6-1 
6 
REINFORCED CONCRETE 
Finley A. Charney, Ph.D., P.E. 
In this chapter, a 12-story reinforced concrete office building with some retail shops on the first floor is 
designed for both high and moderate seismic loadings. For the more extreme loading, it is assumed that 
the structure will be located in Berkeley, California, and for the moderate loading, in Honolulu, Hawaii. 
Figure 6-1 shows the basic structural configuration for each location in plan view and Figure 6-2, in 
section. The building, to be constructed primarily from sand-lightweight (LW) aggregate concrete, has 
12 stories above grade and one basement level. The typical bays are 30 ft long in the north-south (N-S) 
direction and either 40 ft or 20 ft long in the east-west (E-W) direction. The main gravity framing system 
consists of seven continuous 30-ft spans of pan joists. These joists are spaced 36 in. on center and have 
an average web thickness of 6 in. and a depth below slab of 16 in. Due to fire code requirements, a 4-in.- 
thick floor slab is used, giving the joists a total depth of 20 in. 
The joists along Gridlines 2 through 7 are supported by variable depth "haunched" girders spanning 40 ft 
in the exterior bays and 20 ft in the interior bays. The girders are haunched to accommodate 
mechanical-electrical systems. The girders are not haunched on exterior Gridlines 1 and 8, and the 40-ft 
spans have been divided into two equal parts forming a total of five spans of 20 ft. The girders along all 
spans of Gridlines A and D are of constant depth, but along Gridlines B and C, the depth of the end bay 
girders has been reduced to allow for the passage of mechanical systems. 
Normal weight (NW) concrete walls are located around the entire perimeter of the basement level. NW 
concrete also is used for the first (ground) floor framing and, as described later, for the lower levels of the 
structural walls in the Berkeley building. 
For both locations, the seismic-force-resisting system in the N-S direction consists of four 7-bay momentresisting 
frames. The interior frames differ from the exterior frames only in the end bays where the 
girders are of reduced depth. At the Berkeley location, these frames are detailed as special momentresisting 
frames. Due to the lower seismicity and lower demand for system ductility, the frames of the 
Honolulu building are detailed as intermediate moment-resisting frames. 
In the E-W direction, the seismic-force-resisting system for the Berkeley building is a dual system 
composed of a combination of frames and frame-walls (walls integrated into a moment-resisting frame). 
Along Gridlines 1 and 8, the frames have five 20-ft bays with constant depth girders. Along Gridlines 2 
and 7, the frames consist of two exterior 40-ft bays and one 20-ft interior bay. The girders in each span 
are of variable depth as described earlier. At Gridlines 3, 4, 5 and 6, the interior bay has been filled with 
a shear panel and the exterior bays consist of 40-ft-long haunched girders. For the Honolulu building, the 
structural walls are not necessary so E-W seismic resistance is supplied by the moment frames along 
Gridlines 1 through 8. The frames on Gridlines 1 and 8 are five-bay frames and those on Gridlines 2 
through 7 are three-bay frames with the exterior bays having a 40-ft span and the interior bay having a 
20-ft span. Hereafter, frames are referred to by their gridline designation (e.g., Frame 1 is located on
FEMA 451, NEHRP Recommended Provisions: Design Examples 
6-2 
Figure 6-2A 
102'-6" 
5 at 20'-0" 
216'-0" 
7 at 30'-0" 
Figure 6-2B 
N 
' ' 
Figure 6-1 Typical floor plan of the Berkeley building. The Honolulu building is 
similar but without structural walls (1.0 ft = 0.3048 m). 
Gridline 1). It is assumed that the structure for both the Berkeley and Honolulu locations is founded on 
very dense soil (shear wave velocity of approximately 2000 ft/sec).
Chapter 6, Reinforced Concrete 
6-3 
R
12 
11 
10 
9 
8 
7 
6 
5 
4 
3 
2 
G 
Story Level 
B 
1 
2 
3 
4 
5 
6 
7 
8 
9 
10 
11 
12 
11 at 12'-6" 18'-0" 15'-0" 
40'-0" 20'-0" 40'-0" 
' ' 
R
12 
11 
10 
9 
8 
7 
6 
5 
4 
3 
2 
G 
Story Level 
B 
1 
2 
3 
4 
5 
6 
7 
8 
9 
10 
11 
12 
11 at 12'-6" 18'-0" 15'-0" 
40'-0" 20'-0" 40'-0" 
' ' 
A. Section at Wall B. Section at Frame 
Figure 6-2 Typical elevations of the Berkeley building; the Honolulu building is 
similar but without structural walls (1.0 ft = 0.3048 m).
FEMA 451, NEHRP Recommended Provisions: Design Examples 
6-4 
The calculations herein are intended to provide a reference for the direct application of the design 
requirements presented in the 2000 NEHRP Recommended Provisions (hereafter, the Provisions) and to 
assist the reader in developing a better understanding of the principles behind the Provisions. 
Because a single building configuration is designed for both high and moderate levels of seismicity, two 
different sets of calculations are required. Instead of providing one full set of calculations for the 
Berkeley building and then another for the Honolulu building, portions of the calculations are presented 
in parallel. For example, the development of seismic forces for the Berkeley and Honolulu buildings are 
presented before structural design is considered for either building. The full design then is given for the 
Berkeley building followed by the design of the Honolulu building. Each major section (development of 
forces, structural design, etc.) is followed by discussion. In this context, the following portions of the 
design process are presented in varying amounts of detail for each structure: 
1. Development and computation of seismic forces; 
2. Structural analysis and interpretation of structural behavior; 
3. Design of structural members including typical girder in Frame 1, typical interior column in Frame 1, 
typical beam-column joint in Frame 1, typical girder in Frame 3, typical exterior column in Frame 3, 
typical beam-column joint in Frame 3, boundary elements of structural wall (Berkeley building only) 
and panel of structural wall (Berkeley building only). 
The design presented represents the first cycle of an iterative design process based on the equivalent 
lateral force (ELF) procedure according to Provisions Chapter 5. For final design, the Provisions may 
require that a modal response spectrum analysis or time history analysis be used. The decision to use 
more advanced analysis can not be made a priori because several calculations are required that cannot be 
completed without a preliminary design. Hence, the preliminary design based on an ELF analysis is a 
natural place to start. The ELF analysis is useful even if the final design is based on a more sophisticated 
analysis (e.g., forces from an ELF analysis are used to apply accidental torsion and to scale the results 
from the more advanced analysis and are useful as a check on a modal response spectrum or time-history 
analysis). 
In addition to the Provisions, ACI 318 is the other main reference in this example. Except for very minor 
exceptions, the seismic-force-resisting system design requirements of ACI 318 have been adopted in their 
entirety by the Provisions. Cases where requirements of the Provisions and ACI 318 differ are pointed 
out as they occur. ASCE 7 is cited when discussions involve live load reduction, wind load, and load 
combinations. 
Other recent works related to earthquake resistant design of reinforced concrete buildings include: 
ACI 318 American Concrete Institute. 1999 [2002]. Building Code Requirements and 
Commentary for Structural Concrete. 
ASCE 7 American Society of Civil Engineers. 1998 [2002]. Minimum Design Loads for 
Buildings and Other Structures. 
Fanella Fanella, D.A., and M. Munshi. 1997. Design of Low-Rise Concrete Buildings for 
Earthquake Forces, 2nd Edition. Portland Cement Association, Skokie, Illinois. 
ACI 318 Notes Fanella, D.A., J. A. Munshi, and B. G. Rabbat, Editors. 1999. Notes on ACI 318-99 
Building Code Requirements for Structural Concrete with Design Applications. Portland 
Cement Association, Skokie, Illinois.
Chapter 6, Reinforced Concrete 
6-5 
ACI SP127 Ghosh, S. K., Editor. 1991. Earthquake-Resistant Concrete Structures Inelastic 
Response and Design, ACI SP127. American Concrete Institute, Detroit, Michigan. 
Ghosh Ghosh, S. K., A. W. Domel, and D. A. Fanella. 1995. Design of Concrete Buildings for 
Earthquake and Wind Forces, 2nd Edition. Portland Cement Association, Skokie, Illinois. 
Paulay Paulay, T., and M. J. N. Priestley. 1992. Seismic Design of Reinforced Concrete and 
Masonry Buildings. John Wiley & Sons, New York. 
The Portland Cement Associations notes on ACI 318 contain an excellent discussion of the principles 
behind the ACI 318 design requirements and an example of the design and detailing of a frame-wall 
structure. The notes are based on the requirements of the 1997 Uniform Building Code (International 
Conference of Building Officials) instead of the Provisions. The other publications cited above provide 
additional background for the design of earthquake-resistant reinforced concrete structures. 
Most of the large-scale structural analysis for this chapter was carried out using the ETABS Building 
Analysis Program developed by Computers and Structures, Inc., Berkeley, California. Smaller portions 
of the structure were modeled using the SAP2000 Finite Element Analysis Program, also developed by 
Computers and Structures. Column capacity and design curves were computed using Microsoft Excel, 
with some verification using the PCACOL program created and developed by the Portland Cement 
Association. 
Although this volume of design examples is based on the 2000 Provisions, it has been annotated to reflect 
changes made to the 2003 Provisions. Annotations within brackets, [ ], indicate both organizational 
changes (as a result of a reformat of all of the chapters of the 2003 Provisions) and substantive technical 
changes to the 2003 Provisions and its primary reference documents. While the general concepts of the 
changes are described, the design examples and calculations have not been revised to reflect the changes 
to the 2003 Provisions. 
The changes related to reinforced concrete in Chapter 9 of the 2003 Provisions are generally intended to 
maintaining compatibility between the Provisions and the ACI 318-02. Portions of the 2000 Provisions 
have been removed because they were incorporated into ACI 318-02. Other chances to Chapter 9 are 
related to precast concrete (as discussed in Chapter 7 of this volume of design examples). 
Some general technical changes in the 2003 Provisions that relate to the calculations and/or design in this 
chapter include updated seismic hazard maps, revisions to the redundancy requirements, revisions to the 
minimum base shear equation, and revisions several of the system factors (R, O0, Cd) for dual systems. 
Where they affect the design examples in this chapter, other significant changes to the 2003 Provisions 
and primary reference documents are noted. However, some minor changes to the 2003 Provisions and 
the reference documents may not be noted. 
Note that these examples illustrate comparisons between seismic and wind loading for illustrative 
purposes. Wind load calculations are based on ASCE 7-98 as referenced in the 2000 Provisions, and 
there have not been any comparisons or annotations related to ASCE 7-02.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
6-6 
6.1 DEVELOPMENT OF SEISMIC LOADS AND DESIGN REQUIREMENTS 
6.1.1 Seismicity 
Using Provisions Maps 7 and 8 [Figures 3.3-3 and 3.3-4] for Berkeley, California, the short period and 
one-second period spectral response acceleration parameters SS and S1 are 1.65 and 0.68, respectively. 
[The 2003 Provisions have adopted the 2002 USGS probabilistic seismic hazard maps, and the maps have 
been added to the body of the 2003 Provisions as figures in Chapter 3 (instead of the previously used 
separate map package.] For the very dense soil conditions, Site Class C is appropriate as described in 
Provisions Sec. 4.1.2.1 [3.5.1]. Using SS = 1.65 and Site Class C, Provisions Table 4.1.2.4a [3.3-1] lists a 
short period site coefficient Fa of 1.0. For S1 > 0.5 and Site Class C, Provisions Table 4.1.2.4b [3.3-2] 
gives a velocity based site coefficient Fv of 1.3. Using Provisions Eq. 4.1.2.4-1 and 4.1.2.4-2 [3.3-1 and 
3.3-2], the maximum considered spectral response acceleration parameters for the Berkeley building are: 
SMS = FaSS = 1.0 x 1.65 = 1.65 
SM1 = FvS1 = 1.3 x 0.68 = 0.884 
The design spectral response acceleration parameters are given by Provisions Eq. 4.1.2.5-1 and 4.1.2.5-2 
[3.3-3 and 3.3-4]: 
SDS = (2/3) SMS = (2/3) 1.65 = 1.10 
SD1 = (2/3) SM1 = (2/3) 0.884 = 0.589 
The transition period (Ts) for the Berkeley response spectrum is: 
0.589 0.535 sec 
1.10 
D1 
s 
DS 
T S
S 
= = = 
Ts is the period where the horizontal (constant acceleration) portion of the design response spectrum 
intersects the descending (constant velocity or acceleration inversely proportional to T) portion of the 
spectrum. It is used later in this example as a parameter in determining the type of analysis that is 
required for final design. 
For Honolulu, Provisions Maps 19 and 20 [Figure 3.3-10] give the short-period and 1-sec period spectral 
response acceleration parameters of 0.61 and 0.178, respectively. For the very dense soil/firm rock site 
condition, the site is classified as Site Class C. Interpolating from Provisions Table 4.1.4.2a [3.3-1], the 
short-period site coefficient (Fa) is 1.16 and, from Provisions Table 4.1.2.4b [3.3-2], the interpolated 
long-period site coefficient (Fv) is 1.62. The maximum considered spectral response acceleration 
parameters for the Honolulu building are: 
SMS = FaSS = 1.16 x 0.61 = 0.708 
SM1 = FvS1 = 1.62 x 0.178 = 0.288 
and the design spectral response acceleration parameters are: 
SDS = (2/3) SMS = (2/3) 0.708 = 0.472 
SD1 = (2/3) SM1 = (2/3) 0.288 = 0.192 
The transition period (Ts) for the Honolulu response spectrum is:
Chapter 6, Reinforced Concrete 
6-7 
0.192 0.407 sec 
0.472 
D1 
s 
DS 
T S
S 
= = = 
6.1.2 Structural Design Requirements 
According to Provisions Sec. 1.3 [1.2], both the Berkeley and the Honolulu buildings are classified as 
Seismic Use Group I. Provisions Table 1.4 [1.3] assigns an occupancy importance factor (I) of 1.0 to all 
Seismic Use Group I buildings. 
According to Provisions Tables 4.2.1a and 4.2.1b [Tables 1.4-1 and 1.4-2], the Berkeley building is 
classified as Seismic Design Category D. The Honolulu building is classified as Seismic Design 
Category C because of the lower intensity ground motion. 
The seismic-force-resisting systems for both the Berkeley and the Honolulu buildings consist of momentresisting 
frames in the N-S direction. E-W loading is resisted by a dual frame-wall system in the 
Berkeley building and by a set of moment-resisting frames in the Honolulu building. For the Berkeley 
building, assigned to Seismic Design Category D, Provisions Sec. 9.1.1.3 [9.2.2.1.3] (which modifies 
language in the ACI 318 to conform to the Provisions) requires that all moment-resisting frames be 
designed and detailed as special moment frames. Similarly, Provisions Sec. 9.1.1.3 [9.2.2.1.3] requires 
the structural walls to be detailed as special reinforced concrete shear walls. For the Honolulu building 
assigned to Seismic Design Category C, Provisions Sec. 9.1.1.3 [9.2.2.1.3] allows the use of intermediate 
moment frames. According to Provisions Table 5.2.2 [4.3-1], neither of these structures violate height 
restrictions. 
Provisions Table 5.2.2 [4.3-1] provides values for the response modification coefficient (R), the system 
over strength factor (O0), and the deflection amplification factor (Cd) for each structural system type. The 
values determined for the Berkeley and Honolulu buildings are summarized in Table 6-1. 
Table 6-1 Response Modification, Overstrength, and Deflection Amplification Coefficients 
for Structural Systems Used 
Location 
Response 
Direction Building Frame Type R O0 Cd 
Berkeley N-S Special moment frame 8 3 5.5 
E-W Dual system incorporating special moment 
frame and structural wall 
8 2.5 6.5 
Honolulu N-S Intermediate moment frame 5 3 4.5 
E-W Intermediate moment frame 5 3 4.5 
[For a dual system consisting of a special moment frame and special reinforced concrete shear walls, R = 
7, O0 = 2.5, and Cd = 5.5 in 2003 Provisions Table 4.3-1.] 
For the Berkeley building dual system, the Provisions requires that the frame portion of the system be 
able to carry 25 percent of the total seismic force. As discussed below, this requires that a separate 
analysis of a frame-only system be carried out for loading in the E-W direction. 
With regard to the response modification coefficients for the special and intermediate moment frames, it 
is important to note that R = 5.0 for the intermediate frame is 0.625 times the value for the special frame. 
This indicates that intermediate frames can be expected to deliver lower ductility than that supplied by the 
more stringently detailed special moment frames.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
6-8 
For the Berkeley system, the response modification coefficients are the same (R = 8) for the frame and 
frame-wall systems but are higher than the coefficient applicable to a special reinforced concrete 
structural wall system (R = 6). This provides an incentive for the engineer to opt for a frame-wall system 
under conditions where a frame acting alone may be too flexible or a wall acting alone cannot be 
proportioned due to excessively high overturning moments. 
6.1.3 Structural Configuration 
Based on the plan view of the building shown in Figure 6-1, the only possibility of a plan irregularity is a 
torsional irregularity (Provisions Table 5.2.3.2 [4.3-2]) of Type 1a or 1b. While the actual presence of 
such an irregularity cannot be determined without analysis, it appears unlikely for both the Berkeley and 
the Honolulu buildings because the lateral-force-resisting elements of both buildings are distributed 
evenly over the floor. For the purpose of this example, it is assumed (but verified later) that torsional 
irregularities do not exist. 
As for the vertical irregularities listed in Provisions Table 5.2.3.3 [4.3-3], the presence of a soft or weak 
story cannot be determined without calculations based on an existing design. In this case, however, the 
first story is suspect, because its height of 18 ft is well in excess of the 12.5-ft height of the story above. 
As with the torsional irregularity, it is assumed (but verified later) that a vertical irregularity does not 
exist. 
6.2 DETERMINATION OF SEISMIC FORCES 
The determination of seismic forces requires knowledge of the magnitude and distribution of structural 
mass, the short period and long period response accelerations, the dynamic properties of the system, and 
the system response modification factor (R). Using Provisions Eq. 5.4.1 [5.2-1], the design base shear for 
the structure is: 
V = CSW 
where W is the total (seismic) weight of the building and CS is the seismic response coefficient. The upper 
limit on CS is given by Provisions Eq. 5.4.1.1-1 [5.2-2]: 
/DS 
S 
C S
R I 
= 
For intermediate response periods, Eq. 5.4.1.1-2 [5.2-3] controls: 
( / ) 
D1 
S 
C S 
T R I 
= 
However, the response coefficient must not be less than that given by Eq. 5.4.1.1-3 [changed in the 2003 
Provisions]: 
CS = 0.044SDSI 
Note that the above limit will apply when the structural period is greater than SD1/0.044RSDS. This limit is 
(0.589)/(0.044 x 8 x 1.1) = 1.52 sec for the Berkeley building and (0.192)/(0.044 x 5 x 0.472) = 1.85 sec 
for the Honolulu building. [The minimum Cs value is simply 0.01in the 2003 Provisions, which would 
not be applicable to this example as discussed below.]
Chapter 6, Reinforced Concrete 
6-9 
In each of the above equations, the importance factor (I) is taken as 1.0. With the exception of the period 
of vibration (T), all of the other terms in previous equations have been defined and/or computed earlier in 
this chapter. 
6.2.1 Approximate Period of Vibration 
Requirements for the computation of building period are given in Provisions Sec. 5.4.2 [5.2.2]. For the 
preliminary design using the ELF procedure, the approximate period (Ta) computed in accordance with 
Provisions Eq. 5.4.2.1-1 [5.2-6] could be used: 
x 
Ta=Crhn 
Because this formula is based on lower bound regression analysis of measured building response in 
California, it will generally result in periods that are lower (hence, more conservative for use in predicting 
base shear) than those computed from a more rigorous mathematical model. This is particularly true for 
buildings located in regions of lower seismicity. If a more rigorous analysis is carried out (using a 
computer), the resulting period may be too high due to a variety of possible modeling errors. 
Consequently, the Provisions places an upper limit on the period that can be used for design. The upper 
limit is T = CuTa where Cu is provided in Provisions Table 5.4.2 [5.2-1]. 
For the N-S direction of the Berkeley building, the structure is a reinforced concrete moment-resisting 
frame and the approximate period is calculated according to Provisions Eq. 5.4.2.1-1 [5.2-6]. Using 
Provisions Table 5.4.2.1 [5.2-2], Cr = 0.016 and x = 0.9. With hn = 155.5 ft, Ta = 1.50 sec. With SD1 > 
0.40 for the Berkeley building, Cu = 1.4 and the upper limit on the analytical period is T = 1.4(1.5) = 2.1 
sec. 
For E-W seismic activity in Berkeley, the structure is a frame-wall system with Cr = 0.020 and x =0.75. 
Substituting the appropriate values in Provisions Eq. 5.4.2.1-1 [5.2-6], the E-W period Ta = 0.88 sec. The 
upper limit on the analytical period is (1.4)0.88 = 1.23 sec. 
For the Honolulu building, the Ta = 1.5 sec period computed above for concrete moment frames is 
applicable in both the N-S and E-W direction. For Honolulu, SD1 is 0.192g and, from Provisions Table 
5.4.2 [5.2-1], Cu can be taken as 1.52. The upper limit on the analytical period is T = 1.52(1.5) = 2.28 sec. 
The period to be used in the ELF analysis will be in the range of Ta to CuTa. If an accurate analysis 
provides periods greater than CuTa, CuTa should be used. If the accurate analysis produces periods less 
than CuTa but greater than Ta, the period from the analysis should be used. Finally, if the accurate analysis 
produces periods less than Ta, Ta may be used. 
Later in this chapter, the more accurate periods will be computed using a finite element analysis program. 
Before this can be done, however, the building mass must be determined. 
6.2.2 Building Mass 
Before the building mass can be determined, the approximate size of the different members of the 
seismic-force-resisting system must be established. For special moment frames, limitations on 
beam-column joint shear and reinforcement development length usually control. This is particularly true 
when lightweight (LW) concrete is used. An additional consideration is the amount of vertical 
reinforcement in the columns. ACI 318 Sec. 21.4.3.1 limits the vertical steel reinforcing ratio to 6 percent 
for special moment frame columns; however, 4 percent vertical steel is a more practical limit.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
1ACI 318 Sec. 21.6.4 [21.7.4] gives equations for the shear strength of the panels of structural walls. In the equations, the term 
c appears, but there is no explicit requirement to reduce the shear strength of the concrete when LW aggregate is used. f ' 
However, ACI 318 Sec. 11.2 states that wherever the term c appears in association with shear strength, it should be f ' 
multiplied by 0.75 when all-LW concrete is used and by 0.85 when sand-LW concrete is used. In this example, which utilizes 
sand-LW concrete, the shear strength of the concrete will be multiplied by 0.85 as specified in ACI 318 Chapter 11. 
6-10 
Based on a series of preliminary calculations (not shown here), it is assumed that all columns and 
structural wall boundary elements are 30 in. by 30 in., girders are 22.5 in. wide by 32 in. deep, and the 
panel of the structural wall is 16 in. thick. It has already been established that pan joists are spaced 36 in. 
o.c., have an average web thickness of 6 in., and, including a 4-in.-thick slab, are 20 in. deep. For the 
Berkeley building, these member sizes probably are close to the final sizes. For the Honolulu building 
(which has no structural wall and ultimately ends up with slightly smaller elements), the masses computed 
from the above member sizes are on the conservative (heavy) side. 
In addition to the building structural weight, the following superimposed dead loads (DL) were assumed: 
Partition DL (and roofing) = 10 psf 
Ceiling and mechanical DL = 15 psf 
Curtain wall cladding DL = 10 psf 
Based on the member sizes given above and on the other dead load, the individual story weights, masses, 
and mass moments of inertia are listed in Table 6-2. These masses were used for both the Berkeley and 
the Honolulu buildings. 
As discussed below, the mass and mass moments of inertia are required for the determination of modal 
properties using the ETABS program. Note from Table 6-2 that the roof and lowest floor have masses 
slightly different from the typical floors. It is also interesting to note that the average density of this 
building is 11.2 pcf. A normal weight (NW) concrete building of the same configuration would have a 
density of approximately 14.0 pcf. 
The use of LW instead of NW concrete reduces the total building mass by more than 20 percent and 
certainly satisfies the minimize mass rule of earthquake-resistant design. However, there are some 
disadvantages to the use of LW concrete. In general, LW aggregate reinforced concrete has a lower 
toughness or ductility than NW reinforced concrete and the higher the strength, the larger the reduction in 
available ductility. For this reason and also the absence of pertinent test results, ACI 318 Sec. 21.2.4.2 
allows a maximum compressive strength of 4,000 psi for LW concrete in areas of high seismicity. [Note 
that in ACI 318-02 Sec. 21.2.4.2, the maximum compressive strength for LW concrete has been increased 
to 5,000 psi.] A further penalty placed on LW concrete is the reduction of shear strength. This primarily 
affects the sizing of beam-column joints (ACI 318 Sec. 21.5.3.2) but also has an effect on the amount of 
shear reinforcement required in the panels of structural walls.1 For girders, the reduction in shear strength 
of LW aggregate concrete usually is of no concern because ACI 318 disallows the use of the concrete in 
determining the shear resistance of members with significant earthquake shear (ACI 318 Sec. 21.4.5.2). 
Finally, the required tension development lengths for bars embedded in LW concrete are significantly 
greater than those required for NW concrete. 
Table 6-2 Story Weights, Masses, and Moments of Inertia 
Story Level Weight (kips) 
Mass 
(kips-sec2/in.) 
Mass Moment of Inertia 
(in.-kip-sec2/rad)
Chapter 6, Reinforced Concrete 
6-11 
Roof 
12 
11 
10
98765432 
Total 
2,783 
3,051 
3,051 
3,051 
3,051 
3,051 
3,051 
3,051 
3,051 
3,051 
3,051 
3,169 
36,462 
7.202 
7.896 
7.896 
7.896 
7.896 
7.876 
7.896 
7.896 
7.896 
7.896 
7.896 
8.201 
4,675,000 
5,126,000 
5,126,000 
5,126,000 
5,126,000 
5,126,000 
5,126,000 
5,126,000 
5,126,000 
5,126,000 
5,126,000 
5,324,000 
1.0 kip = 4.45 kN, 1.0 in. = 25.4 mm. 
6.2.3 Structural Analysis 
Structural analysis is used primarily to determine the forces in the elements for design purposes, compute 
story drift, and assess the significance of P-delta effects. The structural analysis also provides other 
useful information (e.g., accurate periods of vibration and computational checks on plan and vertical 
irregularities). The computed periods of vibration are addressed in this section and the other results are 
presented and discussed later. 
The ETABS program was used for the analysis of both the Berkeley and Honolulu buildings. Those 
aspects of the model that should be noted are: 
1. The structure was modeled with 12 levels above grade and one level below grade. The perimeter 
basement walls were modeled as shear panels as were the main structural walls. It was assumed that 
the walls were "fixed" at their base. 
2. As automatically provided by the ETABS program, all floor diaphragms were assumed to be 
infinitely rigid in plane and infinitely flexible out-of-plane. 
3. Beams, columns, and structural wall boundary members were represented by two-dimensional frame 
elements. Each member was assumed to be uncracked, and properties were based on gross area for 
the columns and boundary elements and on effective T-beam shapes for the girders. (The effect of 
cracking is considered in a simplified manner.) The width of the flanges for the T-beams is based on 
the definition of T-beams in ACI 318 Sec. 8.10. Except for the slab portion of the joists which 
contributed to T-beam stiffness of the girders, the flexural stiffness of the joists was ignored. For the 
haunched girders, an equivalent depth of stem was used. The equivalent depth was computed to 
provide a prismatic member with a stiffness under equal end rotation identical to that of the 
nonprismatic haunched member. Axial, flexural, and shear deformations were included for all 
members. 
4. The structural walls of the Berkeley building are modeled as a combination of boundary elements and 
shear panels. 
5. Beam-column joints are modeled as 50 percent rigid. This provides effective stiffness for 
beam-column joints halfway between a model with fully rigid joints (clear span analysis) and fully 
flexible joints (centerline analysis).
FEMA 451, NEHRP Recommended Provisions: Design Examples 
6-12 
6. P-delta effects are ignored. An evaluation of the accuracy of this assumption is provided later in this 
example. 
6.2.4 Accurate Periods from Finite Element Analysis 
The computed periods of vibration and a description of the associated modes of vibration are given for the 
first 11 modes of the Berkeley building in Table 6-3. With 11 modes, the accumulated modal mass in 
each direction is more than 90 percent of the total mass. Provisions Sec. 5.5.2 [5.3.2] requires that a 
dynamic analysis must include at least 90 percent of the actual mass in each of the two orthogonal 
directions. Table 6-4 provides the computed modal properties for the Honolulu building. In this case, 90 
percent of the total mass was developed in just eight modes. 
For the Berkeley building, the computed N-S period of vibration is 1.77 sec. This is between the 
approximate period, Ta = 1.5 sec, and CuTa = 2.1 sec. In the E-W direction, the computed period is 1.40 
sec, which is greater than both Ta = 0.88 sec and CuTa = 1.23 sec. 
If cracked section properties were used, the computed period values for the Berkeley building would be 
somewhat greater. For preliminary design, it is reasonable to assume that each member has a cracked 
moment of inertia equal to one-half of the gross uncracked moment of inertia. Based on this assumption, 
and the assumption that flexural behavior dominates, the cracked periods would be approximately 1.414 
(the square root of 2.0) times the uncracked periods. Hence, for Berkeley, the cracked N-S and E-W 
periods are 1.414(1.77) = 2.50 sec, and 1.414(1.4) = 1.98 sec, respectively. Both of these cracked periods 
are greater than CuTa, so CuTa can be used in the ELF analysis. 
For the Honolulu building, the uncracked periods in the N-S and E-W directions are 1.78 and 1.87 sec, 
respectively. The N-S period is virtually the same as for the Berkeley building because there are no walls 
in the N-S direction of either building. In the E-W direction, the increase in period from 1.4 sec to 1.87 
sec indicates a significant reduction in stiffness due to the loss of the walls in the Honolulu building. For 
both the E-W and the N-S directions, the approximate period (Ta) for the Honolulu building is 1.5 sec, 
and CuTa is 2.28 sec. Both of the computed periods fall within these bounds. However, if cracked section 
properties were used, the computed periods would be 2.52 sec in the N-S direction and 2.64 sec in the 
E-W direction. For the purpose of computing ELF forces, therefore, a period of 2.28 sec can be used for 
both the N-S and E-W directions in Honolulu. 
A summary of the approximate and computed periods is given in Table 6-5.
Chapter 6, Reinforced Concrete 
6-13 
Table 6-3 Periods and Modal Response Characteristics for the Berkeley Building 
Mode 
Period* 
(sec) 
% of Effective Mass Represented by Mode** 
N-S E-W Description 
1 
2 
3 
4 
5 
6 
7 
8 
9 
10 
11 
1.77 
1.40 
1.27 
0.581 
0.394 
0.365 
0.336 
0.230 
0.210 
0.171 
0.135 
80.23 (80.2) 
0.0 (80.2) 
0.0 (80.2) 
8.04 (88.3) 
0.00 (88.3) 
0.00 (88.3) 
2.24 (90.5) 
0.88 (91.4) 
0.00 (91.4) 
0.40 (91.8) 
0.00 (91.8) 
00.00 (0.00) 
71.48 (71.5) 
0.00 (71.5) 
0.00 (71.5) 
0.00 (71.5) 
14.17 (85.6) 
0.00 (85.6) 
0.00 (85.6) 
0.00 (85.6) 
0.00 (85.6) 
4.95 (90.6) 
First Mode N-S 
First Mode E-W 
First Mode Torsion 
Second Mode N-S 
Second Mode Torsion 
Second Mode E-W 
Third Mode N-S 
Fourth Mode N-S 
Third Mode Torsion 
Fifth Mode N-S 
Third Mode E-W 
* Based on gross section properties. 
** Accumulated mass in parentheses. 
Table 6-4 Periods and Modal Response Characteristics for the Honolulu Building 
Mode 
Period* 
(sec) 
% of Effective Mass Represented by Mode** 
N-S E-W Description 
1 
2 
3 
4 
5 
6 
7 
8 
9 
10 
11 
1.87 
1.78 
1.38 
0.610 
0.584 
0.452 
0.345 
0.337 
0.260 
0.235 
0.231 
79.7 (79.7) 
0.00 (79.7) 
0.00 (79.7) 
8.79 (88.5) 
0.00 (88.5) 
0.00 (88.5) 
2.27 (90.7) 
0.00 (90.7) 
0.00 (90.7) 
0.89 (91.6) 
0.00 (91.6) 
0.00 (0.00) 
80.25 (80.2) 
0.00 (80.2) 
0.00 (80.2) 
8.04 (88.3) 
0.00 (88.3) 
0.00 (88.3) 
2.23 (90.5) 
0.00 (90.5) 
0.00 (90.5) 
0.87 (91.4) 
First Mode E-W 
First Mode N-S 
First Mode Torsion 
Second Mode E-W 
Second Mode N-S 
Second Mode Torsion 
Third Mode E-W 
Third Mode N-S 
Third Mode Torsion 
Fourth Mode E-W 
Fourth Mode N-S 
* Based on gross section properties. 
** Accumulated mass in parentheses. 
Table 6-5 Comparison of Approximate and "Exact" Periods (in seconds) 
Method of Period 
Computation 
Berkeley Honolulu 
N-S E-W N-S E-W 
Approximate Ta 1.50 0.88 1.50 1.50 
Approximate  Cu 2.10* 1.23 2.28 2.28 
ETABS (gross) 1.77 1.40 1.78 1.87 
ETABS (cracked) 2.50 1.98 2.52 2.64 
* Values in italics should be used in the ELF analysis. 
6.2.5 Seismic Design Base Shear
FEMA 451, NEHRP Recommended Provisions: Design Examples 
6-14 
The seismic design base shear for the Berkeley is computed below. 
In the N-S direction with W = 36,462 kips (see Table 6-2), SDS = 1.10, SD1 = 0.589, R = 8, I = 1, and T = 
2.10 sec: 
, 
1.10 0.1375 
/ 8/1 
DS 
S max 
C S
R I 
= = = 
0.589 0.0351 
( / ) 2.10(8/1) 
D1 
S 
C S 
T R I 
= = = 
CS,min=0.044SDSI=0.044(1.1)(1)=0.0484 
[As noted previously in Sec. 6.2, the minimum Cs value is 0.01 in the 2003 Provisions.] 
CS,min = 0.0484 controls, and the design base shear in the N-S direction is V = 0.0484 (36,462) = 1,765 
kips. 
In the stiffer E-W direction, CS,max and CS,min are as before, T = 1.23 sec, and 
0.589 0.0598 
( / ) 1.23(8/1) 
D1 
S 
C S 
T R I 
= = = 
In this case, CS = 0.0598 controls and V = 0.0598 (36,462) = 2,180 kips 
For the Honolulu building, base shears are computed in a similar manner and are the same for the N-S and 
the E-W directions. With W = 36,462 kips, SDS = 0.474, SD1 = 0.192, R = 5, I = 1, and T = 2.28 sec: 
, 
0.472 0.0944 
/ 5/1 
DS 
S max 
C S
R I 
= = = 
0.192 0.0168 
( / ) 2.28(5/1) 
D1 
S 
C S 
T R I 
= = = 
CS,min=0.044SDSI=0.044(0.472)(1.0)=0.0207 
CS = 0.0207 controls and V = 0.0207 (36,462) = 755 kips 
A summary of the Berkeley and Honolulu seismic design parameters are provided in Table 6-6. 
Note that Provisions Sec. 5.4.6 [5.2.6.1] states that for the purpose of computing drift, a base shear 
computed according to Provisions Eq. 5.4.1.1-2 [5.2-3] (used to compute CS above) may be used in lieu 
of the shear computed using Provisions Eq. 5.4.1.1-3 [5.2-4] (used to compute CS,min above). 
Table 6-6 Comparison of Periods, Seismic Shears Coefficients, and Base Shears 
for the Berkeley and Honolulu Buildings 
Location 
Response 
Direction Building Frame Type 
T 
(sec) Cs 
V 
(kips)
Chapter 6, Reinforced Concrete 
6-15 
Berkeley N-S Special moment frame 2.10 0.0485 1,765 
E-W Dual system incorporating special moment 
frame and structural wall 
1.23 0.0598 2,180 
Honolulu N-S Intermediate moment frame 2.28 0.0207 755 
E-W Intermediate moment frame 2.28 0.0207 755 
1.0 kip = 4.45 kN. 
6.2.6 Development of Equivalent Lateral Forces 
The vertical distribution of lateral forces is computed from Provisions Eq. 5.4.3-1 and 5.4.3-2 [5.2-10 and 
5.2-11]: 
Fx = CvxV 
k 
x x 
vx n 
k 
i i 
i = 1 
C = w h 
S w h 
where 
k = 1.0 for T < 0.5 sec 
k = 2.0 for T > 2.5 sec 
k = 0.75 + 0.5T for 1.0 < T < 2.5 sec 
Based on the equations above, the seismic story forces, shears, and overturning moments are easily 
computed using an Excel spreadsheet. The results of these computations are shown in Tables 6-7a and 
6-7b for the Berkeley buildings and in Table 6-8 for the Honolulu building. A note at the bottom of each 
table gives the calculated vertical force distribution factor (k). The tables are presented with as many 
significant digits to the left of the decimal as the spreadsheet generates but that should not be interpreted 
as real accuracy; it is just the simplest approach. Also, some of the sums are not exact due to truncation 
error.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
6-16 
Table 6-7a Vertical Distribution of N-S Seismic Forces for the Berkeley Building* 
Level 
Height h 
(ft) 
Weight W 
(kips) Whk Whk/S 
Force Fx 
(kips) 
Story 
Shear Vx 
(kips) 
Overturning 
Moment 
Mx (ft-k) 
R 
12 
11 
10 
9 
8 
7 
6 
5 
4 
3 
2 
Total 
155.5 
143.0 
130.5 
118.0 
105.5 
93.0 
80.5 
68.0 
55.5 
43.0 
30.5 
18.0 
2,783 
3,051 
3,051 
3,051 
3,051 
3,051 
3,051 
3,051 
3,051 
3,051 
3,051 
3,169 
36,462 
24,526,067 
23,123,154 
19,612,869 
16,361,753 
13,375,088 
10,658,879 
8,220,056 
6,066,780 
4,208,909 
2,658,799 
1,432,788 
575,987 
130,821,129 
0.187 
0.177 
0.150 
0.125 
0.102 
0.081 
0.063 
0.046 
0.032 
0.020 
0.011 
0.004 
0.998 
330.9 
311.9 
264.6 
220.7 
180.4 
143.8 
110.9 
81.8 
56.8 
35.9 
19.3 
7.8 
1764.8 
330.9 
642.8 
907.4 
1,128.1 
1,308.5 
1,452.3 
1,563.2 
1,645.0 
1,701.8 
1,737.7 
1,757.0 
1,764.8 
4,136 
12,170 
23,512 
37,613 
53,970 
72,123 
91,663 
112,226 
133,498 
155,219 
177,181 
208,947 
* Table based on T = 2.1 sec and k = 1.8. 
1.0 ft = 0.3048 m, 1.0 kip = 4.45 kN, 1.0 ft-kip = 1.36 kN-m. 
Table 6-7b Vertical Distribution of E-W Seismic Forces for the Berkeley Building* 
Level 
Height h 
(ft) 
Weight W 
(kips) Whk Whk/S 
Force Fx 
(kips) 
Story 
Shear Vx 
(kips) 
Overturning 
Moment 
Mx (ft-k) 
R 
12 
11 
10 
9 
8 
7 
6 
5 
4 
3 
2 
Total 
155.5 
143.0 
130.5 
118.0 
105.5 
93.0 
80.5 
68.0 
55.5 
43.0 
30.5 
18.0 
2,783 
3,051 
3,051 
3,051 
3,051 
3,051 
3,051 
3,051 
3,051 
3,051 
3,051 
3,169 
36,462 
2,730,393 
2,669,783 
2,356,408 
2,053,814 
1,762,714 
1,483,957 
1,218,579 
967,870 
733,503 
517,758 
323,975 
163,821 
16,982,575 
0.161 
0.157 
0.139 
0.121 
0.104 
0.087 
0.072 
0.057 
0.043 
0.030 
0.019 
0.010 
1.000 
350.6 
342.8 
302.5 
263.7 
226.3 
190.5 
156.5 
124.3 
94.2 
66.5 
41.6 
21.0 
2180.5 
351 
693 
996 
1,260 
1,486 
1,676 
1,833 
1,957 
2,051 
2,118 
2,159 
2,180 
4,382 
13,049 
25,497 
41,242 
59,816 
80,771 
103,682 
128,146 
153,788 
180,260 
207,253 
246,500 
* Table based on T = 1.23 sec and k = 1.365. 
1.0 ft = 0.3048 m, 1.0 kip = 4.45 kN, 1.0 ft-kip = 1.36 kN-m.
Chapter 6, Reinforced Concrete 
6-17 
Table 6-8 Vertical Distribution of N-S and E-W Seismic Forces for the Honolulu Building* 
Level 
Height h 
(ft) 
Weight W 
(kips) Whk Whk/S 
Force Fx 
(kips) 
Story 
Shear Vx 
(kips) 
Overturning 
Moment 
Mx (ft-k) 
R 
12 
11 
10 
9 
8 
7 
6 
5 
4 
3 
2 
Total 
155.5 
143.0 
130.5 
118.0 
105.5 
93.0 
80.5 
68.0 
55.5 
43.0 
30.5 
18.0 
2,783 
3,051 
3,051 
3,051 
3,051 
3,051 
3,051 
3,051 
3,051 
3,051 
3,051 
3,169 
36,462 
38,626,348 
36,143,260 
30,405,075 
25,136,176 
20,341,799 
16,027,839 
12,210,028 
8,869,192 
6,041,655 
3,729,903 
1,948,807 
747,115 
200,218,197 
0.193 
0.181 
0.152 
0.126 
0.102 
0.080 
0.061 
0.044 
0.030 
0.019 
0.010 
0.004 
1.002 
145.6 
136.2 
114.6 
94.8 
76.7 
60.4 
46.0 
33.4 
22.8 
14.1 
7.3 
2.8 
754.7 
145.6 
281.9 
396.5 
491.2 
567.9 
628.3 
674.3 
707.8 
730.5 
744.6 
751.9 
754.8 
1,820 
5,343 
10,299 
16,440 
23,539 
31,393 
39,822 
48,669 
57,801 
67,108 
76,508 
90,093 
* Table based on T = 2.28 sec and k = 1.89. 
1.0 ft = 0.3048 m, 1.0 kip = 4.45 kN, 1.0 ft-kip = 1.36 kN-m 
The computed seismic story shears for the Berkeley and Honolulu buildings are shown graphically in 
Figures 6-3 and 6-4, respectively. Also shown in the figures are the story shears produced by ASCE 7 
wind loads. For Berkeley, a 3-sec gust of 85 mph was used and, for Honolulu, a 3-sec gust of 105 mph. 
In each case, an Exposure B classification was assumed. The wind shears have been factored by a value 
of 1.36 (load factor of 1.6 times directionality factor 0.85) to bring them up to the ultimate seismic 
loading limit state represented by the Provisions. 
As can be seen from the figures, the seismic shears for the Berkeley building are well in excess of the 
wind shears and will easily control the design of the members of the frames and walls. For the Honolulu 
building, the N-S seismic shears are significantly greater than the corresponding wind shears, but the E-W 
seismic and wind shears are closer. In the lower stories of the building, wind controls the strength 
demands and, in the upper levels, seismic forces control the strength demands. (A somewhat more 
detailed comparison is given later when the Honolulu building is designed.) With regards to detailing the 
Honolulu building, all of the elements must be detailed for inelastic deformation capacity as required by 
ACI 318 rules for intermediate moment frames.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
6-18 
0 
20 
40 
60 
80 
100 
120 
140 
160 
0 500 1,000 1,500 2,000 2,500 
Shear, kips 
Height, ft 
E-W seismic 
N-S seismic 
E-W wind 
N-S wind 
Figure 6-3 Comparison of wind and seismic story shears for the Berkeley building (1.0 
ft = 0.3048 m, 1.0 kip = 4.45 kN).
Chapter 6, Reinforced Concrete 
6-19 
0 
20 
40 
60 
80 
100 
120 
140 
160 
0 200 400 600 800 1,000 1,200 
Shear, kips 
Height, ft 
Seismic 
E-W wind 
N-S wind 
Figure 6-4 Comparison of wind and seismic story shears for the Honolulu building (1.0 ft 
= 0.3048 m, 1.0 kip = 4.45 kN). 
6.3 DRIFT AND P-DELTA EFFECTS 
6.3.1 Direct Drift and P-Delta Check for the Berkeley Building 
Drift and P-delta effects are checked according to Provisions Sec. 5.2.8 [5.2.6.1] and 5.4.6 [5.2.6.2], 
respectively. According to Provisions Table 5.2.8 [4.5-1], the story drift limit for this Seismic Use Group 
I building is 0.020hsx where hsx is the height of story x. This limit may be thought of as 2 percent of the 
story height. Quantitative results of the drift analysis for the N-S and E-W directions are shown in Tables 
6-9a and 6-9b, respectively. 
With regards to the values shown in Table 6-9a , it must be noted that cracked section properties were 
used in the structural analysis and that 0.0351/0.0484=0.725 times the story forces shown in Table 6-7a 
were applied. This adjusts for the use of Provisions Eq. 5.4.1.1-3 [not applicable in the 2003 Provisions], 
which governed for base shear, was not used in computing drift. In Table 6-9b, cracked section
FEMA 451, NEHRP Recommended Provisions: Design Examples 
6-20 
0 
20 
40 
60 
80 
100 
120 
140 
160
0.0 5.0 10.0 15.0 20.0 25.0 30.0 35.0 40.0 
Total drift, in. 
Height, ft 
E-W* 
6.5 x E-W 
N-S* 
5.5 x N-S 
2% limit 
* Elasticlly computed under code-prescribed seismic forces 
Figure 6-5 Drift profile for Berkeley building (1.0 ft = 
0.3048 m, 1.0 in. = 25.4 mm). 
properties were also used, but the modifying factor does not apply because Provisions Eq. 5.4.1.1-2 [5.2- 
3] controlled in this direction. 
In neither case does the computed drift ratio (magnified story drift/hsx) exceed 2 percent of the story 
height. Therefore, the story drift requirement is satisfied. A plot of the total drift resulting from both the 
N-S and E-W equivalent lateral seismic forces is shown in Figure 6-5. 
An example calculation for drift in Story 5 loaded in the E-W direction is given below. Note that the 
relevant row is highlighted in Table 6-9b. 
Deflection at top of story = d5e = 1.812 in. 
Deflection at bottom of story = d4e = 1.410 in. 
Story drift = .5e = d5e - d4e = 1.812 - 1.410 = 0.402 in. 
Deflection amplification factor, Cd = 6.5 
Importance factor, I = 1.0 
Magnified story drift = .5 = Cd .5e/I = 6.5(0.402)/1.0 = 2.613 in. 
Magnified drift ratio = .5/h5 = (2.613/150) = 0.01742 = 1.742% < 2.0% OK
Chapter 6, Reinforced Concrete 
6-21 
Table 6-9a Drift Computations for the Berkeley Building Loaded in the N-S Direction 
Story 
Total Deflection 
(in.) 
Story Drift 
(in.) 
Story Drift  Cd * 
(in.) 
Drift Ratio 
(%) 
12 
11 
10
987654321 
3.640 
3.533 
3.408 
3.205 
2.973 
2.697 
2.393 
2.059 
1.711 
1.363 
0.999 
0.618 
0.087 
0.145 
0.203 
0.232 
0.276 
0.305 
0.334 
0.348 
0.348 
0.364 
0.381 
0.618 
0.478 
0.798 
1.117 
1.276 
1.515 
1.675 
1.834 
1.914 
1.914 
2.002 
2.097 
3.397 
0.319 
0.532 
0.744 
0.851 
1.010 
1.117 
1.223 
1.276 
1.276 
1.334 
1.398 
1.573 
* Cd = 5.5 for loading in this direction; total drift is at top of story, story height = 150 in. for Levels 3 
through roof and 216 in. for Level 2. 
1.0 in. = 25.4 mm. 
Table 6-9b Drift Computations for the Berkeley Building Loaded in the E-W Direction 
Story 
Total Drift 
(in.) 
Story Drift 
(in.) 
Story Drift  Cd 
* 
(in.) 
Drift Ratio 
(%) 
12 
11 
10
98765
4321 
4.360 
4.060 
3.720 
3.380 
3.020 
2.620 
2.220 
1.812 
1.410 
1.024 
0.670 
0.362 
0.300 
0.340 
0.340 
0.360 
0.400 
0.400 
0.408 
0.402 
0.386 
0.354 
0.308 
0.362 
1.950 
2.210 
2.210 
2.340 
2.600 
2.600 
2.652 
2.613 
2.509 
2.301 
2.002 
2.353 
1.300 
1.473 
1.473 
1.560 
1.733 
1.733 
1.768 
1.742 
1.673 
1.534 
1.335 
1.089 
* Cd = 6.5 for loading in this direction; total drift is at top of story, story height = 150 in. for Levels 3 
through roof and 216 in. for Level 2. 
1.0 in. = 25.4 mm. 
When a soft story exists in a Seismic Design Category D building, Provisions Table 5.2.5.1 [4.4-1] 
requires that a modal analysis be used. However, Provisions Sec. 5.2.3.3 [4.3.2.3] lists an exception: 
Structural irregularities of Types 1a, 1b, or 2 in Table 5.2.3.3 [4.3-2] do not apply where no story drift ratio 
under design lateral load is less than or equal to 130 percent of the story drift ratio of the next story above. . 
. . The story drift ratios of the top two stories of the structure are not required to be evaluated.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
6-22 
For the building responding in the N-S direction, the ratio of first story to second story drift ratios is 
1.573/1.398 = 1.13, which is less than 1.3. For E-W response, the ratio is 1.089/1.335 = 0.82, which also 
is less than 1.3. Therefore, a modal analysis is not required and the equivalent static forces from Tables 
6-7a and 6-7b may be used for design. 
The P-delta analysis for each direction of loading is shown in Tables 6-10a and 6-10b. The upper limit on 
the allowable story stability ratio is given by Provisions Eq. 5.4.6.2-2 [changed in the 2003 Provisions] 
as: 
max 0.5 0.50 
Cd 
. 
 
= = 
Taking  as 1.0 (see Provisions Sec. 5.4.6.2 [not applicable in the 2003 Provisions]), the stability ratio 
limit for the N-S direction is 0.5/(1.0)5.5 = 0.091, and for the E-W direction the limit is 0.5/(1.0)6.5 = 
0.077. 
[In the 2003 Provisions, the maximum limit on the stability coefficient has been replaced by a 
requirement that the stability coefficient is permitted to exceed 0.10 if and only if the resistance to lateral 
forces is determined to increase in a monotonic nonlinear static (pushover) analysis to the target 
displacement as determined in Sec. A5.2.3. P-delta effects shall be included in the analysis. Therefore, 
in this example, the stability coefficient should be evaluated directly using 2003 Provisions Eq. 5.2.-16.] 
For this P-delta analysis a (reduced) story live load of 20 psf was included in the total story weight 
calculations. Deflections are based on cracked sections, and story shears are adjusted as necessary for use 
of Provisions Eq. 5.4.1.1-3 [5.2-3]. As can be seen in the last column of each table, the stability ratio (.) 
does not exceed the maximum allowable value computed above. Moreover, since the values are less than 
0.10 at all levels, P-delta effects can be neglected for both drift and strength computed limits according to 
Provisions Sec. 5.4.6.2 [5.2.6.2]. 
An example P-delta calculation for the Level 5 under E-W loading is shown below. Note that the relevant 
row is highlighted in Table 6-10b. 
Magnified story drift = .5 = 2.613 in. 
Story shear = V5 = 1,957 kips 
Accumulated story weight P5 = 27,500 kips 
Story height = hs5= 150 in. 
Cd = 6.5 
. = (P5 (.5/Cd)) /(V5hs5) = 27,500(2.613/6.5)/(1957.1)(150) = 0.0377 < 0.077 OK 
[Note that the equation to determine the stability coefficient has been changed in the 2003 Provisions. 
The importance factor, I, has been added to 2003 Provisions Eq. 5.2-16. However, this does not affect 
this example because I = 1.0.]
Chapter 6, Reinforced Concrete 
6-23 
Table 6-10a P-Delta Computations for the Berkeley Building Loaded in the N-S Direction 
Level 
Story Drift 
(in.) 
Story Shear * 
(kips) 
Story Dead 
Load 
(kips) 
Story Live 
Load 
(kips) 
Total Story 
Load 
(kips) 
Accum. Story 
Load 
(kips) 
Stability 
Ratio 
. 
12 
11 
10
987654321 
0.478 
0.798 
1.117 
1.276 
1.515 
1.675 
1.834 
1.914 
1.914 
2.002 
2.097 
3.397 
239.9 
466.0 
657.8 
817.9 
948.7 
1052.9 
1133.3 
1192.6 
1233.8 
1259.8 
1273.8 
1279.5 
2783 
3051 
3051 
3051 
3051 
3051 
3051 
3051 
3051 
3051 
3051 
3169 
420 
420 
420 
420 
420 
420 
420 
420 
420 
420 
420 
420 
3203 
3471 
3471 
3471 
3471 
3471 
3471 
3471 
3471 
3471 
3471 
3589 
3203 
6674 
10145 
13616 
17087 
20558 
24029 
27500 
30971 
34442 
37913 
41502 
0.0077 
0.0138 
0.0209 
0.0257 
0.0331 
0.0396 
0.0471 
0.0535 
0.0582 
0.0663 
0.0757 
0.0928 
* Story shears in Table 6-7a factored by 0.725. See Sec. 6.3.1. 
1.0 in. = 25.4 mm, 1.0 kip = 4.45 kN. 
Table 6-10b P-Delta Computations for the Berkeley Building Loaded in the E-W Direction 
Level 
Story Drift 
(in.) 
Story Shear 
(kips) 
Story Dead 
Load 
(kips) 
Story Live 
Load 
(kips) 
Total Story 
Load 
(kips) 
Accum. Story 
Load 
(kips) 
Stability 
Ratio 
. 
12 
11 
10
9876 
5
4321 
1.950 
2.210 
2.210 
2.340 
2.600 
2.600 
2.652 
2.613 
2.509 
2.301 
2.002 
2.353 
350.6 
693.3 
995.9 
1259.6 
1485.9 
1676.4 
1832.9 
1957.1 
2051.3 
2117.8 
2159.4 
2180.4 
2783 
3051 
3051 
3051 
3051 
3051 
3051 
3051 
3051 
3051 
3051 
3169 
420 
420 
420 
420 
420 
420 
420 
420 
420 
420 
420 
420 
3203 
3471 
3471 
3471 
3471 
3471 
3471 
3471 
3471 
3471 
3471 
3589 
3203 
6674 
10145 
13616 
17087 
20558 
24029 
27500 
30971 
34442 
37913 
41502 
0.0183 
0.0218 
0.0231 
0.0259 
0.0307 
0.0327 
0.0357 
0.0377 
0.0389 
0.0384 
0.0361 
0.0319 
1.0 in. = 25.4 mm, 1.0 kip = 4.45 kN. 
6.3.2 Test for Torsional Irregularity for Berkeley Building 
In Sec. 6.1.3 it was mentioned that torsional irregularities are unlikely for the Berkeley building because 
the elements of the seismic-force-resisting system were well distributed over the floor area. This will now 
be verified by applying the story forces of Table 6-3a at an eccentricity equal to 5 percent of the building 
dimension perpendicular to the direction of force (accidental torsion requirement of Provisions Sec. 
5.4.4.2 [5.2.4.2]). This test is required per Provisions Sec. 5.2.3.2 [4.3.2.2]. Analysis was performed 
using the ETABS program.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
6-24 
The eccentricity is 0.05(102.5) = 5.125 ft for forces in the N-S direction and 0.05(216) = 10.8 ft in the EW 
direction. 
For forces acting in the N-S direction: 
Total displacement at center of mass = davg = 3.640 in. (see Table 6-9a) 
Rotation at center of mass = 0.000189 radians 
Maximum displacement at corner of floor plate = dmax = 3.640 + 0.000189(102.5)(12)/2 = 3.756 in. 
Ratio dmax/davg = 3.756/3.640 = 1.03 < 1.20, so no torsional irregularity exists. 
For forces acting in the E-W direction: 
Total displacement at center of mass = davg = 4.360 in. (see Table 6-9b) 
Rotation at center of mass = 0.000648 radians 
Maximum displacement at corner of floor plate = dmax = 4.360 + 0.000648(216)(12)/2 = 5.200 in. 
Ratio dmax/davg = 5.200/4.360 = 1.19 < 1.20, so no torsional irregularity exists. 
It is interesting that this building, when loaded in the E-W direction, is very close to being torsionally 
irregular (irregularity Type 1a of Provisions Table 5.2.3.2 [4.3-2]), even though the building is extremely 
regular in plan. The torsional flexibility of the building arises from the fact that the walls exist only on 
interior Gridlines 3, 4, 5, and 6. 
6.3.3 Direct Drift and P-Delta Check for the Honolulu Building 
The interstory drift computations for the Honolulu building deforming under the N-S and E-W equivalent 
static forces are shown in Tables 6-11a and 6-11b. As with the Berkeley building, the analysis used 
cracked section properties. The applied seismic forces, shown previously in Table 6-3b were multiplied 
by the ratio 0.0168/0.0207 = 0.808 to adjust for the use of Provisions Eq. 5.4.1.1-3. [As noted previously 
in Sec. 6.2, the minimum Cs value has been removed in the 2003 Provisions.] 
These tables, as well as Figure 6-6, show that the story drift at each level is less than the allowable 
interstory drift of 0.020hsx (Provisions Table 5.2.8 [4.5-1]). Even though it is not pertinent for Seismic 
Design Category C buildings, a soft first story does not exist for the Honolulu building because the ratio 
of first story to second story drift does not exceed 1.3.
Chapter 6, Reinforced Concrete 
6-25 
E-W* 
6.5 x E-W 
N-S* 
5.5 x N-S 
2% limit 
0 
20 
40 
60 
80 
100 
120 
140 
160
0.0 5.0 10.0 15.0 20.0 25.0 30.0 35.0 40.0 
Total drift, in. 
Height, ft 
* Elasticlly computed under code-prescribed seismic forces 
Figure 6-6 Drift profile for the Honolulu building (1.0 ft = 0.3048 m, 
1.0 in. = 25.4 mm).
FEMA 451, NEHRP Recommended Provisions: Design Examples 
6-26 
Table 6-11a Drift Computations for the Honolulu Building Loaded in the N-S Direction 
Story 
Total Drift 
(in.) 
Story Drift 
(in.) 
Story Drift  Cd 
* 
(in.) 
Drift Ratio 
(%) 
12 
11 
10 
9 
8 
7 
6 
5 
4 
3 
2 
1 
1.766 
1.726 
1.656 
1.559 
1.441 
1.306 
1.157 
0.997 
0.829 
0.658 
0.482 
0.297 
0.040 
0.069 
0.097 
0.118 
0.136 
0.149 
0.160 
0.168 
0.171 
0.176 
0.184 
0.297 
0.182 
0.313 
0.436 
0.531 
0.611 
0.669 
0.720 
0.756 
0.771 
0.793 
0.829 
1.338 
0.121 
0.208 
0.291 
0.354 
0.407 
0.446 
0.480 
0.504 
0.514 
0.528 
0.553 
0.619 
* Cd = 4.5 for loading in this direction; total drift is at top of story, story height = 150 in. for Levels 3 
through roof and 216 in. for Level 2. 
1.0 in. = 25.4 mm. 
Table 6-11b Drift Computations for the Honolulu Building Loaded in the E-W Direction 
Story 
Total Drift 
(in.) 
Story Drift 
(in.) 
Story Drift  Cd * 
(in.) 
Drift Ratio 
(%) 
12 
11 
10 
9 
8 
7 
6 
5 
4 
3 
2 
1 
2.002 
1.941 
1.850 
1.734 
1.597 
1.440 
1.269 
1.089 
0.903 
0.713 
0.522 
0.325 
0.061 
0.090 
0.116 
0.137 
0.157 
0.171 
0.179 
0.186 
0.191 
0.191 
0.197 
0.325 
0.276 
0.407 
0.524 
0.618 
0.705 
0.772 
0.807 
0.836 
0.858 
0.858 
0.887 
1.462 
0.184 
0.271 
0.349 
0.412 
0.470 
0.514 
0.538 
0.558 
0.572 
0.572 
0.591 
0.677 
* Cd = 4.5 for loading in this direction; total drift is at top of story, story height = 150 in. for Levels 3 
through roof and 216 in. for Level 2. 
1.0 in. = 25.4 mm. 
A sample calculation for Level 5 of Table 6-11b (highlighted in the table) is as follows: 
Deflection at top of story = d5e =1.089 in. 
Deflection at bottom of story = d4e = 0.903 in. 
Story drift = .5e = d5e - d4e = 1.089-0.0903 = 0.186 in. 
Deflectiom amplification factor, Cd = 4.5 
Importance factor, I = 1.0 
Magnified story drift = .5 = Cd .5e/I = 4.5(0.186)/1.0 = 0.836 in. 
Magnified drift ratio = .5 / h5 = (0.836/150) = 0.00558 = 0.558% < 2.0% OK
Chapter 6, Reinforced Concrete 
6-27 
Therefore, story drift satisfies the drift requirements. 
Calculations for P-delta effects are shown in Tables 6-12a and 6-12b for N-S and E-W loading, 
respectively. The stability ratio at the 5th story from Table 6-12b is computed: 
Magnified story drift = .5 = 0.836 in. 
Story shear = V5 = 571.9 = kips 
Accumulated story weight P5 = 27500 kips 
Story height = hs5 = 150 in. 
Cd = 4.5 
. = [P5 (.5/Cd)]/(V5hs5) = 27500(0.836/4.5)/(571.9)(150) = 0.0596 
[Note that the equation to determine the stability coefficient has been changed in the 2003 Provisions. 
The importance factor, I, has been added to 2003 Provisions Eq. 5.2-16. However, this does not affect 
this example because I = 1.0.] 
The requirements for maximum stability ratio (0.5/Cd = 0.5/4.5 = 0.111) are satisfied. Because the 
stability ratio is less than 0.10 at all floors, P-delta effects need not be considered (Provisions Sec. 5.4.6.2 
[5.2.6.2]). (The value of 0.1023 in the first story for the E-W direction is considered by the author to be 
close enough to the criterion.) 
Table 6-12a P-Delta Computations for the Honolulu Building Loaded in the N-S Direction 
Level 
Story Drift 
(in.) 
Story Shear * 
(kips) 
Story Dead 
Load 
(kips) 
Story Live 
Load 
(kips) 
Total Story 
Load 
(kips) 
Accum. Story 
Load 
(kips) 
Stability 
Ratio 
. 
12 
11 
10 
9 
8 
7 
6 
5 
4 
3 
2 
1 
0.182 
0.313 
0.436 
0.531 
0.611 
0.669 
0.720 
0.756 
0.771 
0.793 
0.829 
1.338 
117.7 
227.7 
320.4 
396.9 
458.9 
507.7 
544.9 
571.9 
590.3 
601.6 
607.6 
609.8 
2783 
3051 
3051 
3051 
3051 
3051 
3051 
3051 
3051 
3051 
3051 
3169 
420 
420 
420 
420 
420 
420 
420 
420 
420 
420 
420 
420 
3203 
3471 
3471 
3471 
3471 
3471 
3471 
3471 
3471 
3471 
3471 
3589 
3203 
6674 
10145 
13616 
17087 
20558 
24029 
27500 
30971 
34442 
37913 
41502 
0.0073 
0.0136 
0.0205 
0.0270 
0.0337 
0.0401 
0.0470 
0.0539 
0.0599 
0.0672 
0.0766 
0.0937 
* Story shears in Table 6-8 factored by 0.808. See Sec. 6.3.3. 
1.0 in. = 25.4 mm, 1.0 kip = 4.45 kN.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
6-28 
Table 6-12b P-Delta Computations for the Honolulu Building Loaded in the E-W Direction 
Level 
Story Drift 
(in.) 
Story Shear * 
(kips) 
Story Dead 
Load 
(kips) 
Story Live 
Load 
(kips) 
Total Story 
Load 
(kips) 
Accum. Story 
Load 
(kips) 
Stability 
Ratio 
. 
12 
11 
10 
9 
8 
7 
6 
5 
4 
3 
2 
1 
0.276 
0.407 
0.524 
0.618 
0.705 
0.772 
0.807 
0.836 
0.858 
0.858 
0.887 
1.462 
117.7 
227.7 
320.4 
396.9 
458.9 
507.7 
544.9 
571.9 
590.3 
601.6 
607.6 
609.8 
2783 
3051 
3051 
3051 
3051 
3051 
3051 
3051 
3051 
3051 
3051 
3169 
420 
420 
420 
420 
420 
420 
420 
420 
420 
420 
420 
420 
3203 
3471 
3471 
3471 
3471 
3471 
3471 
3471 
3471 
3471 
3471 
3589 
3203 
6674 
10145 
13616 
17087 
20558 
24029 
27500 
30971 
34442 
37913 
41502 
0.0111 
0.0177 
0.0246 
0.0314 
0.0389 
0.0463 
0.0527 
0.0596 
0.0667 
0.0728 
0.0820 
0.1023 
* Story shears in Table 6-8 factored by 0.808. See Sec. 6.3.3. 
1.0 in. = 25.4 mm, 1.0 kip = 4.45 kN. 
6.3.4 Test for Torsional Irregularity for the Honolulu Building 
A test for torsional irregularity for the Honolulu building can be performed in a manner similar to that for 
the Berkeley building. However, it is clear that a torsional irregularity will not occur for the Honolulu 
building if the Berkeley building is not irregular. This will be the case because the walls, which draw the 
torsional resistance towards the center of the Berkeley building, do not exist in the Honolulu building. 
6.4 STRUCTURAL DESIGN OF THE BERKELEY BUILDING 
6.4.1 Material Properties 
For the Berkeley building, sand-LW aggregate concrete of 4,000 psi strength is used everywhere except 
for the lower two stories of the structural walls where 6,000 psi NW concrete is used. All reinforcement 
has a specified yield strength of 60 ksi, except for the panel of the structural walls which contains 40 ksi 
reinforcement. This reinforcement must conform to ASTM A706. According to ACI 318 Sec. 21.2.5, 
however, ASTM A615 reinforcement may be used if the actual yield strength of the steel does not exceed 
the specified strength by more than 18 ksi and the ratio of actual ultimate tensile stress to actual tensile 
yield stress is greater than 1.25. 
6.4.2 Combination of Load Effects 
Using the ETABS program, the structure was analyzed for the equivalent lateral loads shown in Tables 
6-7a and 6-7b. For strength analysis, the loads were applied at a 5 percent eccentricity as required for 
accidental torsion by Provisions Sec. 5.4.4.2 [5.2.4.2]. Where applicable, orthogonal loading effects were 
included per Provisions Sec. 5.2.5.2.3 [4.4.2.3]. The torsional magnification factor (Ax) given by 
Provisions Eq. 5.4.4.3-1 [5.2-13] was not used because the building has no significant plan irregularities. 
Provisions Sec. 5.2.7 [4.2.2.1] and Eq. 5.2.7-1 and 5.2.7-2 [4.2-1 and 4.2-2] require combination of load 
effects be developed on the basis of ASCE 7, except that the earthquake load effect, E, be defined as: 
Chapter 6, Reinforced Concrete 
6-29 
E=.QE+0.2SDSD 
when gravity and seismic load effects are additive and 
E=.QE-0.2SDSD 
when the effects of seismic load counteract gravity. 
The special load combinations given by Provisions Eq. 5.2.7-1 and 5.2.7-2 [4.2-3 and 4.2-4] do not apply 
to the Berkeley building because there are no discontinuous elements supporting stiffer elements above 
them. (See Provisions Sec. 9.6.2 [9.4.1].) 
The reliability factor (.) in Eq. 5.2.7-1 and 5.2.7-2 [not applicable in the 2003 Provisions] should be 
taken as the maximum value of .x defined by Provisions Eq. 5.2.4.2: 
2 20 
x 
x 
rmax Ax 
. = - 
where Ax is the area of the floor or roof diaphragm above the story under consideration and is the rmaxx 
largest ratio of the design story shear resisted by a single element divided by the total story shear for a 
given loading. The computed value for . must be greater than or equal to 1.0, but need not exceed 1.5. 
Special moment frames in Seismic Design Category D are an exception and must be proportioned such 
that . is not greater than 1.25. 
For the structure loaded in the N-S direction, the structural system consists of special moment frames, and 
rix is taken as the maximum of the shears in any two adjacent columns in the plane of a moment frame 
divided by the story shear. For interior columns that have girders framing into both sides, only 70 percent 
of the individual column shear need be included in this sum. In the N-S direction, there are four identical 
frames. Each of these frames has eight columns. Using the portal frame idealization, the shear in an 
interior column will be Vinterior = 0.25 (2/14) V = 0.0357V. 
Similarly, the shear in an exterior column will be Vexterior= 0.25 (1/14) V = 0.0179V. 
For two adjacent interior columns: 
0.7( int int ) 0.7(0.0375 0.0375 ) 0.0525 
ix 
r V V V V 
V V 
+ + 
= = = 
For one interior and one exterior column: 
(0.7 int ext ) 0.7(0.0375 ) 0.0179 ) 0.0441 
ix 
r V V V V 
V V 
+ + 
= = = 
The larger of these values will produce the largest value of .x. Hence, for a floor diaphragm area Ax equal 
to 102.5  216 = 22,140 square ft: 
2 20 0.56 
0.0525 22,140 . x = - = -
FEMA 451, NEHRP Recommended Provisions: Design Examples 
6-30 
As this value is less than 1.0, . will be taken as 1.0 in the N-S direction. 
For seismic forces acting in the E-W direction, the walls carry significant shear, and for the purposes of 
computing ., it will be assumed that they take all the shear. According to the Provisions, rix for walls is 
taken as the shear in the wall multiplied by 10/lw and divided by the story shear. The term lw represents 
the plan length of the wall in feet. Thus, for one wall: 
maxx ix 0.25 (10 / 20) 0.125 
r r V
V 
= = = 
Only 80 percent of the . value based on the above computations need be used because the walls are part 
of a dual system. Hence, in the E-W direction 
0.8 2 20 0.740 
0.125 22,140 . x 
. . 
= ..- ..= 
. . 
and as with the N-S direction, . may be taken as 1.0. Note that . need not be computed for the columns 
of the frames in the dual system, as this will clearly not control. 
[The redundancy requirements have been substantially changed in the 2003 Provisions. For a building 
assigned to Seismic Design Category D, . = 1.0 as long as it can be shown that failure beam-to-column 
connections at both ends of a single beam (moment frame system) or failure of a single shear wall with 
aspect ratio greater than 1.0 (shear wall system) would not result in more than a 33 percent reduction in 
story strength or create an extreme torsional irregularity. Alternatively, if the structure is regular in plan 
and there are at least 2 bays of perimeter framing on each side of the structure in each orthogonal 
direction, it is permitted to use, . = 1.0. Per 2003 Provisions Sec. 4.3.1.4.3 special moment frames in 
Seismic Design Category D must be configured such that the structure satisfies the criteria for . = 1.0. 
There are no reductions in the redundancy factor for dual systems. Based on the preliminary design, . = 
1.0 for because the structure has a perimeter moment frame and is regular.] 
For the Berkeley structure, the basic ASCE 7 load combinations that must be considered are: 
1.2D + 1.6L 
1.2D + 0.5L  1.0E 
0.9D  1.0E 
The ASCE 7 load combination including only 1.4 times dead load will not control for any condition in 
this building. 
Substituting E from the Provisions, with . taken as 1.0, the following load combinations must be used for 
earthquake: 
(1.2 + 0.2SDS)D + 0.5L + E 
(1.2 + 0.2SDS)D + 0.5L - E 
(0.9 - 0.2 SDS)D + E 
(0.9 - 0.2SDS)D - E 
Finally, substituting 1.10 for SDS, the following load combinations must be used for earthquake: 
1.42D + 0.5L + E 
1.42D + 0.5L - E
Chapter 6, Reinforced Concrete 
2The analysis used to create Figures 6-7 and 6-8 did not include the 5 percent torsional eccentricity or the 30 percent orthogonal 
loading rules specified by the Provisions. The eccentricity and orthogonal load were included in the analysis carried out for 
member design. 
6-31 
0.68D + E 
0.68D - E 
It is very important to note that use of the ASCE 7 load combinations in lieu of the combinations given in 
ACI Chapter 9 requires use of the alternate strength reduction factors given in ACI 318 Appendix C: 
Flexure without axial load f = 0.80 
Axial compression, using tied columns f = 0.65 (transitions to 0.8 at low axial loads) 
Shear if shear strength is based on nominal axial-flexural capacity f = 0.75 
Shear if shear strength is not based on nominal axial-flexural capacity f = 0.55 
Shear in beam-column joints f = 0.80 
[The strength reduction factors in ACI 318-02 have been revised to be consistent with the ASCE 7 load 
combinations. Thus, the factors that were in Appendix C of ACI 318-99 are now in Chapter 9 of ACI 
318-02, with some modification. The strength reduction factors relevant to this example as contained in 
ACI 318-02 Sec. 9.3 are: 
Flexure without axial load f = 0.9 (tension-controlled sections) 
Axial compression, using tied columns f = 0.65 (transitions to 0.9 at low axial loads) 
Shear if shear strength is based o nominal axial-flexural capacity f = 0.75 
Shear if shear strength is not based o nominal axial-flexural capacity f = 0.60 
Shear in beam-column joints f = 0.85] 
6.4.3 Comments on the Structures Behavior Under E-W Loading 
Frame-wall interaction plays an important role in the behavior of the structure loaded in the E-W 
direction. This behavior is beneficial to the design of the structure because: 
1. For frames without walls (Frames 1, 2, 7, and 8), the shears developed in the girders (except for the 
first story) do not differ greatly from story to story. This allows for a uniformity in the design of the 
girders. 
2. For frames containing structural walls (Frames 3 through 6), the overturning moments in the 
structural walls are reduced significantly as a result of interaction with the remaining frames (Frames 
1, 2, 7, and 8). 
3. For the frames containing structural walls, the 40-ft-long girders act as outriggers further reducing the 
overturning moment resisted by the structural walls. 
The actual distribution of story forces developed in the different frames of the structure is shown in 
Figure 6-7.2 This figure shows the response of Frames 1, 2, and 3 only. By symmetry, Frame 8 is similar 
to Frame 1, Frame 7 is similar to Frame 2, and Frame 6 is similar to Frame 3. Frames 4 and 5 have a 
response that is virtually identical to that of Frames 3 and 6. 
As may be observed from Figure 6-7, a large reverse force acts at the top of Frame 3 which contains a 
structural wall. This happens because the structural wall pulls back on (supports) the top of Frame 1. The 
deflected shape of the structure loaded in the E-W direction (see Figure 6-5) also shows the effect of 
frame-wall interaction because the shape is neither a cantilever mode (wall alone) nor a shear mode
FEMA 451, NEHRP Recommended Provisions: Design Examples 
6-32 
-58.4 
-8.58 
-8.68 
-2.88 
0.96 
4.26 
6.76 
8.62 
8.72 
13.28 
-18.96 
107.8 
1
2
3
4
5
6
7
8
9 
10 
1 
12 
Story force, kips 
-114.9
-30.1 
-20.18 
-8.18 
1.64 
9.56 
15.74 
19.98 
21.48 
26.56 
-29.12 
255.7 
1
2
3
4
5
6
7
8
9 
10 
11 
12 
91.96 
31.79 
34.77 
34.37 
36.26 
39.88 
45.14 
52.14 
61.29 
66.71 
120.88 
-77.67 
1
2
3
4
5
6
7
8
9 
10 
1 
12 
Story force, kips Story force, kips 
Frame 1 Frame 2 Frame 3 (includes wall) 
Figure 6-7 Story forces in the E-W direction (1.0 kip = 4.45 kN). 
(frame alone). It is the straightening out of the deflected shape of the structure that causes the story 
shears in the frames without walls to be relatively equal. 
A plot of the story shears in Frames 1, 2, and 3 is shown in Figure 6-8. The distribution of overturning 
moments is shown in Figure 6-9 and indicates that the relatively stiff Frames 1 and 3 resist the largest 
portion of the total overturning moment. The reversal of moment at the top of Frame 3 is a typical 
response characteristic of frame-wall interaction. 
6.4.4 Analysis of Frame-Only Structure for 25 Percent of Lateral Load 
When designing a dual system, Provisions Sec. 5.2.2.1 [4.3.1.1] requires the frames (without walls) to 
resist at least 25 percent of the total base shear. This provision ensures that the dual system has sufficient 
redundancy to justify the increase from R = 6 for a special reinforced concrete structural wall to R = 8 for 
a dual system (see Provisions Table 5.2.2 [4.3-1]). [Note that R = 7 per 2003 Provisions Table 4.3-1.] 
The 25 percent analysis was carried out using the ETABS program with the mathematical model of the 
building being identical to the previous version except that the panels of the structural wall were removed. 
The boundary elements of the walls were retained in the model so that behavior of the interior frames 
(Frames 3, 4, 5, and 6) would be analyzed in a rational way. 
The results of the analysis are shown in Figures 6-10, 6-11, and 6-12. In these figures, the original 
analysis (structural wall included) is shown by a solid line and the 25 percent (backup frame) analysis 
(structural wall removed) is shown by a dashed line. As can be seen, the 25 percent rule controls only at 
the lower level of the building.
Chapter 6, Reinforced Concrete 
6-33 
0 
20 
40 
60 
80 
100 
120 
140 
160 
-200 -100 0 100 200 300 400 500 600 
Shear, kips 
Height, ft 
Frame 1 
Frame 2 
Frame 3 
Figure 6-8 Story shears in the E-W direction (1.0 ft = 0.3048 m, 1.0 kip = 4.45 kN). 
0 
20 
40 
60 
80 
100 
120 
140 
160
-10,000 0 10,000 20,000 30,000 40,000 50,000 
Bending moment, ft-kips 
Height, ft 
Frame 1 
Frame 2 
Frame 3 
Figure 6-9 Story overturning moments in the E-W direction (1.0 ft = 0.3048 m, 1.0 ft-kip = 1.36 kN-m).
FEMA 451, NEHRP Recommended Provisions: Design Examples 
6-34 
0 
20 
40 
60 
80 
100 
120 
140 
160 
0 50 100 150 200 250 300 350 
Shear, kips 
Height, ft 
25% V analysis for Frame 1 
Frame 1 
Figure 6-10 25 percent story shears, Frame 1 E-W direction (1.0 ft = 0.3048 
m, 1.0 kip = 4.45 kN).
Chapter 6, Reinforced Concrete 
6-35 
0 
20 
40 
60 
80 
100 
120 
140 
160 
0 50 100 150 200 250 300 350 
Shear, kips 
Height, ft 
25% V analysis for Frame 2 
Frame 2 
Figure 6-11 25 percent story shears, Frame 2 E-W direction (1.0 ft = 0.3048 
m, 1.0 kip = 4.45 kN).
FEMA 451, NEHRP Recommended Provisions: Design Examples 
6-36 
0 
20 
40 
60 
80 
100 
120 
140 
160 
0 50 100 150 200 250 300 350 
Shear, kips 
Height, ft 
25% V analysis for Frame 3 
Frame 3 (without panels) 
Figure 6-12 25 percent story shear, Frame 3 E-W direction (1.0 ft = 0.3048 m, 
1.0 kip = 4.45 kN).. 
6.4.5 Design of Frame Members for the Berkeley Building 
A sign convention for bending moments is required in flexural design. In this example, when the steel at 
the top of a beam section is in tension, the moment is designated as a negative moment. When the steel at 
the bottom is in tension, the moment is designated as a positive moment. All moment diagrams are drawn 
using the reinforced concrete or tension-side convention. For beams, this means negative moments are 
plotted on the top and positive moments are plotted on the bottom. For columns, moments are drawn on 
the tension side of the member. 
6.4.5.1 Initial Calculations 
Before the quantity and placement of reinforcement is determined, it is useful to establish, in an overall 
sense, how the reinforcement will be distributed. The preliminary design established that beams would 
have a maximum depth of 32 in. and columns would be 30 in. by 30 in. In order to consider the 
beam-column joints confined per ACI 318 Sec. 21.5, it was necessary to set the beam width to 22.5 in., 
which is 75 percent of the column width. 
In order to determine the effective depth used for the design of the beams, it is necessary to estimate the 
size and placement of the reinforcement that will be used. In establishing this depth, it is assumed that #8 
bars will be used for longitudinal reinforcement and that hoops and stirrups will be constructed from #3 
deformed bars. In all cases, clear cover of 1.5 in. is assumed. Since this structure has beams spanning in
Chapter 6, Reinforced Concrete 
6-37 
2'-6" 
1.5" cover 
#8 bar 
#3 hoop 
East-west 
spanning beam 
2'-8" 
2'-51
2" 
2'-41
2" 
North-south 
spanning beam 
Figure 6-13 Layout for beam reinforcement (1.0 ft = 0.3048 m, 1.0 in = 25.4 
mm). 
two orthogonal directions, it is necessary to layer the flexural reinforcement as shown in Figure 6-13. The 
reinforcement for the E-W spanning beams was placed in the upper and lower layers because the strength 
demand for these members is somewhat greater than that for the N-S beams. 
Given Figure 6-13, compute the effective depth for both positive and negative moment as: 
Beams spanning in the E-W direction, d = 32 - 1.5 - 0.375 - 1.00/2 = 29.6 in. 
Beams spanning in the N-S direction, d = 32 - 1.5 - 0.375 - 1.0 - 1.00/2 = 28.6 in. 
For negative moment bending, the effective width is 22.5 in. for all beams. For positive moment, the slab 
is in compression and the effective T-beam width varies according to ACI 318 Sec. 8.10. The effective 
widths for positive moment are as follows (with the parameter controlling effective width shown in 
parentheses): 
20-ft beams in Frames 1 and 8 b = 22.5 + 20(12)/12 = 42.5 in. (span length) 
Haunched beams b = 22.5 + 2[8(4)] = 86.5 in. (slab thickness) 
30-ft beams in Frames A, B, C, and D b = 22.5 + [6(4)] = 46.5 in. (slab thickness) 
ACI 318 Sec. 21.3.2 controls the longitudinal reinforcement requirements for beams. The minimum 
reinforcement to be provided at the top and bottom of any section is: 
2 
, 
200 200(22.5)29.6 2.22 in. 
60,000 
w 
s min 
y 
A b d 
f 
= = =
FEMA 451, NEHRP Recommended Provisions: Design Examples 
6-38 
This amount of reinforcement can be supplied by three #8 bars with As = 2.37 in.2 Since the three #8 bars 
will be provided continuously top and bottom, reinforcement required for strength will include these #8 
bars. 
Before getting too far into member design, it is useful to check the required tension development length 
for hooked bars since the required length may control the dimensions of the columns and the boundary 
elements of the structural walls. 
From Eq. 21-6 of ACI 318 Sec. 21.5.4.1, the required development length is: 
65
y b 
dh 
c 
f d 
l 
f 
= 
' 
For NW concrete, the computed length should not be less than 6 in. or 8db. For LW concrete, the 
minimum length is the larger of 1.25 times that given by ACI 318 Eq. 21-6, 7.5 in., or 10db. For fc' = 
4,000 psi LW concrete, ACI 318 Eq. 21-6 controls for #3 through #11 bars. 
For straight top bars, ld = 3.5ldh and for straight bottom bars, ld = 2.5ldh. These values are applicable 
only when the bars are anchored in well confined concrete (e.g., column cores and plastic hinge regions 
with confining reinforcement). The development length for the portion of the bar extending into 
unconfined concrete must be increased by a factor of 1.6. Development length requirements for hooked 
and straight bars are summarized in Table 6-13. 
Where hooked bars are used, the hook must be 90 degrees and be located within the confined core of the 
column or boundary element. For bars hooked into 30-in.-square columns with 1.5 in. of cover and #4 
ties, the available development length is 30 - 1.50 - 0.5 = 28.0 in. With this amount of available length, 
there will be no problem developing hooked bars in the columns. As required by ACI 318 Sec. 12.5, 
hooked bars have a 12db extension beyond the bend. ACI 318 Sec. 7.2 requires that #3 through #8 bars 
have a 6db bend diameter and #9 through #11 bars have a 8db diameter. 
Table 6-13 is applicable to bars anchored in joint regions only. For development of bars outside of joint 
regions, ACI 318 Chapter 12 should be used. 
Table 6-13 Tension Development Length Requirements for Hooked Bars 
and Straight Bars in 4,000 psi LW Concrete 
Bar Size db (in.) ldh hook (in.) ld top (in.) ld bottom (in.) 
#4 
#5 
#6 
#7 
#8 
#9 
#10 
#11 
0.500 
0.625 
0.750 
0.875 
1.000 
1.128 
1.270 
1.410 
9.1 
11.4 
13.7 
16.0 
18.2 
20.6 
23.2 
25.7 
31.9 
39.9 
48.0 
56.0 
63.7 
72.1 
81.2 
90.0 
22.8 
28.5 
34.3 
40.0 
45.5 
51.5 
58.0 
64.2 
1.0 in. = 25.4 mm.
Chapter 6, Reinforced Concrete 
3See Chapter 1 of the 2nd Edition of the Handbook of Concrete Engineering edited by Mark Fintel (New York: Van Nostrand 
Reinhold Company, 1984). 
6-39 
6.4.5.2 Design of Members of Frame 1 for E-W Loading 
For the design of the members of Frame 1, the equivalent lateral forces of Table 6-7b were applied at an 
eccentricity of 10.5 ft together with 30 percent of the forces of Table 6-7a applied at an eccentricity of 5.0 
ft. The eccentricities were applied in such a manner as to maximize torsional response and produce the 
largest shears in Frame 1. 
For this part of the example, the design and detailing of all five beams and one interior column of Level 5 
are presented in varying amounts of detail. The beams are designed first because the flexural capacity of 
the as-designed beams is a factor in the design and detailing of the column and the beam-column joint. 
The design of a corner column will be presented later. 
Before continuing with the example, it should be mentioned that the design of ductile reinforced concrete 
moment frame members is dominated by the flexural reinforcement in the beams. The percentage and 
placement of beam flexural reinforcement governs the flexural rebar cutoff locations, the size and spacing 
of beam shear reinforcement, the cross-sectional characteristics of the column, the column flexural 
reinforcement, and the column shear reinforcement. The beam reinforcement is critical because the basic 
concept of ductile frame design is to force most of the energy-absorbing deformation to occur through 
inelastic rotation in plastic hinges at the ends of the beams. 
In carrying out the design calculations, three different flexural strengths were used for the beams. These 
capacities were based on: 
Design strength f = 0.8, tensile stress in reinforcement at 1.00 fy 
Nominal strength f = 1.0, tensile stress in reinforcement at 1.00 fy 
Probable strength f = 1.0, tensile stress in reinforcement at 1.25 fy 
Various aspects of the design of the beams and other members depend on the above capacities as follows: 
Beam rebar cutoffs Design strength 
Beam shear reinforcement Probable strength of beam 
Beam-column joint strength Probable strength of beam 
Column flexural strength 6/5  nominal strength of beam 
Column shear strength Probable strength of column 
In addition, beams in ductile frames will always have top and bottom longitudinal reinforcement 
throughout their length. In computing flexural capacities, only the tension steel will be considered. This 
is a valid design assumption because reinforcement ratios are quite low, yielding a depth to the neutral 
axis similar to the depth of the compression reinforcement (d'/d is about 0.08, while the neutral axis depth 
at ultimate ranges from 0.07 to 0.15 times the depth) .3 
The preliminary design of the girders of Frame 1 was based on members with a depth of 32 in. and a 
width of 22.5 in. The effective depth for positive and negative bending is 29.6 in. and the effective 
widths for positive and negative bending are 42.5 and 22.5 in., respectively. This assumes the stress 
block in compression is less than the 4.0-inch flange thickness. 
The layout of the geometry and gravity loading on the three eastern-most spans of Level 5 of Frame 1 as 
well as the unfactored gravity and seismic moments are illustrated in Figure 6-14. The seismic moments 
are taken directly from the ETABS program output and the gravity moments were computed by hand
FEMA 451, NEHRP Recommended Provisions: Design Examples 
6-40 
4,515 
4,515 
4,708 
4,635 
4,457 
3,988 
492 
786 
562 492 
715 715 715 715 
492 
152 
173 
242 221 221 
152 152 
221 221 
5,232 5,225 
951 
4,122 3,453 
834 
5,834 5,761 
4,222 4,149 
834 
5,641 
4,028 
5,641 
4,028 
1.42D + 0.5L + E 
0.68D - E 
1.2D+1.6L - midspan 
(a) 
Span layout 
and loading 
(b) 
Earthquake moment 
(in.-kips) 
(c) 
Unfactored DL moment 
(in.-kips) 
(d) 
Unfactored LL moment 
(in.-kips) 
(e) 
Required strength 
envelopes (in.-kips) 
17'-6" 
20'-0" 20'-0" 
20'-0" 
' 
Figure 6-14 Bending moments for Frame 1 (1.0 ft = 0.3048 m, 1.0 in.-kip = 0.113 kN-m). 
using the coefficient method of ACI 318 Chapter 8. Note that all moments (except for midspan positive 
moment) are given at the face of the column and that seismic moments are considerably greater than those 
due to gravity. 
Factored bending moment envelopes for all five spans are shown in Figure 6-14. Negative moment at the 
supports is controlled by the 1.42D + 0.5L + 1.0E load combination, and positive moment at the support 
is controlled by 0.68D - 1.0E. Midspan positive moments are based on the load combination 1.2D + 1.6L. 
The design process is illustrated below starting with Span B-C. 
6.4.5.2.1 Span B-C 
1. Design for Negative Moment at the Face of the Support 
Mu = 1.42(-715) + 0.5(-221) + 1.0(-4515) = -5,641 in.-kips 
Try two #9 bars in addition to the three #8 bars required for minimum steel: 
As = 2(1.0) + 3(0.79) = 4.37 in.2 
fc' = 4,000 psi 
fy = 60 ksi
Chapter 6, Reinforced Concrete 
6-41 
Width b for negative moment = 22.5 in. 
d = 29.6 in. 
Depth of compression block, a = Asfy/.85fc'b 
a = 4.37 (60)/[0.85 (4) 22.5] = 3.43 in. 
Design strength, fMn = fAsfy(d - a/2) 
fMn = 0.8(4.37)60(29.6 - 3.43/2) = 5,849 in.-kips > 5,641 in.-kips OK 
2. Design for Positive Moment at Face of Support 
Mu = [-0.68(715)] + [1.0(4,515)] = 4,028 in.-kips 
Try two #7 bars in addition to the three #8 bars already provided as minimum steel: 
As = [2(0.60)] + [3(0.79)] = 3.57 in. 2 
Width b for positive moment = 42.5 in. 
d = 29.6 in. 
a = [3.57(60)]/[0.85(4)42.5] = 1.48 in. 
fMn = 0.8(3.57) 60(29.6 - 1.48/2) = 4,945 in.-kips > 4,028 in.-kips OK 
3. Positive Moment at Midspan 
Mu = [1.2(492)] + [1.6(152)] = 833.6 in.-kips 
Minimum reinforcement (three #8 bars) controls by inspection. This positive moment reinforcement will 
also work for Spans A'-B and A-A'. 
6.4.5.2.2 Span A'-B 
1. Design for Negative Moment at the Face of Support A' 
Mu = [1.42(-715)] + [0.5(-221)] + [1.0(-4,708)] = -5,834 in.-kips 
Three #8 bars plus two #9 bars (capacity = 5,849 in.-kips) will work as shown for Span B-C. 
2. Design for Negative Moment at the Face of Support B 
Mu = [1.42(-715)] + [0.5(-221)] + [1.0(-4,635)] = -5,761 in.-kips 
As before, use three #8 bars plus two #9 bars. 
3. Design for Positive Moment at Face of Support A' 
Mu = [-0.68(715)] + [1.0(4708)] = 4,222 in.-kips 
Three #8 bars plus two #7 bars (capacity = 4,945 in.-kips) works as shown for Span B-C. 
4. Design for Positive Moment at Face of Support B' 
Mu = [-0.68(715)] + [1.0(4,635)] = 4,149 in.-kips 
As before, use three #8 bars plus two #7 bars.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
6-42 
6.4.5.2.3 Span A-A' 
1. Design for Negative Moment at the Face of Support A 
Mu = [1.42(-492)] + [0.5(-152)] + [1.0(-4,457)] = -5,232 in.-kips 
Try three #8 bars plus two #8 bars: 
As = 5  0.79 = 3.95 in.2 
Width b for negative moment = 22.5 in. 
d = 29.6 in. 
a = [3.95(60)/[0.85(4)22.5] = 3.10 in. 
fMn =[0.8(3.95)60] (29.6 - 3.10/2) = 5,318 in.-kips > 5,232 in.-kips OK 
2. Design for Negative Moment at the Face of Support A' 
Mu = [1.42(-786)] + [0.5(-242)] + [1.0(-3,988)] = -5,225 in.-kips 
Use three #8 bars plus two #9 bars as required for Support B of Span A'-B. 
3. Design for Positive Moment at Face of Support A 
Mu = [-0.68(492)] + [1.0(4,457)] = 4,122 in.-kips 
Three #8 bars plus two #7 bars will be sufficient. 
4. Design for Positive Moment at Face of Support A' 
Mu = [-0.68(786)] + [1.0(3,988)] = 3,453 in.-kips 
As before, use three #8 bars plus two #7 bars. 
6.4.5.2.4 Spans C-C' and C'-D 
Reinforcement requirements for Spans C-C' and C'-D are mirror images of those computed for Spans 
A'-B and A-A', respectively. 
In addition to the computed strength requirements and minimum reinforcement ratios cited above, the 
final layout of reinforcing steel also must satisfy the following from ACI 318 Sec. 21.3.2: 
Minimum of two bars continuous top and bottom OK (three #8 bars continuous top and bottom) 
Positive moment strength greater than OK (at all joints) 
50 percent negative moment strength at a joint 
Minimum strength along member greater OK (As provided = three #8 bars is more than 
than 0.25 maximum strength 25 percent of reinforcement provided at joints) 
The preliminary layout of reinforcement is shown in Figure 6-15. The arrangement of bars actually 
provided is based on the above computations with the exception of Span B-C where a total of six #8 top 
bars were used instead of the three #8 bars plus two #9 bars combination. Similarly, six #8 bars are used 
at the bottom of Span B-C. The use of six #8 bars is somewhat awkward for placing steel, but it allows
Chapter 6, Reinforced Concrete 
6-43 
' 
Note: 
Drawing not to scale 
' 
(2) #8 
(3) #8 
(2) #9 
(2) #7 
(3) #8 
(2) #7 
(3) #8 (3) #8 
(2) #8 
(2) #9 
(2) #7 
20'-0" 
(2) #7 
(3) #8 
2'-8" 
2'-6" 
(typical) 
Figure 6-15 Preliminary rebar layout for Frame 1 (1.0 ft = 03.048 m). 
for the use of three #8 continuous top and bottom at all spans. An alternate choice would have been to 
use two #9 continuous across the top of Span B-C instead of the three of the #8 bars. However, the use of 
two #9 bars (. = 0.00303) does not meet the minimum reinforcement requirement .min = 0.0033. 
As mentioned above, later phases of the frame design will require computation of the design strength and 
the maximum probable strength at each support. The results of these calculations are shown in Table 
6-14. 
Table 6-14 Design and Maximum Probable Flexural Strength For Beams in Frame 1 
Item 
Location 
A A' B C C' D 
Negative 
Moment 
Reinforcement five #8 three #8 + 
two #9 
six #8 six #8 three #8 + 
two #9 
five #8 
Design Strength 
(in.-kips) 5,318 5,849 6,311 7,100 5,849 5,318 
Probable Strength 
(in.-kips) 8,195 8,999 9,697 9,697 8,999 8,195 
Positive 
Moment 
Reinforcement three #8 + 
two #7 
three #8 + 
two #7 
six #8 six #8 three #8 + 
two #7 
three #8 + 
two #7 
Design Strength 
(in.-kips) 4,945 4,945 6,510 6,510 4,945 4,945 
Probable Strength 
(in.-kips) 7,677 7,677 10,085 10,085 7,655 7,677 
1.0 in.-kip = 0.113 kN-m. 
As an example of computation of probable strength, consider the case of six #8 top bars: 
As = 6(0.79) = 4.74 in.2 
Width b for negative moment = 22.5 in. 
d = 29.6 in. 
Depth of compression block, a = As(1.25fy)/0.85fc'b 
a = 4.74(1.25)60/[0.85(4)22.5] = 4.65 in. 
Mpr = 1.0As(1.25fy)(d - a/2)
FEMA 451, NEHRP Recommended Provisions: Design Examples 
6-44
Mpr = 1.0(4.74)1.25(60)(29.6 - 4.65/2) = 9,697 in.-kips 
For the case of six #8 bottom bars: 
As = 6(0.79) = 4.74 in.2 
Width b for positive moment = 42.5 in. 
d = 29.6 in. 
a = 4.74(1.25)60/(0.85  4  42.5 ) = 2.46 in. 
Mpr = 1.0(4.74)1.25(60)(29.6 - 2.46/2) = 10,085 in.-kips 
6.4.5.2.5 Adequacy of Flexural Reinforcement in Relation to the Design of the Beam-Column Joint 
Prior to this point in the design process, the layout of reinforcement has been considered preliminary 
because the quantity of reinforcement placed in the girders has a direct bearing on the magnitude of the 
stresses developed in the beam-column joint. If the computed joint stresses are too high, the only 
remedies are increasing the concrete strength, increasing the column area, changing the reinforcement 
layout, or increasing the beam depth. The option of increasing concrete strength is not viable for this 
example because it is already at the maximum (4,000 psi) allowed for LW concrete. If absolutely 
necessary, however, NW concrete with a strength greater than 4,000 psi may be used for the columns and 
beam-column joint region while the LW concrete is used for the joists and beams. 
The design of the beam-column joint is based on the requirements of ACI 318 Sec. 21.5.3. The 
determination of the forces in the joint of the column on Gridline C of Frame 1 is based on Figure 6-16a, 
which shows how plastic moments are developed in the various spans for equivalent lateral forces acting 
to the east. An isolated subassemblage from the frame is shown in Figure 6-16b. The beam shears shown 
in Figure 6-16c are based on the probable moment strengths shown in Table 6-14. 
For forces acting from west to east, compute the earthquake shear in Span B-C: 
VE = (Mpr
- + Mpr
+ )/lclear = (9,697 + 10,085)/(240 - 30) = 94.2 kips 
For Span C-C': 
VE = (10,085 + 8,999)/(240 - 30) = 90.9 kips 
With the earthquake shear of 94.2 and 90.9 kips being developed in the beams, the largest shear that 
theoretically can be developed in the column above Level 5 is 150.5 kips. This is computed from 
equilibrium as shown at the bottom of Figure 6-16: 
94.2(9.83) + 90.9(10.50) = 2Vc(12.5/2) 
Vc = 150.4 kips 
With equal spans, gravity loads do not produce significant column shears, except at the end column, 
where the seismic shear is much less. Therefore, gravity loads are not included in this computation. 
The forces in the beam reinforcement for negative moment are based on six #8 bars at 1.25 fy: 
T = C = 1.25(60)[(6(0.79)] = 355.5 kips
Chapter 6, Reinforced Concrete 
6-45 
' 
- 
+ 
- 
+ + 
- 
+
- - 
9,697 8,999 
10,085 10,085 
150.4
150.4 
90.9 
94.2 
(a) 
Plastic 
mechanism 
(b) 
Plastic 
moments (in.- 
kips) in spans 
B-C and C-C' 
(c) 
Girder and 
column shears 
(kips) 
' 
' 
20'-0" 20'-0" 
9'-10" 10'-6" 
Figure 6-16 Diagram for computing column shears (1.0 ft = 
0.3048 m, 1.0 kip = 4.45kN, 1.0 in.-kip = 0.113 kN-m). 
For positive moment, six #8 bars also are used, assuming C = T, C = 355.5 kips. 
As illustrated in Figure 6-17, the joint shear force Vj is computed as: 
Vj = T + C - VE 
= 355.5 + 355.5 - 150.4 
= 560.6 kips 
The joint shear stress is: 
2 2 
560.5 623 psi 
30 
j 
j 
c 
V 
v 
d 
= = =
FEMA 451, NEHRP Recommended Provisions: Design Examples 
6-46 
C = 355.5 kips 
C = 355.5 kips T = 355.5 kips 
T = 355.5 kips 
Vc = 150.5 kips 
V = 2(355.5)-150.5 
= 560.5 kips 
g 
Figure 6-17 Computing joint shear stress (1.0 kip = 4.45kN). 
For joints confined on three faces or on two opposite faces, the allowable shear stress for LW concrete is 
based on ACI 318 Sec. 21.5.3. Using f = 0.80 for joints (from ACI Appendix C) and a factor of 0.75 as a 
modifier for LW concrete: 
v j,allowable=0.80(0.75)(15 4,000)=569 psi 
[Note that for joints, f = 0.85 per ACI 318-02 Sec 9.3 as referenced by the 2003 Provisions. See Sec 
6.4.2 for discussion.] 
Since the actual joint stress (623 psi) exceeds the allowable stress (569 psi), the joint is overstressed. One 
remedy to the situation would be to reduce the quantity of positive moment reinforcement. The six #8 
bottom bars at Columns B and C could be reduced to three #8 bars plus two #7 bars. This would require a 
somewhat different arrangement of bars than shown in Figure 6-15. It is left to the reader to verify that 
the joint shear stress would be acceptable under these circumstances. Another remedy would be to 
increase the size of the column. If the column is increased in size to 32 in. by 32 in., the new joint shear 
stress is: 
2 2 
560.5 547psi < 569 psi 
32 
j 
j 
c 
V 
v 
d 
= = = 
which is also acceptable. For now we will proceed with the larger column, but as discussed later, the final 
solution will be to rearrange the bars to three #8 plus two #7. 
Joint stresses would be checked for the other columns in a similar manner. Because the combined area of 
top and bottom reinforcement used at Columns A, A', C', and D is less than that for Columns B and C, 
these joints will not be overstressed. 
Given that the joint stress is acceptable, ACI 318 Sec. 21.5.2.3 controls the amount of reinforcement 
required in the joint. Since the joint is not confined on all four sides by a beam, the total amount of 
transverse reinforcement required by ACI 318 Sec. 21.4.4 will be placed within the depth of the joint. As 
shown later, this reinforcement consists of four-leg #4 hoops at 4 in. on center.
Chapter 6, Reinforced Concrete 
6-47 
Because the arrangement of steel is acceptable from a joint strength perspective, the cutoff locations of 
the various bars may be determined (see Figure 6-15 for a schematic of the arrangement of 
reinforcement). The three #8 bars (top and bottom) required for minimum reinforcement are supplied in 
one length that runs continuously across the two end spans and are cut off in the center span. An 
additional three #8 bars are placed top and bottom in the center span; these bars are cut off in Spans A'-B 
and C-C'. At Supports A, A', C' and D, shorter bars are used to make up the additional reinforcement 
required for strength. 
To determine where bars should be cut off in each span, it is assumed that theoretical cutoff locations 
correspond to the point where the continuous top and bottom bars develop their design flexural strength. 
Cutoff locations are based on the members developing their design flexural capacities (fy = 60 ksi and f = 
0.8). Using calculations similar to those above, it has been determined that the design flexural strength 
supplied by a section with only three #8 bars is 3,311 in.-kips for positive moment and 3,261 in.-kips for 
negative moment. 
Sample cutoff calculations are given first for Span B-C. To determine the cutoff location for negative 
moment, it is assumed that the member is subjected to earthquake plus 0.68 times the dead load forces. 
For positive moment cutoffs, the loading is taken as earthquake plus 1.42 times dead load plus 0.5 times 
live load. Loading diagrams for determining cut off locations are shown in Figure 6-18. 
For negative moment cutoff locations, refer to Figure 6-19a, which is a free body diagram of the west end 
of the member. Since the goal is to develop a negative moment capacity of 3,261 in.-kips in the 
continuous #8 bars summing moments about Point A in Figure 6-19a: 
6,311 0.121 2 73.7 3,261 
2 
+ x - x= 
In the above equation, 6,311 (in.-kips) is the negative moment capacity for the section with six #8 bars, 
0.121 (kips/in.) is 0.68 times the uniform dead load, 73.3 kips is the end shear, and 3,261 in.-kips is the 
design strength of the section with three #8 bars. Solving the quadratic equation results in x = 42.9 in. 
ACI 318 Sec. 12.10.3 requires an additional length equal to the effective depth of the member or 12 bar 
diameters (whichever is larger). Consequently, the total length of the bar beyond the face of the support 
is 42.9 + 29.6 = 72.5 in. and a 6 ft-1 in. extension beyond the face of the column could be used. 
For positive moment cutoff, see Figure 6-14 and Figure 6-19b. The free body diagram produces an 
equilibrium equation as: 
6,510 0.281 2 31.6 3,311 
2 
- x - x= 
where the distance x is computed to be 75.7 in. Adding the 29.6 in. effective depth, the required 
extension beyond the face of the support is 76.0 + 29.6 = 105.3 in, or 8 ft-9 in. Note that this is exactly at 
the midspan of the member.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
6-48 
E
E +0.68D 
E +1.42D+0.68L 
X- X+ 
WL = 0.66 klf 
WD = 2.14 klf 
Face of 
support 
17'-6" 
6,311 
3,261 
3,311 
6,510 
Bending moment 
(in.-kips) 
Figure 6-18 Loading for determination of rebar cutoffs 
(1.0 ft = 0.3048 m, 1.0 klf = 14.6 kN/m, 1.0 in.-kip = 
0.113 kN-m). 
6,311 in.-kips 
XA 
0.68 WD = 0.121 kip/in. 
73.7 kips 
X+ 
A 
31.6 kips 
1.42D+0.5L = 0.281 kip/in. 6,510 in.-kips 
(b) 
(a) 
Figure 6-19 Free body diagrams (1.0 kip = 
4.45kN, 1.0 klf = 14.6 kN/m, 1.0 in.-kip = 
0.113 kN-m). 
Clearly, the short bottom bars shown in Figure 6-15 are impractical. Instead, the bottom steel will be 
rearranged to consist of three #8 plus two #7 bars continuous. Recall that this arrangement of 
reinforcement will satisfy joint shear requirements, and the columns may remain at 30 in. by 30 in. 
As shown in Figure 6-20, another requirement in setting cutoff length is that the bar being cut off must 
have sufficient length to develop the strength required in the adjacent span. From Table 6-13, the 
required development length of the #9 top bars in tension is 72.1 in. if the bar is anchored in a confined 
joint region. The confined length in which the bar is developed is shown in Figure 6-20 and consists of
Chapter 6, Reinforced Concrete 
6-49 
dh 
F 
c 
d = 2'-8" 
d = 2'-6" 
7'-10" 
6'-1" 
Confined 
region 
Must also check 
for force F. Required 
length = 3.5 l = 72.1" 
#9 bar 
Cut off length based 
on moments in span 
A'-B 
b ' 
Figure 6-20 Development length for top bars (1.0 ft = 0.3048 m, 1.0 in = 25.4 mm). 
the column depth plus twice the depth of the girder. This length is 30 + 32 + 32 = 94 in., which is greater 
than the 72.1 in. required. The column and girder are considered confined because of the presence of 
closed hoop reinforcement as required by ACI 318 Sec. 21.3.3 and 21.4.4. 
The bottom bars are spliced at the center of Spans A'-B and C-C' as shown in Figure 6-21. The splice 
length is taken as the bottom bar Class B splice length for #8 bars. According to ACI 318 Sec. 12.15, the 
splice length is 1.3 times the development length. From ACI 318 Sec. 12.2.2, the development length (ld) 
is computed from: 
' 
3 
40 
d y 
b c tr 
b 
l f 
d f c K 
d 
a.. 
= 
. + . 
. . 
. . 
using a = 1.0 (bottom bar),  =1.0 (uncoated), . = 1.0 (#9 bar), . = 1.3 (LW concrete), taking c as the 
cover (1.5 in.) plus the tie dimension (0.5 in.) plus 1/2 bar diameter (0.50 in.) = 2.50 in., and using 
Ktr = 0, the development length for one #9 bar is: 
3 60,000 1 1 1.0 1.3 (1.0) 37.0 in. 
40 4,000 2.5 0 
1.0 
ld 
. .    
=... ... .. + .. = 
. . 
The splice length = 1.3  37.0 = 48.1 in. Therefore, use a 48-in. contact splice. According to ACI 318 
Sec. 21.3.2.3, the entire region of the splice must be confined by closed hoops spaced no closer than d/4 
or 4 in. 
The final bar placement and cutoff locations for all five spans are shown in Figure 6-21. Due to the 
different arrangement of bottom steel, the strength at the supports must be recomputed. The results are 
shown in Table 6-15.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
6-50 
(3) #8 
' ' 
(2) #9 (2) #8 
(3) #8 
5'-0" 
4'-0" 
(3) #8 + 
(2) #7 
Hoop spacing (from each end): 
Typical spans A-A', B-C, C'-D 
(4) #3 leg 1 at 2", 19 at 5.5" 
Typical spans A'-B,C-C', 
(4) #3 leg 1 at 2", 
15 at 5.5", 6 at 4" 
Figure 6-21 Final bar arrangement (1.0 ft = 0.3048 m, 1.0 in = 25.4 mm). 
Table 6-15 Design and Maximum Probable Flexural Strength For Beams in Frame 1 (Revised) 
Item 
Location 
A A' B C C' D 
Negative 
Moment 
Reinforcement five #8 three #8 + 
two #9 
six #8 six #8 three #8 + 
two #9 
five #8 
Design Moment 
(in.-kips) 5,318 5,849 6,311 6,311 5,849 5,318 
Probable moment 
(in.-kips) 8,195 8,999 9,696 9,696 8,999 8,195 
Positive 
Moment 
Reinforcement three #8 + 
two #7 
three #8 + 
two #7 
three #8 + 
two #7 
three #8 
+ two #7 
three #8 + 
two #7 
three #8 + 
two #7 
Design Moment 
(in.-kips) 4,944 4,944 4,944 4,944 4,944 4,944 
Probable moment 
(in.-kips) 7,677 7,677 7,677 7,677 7,677 7,677 
1.0 in.-kip = 0.113 kN-m. 
6.4.5.2.6 Transverse Reinforcement 
Transverse reinforcement requirements are covered in ACI 318 Sec. 21.3.3 (minimum reinforcement) and 
21.3.4 (shear strength). 
To avoid nonductile shear failures, the shear strength demand is computed as the sum of the factored 
gravity shear plus the maximum probable earthquake shear. The maximum probable earthquake shear is 
based on the assumption that f = 1.0 and the flexural reinforcement reaches a tensile stress of 1.25fy. The 
probable moment strength at each support is shown in Table 6-15.
Chapter 6, Reinforced Concrete 
6-51 
Figure 6-22 illustrates the development of the design shear strength envelopes for Spans A-A', A'-B, and 
B-C. In Figure 6-22a, the maximum probable earthquake moments are shown for seismic forces acting to 
the east (solid lines) and to the west (dashed lines). The moments shown occur at the face of the supports. 
The earthquake shears produced by the maximum probable moments are shown in Figure 6-22b. For 
Span A-B, the values shown in the figure are: 
pr pr 
E 
clear 
M M 
V 
l
- + + 
= 
where lclear = 17 ft-6 in. = 210 in. 
Note that the earthquake shears act in different directions depending on the direction of load. 
For forces acting to the east, VE = (9696 + 7677) / 210 = 82.7 kips. 
For forces acting to the west, VE = (8999 + 7677) / 210 = 79.4 kips.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
6-52 
' 
9,696 9,696 
7,677 7,677 7,677 7,677 
8,195 8,999 
79.4 82.7 82.7 
75.6 79.4 82.7 
29.5 
29.5 29.5 
29.5 
29.5 
29.5 
112.2 
53.2 
112.2 
53.2 
53.2 
112.2 
49.9 
108.9 
49.9 
108.9 
46.1 
105.1 
Loading 
(a) 
Seismic moment 
(tension side) 
in.-kips 
kips
positive 
kips
positive 
kips
positive 
(d) 
Design shear 
seismic + gravity 
240" 
210" 
15" 15" 
(c) 
Gravity shear 
(1.42D + 0.5L) 
(b) 
Seismic shear 
Figure 6-22 Shear forces for transverse reinforcement (1.0 in = 25.4 mm, 1.0 kip 
= 4.45kN, 1.0 in.-kip = 0.113 kN-m).
Chapter 6, Reinforced Concrete 
6-53 
9.3" 54.4" 41.3" 
f V s = 132.6 kips s = 5" 
s = 6" 
s = 7" 
f V s = 110.5 kips 
f V s = 94.7 kips 
112.2 kips 
53.2 kips 
112.2 kips 
53.2 kips 
Figure 6-23 Detailed shear force envelope in Span B-C (1.0 in = 
25.4 mm, 1.0 kip = 4.45kN). 
The gravity shears shown in Figure 6-22c are: 
Factored gravity shear = VG = 1.42Vdead + 0.5Vlive 
Vdead = 2.14  17.5/2 = 18.7 kips 
Vlive = 0.66  17.5/2 = 5.8 kips 
VG = 1.42(18.7) + 0.5(5.8) = 29.5 kips 
Total design shears for each span are shown in Figure 6-22d. The strength envelope for Span B-C is 
shown in detail in Figure 6-23, which indicates that the maximum design shears is 82.7 + 29.5 = 112.2 
kips. While this shear acts at one end, a shear of 82.7 - 29.5 = 53.2 kips acts at the opposite end of the 
member. 
In designing shear reinforcement, the shear strength can consist of contributions from concrete and from 
steel hoops or stirrups. However, according to ACI 318 Sec. 21.3.4.2, the design shear strength of the 
concrete must be taken as zero when the axial force is small (Pu/Agf !c 
< 0.05) and the ratio VE/Vu is greater 
than 0.5. From Figure 6-22, this ratio is VE/Vu = 82.7/112.2 = 0.73, so concrete shear strength must be 
taken as zero. Using the ASCE 7 compatible f for shear = 0.75, the spacing of reinforcement required is 
computed as described below. [Note that this is the basic strength reduction factor for shear per ACI 318- 
02 Sec 9.3. See Sec 6.4.2 for discussion.] 
Compute the shear at d = 29.6 in. from the face of the support: 
Vu = fVs = 112.2 - (29.6/210)(112.2 - 53.2) = 103.9 kips 
Vs = Avfyd/s 
Assuming four #3 vertical legs (Av = 0.44 in.2), fv = 60 ksi and d = 29.6 in., compute the required spacing: 
s = fAvfyd/Vu = 0.75[4(0.11)](60)(29.6/103.9) = 5.65 in., say 5.5 in. 
At midspan, the design shear Vu = (112.2 + 53.2)/2 = 82.7 kips. Compute the required spacing: 
s = 0.75[4(0.11)](60)(29.6/82.7) = 7.08 in., say 7.0 in. 
Check maximum spacing per ACI 318 Sec. 21.3.3.2:
FEMA 451, NEHRP Recommended Provisions: Design Examples 
6-54
d/4 = 29.6/4 = 7.4 in. 
8db = 8(1.0) = 8.0 in. 
24dh = 24(3/8) = 9.0 in. 
The spacing must vary between 5.5 in. at the support and 7.0 in. at midspan. Due to the relatively flat 
shear force gradient, a spacing of 5.5 in. will be used for the full length of the beam. The first hoop must 
be placed 2 in. from the face of the support. This arrangement of hoops will be used for Spans A-A', B-C, 
and C'-D. In Spans A'-B and C-C', the bottom flexural reinforcement is spliced and hoops must be placed 
over the splice region at d/4 or a maximum of 4 in. on center. 
ACI 318 Sec. 21.3.3.1 states that closed hoops are required over a distance of twice the member depth 
from the face of the support. From that point on, stirrups may be used. For the girders of Frame 1, 
however, stirrups will not be used, and the hoops will be used along the entire member length. This is 
being done because the earthquake shear is a large portion of the total shear, the girder is relatively short, 
and the economic premium is negligible. 
Where hoops are required (first 64 in. from face of support), longitudinal reinforcing bars should be 
supported as specified in ACI 318 Sec. 7.10.5.3. Hoops should be arranged such that every corner and 
alternate longitudinal bar is supported by a corner of the hoop assembly and no bar should be more than 6 
in. clear from such a supported bar. Details of the transverse reinforcement layout for all spans of Level 5 
of Frame 1 are shown in Figure 6-21. 
6.4.5.3 Design of a Typical Interior Column of Frame 1 
This section illustrates the design of a typical interior column on Gridline A'. The column, which 
supports Level 5 of Frame 1, is 30 in. square and is constructed from 4,000 psi LW concrete, 60 ksi 
longitudinal reinforcement, and 60 ksi transverse reinforcement. An isolated view of the column is 
shown in Figure 6-24. The flexural reinforcement in the beams framing into the column is shown in 
Figure 6-21. Using simple tributary area calculations (not shown), the column supports an unfactored 
axial dead load of 528 kips and an unfactored axial live load of 54 kips. The ETABS analysis indicates 
that the maximum axial earthquake force is 84 kips, tension or compression. The load combination used 
to compute this force consists of full earthquake force in the E-W direction, 30 percent of the N-S force, 
and accidental torsion. Because no beams frame into this column along Gridline A', the column bending 
moment for N-S forces can be neglected. Hence, the column is designed for axial force plus uniaxial 
bending.
Chapter 6, Reinforced Concrete 
6-55 
See Figure 6-21 
for girder 
reinforcement 
L 
30" 
' 
Level 5 
Level 4 
20'-0" 20'-0" 
32" 32" 
P = 54 kips Includes 
PD = 528 kips level 5 
12'-6" 
Figure 6-24 Layout and loads on column of Frame A (1.0 ft = 0.3048 m, 
1.0 in = 25.4 mm, 1.0 kip = 4.45kN). 
6.4.5.3.1 Longitudinal Reinforcement 
To determine the axial design loads, use the basic load combinations: 
1.42D + 0.5L + 1.0E 
0.68D - 1.0E. 
The combination that results in maximum compression is: 
Pu = 1.42(528) + 0.5(54) + 1.0(84) = 861 kips (compression) 
The combination for minimum compression (or tension) is: 
Pu = 0.68(528) - 1.0(84) = 275 kips (compression) 
The maximum axial compression force of 861 kips is greater than 0.1fc'Ag = 0.1(4)(302) = 360 kips. Thus, 
according to ACI 318 Sec. 21.4.2, the nominal column flexural strength must be at least 6/5 of the 
nominal flexural strength of the beams framing into the column. Beam moments at the face of the support 
are used for this computation. These capacities are provided in Table 6-15. 
Nominal (negative) moment strength at end A' of Span A-A' = 5,849/0.8 = 7,311 in.-kips 
Nominal (positive) moment strength at end A' of Span A' B = 4,945/0.8 = 6,181 in.-kips 
Average nominal moment framing into joint = 6,746 in.-kips 
Nominal column design moment = 6/5  6746 = 8,095 in.-kips. 
Knowing the factored axial load and the required design flexural strength, a column with adequate 
capacity must be selected. Figure 6-25 gives design curves for 30 in. by 30 in. columns of 4,000 psi 
concrete and reinforcement consisting of 12 #8, #9, or #10 bars. These curves, computed with a
FEMA 451, NEHRP Recommended Provisions: Design Examples 
6-56 
(12) #10 
(12) #9 
(12) #8 
0 
0 
500 1,000 1,500 2,000 
1,000 
2,000 
3,000 
4,000 
5,000 
-1,000 
-2,000 
M u (ft-kips) 
Pu (kips) 
Figure 6-25 Design interaction diagram for column on 
Gridline A' (1.0 kip = 4.45kN, 1.0 ft-kip = 1.36 kN-m). 
Microsoft Excel spreadsheet, are based on a f factor of 1.0 as required for nominal strength. At axial 
forces of 275 kips and 861 kips, solid horizontal lines are drawn. The dots on the lines represent the 
required nominal flexural strength (8095 in.-kips) at each axial load level. These dots must lie to the left 
of the curve representing the design columns. For both the minimum and maximum axial forces, a 
column with 12 #8 bars (with As = 9.48 in.2 and 1.05 percent of steel) is clearly adequate. 
6.4.5.3.2 Transverse Reinforcement 
ACI 318 Sec. 21.4.4 gives the requirements for minimum transverse reinforcement. For rectangular 
sections with hoops, ACI 318 Eq. 21-3 and 21-4 are applicable: 
0.3 c c g 1 
sh 
yh ch 
A sh f A 
f A 
. '.. . 
= ... ..... -.. 
0.09 c 
sh c 
yh 
f A shf
' 
= 
The first of these equations controls when Ag/Ach > 1.3. For the 30-in.-by-30-in. columns: 
Ach = (30 - 1.5 - 1.5)2 = 729 in.2 
Ag = 30 (30) = 900 in.2 
Ag/Ach = 900/729 = 1.24 
ACI 318 Eq. 21-4 therefore controls.
Chapter 6, Reinforced Concrete 
6-57 
For LW concrete, try hoops with four #4 legs and fc' = 4,000 psi: 
hc = 30 - 1.5 - 1.5 - 0.25 - 0.25 = 26.5 in. 
s = [4 (0.2) 60,000]/[0.09 (26.5) 4000] = 5.03 in. 
However, the maximum spacing of transverse reinforcement is the lesser of one-fourth the maximum 
column dimension (30/4=7.5 in.), six bar diameters (6  1.0 = 6.0 in.), or the dimension sx where: 
4 14 
3 
x 
x 
s h 
- 
= + 
and where hx is the maximum horizontal spacing of hoops or cross ties. For the column with twelve #8 
bars and #4 hoops and cross ties, hx = 8.833 in. and sx = 5.72 in. The 5.03-in. spacing required by ACI Eq. 
21-4 controls, so a spacing of 5 in. will be used. This transverse reinforcement must be spaced over a 
distance lo = 30 in. at each end of the member and, according to ACI 318 Sec. 21.5.2, must extend 
through the joint at (at most) the same spacing. 
ACI 318 Sec. 21.4.4.6 requires a maximum spacing of transverse reinforcement in the region of the 
column not covered by Sec. 21.4.4.4. The maximum spacing is the smaller of 6.0 in. or 6db, which for #8 
bars is also 6 in. ACI 318 requires transverse steel at this spacing, but it does not specify what the details 
of reinforcement should be. In this example, hoops and crossties with the same details as those placed in 
the critical regions of the column are used. 
6.4.5.3.3 Transverse Reinforcement Required for Shear 
The amount of transverse reinforcement computed above is the minimum required. The column also 
must be checked for shear with the column shears being based on the maximum probable moments in the 
beams that frame into the column. The average probable moment is roughly 1.25/0.8 (f = 0.8) times the 
average design moment = (1.25/0.8)(5397) = 8,433 in.-kips. With a clear height of 118 in., the column 
shear can be estimated at 8433/(0.5x118) = 143 kips. This shear will be compared to the capacity 
provided by the 4-leg #4 hoops spaced at 6 in. on center. If this capacity is well in excess of the demand, 
the columns will be acceptable for shear. 
For the design of column shear capacity, the concrete contribution to shear strength may be considered 
because Pu > Agf !c
/20. Using a shear strength reduction factor of 0.85 for sand-LW concrete (ACI 318 
Sec. 11.2.1.2) in addition to the capacity reduction factor for shear, the design shear strength contributed 
by concrete is: 
fVc=f0.75fc'bcdc=0.75(0.85)( 4,000(30)(27.5)= 33.2 kips 
fVs=fAvfyd/s=0.75(4)(0.2)(60)(27.5) / 6=165 kips 
fVn=fVc+fVs=33.2+165=198.2 kips > 143 kips OK 
The column with the minimum transverse steel is therefore adequate for shear. The final column detail 
with both longitudinal and transverse reinforcement is given in Figure 6-26. The spacing of 
reinforcement through the joint has been reduced to 4 in. on center. This is done for practical reasons 
only. Column bar splices, where required, should be located in the center half of the of the column and 
must be proportioned as (Class B) tension splices.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
6-58 
' 
Level 7 
Level 6 
30" 
32" 32" 
30" 
30" 
(12) #8 bars 
#4 hoops 
+ + 
2" 7 at 4" 2" 6 at 5" 9 at 6" 6 at 5" 2" 
Figure 6-26 Details of reinforcement for column (1.0 in = 25.4 mm). 
6.4.5.4 Design of Haunched Girder 
The design of a typical haunched girder of Level 5 of Frame 3 is now illustrated. This girder is of 
variable depth with a maximum depth of 32 in. at the support and a minimum depth of 20 in. for the 
middle half of the span. The length of the haunch at each end (as measured from the face of the support) 
is 8 ft-9 in. The width of the web of girder is 22.5 in. throughout. 
Based on a tributary gravity load analysis, this girder supports an average of 3.375 kips/ft of dead load 
and 0.90 kips/ft of reduced live load. For the purpose of estimating gravity moments, a separate analysis 
of the girder was carried out using the SAP2000 program. End A of the girder was supported with 
half-height columns pinned at midstory and End B, which is supported by a shear wall, was modeled as 
fixed. Each haunch was divided into four segments with nonprismatic section properties used for each 
segment. The loading and geometry of the girder is shown in Figure 6-27a. 
For determining earthquake forces, the entire structure was analyzed using the ETABS program. This 
analysis included 100 percent of the earthquake forces in the E-W direction and 30 percent of the
Chapter 6, Reinforced Concrete 
6-59 
Negative moment hinge 
-2,000
0 
-4,000 
-6,000 
-8,000 
-10,000 
-12,000 
-14,000 
8,000 
6,000 
4,000 
2,000 
Range of possible positive 
moment hinges 
Level 5 
4'-0" 6'-0" 6'-0" 4'-0" 
(d) Potential plastic 
hinge locations 
Level 5 
(c) Flexural 
reinforcement 
details 
(7) #11 
(5) #9 
6,982 = fMn 4,102 = fMn 
-13,167 = fMn -6,824 = fMn 
1.42D + 0.5L + 0.5E 
1.2D + 1.6L 
1.42D + 0.5L - 0.5E 
0.68D + 0.5E 
0.68D - E 
Strength envelope 
Level 5 
(a) Span geometry 
and loading 
(b) Moment envelope 
(in.-kips) 
1'-3" 8'-9" 10'-0" 10'-0" 8'-9" 1'-3" 
WL = 0.90 kips/ft 
WD = 3.38 kips/ft 
Figure 6-27 Design forces and detailing of haunched girder (1.0 ft = 0.3048 m, 1.0 
k/ft = 14.6 kN/m, 1.0 in.-kip = 0.113 kN-m). 
earthquake force in the N-S direction, and accidental torsion. Each of these systems of lateral forces was 
placed at a 5 percent eccentricity with the direction of the eccentricity set to produce the maximum 
seismic shear in the member.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
6-60 
6.4.5.4.1 Design and Detailing of Longitudinal Reinforcement 
The results of the analysis for five different load combinations are shown in Figure 6-27b. Envelopes of 
maximum positive and negative moment are shown on the figure indicate that 1.42D + 0.5L  E controls 
negative moment at the support, 0.68D  E controls positive moment at the support, and 1.2D + 1.6L 
controls positive moment at midspan. The maximum positive moment at the support is less than 50 
percent of the maximum negative moment and the positive and negative moment at midspan is less than 
25 percent of the maximum negative moment; therefore, the design for negative moment controls the 
amount of reinforcement required at all sections per ACI 318 Sec. 21.3.2.2. 
For a factored negative moment of 12,600 in.-kips at Support B, try seven #11 bars, and assuming #3 
hoops: 
As = 7  1.54 = 10.92 in.2 
d = 32- 1.5 - 3/8 - 1.41/2 = 29.4 in. 
. = 10.92/(29.4  22.5) = 0.0165 < 0.025, O.K. 
b = 22.5 in. 
Depth of compression block, a = [10.92 (60)]/[0.85 (4) 22.5] = 8.56 in. 
Design strength, fMn = [0.8 (10.92) 60](29.4 - 8.56/2) = 13,167 in.-kips > 12,600 in.-kips OK 
For positive moment at the support, try five #9 bars, which supplies about half the negative moment 
reinforcement: 
As = 5 (1.0) = 5.00 in.2 
d = 32 - 1.5 - 3/8 - 1.128/2 = 29.6 in. 
. = 5.00/(29.6  22.5) = 0.0075 > 0.033, O.K. 
b = 86.5 in. (assuming stress block in flange) 
a = [5.00 (60)]/(0.85 (4) 86.5] = 1.02 in. 
fMn = [0.8 (5.00) 60] (29.6 - 1.02/2) = 6,982 in.-kips. 
This moment is larger than the design moment and, as required by ACI 318 Sec. 21.3.2.2, is greater than 
50 percent of the negative moment capacity at the face of the support. 
For positive moment at midspan the same five #9 bars used for positive moment at the support will be 
tried: 
As = 5 (1.0) = 5.00 in.2 
d = 20 - 1.5 - 3/8 - 1.128/2 = 17.6 in. 
. =5.00/(17.6  22.5) = 0.0126 
b = 86.5 in. 
a = [5.00 (60)]/[0.85 (4) 86.5] = 1.02 in. 
fMn = [0.8 (5.00) 60] (17.6 - 1.02/2) = 4,102 in.-kips > 3,282 in.-kips. OK 
The five #9 bottom bars are adequate for strength and satisfy ACI 318 Sec. 21.3.2.2, which requires that 
the positive moment capacity be not less than 25 percent of the negative moment capacity at the face of 
the support. 
For negative moment in the 20-ft span between the haunches, four #11 bars (. = 0.016) could be used at 
the top. These bars provide a strength greater than 25 percent of the negative moment capacity at the 
support. Using four bars across the top also eliminates the possibility that a negative moment hinge will 
form at the end of the haunch (8 ft-9 in. from the face of the support) when the 0.68D - E load 
combination is applied. These four top bars are part of the negative moment reinforcement already sized
Chapter 6, Reinforced Concrete 
6-61 
for negative moment at the support. The other three bars extending from the support are not needed for 
negative moment in the constant depth region and would be cut off approximately 6 ft beyond the 
haunch; however, this detail results in a possible bar cutoff in a plastic hinge region (see below) that is not 
desirable. Another alternative would be to extend all seven #11 bars across the top and thereby avoid the 
bar cutoff in a possible plastic hinge region; however, seven #11 bars in 20-in. deep portion of the girder 
provide . = 0.028, which is a violation of ACI 21.3.2.1 (.max = 0.025). The violation is minor and will be 
accepted in lieu of cutting off the bars in a potential plastic hinge region. Note that these bars provide a 
negative design moment capacity of 6,824 in.-kips in the constant depth region of the girder. 
The layout of longitudinal reinforcement used for the haunched girder is shown in Figure 6-27c, and the 
flexural strength envelope provided by the reinforcement is shown in Figure 6-27b. As noted in Table 
6-13, the hooked #11 bars can be developed in the confined core of the columns. Finally, where seven 
#11 top bars are used, the spacing between bars is approximately 1.4 in., which is greater than the 
diameter of a #11 bar and is therefore acceptable. This spacing should accommodate the vertical column 
reinforcement. 
Under combined gravity and earthquake load, a negative moment plastic hinge will form at the support 
and, based on the moment envelopes from the loading (Figure 6-27b), the corresponding positive moment 
hinge will form in the constant depth portion of the girder. As discussed in the following sections, the 
exact location of plastic hinges must be determined in order to design the transverse reinforcement. 
6.4.5.4.2 Design and Detailing of Transverse Reinforcement 
The design for shear of the haunched girder is complicated by its variable depth; therefore, a tabular 
approach is taken for the calculations. Before the table may be set up, however, the maximum probable 
strength must be determined for negative moment at the support and for positive moment in the constant 
depth region, 
For negative moment at the face of the support and using seven #11 bars: 
As = 7 (1.56) = 10.92 in.2 
d = 32 - 1.5 - 3/8 - 1.41/2 = 29.4 in. 
b = 22.5 in. 
a = [10.92 (1.25) 60]/[0.85 (4) 22.5] = 10.71 in. 
Mpr = 1.0(10.92)(1.25)(60)(29.4 - 10.71/2) = 19,693 in.-kips. 
For positive moment in the constant depth region and using five #9 bars: 
As = 5 (1.0) = 5.00 in.2 
d = 20 - 1.5 - 3/8 - 1.128/2 = 17.6 in. 
b = 86.5 in. 
a = [5.00 (1.25) 60]/[0.85 (4) 86.5] = 1.28 in. 
Mpr = [1.0 (5.00) 1.25 (60)] (17.6 - 1.28/2) = 6,360 in.-kips 
Before the earthquake shear may be determined, the location of the positive moment hinge that will form 
in the constant depth portion of the girder must be identified. To do so, consider the free-body diagram of 
Figure 6-28a. Summing moments (clockwise positive) about point B gives: 
2 
0 
pr pr 2 
M+ +M- +Rx-wx = 
At the positive moment hinge the shear must be zero, thus R  wx = 0
FEMA 451, NEHRP Recommended Provisions: Design Examples 
4The equation for the location of the plastic hinge is only applicable if the hinge forms in the constant depth region of the girder. 
If the computed distance x is greater than 28 ft - 9 in. (345 in.), the result is erroneous and a trial and error approach is required to 
find the actual hinge location. 
6-62 
By combining the above equations: 
2(Mpr Mpr ) 
x 
w 
+ + - 
= 
Using the above equation with Mpr as computed and w = 1.42(3.38) + 0.5(0.90) = 5.25 k/ft = 0.437 k/in., 
x = 345 in., which is located exactly at the point where the right haunch begins.4 
The reaction is computed as R = 345 (0.437) = 150.8 kips. 
The earthquake shear is computed as VE = R = wL/2 = 150.8-(0.437)(450)/2 = 52.5 kips 
This earthquake shear is smaller than would have been determined if the positive moment hinge had 
formed at the face of support. 
The earthquake shear is constant along the span but changes sign with the direction of the earthquake. In 
Figure 6-28a, this shear is shown for the equivalent lateral seismic forces acting to the west. The factored 
gravity load shear (1.42VD + 0.5VL) varies along the length of the span as shown in Figure 6-28b. At 
Support A, the earthquake shear and factored gravity shear are additive, producing a design ultimate shear 
of 150.8 kips. At midspan, the shear is equal to the earthquake shear acting alone and, at Support C, the 
ultimate design shear is -45.8 kips. Earthquake, gravity, and combined shears are shown in Figures 6-28a 
through 6-28c and are tabulated for the first half of the span in Table 6-16. For earthquake forces acting 
to the east, the design shears are of the opposite sign of those shown in Figure 6-28. 
According to ACI 318 Sec. 21.3.4.2, the contribution of concrete to member shear strength must be taken 
as zero when VE/VU is greater than 0.5 and Pu/Agf !c 
is less than 0.05. As shown in Table 6-16, the VE/VU 
ratio is less than 0.5 within the first three-fourths of the haunch length but is greater than 0.50 beyond this 
point. In this example, it is assumed that if VE/VU is less than 0.5 at the support, the concrete strength can 
be used along the entire length of the member. 
The concrete contribution to the design shear strength is computed as: 
fVc=f(0.85)2fc'bwd 
where the ASCE 7 compatible f = 0.75 for shear, and the 0.85 term is the shear strength reduction factor 
for sand-LW concrete. [Note that this is the basic strength reduction factor for shear per ACI 318-02 Sec 
9.3. See Sec 6.4.2 for discussion.] The remaining shear, fVs = Vu - fVc, must be resisted by closed hoops 
within a distance 2d from the face of the support and by stirrups with the larger of 6dh or 3.0 in. hook 
extensions elsewhere. The 6dh or 3.0 in. seismic hook extension is required by ACI 318 Sec. 21.3.3.3.
Chapter 6, Reinforced Concrete 
6-63 
45.8 kips 
(a) Location of 
plastic hinge 
+
PR 
' 
(b) Earthquake 
shear (kips) 
(c) Factored gravity 
shear (kips) 
pos 
pos 
(d) Earthquake + 
factored gravity 
shear (kips) 
(e) 
150.8 kips 98.3 kips 
98.3 kips 
52.5 kips 
face of support 
M 
"B" 
Provide two additional 
hoops (detail B) at kink 
A B 
A B 
6 at 6" 13 at 5" 30 at 4" 
#3 hoops 
6db 
6db 
R 
W
x 
Figure 6-28 Computing shear in haunched girder (1.0 in = 25.4 mm, 1.0 kip = 4.45kN).
FEMA 451, NEHRP Recommended Provisions: Design Examples 
6-64 
Table 6-16 Design of Shear Reinforcement for Haunched Girder 
Item 
Distance from Center of Support (in.) 
Units 
15 42.25 67.5 93.75 120 180 240 
Ve 52.5 52.5 52.5 52.5 52.5 52.5 52.5 
kips 
1.42VD + 0.5VL 98.3 86.4 75.4 63.9 52.4 26.2 0.0 
Vu 150.8 139.2 127.9 116.6 104.9 78.7 52.5 
VE/VU 0.35 0.38 0.41 0.45 0.50 0.67 1.00 
d 29.4 26.5 23.5 20.5 17.6 17.6 17.6 in. 
fVC 53.3 48.1 42.6 37.2 0.0 0.0 0.0 
kips 
fVS 97.5 91.2 85.3 79.4 104.9 78.7 52.5 
s 5.97 5.78 5.46 5.12 3.32 4.43 6.64 
d/4 7.35 6.63 5.88 5.13 4.40 4.40 4.40 in. 
Spacing #3 at 6 #3 at 5 #3 at 5 #3 at 5 #3 at 4 #3 at 4 #3 at 4 
1.0 in. = 25.4 mm, 1.0 kip = 4.45 kN. 
In Table 6-16, spacings are computed for four #3 vertical leg hoops or stirrups. As an example, consider 
four #3 vertical legs at the section at the face of the support: 
fVc = f(0.85)2 fc' db = 0.75(0.85)2(4000)0.529.4(22.5) = 53,300 lb = 53.3 kips 
fVs = Vu - fVc = 150.8 - 53.3 = 97.5 kips 
fVs = fAvfyd/s = 97.5 kips 
s = [0.75(4)0.11(60)29.4]/97.5 = 5.97 in. 
The maximum spacing allowed by ACI 318 is shown in Table 6-16. These spacings govern only in the 
center portion of the beam. In the last line of the table, the hoop and stirrup spacing as actually used is 
shown. This spacing, together with hoop and stirrup details, is illustrated in Figure 6-28d. The double 
U-shaped stirrups (and cap ties) in the central portion of the beam work well with the #11 top bars and 
with the #9 bottom bars. 
6.4.5.4.3 Design of Beam-Column Joint 
The design of the beam-column joint at Support A of the haunched girder is controlled by seismic forces 
acting to the west, which produces negative moment at Support A. ACI 318 Sec. 21.5 provides 
requirements for the proportioning and detailing of the joint. 
A plastic mechanism of the beam is shown in Figure 6-29a. Plastic hinges have formed at the support 
and at the location of the far haunch transition. With a total shear at the face of the support of 150.8 kips, 
the moment at the centerline of the column may be estimated as 
MCL = Mpr + 15(150.6) = 19,693 + 15(150.6) = 21,955 in.-kips. 
The total shear in the columns above and below the joint is estimated as 21,955/(150) = 146.3 kips.
Chapter 6, Reinforced Concrete 
6-65 
(a) 
Plastic mechanism 
- + 
(b) 
Plastic moment 
(in.-kips) 
19,693 
10,800 
6,360 
(c) 
Column shears 
(kips) 
61.1 kips 
146.3 kips 
146.3 kips 
288" 
75" 75" 
15" 450" 15" 
Figure 6-29 Computation of column shears for use in joint 
design (1.0 in = 25.4 mm, 1.0 kip = 4.45 kN). 
The stresses in the joint are computed from equilibrium considering the reinforcement in the girder to be 
stressed at 1.25fy. A detail of the joint is shown in Figure 6-30. Compute the joint shear Vj: 
Force in the top reinforcement = 1.25Asfy = 1.25(7)1.56(60) = 819 kips 
Joint shear = Vj = 819.0 - 146.3 = 672.7 kips 
The joint shear stress vj = Vj/dc
2 = 672.7/[30 (30)] = 0.819 ksi 
In the case being considered, all girders framing into the joint have a width equal to 0.75 times the 
column dimension so confinement is provided on three faces of the joint. According to ACI 318 Sec. 
21.5.3, the allowable joint shear stress = 0.75f(15)2pfc'. The 0.75 term is the strength reduction factor for 
LW concrete. Compute the allowable joint shear stress: 
vj,allowable = 0.75(0.80)15(4,000)0.5 
= 569 psi = 0.569 ksi
FEMA 451, NEHRP Recommended Provisions: Design Examples 
6-66 
T = 819 kips 
146 kips 
V = 819-146 = 673 kips 
30" 
J 
C = 819 kips 
Figure 6-30 Computing joint shear force (1.0 kip = 
4.45kN). 
This allowable stress is significantly less than the applied joint shear stress. There are several ways to 
remedy the situation: 
1. Increase the column size to approximately 35  35 (not recommended) 
2. Increase the depth of the haunch so that the area of reinforcement is reduced to seven #10 bars. This 
will reduce the joint shear stress to a value very close to the allowable stress. 
2. Use 5000 psi NW concrete for the column. This eliminates the 0.75 reduction factor on allowable 
joint stress, and raises the allowable stress to 848 psi. 
For the remainder of this example, it is assumed that the lower story columns will be constructed from 
5000 psi NW concrete. 
Because this joint is confined on three faces, the reinforcement within the joint must consist of the same 
amount and spacing of transverse reinforcement in the critical region of the column below the joint. This 
reinforcement is detailed in the following section. 
6.4.5.5 Design and Detailing of Typical Interior Column of Frame 3 
The column supporting the west end of the haunched girder between Gridlines A and B is shown in 
Figure 6-31. This column supports a total unfactored dead load of 804 kips and a total unfactored live 
load of 78 kips. From the ETABS analysis, the axial force on the column from seismic forces is 129 
kips. The design axial force and bending moment in the column are based on one or more of the load 
combinations presented below. 
Earthquake forces acting to the west are: 
Pu = 1.42(804) + 0.5(78) + 1.0(129) 
= 1310 kips (compression)
Chapter 6, Reinforced Concrete 
6-67 
Level 5 
Level 4 
32" 32" 
12'-6" 
20" 
P L = 78.4 kips Includes 
PD = 803.6 kips level 5 
Figure 6-31 Column loading (1.0 ft = 0.3048 m, 
1.0 in = 25.4 mm, 1.0 kip = 4.45kN). 
This axial force is greater than 0.1fc'Ag = 360 kips; therefore, according to ACI 318 Sec. 21.4.2.1, the 
column flexural strength must be at least 6/5 of the nominal strength (using f = 1.0 and 1.0 fy) of the beam 
framing into the column. The nominal beam moment capacity at the face of the column is 16,458 
in.-kips. The column must be designed for six-fifths of this moment, or 19,750 in-kips. Assuming a 
midheight inflection point for the column above and below the beam, the column moment at the 
centerline of the beam is 19,750/2 = 9,875 in.-kips, and the column moment corrected to the face of the 
beam is 7,768 in.-kips. 
Earthquake forces acting to the east are: 
Pu = 0.68(804) - 1.0(129) = 424 kips (compression) 
This axial force is greater than 0.1fc'Ag = 360 kips. For this loading, the end of the beam supported by the 
column is under positive moment, with the nominal beam moment at the face of the column being 8,715 
in.-kips. Because Pu > 0.1fc'Ag, the column must be designed for 6/5 of this moment, or 10,458 in.-kips. 
Assuming midheight inflection points in the column, the column moment at the centerline and the face of 
the beam is 5,229 and 4,113 in.-kips, respectively. 
Axial force for gravity alone is: 
Pu = 1.6(804) + 1.2(78) = 1,380 kips (compression)
FEMA 451, NEHRP Recommended Provisions: Design Examples 
6-68 
(12) #10 
(12) #9 
(12) #8 
M u (ft-kips) 
0 
0 
500 1,000 1,500 2,000 
1,000 
2,000 
3,000 
4,000 
5,000 
-1,000 
-2,000 
Pu (kips) 
Figure 6-32 Interaction diagram and column design forces 
(1.0 kip = 4.45kN, 1.0 ft-kip = 1.36 kN-m). 
This is approximately the same axial force as designed for earthquake forces to the west, but as can be 
observed from Figure 6-25, the design moment is significantly less. Hence, this loading will not control. 
6.4.5.5.1 Design of Longitudinal Reinforcement 
Figure 6-32 shows an axial force-bending moment interaction diagram for a 30 in. by 30 in. column with 
12 bars ranging in size from #8 to #10. A horizontal line is drawn at each of the axial load levels 
computed above, and the required flexural capacity is shown by a solid dot on the appropriate line. The 
column with twelve #8 bars provides more than enough strength for all loading combinations. 
6.4.5.5.2 Design of Transverse Reinforcement 
In Sec. 6.4.5.3, an interior column supporting Level 5 of Frame 1 was designed. This column has a shear 
strength of 198.2 kips, which is significantly greater than the imposed seismic plus gravity shear of 146.3 
kips. For details on the computation of the required transverse reinforcement for this column, see the 
Transverse Reinforcement and Transverse Reinforcement Required for Shear subsections in Sec. 
6.4.5.3. A detail of the reinforcement of the column supporting Level 5 of Frame 3 is shown in Figure 6- 
33. The section of the column through the beams shows that the reinforcement in the beam-column joint 
region is relatively uncongested.
Chapter 6, Reinforced Concrete 
6-69 
Level 6 
Level 7 
30" 
30" 
30" 
(12) #8 bars 
#4 hoops 
+ + 
6 at 4" 4" 2" 7 at 4" 5" 8 at 6" 5" 7 at 4" 2" 
Figure 6-33 Column detail (1.0 in = 25.4 mm). 
6.4.5.6 Design of Structural Wall of Frame 3 
The factored forces acting on the structural wall of Frame 3 are summarized in Table 6-17. The axial 
compressive forces are based on a tributary area of 1,800 square ft for the entire wall, an unfactored dead 
load of 160 psf, and an unfactored (reduced) live load of 20 psf. For the purposes of this example it is 
assumed that these loads act at each level, including the roof. The total axial force for a typical floor is: 
Pu = 1.42D + 0.5L = 1,800((1.420.16) + 0.50x0.02)) = 427 kips for maximum compression 
Pu = 0.68D = 1,800(0.680.16) = 196 kips for minimum compression 
The bending moments come from the ETABS analysis. Note the reversal in the moment sign due to the 
effects of frame-wall interaction. Each moment contains two parts: the moment in the shear panel and the 
couple resulting from axial forces in the boundary elements. For example, at the base of Level 2:
FEMA 451, NEHRP Recommended Provisions: Design Examples 
6-70
ETABS panel moment =162,283 in.-kips 
ETABS column force = 461.5 kips 
Total moment, Mu = 162,283 + 240(461.5) = 273,043 in.-kips 
The shears in Table 6-17 also consist of two parts, the shear in the panel and the shear in the column. 
Using Level 2 as an example: 
ETABS panel shear = 527 kips 
ETABS column shear = 5.90 kips 
Total shear, Vu = 527 + 2(5.90) = 539 kips 
As with the moment, note the reversal in wall shear, not only at the top of the wall but also at Level 1 
where the first floor slab acts as a support. If there is some in-plane flexibility in the first floor slab, or if 
some crushing were to occur adjacent to the wall, the shear reversal would be less significant, or might 
even disappear. For this reason, the shear force of 539 kips at Level 2 will be used for the design of 
Level 1 as well. 
Recall from Sec. 6.2.2 that the structural wall boundary elements are 30 in. by 30 in. in size. The basic 
philosophy of this design will be to use these elements as special boundary elements where a close 
spacing of transverse reinforcement is used to provide extra confinement. This avoids the need for 
confining reinforcement in the wall panel. Note, however, that there is no code restriction on extending 
the special boundary elements into the panel of the wall. 
It should also be noted that preliminary calculations (not shown) indicate that a 12-in. thickness of the 
wall panel is adequate for this structure. This is in lieu of the 18-in. thickness assumed when computing 
structural mass. 
Table 6-17 Design Forces for Structural Wall 
Supporting 
Level 
Axial Compressive Force Pu (kips) Moment Mu 
(in.-kips) 
Shear Vu 
1.42D + 0.5L 0.68D (kips) 
R 
12 
11 
10 
987654321 
427 
854 
1,281 
1,708 
2,135 
2,562 
2,989 
3,416 
3,843 
4,270 
4,697 
5,124 
5,550 
196 
392 
588 
783 
979 
1,175 
1,371 
1,567 
1,763 
1,958 
2,154 
2,350 
2,546 
-30,054 
-39,725 
-49,954 
-51,838 
-45,929 
-33,817 
17,847 
45,444 
78,419 
117,975 
165,073 
273,043 
268,187 
-145 
-4 
62 
118 
163 
203 
240 
274 
308 
348 
390 
539 
-376 (use 539) 
1.0 kip = 4.45 kN, 1.0 in.-kip = 0.113 kN-m.
Chapter 6, Reinforced Concrete 
6-71 
6.4.5.6.1 Design of Panel Shear Reinforcement 
First determine the required shear reinforcement in the panel and then design the wall for combined 
bending and axial force. The nominal shear strength of the wall is given by ACI 318 Eq. 21-7: 
Vn=Acv(ac fc'+.nfy) 
where ac = 2.0 because hw/lw = 155.5/22.5 = 6.91 > 2.0. Note that the length of the wall was taken as the 
length between boundary element centerlines (20 ft) plus one-half the boundary element length (2.5 ft) at 
each end of the wall. 
Using fc' = 4000 psi, fy = 40 ksi, Acv = (270)(12) = 3240 in.2, and taking f for shear = 0.55, the ratio of 
horizontal reinforcement is computed: 
Vu = fVn 
539.000 (0.85 2 4,000)3,240 
0.55 0.0049 
. n 3,240(40,000) 
.. ..-  
=. . = 
Note that the factor of 0.85 on concrete strength accounts for the use of LW concrete. Reinforcement 
ratios for the other stories are given in Table 6-18. This table gives requirements using fc' = 4,000 psi, as 
well as 6,000 psi NW concrete. As shown later, the higher strength NW concrete is required to manage 
the size of the boundary elements of the wall. Also shown in the table is the required spacing of 
horizontal reinforcement assuming that two curtains of #4 bars will be used. If the required steel ratio is 
less than 0.0025, a ratio of 0.0025 is used to determine bar spacing. 
Table 6-18 Design of Structural Wall for Shear 
Level fc' = 4,000 psi (lightweight) fc' = 6,000 psi (normal weight) 
Reinforcement 
ratio 
Spacing1 
(in.) 
Reinforcement 
ratio 
Spacing * 
(in.) 
R 
12 
11 
10 
9 
8 
7 
6 
5 
4 
3 
2 
1 
0.00250 
0.00250 
0.00250 
0.00250 
0.00250 
0.00250 
0.00250 
0.00250 
0.00250 
0.00250 
0.00278 
0.00487 
0.00487 
13.33 (12.0) 
13.33 (12.0) 
13.33 (12.0) 
13.33 (12.0) 
13.33 (12.0) 
13.33 (12.0) 
13.33 (12.0) 
13.33 (12.0) 
13.33 (12.0) 
13.33 (12.0) 
12.00 (6.0) 
6.84 (6.0) 
6.84 (6.0) 
0.00250 
0.00250 
0.00250 
0.00250 
0.00250 
0.00250 
0.00250 
0.00250 
0.00250 
0.00250 
0.00250 
0.00369 
0.00369 
13.33 (12.0) 
13.33 (12.0) 
13.33 (12.0) 
13.33 (12.0) 
13.33 (12.0) 
13.33 (12.0) 
13.33 (12.0) 
13.33 (12.0) 
13.33 (12.0) 
13.33 (12.0) 
13.33 (9.0) 
9.03 (9.0) 
9.03 (9.0) 
* Values in parentheses are actual spacing used. 
1.0 in. = 25.4 mm. 
For LW concrete, the required spacing is 6.84 in. at Levels 1 and 2. Minimum reinforcement 
requirements control all other levels. For the final design, it is recommended to use a 6-in. spacing at
FEMA 451, NEHRP Recommended Provisions: Design Examples 
6-72 
Levels 1, 2, and 3 and a 12-in. spacing at all levels above. The 6-in. spacing is extended one level higher 
that required because it is anticipated that an axial-flexural plastic hinge could propagate this far. 
For the NW concrete, the required spacing is 9.03 in. at Levels 1 and 2 and minimum reinforcement 
requirements control elsewhere. For the final design, a 9-in. spacing would be used at Levels 1, 2, and 3 
with a 12-in. spacing at the remaining levels. 
ACI 318 Sec. 21.6.4.3 [21.7.4.3] requires the vertical steel ratio to be greater than or equal to the 
horizontal steel ratio if hwl/lw is less than 2.0. As this is not the case for this wall, the minimum vertical 
reinforcement ratio of 0.0025 is appropriate. Vertical steel consisting of two curtains of #4 bars at 12 in. 
on center provides a reinforcement ratio of 0.0028, which ill be used at all levels. 
6.4.5.6.2 Design for Flexure and Axial Force 
The primary consideration in the axial-flexural design of the wall is determining whether or not special 
boundary elements are required. ACI 318 provides two methods for this. The first approach, specified in 
ACI 318 Sec. 21.6.6.2 [21.7.6.2], uses a displacement based procedure. The second approach, described 
in ACI 318 Sec. 21.6.6.3 [21.7.6.3], is somewhat easier to implement but, due to its empirical nature, is 
generally more conservative. In the following presentation, only the displacement based method will be 
used for the design of the wall. 
Using the displacement based approach, boundary elements are required if the length of the compression 
block, c, satisfies ACI 318 Eq. 21-8: 
600( ) 
w
u w 
c l 
d h 
= 
where du is the total elastic plus inelastic deflection at the top of the wall. From Table 6-9b, the total 
elastic roof displacement is 4.36 in., and the inelastic drift is Cd times the elastic drift, or 6.5(4.36) = 28.4 
in. or 2.37 feet. Recall that this drift is based on cracked section properties assuming Icracked = 0.5 Igross and 
assuming that flexure dominates. Using this value together with lw = 22.5 ft, and hw = 155.5 ft: 
22.5 2.46 ft = 29.52 in. 
600( ) 600(2.37 155.5) 
w
u w 
l 
d h 
= = 
To determine if c is greater than this value, a strain compatibility analysis must be performed for the wall. 
In this analysis, it is assumed that the concrete reaches a maximum compressive strain of 0.003 and the 
wall reinforcement is elastic-perfectly plastic and yields at the nominal value. A rectangular stress block 
was used for concrete in compression, and concrete in tension was neglected. A straight line strain 
distribution was assumed (as allowed by ACI 318 Sec. 21.6.5.1 [21.7.5.1]). Using this straight line 
distribution, the extreme fiber compressive strain was held constant at 0.003, and the distance c was 
varied from 100,000 in. (pure compression) to 1 in. (virtually pure tension). For each value of c, a total 
cross sectional nominal axial force (Pn) and nominal bending moment (Mn) were computed. Using these 
values, a plot of the axial force (Pn) versus neutral axis location (c) was produced. A design value axial 
force-bending moment interaction diagram was also produced. 
The analysis was performed using an Excel spreadsheet. The concrete was divided into 270 layers, each 
with a thickness of 1 in. The exact location of the reinforcement was used. When the reinforcement was 
in compression, an adjustment was made to account for reinforcement and concrete sharing the same 
physical volume.
Chapter 6, Reinforced Concrete 
6-73 
-4,000 
-2,000
0 
2,000 
4,000 
6,000 
8,000 
10,000 
12,000 
14,000 
16,000 
18,000 
0 100,000 200,000 300,000 400,000 500,000 600,000 700,000 800,000 
Bending moment, in.-kips 
Axial force, kips 
1.42D + 0.5L 
0.68 D 
6 ksi NW 
4 ksi LW 
Figure 6-34 Interaction diagram for structural wall (1.0 kip = 4.45kN, 1.0 in.-kip = 0.113 kN-m). 
Two different sections were analyzed: one with fc' = 4,000 psi (LW concrete) and the other with fc' = 
6,000 psi (NW concrete). In each case, the boundary elements were assumed to be 30 in. by 30 in. and 
the panel was assumed to be 12 in. thick. Each analysis also assumed that the reinforcement in the 
boundary element consisted of twelve #9 bars, producing a reinforcement ratio in the boundary element 
of 1.33 percent. Panel reinforcement consisted of two curtains of #4 bars spaced at approximately 12 in. 
on center. For this wall the main boundary reinforcement has a yield strength of 60 ksi, and the vertical 
panel steel yields at 40 ksi. 
The results of the analysis are shown in Figures 6-34 and 6-35. The first of these figures is the nominal 
interaction diagram multiplied by f = 0.65 for tied sections. Also plotted in the figure are the factored 
P-M combinations from Table 6-17. The section is clearly adequate for both 4,000 psi and 6,000 psi 
concrete because the interaction curve fully envelopes the design values.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
6-74 
-3,000 
-1,000 
1,000 
3,000 
5,000 
7,000 
9,000 
11,000 
13,000 
15,000 
0 50 100 150 200 250 
Neutral axis location, in. 
Factored axial force, kips 
6 ksi NW 
4 ksi LW 
0 
Figure 6-35 Variation of neutral axis depth with compressive force (1.0 in = 25.4 mm, 1.0 kip = 4.45kN). 
Figure 6-35 shows the variation in neutral axis depth with axial force. For a factored axial force of 5,550 
kips, the distance c is approximately 58 in. for the 6,000 psi NW concrete and c is in excess of 110 in. for 
the 4,000 psi LW concrete. As both are greater than 29.52 in., special boundary elements are clearly 
required for the wall. 
According to ACI 318 Sec. 21.6.6.4 [21.7.6.4], the special boundary elements must have a plan length of 
c - 0.1lw, or 0.5c, whichever is greater. For the 4,000 psi concrete, the first of these values is 110 - 
0.1(270) = 83 in., and the second is 0.5(110) = 55 in. Both of these are significantly greater than the 30 
in. assumed in the analysis. Hence, the 30-in. boundary element is not adequate for the lower levels of 
the wall if fc' = 4,000 psi. For the 6,000 psi concrete, the required length of the boundary element is 
58-0.1(270) = 31 in., or 0.5(58) = 29 in. The required value of 31 in. is only marginally greater than the 
30 in. provided and will be deemed acceptable for the purpose of this example. 
The vertical extent of the special boundary elements must not be less than the larger of lw or Mu/4Vu. The 
wall length lw = 22.5 ft and, of the wall at Level 1, Mu/4Vu = 273,043/4(539) =126.6 in., or 10.6 ft. 22.5 ft 
controls and will be taken as the required length of the boundary element above the first floor. The 
special boundary elements will begin at the basement level, and continue up for the portion of the wall 
supporting Levels 2 and 3. Above that level, boundary elements will still be present, but they will not be 
reinforced as special boundary elements. 
Another consideration for the boundary elements is at what elevation the concrete may change from 6,000 
psi NW to 4,000 LW concrete. Using the requirement that boundary elements have a maximum plan 
dimension of 30 in., the neutral axis depth (c) must not exceed approximately 57 in. As may be seen from 
Figure 6-35, this will occur when the factored axial force in the wall falls below 3,000 kips. From Table 
6-17, this will occur between Levels 6 and 7. Hence, 6,000 psi concrete will be continued up through 
Level 7. Above Level 7, 4,000 psi LW concrete may be used.
Chapter 6, Reinforced Concrete 
6-75 
(12) #9 
#5 
at 5" o.c. Alternate 
location of 90 bend 
4" 
4" 
#5 at 5" o.c. 
4" 
#5 x 
developed in wall 
Figure 6-36 Details of structural wall boundary element (1.0 in = 25.4 mm). 
Where special boundary elements are required, transverse reinforcement must conform to ACI 318 Sec. 
21.6.6.4(c) [21.7.6.4(c)], which refers to Sec. 21.4.4.1 through 21.4.4.3. If rectangular hoops are used, 
the transverse reinforcement must satisfy ACI 318 Eq. 21-4: 
0.09 c 
sh c 
yh 
f A shf
' 
= 
If #5 hoops are used in association with two crossties in each direction, Ash = 4(0.31) = 1.24 in.2, and hc = 
30 - 2(1.5) - 0.525 = 26.37 in. With fc' = 6 ksi and fyh = 60 ksi: 
0.09(12.62.437) 6 5.22 
60 
s= = 
If 4,000 psi concrete is used, the required spacing increases to 7.83 in. 
Maximum spacing is the lesser of h/4, 6db, or sx where sx = 4 + (14-hx)/3. With hx = 8.83 in., the third of 
these spacings controls at 5.72 in. The 5.22-in. spacing required by ACI 318 Eq. 21-4 is less than this, so 
a spacing of 5 in. on center will be used wherever the special boundary elements are required. 
Details of the panel and boundary element reinforcement are shown in Figures 6-36 and 6-37, 
respectively. The vertical reinforcement in the boundary elements will be spliced as required using Type 
2 mechanical splices at all locations. According to Table 6-13 (prepared for 4,000 psi LW concrete), 
there should be no difficulty in developing the horizontal panel steel into the 30-in.-by-30-in. boundary 
elements.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
6-76 
#5 at 4" 
(24) #11 
#4 at 12" EF 
#4 at 6" EF 
Class 
B 
Class 
B 
#5 at 4"
#4 at 12" EF 
#4 at 6" EF 
#5 at 4" 
(24) #11 
#4 at 12" 
See figure EW-EF 
6-26 
#4 at 12" 
EW-EF 
#4 at 4" 
(24) #9 
#4 at 12" 
EW-EF 
#4 at 4" 
#4 at 12" 
EW-EF 
#4 at 4" 
(24) #10 
#4 at 12" 
EW-EF 
#4 at 4" 
#4 at 12" 
EW-EF 
Class 
B 
Class 
B 
See figure 
6-26 
(12) #9 
#4 at 12" EF 
#4 at 12" EF 
See figure 
6-26 
(12) #9 
#4 at 12" EF 
#4 at 12" EF 
See figure 
6-26 
#4 at 12" EF 
#4 at 12" EF 
See figure 
6-26 
#4 at 12" EF 
#4 at 12" EF 
See figure 
6-26 
#4 at 12" EF 
#4 at 12" EF 
f'c = 4.0ksi 
(LW) 
f'c = 
6.0ksi 
(NW) 
f'c = 4.0ksi 
(LW) 
Figure 6-37 Overall details of structural wall (1.0 in = 25.4 mm). 
ACI 318 Sec. 21.6.6.4(d) [21.7.6.4(d)] also requires that the boundary element transverse reinforcement 
be extended into the foundation tie beam a distance equal to the tension development length of the #9 bars 
used as longitudinal reinforcement in the boundary elements. Assuming the tie beam consists of 6,000 psi 
NW concrete, the development length for the #9 bar is 2.5 times the value given by ACI 318 Eq. 21-6: 
2.5 2.5 60,000(1.128) 33.6 in. 
65 65 6,000 
y b 
d 
c 
f d 
l 
f 
. . 
= . .= = 
.. ' .. 
Hence, the transverse boundary element reinforcement consisting of #5 hoops with two crossties in each 
direction, spaced at 5 in. on center, will extend approximately 3 ft into the foundation tie beam. 
6.5 STRUCTURAL DESIGN OF THE HONOLULU BUILDING 
The structure illustrated in Figure 6-1 and 6-2 is now designed and detailed for the Honolulu building. 
Because of the relatively moderate level of seismicity, the lateral load resisting system will consist of a 
series of intermediate moment-resisting frames in both the E-W and N-S directions. This is permitted for 
Seismic Design Category C buildings under Provisions Sec. 9.6 [9.4]. Design guidelines for the 
reinforced concrete framing members are provided in ACI 318 Sec. 21.10 [21.12].
Chapter 6, Reinforced Concrete 
6-77 
Preliminary design for the Honolulu building indicated that the size of the perimeter frame girders could 
be reduced to 30 in. deep by 20 in. wide (the Berkeley building has girders that are 32 in. deep by 22.5 in. 
wide) and that the columns could be decreased to 28 in. square (the Berkeley building uses 30-in.-by-30- 
in. columns). The haunched girders along Frames 2 through 7 have a maximum depth of 30 in. and a 
width of 20 in. in the Honolulu building (the Berkeley building had haunches with a maximum depth of 
32 in. and a width of 22.5 in.). The Frame 2 through Frame 7 girders in Bays B-C have a constant depth 
of 30 in. Using these reduced properties, the computed drifts will be increased over those shown in 
Figure 6-6, but will clearly not exceed the drift limits. 
6.5.1 Material Properties 
ACI 318 has no specific limitations for materials used in structures designed for moderate seismic risk. 
For the Honolulu building, 4,000 psi sand-LW concrete is used with ASTM A615 Grade 60 rebar for 
longitudinal reinforcement and Grade 60 or Grade 40 rebar for transverse reinforcement. 
6.5.2 Combination of Load Effects 
For the design of the Honolulu building, all masses and superimposed gravity loads generated for the 
Berkeley building are used. This is conservative because the members for the Honolulu building are 
slightly smaller than the corresponding members for the Berkeley building. Also, the Honolulu building 
does not have reinforced concrete walls on Gridlines 3, 4, 5, and 6 (these walls are replaced by infilled, 
nonstructural masonry designed with gaps to accommodate frame drifts in the Honolulu building). 
Provisions Sec. 5.2.7 [4.2.2] and Eq. 5.2.7-1 and 5.2.7-2 [4.2-1 and 4.2-2] require a combination of load 
effects to be developed on the basis of ASCE 7, except that the earthquake load (E) is defined as: 
E=.QE+0.2SDSD 
when gravity and seismic load effects are additive and as: 
E=.QE-0.2SDSD 
when the effects of seismic load counteract gravity. 
For Seismic Design Category C buildings, Provisions Sec. 5.2.4.1 [4.3.3.1] permits the reliability factor 
(.) to be taken as 1.0. The special load combinations of Provisions Eq. 5.2.7-1 and 5.2.7-2 [4.2-3 and 
4.2-4] do not apply to the Honolulu building because there are no discontinuous elements supporting 
stiffer elements above them. (See Provisions Sec. 9.6.2 [9.4.1].) 
For the Honolulu structure, the basic ASCE 7 load combinations that must be considered are: 
1.2D + 1.6L 
1.2D + 0.5L  1.0E 
0.9D  1.0E 
The ASCE 7 load combination including only 1.4 times dead load will not control for any condition in 
this building.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
6-78 
Substituting E from the Provisions and with . taken as 1.0, the following load combinations must be used 
for earthquake: 
(1.2 + 0.2SDS)D + 0.5L + E 
(1.2 + 0.2SDS)D + 0.5L - E 
(0.9 - 0.2SDS)D + E 
(0.9 - 0.2SDS)D -E 
Finally, substituting 0.472 for SDS (see Sec. 6.1.1), the following load combinations must be used for 
earthquake: 
1.30D + 0.5L + E 
1.30D + 0.5L - E 
0.80D + E 
0.80D - E 
Note that the coefficients on dead load have been slightly rounded to simplify subsequent calculations. 
As E-W wind loads apparently govern the design at the lower levels of the building (see Sec. 6.2.6 and 
Figure 6-4), the following load combinations should also be considered: 
1.2D + 0.5L + 1.6W 
1.2D + 0.5L - 1.6W 
0.9D - 1.6W 
The wind load (W) from ASCE 7 includes a directionality factor of 0.85. 
It is very important to note that use of the ASCE 7 load combinations in lieu of the combinations given in 
ACI 318 Chapter 9 requires use of the alternate strength reduction factors given in ACI 318 Appendix C: 
Flexure without axial load f = 0.80 
Axial compression, using tied columns f = 0.65 (transitions to 0.8 at low axial loads) 
Shear if shear strength is based on nominal axial-flexural capacity f = 0.75 
Shear if shear strength is not based on nominal axial-flexural capacity f = 0.55 
Shear in beam-column joints f = 0.80 
[The strength reduction factors in ACI 318-02 have been revised to be consistent with the ASCE 7 load 
combinations. Thus, the factors that were in Appendix C of ACI 318-99 are now in Chapter 9 of ACI 
318-02, with some modification. The strength reduction factors relevant to this example as contained in 
ACI 318-02 Sec. 9.3 are: 
Flexure without axial load f = 0.9 (tension-controlled sections) 
Axial compression, using tied columns f = 0.65 (transitions to 0.9 at low axial loads) 
Shear if shear strength is based o nominal axial-flexural capacity f = 0.75 
Shear if shear strength is not based o nominal axial-flexural capacity f = 0.60 
Shear in beam-column joints f = 0.85] 
6.5.3 Accidental Torsion and Orthogonal Loading (Seismic Versus Wind) 
As has been discussed and as illustrated in Figure 6-4, wind forces appear to govern the strength 
requirements of the structure at the lower floors, and seismic forces control at the upper floors. The 
seismic and wind shears, however, are so close at the midlevels of the structure that a careful evaluation
Chapter 6, Reinforced Concrete 
6-79 
PW 
L 
PL 
PW 
Case 1 
0.75 PL 
Case 3 
0.75 PL 
0.75 PW 
0.75 PW 
Case 2 Case 4 
0.75 PW 
0.56 PW 
0.75 PL 
0.56 PW 0.56 PL 
0.75 PW 
L 
0.56 PL 
PW 
0.75 PW 0.75 PL 
0.75 PL 
0.75 PW 
0.75 P 
PL 
P 
Figure 6-38 Wind loading requirements from ASCE 7. 
must be made to determine which load governs for strength. This determination is complicated by the 
differing (wind versus seismic) rules for applying accidental torsion and for considering orthogonal 
loading effects. 
Because the Honolulu building is in Seismic Design Category C and has no plan irregularities of Type 5 
in Provisions Table 5.2.3.2 [4.3-2], orthogonal loading effects need not be considered per Provisions Sec. 
5.2.5.2.2 [4.4.2.2]. However, as required by Provisions Sec. 5.4.4.2 [5.2.4.2], seismic story forces must 
be applied at a 5 percent accidental eccentricity. Torsional amplification is not required per Provisions 
Sec. 5.4.4.3 [5.2.4.3] because the building does not have a Type 1a or 1b torsional irregularity. (See Sec. 
6.3.2 and 6.3.4 for supporting calculations and discussion.) 
For wind, ASCE 7 requires that buildings over 60 ft in height be checked for four loading cases. The 
required loads are shown in Figure 6-38, which is reproduced directly from Figure 6-9 of ASCE 7. In 
Cases 1 and 2, load is applied separately in the two orthogonal directions. Case 2 may be seen to produce 
torsional effects because 7/8 of the total force is applied at an eccentricity of 3.57% the building width. 
This is relatively less severe than required for seismic effects, where 100 percent of the story force is 
applied at a 5 percent eccentricity. 
For wind, Load Cases 3 and 4 require that 75 percent of the wind pressures from the two orthogonal 
directions be applied simultaneously. Case 4 is similar to Case 2 because of the torsion inducing pressure 
unbalance. As mentioned earlier, the Honolulu building has no orthogonal seismic loading requirements.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
6-80 
In this example, only loading in the E-W direction is considered. Hence, the following lateral load 
conditions were applied to the ETABS model: 
100% E-W Seismic applied at 5% eccentricity 
ASCE 7 Wind Case 1 applied in E-W direction only 
ASCE 7 Wind Case 2 applied in E-W direction only 
ASCE 7 Wind Case 3 
ASCE 7 Wind Case 4 
All cases with torsion are applied in such a manner as to maximize the shears in the elements of Frame 1. 
6.5.4 Design and Detailing of Members of Frame 1 
In this section, the girders and a typical interior column of Level 5 of Frame 1 are designed and detailed. 
For the five load cases indicated above, the girder shears produced from seismic effects control at the fifth 
level, with the next largest forces coming from direct E-W wind without torsion. This is shown 
graphically in Figure 6-39, where the shears in the exterior bay of Frame 1 are plotted vs. story height. 
Wind controls at the lower three stories and seismic controls for all other stories. This is somewhat 
different from that shown in Figure 6-4, wherein the total story shears are plotted and where wind 
controlled for the lower five stories. The basic difference between Figures 6-4 and 6-39 is that Figure 6- 
39 includes accidental torsion and, hence, Frame 1 sees a relatively larger seismic shear.
Chapter 6, Reinforced Concrete 
6-81 
0 
20 
40 
60 
80 
100 
120 
140 
160 
0 5 10 15 20 25 30 35 40 
Girder shear, kips 
Height, ft 
Seismic (with torsion) 
Wind (without torsion) 
Figure 6-39 Wind vs. seismic shears in exterior bay of Frame 1 (1.0 ft = 0.3048 m, 1.0 kip = 4.45kN). 
6.5.4.1 Initial Calculations 
The girders of Frame 1 are 30 in. deep and 20 in. wide. For positive moment bending, the effective width 
of the compression flange is taken as 20 + 20(12)/12 = 40.0 in. Assuming 1.5 in. cover, #3 stirrups and 
#8 longitudinal reinforcement, the effective depth for computing flexural and shear strength is 27.6 in. 
6.5.4.2 Design of Flexural Members 
ACI 318 Sec. 21.10.4 [21.12.4] gives the minimum requirements for longitudinal and transverse 
reinforcement in the beams of intermediate moment frames. The requirements for longitudinal steel are 
as follows: 
1. The positive moment strength at the face of a joint shall be at least one-third of the negative moment 
strength at the same joint. 
2. Neither the positive nor the negative moment strength at any section along the length of the member 
shall be less than one-fifth of the maximum moment strength supplied at the face of either joint.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
6-82 
2,835 
2,835 
2,886 
2,852 
2,796 
2,492 
502 
802 
573 502 
729 729 729 729 
502 
155 
176 
247 225 225 
155 155 
225 225 
3,526 3,658 
969 
2,394 1,850 
850 
3,946 3,927 
2,302 2,269 
850 
3,910 
2,252 
3,910 
2,252 1.3D+0.5L+E 
0.8D - E 
1.2D+1.6L 
(a) 
Span layout 
and loading 
(b) 
Earthquake moment 
(in.-kips) 
(c) 
Unfactored DL moment 
(in.-kips) 
(d) 
Unfactored LL moment 
(in.-kips) 
(e) 
Required strength 
envelopes (in.-kips) 
17.67' 
20.0' 20.0' 20.0' 
' 
W = 0.66 kips/ft 
W = 2.14 kips/ft 
L
D 
Figure 6-40 Bending moment envelopes at Level 5 of Frame 1 (1.0 ft = 0.3048 m, 1.0 kip/ft 
= 14.6 kN/m, 1.0 in.-kip = 0.113 kN-m). 
The second requirement has the effect of requiring top and bottom reinforcement along the full length of 
the member. The minimum reinforcement ratio at any section is taken from ACI 318 Sec. 10.5.1 as 200/fy 
or 0.0033 for fy = 60 ksi. However, according to ACI 318 Sec. 10.5.3, the minimum reinforcement 
provided need not exceed 1.3 times the amount of reinforcement required for strength. 
The gravity loads and design moments for the first three spans of Frame 1 are shown in Figure 6-40. The 
seismic moments are taken directly from the ETABS analysis, and the gravity moments were computed 
by hand using the ACI coefficients. All moments are given at the face of the support. The gravity 
moments shown in Figures 6-40c and 6-40d are slightly larger than those shown for the Berkeley building 
(Figure 6-14) because the clear span for the Honolulu building increases due to the reduction in column 
size from 30 in. to 28 in. 
Based on preliminary calculations, the reinforcement layout of Figure 6-41 will be checked. Note that the 
steel clearly satisfies the detailing requirements of ACI 318 Sec. 21.10.4 [21.12.4].
Chapter 6, Reinforced Concrete 
6-83 
' ' 
(1) #7 
(2) #8 
20'-0" 
30" 
28" 48" 
#3x stirrups spaced from 
each support: 1 at 2", 10 at 6", 
5 at 8" (typical each span). 
(3) #8 
(2) #8 
(3) #8 
(2) #8 
(3) #8 
(2) #8 
(1) #7 (1) #7 
(2) #8 
(1) #7 
(3) #7 (3) #7 
5'-0" 
(typical) 
Figure 6-41 Preliminary reinforcement layout for Level 5 of Frame 1 (1.0 in = 25.4 
mm, 1.0 ft = 0.3048 m). 
6.5.4.2.1 Design for Negative Moment at Face of Support A 
Mu = -1.3 (502) - 0.5 (155) - 1.0 (2,796) = -3,526 in.-kips 
Try three #7 short bars and two #8 long bars. 
As = 3 (0.60) + 2 (0.79) = 3.38 in.2 
. = 0.0061 
Depth of compression block, a = [3.38 (60)]/[0.85 (4) 20] = 2.98 in. 
Nominal moment capacity, Mn = Asfy(d - a/2) = [3.38 (60.0)] [27.6 - 2.98/2] = 5,295 in.-kips 
Design capacity, fMn = 0.8(5,295) = 4,236 in.-kips > 3,526 in.-kips OK 
6.5.4.2.2 Design for Positive Moment at Face of Support A 
Mu = -0.8 (502) + 1.0 (2,796) = 2,394 in.-kips 
Try three #8 long bars. 
Asfy = 3 (0.79) = 2.37 in.2 
. = 0.0043 
a = 2.37 (60)/[0.85 (4) 40] = 1.05 in. 
Mn = Asfy(d - a/2) = [2.37 (60.0)][27.6 - 1.05/2] = 3,850 in.-kips 
fMn = 0.8(3850) = 3,080 in.-kips > 2,394 in.-kips OK 
This reinforcement also will work for positive moment at all other supports. 
6.5.4.2.3 Design for Negative Moment at Face of Support A' 
Mu = -1.3 (729) - 0.5 (225) - 1.0 (2,886) = 3,946 in.-kips 
Try four #8 long bars and one #7 short bar: 
As = 4 (0.79) + 1 (0.6) = 3.76 in.2 
. = 0.0068
FEMA 451, NEHRP Recommended Provisions: Design Examples 
6-84
a = [3.76 (60)]/[0.85 (4) 20] = 3.32 in. 
Mn = Asfy(d - a/2) = [3.76 (60.0)][27.6 - 3.32/2] = 5,852 in.-kips 
fMn = 0.8(5,852) = 4,681 in.-kips > 3,946 in.-kips OK 
This reinforcement will also work for negative moment at Supports B and C. Therefore, the flexural 
reinforcement layout shown in Figure 6-41 is adequate. The top short bars are cut off 5 ft-0 in. from the 
face of the support. The bottom bars are spliced in Spans A'-B and C-C' with a Class B lap length of 48 
in. Unlike special moment frames, there are no requirements that the spliced region of the bars in 
intermediate moment frames be confined by hoops over the length of the splice. 
6.5.4.2.4 Design for Shear Force in Span A'-B: 
ACI 318 Sec. 21.10.3 [21.12.3] provides two choices for computing the shear strength demand in a 
member of an intermediate moment frame: 
1. The first option requires that the design shear force for earthquake be based on the nominal moment 
strength at the ends of the members. Nominal moment strengths are computed with a flexural 
reinforcement tensile strength of 1.0fy and a flexural f factor of 1.0. The earthquake shears computed 
from the nominal flexural strength are added to the factored gravity shears to determine the total 
design shear. 
2. The second option requires that the design earthquake shear force be 2.0 times the factored 
earthquake shear taken from the structural analysis. This shear is used in combination with the 
factored gravity shears. 
For this example, the first option is used. The nominal strengths at the ends of the beam were computed 
earlier as 3850 in.-kips for positive moment at Support A' and 5,852 in.-kips for negative moment at 
Support B. Compute the design earthquake shear VE: 
5,852 3,850 45.8 kips 
VE 212 
+ 
= = 
where 212 in. is the clear span of the member. For earthquake forces acting in the other direction, the 
earthquake shear is 43.1 kips. 
The gravity load shears at the face of the supports are: 
2.14(20 2.33) 18.9 kips 
VD 2
- 
= = 
0.66(20 2.33) 5.83 kips 
VL 2
- 
= = 
The factored design shear Vu = 1.3(18.9) + 0.5(5.8) + 1.0(45.8) = 73.3 kips. This shear force applies for 
earthquake forces coming from either direction as shown in the shear strength design envelope in Figure 
6-42. 
The design shear force is resisted by a concrete component (Vc) and a steel component (Vs). Note that the 
concrete component may be used regardless of the ratio of earthquake shear to total shear. The required 
design strength is:
Chapter 6, Reinforced Concrete 
6-85 
Vu # fVc + fVs 
where f = 0.75 for shear. 
(0.85) (2 4,000)20(27.6) 59.3 kips 
Vc= 1,000 = 
The factor of 0.85 above reflects the reduced shear capacity of sand-LW concrete. 
The shear to be resisted by steel, assuming stirrups consist of two #3 legs (Av = 0.22) and fy = 40 ksi is: 
73.3 0.75(59.3) 38.4 kips 
0.75 
u c 
s 
V V V
f 
f 
- - 
= = = 
Using VS = Av fyd/s: 
(0.22)(40)(27.6) 6.32 in. 
38.4 
s= = 
Minimum transverse steel requirements are given in ACI 318 Sec. 21.10.4.2 [21.12.4.2]. The first stirrup 
should be placed 2 in. from the face of the support, and within a distance 2h from the face of the support, 
the spacing should be not greater than d/4, eight times the smallest longitudinal bar diameter, 24 times the 
stirrup diameter, or 12 in. For the beam under consideration d/4 controls minimum transverse steel, with 
the maximum spacing being 27.6/4 = 6.9 in. This is slightly greater, however, than the 6.32 in. required 
for strength. In the remainder of the span, stirrups should be placed at a maximum of d/2 (ACI 318 Sec. 
21.10.4.3 [21.12.4.3]). 
Because the earthquake shear (at midspan) is greater than 50 percent of the shear strength provided by 
concrete alone, the minimum requirements of ACI 318 Sec. 11.5.5.3 must be checked: 
0.2(40,000) 8.0 in. 
smax= 50(20) = 
This spacing controls over the d/2 requirement. The final spacing used for the beam is shown in Figure 6- 
41. This spacing is used for all other spans as well. The stirrups may be detailed according to ACI 318 
Sec. 7.1.3, which requires a 90-degree hook with a 6db extension. This is in contrast to the details of the 
Berkeley building where full hoops with 135-degree hooks are required in the critical region (within 2d 
from the face of the support) and stirrups with 135-degree hooks are required elsewhere.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
6-86 
' 
5,852 5,852 
3,850 3,850 3,850 3,850 
5,295 5,852 
45.8 45.8 45.8 
43.1 45.8 45.8 
27.5 
27.5 27.5 
27.5 
27.5 
27.5 
73.3 
18.3 
73.3 
18.3 
Loading
18.3 
73.3 
18.3 
73.3 73.3 
18.3 
70.6 
15.6 
(a) 
Seismic moment 
(tension side) 
in.-kips 
kips 
positive 
kips 
positive 
kips 
positive 
(b) 
Seismic shear 
(c) 
Gravity shear 
(1.175D + 1.0L) 
(d) 
Design shear 
seismic + gravity 
14" 212" 14" 
240" 
Figure 6-42 Shear strength envelopes for Span A'-B of Frame 1 (1.0 in = 
25.4 mm, 1.0 kip = 4.45kN, 1.0 in.-kip = 0.113 kN-m).
Chapter 6, Reinforced Concrete 
6-87 
' 
20'-0" 20'-0" 
12'-6" 
Level 4 
Level 5 
30" 30" 
28" 
See Figure 6-41 
for girder 
reinforcement 
P = 54 kips Includes 
P = 528 kips level 5 
Figure 6-43 Isolated view of column A' (1.0 ft = 0.3048 m, 1.0 kip = 
4.45kN). 
6.5.4.3 Design of Typical Interior Column of Frame 1 
This section illustrates the design of a typical interior column on Gridline A'. The column, which 
supports Level 5 of Frame 1, is 28 in. square and is constructed from 4,000 psi LW concrete, 60 ksi 
longitudinal reinforcement, and 40 ksi transverse reinforcement. An isolated view of the column is 
shown in Figure 6-43. 
The column supports an unfactored axial dead load of 528 kips and an unfactored axial live load of 54 
kips. The ETABS analysis indicates that the axial earthquake force is 33.2 kips, the earthquake shear 
force is 41.9 kips, and the earthquake moments at the top and the bottom of the column are 2,137 and 
2,708 in.-kips, respectively. Moments and shears due to gravity loads are assumed to be negligible. 
6.5.4.3.1 Design of Longitudinal Reinforcement 
The factored gravity force for maximum compression (without earthquake) is: 
Pu = 1.2(528) + 1.6(54) = 720 kips 
This force acts with no significant gravity moment. 
The factored gravity force for maximum compression (including earthquake) is: 
Pu = 1.3(528) + 0.5(54) + 33.2 = 746.6 kips 
The factored gravity force for minimum compression (including earthquake) is: 
Pu = 0.8(528) - 33.2 = 389.2 kips
FEMA 451, NEHRP Recommended Provisions: Design Examples 
6-88 
Since the frame being designed is unbraced in both the N-S and E-W directions, slenderness effects 
should be checked. For a 28-in.-by-28-in. column with a clear unbraced length. lu = 120 in., r = 0.3(28) 
= 8.4 in. (ACI 318 Sec. 10.11.3) and lu/r = 120/8.4 = 14.3. 
ACI 318 Sec. 10.11.4.2 states that the frame may be considered braced against sidesway if the story 
stability factor is less than 0.05. This factor is given as: 
u 0 
u c 
Q P 
V l
= S d 
which is basically the same as Provisions Eq. 5.4.6.2-1 [5.2-16] except that in the ACI equation, the 
gravity forces are factored. [Note also that the equation to determine the stability coefficient has been 
changed in the 2003 Provisions. The importance factor, I, has been added to 2003 Provisions Eq. 5.2-16. 
However, this does not affect this example because I = 1.0.] ACI is silent on whether or not d0 should 
include Cd. In this example, d0 does not include Cd, and is therefore consistent with the Provisions. As 
can be seen from earlier calculations shown in Table 6-12b, the ACI story stability factor will be in excess 
of 0.05 for Level 5 of the building responding in the E-W direction. Hence, the structure must be 
considered unbraced. 
Even though the frame is defined as unbraced, ACI 318 Sec. 10.13.2 allows slenderness effects to be 
neglected when klu/r < 22. This requires that the effective length factor k for this column be less than 
1.54. For use with the nomograph for unbraced columns (ACI 318 Figure R10.12.1b): 
(45,000) 187.5 
Girder 240 
EI E E 
L 
.. .. = = 
. . 
According to ACI 318 Sec. 10.12.3: 
0.4
(1 ) 
150 
Column 
d 
Column 
EI 
EI 
L 
 
. . 
.. .. =.. + .. 
. . 
Using the 1.2 and 1.6 load factors on gravity load: 
1.2(528) 0.88 
 d= 720 = 
3 
28 (28) 51,221 in.4 
IColumn= 12 = 
0.4(51, 221 ) 
1 0.88 72.7 
Column 150 
E 
EI E 
L 
.. .. = + = 
. . 
Because there is a column above and below as well as a beam on either side: 
72.7 0.39 
.Top =. Bottom=187.5 =
Chapter 6, Reinforced Concrete 
5For loading in the N-S direction, the column under consideration has no beam framing into it in the direction of loading. If the 
stiffness contributed by the joists and the spandrel beam acting in torsion is ignored, the effective length factor for the column in 
the N-S direction is effectively infinity. However, this column is only one of four in a story containing a total of 36 columns. 
Since each of the other 32 columns has a lateral stiffness well in excess of that required for story stability in the N-S direction, 
the columns on Lines A' and C' can be considered to be laterally supported by the other 32 columns and therefore can be 
designed using an effective length factor of 1.0. A P-delta analysis carried out per the ACI Commentary would be required to 
substantiate this. 
6-89 
500 
1,500 
2,500 
-500 
200 400 600 800 1,000 
P (kips) 
1,000 
2,000 
3,000
0 
M x (ft-kips) 
0 
Figure 6-44 Interaction diagram for column (1.0 kip = 
4.45kN, 1.0 ft-kip = 1.36 kN-m). 
and the effective length factor k = 1.15 (ACI 318 Figure R10.12.1b). As the computed effective length 
factor is less than 1.54, slenderness effects need not be checked for this column.5 
Continuing with the design, an axial-flexural interaction diagram for a 28-in.-by-28-in. column with 12 
#8 bars (. = 0.0121) is shown in Figure 6-44. The column clearly has the strength to support the applied 
loads (represented as solid dots in the figure). 
6.5.4.3.2 Design and Detailing of Transverse Reinforcement 
ACI 318 Sec. 21.10.3 [21.12.3] allows the column to be checked for 2.0 times the factored shear force as 
derived from the structural analysis. The ETABS analysis indicates that the shear force is 41.9 kips and 
the design shear is 2.0(41.9) = 83.8 kips. 
The concrete supplies a capacity of: 
Vc=0.85(2)fc'bwd=0.85(2) 4,000(28)(25.6)= 77.1 kips
FEMA 451, NEHRP Recommended Provisions: Design Examples 
6-90 
The requirement for steel reinforcement is: 
83.8 0.75(77.1) 34.6 kips 
0.75 
u c 
s 
V V V
f 
f
- - 
= = = 
Using ties with four #3 legs, s = [4(0.11)] [40.0 (25.6/34.6)] = 13.02 in. 
ACI 318 Sec. 21.10.5 [21.12.5] specifies the minimum reinforcement required. Within a region lo from 
the face of the support, the tie spacing should not exceed: 
8.0db = 8.0 (1.008) = 8.00 in. (using #8 longitudinal bars) 
24dtie = 24 (3/8) = 9.0 in. (using #3 ties) 
1/2 the smallest dimension of the frame member = 28/2 = 14 in. 
12 in. 
The 8.0 in. maximum spacing controls. Ties at this spacing are required over a length lo of: 
1/6 clearspan of column = 120/6 = 20 in. 
maximum cross section dimension = 28 in. 
18.0 in. 
Given the above, a four-legged #3 tie spaced at 8 in. over a depth of 28 in. will be used. One tie will be 
provided at 4 in. below the beam soffit, the next tie is placed 4 in. above the floor slab, and the remaining 
ties are spaced at 8 in. on center. The final spacing is as shown in Figure 6-45. Note that the tie spacing 
is not varied beyond lo.
Chapter 6, Reinforced Concrete 
6-91 
' 
Level 7 
Level 6 
28" 
30" 30" 
28" 
28" 
(12) #8 bars 
4" 14 spaces at 8" o.c. 4" 
3" 3 at 8" 3" 
Figure 6-45 Column reinforcement (1.0 in = 25.4 mm). 
6.5.4.4 Design of Beam-Column Joint 
Joint reinforcement for intermediate moment frames is addressed in ACI 318 Sec. 21.10.5.3 [21.12.5.5], 
which refers to Sec. 11.11.2. ACI 318 Sec. 11.11.2 requires that all beam-column connections have a 
minimum amount of transverse reinforcement through the beam-column joints. The only exception is in 
nonseismic frames where the column is confined on all four sides by beams framing into the column. The 
amount of reinforcement required is given by ACI 318 Eq. 11-13: 
50 w 
v 
y 
A b s 
f 
. . 
= .. .. 
. . 
This is the same equation used to proportion minimum transverse reinforcement in beams. Assuming Av 
is supplied by four #3 ties and fy = 40 ksi: 
4(0.11)(40,000) 12.6 in. 
50(28) 
s= =
FEMA 451, NEHRP Recommended Provisions: Design Examples 
6-92 
1.3D + 0.5L + E 
1.2D + 1.6L 
1.3D + 0.5L - E 
0.8D + E 
0.8D - E 
Strength envelope 
(b) Moment envelope 
(in.-kips) 
L
D 
-2,000 
0 
-4,000 
-6,000 
-8,000 
-10,000 
-12,000 
8,000 
6,000 
4,000 
2,000 
Level 5 
Level 7 
(a) Span geometry 
and loading 
48" 48" 
1'-2" 
8'-10" 10'-0" 
1'-2" 
10'-0" 8'-10" 
W = 0.90 kips/ft 
W = 0.90 kips/ft 
30" 
20" 
48" 
(3) #10 (3) #10 (2) #10 (2) #10 
(4) #9 
(4) #9 
Figure 6-46 Loads, moments, and reinforcement for haunched girder (1.0 in = 25.4 mm, 1.0 ft = 0.3048 
m, 1.0 kip/ft = 14.6 kN/m, 1.0 in.-kip = 0.113 kN-m). 
This effectively requires only two ties within the joint. However, the first tie will be placed 3 in. below 
the top of the beam and then three additional ties will be placed below this hoop at a spacing of 8 in. The 
final arrangement of ties within the beam-column joint is shown in Figure 6-45. 
6.5.5 Design of Members of Frame 3 
6.5.5.1 Design of Haunched Girder 
A typical haunched girder supporting Level 5 of Frame 3 is now illustrated. This girder, located between 
Gridlines A and B, has a variable depth with a maximum depth of 30 in. at the support and a minimum 
depth of 20 in. for the middle half of the span. The length of the haunch at each end (as measured from 
the face of the support) is 106 in. The width of the girder is 20 in. throughout. The girder frames into 28- 
in.-by-28-in. columns on Gridlines A and B. As illustrated in Figure 6-46c, the reinforcement at Gridline 
B is extended into the adjacent span (Span B-C) instead of being hooked into the column.
Chapter 6, Reinforced Concrete 
6-93 
Based on a tributary gravity load analysis, this girder supports an average of 3.38 kips/ft of dead load and 
0.90 kips/ft of reduced live load. A gravity load analysis of the girder was carried out in a similar manner 
similar to that described above for the Berkeley building. 
For determining earthquake forces, the entire structure was analyzed using the ETABS program. This 
analysis included 100 percent of the earthquake forces in the E-W direction placed at a 5 percent 
eccentricity with the direction of the eccentricity set to produce the maximum seismic shear in the 
member. 
6.5.5.2 Design of Longitudinal Reinforcement 
The results of the analysis are shown in Figure 6-46b for five different load combinations. The envelopes 
of maximum positive and negative moment indicate that 1.2D + 1.6L and 1.3D + 0.5L  E produce 
approximately equal negative end moments. Positive moment at the support is nearly zero under 0.8D - 
E, and gravity controls midspan positive moment. Since positive moment at the support is negligible, a 
positive moment capacity of at least one-third of the negative moment capacity will be supplied per ACI 
318 Sec. 21.10.4.1 [21.12.4.1]. The minimum positive or negative moment strength at any section of the 
span will not be less than one-fifth of the maximum negative moment strength. 
For a factored negative moment of 8,106 in.-kips on Gridline A, try six #10 bars. Three of the bars are 
short, extending just past the end of the haunch. The other three bars are long and extend into Span B-C. 
As = 6 (1.27) = 7.62 in.2 
d = 30 - 1.5 - 0.375 - 1.27/2 = 27.49 in. 
. = 7.62/[20 (27.49)] = 0.0139 
Depth of compression block, a = [7.62 (60)]/[0.85 (4) 20.0] = 6.72 in. 
Nominal capacity, Mn = [7.62 (60)](27.49 - 6.72/2) = 11,031 in-kips 
Design capacity, fMn = 0.8(11,031) = 8,824 in.-kips > 8,106 in.-kips OK 
The three #10 bars that extend across the top of the span easily supply a minimum of one-fifth of the 
negative moment strength at the face of the support. 
For a factored negative moment of 10,641 in.-kips on Gridline B, try eight #10 bars. Three of the bars 
extend from Span A-B, three extend from Span B-C, and the remaining two are short bars centered over 
Support B. 
As = 8 (1.27) = 10.16 in.2 
d = 30 - 1.5 - 0.375 - 1.27/2 = 27.49 in. 
. = 10.16/[20 (27.49)] = 0.0185 
a = [10.16 (60)]/[0.85 (4) 20.0] = 8.96 in. 
Mn = [10.16 (60)](27.49 - 8.96/2) = 13,996 in.-kips 
fMn = 0.8(13,996) = 11,221 in.-kips > 10,641 in.-kips OK 
For the maximum factored positive moment at midspan of 2,964 in-kips., try four #9 bars: 
As = 4 (1.0) = 4.00 in.2 
d = 20 - 1.5 - 0.375 - 1.128/2 = 17.56 in. 
. = 4.0/[20 (17.56)] = 0.0114 
a = [4.00 (60)]/[0.85 (4) 84] = 0.84 in. (effective flange width = 84 in.) 
Mn = [4.00 (60)](17.56 - 0.84/2) = 4,113 in.-kips 
fMn = 0.8(4,113) = 3,290 in.-kips > 2,964 OK
FEMA 451, NEHRP Recommended Provisions: Design Examples 
6-94 
Even though they provide more than one-third of the negative moment strength at the support, the four #9 
bars will be extended into the supports as shown in Figure 6-46. The design positive moment strength for 
the 30-in.-deep section with four #9 bars is computed as follows: 
As = 4 (1.00) = 1.00 in.2 
d = 30 - 1.5 - 0.375 - 1.128/2 = 27.56 in. 
. = 4.00/[20 (27.56)] = 0.0073 
a = [4.0 (60)]/[0.85 (4) 20.0] = 0.84 in. 
Mn = [4.00 (60)] (27.56 - 0.84/2) = 6,514 in.-kips 
fMn = 0.8(6,514) = 5,211 in.-kips 
The final layout of longitudinal reinforcement used is shown in Figure 6-46. Note that the supplied 
design strengths at each location exceed the factored moment demands. The hooked #10 bars can easily 
be developed in the confined core of the columns. Splices shown are Class B and do not need to be 
confined within hoops. 
6.5.5.3 Design of Transverse Reinforcement 
For the design for shear, ACI 318 Sec. 21.10.3 [21.12.3] gives the two options discussed above. For the 
haunched girder, the approach based on the nominal flexural capacity (f = 1.0) of the girder will be used 
as follows: 
For negative moment and six #10 bars, the nominal moment strength = 11,031 in.-kips 
For negative moment and eight #10 bars, the nominal strength =13,996 in.-kips 
For positive moment and four #9 bars, the nominal moment strength = 6,514 in.-kips 
Earthquake shear when Support A is under positive seismic moment is: 
VE = (13,996 + 6,514)/(480 - 28) = 45.4 kips 
Earthquake shear when Support B is under positive seismic moment is: 
VE = (11,031 + 6,514)/(480 - 28) = 38.8 kips 
VG = 1.3VD + 0.5VL = 1.3 (63.6) + 0.5(16.9) = 91.1 kips 
Maximum total shear occurs at Support B: 
Vu = 45.4+91.1 = 136.5 kips 
The shear at Support A is 38.8 + 91.9 = 130.1 kips. The complete design shear (demand) strength 
envelope is shown in Figure 6-47a. Due to the small difference in end shears, use the larger shear for 
designing transverse reinforcement at each end. 
Stirrup spacing required for strength is based on two #4 legs with fy = 60 ksi. 
(0.85)(2) 4,000)(20)(27.6) 59.3 kips 
Vc= 1,000 = 
136.5 0.75(59.3) 122.7 kips 
0.75 
u c 
s 
V V V 
f 
f
- - 
= = =
Chapter 6, Reinforced Concrete 
6-95 
8'-10" 20'-0" 8'-10" 
5" o.c. 
8" o.c. 
6" o.c. 
130.1 kips 
136.4 kips 
42.0 kips 
#4 stirrups 
(a) 
Required shear 
strength envelope 
(b) 
Spacing of 
transverse 
reinforcement 
2"
46.0 kips 52.5 kips 
2" 
5" o.c. 
6" o.c. 
Figure 6-47 Shear force envelope for haunched girder (1.0 ft = 
0.3048m, 1.0 in = 25.4 mm, 1.0 kip = 4.45kN). 
Using Vs = Av fyd/s: 
(0.4)(60)(27.6) 5.39 in. 
122.7 
s= = 
Following the same procedure as shown above, the spacing required for other stations is: 
At support, h = 30 in., VU = 136.4 kips s = 5.39 in. 
Middle of haunch, h = 25 in., VU = 114.9 kips s = 6.67 in. 
End of haunch, h =20 in., VU =93.4 kips s = 7.61 in. 
Quarter point of region of 20-in. depth, VU = 69.2 kips s = 12.1 in. 
Midspan, Vu = 45.1 kips s = 29.7 in. 
Within a region 2h from the face of the support, the allowable maximum spacing is d/4 = 6.87 in. at the 
support and approximately 5.60 in. at midhaunch. Outside this region, the maximum spacing is d/2 = 
11.2 in. at midhaunch and 8.75 in. at the end of the haunch and in the 20-in. depth region. At the 
haunched segments at either end of the beam, the first stirrup is placed 2 in. from the face of the support 
followed by four stirrups at a spacing of 5 in, and then 13 stirrups at 6 in. through the remainder of the 
haunch. For the constant 20-in.-deep segment of the beam, a constant spacing of 8 in. is used. The final 
spacing of stirrups used is shown in Figure 6-47b. Three additional stirrups should be placed at each 
bend or kink in the bottom bars. One should be located at the kink and the others approximately 2 in. 
on either side of the kink.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
6-96 
6.5.5.4 Design of Supporting Column 
The column on Gridline A which supports Level 5 of the haunched girder is 28 in. by 28 in. and supports 
a total unfactored dead load of 803.6 kips and an unfactored reduced live load of 78.4 kips. The layout of 
the column is shown in Figure 6-48. Under gravity load alone, the unfactored dead load moment is 2,603 
in.-kips and the corresponding live load moment is 693.0 in.-kips. The corresponding shears are 43.4 and 
11.5 kips, respectively. The factored gravity load combinations for designing the column are as follows: 
Bending moment, M = 1.2(2,603) + 1.6(693) 
= 4,232 in.-kips 
This moment causes tension on the outside face of the top of the column and tension on the inside face of 
the bottom of the column. 
Shear, V = 1.2(43.4) + 0.5(11.5) = 57.8 kips 
Axial compression, P = 1.2(803.6) + 1.6(78.4) 
= 1,090 kips 
For equivalent static earthquake forces acting from west to east, the forces in the column are obtained 
from the ETABS analysis as follows: 
Moment at top of column = 690 in.-kips (tension on inside face subtracts from gravity) 
Moment at bottom of column = 874 in.-kips (tension on outside face subtracts from gravity) 
Shear in column = 13.3 kips (opposite sign of gravity shear) 
Axial force = 63.1 kips tension 
The factored forces involving earthquake from west to east are: 
Moment at top 0.80(2603) - 690 = 1,392 in.-kips 
Moment at bottom = 0.80(2603) - 874 = 1,208 in.-kips 
Shear = 0.80(43.4) - 2(13.3) = 8.1 kips (using the second option for computing EQ shear) 
Axial force = 0.80(803.6) - 63.1 = 580 kips 
For earthquake forces acting from east to west, the forces in the column are obtained from the ETABS 
analysis as follows: 
Moment at top of column = 690 in.-kips (tension on outside face adds to gravity) 
Moment at bottom of column = 874 in.-kips (tension on inside face adds to gravity) 
Shear in column = 13.3 kips (same sign of gravity shear) 
Axial force = 63.1 kips compression
Chapter 6, Reinforced Concrete 
6-97 
P = 78.4 kips Includes 
P = 803.6 kips level 5 
Level 4 
Level 5 
L
D 
20" 
30" 30" 
12'-6" 
28" 
Figure 6-48 Loading for Column A, Frame 3 (1.0 ft = 
0.3048 m, 1.0 in = 25.4 mm, 1.0 kip = 4.45kN). 
The factored forces involving earthquake from east to west are: 
Moment at top 1.3(2,603) + 0.5(693) + 690 = 4,420 in.-kips 
Moment at bottom = 1.3(2,603) + 0.5(693) + 874 = 4,604 in.-kips 
Shear = 1.3(43.4) + 0.5(11.5) + 2(13.3) = 94.6 kips (using second option for computing EQ shear) 
Axial force = 1.3(803.6) + 0.5(78.4) + 63.1 = 1,147 kips 
As may be observed from Figure 6-49, the column with 12 #8 bars is adequate for all loading 
combinations. Since the maximum design shear is less than that for the column previously designed for 
Frame 1 and since minimum transverse reinforcement controlled that column, the details for the column 
currently under consideration are similar to those shown in Figure 6-45. The actual details for the column 
supporting the haunched girder of Frame 3 are shown in Figure 6-50.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
6-98 
500 
1,500 
2,500 
-500 
200 400 600 800 1,000 
P (kips) 
2,000 
1,000
0 
3,000 
M x (ft-kips) 
0 
Figure 6-49 Interaction diagram for Column A, Frame 3 
(1.0 kip = 4.45kN, 1.0 ft-kip = 1.36 kN-m). 
6.5.5.5 Design of Beam-Column Joint 
The detailing of the joint of the column supporting Level 5 of the haunched girder is the same as that for 
the column interior column of Frame A. The joint details are shown in Figure 6-50.
Chapter 6, Reinforced Concrete 
6-99 
Level 5 
Level 4 
28" 
28" 
28" 
(12) #8 bars 
4" 14 spaces at 8" o.c. 4" 
Figure 6-50 Details for Column A, Frame 3 (1.0 
in = 25.4 mm).
7-1 
7 
PRECAST CONCRETE DESIGN 
Gene R. Stevens, P.E. and James Robert Harris, P.E., Ph.D. 
This chapter illustrates the seismic design of precast concrete members using the NEHRP Recommended 
Provisions (referred to herein as the Provisions) for buildings in several different seismic design 
categories. Very briefly, for precast concrete structural systems, the Provisions: 
1. Requires the system (even if the precast carries only gravity loads) to satisfy one of the following two 
sets of provisions: 
a. Resist amplified chord forces in diaphragms and, if moment-resisting frames are used as the 
vertical system, provide a minimum degree of redundancy measured as a fraction of available 
bays, or 
b. Provide a moment-resisting connection at all beam-to-column joints with positive lateral support 
for columns and with special considerations for bearing lengths. 
(In the authors opinion this does not apply to buildings in Seismic Design Category A.) 
2. Requires assurance of ductility at connections that resist overturning for ordinary precast concrete 
shear walls. (Because ordinary shear walls are used in lower Seismic Design Categories, this 
requirement applies in Seismic Design Categories B and C.) 
3. Allows special moment frames and special shear walls of precast concrete to either emulate the 
behavior of monolithic concrete or behave as jointed precast systems. Some detail is given for special 
moment frame designs that emulate monolithic concrete. To validate designs that do not emulate 
monolithic concrete, reference is made to a new ACI testing standard (ACI T1.1-01). 
4. Defines that monolithic emulation may be achieved through the use of either: 
a. Ductile connections, in which the nonlinear response occurs at a connection between a precast 
unit and another structural element, precast or not, or 
b. Strong connections, in which the nonlinear response occurs in reinforced concrete sections 
(generally precast) away from connections that are strong enough to avoid yield even as the 
forces at the nonlinear response location increase with strain hardening. 
5. Defines both ductile and strong connections can be either: 
a. Wet connections where reinforcement is spliced with mechanical couplers, welds, or lap splices 
(observing the restrictions regarding the location of splices given for monolithic concrete) and the 
connection is completed with grout, or
FEMA 451, NEHRP Recommended Provisions: Design Examples 
7-2 
b. Dry connections, which are defined as any connection that is not a wet connection. 
6. Requires that ductile connections be either: 
a. Type Y, with a minimum ductility ratio of 4 and specific anchorage requirements, or 
b. Type Z, with a minimum ductility ratio of 8 and stronger anchorage requirements. 
Many of these requirements have been adopted into the 2002 edition of ACI 318, but some differences 
remain. Where those differences are pertinent to the examples illustrated here, they are explained. 
The examples in Sec. 7.1 illustrate the design of untopped and topped precast concrete floor and roof 
diaphragms of the five-story masonry buildings described in Sec. 9.2 of this volume of design examples. 
The two untopped precast concrete diaphragms of Sec. 7.1.1 show the requirements for Seismic Design 
Categories B and C using 8-in.-thick hollow core precast, prestressed concrete planks. Sec. 7.1.2 shows 
the same precast plank with a 2  in.-thick composite lightweight concrete topping for the five-story 
masonry building in Seismic Design Category D described in Sec. 9.2. Although untopped diaphragms 
are commonly used in regions of low seismic hazard, the only place they are addressed in the Provisions 
is the Appendix to Chapter 9. The reader should bear in mind that the appendices of the Provisions are 
prepared for trial use and comment, and future changes should be expected. 
The example in Sec. 7.2 illustrates the design of an ordinary precast concrete shear wall building in a 
region of low or moderate seismicity, which is where most precast concrete seismic-force-resisting 
systems are constructed. The precast concrete walls in this example resist the seismic forces for a threestory 
office building, located in southern New England (Seismic Design Category B). There are very few 
seismic requirements for such walls in the Provisions. One such requirement qualifies is that overturning 
connections qualify as the newly defined Type Y or Z. ACI 318-02 identifies this system as an 
intermediate precast concrete shear wall and does not specifically define the Type Y or Z connections. 
Given the brief nature of the requirements in both the Provisions and ACI 318, the authors offer some 
interpretation. This example identifies points of yielding for the system and connection features that are 
required to maintain stable cyclic behavior for yielding. 
The example in Sec. 7.3 illustrates the design of a special precast concrete shear wall for a single-story 
industrial warehouse building in the Los Angeles. For buildings in Seismic Design Category D, 
Provisions Sec. 9.1.1.12 [9.2.2.4] requires that the precast seismic-force-resisting system emulate the 
behavior of monolithic reinforced concrete construction or that the systems cyclic capacity be 
demonstrated by testing. The Provisions describes methods specifically intended to emulate the behavior 
of monolithic construction, and dry connections are permitted. Sec. 7.3 presents an interpretation of 
monolithic emulation of precast shear wall panels with ductile, dry connections. Whether this connection 
would qualify under ACI 318-02 is a matter of interpretation. The design is computed using the 
Provisions rules for monolithic emulation; however, the system probably would behave more like a 
jointed precast system. Additional clarity in the definition and application of design provisions of such 
precast systems is needed. 
Tilt-up concrete wall buildings in all seismic zones have long been designed using the precast wall panels 
as shear walls in the seismic-force-resisting system. Such designs have usually been performed using 
design force coefficients and strength limits as if the precast walls emulated the performance of cast-inplace 
reinforced concrete shear walls, which they usually do not. In tilt-up buildings subject to strong 
ground shaking, the in-plane performance of the precast panels has rarely been a problem, primarily 
because there has been little demand for post-elastic performance in that direction. Conventional tilt-up 
buildings may deserve a unique treatment for seismic-resistant design, and they are not the subject of any 
of the examples in this chapter, although tilt-up panels with large height-to-width ratios could behave in 
the fashion described in design example 7.3.
Chapter 7, Precast Concrete Design 
7-3 
In addition to the Provisions, the following documents are either referred to directly or are useful design 
aids for precast concrete construction: 
ACI 318-99 American Concrete Institute. 1999. Building Code Requirements and 
Commentary for Structural Concrete. 
ACI 318-02 American Concrete Institute. 2002. Building Code Requirements and 
Commentary for Structural Concrete. 
AISC LRFD American Institute of Steel Construction. 2002. Manual of Steel Construction, 
Load & Resistance Factor Design, Third Edition. 
ASCE 7 American Society of Civil Engineers. 1998 [2002]. Minimum Design Loads for 
Buildings and Other Structures. 
Hawkins Hawkins, Neil M., and S. K. Ghosh. 2000. Proposed Revisions to 1997 
NEHRP Recommended Provisions for Seismic Regulations for Precast Concrete 
Structures, Parts 1, 2, and 3. PCI Journal, Vol. 45, No. 3 (May-June), No. 5 
(Sept.-Oct.), and No. 6 (Nov.-Dec.). 
Moustafa Moustafa, Saad E. 1981 and 1982. Effectiveness of Shear-Friction 
Reinforcement in Shear Diaphragm Capacity of Hollow-Core Slabs. PCI 
Journal, Vol. 26, No. 1 (Jan.-Feb. 1981) and the discussion contained in PCI 
Journal, Vol. 27, No. 3 (May-June 1982). 
PCI Handbook Precast/Prestressed Concrete Institute. 1999. PCI Design Handbook, Fifth 
Edition. 
PCI Details Precast/Prestressed Concrete Institute. 1988. Design and Typical Details of 
Connections for Precast and Prestressed Concrete, Second Edition. 
SEAA Hollow Core Structural Engineers Association of Arizona, Central Chapter. Design and 
Detailing of Untopped Hollow-Core Slab Systems for Diaphragm Shear. 
The following style is used when referring to a section of ACI 318 for which a change or insertion is 
proposed by the Provisions: Provisions Sec. xxx (ACI Sec. yyy) where xxx is the section in the 
Provisions and yyy is the section proposed for insertion into ACI 318-99. 
Although this volume of design examples is based on the 2000 Provisions, it has been annotated to reflect 
changes made for the 2003 Provisions. Annotations within brackets, [ ], indicate both organizational 
changes (as a result of a reformatting of all chapters for the 2003 Provisions) and substantive technical 
changes to the Provisions and its primary reference documents. Although the general conepts of the 
changes are described, the design examples and calculations have not been revised to reflect the changes 
made for the 2003 Provisions. 
The most significant change related to precast concrete in the 2003 Provisions is that precast shear wall 
systems are now recognized separately from cast-in-place systems. The 2003 Provisions recognizes 
ordinary and intermediate precast concrete shear walls. The design of ordinary precast shear walls is 
based on ACI 318-02 excluding Chapter 21 and the design of intermediate shear walls is based on ACI 
318-02 Sec. 21.13 (with limited modifications in Chapter 9 of the 2003 Provisions). The 2003 Provisions 
does not distinguish between precast and cast-in-place concrete for special shear walls. Special precast 
shear walls either need to satisfy the design requirements for special cast-in-place concrete shear walls
FEMA 451, NEHRP Recommended Provisions: Design Examples 
7-4 
(ACI 318-02 Sec. 21.7) or most be substantiated using experimental evidence and analysis (2003 
Provisions Sec. 9.2.2.4 and 9.6). Many of the design provisions for precast shear walls in the 2000 
Provisions have been removed, and the requirements in ACI 318-02 are in some ways less specific. 
Where this occurs, the 2000 Provisions references in this chapter are simply annotated as [not applicable 
in the 2003 Provisions]. Commentary on how the specific design provision was incorporated into ACI 
318-02 is included where appropriate. 
Some general technical changes for the 2003 Provisions that relate to the calculations and/or designs in 
this chapter include updated seismic hazard maps, revisions to the redundancy requirements, and 
revisions to the minimum base shear equation. Where they affect the design examples in the chapter, 
other significant changes for the 2003 Provisions and primary reference documents are noted. However, 
some minor changes may not be noted.
Chapter 7, Precast Concrete Design 
1Note that this equation is incorrectly numbered as 5.2.5.4 in the first printing of the 2000 Provisions. 
7-5 
7.1 HORIZONTAL DIAPHRAGMS 
Structural diaphragms are horizontal or nearly horizontal elements, such as floors and roofs, that transfer 
seismic inertial forces to the vertical seismic-force-resisting members. Precast concrete diaphragms may 
be constructed using topped or untopped precast elements depending on the Seismic Design Category of 
the building. Reinforced concrete diaphragms constructed using untopped precast concrete elements are 
addressed in the Appendix to Chapter 9 of the Provisions. Topped precast concrete elements, which act 
compositely or noncompositely for gravity loads, are designed using the requirements of ACI 318-99 Sec. 
21.7 [ACI 318-02 Sec. 21.9]. 
7.1.1 Untopped Precast Concrete Units for Five-Story Masonry Buildings Located in 
Birmingham, Alabama, and New York, New York 
This example illustrates floor and roof diaphragm design for the five-story masonry buildings located in 
Birmingham, Alabama, on soft rock (Seismic Design Category B) and in New York, New York (Seismic 
Design Category C). The example in Sec. 9.2 provides design parameters used in this example. The 
floors and roofs of these buildings are to be untopped 8-in.-thick hollow core precast, prestressed concrete 
plank. Figure 9.2-1 shows the typical floor plan of the diaphragms. 
7.1.1.1 General Design Requirements 
In accordance with the Provisions and ACI 318, untopped precast diaphragms are permitted only in 
Seismic Design Categories A through C. The Appendix to Chapter 9 provides design provisions for 
untopped precast concrete diaphragms without limits as to the Seismic Design Category. Diaphragms 
with untopped precast elements are designed to remain elastic, and connections are designed for limited 
ductility. No out-of-plane offsets in vertical seismic-force-resisting members (Type 4 plan irregularities) 
are permitted with untopped diaphragms. Static rational models are used to determine shears and 
moments on joints as well as shear and tension/compression forces on connections. Dynamic modeling of 
seismic response is not required. 
The design method used here is that proposed by Moustafa. This method makes use of the shear friction 
provisions of ACI 318 with the friction coefficient, , being equal to 1.0. To use  = 1.0, ACI 318 
requires grout or concrete placed against hardened concrete to have clean, laitance free, and intentionally 
roughened surfaces with a total amplitude of about 1/4 in. (peak to valley). Roughness for formed edges 
is provided either by sawtooth keys along the length of the plank or by hand roughening with chipping 
hammers. Details from the SEAA Hollow Core reference are used to develop the connection details. 
The terminology used is defined in ACI 318 Chapter 21 and Provisions Chapter 9. These two sources 
occasionally conflict (such as the symbol  used above), but the source is clear from the context of the 
discussion. Other definitions (e.g., chord elements) are provided as needed for clarity in this example. 
7.1.1.2 General In-Plane Seismic Design Forces for Untopped Diaphragms 
The in-plane diaphragm seismic design force (F!px) for untopped precast concrete in Provisions Sec. 
9A.3.3 [A9.2.2] shall not be less than the forcee calculated from either of the following two criteria: 
1. .O0Fpx but not less than .O0Cswpx where 
F px is calculated from Provisions Eq. 5.2.6.4.41 [4.6-3], which also bounds Fpx to be not less than 
0.2SDSIwpx and not more than 0.4SDSIwpx. This equation normally is specified for Seismic Design
FEMA 451, NEHRP Recommended Provisions: Design Examples 
7-6 
Categories D and higher; it is intended in the Provisions Appendix to Chapter 9 that the same 
equation be used for untopped diaphragms in Seismic Design Categories B and C. 
. is the reliability factor, which is 1.0 for Seismic Design Categories A through C per Provisions Sec. 
5.2.4.1 [4.3.3.1]. 
O0 is the overstrength factor (Provisions Table 5.2.2 [4.3-1]) 
Cs is the seismic response coefficient (Provisions Sec. 5.4.1.1 [5.2.1.1]) 
wpx is the weight tributary to the diaphragm at Level x 
SDS is the spectral response acceleration parameter at short periods (Provisions Sec. 4.1.2 [3.3.3]) 
I is the occupancy importance factor (Provisions Sec. 1.4 [1.3]) 
2. 1.25 times the shear force to cause yielding of the vertical seismic-force-resisting system. 
For the five-story masonry buildings of this example, the shear force to cause yielding is first 
estimated to be that force associated with the development of the nominal bending strength of the 
shear walls at their base. This approach to yielding uses the first mode force distribution along the 
height of the building and basic pushover analysis concepts, which can be approximated as: 
F!px = 1.25KFpx
* where 
K is the ratio of the yield strength in bending to the demand, My/Mu. (Note that f = 1.0) 
Fpx
* is the seismic force at each level for the diaphragm as defined above by Provisions Eq. 5.2.6.4.4 
[4.6-2] and not limited by the minima and maxima for that equation. 
This requirement is different from similar requirements elsewhere in the Provisions. For components 
thought likely to behave in a brittle fashion, the designer is required to apply the overstrength factor and 
then given an option to check the maximum force that can be delivered by the remainder of the structural 
system to the element in question. The maximum force would normally be computed from a plastic 
mechanism analysis. If the option is exercised, the designer can then use the smaller of the two forces. 
Here the Provisions requires the designer to compute both an overstrength level force and a yield level 
force and then use the larger. This appears to conflict with the Commentary. 
For Seismic Design Categories B and C, Provisions Sec. 5.2.6.2.6 [4.6.1.9] defines a minimum diaphragm 
seismic design force that will always be less than the forces computed above. 
For Seismic Design Category C, Provisions Sec. 5.2.6.3.1 [4.6.2.2] requires that collector elements, 
collector splices, and collector connections to the vertical seismic-force-resisting members be designed in 
accordance with Provisions Sec. 5.2.7.1 [4.2.2.2], which places the overstrength factor on horizontal 
seismic forces and combines the horizontal and vertical seismic forces with the effects of gravity forces. 
Because vertical forces do not normally affect diaphragm collector elements, splices, and connections, the 
authors believe that Provisions Sec. 5.2.7.1 [4.2.2.2] is satisfied by the requirements of Provisions Sec. 
9A.3.3 [A9.2.2], which requires use of the overstrength factor. 
Parameters from the example in Sec. 9.2 used to calculate in-plane seismic design forces for the 
diaphragms are provided in Table 7.1-1.
Chapter 7, Precast Concrete Design 
7-7 
Table 7.1-1 Design Parameters from Example 9.2 
Design Parameter Birmingham 1 New York City 
. 1.0 1.0 
Oo 2.5 2.5 
Cs 0.12 0.156 
wi (roof) 861 kips 869 kips 
wi (floor) 963 kips 978 kips 
SDS 0.24 0.39 
I 1.0 1.0 
1.0 kip = 4.45 kN, 1.0 ft-kip = 1.36 kN-m. 
The Provisions Appendix to Chapter 9 does not give the option of using the overstrength factor O0 to 
estimate the yield of the vertical system, so Mn for the wall is computed from the axial load moment 
interaction diagram data developed in Sec. 9.2. The shape of the interaction diagram between the 
balanced point and pure bending is far enough from a straight line (see Figure 9.2-6) in the region of 
interest that simply interpolating between the points for pure bending and balanced conditions is 
unacceptably unconservative for this particular check. An intermediate point on the interaction diagram 
was computed for each wall in Sec. 9.2, and that point is utilized here. Yielding begins before the 
nominal bending capacity is reached, particularly when the reinforcement is distributed uniformly along 
the wall rather than being concentrated at the ends of the wall. For lightly reinforced walls with 
distributed reinforcement and with axial loads about one-third of the balanced load, such as these, the 
yield moment is on the order of 90 to 95 percent of the nominal capacity. It is feasible to compute the 
moment at which the extreme bar yields, but that does not appear necessary for design. A simple factor 
of 0.95 was applied to the nominal capacity here. Thus, Table 7.1-2 shows the load information from 
Sec. 9.2 (the final numbers in this section may have changed, because this example was completed first). 
The factor K is large primarily due to consideration of axial load. The strength for design is controlled by 
minimum axial load, whereas K is maximum for the maximum axial load, which includes some live load 
and a vertical acceleration on dead load.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
7-8 
Table 7.1-2 Shear Wall Overstrength 
Birmingham 1 New York City 
Pure Bending, Mn0 963 ft-kips 1,723 ft-kips 
Intermediate Load, MnB 5,355 ft-kips 6,229 ft-kips 
Intermediate Load, PnB 335 kips 363 kips 
Maximum Design Load, Pu 315 kips 327 kips 
Interpolated Mn 5,092 ft-kips 5,782 ft-kips 
Approximate My 4,837 ft-kips 5,493 ft-kips 
Design Mu 2,640 ft-kips 3,483 ft-kips 
Factor K = My/Mu 1.83 1.58 
7.1.1.3 Diaphragm Forces for Birmingham Building 1 
The weight tributary to the roof and floor diaphragms (wpx) is the total story weight (wi) at Level i minus 
the weight of the walls parallel to the direction of loading. 
Compute diaphragm weight (wpx) for the roof and floor as follows: 
Roof
Total weight = 861 kips 
Walls parallel to force = (45 psf)(277 ft)(8.67 ft/2) = -54 kips 
wpx = 807 kips 
Floors 
Total weight = 963 kips 
Walls parallel to force = (45 psf)(277 ft)(8.67 ft) = -108 kips 
wpx = 855 kips 
Compute diaphragm demands in accordance with Provisions Eq. 5.2.6.4.4 [4.6.3.4]: 
n 
i 
i x 
px n px 
i 
i x 
F 
F w
w 
=
=
S 
= 
S 
Calculations for Fpx are provided in Table 7.1-3.
Chapter 7, Precast Concrete Design 
7-9 
Table 7.1-3 Birmingham 1 Fpx Calculations 
Level 
wi 
(kips) 
n 
i 
i x
w 
= S 
(kips) 
Fi 
(kips) 
n 
i i 
i x
F V 
= 
S = 
(kips) 
wpx 
(kips) 
Fpx 
(kips) 
Roof 
4321 
861 
963 
963 
963 
963 
861 
1,820 
2,790 
3,750 
4,710 
175 
156 
117 
78 
39 
175 
331 
448 
527 
566 
807 
855 
855 
855 
855 
164 
155 
137 
120 
103 
1.0 kip = 4.45 kN. 
The values for Fi and Vi used in Table 7.1-3 are listed in Table 9.2-2. 
The minimum value of Fpx = 0.2SDSIwpx = 0.2(0.24)1.0(807 kips) = 38.7 kips (at the roof) 
= 0.2(0.24)1.0(855 kips) = 41.0 kips (at floors) 
The maximum value of Fpx = 0.4SDSIwpx = 2(38.7 kips) = 77.5 kips (at the roof) 
= 2(41.0 kips) = 82.1 kips (at floors) 
Note that Fpx by Table 7.1-3 is substantially larger than the maximum Fpx. This is generally true at upper 
levels if the R factor is less than 5. The value of Fpx used for the roof diaphragm is 82.1 kips. Compare 
this value to Cswpx to determine the minimum diaphragm force for untopped diaphragms as indicated 
previously. 
Cswpx = 0.12(807 kips) = 96.8 kips (at the roof) 
Cswpx = 0.12(855 kips) = 103 kips (at the floors) 
Since Cswpx is larger than Fpx, the controlling force is Cswpx. Note that this will always be true when I = 
1.0 and R is less than or equal to 2.5. Therefore, the diaphragm seismic design forces are as follows: 
F!px = .O0Cswpx = 1.0(2.5)(96.8 kips) = 242 kips (at the roof) 
F!px = .O0Cswpx = 1.0(2.5)(103 kips) = 256 kips (at the floors) 
The second check on design force is based on yielding of the shear walls: 
F!px = 1.25KFpx* = 1.25(1.85)164 kips = 379 kips (at the roof) 
F!px = 1.25KFpx* = 1.25(1.85)155 kips = 358 kips (at the floors) 
For this example, the force to yield the walls clearly controls the design. To simplify the design, the 
diaphragm design force used for all levels will be the maximum force at any level, 379 kips. 
7.1.1.4 Diaphragm Forces for New York Building 
The weight tributary to the roof and floor diaphragms (wpx) is the total story weight (wi) at Level i minus 
the weight of the walls parallel to the force. 
FEMA 451, NEHRP Recommended Provisions: Design Examples 
7-10 
Compute diaphragm weight (wpx) for the roof and floor as follows: 
Roof
Total weight = 870 kips 
Walls parallel to force = (48 psf)(277 ft)(8.67 ft/2) = -58 kips 
wpx = 812 kips 
Floors 
Total weight = 978 kips 
Walls parallel to force = (48 psf)(277 ft)(8.67 ft) = -115 kips 
wpx = 863 kips 
Calculations for Fpx using Provisions Eq. 5.2.6.4.4 [4.6.3.4] are not required for the first set of forces 
because Cswpx will be greater than or equal to the maximum value of Fpx = 0.4SDSIwpx when I = 1.0 and R 
is less than or equal to 2.5. Compute Cswpx as: 
Cswpx = 0.156(812 kips) = 127 kips (at the roof) 
Cswpx = 0.156(863 kips) = 135 kips (at the floors) 
The diaphragm seismic design forces are: 
F!px = .O0Cswpx = 1.0(2.5)(127 kips) = 318 kips (at the roof) 
F!px = .O0Cswpx = 1.0(2.5)(135 kips) = 337 kips (at the floors) 
Calculations for Fpx using Provisions Eq. 5.2.6.4.4 [4.6.3.4] are required for the second check F!px = 
1.25KFpx. Following the same procedure as illustrated in the previous section, the maximum Fpx is 214 
kips at the roof. Thus, 
1.25KFpx* = 1.25(1.58)214 kips = 423 kips (at the roof) 
To simplify the design, the diaphragm design force used for all levels will be the maximum force at any 
level. The diaphragm seismic design force (423 kips) is controlled by yielding at the base of the walls, 
just as with the Birmingham 1 building. 
7.1.1.5 Static Analysis of Diaphragms 
The balance of this example will use the controlling diaphragm seismic design force of 423 kips for the 
New York building. In the transverse direction, the loads will be distributed as shown in Figure 7.1-1.
Chapter 7, Precast Concrete Design 
7-11 
W2 
W1 W1 
F F F F 
40'-0" 3 at 24'-0" = 72'-0" 40'-0" 
152'-0" 
Figure 7.1-1 Diaphragm force distribution and analytical model (1.0 ft = 
0.3048 m). 
Assuming the four shear walls have the same stiffness and ignoring torsion, the diaphragm reactions at 
the transverse shear walls (F as shown in Figure 7.1-1) are computed as follows: 
F = 423 kips/4 = 105.8 kips 
The uniform diaphragm demands are proportional to the distributed weights of the diaphragm in different 
areas (see Figure 7.1-1). 
W1 = [67 psf(72 ft) + 48 psf(8.67 ft)4](423 kips / 863 kips) = 3,180 lb/ft 
W2 = [67 psf(72 ft)](423 kips / 863 kips) = 2,364 lb/ft 
Figure 7.1-2 identifies critical regions of the diaphragm to be considered in this design. These regions 
are: 
Joint 1  maximum transverse shear parallel to the panels at panel-to-panel joints 
Joint 2  maximum transverse shear parallel to the panels at the panel-to-wall joint 
Joint 3  maximum transverse moment and chord force 
Joint 4  maximum longitudinal shear perpendicular to the panels at the panel-to-wall connection 
(exterior longitudinal walls) and anchorage of exterior masonry wall to the diaphragm for out-ofplane 
forces 
Joint 5  collector element and shear for the interior longitudinal walls
FEMA 451, NEHRP Recommended Provisions: Design Examples 
7-12 
72'-0" 
4 1 2 3 
5 
36'-0" 
4'-0" 
24'-0" 
Figure 7.1-2 Diaphragm plan and critical design regions (1.0 ft = 0.3048 m). 
Provisions Sec. 9.1.1.4 [not applicable in 2003 Provisions] defines a chord amplification factor for 
diaphragms in structures having precast gravity-load systems. [The chord amplification factor has been 
dropped in the 2003 Provisions and does not occur in ASC 318-02. See the initial section of this chapter 
for additional discussion on changes for the 2003 Provisions.] This amplification factor appears to apply 
to buildings with vertical seismic-force-resisting members constructed of precast or monolithic concrete. 
Because these masonry wall buildings are similar to buildings with concrete walls, this amplification 
factor has been included in calculating the chord forces. The amplification factor is: 
2 
1 0.4 
1.0 
12 
eff 
d 
d 
s
L
b 
b 
h 
. . .. 
.+ . .. 
.. . ...= 
where 
Leff = length of the diaphragm between inflection points. Since the diaphragms have no infection 
points, twice the length of the 40-ft-long cantilevers is used for Leff = 80 ft 
hs = story height = 8.67 ft 
bd = diaphragm width = 72 ft 
The amplification factor = ( ) ( ) = 1.03 
2 1 0.4 80 
72 
72 
12 8.67 
..+ .. .... 
.. . ...
Chapter 7, Precast Concrete Design 
7-13 
Joint forces are: 
Joint 1  Transverse forces 
Shear, Vu1 = 3.18 kips/ft (36 ft) = 114.5 kips 
Moment, Mu1 = 114.5 kips (36 ft/2) = 2,061 ft-kips 
Chord tension force, Tu1 = M/d = 1.03(2,061 ft-kips/71 ft) = 29.9 kips 
Joint 2  Transverse forces 
Shear, Vu2 = 3.18 kips/ft (40 ft) = 127 kips 
Moment, Mu2 = 127 kips (40 ft/2) = 2,540 ft-kips 
Chord tension force, Tu2 = M/d = 1.03(2,540 ft-kips/71 ft) = 36.9 kips 
Joint 3  Transverse forces 
Shear, Vu3 = 127 kips + 2.36 kips/ft (24 ft) - 105.8 kips = 78.1 kips 
Moment, Mu3 = 127 kips (44 ft) + 56.7 kips (12 ft) - 105.8 kips (24 ft) = 3,738 ft-kips 
Chord tension force, Tu3 = M/d = 1.03(3,738 ft-kips/71 ft) = 54.2 kips 
Joint 4  Longitudinal forces 
Wall Force, F = 423 kips/8 = 52.9 kips 
Wall shear along wall length, Vu4 = 52.9 kips (36 ft)/(152 ft /2) = 25.0 kips 
Collector force at wall end, Tu4 = Cu4 = 52.9 kips - 25.0 kips = 27.9 kips 
Joint 4  Out-of-plane forces 
The Provisions have several requirements for out-of-plane forces. None are unique to precast 
diaphragms and all are less than the requirements in ACI 318 for precast construction regardless 
of seismic considerations. Assuming the planks are similar to beams and comply with the 
minimum requirements of Provisions Sec. 5.2.6.1.1 [4.6.1.1] (Seismic Design Category A and 
greater) [In the 2003 Provisions, all requirements for Seismic Design Category A are in Sec. 1.5 
but they generally are the same as those in the 2000 Provisions. The design and detailing 
requirements in 2003 Provisions Sec. 4.6 apply to Seismic Design Category B and greater], the 
required out-of-plane horizontal force is: 
0.05(D + L)plank = 0.05(67 psf + 40 psf)(24 ft/2) = 64.2 plf 
According to Provisions Sec. 5.2.6.1.2 [4.6.1.2] (Seismic Design Category A and greater), the 
minimum anchorage for masonry walls is: 
Fp = 400(SDS)I = 400(0.39)1.0 = 156 plf 
According to Provisions Sec. 5.2.6.2.7 [4.6.1.3] (Seismic Design Category B and greater), 
bearing wall anchorage shall be designed for a force computed as: 
0.4(SDS)Wwall = 0.4(0.39)(48 psf)(8.67 ft) = 64.9 plf 
Provisions Sec. 5.2.6.3.2 [4.6.2.1] (Seismic Design Category C and greater) requires masonry 
wall anchorage to flexible diaphragms to be designed for a larger force. This diaphragm is
FEMA 451, NEHRP Recommended Provisions: Design Examples 
7-14 
considered rigid with respect to the walls, and considering that it is designed to avoid yield under 
the loads that will yield the walls, this is a reasonable assumption. 
Fp = 1.2(SDS)Iwp = 1.2(0.39)1.0[(48 psf)(8.67 ft)] = 195 plf 
[In the 2003 Provisions, Eq. 4.6-1 in Sec. 4.6.2.1 has been changed to 0.85SDSIWp.] 
The force requirements in ACI 318 Sec. 16.5 will be described later. 
Joint 5  Longitudinal forces 
Wall force, F = 423 kips/8 = 52.9 kips 
Wall shear along each side of wall, Vu4 = 52.9 kips [2(36 ft)/152 ft]/2 = 12.5 kips 
Collector force at wall end, Tu5 = Cu5 = 52.9 kips - 25.0 kips = 27.9 kips 
ACI 318 Sec. 16.5 also has minimum connection force requirements for structural integrity of precast 
concrete bearing wall building construction. For buildings over two stories there are force requirements 
for horizontal and vertical members. This building has no vertical precast members. However, ACI 318 
Sec. 16.5.1 specifies that the strengths . . . for structural integrity shall apply to all precast concrete 
structures. This is interpreted to apply to the precast elements of this masonry bearing wall structure. 
The horizontal tie force requirements are: 
1. 1,500 lb/ft parallel and perpendicular to the span of the floor members. The maximum spacing of ties 
parallel to the span is 10 ft. The maximum spacing of ties perpendicular to the span is the distance 
between supporting walls or beams. 
2. 16,000 lb parallel to the perimeter of a floor or roof located within 4 ft of the edge at all edges. 
ACIs tie forces are far greater than the minimum tie forces given in the Provisions for beam supports and 
anchorage for of masonry walls. They do control some of the reinforcement provided, but most of the 
reinforcement is controlled by the computed connections for diaphragm action. 
7.1.1.6 Diaphragm Design and Details 
Before beginning the proportioning of reinforcement, a note about ACIs f factors is necessary. The 
Provisions cites ASCE 7 for combination of seismic load effects with the effects of other loads. Both 
ASCE 7 and the Provisions make it clear that the appropriate f factors within ACI 318 are those 
contained within Appendix C of ACI 318-99. These factors are about 10% less than the comparable 
factors within the main body of the standard. The 2002 edition of ACI 318 has placed the ASCE 7 load 
combinations within the main body of the standard and revised the f factors accordingly. This example 
uses the f factors given in the 2002 edition of ACI 318, which are the same as those given in Appendix C 
of the 1999 edition with one exception. Thus, the f factors used here are: 
Tension control (bending and ties) f = 0.90 
Shear f = 0.75 
Compression control in tied members f = 0.65. 
The minimum tie force requirements given in ACI 318 Sec. 16.5 are specified as nominal values, meaning 
that f = 1.00 for those forces.
Chapter 7, Precast Concrete Design 
7-15 
Splice bars 
(2) #7 bars 
(chord bars) 
3" 
2" 3" 3" 
3" 
Grouted 
chord / collector 
element along exterior 
edge of precast plank 
Contact 
lap splice 
Prestressed 
hollow core 
plank 
Artificially roughened 
surfaces of void as 
required 
4" spiral of 1
4" wire 
with 2" pitch over each 
lap splice may be required 
depending on geometry 
of specific voids in plank. 
Figure 7.1-3 Joint 3 chord reinforcement at the exterior edge (1.0 in. = 25.4 mm). 
7.1.1.6.1 Design and Detailing at Joint 3 
Joint 3 is designed first to check the requirements of Provisions Sec. 9A.3.9 [A9.2.4], which references 
ACI 318 Sec. 21.7.8.3 [21.9.8.3], which then refers to ACI 318 Sec. 21.7.5.3 [21.9.5.3]. This section 
provides requirements for transverse reinforcement in the chords of the diaphragm. The compressive 
stress in the chord is computed using the ultimate moment based on a linear elastic model and gross 
section properties. To determine the in-plane section modulus (S) of the diaphragm, an equivalent 
thickness (t) based on the cross sectional area is used for the hollow core precast units as follows. 
t = area/width = 215/48 = 4.5 in. 
S = td2/6 
Chord compressive stress is computed as: 
Mu/S = 6Mu3/td2 = 6(3,738  12)/(4.5)(72  12)2 = 80.1 psi 
The design 28-day compressive strength of the grout is 4,000 psi. Since the chord compressive stress is 
less than 0.2 fc' = 0.2(4,000) = 800 psi, the transverse reinforcement indicated in ACI 318 Sec. 21.4.4.1 
through 21.4.4.3 is not required. 
Compute the required amount of chord reinforcement as: 
Chord reinforcement, As3 = Tu3/ffy = (54.2 kips)/[0.9(60 ksi)] = 1.00 in.2 
Use two #7 bars, As = 2(0.60) = 1.20 in.2 along the exterior edges (top and bottom of the plan in Figure 
7.1- 2). Require cover for chord bars and spacing between bars at splices and anchorage zones by ACI 
318 Sec. 21.7.8.3 [21.9.8.3]. 
Minimum cover = 2.5(7/8) = 2.19 in., but not less than 2.0 in. 
Minimum spacing = 3(7/8) = 2.63 in., but not less than 1-1/2 in. 
Figure 7.1-3 shows the chord element at the exterior edges of the diaphragm. The chord bars extend 
along the length of the exterior longitudinal walls and act as collectors for these walls in the longitudinal 
direction (see Joint 4 collector reinforcement and Figure 7.1-7).
FEMA 451, NEHRP Recommended Provisions: Design Examples 
7-16 
(2) #6 
(collector bars) 
33
4" 
21 2" 
31 2" 
2" 
#3 x 4'-0" (behind) 
at each joint 
between planks 
Figure 7.1-4 Interior joint reinforcement at the ends of plank 
and the collector reinforcement at the end of the interior 
longitudinal walls - Joints 1 and 5 (1.0 in. = 25.4 mm). 
Joint 3 must also be checked for the minimum ACI tie forces. The chord reinforcement obviously 
exceeds the 16 kip perimeter force requirement. The 1.5 kips per foot requirement requires a 6 kip tie at 
each joint between the planks, which is satisfied with a #3 bar in each joint (0.11 in.2 at 60 ksi = 6.6 kips). 
This bar is required at all bearing walls and is shown in subsequent details. 
7.1.1.6.2 Joint 1 Design and Detailing 
The design must provide sufficient reinforcement for chord forces as well as shear friction connection 
forces as follows: 
Chord reinforcement, As1 = Tu1/ffy = (29.9 kips)/[0.9(60 ksi)] = 0.55 in.2 (collector force from Joint 4 
calculations at 27.9 kips is not directly additive). 
Shear friction reinforcement, Avf1 = Vu1/ffy = (114.5 kips)/[(0.75)(1.0)(60 ksi)] = 2.54 in.2 
Total reinforcement required = 2(0.55 in.2) + 2.54 in.2 = 3.65 in.2 
ACI tie force = (3 kips/ft)(72 ft) = 216 kips; reinforcement = (216 kips)/(60 ksi) = 3.60 in.2 
Provide four #7 bars (two at each of the outside edges) plus four #6 bars (two each at the interior joint at 
the ends of the plank) for a total area of reinforcement of 4(0.60 in2) + 4(0.44 in.2) = 4.16 in.2 
Because the interior joint reinforcement acts as the collector reinforcement in the longitudinal direction 
for the interior longitudinal walls, the cover and spacing of the two # 6 bars in the interior joints will be 
provided to meet the requirements of ACI 318 Sec. 21.7.8.3 [21.9.8.3]: 
Minimum cover = 2.5(6/8) = 1.88 in., but not less than 2.0 in. 
Minimum spacing = 3(6/8) = 2.25 in., but not less than 1-1/2 in. 
Figure 7.1-4 shows the reinforcement in the interior joints at the ends of the plank, which is also the 
collector reinforcement for the interior longitudinal walls (Joint 5). The two #6 bars extend along the 
length of the interior longitudinal walls as shown in Figure 7.1-8.
Chapter 7, Precast Concrete Design 
7-17 
112" 212" 2" 
2" (2) #6 anchored 4'-0" 
into plank at ends. 
Figure 7.1-5 Anchorage region of shear reinforcement for Joint 1 and collector 
reinforcement for Joint 5 (1.0 in. = 25.4 mm). 
Figure 7.1-5 shows the extension of the two #6 bars of Figure 7.1-4 into the region where the plank is 
parallel to the bars. The bars will need to be extended the full length of the diaphragm unless 
supplemental plank reinforcement is provided. This detail makes use of this supplement plank 
reinforcement (two #6 bars or an equal area of strand per ACI 318-99 Sec. 21.7.5.2 [21.9.5.2]) and shows 
the bars anchored at each end of the plank. The anchorage length of the #6 bars is calculated using ACI 
318-99 Sec. 21.7.5.4 [21.9.5.4] which references ACI 318 Sec. 21.5.4: 
60,000( ) 
1.6(2.5) 1.6(2.5) 58.2 
65 65 4,000 
y b b 
d b 
c 
f d d 
l d 
f 
. . . . 
= ... '...= .. ..= 
The 2.5 factor is for the difference between straight and hooked bars, and the 1.6 factor applies when the 
development length is not within a confined core. Using #6 bars, the required ld = 58.2(0.75 in.) = 43.7 
in. Therefore, use ld = 4 ft, which is the width of the plank. 
7.1.1.6.3 Joint 2 Design and Detailing 
The chord design is similar to the previous calculations: 
Chord reinforcement, As2 = Tu2/ffy = (36.9 kips)/[0.9(60 ksi)] = 0.68 in.2 
The shear force may be reduced along Joint 2 by the shear friction resistance provided by the 
supplemental chord reinforcement (2Achord - As2) and by the four #6 bars projecting from the interior 
longitudinal walls across this joint. The supplemental chord bars, which are located at the end of the 
walls, are conservatively excluded here. The shear force along the outer joint of the wall where the plank 
is parallel to the wall is modified as: 
( ) ( )( )( ) 2 2 4#6 Mod 127 0.75 60 1.0 4 0.44 47.8 kips 
Vu =Vu-..ffyA ..= -..  ..= 
This force must be transferred from the planks to the wall. Using the arrangement shown in Figure 7.1-6, 
the required shear friction reinforcement (Avf2) is computed as: 
= = 0.79 in.2 ( ) 
2 
2 sin cos 
Mod 
u 
vf 
y f f 
A V 
f f  a a 
= 
+ ( )( ) 
47.8 
0.75 60 1.0 sin 26.6 + cos 26.6
FEMA 451, NEHRP Recommended Provisions: Design Examples 
7-18 
Use two #3 bars placed at 26.6 degrees (2-to-1 slope) across the joint at 4 ft from the ends of the plank 
and at 8 ft on center (three sets per plank). The angle (af) used above provides development of the #3 bars 
while limiting the grouting to the outside core of the plank. The total shear reinforcement provided is 
9(0.11 in.2) = 0.99 in.2 
The shear force between the other face of this wall and the diaphragm is: 
Vu2 - F = 127 - 106 = 21 kips 
The shear friction resistance provided by #3 bars in the grout key between each plank (provided for the 
1.5 klf requirement of the ACI) is computed as: 
fAvffy = (0.75)(10 bars)(0.11 in.2)(60 ksi)(1.0) = 49.5 kips 
The development length of the #3 and #4 bars will now be checked. For the 180 degree standard hook 
use ACI 318 Sec. 12.5, ldh = lhb times the factors of ACI 318 Sec. 12.5.3, but not less than 8db or 6 in. 
Side cover exceeds 2-1/2 in. and cover on the bar extension beyond the hook is provided by the grout and 
the planks, which is close enough to 2 in. to apply the 0.7 factor of ACI 318 Sec. 12.5.3.2. The 
continuous #5 provides transverse reinforcement, but it is not arranged to take advantage of ACI 318s 
0.8 factor. For the #3 hook: 
0.7(1,200) 0.7(1,200)0.375 = 4.95 in. (6" minimum) 
4,000 
b 
dh 
c 
d 
l 
f 
= = 
' 
The available distance for the perpendicular hook is about 5-1/2 in. The bar will not be fully developed at 
the end of the plank because of the 6 in. minimum requirement. The full strength is not required for shear 
transfer. By inspection, the diagonal #3 hook will be developed in the wall as required for the computed 
diaphragm-to-shear-wall transfer. The straight end of the #3 will now be checked. The standard 
development length of ACI 318 Sec. 12.2 is used for ld. 
= 14.2 in. 60,000(0.375) 
25 25 4,000 
y b 
d 
c 
f d 
l 
f 
= = 
' 
Figure 7.1-6 shows the reinforcement along each side of the wall on Joint 2.
Chapter 7, Precast Concrete Design 
7-19 
#3x 2'-6" 
standard hook 
grouted into 
each key joint 
(1) #5 
continuous 
in joint to 
anchor hooks 
(2) #5 in 
masonry 
bond beam 
#3 x standard hooks 
embedded in grouted 
edge cell of plank. Provide 
3 sets for each plank. 
2 
1 
71
2" 
2'-2" 
2'-2" 
2" cover 
Vertical 
reinforcement 
in wall 
Figure 7.1-6 Joint 2 transverse wall joint reinforcement (1.0 in. = 25.4 mm, 
1.0 ft = 0.3048 m). 
7.1.1.6.4 Joint 4 Design and Detailing 
The required shear friction reinforcement along the wall length is computed as: 
Avf4 = Vu4/ffy = (25.0 kips)/[(0.75)(1.0)(60 ksi)] = 0.56 in.2 
Based upon the ACI tie requirement, provide #3 bars at each plank-to-plank joint. For eight bars total, the 
area of reinforcement is 8(0.11) = 0.88 in.2, which is more than sufficient even considering the marginal 
development length, which is less favorable at Joint 2. The bars are extended 2 ft into the grout key, 
which is more than the development length and equal to half the width of the plank. 
The required collector reinforcement is computed as: 
As4 = Tu4/ffy = (27.9 kips)/[0.9(60 ksi)] = 0.52 in.2 
The two #7 bars, which are an extension of the transverse chord reinforcement, provide an area of 
reinforcement of 1.20 in.2 
The reinforcement required by the Provisions for out-of-plane force is (195 plf) is far less than the ACI 
318 requirement. 
Figure 7.1-7 shows this joint along the wall.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
7-20 
#3x 2'-6" 
standard hook 
grouted into 
each key joint 
(2) #5 in 
bond beam 
(2) #7 bars 
in joint 
(chord bars) 
Vertical wall 
reinforcement 
beyond 
2" 
cover 
Figure 7.1-7 Joint 4 exterior longitudinal walls to diaphragm 
reinforcement and out-of-plane anchorage (1.0 in. = 25.4 mm, 
1.0 ft = 0.3048 m). 
7.1.1.6.5 Joint 5 Design and Detailing 
The required shear friction reinforcement along the wall length is computed as: 
Avf5 = Vu5/ffy = (12.5 kips)/[(0.75)(1.0)(60 ksi)] = 0.28 in.2 
Provide #3 bars at each plank-to-plank joint for a total of 8 bars. 
The required collector reinforcement is computed as: 
As5 = Tu5/ffy = (27.9 kips)/[0.9(60 ksi)] = 0.52 in.2 
Two #6 bars specified for the design of Joint 1 above provide an area of reinforcement of 0.88 in.2 Figure 
7.1-8 shows this joint along the wall.
Chapter 7, Precast Concrete Design 
7-21 
4" 
#3 x 4'-8" 
grouted into 
each key joint 
(2) #5 in 
bond beam 
(2) #6 bars 
in joint 
(collector bars) 
Vertical wall 
reinforcement 
beyond 
Figure 7.1-8 Wall-to-diaphragm reinforcement along interior 
longitudinal walls - Joint 5 (1.0 in. = 25.4 mm, 1.0 ft = 0.3048 m). 
7.1.2 Topped Precast Concrete Units for Five-Story Masonry Building, Los Angeles, 
California (see Sec. 9.2) 
This design shows the floor and roof diaphragms using topped precast units in the five-story masonry 
building in Los Angeles, California. The topping thickness exceeds the minimum thickness of 2 in. as 
required for composite topping slabs by ACI 318 Sec. 21.7.4 [21.9.4]. The topping shall be lightweight 
concrete (weight = 115 pcf) with a 28-day compressive strength (fc' ) of 4,000 psi and is to act 
compositely with the 8-in.-thick hollow-core precast, prestressed concrete plank. Design parameters are 
provided in Sec. 9.2. Figure 9.2-1 shows the typical floor and roof plan. 
7.1.2.1 General Design Requirements 
Topped diaphragms may be used in any Seismic Design Category. ACI 318 Sec. 21.7 [21.9]provides 
design provisions for topped precast concrete diaphragms. Provisions Sec. 5.2.6 [4.6] specifies the forces 
to be used in designing the diaphragms. The amplification factor of Provisions Sec. 9.1.1.4 [not 
applicable in the 2003 Provisions] is 1.03, the same as previously computed for the untopped diaphragm. 
FEMA 451, NEHRP Recommended Provisions: Design Examples 
7-22 
[As noted above, the chord amplification factor has been dropped for the 2003 Provisions and does not 
occur in ASC 318-02.] 
7.1.2.2 General In-Plane Seismic Design Forces for Topped Diaphragms 
The in-plane diaphragm seismic design force (Fpx) is calculated using Provisions Eq. 5.2.6.4.4 [4.6-2] but 
must not be less than 0.2SDSIwpx and need not be more than 0.4SDSIwpx. Vx must be added to Fpx calculated 
using Eq. 5.2.6.4.4 [4.6-2] where: 
wpx = the weight tributary to the diaphragm at Level x 
SDS = the spectral response acceleration parameter at short periods (Provisions Sec. 4.1.2 [3.3.5]) 
I = occupancy importance factor (Provisions Sec. 1.4 [1.3]) 
Vx = the portion of the seismic shear force required to be transferred to the components of the 
vertical seismic-force-resisting system due to offsets or changes in stiffness of the vertical 
resisting member at the diaphragm being designed. 
For Seismic Design Category C and higher, Provisions Sec. 5.2.6.3.1 [4.6.2.2] requires that collector 
elements, collector splices, and collector connections to the vertical seismic-force-resisting members be 
designed in accordance with Provisions Sec. 5.2.7.1 [4.2.2.2], which combines the diaphragm forces 
times the overstrength factor ( O0
) and the effects of gravity forces. The parameters from example in Sec. 
9.2 used to calculate in-plane seismic design forces for the diaphragms are provided in Table 7.1-4. 
Table 7.1-4 Design Parameters from Sec. 9.2 
Design Parameter Value 
Oo
2.5 
wi (roof) 1,166 kips 
wi (floor) 1,302 kips 
SDS 1.0 
I 1.0 
Seismic Design Category D 
1.0 kip = 4.45 kN. 
7.1.2.3 Diaphragm Forces 
As indicated previously, the weight tributary to the roof and floor diaphragms (wpx) is the total story 
weight (wi) at Level i minus the weight of the walls parallel to the force. 
Compute diaphragm weight (wpx) for the roof and floor as: 
Roof
Total weight = 1,166 kips 
Walls parallel to force = (60 psf)(277 ft)(8.67 ft/2) = -72 kips 
wpx = 1,094 kips
Chapter 7, Precast Concrete Design 
7-23 
Floors 
Total weight = 1,302 kips 
Walls parallel to force = (60 psf)(277 ft)(8.67 ft) = -144 kips 
wpx = 1,158 kips 
Compute diaphragm demands in accordance with Provisions Eq. 5.2.6.4.4 [4.6-2]: 
n 
i 
i x 
px n px 
i 
i x 
F 
F w
w 
=
=
S 
= 
S 
Calculations for Fpx are provided in Table 7.1-5. The values for Fi and Vi are listed in Table 9.2-17. 
Table 7.1-5 Fpx Calculations from Sec. 9.2 
Level 
wi 
(kips) 
n 
i 
i x
w 
= S 
(kips) 
Fi 
(kips) 
n 
i i 
i x
F V 
= 
S = 
(kips) 
wpx 
(kips) 
Fpx 
(kips) 
Roof 
4321 
1,166 
1,302 
1,302 
1,302 
1,302 
1,166 
2,468 
3,770 
5,072 
6,384 
564 
504 
378 
252 
126 
564 
1,068 
1,446 
1,698 
1,824 
1,094 
1,158 
1,158 
1,158 
1,158 
529 
501 
444 
387 
331 
1.0 kip = 4.45 kN. 
The minimum value of Fpx = 0.2SDSIwpx = 0.2(1.0)1.0(1,094 kips) = 219 kips (at the roof) 
= 0.2(1.0)1.0(1,158 kips) = 232 kips (at floors) 
The maximum value of Fpx = 0.4SDSIwpx = 2(219 kips) = 438 kips (at the roof) 
= 2(232 kips) = 463 kips (at floors) 
The value of Fpx used for design of the diaphragms is 463 kips, except for collector elements where forces 
will be computed below. 
7.1.2.4 Static Analysis of Diaphragms 
The seismic design force of 463 kips is distributed as in Sec. 7.1.1.6 (Figure 7.1-1 shows the distribution). 
The force is only 9.5 percent higher than that used to design the untopped diaphragm for the New York 
design due to the intent to prevent yielding in the untopped diaphragm. Figure 7.1-2 shows critical 
regions of the diaphragm to be considered in this design. Collector elements will be designed for 2.5 
times the diaphragm force based on the overstrength factor (O0). 
Joint forces taken from Sec. 7.1.1.5 times 1.095 are as:
FEMA 451, NEHRP Recommended Provisions: Design Examples 
7-24 
Joint 1  Transverse forces 
Shear, Vu1 = 114.5 kips  1.095 = 125 kips 
Moment, Mu1 = 2,061 ft-kips  1.095 = 2,250 ft-kips 
Chord tension force, Tu1 = M/d = 1.03  2,250 ft-kips / 71 ft = 32.6 kips 
Joint 2  Transverse forces 
Shear, Vu2 = 127 kips  1.095 = 139 kips 
Moment, Mu2 = 2,540 ft-kips  1.095 = 2,780 ft-kips 
Chord tension force, Tu2 = M/d = 1.03  2,780 ft-kips / 71 ft = 39.3 kips 
Joint 3  Transverse forces 
Shear, Vu3 = 78.1 kips  1.095 = 85.5 kips 
Moment, Mu2 = 3,738 ft-kips  1.095 = 4,090 ft-kips 
Chord tension force, Tu3 = M/d = 1.03  4,090 ft-kips/71 ft = 59.3 kips 
Joint 4  Longitudinal forces 
Wall Force, F = 52.9 kips  1.095 = 57.9 kips 
Wall shear along wall length, Vu4 =25 kips  1.095 = 27.4 kips 
Collector force at wall end, O0Tu4 = 2.5(27.9 kips)(1.095) = 76.4 kips 
Out-of-Plane forces 
Just as with the untopped diaphragm, the out-of-plane forces are controlled by ACI 318 Sec. 16.5, 
which requires horizontal ties of 1.5 kips per foot from floor to walls. 
Joint 5  Longitudinal forces 
Wall Force, F = 463 kips / 8 walls = 57.9 kips 
Wall shear along each side of wall, Vu4 = 12.5 kips  1.095 = 13.7 kips 
Collector force at wall end, O0Tu4 = 2.5(27.9 kips)(1.095) = 76.4 kips 
7.1.2.5 Diaphragm Design and Details 
7.1.2.5.1 Minimum Reinforcement for 2.5 in. Topping 
ACI 318 Sec. 21.7.5.1 [21.9.5.1] references ACI 318 Sec. 7.12, which requires a minimum As = 0.0018bd 
for welded wire fabric. For a 2.5 in. topping, the required As = 0.054 in.2/ft. WWF 1010 - W4.5W4.5 
provides 0.054 in.2/ft. The minimum spacing of wires is 10 in. and the maximum spacing is 18 in. Note 
that the ACI 318 Sec. 7.12 limit on spacing of five times thickness is interpreted such that the topping 
thickness is not the pertinent thickness. 
7.1.2.5.2 Boundary Members 
Joint 3 has the maximum bending moment and is used to determine the boundary member reinforcement 
of the chord along the exterior edge. The need for transverse boundary member reinforcement is 
reviewed using ACI 318 Sec. 21.7.5.3 [21.9.5.3]. Calculate the compressive stress in the chord with the 
ultimate moment using a linear elastic model and gross section properties of the topping. It is
Chapter 7, Precast Concrete Design 
7-25 
Figure 7.1-9 Diaphragm plan and section cuts. 
conservative to ignore the precast units, but not necessary. As developed previously, the chord 
compressive stress is: 
6Mu3/td2 = 6(4,090  12)/(2.5)(72  12)2 = 158 psi 
The chord compressive stress is less than 0.2fc' = 0.2(4,000) = 800 psi. Transverse reinforcement in the 
boundary member is not required. 
The required chord reinforcement is: 
As3 = Tu3/ffy = (59.3 kips)/[0.9(60 ksi)] = 1.10 in.2 
7.1.2.5.3 Collectors 
The design for Joint 4 collector reinforcement at the end of the exterior longitudinal walls and for Joint 5 
at the interior longitudinal walls is the same. 
As4 = As5 = O0Tu4/ffy = (76.4 kips)/[0.9(60 ksi)] = 1.41 in.2 
Use two #8 bars (As = 2  0.79 = 1.58 in.2) along the exterior edges, along the length of the exterior 
longitudinal walls, and along the length of the interior longitudinal walls. Provide cover for chord and 
collector bars and spacing between bars per ACI 318 Sec. 21.7.8.3 [21.9.8.3]. 
Minimum cover = 2.5(8/8) = 2.5 in., but not less than 2.0 in. 
Minimum spacing = 3(8/8) = 3.0 in., but not less than 1-1/2 in. 
Figure 7.1-9 shows the diaphragm plan and section cuts of the details and Figure 7.1-10, the boundary 
member and chord/collector reinforcement along the edge. Given the close margin on cover, the 
transverse reinforcement at lap splices also is shown.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
7-26 
Splice bars 
(2) #8 bars 
(chord bars) 
31
2" 
3" 3" 
21
2" 
Grouted 
chord / collector 
element along exterior 
edge of precast plank 
Contact 
lap splice 
Prestressed 
hollow core 
plank with 
roughened 
top surface 
Artificially 
roughened 
edge 
WWF bend 
down into 
chord 
21
2" min 
(concrete 
topping) 
41
2" spiral of 1
4" wire 
with 2" pitch over each 
lap splice. 
Figure 7.1-10 Boundary member, and chord and collector reinforcement (1.0 in. = 25.4 
mm). 
(2) #8 
(collector bars) 
3" 
21
21 3" 2" 
2" 
21
2" min 
topping 
WWF 
Figure 7.1-11 Collector reinforcement at the end 
of the interior longitudinal walls - Joint 5 (1.0 in. 
= 25.4 mm, 1.0 ft = 0.3048 m). 
Figure 7.1-11 shows the collector reinforcement for the interior longitudinal walls. The side cover of 
2-1/2 in. is provided by casting the topping into the cores and by the stems of the plank. A minimum 
space of 1 in. is provided between the plank stems and the sides of the bars. 
7.1.2.5.4 Shear Resistance
Chapter 7, Precast Concrete Design 
7-27 
Cut out alternate face shells 
(16" o.c. each side) and place 
topping completely through 
wall and between planks 
(2) #5 in 
masonry 
bond beam 
#3x4'-0" at 16" to 
lap with WWF 
(2) #8 
collector bars 
1" clear 
Vertical 
reinforcement 
WWF 10 x 10 
W4.5 x W4.5 
Figure 7.1-12 Wall-to-diaphragm reinforcement along interior 
longitudinal walls - Joint 5 (1.0 in. = 25.4 mm, 1.0 ft = 0.3048 m). 
Thin composite and noncomposite topping slabs on precast floor and roof members may not have reliable 
shear strength provided by the concrete. In accordance with ACI 318 Sec. 21.7.7.2 [21.9.7.2], all of the 
shear resistance must be provided by the reinforcement (that is, Vc = 0). 
fVn = fAcv.nfy = 0.75(0.054 in.2/ft)60 ksi = 2.43 kips/ft 
The shear resistance in the transverse direction is: 
2.43 kips/ft (72 ft) = 175 kips 
which is greater than the Joint 2 shear (maximum transverse shear) of 139 kips. No. 3 dowels are used to 
make the welded wire fabric continuous across the masonry walls. The topping is to be cast into the 
masonry walls as shown in Figure 7.1-12, and the spacing of the No. 3 bars is set to be modular with the 
CMU. 
The required shear reinforcement along the exterior longitudinal wall (Joint 4) is: 
Avf4 = Vu4/ffy = (27.4 kips)/[(0.75)(1.0)(60 ksi)] = 0.61 in.2 
7.1.2.5.5 Check Out-of-Plane Forces 
At Joint 4 with bars at 2 ft on center, Fp = 624 plf = 2 ft(624 plf) = 1.25 kips. The required reinforcement, 
As = 1.25/(0.9)(60ksi) = 0.023 in.2. Provide #3 bars at 2 ft on center, which provides a nominal strength 
of 0.11 x 60 / 2 = 3.3 klf. The detail provides more than required by ACI 318 Sec. 16.5 for the 1.5 klf tie 
force. The development length was checked in the prior example. Using #3 bars at 2 ft on center will be 
adequate, and the detail is shown in Figure 7.1-13. The detail at joint 2 is similar.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
7-28 
(2) #5 in 
masonry 
bond beam 
(2) #8 
(collector bars) 
WWF 10 x10 
W4.5 x W4.5 
2" 
Vertical wall 
reinforcement 
beyond 
#3x STD HK 
2'-6" 
at 2'-0" o.c. 
Cut out face shells 
@ 2'-0" and place 
topping into wall 
Figure 7.1-13 Exterior longitudinal wall-to-diaphragm 
reinforcement and out-of-plane anchorage - Joint 4 (1.0 in. = 
25.4 mm, 1.0 ft = 0.3048 m).
Chapter 7, Precast Concrete Design 
7-29 
7.2 THREE-STORY OFFICE BUILDING WITH PRECAST CONCRETE SHEAR 
WALLS 
This example illustrates the seismic design of ordinary precast concrete shear walls that may be used in 
regions of low to moderate seismicity. The Provisions has one requirement for detailing such walls: 
connections that resist overturning shall be Type Y or Z. ACI 318-02 has incorporated a less specific 
requirement, renamed the system as intermediate precast structural walls, and removed some of the detail. 
This example shows an interpretation of the intent of the Provisions for precast shear wall systems in 
regions of moderate and low seismicity, which should also meet the cited ACI 318-02 requirements. 
[As indicated at the beginning of this chapter, the requirements for precast shear wall systems in the 2003 
Provisions have been revised  primarily to point to ACI 318-02 by reference. See also Sec. 7.2.2.1 for 
more discussion of system requirements.] 
7.2.1 Building Description 
This precast concrete building is a three-story office building (Seismic Use Group I) in southern New 
England on Site Class D soils. The structure utilizes 10-ft-wide by 18-in.-deep prestressed double tees 
(DTs) spanning 40 ft to prestressed inverted tee beams for the floors and the roof. The DTs are to be 
constructed using lightweight concrete. Each of the above-grade floors and the roof are covered with a 2- 
in.-thick (minimum), normal weight cast-in-place concrete topping. The vertical seismic-force-resisting 
system is to be constructed entirely of precast concrete walls located around the stairs and 
elevator/mechanical shafts. The only features illustrated in this example are the rational selection of the 
seismic design parameters and the design of the reinforcement and connections of the precast concrete 
shear walls. The diaphragm design is not illustrated. 
As shown in Figure 7.2-1, the building has a regular plan. The precast shear walls are continuous from 
the ground level to 12 ft above the roof. Walls of the elevator/mechanical pits are cast-in-place below 
grade. The building has no vertical irregularities. The story-to-story height is 12 ft.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
7-30 
25'-0" 25'-0" 25'-0" 25'-0" 25'-0" 25'-0" 
150'-0" 
40'-0" 40'-0" 40'-0" 
120'-0" 
15'-0" 
8'-0" 
26 IT 28 
precast 
beams 
18" DT roof and floor 
slabs (10 DT 18) 
8" precast 
shear walls
8'-0" 
Figure 7.2-1 Three-story building plan (1.0 in. = 25.4 mm, 1.0 ft = 0.3048 m). 
The precast walls are estimated to be 8 in. thick for building mass calculations. These walls are normal 
weight concrete with a 28-day compressive strength, fc' = 5,000 psi. Reinforcing bars used at the ends of 
the walls and in welded connectors are ASTM A706 (60 ksi yield strength). The concrete for the 
foundations and below-grade walls has a 28-day compressive strength, fc' = 4,000 psi. 
7.2.2 Design Requirements 
7.2.2.1 Seismic Parameters of the Provisions 
The basic parameters affecting the design and detailing of the building are shown in Table 7.2-1.
Chapter 7, Precast Concrete Design 
7-31 
Table 7.2-1 Design Parameters 
Design Parameter Value 
Seismic Use Group I I = 1.0 
SS (Map 1 [Figure 3.3-1]) 0.266 
S1 (Map 2 [Figure 3.3-2]) 0.08 
Site Class D 
Fa 1.59 
Fv 2.4 
SMS = FaSS 0.425 
SM1 = FvS1 0.192 
SDS = 2/3 SMS 0.283 
SD1 = 2/3 SM1 0.128 
Seismic Design Category B 
Basic Seismic-Force-Resisting System Bearing Wall System 
Wall Type * Ordinary Reinforced Concrete Shear Walls 
R 4 
O0 2.5 
Cd 4 
* Provisions Sec. 9.1.1.3 [9.2.2.1.3] provides for the use of ordinary reinforced concrete shear 
walls in Seismic Design Category B, which does not require adherence to the special seismic 
design provisions of ACI 318 Chapter 21. 
[The 2003 Provisions have adopted the 2002 U.S. Geological Survey probabilistic seismic hazard maps 
and the maps have been added to the body of the 2003 Provisions as figures in Chapter 3. These figures 
replace the previously used separate map package.] 
[Ordinary precast concrete shear walls is recognized as a system in Table 4.3-1 of the 2003 Provisions. 
Consistent with the philosophy that precast systems are not expected to perform as well as cast-in-place 
systems, the design factors for the ordinary precast concrete shear walls per 2003 Provisions Table 4.3-1 
are: R = 3, O0 = 2.5, and Cd = 3. Note that while this system is permitted in Seismic Design Category B, 
unline ordinary reinforced concrete shear walls, it is not permitted in Seismic Design Category C. 
Alternatively, as this example indicates conceptually, this building could be designed incorporating 
intermediate precast concrete shear walls with the following design values per 2003 Provisions Table 4.3- 
1: R = 4, O0 = 2.5, and Cd = 4.] 
7.2.2.2 Structural Design Considerations 
7.2.2.2.1 Precast Shear Wall System 
This system is designed to yield in bending at the base of the precast shear walls without shear slippage at 
any of the joints. Although not a stated design requirement of the Provisions or ACI 318-02 for this 
Seismic Design Category, shear slip could kink the vertical rebar at the connection and sabotage the 
intended performance, which counts on an R factor of 4. The flexural connections at the ends of the
FEMA 451, NEHRP Recommended Provisions: Design Examples 
7-32 
walls, which are highly stressed by seismic forces, are designed to be the Type Y connection specified in 
the Provisions. See Provisions Sec. 9.1.1.2 [9.2.2.1.1] (ACI Sec. 21.1 [21.1]) for the definitions of 
ordinary precast concrete structural walls and Provisions Sec. 9.1.1.12 [not applicable for the 2003 
Provisions] (ACI Sec. 21.11.6) for the connections. The remainder of the connections (shear connectors) 
are then made strong enough to ensure that the inelastic straining is forced to the intended location. 
[Per 2003 Provisions Sec. 9.2.2.1.1 (ACI 318-02 Sec. 21.1), ordinary precast concrete shear walls need 
only satisfy the requirements of ACI 318-02 Chapters 1-18 (with Chapter 16 superceding Chapter 14). 
Therefore, the connections are to be designed in accordance with ACI 318-02 Sec. 16.6.] 
Although it would be desirable to force yielding to occur in a significant portion of the connections, it 
frequently is not possible to do so with common configurations of precast elements and connections. The 
connections are often unavoidable weak links. Careful attention to detail is required to assure adequate 
ductility in the location of first yield and that no other connections yield prematurely. For this particular 
example, the vertical bars at the ends of the shear walls act as flexural reinforcement for the walls and are 
selected as the location of first yield. The yielding will not propagate far into the wall vertically due to 
the unavoidable increase in flexural strength provided by unspliced reinforcement within the panel. The 
issue of most significant concern is the performance of the shear connections at the same joint. The 
connections are designed to provide the necessary shear resistance and avoid slip without unwittingly 
increasing the flexural capacity of the connection because such an increase would also increase the 
maximum shear force on the joint. At the base of the panel, welded steel angles are designed to be 
flexible for uplift but stiff for in-plane shear. 
7.2.2.2.2 Building System 
No height limitations are imposed (Provisions Table 5.2.2 [4.3-1]). 
For structural design, the floors are assumed to act as rigid horizontal diaphragms to distribute seismic 
inertial forces to the walls parallel to the motion. The building is regular both in plan and elevation, for 
which, according to Provisions Table 5.2.5.1 [4.4-4], use of the ELF procedure (Provisions Sec. 5.4 [5.2]) 
is permitted. 
Orthogonal load combinations are not required for this building (Provisions Sec. 5.2.5.2.1 [4.4.2.1]). 
Ties, continuity, and anchorage (Provisions Sec. 5.2.6.1 and 5.2.6.2 [4.6.1.1 and 4.6.1.2]) must be 
explicitly considered when detailing connections between the floors and roof, and the walls and columns. 
This example does not include consideration of nonstructural elements. 
Collector elements are required due to the short length of shear walls as compared to the diaphragm 
dimensions, but are not designed in this example. 
Diaphragms need to be designed for the required forces (Provisions Sec. 5.2.6.2.6 [4.6.1.9]), but that 
design is not illustrated here. 
The bearing walls must be designed for a force perpendicular to their plane (Provisions Sec. 5.2.6.2.7 
[4.6.1.3]), but this requirement is of no real consequence for this building. 
The drift limit is 0.025hsx (Provisions Table 5.2.8 [4.5-1]), but drift is not computed here.
Chapter 7, Precast Concrete Design 
7-33 
ACI 318 Sec. 16.5 requires minimum strengths for connections between elements of precast building 
structures. The horizontal forces were described in Sec. 7.1; the vertical forces will be described in this 
example. 
7.2.3 Load Combinations 
The basic load combinations (Provisions Sec. 5.2.7 [4.2.2]) require that seismic forces and gravity loads 
be combined in accordance with the factored load combinations presented in ASCE 7 except that the 
factors for seismic loads (E) are defined by Provisions Eq. 5.2.7-1 and 5.2.7-2 [4.2-1 and 4.2-2]: 
E = .QE  0.2SDSD = (1.0)QE  (0.2)(0.283)D = QE  0.0567D 
According to Provisions Sec. 5.2.4.1 [4.3.3.1], . = 1.0 for structures in Seismic Design Categories A, B, 
and C, even though this seismic resisting system is not particularly redundant. 
The relevant load combinations from ASCE 7 are: 
1.2D  1.0E + 0.5L 
0.9D  1.0E 
Into each of these load combinations, substitute E as determined above: 
1.26D + QE + 0.5L 
1.14D - QE + 0.5L (will not control) 
0.96D + QE (will not control) 
0.843D - QE 
These load combinations are for loading in the plane of the shear walls. 
7.2.4 Seismic Force Analysis 
7.2.4.1 Weight Calculations 
For the roof and two floors 
18 in. double tees (32 psf) + 2 in. topping (24 psf) = 56.0 psf 
Precast beams at 40 ft = 12.5 psf 
16 in. square columns = 4.5 psf 
Ceiling, mechanical, miscellaneous = 4.0 psf 
Exterior cladding (per floor area) = 5.0 psf 
Partitions = 10.0 psf 
Total = 92.0 psf 
The weight of each floor including the precast shear walls is: 
(120 ft)(150 ft)(92 psf/1,000) + [15 ft(4) + 25 ft(2)](12 ft)(0.10 ksf) = 1,790 kips 
Considering the roof to be the same weight as a floor, the total building weight is W = 3(1,790 kips) = 
5,360 kips. 
7.2.4.2 Base Shear
FEMA 451, NEHRP Recommended Provisions: Design Examples 
7-34 
The seismic response coefficient (Cs) is computed using Provisions Eq. 5.4.1.1-1 [5.2-2]: 
0.283 0.0708 
/ 41 
DS 
S 
C S
R I 
= = = 
except that it need not exceed the value from Provisions Eq. 5.4.1.1-2 [5.2-3] computed as: 
C 
S 
S T R I 
= D1 = = 0128 
029 4 1 
0110 
( / ) 
. 
. ( / ) 
. 
where T is the fundamental period of the building computed using the approximate method of Provisions 
Eq. 5.4.2.1-1 [5.2-6]: 
T Ch a r n x 
= =(0.02)(36)0.75=0.29sec 
Therefore, use Cs = 0.0708, which is larger than the minimum specified in Provisions Eq. 5.4.1.1-3 [not 
applicable in the 2003 Provisions]: 
Cs = 0.044ISDS = (.044)(1.0)(0.283) = 0 .0125 
[The minimum Cs has been changed to 0.01 in the 2003 Provisions.] 
The total seismic base shear is then calculated using Provisions Eq. 5.4-1 [5.2-1] as: 
V = CsW = (0.0708)(5,370) = 380 kips 
Note that this force is substantially larger than a design wind would be. If a nominal 20 psf were applied 
to the long face and then amplified by a load factor of 1.6, the result would be less than half this seismic 
force already reduced by an R factor of 4. 
7.2.4.3 Vertical Distribution of Seismic Forces 
The seismic lateral force (Fx) at any level is determined in accordance with Provisions Sec. 5.4.3 [5.2.3]: 
Fx =CvxV 
where 
1 
k 
x x 
vx n k 
i i 
i 
C w h 
w h 
= 
= 
S 
Since the period, T < 0.5 sec, k = l in both building directions. With equal weights at each floor level, the 
resulting values of Cvx and Fx are as follows: 
Roof Cvr = 0.50 Fr = 190 kips 
Third Floor Cv3 = 0.33 F3 = 127 kips 
Second Floor Cv2 = 0.17 F2 = 63.0 kips
Chapter 7, Precast Concrete Design 
7-35 
12'-0" 12'-0" 12'-0" 
95 kips 
63.5 kips 
31.5 kips 
Grade 
25'-0" 
V = SF = 190 kips 
Figure 7.2-2 Forces on the longitudinal walls 
(1.0 kip = 4.45 kN, 1.0 ft = 0.3048 m). 
7.2.4.4 Horizontal Shear Distribution and Torsion 
7.2.4.4.1 Longitudinal Direction 
Design each of the 25-ft-long walls at the elevator/mechanical shafts for half the total shear. Since the 
longitudinal walls are very close to the center of rigidity, assume that torsion will be resisted by the 15-ftlong 
stairwell walls in the transverse direction. The forces for each of the longitudinal walls are shown in 
Figure 7.2-2. 
7.2.4.4.2 Transverse Direction 
Design the four 15-ft-long stairwell walls for the total shear including 5 percent accidental torsion 
(Provisions Sec. 5.4.4.2 [5.2.4.2]). A rough approximation is used in place of a more rigorous analysis 
considering all of the walls. The maximum force on the walls is computed as: 
V = 380/4 + 380(0.05)(150)/[(100 ft moment arm)  (2 walls in each set)] = 109 kips 
Thus
Fr = 109(0.50) = 54.5 kips 
F3 = 109(0.33) = 36.3 kips 
F2 = 109(0.167) = 18.2 kips 
Seismic forces on the transverse walls of the stairwells are shown in Figure 7.2-3.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
7-36 
12'-0" 12'-0" 12'-0" 
54.5 kips 
36.3 kips 
18.2 kips 
Grade 
15'-0" 
V = SF = 109 kips 
Figure 7.2-3 Forces on the transverse walls 
(1.0 kip = 4.45 kN, 1.0 ft = 0.3048 m). 
7.2.5 PROPORTIONING AND DETAILING 
The strength of members and components is determined using the strengths permitted and required in 
ACI 318 excluding Chapter 21 (see Provisions Sec. 9.1.1.3 [9.2.2.1.3]). 
7.2.5.1 Overturning Moment and End Reinforcement 
Design shear panels to resist overturning by means of reinforcing bars at each end with a direct tension 
coupler at the joints. A commonly used alternative is a threaded post-tensioning bar inserted through the 
stack of panels, but the behavior is different, and the application of the rules for a Type Y connection to 
such a design is not clear. 
7.2.5.1.1 Longitudinal Direction 
The free-body diagram for the longitudinal walls is shown in Figure 7.2-4. The tension connection at the 
base of the precast panel to the below grade wall is governed by the seismic overturning moment and the 
dead loads of the panel and supported floors and roof. In this example, the weights for an elevator 
penthouse, with a floor and equipment at 180 psf between the shafts and a roof at 20 psf, are included. 
The weight for the floors includes double tees, ceiling and partition (total of 70 psf), but not beams and 
columns. Floor live load is 50 psf, except 100 psf is used in the elevator lobby. Roof snow load is 30 psf. 
(The elevator penthouse is so small that it was ignored in computing the gross seismic forces on the 
building, but it is not ignored in the following calculations.)
Chapter 7, Precast Concrete Design 
7-37 
12'-0" 12'-0" 12'-0" 
95 kips 
63.5 kips 
31.5 kips 
23'-6" 
V 
12'-0" 
9" 9" 
T 
C 
12'-0" 
D 
D 
D 
D 
Figure 7.2-4 Free-body diagram for 
longitudinal walls (1.0 kip = 4.45 kN, 
1.0 ft = 0.3048 m). 
At the base 
ME = (95 kips)(36 ft) + (63.5 kips)(24 ft) + (31.5 kips)(12 ft) = 5,520 ft-kips 
3D = wall + exterior floors (& roof) + lobby floors + penthouse floor + penthouse roof 
= (25 ft)(48 ft)(0.1 ksf) + (25 ft)(48 ft/2)(0.070 ksf)(3) + (25 ft)(8 ft/2)(0.070 ksf)(2) + 
(25 ft)(8 ft/2)(0.18 ksf) + (25ft )(24 ft/2)(0.02 ksf) 
= 120 + 126 + 14 + 18 + 6 = 284 kips 
3L = (25 ft)(48 ft/2)(0.05 ksf)(2) + (25 ft)(8 ft/2)(0.1 ksf) = 60 + 10 = 70 kips 
3S = (25ft)(48 ft + 24 ft)(0.03 ksf)/2 = 27 kips 
Using the load combinations described above, the vertical loads for combining with the overturning 
moment are computed as: 
Pmax = 1.26 D + 0.5 L + 0.2 S = 397 kips 
Pmin = 0.843 D = 239 kips 
The axial load is quite small for the wall panel. The average compression Pmax / Ag = 0.165 ksi (3.3 
percent of f'c). Therefore, the tension reinforcement can easily be found from the simple couple shown on 
Figure 7.2-4. 
The effective moment arm is: 
jd = 25 - 1.5 = 23.5 ft
FEMA 451, NEHRP Recommended Provisions: Design Examples 
7-38 
and the net tension on the uplift side is: 
min 5320 239 107 kips 
u 2 23.5 2 
T M P 
jd 
= - = - = 
The required reinforcement is: 
As = Tu/ffy = (107 kips)/[0.9(60 ksi)] = 1.98 in.2 
Use two #9 bars (As = 2.0 in.2 ) at each end with direct tension couplers for each bar at each panel joint. 
Since the flexural reinforcement must extend a minimum distance d (the flexural depth)beyond where it is 
no longer required, use both #9 bars at each end of the panel at all three levels for simplicity. 
At this point a check of ACI 318 Sec. 16.5 will be made. Bearing walls must have vertical ties with a 
nominal strength exceeding 3 kips/ft, and there must be at least two ties per panel. With one tie at each 
end of a 25 ft panel, the demand on the tie is: 
Tn = (3 kip/ft)(25 ft)/2 = 37.5 kip 
The two #9 bars are more than adequate for the ACI requirement. 
Although no check for confinement of the compression boundary is required for ordinary precast concrete 
shear walls, it is shown here for interest. Using the check from ACI 318-99 Sec. 21.6.6.2 [21.7.6.2], the 
depth to the neutral axis is: 
Total compression force = As fy + Pmax = (2.0)(60) + 397 = 517 kips 
Compression block a = (517 kips)/[(0.85)(5 ksi)(8 in. width)] = 15.2 in. 
Neutral axis depth c = a/(0.80) = 19.0 in. 
The maximum depth (c) with no boundary member per ACI 318-99 Eq. 21-8 [21-8] is: 
c ( ) 
l 
h u w 
= 
600 d / 
where the term (du/hw) shall not be taken less than 0.007. Once the base joint yields, it is unlikely that 
there will be any flexural cracking in the wall more than a few feet above the base. An analysis of the 
wall for the design lateral forces using 50% of the gross moment of inertia, ignoring the effect of axial 
loads, and applying the Cd factor of 4 to the results gives a ratio (du/hw) far less than 0.007. Therefore, 
applying the 0.007 in the equation results in a distance c of 71 in., far in excess of the 19 in. required. 
Thus, ACI 318-99 would not require transverse reinforcement of the boundary even if this wall were 
designed as a special reinforced concrete shear wall. For those used to checking the compression stress as 
an index: 
= 742 psi ( ) 
( ) 
( )2 ( ) 
389 6 5,520 
8 25 12 8 25 12 
P M 
A S 
s = + = + 
The limiting stress is 0.2fc' , which is 1000 psi, so no transverse reinforcement is required at the ends of 
the longitudinal walls. 
7.2.5.1.2 Transverse Direction
Chapter 7, Precast Concrete Design 
7-39 
12'-0" 12'-0" 12'-0" 
18.2 kips 
36.3 kips 
54.5 kips 
12'-0" 
9" 13'-6" 9" 
7'-0" 
V C 
T 
D
D
D
D 
Figure 7.2-5 Free-body diagram of the transverse walls 
(1.0 kip = 4.45 kN, 1.0 ft = 0.3048 m). 
The free-body diagram of the transverse walls is shown in Figure 7.2-5. The weight of the precast 
concrete stairs is 100 psf and the roof over the stairs is 70 psf. 
The transverse wall is similar to the longitudinal wall. 
At the base 
ME = (54.5 kips)(36 ft) + (36.3 kips)(24 ft) + (18.2 kips)(12 ft) = 3,052 ft-kips 
3D = (15 ft)(48 ft)(0.1 ksf) + 2(12.5 ft/2)(10 ft/2)(0.07 ksf)(3) + (15 ft)(8 ft/2)[(0.1 ksf)(3) + 
(0.07 ksf)] = 72 + 13 + 18 + 4 = 107 kips 
3L = 2(12.5 ft/2)(10 ft/2)(0.05 ksf)(2) + (15 ft)(8 ft/2)(0.1 ksf)(3) = 6 + 18 = 24 kips 
3S = [2(12.5 ft/2)(10 ft/2) + (15 ft)(8 ft/2)](0.03 ksf) = 3.7 kips 
Pmax = 1.26(107) + 0.5(24) + 0.2(4) = 148 kips 
Pmin = 0.843(107) = 90.5 kips 
jd = 15 - 1.5 = 13.5 ft 
Tu = (Mnet/jd) - Pmin/2 = (3,052/13.5) - 90.5/2 = 181 kips 
As = Tu/ffy = (181 kips)/[0.9(60 ksi)] = 3.35 in.2
FEMA 451, NEHRP Recommended Provisions: Design Examples 
7-40 
Use two #10 and one #9 bars (As = 3.54 in.2 ) at each end of each wall with a direct tension coupler at each 
bar for each panel joint. All three bars at each end of the panel will also extend up through all three levels 
for simplicity. Following the same method for boundary member check as on the longitudinal walls: 
Total compression force = As fy + Pmax = (3.54)(60) + 148 = 360 kips 
Compression block a = (360 kips)/[(0.85)(5 ksi)(8 in. width)] = 10.6 in. 
Neutral axis depth c = a/(0.80) = 13.3 in. 
Even though this wall is more flexible and the lateral loads will induce more flexural cracking, the 
computed deflections are still small and the minimum value of 0.007 is used for the ratio (du/hw). This 
yields a maximum value of c = 42.9 in., thus confinement of the boundary would not be required. The 
check of compression stress as an index gives: 
= 951 psi ( ) 
( ) 
( )2 ( ) 
140 6 2,930 
8 15 12 8 15 12 
P M 
A S 
s = + = + 
Since s < 1,000 psi, no transverse reinforcement is required at the ends of the transverse walls. Note how 
much closer to the criterion this transverse wall is by the compression stress check. 
The overturning reinforcement and connection are shown in Figures 7.2-6. Provisions Sec. 9.1.1.12 [not 
applicable in the 2003 Provisions] (ACI 21.11.6.4) requires that this Type Y connection develop a 
probable strength of 125% of the nominal strength and that the anchorage on either side of the connection 
develop 130% of the defined probable strength. [As already noted, the connection requirements for 
ordianry precast concrete shear walls have been removed in the 2003 Provisions and the ACI 318-02 
requirements are less specific.] The 125% requirement applies to the grouted mechanical splice, and the 
requirement that a mechanical coupler develop 125% of specified yield strength of the bar is identical to 
the Type 1 coupler defined by ACI 318 Sec. 21.2.6.1. Some of the grouted splices on the market can 
qualify as the Type 2 coupler defined by ACI, which must develop the specified tensile strength of the 
bar. The development length, ld, for the spliced bars is multiplied by both the 1.25 and the 1.3 factors to 
satisfy the Provisions requirement. The bar in the panel is made continuous to the roof, therefore no 
calculation of development length is necessary in the panel. The dowel from the foundation will be 
hooked, otherwise the depth of the foundation would be more than required for structural reasons. The 
size of the foundation will provide adequate cover to allow the 0.7 factor on ACIs standard development 
length for hooked bars. For the # 9 bar: 
' 
1.3(1.25) (1.6.25)0.7(1200) 1365(1.128) 24.3 in. 
4000 
b 
dh 
c 
l d 
f 
= = = 
Similarly, for the #10 bar, the length is 27.4 in. 
Like many shear wall designs, this design does concentrate a demand for overturning resistance on the 
foundation. In this instance the resistance may be provided by a large footing (on the order of 20 ft by 28 
ft by 3 ft thick) under the entire stairwell, or by deep piers or piles with an appropriate cap for load 
transfer. Refer to Chapter 4 for examples of design of each type of foundation, although not for this 
particular example.
Chapter 7, Precast Concrete Design 
7-41 
(2) #10 & (1) #9 ea. end, 
full ht. of 15' transverse 
wall panel 
(2) #9 ea. end, full height 
of 25' longitudinal 
wall panel 
8" precast wall 
Direct tension 
coupler-(typical) 
8" 8" 
Longitudinal Wall Transverse Wall 
1" shim and drypack 
(typical) 
Reinforced foundation 
not designed in the 
example 
3" min 25" min for #9 
Development at Foundation 
Standard hook to develop 
overturning reinforcement 
28" min for #10 
Figure 7.2-6 Overturning connection detail at the base of the 
walls (1.0 in = 25.4 mm, 1.0 ft = 0.3048 m). 
7.2.5.2 Shear Connections and Reinforcement 
Panel joints often are designed to resist the shear force by means of shear friction but that technique is not 
used for this example because the joint at the foundation will open due to flexural yielding. This opening 
would concentrate the shear stress on the small area of the dry-packed joint that remains in compression. 
This distribution can be affected by the shims used in construction. Tests have shown that this often leads 
to slip of the joint, which could lead to a kink in the principal tension reinforcement at or near its splice 
and destroy the integrity of the system. Therefore, the joint will be designed with direct shear connectors 
that will prevent slip along the joint. This is the authors interpretation of the Provisions text indicating 
that Type Y connections shall develop under flexural, shear, and axial load actions, as required, a 
probable strength. . . . based upon 125 percent of the specified yield in the connection. It would not be 
required by the ACI 318-02 rules for intermediate precast walls.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
7-42 
Welded wire 
fabric Plate 3
8x4x1'-0" 
L4x3x5
16x0'-8" 
LLH 
Plate 1
2x12x1'-0" 
1
4 8 
Drypack 
(a) Section through connection 
(b) Side elevation 
(c) Section through embedded assembly 
#5, see (c) 
3
4" H.A.S. 
1
4 
4 
C8x18.75 
1
4 
Figure 7.2-7 Shear connection at base (1.0 in = 25.4 mm, 1.0 ft = 0.3048 m). 
7.2.5.2.1 Longitudinal Direction 
The shear amplification factor is determined as: 
2 
(1.25) max / 2 (2.0 in. )(1.25)(60 ksi)(23.5 ft) (397 kip)(23.5 ft / 2) 
5320 ft-kip 
capacity s y 
demand u 
M A fjd P jd 
M M
+ + 
= = 
=1.54 
Therefore, the design shear (Vu) at the base is 1.54(190 kips) = 292 kips 
The base shear connection is shown in Figure 7.2-7 and is to be flexible vertically but stiff horizontally in 
the plane of the panel. The vertical flexibility is intended to minimize the contribution of these 
connections to overturning resistance, which would simply increase the shear demand. 
In the panel, provide an assembly with two face plates 3/8 in.  4 in.  12 in. connected by a C8x18.75 
and with diagonal #5 bars as shown in the figure. In the foundation provide an embedded plate 1/2  12  
1'-6" with six 3/4 in. diameter headed anchor studs. In the field, weld an L 4  3  5/16  0'-8", long leg 
horizontal, on each face. The shear capacity of this connection is checked:
Chapter 7, Precast Concrete Design 
7-43 
Shear in the two loose angles 
fVn = f(0.6Fu)tl(2) = (0.75)(0.6)(58 ksi)(0.3125 in.)(8 in.)(2) = 130.5 kip 
Weld at toe of loose angles 
fVn = f(0.6Fu)tel(2) = (0.75)(0.6)(70 ksi)(0.25 in. / o 2)(8 in.)(2) = 89.1 kip 
Weld at face plates, using Table 8-9 in AISC Manual (3rd edition; same table is 8-42 in 2nd edition) 
fVn = CC1Dl(2 sides) 
C1 = 1.0 for E70 electrodes 
l = 8 in. 
D = 4 (sixteenths of an inch) 
k = 2 in. / 8 in. = 0.25 
` a = eccentricity, summed vectorially: horizontal component is 4 in.; vertical component is 2.67 
in.; thus, al = 4.80 in. and a = 4.8 in./8 in. = 0.6 from the table. By interpolation, C = 1.29 
fVn = (1.29)(1.0)(4)(8)(2) = 82.6 kip 
Weld from channel to plate has at least as much capacity, but less demand. 
Bearing of concrete at steel channel 
fc = f(0.85f'c) = 0.65(0.85)(5 ksi) = 2.76 ksi 
The C8 has the following properties: 
tw = 0.487 in. 
bf = 2.53 in. 
tf = 0.39 in. (average) 
The bearing will be controlled by bending in the web (because of the tapered flange, the critical 
flange thickness is greater than the web thickness). Conservatively ignoring the concretes 
resistance to vertical deformation of the flange, compute the width (b) of flange loaded at 2.76 ksi 
that develops the plastic moment in the web: 
Mp = fFytw
2/4 = (0.9)(50 ksi)(0.4872 in.2)/4 = 2.67 in.-kip/in. 
Mu = fc[(b-tw)2/2 - (tw/2)2/2] = 2.76[(b - 0.243 in.)2 - (0.243 in.)2]/2 
setting the two equal results in b = 1.65 in. 
Therefore bearing on the channel is 
fVc = fc(2 - tw)(l) = (2.76 ksi)[(2(1.65) - 0.487 in.](6 in.) = 46.6 kip 
To the bearing capacity on the channel is added the 4 - #5 diagonal bars, which are effective in 
tension and compression; f = 0.75 for shear is used here: 
fVs = ffyAscosa = (0.75)(60 ksi)(4)(0.31 in.2)(cos 45E) = 39.5 kip 
Thus, the total capacity for transfer to concrete is: 
fVn = fVc + fVs = 46.6 + 39.6 = 86.1 kip
FEMA 451, NEHRP Recommended Provisions: Design Examples 
7-44
The capacity of the plate in the foundation is governed by the headed anchor studs. The Provisions 
contain the new anchorage to concrete provisions that are in ACI 318-02 Appendix D. [In the 2003 
Provisions, the anchorage to concrete provisions have been removed and replaced by the reference to 
ACI 318-02.] Capacity in shear for anchors located far from an edge of concrete, such as these, and 
with sufficient embedment to avoid the pryout failure mode is governed by the capacity of the steel: 
fVs = f n Ase fut = (0.65)(6 studs)(0.44 in.2 per stud)(60 ksi) = 103 kip 
Provisions Sec 9.2.3.3.2 (ACI 318-02 Sec. D.3.3.3) specifies an additional factor of 0.75 to derate 
anchors in structures assigned to Seismic Design Categories C and higher. 
In summary the various shear capacities of the connection are: 
Shear in the two loose angles: 130.5 kip 
Weld at toe of loose angles: 89.1 kip 
Weld at face plates: 82.6 kip 
Transfer to concrete: 86.1 kip 
Headed anchor studs at foundation: 103 kip 
The number of embedded plates (n) required for a panel is: 
n = 292/82.6 = 3.5 
Use four connection assemblies, equally spaced along each side (5'-0" on center works well to avoid the 
end reinforcement). The plates are recessed to position the #5 bars within the thickness of the panel and 
within the reinforcement of the panel. 
It is instructive to consider how much moment capacity is added by the resistance of these connections to 
vertical lift at the joint. The vertical force at the tip of the angle that will create the plastic moment in the 
leg of the angle is: 
T = Mp / x = Fylt2/4 / (l-k) = (36 ksi)(8 in)(0.31252 in.2)/4]/(4 in. - 0.69 in.) = 2.12 kips 
There are four assemblies with two loose angles each, giving a total vertical force of 17 kips. The 
moment resistance is this force times half the length of the panel, which yields 212 ft-kips. The total 
demand moment, for which the entire system is proportioned, is 5320 ft - kips. Thus, these connections 
will add about 4% to the resistance and ignoring this contribution is reasonable. If a straight plate 1/4 in. 
x 8 in., which would be sufficient, were used and if the welds and foundation embedment did not fail first, 
the tensile capacity would be 72 kips each, a factor of 42 increase over the angles, and the shear 
connections would have the unintended effect of more than doubling the flexural resistance, which could 
easily cause failures in the system. 
Using ACI 318 Sec. 11.10, check the shear strength of the precast panel at the first floor: 
fVc=f2Acv fc'hd=0.85(2) 5,000(8)(23.5)(12) = 271 kips 
Because fVc $Vu = 190 kips, the wall is adequate for shear without even considering the reinforcement. 
Note that the shear strength of wall itself is not governed by the overstrength required for the connection. 
However, since Vu $ fVc/2 = 136 kips, ACI Sec. 11.10.8 requires minimum wall reinforcement in 
accordance with ACI 318 Sec. 11.10.9.4 rather than Chapter 14 or 16. For the minimum required .h = 
0.0025, the required reinforcement is:
Chapter 7, Precast Concrete Design 
7-45 
Plate 5
16x5"x0'-8" 
1
4 
See Figure 7.2-7 
for embedded 
plates 
Shim and 
drypack 
Horizontal and 
vertical edges 
Figure 7.2-8 Shear connections on each side of the wall at the 
second and third floors (1.0 in = 25.4 mm). 
Av = 0.0025(8)(12) = 0.24 in.2/ft 
As before, use two layers of welded wire fabric, WWF 44 - W4.0W4.0, one on each face. Shear 
reinforcement provided, Av = 0.12(2) = 0.24 in.2/ft 
Next, compute the shear strength at Level 2. Since the end reinforcement at the base extends to the top of 
the shear wall, bending is not a concern. Yield of the vertical bars will not occur, the second floor joint 
will not open (unlike at the base) and, therefore, shear friction could rationally be used to design the 
connections at this level and above. Shear keys in the surface of both panels would be advisable. Also, 
because of the lack of flexural yield at the joint, it is not necessary to make the shear connection be 
flexible with respect to vertical movement. To be consistent with the seismic force increase from yielding 
at the base, the shear at this level will be increased using the same amplification factor as calculated for 
the first story. 
The design shear, Vu2 = 1.54(95 + 63.5) = 244 kips. 
Using the same recessed embedded plate assemblies in the panel as at the base, but welded with a straight 
plate, the number of plates, n = 244/82.6 = 2.96. Use three plates, equally spaced along each side. 
Figure 7.2-8 shows the shear connection at the second and third floors of the longitudinal precast concrete 
shear wall panels.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
7-46 
7.2.5.2.2 Transverse Direction 
Use the same procedure as for the longitudinal walls: 
2 
(1.25) max / 2 (3.54 in. )(1.25)(60 ksi)(13.5 ft) (148 kip)(23.5 ft / 2) 
3052 ft-kip 
capacity s y 
demand u 
M A fjdP jd 
M M
+ + 
= = 
= 1.50 
Design shear, Vu at base is 1.50(105 kips) = 157.5 kips. 
Use the same shear connections as at the base of the longitudinal walls (Figure 7.2-7). The connection 
capacity is 82.6 kips and the number of connections required is n = 157.5/82.6 = 1.9. Provide two 
connections on each panel. 
Check the shear strength of the first floor panel as described previously: 
fVc=f2fc'hd=0.85(2) 5,000(8)(13.5)(12) = 156 kips 
Similar to the longitudinal direction, fVc $Vu = 142 kips, but Vu $ fVc/2 so provide two layers of welded 
wire fabric, WWF 44 - W4.0W4.0, one on each face as in the longitudinal walls. 
Compute the shear demand at the second floor level joint as indicated below. 
The design shear, Vu = 1.50(52.3+ 34.9) = 130.8 kips. 
Use the same plates as in the longitudinal walls. The number of plates, n = 130.8/82.6 = 1.6. Use two 
plates, equally spaced. Use the same shear connections for the transverse walls as for the longitudinal 
walls as shown in Figures 7.2-7 and 7.2-8.
Chapter 7, Precast Concrete Design 
7-47 
7.3 ONE-STORY PRECAST SHEAR WALL BUILDING 
This example illustrates precast shear wall seismic design using monolithic emulation as defined in the 
Provisions Sec. 9.1.1.12 [not applicable in the 2003 Provisions] (ACI Sec. 21.11.3) for a single-story 
building in a region of high seismicity. For buildings in Seismic Design Category D, Provisions Sec. 
9.1.1.12 [not applicable in the 2003 Provisions] (ACI Sec. 21.11.2.1) requires that the precast seismicforce-
resisting system emulate the behavior of monolithic reinforced concrete construction or that the 
systems cyclic capacity be demonstrated by testing. This example presents an interpretation of 
monolithic emulation design with ductile connections. Here the connections in tension at the base of the 
wall panels yield by bending steel angles out-of-plane. The same connections at the bottom of the panel 
are detailed and designed to be very strong in shear and to resist the nominal shear strength of the 
concrete panel. 
[Many of the provisions for precast concrete shear walls in areas of high seismicity have been moved out 
of the 2003 Provisions and into ACI 318-02. For structures assigned to Seismic Design Category D, 
2003 Provisions Sec. 9.2.2.1.3 (ACI 318-02 Sec. 21.21.1.4) permits special precast concrete shear walls 
(ACI 318-02 Sec. 21.8) or intermediate precast concrete shear walls (ACI 318-02 Sec. 21.13). The 2003 
Provisions does not differentiate between precast or cast-in-place concrete for special shear walls. This is 
because ACI 318-02 Sec. 21.8 essentially requires special precast concrete shear walls to satisfy the same 
design requirements as special reinforced concrete shear walls (ACI 318-02 Sec. 21.7). Alternatively, 
special precast concrete shear walls are permitted if they satisfy experimental and analytical requirements 
contained in 2003 Provisions Sec. 9.2.2.4 and 9.6.] 
7.3.1 Building Description 
The precast concrete building is a single-story industrial warehouse building (Seismic Use Group I) 
located in the Los Angeles area on Site Class C soils. The structure has 8-ft-wide by 12-1/2-in.-deep 
prestressed double tee (DT) wall panels. The roof is light gage metal decking spanning to bar joists that 
are spaced at 4 ft on center to match the location of the DT legs. The center supports for the joists are 
joist girders spanning 40 ft to steel tube columns. The vertical seismic-force-resisting system is the 
precast/prestressed DT wall panels located around the perimeter of the building. The average roof height 
is 20 ft, and there is a 3 ft parapet. The building is located in the Los Angeles area on Site Class C soils. 
Figure 7.3-1 shows the plan of the building, which is regular.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
7-48 
15 DT at 8'-0" = 120'-0" 
48'-0" 48'-0" 
12 DT at 8'-0" = 96'-0" 
Steel tube 
columns 
Joist girder 
(typical) 
24LH03 
at 4'-0" o.c. 
24LH03 
at 4'-0" o.c. 
3 DT at 8'-0" = 
24'-0" 
16'-0" 
O.H. 
door 
5 DT at 8'-0" = 40'-0" 3 DT at 8'-0" = 
24'-0" 
16'-0" 
O.H. 
door 
Figure 7.3-1 Single-story industrial warehouse building plan (1.0 ft = 0.3048 m). 
The precast wall panels used in this building are typical DT wall panels commonly found in many 
locations but not normally used in Southern California. For these wall panels, an extra 1/2 in. has been 
added to the thickness of the deck (flange). This extra thickness is intended to reduce cracking of the 
flanges and provide cover for the bars used in the deck at the base. The use of thicker flanges is 
addressed later. 
Provisions Sec. 9.1.1.5 [9.2.2.1.5.4] (ACI Sec. 21.2.5.1 [21.2.5.1]) limits the grade and type of 
reinforcement in boundary elements of shear walls and excludes the use of bonded prestressing tendons 
(strand) due to seismic loads. ACI 318-99 Sec. 21.7.5.2 [21.9.5.2] permits the use of strand in boundary 
elements of diaphragms provided the stress is limited to 60,000 psi. This design example uses the strand 
as the reinforcement based on that analogy. The rationale for this is that the primary reinforcement of the 
DT, the strand, is not working as the ductile element of the wall panel and is not expected to yield in an 
earthquake. 
The wall panels are normal-weight concrete with a 28-day compressive strength, fc ' = 5,000 psi. 
Reinforcing bars used in the welded connections of the panels and footings are ASTM A706 (60 ksi). 
The concrete for the foundations has a 28-day compressive strength, fc ' = 4,000 psi.
Chapter 7, Precast Concrete Design 
7-49 
7.3.2 Design Requirements 
7.3.2.1 Seismic Parameters of the Provisions 
The basic parameters affecting the design and detailing of the building are shown in Table 7.3-1. 
Table 7.3-1 Design Parameters 
Design Parameter Value 
Seismic Use Group I I = 1.0 
SS (Map 1 [Figure 3.3-1]) 1.5 
S1 (Map 2 [Figure 3.3-2]) 0.60 
Site Class C 
Fa 1.0 
Fv 1.3 
SMS = FaSS 1.5 
SM1 = FvS1 0.78 
SDS = 2/3 SMS 1.0 
SD1 = 2/3 SM1 0.52 
Seismic Design Category D 
Basic Seismic-Force-Resisting System Bearing Walls System 
Wall Type * Special Reinforced Concrete Shear Wall 
R 5 
O0 2.5 
Cd 5 
* Provisions Sec. 9.7.1.2 [9.2.2.1.3] requires special reinforced concrete shear walls in Seismic 
Design Category D and requires adherence to the special seismic design provisions of ACI 318 
Chapter 21. 
[The 2003 Provisions have adopted the 2002 U.S. Geological Survey seismic hazard maps and the maps 
have been added to the body of the 2003 Provisions as figures in Chapter 3 (instead of the previously 
used separate map package).] 
7.3.2.2 Structural Design Considerations 
7.3.2.2.1 Precast Shear Wall System 
The criteria for the design is to provide yielding in a dry connection for bending at the base of each 
precast shear wall panel while maintaining significant shear resistance in the connection. The flexural 
connection for a wall panel at the base is located in one DT leg while the connection at the other leg is 
used for compression. Per Provisions Sec. 9.1.1.12 (ACI Sec. 21.11.3.1) [not applicable in the 2003 
Provisions], these connections resist the shear force equal to the nominal shear strength of the panel and 
have a nominal strength equal to twice the shear that exists when the actual moment is equal to Mpr
FEMA 451, NEHRP Recommended Provisions: Design Examples 
7-50 
(which ACI defines as f = 1.0 and a steel stress equal to 125% of specified yield). Yielding will develop 
in the dry connection at the base by bending the horizontal leg of the steel angle welded between the 
embedded plates of the DT and footing. The horizontal leg of this angle is designed in a manner to resist 
the seismic tension of the shear wall due to overturning and then yield and deform inelastically. The 
connections on the two legs of the DT are each designed to resist 50 percent of the shear. The anchorage 
of the connection into the concrete is designed to satisfy the Type Z requirements in Provisions Sec. 
9.1.1.12 (ACI Sec. 21.11.6.5) [not applicable in the 2003 Provisions.]. Careful attention to structural 
details of these connections is required to ensure tension ductility and resistance to large shear forces that 
are applied to the embedded plates in the DT and footing. 
[Based on the 2003 Provisions, unless the design of special precast shear walls is substantiated by 
experimental evidence and analysis per 2003 Provisions Sec. 9.2.2.4 (ACI 318-02 Sec. 21.8.2), the design 
must satisfy ACI 318-02 Sec. 21.7 requirements for special structural walls as referenced by ACI 318-02 
Sec. 21.8.1. The connection requirements are not as clearly defined as in the 2000 Provisions.] 
7.3.2.2.2 Building System 
Height limit is 160 ft (Provisions Table 5.2.2 [4.3-1]). 
The metal deck roof acts as a flexible horizontal diaphragm to distribute seismic inertia forces to the walls 
parallel to the earthquake motion (Provisions Sec. 5.2.3.1 [4.3.2.1]). 
The building is regular both in plan and elevation. 
The reliability factor, . is computed in accordance with Provisions Sec. 5.2.4.2 [4.3.3]. The maximum .x 
value is given when maxx is the largest value. is the ratio of design story shear resisted by the r maxx r 
single element carrying the most shear force to the total story shear. All shear wall elements (8-ft-wide 
panels) have the same stiffness. Therefore, the shear in each element is the total shear along a side 
divided by the number of elements (wall panels). The largest maxx value is along the side with the least r 
number of panels. Along the side with 11 panels, maxx is computed as: r 
maxx = = 0.0455 r 
1 2
11 
1.0 
Ax = 96 ft  120 ft = 11,520 ft2 
2 20 2 20 = -2.10 
0.0455 11,520 x 
x 
rmax Ax 
. = - = - 
Therefore, use . = 1.0. 
[The redundancy requirements have been substantially changed for the 2003 Provisions. For a shear wall 
building assigned to Seismic Design Category D, . = 1.0 as long as it can be shown that failure of a single 
shear wall with an aspect ratio greater than 1.0 would not result in more than a 33 percent reduction in 
story strength or create an extreme torsional irregularity. Based on the design procedures for the walls, 
each individual panel should be considered a separate wall with an aspect ratio greater than 1.0. 
Alternatively, if the structure is regular in plan and there are at least two bays of perimeter framing on 
each side of the structure in each orthogonal direction, the exception in 2003 Provisions Sec. 4.3.3.2
Chapter 7, Precast Concrete Design 
7-51 
permits the use of D, . = 1.0. This exception could be interpreted as applying to this example, which is 
regular and has more than two wall panels (bays) in both directions.] 
The structural analysis to be used is the ELF procedure (Provisions Sec. 5.4 [5.2]) as permitted by 
Provisions Table 5.2.5 [4.4-1]. 
Orthogonal load combinations are not required for flexible diaphragms in Seismic Design Category D 
(Provisions Sec. 5.2.5.2.3 [4.4.2.3]). 
This example does not include design of the foundation system, the metal deck diaphragm, or the 
nonstructural elements. 
Ties, continuity, and anchorage (Provisions 5.2.6.1 through 5.2.6.4 [4.6]) must be explicitly considered 
when detailing connections between the roof and the wall panels. This example does not include the 
design of these connections, but sketches of details are provided to guide the design engineer. 
There are no drift limitations for single-story buildings as long as they are designed to accommodate 
predicted lateral displacements (Provisions Table 5.2.8, footnote b [4.5-1, footnote c]). 
7.3.3 Load Combinations 
The basic load combinations (Provisions Sec. 5.2.7) require that seismic forces and gravity loads be 
combined in accordance with the factored load combinations as presented in ASCE 7, except that the load 
factor for earthquake effects (E) is defined by Provisions Eq. 5.2.7-1 and 5.2.7-2 [4.2-1 and 4.2-2]: 
E = .QE  0.2SDSD = (1.0)QE  (0.2)(1.0)D = QE  0.2D 
The relevant load combinations from ASCE 7 are: 
1.2D  1.0E + 0.5L 
0.9D  1.0E 
Note that roof live load need not be combined with seismic loads, so the live load term, L, can be omitted 
from the equation. 
Into each of these load combinations, substitute E as determined above: 
1.4D + QE 
1.0D - QE (will not control) 
1.1D + QE (will not control) 
0.7D - QE 
These load combinations are for the in-plane direction of the shear walls.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
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7.3.4 Seismic Force Analysis 
7.3.4.1 Weight Calculations 
Compute the weight tributary to the roof diaphragm 
Roofing = 2.0 psf 
Metal decking = 1.8 psf 
Insulation = 1.5 psf 
Lights, mechanical, sprinkler system etc. = 3.2 psf 
Bar joists = 2.7 psf 
Joist girder and columns = 0.8 psf 
Total = 12.0 psf 
The total weight of the roof is computed as: 
(120 ft  96 ft)(12 psf/1,000) = 138 kips 
The exterior double tee wall weight tributary to the roof is: 
(20 ft/2 + 3 ft)[42 psf/1,000](120 ft + 96 ft)2 = 236 kips 
Total building weight for seismic lateral load, W = 138 + 236 = 374 kips 
7.3.4.2 Base Shear 
The seismic response coefficient (Cs) is computed using Provisions Eq. 5.4.1.1-1 [5.2-2] as: 
1.0 0.20 
/ 51 
DS 
s 
C S
R I 
= = = 
except that it need not exceed the value from Provisions Eq. 5.4.1.1-2 [5.2-3] as follows: 
( ) ( ) 
1 0.52 0.55 
0.189 5 1 
D 
s 
C S 
T R I 
= = = 
where T is the fundamental period of the building computed using the approximate method of Provisions 
Eq. 5.4.2.1-1 [5.2-6]: 
( )0.75 x (0.02) 20.0 0.189 sec 
Ta=Crhn= = 
Therefore, use Cs = 0.20, which is larger than the minimum specified in Provisions Eq. 5.4.1.1-3 [not 
applicable in the 2003 Provisions]: 
Cs = 0.044ISDS = (0.044)(1.0)(1.0) = 0 .044 
[The minimum Cs value has been changed to 0.01 in. the 2003 Provisions. 
The total seismic base shear is then calculated using Provisions Eq. 5.4-1 [5.2-1] as:
Chapter 7, Precast Concrete Design 
7-53 
8'-0" 
20'-0" 3'-0" 
D1 D1 
Vlu 
D2 
2'-0" 2'-0" 2'-0" 2'-0" 
DT leg 
Foundation 
T C 
Vlu 
Figure 7.3-2 Free-body diagram of a panel in the 
longitudinal direction (1.0 ft = 0.3048 m). 
V = CsW = (0.20)(374) = 74.8 kips 
7.3.4.3 Horizontal Shear Distribution and Torsion 
Torsion is not considered in the shear distribution in buildings with flexible diaphragms. The shear along 
each side of the building will be equal, based on a tributary area force distribution. 
7.3.4.3.1 Longitudinal Direction 
The total shear along each side of the building is V/2 = 37.4 kips. The maximum shear on longitudinal 
panels (at the side with the openings) is: 
Vlu = 37.4/11 = 3.4 kips 
On each side, each longitudinal wall panel resists the same shear force as shown in the free-body diagram 
of Figure 7.3-2, where D1 represents roof joist reactions and D2 is the panel weight.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
7-54 
8'-0" 
20'-0" 3'-0" 
Vtu 
D
2'-0" 2'-0" 
DT leg 
Foundation 
T C 
Vtu 
Figure 7.3-3 Free-body diagram of a panel in the 
transverse direction (1.0 ft = 0.3048 m). 
7.3.4.3.2 Transverse Direction 
Seismic forces on the transverse wall panels are all equal and are: 
Vtu = 37.4/12 = 3.12 kips 
Figure 7.3-3 shows the transverse wall panel free-body diagram. 
Note the assumption of uniform distribution to the wall panels in a line requires that the roof diaphragm 
be provided with a collector element along its edge. The chord designed for diaphragm action in the 
perpendicular direction will normally be capable of fulfilling this function, but an explicit check should 
be made in the design. 
7.3.5 Proportioning and Detailing 
The strength of members and components is determined using the strengths permitted and required in 
ACI 318 including Chapter 21.
Chapter 7, Precast Concrete Design 
7-55 
7.3.5.1 Tension and Shear Forces at the Panel Base 
Design each precast shear panel to resist the seismic overturning moment by means of a ductile tension 
connector at the base of the panel. A steel angle connector will be provided at the connection of each leg 
of the DT panel to the concrete footing. The horizontal leg of the angle is designed to yield in bending as 
needed in an earthquake. Provisions Sec. 9.1.1.12 [not applicable in the 2003 Provisions] requires that 
dry connections at locations of nonlinear action comply with applicable requirements of monolithic 
concrete construction and satisfy the following: 
1. Where the moment action on the connection is assumed equal to Mpr, the co-existing shear on the 
connection shall be no greater than 0.5SnConnection and 
2. The nominal shear strength for the connection shall not be less than the shear strengths of the 
members immediately adjacent to that connection. 
Precisely how ductile dry connections emulate monolithic construction is not clearly explained. The dry 
connections used here do meet the definition of a yielding steel element at a connection contained in ACI 
318-02. For the purposes of this example, these two additional requirements are interpreted as: 
1. When tension from the seismic overturning moment causes 1.25 times the yield moment in the angle, 
the horizontal shear on this connection shall not exceed one-half the nominal shear strength of the 
connection. For this design, one-half the total shear will be resisted by the angle at the DT leg in 
tension and the remainder by the angle at the DT leg in compression. 
2. The nominal shear strength of the connections at the legs need to be designed to exceed the in-plane 
shear strength of the DT. 
Determine the forces for design of the DT connection at the base. 
7.3.5.1.1 Longitudinal Direction 
Use the free-body diagram shown in Figure 7.3-2. The maximum tension for the connection at the base 
of the precast panel to the concrete footing is governed by the seismic overturning moment and the dead 
loads of the panel and the roof. The weight for the roof is 11.2 psf, which excludes the joist girders and 
columns. 
At the base 
ME = (3.4 kips)(20 ft) = 68.0 ft-kips 
Dead loads
( ) = 1.08 kips 1 
11.2 1,000 48 4 
2 
D = .. .. 
. . 
D2 = 0.042(23)(8) = 7.73 kips 
SD = 2(1.08) + 7.73 = 9.89 kips 
1.4D = 13.8 kips 
0.7 D = 6.92 kips 
Compute the tension force due to net overturning based on an effective moment arm, d = 4.0 ft (distance 
between the DT legs). The maximum is found when combined with 0.7D:
FEMA 451, NEHRP Recommended Provisions: Design Examples 
7-56 
21 10" 
2" 
23
8" 
4'-0" 
43 
4" 
average 
M 
23
8" 
Figure 7.3-4 Cross section of the DT 
dry-packed at the footing (1.0 in = 25.4 
mm, 1.0 ft = 0.3048 m). 
Tu = ME/d - 0.7D/2 = 68.0/4 - 6.92/2 = 13.5 kips 
7.3.5.1.2 Transverse Direction 
For the transverse direction, use the free-body diagram of Figure 7.3-3. The maximum tension for 
connection at the base of the precast panel to the concrete footing is governed by the seismic overturning 
moment and the dead loads of just the panel. No load from the roof is included, since it is negligible. 
At the base 
ME = (3.12 kips)(20 ft) = 62.4 ft-kips 
The dead load of the panel (as computed above) is D2 = 7.73 kips, and 0.7D = 5.41. 
The tension force is computed as above for d = 4.0 ft (distance between the DT legs): 
Tu = 62.4/4 - 5.41/2 = 12.9 kips 
This tension force is less than that at the longitudinal wall panels. Use the tension force of the 
longitudinal wall panels for the design of the angle connections. 
7.3.5.2 Panel Reinforcement 
Check the maximum compressive stress in the DT leg for the 
requirement of transverse boundary element reinforcement per ACI 
318 Sec. 21.6.6.3 [21.7.6.3]. Figure 7.3-4 shows the cross section 
used. The section is limited by the area of dry-pack under the DT at 
the footing. 
The reason to limit the area of dry-pack at the footing is to locate the 
boundary elements in the legs of the DT, at least at the bottom of the 
panel. The flange between the legs of the DT is not as susceptible to 
cracking during transportation as are the corners of DT flanges 
outside the confines of the legs. The compressive stress due to the 
overturning moment at the top of the footing and dead load is: 
A = 227 in.2 
S = 3240 in.3 
13,800 12(68,000) 313psi 
227 3,240 
P ME 
A S 
s = + = + = 
Roof live loads need not be included as a factored axial load in the compressive stress check, but the force 
from the prestress steel will be added to the compression stress above because the prestress force will be 
effective a few feet above the base and will add compression to the DT leg. Each leg of the DT will be 
reinforced with one 1/2-in. diameter and one 3/8-in. diameter strand. Figure 7.3-5 shows the location of 
these prestressed strands.
Chapter 7, Precast Concrete Design 
7-57 
Deck mesh 
(1) 1
2" dia. strand 
(1) 3
8" dia. strand 
Leg mesh 
43
4" 
average 
21 
6" 4" 2" 
Figure 7.3-5 Cross section of one DT leg showing the location of the 
bonded prestressing tendons or strand (1.0 in = 25.4 mm). 
Next, compute the compressive stress resulting from these strands. Note the moment at the height of 
strand development above the footing, about 26 in. for the effective stress (fse), is less than at the top of 
footing. This reduces the compressive stress by: 
(3.4)(26) x 1000 27 psi 
3, 240 
= 
In each leg, use 
P = 0.58fpu Aps = 0.58(270 ksi)[0.153 + 0.085] = 37.3 kips 
A = 168 in.2 
e = yb - CGStrand = 9.48 - 8.57 = 0.91 in. 
Sb = 189 in.3 
37,300 0.91(37,300) 402psi 
168 189 
P Pe 
A S 
s = + = + = 
Therefore, the total compressive stress is approximately 313 + 402 - 27 = 688 psi. 
The limiting stress is 0.2 fc', which is 1000 psi, so no special boundary elements are required in the 
longitudinal wall panels. 
Reinforcement in the DT for tension is checked at 26 in. above the footing. The strand reinforcement of 
the DT leg resisting tension is limited to 60,000 psi. The rationale for using this stress is discussed at the 
beginning of this example. 
D2 = (0.042)(20.83)(8) = 7.0 kips
FEMA 451, NEHRP Recommended Provisions: Design Examples 
7-58
Pmin = 0.7(7.0 + 2(1.08)) = 6.41 kips 
ME = (3.4)(17.83) = 60.6 ft-kips 
Tu = Mnet/d - Pmin/2= 12.0 kips 
The area of tension reinforcement required is: 
As = Tu/ffy = (12.0 kips)/[0.9(60 ksi)] = 0.22 in.2 
The area of one  in. diameter and one 3/8 in. diameter strand is 0.153 in.2 + 0.085 in.2 = 0.236 in.2 The 
mesh in the legs is available for tension resistance, but not required in this check. 
To determine the nominal shear strength of the concrete for the connection design, complete the shear 
calculation for the panel in accordance with ACI Sec. 21.6 [21.7]. The demand on each panel is: 
Vu = Vlu = 3.4 kips 
Only the deck between the DT legs is used to resist the in-plane shear (the legs act like flanges, meaning 
that the area effective for shear is the deck between the legs). First, determine the minimum required 
shear reinforcement based on ACI Sec. 21.6.2.1 [21.7.2]. Since 
Acv fc'=2.5(48) 5,000=8.49 kips 
exceeds Vu = 3.4 kips, the reinforcement of the deck is per ACI 318 Sec. 16.4.2. Using welded wire 
fabric, the required areas of reinforcement are: 
Ash = Asv = (0.001)(2.5)(12) = 0.03 in.2/ft 
Provide 6  6 - W2.5  W2.0 welded wire fabric. 
Ash = 0.05 in.2/ft 
Asv = 0.04 in.2/ft 
The nominal shear strength of the wall panel by ACI 318 Sec. 21.6.4.1 is: 
( ) (2.5)(48)2 5,000 0.05(4)(60) 29.0kips 
Vn=Acvac fc'+.nfy = 1,000 + = 
where ac is 2.0 for hw/lw = 23/4 = 5.75, which is greater than 2.0. Given that the connections will be 
designed for a shear of 29 kips, it is obvious that half the nominal shear strength will exceed the seismic 
shear demand, which is 3.4 kips. 
The prestress force and the area of the DT legs are excluded from the calculation of the nominal shear 
strength of the DT wall panel. The prestress force is not effective at the base, where the connection is, 
and the legs are like the flanges of a channel, which are not effective in shear. 
7.3.5.3 Size the Yielding Angle 
The angle, which is the ductile element of the connection, is welded between the plates embedded in the 
DT leg and the footing. This angle is a L5  3-1/2  3/4  0 ft-5 in. with the long leg vertical. The steel 
for the angle and embedded plates will be ASTM A572, Grade 50. The horizontal leg of the angle needs 
to be long enough to provide significant displacement at the roof, although this is not stated as a
Chapter 7, Precast Concrete Design 
7-59 
y 
x 
1" y
4" 
5" 
Mx 
CG 
B 
21
4" 
Tu 
L5x31
2x3 
4x5 
(LLV) 
Fillet 
weld "t" 
Mx 
Fillet 
weld 
k = 11
4" 
Location 
of plastic 
hinge 
t 
t 
Mz 
My 
y 
z 
My
Mz 
Vu 
' 
' 
Vu' 
Tu ' 
Vu' 
Tu ' 
Vu' 
Tu ' 
Figure 7.3-6 Free-body of the angle and the fillet weld connecting the embedded plates in 
the DT and the footing (elevation and section) (1.0 in = 25.4 mm). 
requirement in either the Provisions or ACI 318. This will be examined briefly here. The angle and its 
welds are shown in Figure 7.3-6. 
The bending moment at a distance k from the heel of the angle (location of the plastic hinge in the angle) 
is: 
Mu = Tu(3.5 - k) = 13.5(3.5 - 1.25) = 30.4 in.-kips 
( ) ( )2 5 0.75 
0.9 0.9 50 31.6 in.-kips 
fbMn FyZ 4 
. . 
= = . .= 
.. .. 
Providing a stronger angle (e.g., a shorter horizontal leg) will simply increase the demands on the 
remainder of the assembly. Using Provisions Sec. 9.1.1.12 (ACI Sec. 21.11.6.5) [not applicable in the 
2003 Provisions], the tension force for the remainder of this connection other than the angle is based upon 
a probable strength equal to 140% of the nominal strength. Thus 
(1.4) (50)(5)(0.75)2 / 4 1.4 21.9 kips 
3.5 3.5 1.25 
n 
u 
T M 
k 
'= =  = 
- - 
Check the welds for the tension force of 21.9 kips and a shear force (Vu')of 29.0/2 = 14.5 kips, or the 
shear associated with Tu', whichever is greater. The bearing panel, with its larger vertical load, will give a 
larger shear. 
FEMA 451, NEHRP Recommended Provisions: Design Examples 
7-60
1.4D = 13.8 kips, and V = [Tu'(4) + 1.4D(2)]/20 = [21.9/4 + 13.8(2)]/20 = 5.76 kips. Vn for the panel 
obviously controls. 
But before checking the welds, consider the deformability of the system as controlled by the yielding 
angle. Ignore all sources of deformation except the angle. (This is not a bad assumption regarding the 
double tee itself, but other aspects of the connections, particularly the plate and reinforcement embedded 
in the DT, will contribute to the overall deformation. Also, the diaphragm deformation will overwhelm 
all other aspects of deformation, but this is not the place to address flexible diaphragm issues.) The angle 
deformation will be idealized as a cantilever with a length from the tip to the center of the corner, then 
upward to the level of the bottom of the DT, which amounts to: 
L = 3.5 in. - t/2 + 1 in. - t/2 = 3.75 in. 
Using an elastic-plastic idealization, the vertical deformation at the design moment in the leg is 
dv = TL3/3EI = (13.5 kips)(3.75 in.)3/[3(29000 ksi)(5 in.)(0.75 in.)3/12] = 0.047 in. 
This translates into a horizontal motion at the roof of 0.24 in. (20 ft to the roof, divided by the 4 ft from 
leg to leg at the base of the DT.) With Cd of 4, the predicted total displacement is 0.96 in. These 
displacements are not very large, but now compare with the expectations of the Provisions. The 
approximate period predicted for a 20-ft-tall shear wall building is 0.19 sec. Given a weight of 374 kips, 
as computed previously, this would imply a stiffness from the fundamental equation of dynamics: 
T2 W/gK42W/(gT) 42374 /(386 0.19) 201 kip/in. 
K 
=p . =p =p  = 
Now, given the design seismic base shear of 74.8 kips, this would imply an elastic displacement of 
dh = 74.8 kip / (201 kip/in.) = 0.37 in. 
This is about 50% larger than the simplistic calculation considering only the angle. The bending of angle 
legs about their weak axis has a long history of providing ductility and, thus, it appears that this dry 
connection will provide enough deformability to be in the range of expectation of the Provisions. 
7.3.5.4 Welds to Connection Angle 
Welds will be fillet welds using E70 electrodes. 
For the base metal, fRn = f(Fy)ABM. 
For which the limiting stress is fFy = 0.9(50) = 45.0 ksi. 
For the weld metal, fRn = f(Fy)Aw = 0.75(0.6)70(0.707)Aw. 
For which limiting stress is 22.3 ksi. 
Size a fillet weld, 5 in. long at the angle to embedded plate in the footing: 
Using an elastic approach 
Resultant force = V2+T2= 14.52+ 21.92= 26.3kips
Chapter 7, Precast Concrete Design 
7-61 
Z 
X 
V 
My 
V 
Mz 
Figure 7.3-7 Free-body of angle with welds, top view, 
showing only shear forces and resisting moments. 
Aw = 26.3/22.3 = 1.18 in.2 
t = Aw/l =1.18 in.2/5 in. = 0.24 in. 
For a 3/4 in. angle leg, use a 5/16 in. fillet weld. Given the importance of this weld, increasing the size to 
3/8 in. would be a reasonable step. With ordinary quality control to avoid flaws, increasing the strength 
of this weld by such an amount should not have a detrimental effect elsewhere in the connection. 
Now size the weld to the plate in the DT. Continue to use the conservative elastic method to calculate 
weld stresses. Try a fillet weld 5 in. long across the top and 4 in. long on each vertical leg of the angle. 
Using the free-body diagram of Figure 7.3-6 for tension and Figure 7.3-7 for shear, the weld moments 
and stresses are: 
Mx = Tu'(3.5) = 21.9(3.5) = 76.7 in.-kips 
My = Vu'(3.5) = (14.5)(3.5) = 50.8 in.-kips 
Mz = Vu'(yb + 1.0) 
= 14.5(2.77 + 1.0) = 54.7 in.-kips 
For the weld between the angle and the embedded plate in the DT as shown in Figure 7.3-7 the section 
properties for a weld leg (t) are: 
A = 13t in.2 
Ix = 23.0t in.4 
Iy = 60.4t in.4
FEMA 451, NEHRP Recommended Provisions: Design Examples 
7-62
Ip = Ix + Iy = 83.4t in.4 
yb = 2.77 in. 
xL = 2.5 in. 
To check the weld, stresses are computed at all four ends (and corners). The maximum stress is at the 
lower right end of the inverted U shown in Figure 7.3-6. 
14.5 (54)(2.77) 2.93 ksi 
13 83.4 
21.9 (54.7)(2.5) 0.045 ksi 
13 83.4 
= - (50.8)(2.5) - (76.7)(2.77) = -11.3 ksi 
60.4 23.0t 
u zb 
x 
p 
u z L 
y 
p 
y L x b 
z 
y x 
V M y 
A I t t t 
T M x 
A I t t t 
M x M y 
I I t t 
s
s
s 
= ' + = + =.. .. 
. . 
= - ' + = - + =.. .. 
. . 
= - - .. .. 
. . 
R x2 y2 z21t(2.93)2 (0.045)2 ( 11.3)211.t67ksi 
s = s +s +s = + + - =.. .. 
. . 
Thus, t = 11.67/22.3 = 0.52 in., say 9/16 in. Field welds are conservatively sized with the elastic method 
for simplicity and to minimize construction issues. 
7.3.5.5 Tension and Shear at the Footing Embedment 
Reinforcement to anchor the embedded plates is sized for the same tension and shear, and the 
development lengths are lengthened by an additional 30%, per Provisions Sec. 9.1.1.12 (ACI Sec. 
21.11.6.5) [not applicable in the 2003 Provisions]. Reinforcement in the DT leg and in the footing will be 
welded to embedded plates as shown in Figure 7.3-8. 
The welded reinforcement is sloped to provide concrete cover and to embed the bars in the central region 
of the DT leg and footing. The tension reinforcement area required in the footing is: 
( )( ) 
2 
, 
21.9 0.45 in. 
cos 0.9 60 cos26.5 
u 
s Sloped 
y
A T 
ff . 
' 
= = = 
o 
Use two #5 bars (As = 0.62 in.2 ) at each embedded plate in the footing. 
The shear bars in the footing will be two #4 placed on an angle of two (plus)-to-one. The resultant shear 
resistance is: 
fVn = 0.75(0.2)(2)(60)(cos 26.5E) = 16.1 kips
Chapter 7, Precast Concrete Design 
7-63 
DT 
9 
1 
1 
Interior slab 2 
L6x4x1
2x10" 
(2) #5 
5
16 5 
L5x31
2x3
4x5 
(LLV) 
9
16 4 9
16 5 
Plate 1
2 x 6 x 0'-10" 
(2) #4x24" 
(see Fig 7.3-9) 
Plate 6x41
2x1
2 
C.I.P. 
concrete 
footing 
(2) #4x 1 
2 
2'-6" 
2'-6" 
(2) #4 with 
standard hooks 
(2) #3 with 
standard hooks 
(2) #4x48" 
(See Fig 7.3-9) 
weld 
on #4 
Figure 7.3-8 Section at the connection of the precast/prestressed shear wall panel and 
the footing (1.0 in = 25.4 mm). 
7.3.5.6 Tension and Shear at the DT Embedment 
The area of reinforcement for the welded bars of the embedded plate in the DT, which develop tension as 
the angle bends through cycles is: 
21.9 0.408 in.2 
cos 0.9(60)cos6.3 
u 
s 
y
A T 
ff . 
' 
= = = o 
Two #4 bars are adequate. Note that the bars in the DT leg are required to extend upward 1.3 times the 
development length, which would be 22 in. In this case they will be extended 22 in. past the point of 
development of the effective stress in the strand, which totals about 48 in. 
The same embedded plate used for tension will also be used to resist one-half the nominal shear. This 
shear force is 14.5 kips. The transfer of direct shear to the concrete is easily accomplished with bearing 
on the sides of the reinforcing bars welded to the plate. Two #5 and two #4 bars (explained later) are 
welded to the plate. The available bearing area is approximately Abr = 4(0.5 in.)(5 in.(available)) = 10 in.2 
and the bearing capacity of the concrete is fVn = (0.65)(0.85)(5 ksi)(10 in.2) = 27.6 kips > 14.5 kip 
demand.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
7-64 
The weld of these bars to the plate must develop both the tensile demand and this shear force. The weld 
is a flare bevel weld, with an effective throat of 0.2 times the bar diameter along each side of the bar. 
(Refer to the PCI Handbook.) For the #4 bar, the weld capacity is 
fVn = (0.75)(0.6)(70 ksi)(0.2)(0.5 in.)(2) = 6.3 kips/in. 
The shear demand is prorated among the four bars as (14.5 kip)/4 = 3.5 kip. The tension demand is the 
larger of 1.25 fy on the bar (15 kip) or Tu/2 (11.0 kip). The vectorial sum of shear and tension demand is 
15.4 kip. Thus, the minimum length of weld is 15.4 / 6.3 = 2.4 in. 
7.3.5.7 Resolution of Eccentricities at the DT Embedment 
Check the twisting of the embedded plate in the DT for Mz. 
Use Mz = 54.7 in.-kips. 
( ) ( )( ) 
54.7 0.11 in.2 
0.9 60 9.0 
z 
s 
y
A M
ff jd 
= = = 
Use one #4 bar on each side of the vertical embedded plate in the DT as shown in Figure 7.3-9. This is 
the same bar used to transfer direct shear in bearing. 
Check the DT embedded plate for My (50.8 in.-kips) and Mx (76.7 in.-kips) using the two #4 bars welded 
to the back side of the plate near the corners of the weld on the loose angle and the two #3 bars welded to 
the back side of the plate near the bottom of the DT leg (as shown in Figure 7.3-9). It is relatively 
straightforward to compute the resultant moment magnitude and direction, assume a triangular shaped 
compression block in the concrete, and then compute the resisting moment. It is quicker to make a 
reasonable assumption as to the bars that are effective and then compute resisting moments about the X 
and Y axes. This approximate method is demonstrated here. The #4 bars are effective in resisting Mx, 
and one each of the #3 and #4 bars are effective in resisting My. For My assume that the effective depth 
extends 1 in. beyond the edge of the angle (equal to twice the thickness of the plate). Begin by assigning 
one-half of the corner #4 to each component. 
With Asx = 0.20 + 0.20/2 = 0.30 in.2 , 
fMnx = fAs fy jd = (0.9)(0.3 in.2)(60 ksi)(0.95)(5 in.) = 77 in.-kips (>76.7). 
With Asy = 0.11 + 0.20/2 = 0.21 in.2, 
fMny = fAs fy jd = (0.9)(0.21 in.2)(60 ksi)(0.95)(5 in.) = 54 in.-kips (>50.8). 
Each component is strong enough, so the proposed bars are satisfactory.
Chapter 7, Precast Concrete Design 
7-65 
10" 
3" 
For 1.25 Fy 
Plate 41
2"x6"x1
2" 
with 5
8" slot 
at center 
Plate 10"x6"x1
2" 
1" 
2" 3" 
2" 3" 
(2) #3 with 
standard hook 
#4 
(2) #4 with 
standard hook 
Figure 7.3-9 Details of the embedded plate in the DT at the base (1.0 in = 25.4 mm). 
Metal deck 
L4x3x1
4x 
continuous 
DT 
Plate at each DT leg 
Bar joists 
DT corbel at 
each leg 
Figure 7.3-11 Sketch of connection of load-bearing DT wall panel at the roof (1.0 in = 
25.4 mm).
FEMA 451, NEHRP Recommended Provisions: Design Examples 
7-66 
L4x3x1 Bar joist 
4 
continuous 
2'-0" 
Metal deck 
Deck straps 
as needed 
Plate at each 
DT leg 
#4 continuous 
weld to plates 
Figure 7.3-10 Sketch of connection of non-load-bearing DT wall panel at the 
roof (1.0 in = 25.4 mm, 1.0 ft = 0.3048 m). 
7.3.5.8 Other Connections 
This design assumes that there is no in-plane shear transmitted from panel to panel. Therefore, if 
connections are installed along the vertical joints between DT panels to control the out-of-plane 
alignment, they should not constrain relative movement in-plane. In a practical sense, this means the 
chord for the roof diaphragm should not be a part of the panels. Figures 7.3-10 and 7.3-11 show the 
connections at the roof and DT wall panels. These connections are not designed here. Note that the 
continuous steel angle would be expected to undergo vertical deformations as the panels deform laterally. 
Because the diaphragm supports concrete walls out of their plane, Provisions Sec. 5.2.6.3.2 [4.6.2.1] 
requires specific force minimums for the connection and requires continuous ties across the diaphragm. 
Also, it specifically prohibits use of the metal deck as the ties in the direction perpendicular to the deck 
span. In that direction, the designer may wish to use the top chord of the bar joists, with an appropriate 
connection at the joist girder, as the continuous cross ties. In the direction parallel to the deck span, the 
deck may be used but the laps should be detailed appropriately. 
In precast double tee shear wall panels with flanges thicker than 2-1/2 in., consideration may be given to 
using vertical connections between the wall panels to transfer vertical forces resulting from overturning 
moments and thereby reduce the overturning moment demand. These types of connections are not 
considered here, since the uplift force is small relative to the shear force and cyclic loading of bars in thin 
concrete flanges is not always reliable in earthquakes.
8-1 
8 
COMPOSITE STEEL AND CONCRETE 
James Robert Harris, P.E., Ph.D. and 
Frederick R. Rutz, P.E., Ph.D. 
This chapter illustrates application of the 2000 NEHRP Recommended Provisions to the design of 
composite steel and concrete framed buildings using partially restrained composite connections. This 
system is referred to as a Composite Partially Restrained Moment Frame (C-PRMF) in the Provisions. 
An example of a multistory medical office building in Denver, Colorado, is presented. The Provisions set 
forth a wealth of opportunities for designing composite steel and concrete systems, but this is the only one 
illustrated in this set of design examples. 
The design of partially restrained composite (PRC) connections and their effect on the analysis of frame 
stiffness are the aspects that differ most significantly from a non-composite design. Some types of PRC 
connections have been studied in laboratory tests and a design method has been developed for one in 
particular, which is illustrated in this example. In addition, a method is presented by which a designer 
using readily available frame analysis programs can account for the effect of the connection stiffness on 
the overall frame. 
The example covers only design for seismic forces in combination with gravity, although a check on drift 
from wind load is included. 
The structure is analyzed using three-dimensional static methods. The RISA 3D analysis program, v.4.5 
(Risa Technologies, Foothill Ranch, California) is used in the example. 
Although this volume of design examples is based on the 2000 Provisions, it has been annotated to reflect 
changes made to the 2003 Provisions. Annotations within brackets, [ ], indicate both organizational 
changes (as a result of a reformat of all of the chapters of the 2003 Provisions) and substantive technical 
changes to the 2003 Provisions and its primary reference documents. While the general concepts of the 
changes are described, the design examples and calculations have not been revised to reflect the changes 
to the 2003 Provisions. 
Chapter 10 in the 2003 Provisions has been expanded to include modifications to the basic reference 
document, AISC Seismic, Part II. These modifications are generally related to maintaining compatibility 
between the Provisions and the most recent editions of the ACI and AISC reference documents and to 
incorporate additional updated requirements. Updates to the reference documents, in particular AISC 
Seismic, have some affect on the calculations illustrated herein. 
There are not any general technical changes to other chapters of the 2003 Provisions that have a 
significant effect on the calculations and/or design example in this chapter of the Guide with the possible 
exception of the updated seismic hazard maps. 
FEMA 451, NEHRP Recommended Provisions: Design Examples 
8-2 
Where they affect the design examples in this chapter, significant changes to the 2003 Provisions and 
primary reference documents are noted. However, some minor changes to the 2003 Provisions and the 
reference documents may not be noted. 
In addition to the 2000 NEHRP Recommended Provisions (referred to herein as the Provisions), the 
following documents are referenced: 
ACI 318 American Concrete Institute. 1999. Building Code Requirements for Structural 
Concrete, Standard ACI 318-99. Detroit: ACI. 
AISC LRFD American Institute of Steel Construction. 1999. Load and Resistance Factor Design 
Specification for Structural Steel Buildings. Chicago: AISC. 
AISC Manual American Institute of Steel Construction. 1998. Manual of Steel Construction, Load 
and Resistance Factor Design, Volumes 1 and 2, 2nd Edition. Chicago: AISC. 
AISC Seismic American Institute of Steel Construction. 1997. Seismic Provisions for Structural 
Steel Buildings, including Supplement No. 2 (2000). Chicago: 
AISC SDGS-8 American Institute of Steel Construction. 1996. Partially Restrained Composite 
Connections, Steel Design Guide Series 8. Chicago: AISC. 
ASCE TC American Society of Civil Engineers Task Committee on Design Criteria for 
Composite Structures in Steel and Concrete. October 1998. Design Guide for 
Partially Restrained Composite Connections, Journal of Structural Engineering 
124(10).. 
ASCE 7 American Society of Civil Engineers. 1998. Minimum Design Loads for Buildings 
and Other Structures, ASCE 7-98. Reston: ASCE. 
The short-form designations presented above for each citation are used throughout. 
The symbols used in this chapter are from Chapter 2 of the Provisions, the above referenced documents, 
or are as defined in the text. Customary U.S. units are used.
Chapter 8, Composite Steel and Concrete 
8-3 
W18x35 
(typical for 
E-W beams) 
25'-0" 25'-0" 25'-0" 25'-0" 
12'-6" 12'-6" 25'-0" 25'-0" 25'-0" 12'-6" 12'-6" 
20K5 at 
3'-11
2" o.c. 
(typical) 
25'-0" 
W10 
(typical) 
W21x44 
(typical for 
N-S beams) 
W E 
S
N 
Figure 8-1 Typical floor plan (1.0 ft = 0.3048 m). 
8.1 BUILDING DESCRIPTION 
This four-story medical office building has a structural steel framework (see Figures 8-1 through 8-3). 
The floors and roof are supported by open web steel joists. The floor slab is composite with the floor 
girders and the spandrel beams and the composite action at the columns is used to create moment resisting 
connections. Figure 8-4 shows the typical connection. This connection has been studied in several 
research projects over the past 15 years and is the key to the buildings performance under lateral loads. 
The structure is free of irregularities both in plan and elevation. This is considered a Composite Partially 
Restrained Moment Frame (C-PRMF) per Provisions Table 5.2.2 and in AISC Seismic, and it is an 
appropriate choice for buildings with low-to-moderate seismic demands, which depend on the building as 
well as the ground shaking hazard.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
8-4 
North and South End Elevation 
25'-0" 25'-0" 25'-0" 
4 at 13'-0" = 52'-0" 
25'-0" 25'-0" 
2
3
4
Roof 
W18x35 
(typical) 
Figure 8-2 Building end elevation (1.0 ft = 0.3048 m). 
East and West Side Elevation 
2
3
4
Roof 
12'-6" 12'-6" 25'-0" 25'-0" 25'-0" 12'-6" 12'-6" 
4 at 13'-0" = 52'-0"
W21x44 
(typical) 
Figure 8-3 Building side elevation (1.0 ft = 0.3048 m). 
The building is located in a relatively low hazard region (Denver, Colorado), but some internal storage 
loading and Site Class E are used in this example to provide somewhat higher seismic design forces for 
purposes of illustration, and to push the example into Seismic Design Category C.
Chapter 8, Composite Steel and Concrete 
8-5 
Double angle 
web connection 
Column 
Seat angle 
Girder 
Rebar 
Headed stud 
Concrete 
Figure 8-4 Typical composite connection. 
There are no foundations designed in this example. For this location and system, the typical foundation 
would be a drilled pier and voided grade beam system, which would provide flexural restraint for the 
strong axis of the columns at their base (very similar to the foundation for a conventional steel moment 
frame). The main purpose here is to illustrate the procedures for the partially restrained composite 
connections. The floor slabs serve as horizontal diaphragms distributing the seismic forces, and by 
inspection they are stiff enough to be considered as rigid. 
The typical bay spacing is 25 feet. Architectural considerations allowed an extra column at the end bay of 
each side in the north-south direction, which is useful in what is the naturally weaker direction. The 
exterior frames in the north-south direction have moment-resisting connections at all columns. The 
frames in each bay in the east-west direction have moment-resisting connections at all except the end 
columns. Composite connections to the weak axis of the column are feasible, but they are not required 
for this design. This arrangement is illustrated in the figures. 
Material properties in this example are as follows: 
1. Structural steel beams and columns (ASTM A992): Fy = 50 ksi 
2. Structural steel connection angles and plates (ASTM A36): Fy = 36 ksi 
3. Concrete slab (4.5 inches thick on form deck, normal weight): fc' = 3000 psi 
4. Steel reinforcing bars (ASTM A615): Fy = 60 ksi 
The floor live load is 50 psf, except in 3 internal bays on each floor where medical records storage 
imposes 200 psf, and the roof snow load is taken as 30 psf. Wind loads per ASCE 7 are also checked, and 
the stiffness for serviceability in wind is a factor in the design. Dead loads are relatively high for a steel 
building due to the 4.5" normal weight concrete slab used to control footfall vibration response of the 
open web joist system and the precast concrete panels on the exterior walls. 
This example covers the following aspects of seismic design that are influenced by partially restrained 
composite frame systems: 
1. Load combinations for composite design 
2. Assessing the flexibility of the connections 
3. Incorporating the connection flexibility into the analytical model of the building
FEMA 451, NEHRP Recommended Provisions: Design Examples 
8-6 
4. Design of the connections 
8.2 SUMMARY OF DESIGN PROCEDURE FOR COMPOSITE PARTIALLY 
RESTRAINED MOMENT FRAME SYSTEM 
For buildings with low to moderate seismic demands, the partially restrained composite frame system 
affords an opportunity to create a seismic-force-resisting system in which many of the members are the 
same size as would already be provided for gravity loads. A reasonable preliminary design procedure to 
develop member sizes for a first analysis is as follows: 
1. Proportion composite beams with heavy noncomposite loads based upon the demand for the unshored 
construction load condition. For this example, this resulted in W18x35 beams to support the open 
web steel joists. 
2. Proportion other composite beams, such as the spandrel beams in this example, based upon judgment. 
For this example, the first trial was made using the same W18x35 beam. 
3. Select a connection such that the negative moment strength is about 75 percent of the plastic moment 
capacity of the bare steel beam. 
4. Proportion columns based upon a simple portal analogy for either stiffness or strength. If stiffness is 
selected, keep the columns contribution to story drift to no more than one-third of the target. If 
strength is selected, an approximate effective column length factor of K = 1.5 is suggested for 
preliminary design. Also check that the moment capacity of the column (after adjusting for axial 
loads) is at least as large as that for the beam. 
Those final design checks that are peculiar to the system are explained in detail as the example is 
described. The key difference is that the flexibility of the connection must be taken into account in the 
analysis. There are multiple ways to accomplish this. Some analytical software allows the explicit 
inclusion of linear, or even nonlinear, springs at each end of the beams. Even for software that does not, a 
dummy member can be inserted at each end of each beam that mimics the connection behavior. For this 
example another method is illustrated, which is consistent with the overall requirements of the Provisions 
for linear analysis. The member properties of the composite beam are altered to become an equivalent 
prismatic beam that gives approximately the same flexural stiffness in the sway mode to the entire frame 
as the actual composite beams combined with the actual connections. Prudence in the use of this 
simplification does suggest checking the behavior of the connections under gravity loads to assure that 
significant yielding is confined to the seismic event. 
Once an analytic model is constructed, the member and connection properties are adjusted to satisfy the 
overall drift limits and the individual strength limits. This is much like seismic design for any other frame 
system. Column stability does need to account for the flexibility of the connection, but the AISC LRFD 
and the Provisions approaches considering second order moments from the translation of gravity loads are 
essentially the same. The further checks on details, such as the strong column rule, are also generally 
familiar. Given the nature of the connection, it is also a good idea to examine behavior at service loads, 
but there are not truly standard criteria for this. 
8.3 DESIGN REQUIREMENTS 
8.3.1 Provisions Parameters 
The basic parameters affecting the design and detailing of the buildings are shown in Table 8.1 below.
Chapter 8, Composite Steel and Concrete 
8-7 
Table 8-1 Design Parameters 
Parameter Value 
Ss (Map 1) 0.20 
S1 (Map 2) 0.06 
Site Class E 
Fa 2.5 
Fv 3.5 
SMS = FaSs 0.50 
SM1 = FvS1 0.21 
SDS = 2/3SMS 0.33 
SD1 = 2/3SM1 0.14 
Seismic Design Category C 
Frame Type per 
Provisions Table 5.2.2 
Composite Partially Restrained 
Moment Frame 
R 6 
O0
3 
Cd 5.5 
[The 2003 Provisions have adopted the 2002 USGS probabilistic seismic hazard maps, and the maps have 
been added to the body of the 2003 Provisions as figures in Chapter 3 (instead of the previously used 
separate map package).] 
The frames are designed in accordance with AISC Seismic, Part II, Sec. 8 (Provisions Table 5.2.2). AISC 
SDGS-8 and ASCE TC describe this particular system in detail. Given the need to determine the 
flexibility of the connections, it would be difficult to design such structures without reference to at least 
one of these two documents. 
8.3.2 Structural Design Considerations Per the Provisions 
The building is regular both in plan and elevation. Provisions Table 5.2.5.1 indicates that use of the 
Equivalent Lateral Force procedure in accordance with Provisions Sec. 5.4 is permitted. 
Nonstructural elements (Provisions Chapter 14) are not considered in this example. 
Diaphragms must be designed for the required forces (Provisions Sec. 5.2.6.2.6), however this is not 
unique to this system and therefore is not explained in this example. 
The story drift limit (Provisions Table 5.2.8) is 0.025 times the story height. Although the Cd factor is 
large, 5.5, the seismic forces are low enough that conventional stiffness rules for wind design actually 
control the stiffness. 
Orthogonal effects need not be considered for Seismic Design Category C, provided the structure does 
not have a plan structural irregularity (Provisions Sec. 5.2.5.2.2). 
8.3.3 Building Weight and Base Shear Summary 
The unit weights are as follows:
FEMA 451, NEHRP Recommended Provisions: Design Examples 
8-8 
Non-composite dead load: 
4.5 in. slab on 0.6 in. form deck, plus sag 58 psf 
Joist and beam framing 6 psf 
Columns 2 psf 
66 psf 
Composite dead load: 
Fire insulation 4 psf 
Mechanical and electrical 6 psf 
Ceiling 2 psf 
Partitions 20 psf 
32 psf 
Exterior wall: 
Precast concrete panels: 0.80 klf 
Records storage on 3 bays per floor 120 psf 
(50 percent is used for seismic weight; minimum per the Provisions is 25 percent) 
The building weight, W, is found to be 8,080 kips. The treatment of the dead loads for analysis is 
described in more detail subsequently. 
The Seismic Response Coefficient, Cs, is equal to 0.021: 
0.14 0.021 
1.12 6
1 
D1 
s 
C TS R
I 
= = = 
. . 
. . 
. . 
The methods used to determine W and Cs are similar to those used elsewhere in this volume of design 
examples. The building is somewhat heavy and flexible. The computed periods of vibration in the first 
modes are 2.12 and 2.01 seconds in the north-south and east-west directions, respectively. These are 
much higher than the customary 0.1 second per story rule of thumb, but low-rise frames with small 
seismic force demands typically do have periods substantially in excess of the rule of thumb. The 
approximate period per the Provisions is 0.66 seconds, and the upper bound for this level of ground 
motion is 1.12 seconds. 
The total seismic force or base shear is then calculated as follows: 
V = CsW = (0.021)(8,080) = 170 kips (Provisions Eq. 5.3.2) 
The distribution of the base shear to each floor (again, by methods similar to those used elsewhere in this 
volume of design examples) is found to be: 
Roof (Level 4): 70 kips 
Story 4 (Level 3): 57 kips 
Story 3 (Level 2): 34 kips 
Story 2 (Level 1): 8 kips 
Story 1 (Level 0): 0 kips 
S: 169 kips (difference is rounding; total is 170) 
Without illustrating the techniques, the gross service level wind force following ASCE 7 is 123 kips. 
When including the directionality effect and the strength load factor, the design wind force is somewhat 
less than the design seismic base shear. The wind force is not distributed in the same fashion as the
Chapter 8, Composite Steel and Concrete 
8-9 
seismic force, thus the story shears and the overturning moments for wind are considerably less than for 
seismic. 
8.4 DETAILS OF THE PRC CONNECTION AND SYSTEM 
8.4.1 Connection M-. Relationships 
The composite connections must resist both a negative moment and a positive moment. The negative 
moment connection has the slab rebar in tension and the leg of the seat angle in compression. The 
positive moment connection has the slab concrete in compression (at least the a dimension down from 
the top of the slab) and the seat angle in tension (which results in flexing of the seat angle vertical leg). 
At larger rotations the web angles contribute a tension force that increases the resistance for both negative 
and positive bending. 
Each of these conditions has a moment-rotation relationship available in AISC SDGS-8 and ASCE TC. 
(Unfortunately there are typographical errors in ASCE TC: A + should be replaced by = and the 
symbol for the area of the seat angle is used where the symbol should be that for the area of the web 
angle.) An M-. curve can be developed from these equations: 
Negative moment connection: 
2 (AISC SDGS-8, Eq. 1) 
1(1 C ) 3 
Mn-=C -e- . +C. 
where: 
C1 = 0.18(4  AsFyrb + 0.857ALFy)(d + Y3) 
C2 = 0.775 
C3 = 0.007(AL + AwL)Fy (d + Y3) 
. = girder end rotation, milliradians (radians/1000) 
d = girder depth, in. 
Y3 = distance from top flange of the girder to the centroid of the reinforcement, in. 
As = steel reinforcing area, in.2 
AL = area of seat angle leg, in.2 
AwL = gross area of double web angles for shear calculations, in.2 (For use in these equations AwL is 
limited to 150 percent of AL). 
Fyrb = yield stress of reinforcing, ksi 
Fy = yield stress of seat and web angles, ksi 
Positive moment connection: 
2 (AISC SDGS-8, Eq. 2) 
1(1 C ) (3 4) 
Mn+=C -e- . + C +C . 
where: 
C1 = 0.2400[(0.48AwL ) + AL](d + Y3)Fy 
C2 = 0.0210(d + Y3/2) 
C3 = 0.0100(AwL + AL )(d + Y3)Fy 
C4 = 0.0065 AwL (d + Y3)Fy 
From these equations, curves for M-. can be developed for a particular connection. Figures 8-5 and 8-6 
are M-. curves for the connections associated with the W18x35 girder and the W21x44 spandrel beam
FEMA 451, NEHRP Recommended Provisions: Design Examples 
8-10 
-300 
-200 
-100
0 
100 
200 
300
-25 -20 -15 -10 -5 0 5 10 15 20 25 
Rotation, milliradians 
Moment, ft-kip 
Positive M 
Pos Bilinear 
Neg Bilinear 
Negative M 
Figure 8-5 M-. Curve for W18x35 connection with 6-#5 (1.0 ft-kip = 1.36 kN-m) 
respectively, which are used in this example. The selection of the reinforcing steel, connection angles, 
and bolts are described in the subsequent section, as are the bilinear approximations shown in the figures. 
Among the important features of the connections demonstrated by these curves are: 
1. The substantial ductility in both negative and positive bending, 
2. The differing stiffnesses for negative and positive bending, and 
3. The substantial post-yield stiffness for both negative and positive bending. 
It should be recognized that these curves, and the equations from which they were plotted, do not 
reproduce the line from a single test. They are averages fit to real test data by numerical methods. They 
smear out the slip of bolts into bearing. (There are several articles in the AISC Engineering Journal that 
describe actual test results. They are in Vol. 24, No.2; Vol. 24, No.4; Vol. 27, No.1; Vol. 27, No. 2; and 
Vol 31, No. 2. The typical tests clearly demonstrate the ability of the connection to meet the rotation 
capabilities of AISC Seismic, Section 8.4 - inelastic rotation of 0.015 radians and total rotation capacity 
of 0.030 radians.) 
[Based on the modifications to AISC Seismic, Part II, Sec. 8.4 in 2003 Provisions Sec. 10.5.16, the 
required rotation capabilities are inelastic rotation of 0.025 radians and total rotation of 0.040 radians.]
Chapter 8, Composite Steel and Concrete 
8-11 
-500 
-400 
-300 
-200 
-100
0 
100 
200 
300 
400 
500
-25 -20 -15 -10 -5 0 5 10 15 20 25 
Rotation, milliradians 
Moment, ft-kip 
Positive M 
Pos Bilinear 
Neg Bilinear 
Negative M 
Figure 8-6 M-. Curve for W21x44 connection with 8-#5 (1.0 ft-kip = 1.36 kN-m). 
8.4.2 Connection Design and Connection Stiffness Analysis 
Table 8-2 is taken from a spreadsheet used to compute various elements of the connections for this design 
example. It shows the typical W18x35 girder and the W21x44 spandrel beam with the connections used 
in the final analysis, as well as a W18x35 spandrel beam for the short exterior spans, where a W21x44 
was used in the end. Each major step in the table is described in a line-by-line description following the 
table. [Based on the modifications to AISC Seismic, Part II, Sec. in 2003 Provisions Sec. 10.5.16, the 
nominal strength of the connection must be exceed RyMp for the bare steel beam, where Ry is the ratio of 
expected yield strength to nominal yield strength per AISC Seismic, Part I, Table I-6-1.]
FEMA 451, NEHRP Recommended Provisions: Design Examples 
8-12 
Table 8-2 Partially Restrained Composite Connection Design 
Line Girder Spandrels 
Basic Data 
2 Beam size W18x35 W21x44 W18x35 
3 Span, ft 25 25 12.5 
4 Area of beam, in.2 10.3 13 10.3 
5 I, of beam alone, in.4 510 843 510 
6 Z, plastic modulus of beam, in.3 66.5 95.4 66.5 
7 Beam depth, in. 17.7 20.7 17.7 
8 Slab thickness, in. 7.0 7.0 7.0 
9 Y3 to rebar, in. 5.5 5.5 5.5 
10 Column W10x77 W10x88 W10x77 
11 Flange width, in. 10.2 10.3 10.2 
12 Flange thickness, in. 0.87 0.99 0.87 
13 Flange fillet, k1, in. 0.88 0.94 0.88 
Basic Negative Moment Capacity 
15 Reinforcing bars 6-#5 8-#5 6-#5 
16 As, rebar area, in.2 1.86 2.48 1.86 
17 Tr, rebar tension, kips 111.6 148.8 111.6 
18 Mn
- 
, nominal negative moment, ft-kips 215.8 324.9 215.8 
19 % Mp (Mn
-/beam Mp) 78% 82% 78% 
20 Check: > 50%? (75% per ASCE TC) OK OK OK 
Seat Demands for Negative Moment 
22 Seat angle L7x4x1/2x8 L7x4x5/8x8.5 L7x4x1/2x8 
23 Seat Fy, ksi 36 36 36 
24 Seat thickness, in. 0.5 0.625 0.5 
25 Seat length, in. 8.0 8.5 8.0 
26 Leg area, in.2 4.0 5.3125 4.0 
27 Minimum area = 1.25 Tr /Fy, in.2 3.875 5.167 3.875 
28 Check OK OK OK 
29 Leg yield force, kips 144 191.25 144 
30 Bolts to beam (4) 1"-325X (4) 11/8"-490X (4) 1"-325X 
31 Diameter, in. 1.0 0.875 1.0 
32 Bolt design shear capacity, kips (f = 0.75) 141.2 223.6 141.2 
33 Check Close enough OK Close enough 
Nominal Positive Moment Capacity 
35 Seat k, fillet length, in. 1.000 1.125 1.000 
36 Mp, vertical leg, in.-kips 18.0 29.9 18.0 
37 b' (see Figure 8-7), in. 1.00 0.81 1.00 
38 Seat tension from bending, kips 31.5 63.8 31.5 
39 Seat tension from shear, kips 86.4 114.75 86.4 
40 Tension to bottom flange, kips 31.5 63.8 31.5 
41 Nominal Positive Moment, Mn
+, ft-kips 67.4 149.9 67.4 
42 Percent of Beam Mp 24% 38% 24% 
Demand on Tension Bolts at Nominal Capacity 
44 a' (see Figure 8-7), in. 2.0 2.1 2.0 
45 Q (prying), kips 6.8 10.7 6.8 
46 Bolt tension, kips 38.3 74.5 38.3 
47 Bolts to column (2) 1"-325X (2) 11/8"-490X (2) 1"-325X 
48 Bolt design tension, kips (f = 0.75) 106 168.4 106 
49 Check OK OK OK
Chapter 8, Composite Steel and Concrete 
Line Girder Spandrels 
8-13 
Compute Total Joint Moment to Column based on Nominal Capacities 
51 Connection nominal Mn
- + Mn
+, ft-kips 283 475 283 
52 Minimum column Mp (125% of sum) 177 297 177 
53 Average as percentage of beam 51% 60% 51% 
54 Check OK OK OK 
Concrete Compression Transfer to Column 
56 Rebar Ty + bottom seat Ty, kips 143.10 212.62 143.10 
57 0.85 f'c on two flanges, kips 364.14 367.71 364.14 
58 Projection for flange Mp, in. 2.72 3.10 2.72 
59 Force from flange Mp, kips 225.92 254.88 225.92 
60 Ratio, demand / minimum capacity 0.63 0.83 0.63 
Web Shear Connection (needed for effective stiffness) 
62 Seismic shear demand, kips 11.5 19.9 23.1 
63 Web angles L4x4x1/4x8.5 L4x4x1/4x11.5 L4x4x1/4x8.5 
64 Aw, area of two legs, in.2 4.25 5.75 4.25 
65 Aw, limit based on area of rebar, in.2 2.79 3.72 2.79 
66 Aw, used in M-. calculation, in.2 2.79 3.72 2.79 
Moment Rotation Values for Analysis of Effective Stiffness 
68 Mneg at service level (0.0025 rad), ft-kip -178.0 -267.8 -178.0 
69 Mneg at maximum capacity(0.020 rad), ft-kip -264.5 -397.7 -264.5 
70 Secant stiffness for Mneg at 0.0025 radian 71.2 107.1 71.2 
71 Mpos at service level (0.0025 rad), ft-kip 73.7 117.3 73.7 
72 Mpos at maximum capacity(0.020 rad), ft-kip 208.9 313.9 208.9 
73 Secant stiffness for Mpos at 0.0025 radian 29.5 46.9 29.5 
74 Rotation at nominal Mneg 3.03 3.03 3.03 
75 Rotation at nominal Mpos 2.29 3.70 2.29 
Beam Moments of Inertia 
77 Full composite action force, beam AFy, kips 515.0 650.0 515.0 
78 Y2, to plastic centroid in concrete, in. 5.65 5.30 4.31 
79 Composite beam inertia for pos. bending, in.4 1,593 2,435 1,402 
80 Centroid of all steel for negative bending, in. 6.66 7.81 6.66 
81 Composite beam inertia for neg. bending, in.4 834 1366 834 
82 Equivalent beam for positive and negative, in.4 1,290 2,008 1,175 
83 Weighted connection stiffness, ft-kips/radian 61,263 88,105 61,263 
84 Eff. prismatic inertia, beam and PRCC, in.4 639 955 412 
85 Ratio of eff. prismatic I / I of beam alone 1.25 1.13 0.81 
Check Bottom Bolt Tension at Maximum Deformation 
87 Rotation at f  (rotation at nominal M pos)  Cd 10.7 14.9 10.7 
88 Moment at f  (rot. at nom. M pos)  Cd, ft-kips 152.3 268.2 152.3 
89 Tension demand, kips 80.5 125.1 80.5 
90 Nominal capacity of bolts, kips 141.3 224.5 141.3 
Check Positive Moment Capacity as a Percentage of Beam Mp (50% criterion) 
92 M pos (at 0.020 radians) / Mp beam 75% 79% 75% 
Detailed explanation of the computations in Table 8-2: 
Step 1: Establish nominal negative moment capacity: (This is a step created in this design example; is 
not actually an explicit step in the procedures recommended in the references. It appears to be necessary 
to satisfy the basic Provisions strength requirement. See Provisions Sec. 5.2.1, Sec. 5.2.7, and ASCE 7 
Sec. 2.3.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
8-14 
Lines 15-18: Mn is taken as a simple couple of rebar in slab and force at connection of bottom flange of 
beam; the true maximum moment is larger due to strain hardening in rebar and the bottom connection and 
due to tension force in the web connection, so long as the bottom connection can handle the additional 
demand. The nominal capacity is plotted in Figures 8-5 and 8-6 as the break of the bilinear relation. The 
design capacity, using a resistance factor of 0.85, has two requirements: 
1. f Mn exceeds demand from seismic load combination: basic Provisions requirement 
2. f Mn exceeds demand from total service gravity loads - simply a good idea to maintain reasonable 
initial stiffness for lateral loads; by codes the factored gravity demand can be checked using plastic 
analysis 
Lines 19-20: Mn exceeds 50 percent (by AISC Seismic, Part II, 8.4) of Mp of the bare steel beam. In this 
example, the more stringent recommendation of 75 percent contained within the ASCE TC is followed. 
Note that this check is on nominal strength, not design strength. A larger Mn gives a larger stiffness, thus 
some drift problems can be addressed by increasing connection capacity. 
Step 2: Design bottom seat angle connection for negative moment: 
Lines 22-28: Provide nominal yield of angle leg at least 125 percent of nominal yield of reinforcing steel. 
This allows for increased force due to web shear connection. Strain hardening in the rebar is a factor, but 
strain hardening the angle would probably be as large. AISC SDGS-8 recommends 120 percent. ASCE 
TC recommends 133 percent, but then uses 125 percent to check the bolts. This is a check in 
compression, and the authors elected to use 125 percent. 
Lines 29-33: Provide high strength bolts in normal (not oversized) holes to transfer force between beam 
flange and angle by shear; conventional rules regarding threads in the shear plane apply. The references 
do apply a resistance factor to the bolts, which may be an inconsistent design methodology. A check 
based on overstrength might be more consistent. The capacity at bolt slip could be compared against 
service loads, which would be a good idea for designs subject to strong wind forces. 
Step 3: Establish nominal positive moment capacity: This connection is less stiff and less linear for 
positive moment than for negative moment, and generally weaker. There is not a simple, clear 
mechanism for a nominal positive moment. The authors of this example suggest the following procedure 
which follows the normal methods of structural engineering and yields a point relevant to the results of 
connection tests, in so far as construction of a bilinear approximation is concerned. It significantly 
underestimates the ultimate capacity. 
Lines 35-38: Compute the shear in the vertical leg associated with bending. Figure 8-7 shows the 
mechanics, which is based on methods in the AISC Manual, for computing prying in hanger-type 
connections. Compute the nominal plastic moment of the angle leg bending out of plane (line 36) and 
assume that the location of the maximum moments are at the end of the fillet on the vertical leg (line 35) 
and at the edge of the bolt shaft (line 37). The moment near the bolt is reduced for the material lost at the 
bolt hole. 
Lines 39-40: Check the shear capacity, compare with the shear governed by moment, and use the smaller. 
Shear will control if the angle is thick. 
Line 41: Compute the nominal positive moment as a couple with the force and the distance from the 
bottom of the beam to the center of the compression area of the slab on the column. The concrete 
compression area uses the idealized Whitney stress block (ACI 318). Note that the capacity to transfer
Chapter 8, Composite Steel and Concrete 
8-15 
a' b' k 
a b 
T
Q 
T + Q 
Mp 
M p 
Figure 8-7 Analysis of seat angle for tension. 
concrete compression force to the steel column flange is checked later. The nominal positive moment is 
also shown on Figures 8-5 and 8-6 at the break point in the bilinear relation. 
Step 4: Design the bolts to transfer positive moment tension to the column: 
Lines 44-45: Compute the prying force following AISCs recommended method. The moment in the 
vertical leg is computed as described above, and the moment arm extends from the edge of the bolt shaft 
(closest to the beam) to the bottom edge of the angle. Refer to Figure 8-7. 
Lines 46-48: Add the basic tension to the prying force and compare to the factored design capacity of the 
bolts. Note that the resistance factor is used here to be consistent with step 2. It is common to use the 
same size and grade of bolt as used for the connection to the beam flange, which generally means that 
these bolts have excess capacity. Also, for seismic design, another check at maximum positive moment is 
recommended (see step 9). 
Step 5: Compute the flexural demand on the column: AISC Seismic, Part II, 7 and 8, require that the 
flexural resistance of the column be greater than the demand from the connections, but it does not give 
any particular margin. ASCE TC recommends a ratio of 1.25. 
Lines 51-52: The minimum nominal flexural strength of the column, summed above and below as well as 
adjusted for the presence of axial load, is set to be 125 percent of the demand from the sum of the nominal 
strengths of the connections. 
Lines 53-54: AISC Seismic, Part II, 8.4 requires that the connection capacity exceed 50 percent of the 
plastic moment capacity of the beam. In this example, the negative moment connections are designed for 
75 percent of the beam plastic moment, and this check shows that the average of negative and positive 
nominal moment capacities for the connection exceeds 50 percent of the plastic moment for the beam. A 
later check (step 10) will compare the maximum positive moment resistance to the 50 percent rule. 
Step 6: Check the transfer of force from concrete slab to steel column: The tension in the reinforcing 
steel and the compression couple from positive bending must both transfer. Both flanges provide
FEMA 451, NEHRP Recommended Provisions: Design Examples 
8-16 
resistance if concrete fills the space between the flanges, but full capacity of the second flange has 
probably not been exercised in tests. 
Line 56: Add the yield force of the reinforcement and the tension yield force of the seat angle, both 
previously computed. 
Line 57: Compute an upper bound concrete compression capacity as 0.85f'c times the area of concrete 
bearing on both flanges. 
Lines 58-59: Compute the force that would yield the steel column flanges over the thickness of the slab 
by computing the projection beyond the web fillet that would yield at a load of 0.85fc. This ignores the 
capacity of the flange beyond the slab thickness and is obviously conservative. 
Line 60: Compare the demand with the smaller of the two capacities just computed. 
Step 7: Select the web connection: 
Line 62: The seismic shear is computed by assuming beam end moments equal to the nominal capacity of 
the connections, one in negative moment and one in positive. 
Line 63: The gravity demand must be added, and straight gravity demand must also be checked before 
selecting the actual connection. 
Lines 64-66: The web connection influences the overall stiffness and strength of the connection, 
especially at large rotation angles. The moment-rotation expression include the area of steel in the web 
angles, but also places a limitation based upon 150 percent of the area of the leg of the seat angle for use 
in the computation. 
Step 8: Determine the effective stiffness of the beam and connection system: Determining the equivalent 
stiffness for a prismatic beam involves several considerations. Figure 8-8 shows how the moment along 
the beam varies for gravity and lateral loads as well as composite and non-composite conditions. The 
moment of inertia for the composite beam varies with the sense of the bending moment. The end 
connections can be modeled as regions with their own moments of inertia, as illustrated in the figure. 
Figure 8-9 shows the effective cross section for each of the four stiffnesses: positive and negative bending 
of the composite beam and positive and negative bending of the composite connection. Given a linear 
approximation of each connection stiffness expressed as moment per radian, flexural mechanics leads to a 
simple expression for a moment of inertia of an equivalent prismatic beam. 
Lines 68-73: Compute the negative and positive moments at a rotation of 2.5 milliradians, which is the 
rotation angle that defines the effective stiffness for lateral analysis (per both AISC SDGS-8 and ASCE 
TC). 
Lines 74-75: Using those moments, compute the rotations corresponding to the nominal strength, positive 
and negative. (This is useful when idealizing the behavior as bilinear, which is plotted in Figures 8-5 and 
8-6.) 
Lines 77-79: Compute the moment of inertia of the composite beam in positive bending. Note that the 
system is designed for full composite action, per the recommendations in AISC SDGS-8 and ASCE TC, 
using the criteria in the AISC manual. The positive bending moment of inertia here is computed using 
AISCs lower bound method, which uses an area of steel in the flange adequate to replace the Whitney 
stress block in the concrete flange. This moment of inertia is less than if one used the full concrete area in 
Figure 8-9.
Chapter 8, Composite Steel and Concrete 
8-17 
Lines 80-81: Compute the moment of inertia of the composite beam in negative bending. 
Line 82: Compute an equivalent moment of inertia for the beam recognizing that a portion of the span is 
in positive bending and the remainder is in negative bending. Following the recommendations in AISC 
SDGS-8 and ASCE TC, this is computed as 60 percent of I pos and 40 percent of I neg. 
Lines 83-84: Compute the moment of inertia of a prismatic beam that will give the same total end 
rotation in a sway condition as the actual system. Gravity loads place both connections in negative 
moment, so one will be subject to increasing negative moment while the other will be subject to 
decreasing negative moment. Thus, initially, the negative moment stiffness is the appropriate stiffness, 
which is what is recommended in the AISC SDGS-8 and ASCE TC. For this example the positive and 
negative stiffnesses are combined, weighted by the nominal strengths in positive and negative bending, to 
yield a connection stiffness that is appropriate for analysis up to the nominal strengths defined earlier. 
Defining this weighted stiffness as Kconn and the equivalent composite beam moment of inertia as Icomp, the 
effective moment of inertia is found by: 
I 
I 
EI 
LK 
effective 
comp
comp 
conn 
= 
1+ 
6
FEMA 451, NEHRP Recommended Provisions: Design Examples 
8-18 
(a) 
Typical beam 
(b) 
Moment 
diagram 
(c) 
Combined 
composite 
moment 
(d) 
Variable 
moment 
of inertia 
M
M + M 
Seismic 
LL 
MCombined
MDNC 
I2 I1 
~ ~ 
~ ~ 
~
~ 
+ 
- 
I2 I1 
Klat 4 Klat 3 
DC 
Figure 8-8 Moment diagram for typical beam. 
Line 85: compute the ratio of the moment of inertia of the effective prismatic beam to that for the bare 
steel beam. When using standard computer programs for analysis that have a library of properties of steel 
cross sections, this ratio is a convenient way to adjust the modulus of elasticity and thus easily compute 
the lateral drift of a frame. This adjustment could invalidate routines in programs that automatically 
check various design criteria that depend on the modulus of elasticity. 
Step 9: Check the tension bolts at maximum rotation 
Line 87: Compute the rotation at total drift as Cd times the drift at the design positive moment. 
Line 88: Compute the positive moment corresponding to that drift. 
Line 89: Compute the tension force at the bottom seat angle, ignoring any contribution of the web angles, 
from the moment and a moment arm between the center of the slab thickness and the inflection point in 
the vertical leg of the seat angle, then add the prying force already calculated for a maximum demand on 
the tension bolts. 
Line 90: Compare with the nominal capacity of the bolts (set f = 1.0) 
Step 10: Check the maximum positive moment capacity:
Chapter 8, Composite Steel and Concrete 
8-19 
7" 
3
4" 
clr cover 
6" 
1" clear 
(6) #5 top bars 
in slab 
W18 
Headed studs 
on beam 
(2) rows 3
4"  H.A.S. spaced at 11" o.c. 
4x4x1
4" w/ 
(3) 3
4" A325 bolts 
Bottom flange angle 
with (4) 1" A325X bolts 
to beam flange and 
(2) 1" A325 bolts to 
column flange 
#4x6'-0" @ 10" o.c. transverse. 
Alternate above and below #5's. 
Figure 8-10 Elevation of typical connection (1.0 ft = 0.3048 m, 1.0 in. = 25.4 mm). 
Line 92: The positive moment at 20 milliradians, already calculated, is compared to the plastic moment 
capacity of the steel beam. This is the point at which the 50 percent requirement of AISC Seismic, Part 
II, 8.4 is checked. 
Figure 8-10 shows many of the details of the connection for the W18x35. The headed studs shown 
develop full composite action of the beam between the end and midspan. They do not develop full 
composite action between the column and the inflection point, but it may be easily demonstrated that they 
are more than capable of developing the full force in the reinforcing steel within that distance. The 
transverse reinforcement is an important element of the design, which will be discussed subsequently. 
Alternating the position above and below is simply a preference of the authors. 
8.5 ANALYSIS 
8.5.1 Load Combinations 
A 3D model using Risa 3D was developed. Non-composite dead loads (steel beams, bar joists, form 
deck, and concrete) were input as concentrated loads at the columns on each level rather than uniformly 
distributed to the beams. This was because we want the model for the seismic load combinations to 
address the moments in the PRC connections. The loads subject to composite action are the composite 
dead loads, live loads, and seismic loads, not the non-composite dead loads. But the non-composite dead 
loads still contribute to mass, are subject to ground acceleration, and as such contribute to seismic loads. 
This gets confusing; so a detailed look at the load combinations is appropriate. 
Let us consider four load cases (illustrated in Figures 8-11 and 8-12): 
1. Dc - Composite dead load, which is uniformly distributed and applied to beams (based on 32 psf) 
2. Dnc - Non-composite dead load, which is applied to the columns (based on 66 psf) 
3. L - Composite live load, which is uniformly distributed to beams, using live load reductions
FEMA 451, NEHRP Recommended Provisions: Design Examples 
8-20 
1.2 D nc 1.2 Dc 0.5 L 
1.0 QE 0.067 D nc 0.067 D c 
Figure 8-11 Illustration of input for load combination for 1.2D + 0.5L + 1.0QE + 0.2SDSD. 
4. E - Earthquake load, which is applied laterally to each level of the building and has a vertical 
component applied as a uniformly distributed load to the composite beams 
We will investigate two load combinations. Recall that composite loads are applied to beams, while noncomposite 
loads are applied to columns. But there is an exception: the 0.2SDSD component, which 
represents vertical acceleration from the earthquake is applied to all the dead load on the beams whether it 
is composite or non-composite. This is because even non-composite dead load contributes to mass, and is 
subject to the ground acceleration. Because the non-composite dead load is not distributed on the beams 
in the computer model, an adjustment to the load factor is necessary. The assignment of loads gets a 
little complicated, so pay careful attention: 
Combination 1 = 1.2D + 0.5L + 1.0E 
= 1.2Dnc + 1.2Dc + 0.5L + QE + 0.2SDSD 
= 1.2Dnc + 1.2Dc + 0.5L + QE + 0.067 (Dnc + Dc) 
= 1.2Dnc + 1.2Dc + 0.5L + QE + 0.067Dnc(Dc/Dc) + 0.067Dc 
= 1.2Dnc + 1.2Dc + 0.5L + QE + 0.067Dc(Dnc/Dc) + 0.067Dc 
= 1.2Dnc + 1.2Dc + 0.5L + QE + 0.067Dc(66 psf/32 psf) + 0.067Dc 
= 1.2Dnc + 1.2Dc + 0.5L + QE + 0.138Dc + 0.067Dc 
= 1.2Dnc + 1.405Dc + 0.5L + QE 
QE will be applied in both the north-south and the east-west directions, so this really represents two load 
combinations. 
Dnc = non-composite dead load. 
Dc = composite dead load 
L = live load 
QE = horizontal seismic load 
Now consider at the second load combination: 
Combination 2 = 0.9D + 1.0E 
= 0.9Dnc + 0.9Dc + QE - 0.2SDSD 
= 0.9Dnc + 0.9Dc + QE - 0.067 (Dnc + Dc) 
= 0.9Dnc + 0.9Dc + QE - 0.067 Dnc(Dc/Dc) - 0.067Dc 
= 0.9Dnc + 0.9Dc + QE - 0.067 Dc(Dnc/Dc) - 0.067Dc 
= 0.9Dnc + 0.9Dc + QE - 0.067 Dc(66 psf/32 psf) - 0.067Dc
Chapter 8, Composite Steel and Concrete 
8-21 
0.9 D nc 0.9 Dc 1.0 Q 
0.067 D nc 0.067 D c 
E 
Figure 8-12 Illustration of input for load combination for 0.9D + 1.0QE - 0.2SDSD. 
= 0.9Dnc + 0.9Dc + QE - 0.138Dc - 0.067Dc 
= 0.9Dnc + 0.9Dc + QE - 0.205Dc 
= 0.9Dnc + 0.695Dc + QE 
Again, QE will be applied in both the north-south and the east-west directions, so this represents another 
two load combinations. 
Dnc = non-composite dead load. 
Dc = composite dead load 
L = live load 
QE = horizontal seismic load 
8.5.2 Drift and P-delta 
As defined by the Provisions, torsional irregularity is considered to exist when the maximum 
displacement computed including accidental torsion at one end of the structure transverse to an axis is 
more than 1.2 times the average of the displacements at the two ends of the structure (Provisions Sec. 
5.4.4.3). For this building the maximum displacement at the roof including accidental torsion, is 1.65 in. 
The displacement at the other end of the building in this direction is 1.43 in. The average is 1.54 in. 
Because 1.65 in. < 1.85 in. = (1.2)(1.54 in.), the structure is not torsionally irregular. Consequently, it is 
not necessary to amplify the accidental torsion nor to check the story drift at the corners. A simple check 
at the center of the building suffices. [In the 2003 Provisions, the maximum limit on the stability 
coefficient has been replaced by a requirement that the stability coefficient is permitted to exceed 0.10 if 
and only if the resistance to lateral forces is determined to increase in a monotonic nonlinear static 
(pushover) analysis to the target displacement as determined in Sec. A5.2.3. P-delta effects shall be 
included in the analysis. Therefore, in this example, the stability coefficient should be evaluated directly 
using 2003 Provisions Eq. 5.2.-16.] 
The elastic story drifts were computed by the RISA 3D analysis for the required load combinations. Like 
most modern computer programs for structural analysis, a P-delta amplification can be automatically 
computed, but to illustrate the effect of P-delta in this structure and to check the limit on the stability 
index, two computer runs have been performed, one without the P-delta amplifier and one with it. The 
allowable story drift is taken from Provisions Table 5.2.8. The allowable story drift is 0.025 hsx = 
(0.025)(13 ft  12 in./ft) = 3.9 in. With a Cd of 5.5, this corresponds to a drift 0.71 in. under the 
equivalent elastic forces. At this point design for wind does influence the structure. A drift limit of h/400 
(= 0.39 in.) was imposed, by office practice, to the service level wind load. In order to achieve the 
desired stiffness, the seismic story drift at elastic forces is determined thus:
FEMA 451, NEHRP Recommended Provisions: Design Examples 
8-22 
Elastic story drift limit = (wind drift limit)(total seismic force)/(service level wind force) 
Elastic story drift limit = (0.39 in.)(170 kip)/123 kip = 0.54 in. 
The structure complies with the story drift requirements, but it was necessary to increase the size of the 
spandrel beams from the preliminary W18x35 to W21x44 to meet the desired wind stiffness. This is 
summarized in Table 8-3. The structure also complies with the maximum limit on the stability index 
(Provisions 5.4.6.2-2): 
. 
max 
. .
. . 
= = . . 
* 
= = 
05 05 
05 55 
018 025 
Cd 
 is the ratio of demand to capacity for the story shear, and has not yet been computed. Maximum 
demand and design capacity are tabulated in the following section; the average is about two-thirds. The 
preceding data show that the maximum resistance is higher, especially for positive moment, than the 
value suggested here for design capacity. The average ratio of demand to maximum capacity with a 
resistance factor is well below 0.5, so that value is arbitrarily used to show that the actual stability index is 
within the limits of the Provisions. 
Table 8-3 Story Drift (in.) and P-delta Analysis 
Story 
North-south (X direction) East-west (Z direction) 
without 
P-delta 
with 
P-delta 
P-delta 
amplifier 
Stability 
index 
without 
P-delta 
with 
P-delta 
P-delta 
amplifier 
Stability 
index 
1 0.358 0.422 1.179 0.152 0.312 0.360 1.154 0.133 
2 0.443 0.517 1.167 0.143 0.410 0.471 1.149 0.130 
3 0.449 0.513 1.143 0.125 0.402 0.453 1.127 0.113 
4 0.278 0.304 1.094 0.086 0.239 0.259 1.084 0.077 
8.5.3 Required and Provided Strengths 
The maximum beam end moments from the frame analysis for the seismic load combinations are as 
follows: 
Table 8-4 Maximum Connection Moments and Capacities (ft-kips) 
Quantity 
W18 Girders W21 Spandrels 
Negative Positive Negative Positive 
Demand (level 2), Mu 
Nominal, Mn 
Design capacity, fMn 
143 
216 
184 
36.6 
67.2 
57.1 
118 
325 
276 
103 
149 
127 
The capacities, using a resistance factor of 0.85, are well in excess of the demands. The girder member 
sizes are controlled by gravity load in the construction condition. All other member and connection 
capacities are controlled by the design for drift. The negative moment demands are somewhat larger than 
would result from a more careful analysis, because the use of a prismatic member overestimates the end 
moments due to distributed load (composite gravity load) along the member. The higher stiffness of the 
portion of the beam in positive bending with respect to the connections would result in higher positive 
moments at midspan and lower negative moments at the supports. This conservatism has no real effect on 
this design example. (The above demands and capacities do not include the girders supporting the storage
Chapter 8, Composite Steel and Concrete 
8-23 
bays, which are required to be W18x40 for the gravity load condition. The overall analysis does not take 
that larger member into account.) 
Snow load is not included in the seismic load combinations. (According to the Provisions, snow load 
equal to or less than 30 psf does not have to be included in the mass.) Further, as a designers judgment 
call, it was considered that the moments from 0.2S (= 6 psf) were so small, considering that the roof was 
designed with the same connections as the floors, that it would make no significant difference in the 
design analysis. 
The maximum column forces are shown in Table 8-5; the particular column does support the storage load. 
The effective length of the columns about their weak axis will be taken as 1.0, because they are braced by 
perpendicular frames acting on the strong axes of the columns, and the P-delta analysis captures the 
secondary moments due to the leaning column effect. The effective length about their strong axis will 
exceed 1.0. The ratio of column stiffness to beam stiffness will use the same effective beam stiffness 
computed for the drift analysis, thus for the W10x77 framed into the W18x35 beams: 
Icol / Lcol = 455/(13  12) = 2.92 
Ibeam / Lbeam = 1.25  510 / (25  12) = 2.12 
and the ratio of stiffnesses, G = 2.92 / 2.12 = 1.37 
Although the column in the lowest story has greater restraint at the foundation, and thus a lower K factor, 
it is illustrative to determine K for a column with the same restraint at the top and bottom. From the 
nomographs in the AISC Manual or from equivalent equations, K = 1.45. It turns out that the effective 
slenderness about the strong axis is less than that for the weak axis, so the K factor does not really control 
this design. 
Table 8-5 Column Strength Check, for W10x77 
Seismic Load Combination Gravity Load Combination 
Axial force, Pu 
Moment, Mu 
Interaction equation 
391 kip 
76.3 ft-kip 
0.72 
557 kip 
52.5 ft-kip 
0.89 
8.6 DETAILS OF THE DESIGN 
8.6.1 Overview 
The requirements in AISC Seismic for C-PRMF systems are brief. Some of the requirements are 
references to Part I of AISC Seismic for the purely steel components of the system. A few of those detail 
checks are illustrated here. For this example, more attention is paid to the details of the joint. 
8.6.2 Width-Thickness Ratios 
The width-thickness ratio of the beam flanges, bf /2tf is compared to .p 
given in AISC Seismic, Part I, 
Table I-9-1. Both beam sizes, W18x35 and W21x44 are found to be acceptable. The W21x44 is 
illustrated below:
FEMA 451, NEHRP Recommended Provisions: Design Examples 
8-24 
(AISC Seismic, Table I-9-1) 
52 52 7.35 
50 p 
Fy 
. = = = 
7.22 (AISC Manual) 
2 
f
f 
b
t 
= 
7.22 < 7.35 OK 
The limiting h/t ratios for columns is also given in AISC Seismic, Part I, Table I-9-1. A W10x77 column 
from the lower level of an interior bay with storage load is illustrated (the axial load from the seismic load 
combination is used): 
2 (AISC Seismic, Table I-9-1) 
391 kips = 0.385 > 0.125 
(0.9)(22.6 in. x 50 ksi) 
u 
b y 
P
f P 
= 
191 2.33 191[2.33 0.385] 52.5 (AISC Seismic, Table I-9-1) 
50 
u 
p 
y b y 
P
F P 
. 
f 
. . 
= . - .= - = 
.. .. 
Check: 
OK 
p 52.5 35.7 253
Fy 
. = > = 
13.0 (AISC Manual) 
w 
h 
t 
= 
13.0 < 52.5 OK 
8.6.3 Column Axial Strength 
AISC Seismic, Part I, 8.2 requires that when Pu/fPn > 0.4 (in a seismic load combination), additional 
requirements be met. Selecting the same column as above for our illustration: 
2 (AISC Seismic, Part I, 8.2) 
391 kips 0.53 > 0.4 
(0.85)(22.6 in. )(38.4 ksi) 
u
n 
P
f P
= = 
Therefore the requirements of AISC Seismic, Part I, 8.2a, 8.2b, and 8.2c apply. These necessitate the 
calculation of axial loads using the System Overstrength Factor, O0 
= 3. Analysis needs to be run for two 
additional load combinations: 
1.2D + 0.5L + 0.2S + O0QE (AISC Seismic, Part I, Eq. 4.1) 
and 
0.9D - O0
QE (AISC Seismic, Part I, Eq. 4.2)
Chapter 8, Composite Steel and Concrete 
8-25 
Slab 
edge 
Floor joist 
Seat angle 
12" 
Bottom flange 
angle 
Figure 8-13 Detail at column. 
The axial seismic force in this column is only 7.5 kips, therefore Pu becomes 397 kips, obviously much 
less than fPn. The low seismic axial load is common for a moment-resisting frame system. Given that 
this requirement is a check ignoring bending moment, it does not control the design. 
[The special load combinations have been removed from the 2002 edition of AISC Seismic to eliminate 
inconsistencies with other building codes and standards. Therefore, 2003 Provisions Eq. 4.2-3 and 4.2-4 
should be used in conjunction with the load combinations in ASCE 7.] 
8.6.4 Details of the Joint 
Figure 8-13 shows a plan view at an edge column, concentrating on the arrangement of the steel elements. 
Figure 8-14 shows a section at the same location, showing the arrangement of the reinforcing steel. It is 
not required that the reinforcing bars be equally distributed on the two sides of the column, but it is 
necessary to place at least some of the bars on each side. This means that some overhang of the slab 
beyond the column flange is required. This example shows two of the six bars on the outside face. 
Figure 8-15 shows a plan view at a corner column. U shaped bent bars are used to implement the 
negative moment connection at such a location. Threaded bars directly attached to the column flange are 
also illustrated. Note the close spacing of the headed anchor studs for composite action. The reason for 
the close spacing at this location is that the beam span is half the normal span, yet full composite action is 
still provided.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
8-26 
41 
21 2" 
2" 
12" 
Floor joist 
#4x4'-0" at 10", lap with 
1
2" deformed anchor 
stud on edge plate 
(6) #5 x 15'-0" 
at each column 
L7x4x1
2"x0'-8" LLH 
(2) 1" dia. A325 bolt 
into column 
Figure 8-14 Detail at spandrel. 
W18 spandrel girder 
(2) #5 x 8'-0" with double 
nuts at column flange placed 
2" below finished floor elevation 
Edge of 
concrete 
slab 
(2) rows 3
4" dia. H.A.S. at 
41
2" o.c. (12'-6" spans only) 
W10 column 
(3) #5 in slab 
W21 spandrel beam 
lap with straight bars 
Figure 8-15 Detail at building corner. 
The compressive force in the deck is transferred to both flanges of the column. This is shown in Figure 8- 
16. Note that both flanges can accept compressive forces from the concrete. Also note that the transverse 
reinforcement will carry tension as force is transferred from the principal tension reinforcement through 
the concrete to bearing on the column flange. Strut and tie models can be used to compute the appropriate 
tension.
Chapter 8, Composite Steel and Concrete 
8-27 
Column 
Longitudinal 
Reinforcment 
Transverse 
Reinforcement 
~ ~ ~
~ ~ ~ 
~ ~ ~ ~ ~ ~ 
~ ~ ~ ~ ~ ~ 
Figure 8-16 Force transfer from deck to column. 
AISC SDGS-8 and ASCE TC include the following recommendations regarding the reinforcing steel: 
1. Place the principal tension reinforcement within a strip of width equal to 7 times the width of the 
column flange (or less) 
2. Use at least 6 bars for the principal reinforcement, extend it one quarter of the span from the column, 
but at least 24 bar diameters beyond the inflection point, and extend at least two of the bars over the 
full span 
3. Do not use bars larger than number 6 (0.75 in. diameter) 
4. Provide transverse reinforcement consistent with a strut and tie model to enable the transfer of forces 
(in the authors observation such reinforcement is also necessary to preserve the capacity of the 
headed studs for shear transfer)
9-1 
9 
MASONRY 
James Robert Harris, P.E., Ph.D., Frederick R. Rutz, P.E., 
Ph.D. and Teymour Manzouri, P.E., Ph.D. 
This chapter illustrates application of the 2000 NEHRP Recommended Provisions (herein after the 
Provisions), to the design of a variety of reinforced masonry structures in regions with different levels of 
seismicity. Example 9.1 features a single-story masonry warehouse building with tall, slender walls; 
Example 9.2 presents a five-story masonry hotel building with a bearing wall system designed in areas 
with different seismicities; and Example 9.3 covers a twelve-story masonry building having the same plan 
as the hotel but located in a region of high seismicity. Selected portions of each building are designed to 
demonstrate specific aspects of the design provisions. 
Masonry is a discontinuous and heterogeneous material. The design philosophy of reinforced grouted 
masonry approaches that of reinforced concrete; however, there are significant differences between 
masonry and concrete in terms of restrictions on the placement of reinforcement and the effects of the 
joints. These physical differences create significant differences in the design criteria. 
All structures were analyzed using two-dimensional (2-D) static methods. Examples 9.2 and 9.3 use 
dynamic analyses to determine the structural periods. Example 9.2 employs the SAP 2000 program, 
V6.11 (Computers and Structures, Berkeley, California); Example 9.3 employs the RISA 2D program, 
V.5.5 (Risa Technologies, Foothill Ranch, California). 
Although this volume of design examples is based on the 2000 Provisions, it has been annotated to reflect 
changes made to the 2003 Provisions. Annotations within brackets, [ ], indicate both organizational 
changes (as a result of a reformat of all of the chapters of the 2003 Provisions) and substantive technical 
changes to the 2003 Provisions and its primary reference documents. While the general concepts of the 
changes are described, the design examples and calculations have not been revised to reflect the changes 
to the 2003 Provisions. 
The most significant change to the masonry chapter in the 2003 Provisions is the incorporation by 
reference of ACI 530-02 for strength design in masonry. A significant portion of 2003 Provisions 
Chapter 11 has been replaced by a reference to this standard as well as a limited number of modifications 
to the standard, similar to other materials chapters. This updated chapter, however, does not result in 
significant technical changes as ACI 530-02 is in substantial agreement with the strength design 
methodology contained in the 2000 Provisions. 
Another change to the provisions for masonry structures is the addition of a new lateral system, 
prestressed masonry shear walls. This system is not covered in this volume of design examples. 
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-2 
Some general technical changes in the 2003 Provisions that relate to the calculations and/or design in this 
chapter include updated seismic hazard maps, changes to Seismic Design Category classification for short 
period structures, revisions to the redundancy requirements, revisions to the wall anchorage design 
requirement for flexible diaphragms, and a new Simplified Design Procedure that could be applicable 
to some of the examples in this chapter. 
Where they affect the design examples in this chapter, other significant changes to the 2003 Provisions 
and primary reference documents are noted. However, some minor changes to the 2003 Provisions and 
the reference documents may not be noted. 
In addition to the Provisions, the following documents are referenced in this chapter: 
ACI 318 American Concrete Institute. 1999 [2002]. Building Code Requirements for Concrete 
Structures. 
ACI 530 American Concrete Institute. 1999 [2002]. Building Code Requirements for Masonry 
Structures, ACI 530/ASCE 5/TMS 402. 
ASCE 7 American Society of Civil Engineers. 1998 [2002]. Minimum Design Loads for 
Buildings and Other Structures. 
Amrhein Amrhein, J, and D. Lee. 1994. Tall Slender Walls, 2nd Ed. Masonry Institute of 
America. 
Drysdale Drysdale R., A. Hamid, and L. Baker. 1999. Masonry Structures, Behavior and Design. 
Boulder Colorado: The Boulder Masonry Society. 
IBC International Code Council. 2000. International Building Code. 
UBC International Conference of Building Officials. 1997. Uniform Building Code. 
NCMA National Concrete Masonry Association. A Manual of Facts on Concrete Masonry, 
NCMA-TEK is an information series from the National Concrete Masonry Association, 
various dates. 
SEAOC Seismology Committee, Structural Engineers Association of California. 1999. 
Recommended Lateral Force Requirements and Commentary, 7th Ed. 
The short form designations for each citation are used throughout. The citation to the IBC exists for two 
reasons. One of the designs employees a tall, slender wall that is partially governed by wind loads and 
the IBC provisions are used for that design. Also, the R factors for masonry walls are significantly 
different in the IBC than in the Provisions; this is not true for other structural systems.
Chapter 9, Masonry 
9-3 
5 bays at 20'-0" = 100'-0" 
5 bays at 40'-0" = 200'-0" 
8" concrete 
masonry wall 
Typical glue-lam 
roof beam 
Plywood roof 
sheathing 
Open 
12" concrete 
masory wall 
Figure 9.1-1 Roof plan (1.0 in = 25.4 mm, 1.0 ft = 0.3048 m). 
9.1 WAREHOUSE WITH MASONRY WALLS AND WOOD ROOF, LOS ANGELES, 
CALIFORNIA 
This example features a one-story building with reinforced masonry bearing walls and shear walls. 
9.1.1 Building Description 
This simple rectangular warehouse is 100 ft by 200 ft in plan (Figure 9.1-1). The masonry walls are 30 ft 
high on all sides, with the upper 2 ft being a parapet. The wood roof structure slopes slightly higher 
towards the center of the building for drainage. The walls are 8 in. thick on the long side of the building, 
for which the slender wall design method is adopted, and 12 in. thick on both ends. The masonry is 
grouted in the cells containing reinforcement, but it is not grouted solid. The assumed strength of 
masonry is 2,000 psi. Normal weight concrete masonry units (CMU) with type S mortar are assumed. 
The long side walls are solid (no openings). The end walls are penetrated by several large doors, which 
results in more highly stressed piers between the doors (Figure 9.1-2); thus, the greater thickness for the 
end walls.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-4 
12'-0" 16'-0" 2'-0" 
30'-0" 
12'-0" 8'-0" 
100'-0"
Roof line 
Continuous 
stem wall 
and footing 
Door 
opening 
(typical) 
12" masonry 
4'-0" 
Figure 9.1-2 End wall elevation (1.0 in = 25.4 mm, 1.0 ft = 0.3048 m). 
The floor is concrete slab-on-grade construction. Conventional spread footings are used to support the 
interior steel columns. The soil at the site is a dense, gravelly sand. 
The roof structure is wood and acts as a diaphragm to carry lateral loads in its plane from and to the 
exterior walls. The roofing is ballasted, yielding a total roof dead load of 20 psf. There are no interior 
walls for seismic resistance. This design results in a highly stressed diaphragm with large calculated 
deflections. The design of the wood roof diaphragm and the masonry wall-to-diaphragm connections is 
illustrated in Sec. 10.2. 
In this example, the following aspects of the structural design are considered: 
1. Design of reinforced masonry walls for seismic loads and 
2. Computation of P-delta effects. 
9.1.2 Design Requirements 
[Note that the new Simplified Design Procedure contained in the 2003 Provisions Simplified Alternate 
Chapter 4 as referenced by the 2003 Provisions Sec. 4.1.1 is likely to be applicable to this example, 
subject to the limitations specified in 2003 Provisions Sec. Alt. 4.1.1.] 
9.1.2.1 Provisions Parameters 
Site Class (Provisions Sec. 4.1.2.1 [Sec. 3.5]) = C 
SS (Provisions Map 5 [Figure 3.3-3] ) = 1.50 
S1 (Provisions Map 6 [Figure 3.3-4] ) = 0.60 
Seismic Use Group (Provisions Sec. 1.3[Sec. 1.2]) = I 
[The 2003 Provisions have adopted the 2002 USGS probabilistic seismic hazard maps , and the maps 
have been added to the body of the 2003 Provisions as figures in Chapter 3 (instead of the previously 
used separate map package).] 
Chapter 9, Masonry 
9-5 
The remaining basic parameters depend on the ground motion adjusted for site conditions. 
9.1.2.2 Response Parameter Determination 
The mapped spectral response factors must be adjusted for site class in accordance with Provisions Sec. 
4.1.2.4 [3.3.2]. The adjusted spectral response acceleration parameters are computed according to 
Provisions Eq. 4.1.2.4-1 [3.3-1] and 4.1.2.4-2 [3.3-2] for the short period and one-second period, 
respectively, as follows: 
SMS = FaSS = 1.0(1.50) = 1.50 
SM1 = FvS1 = 1.3(0.60) = 0.78 
Where Fa and Fv are site coefficients defined in Provisions Tables 4.1.2.4a [3.3-1] and 4.1.2.4b [3.3-2], 
respectively. The design spectral response acceleration parameters (Provisions Sec. 4.1.2.5 [Sec. 3.3.3]) 
are determined in accordance with Provisions Eq. 4.1.2.5-1 [Eq. 3.3-3] and 4.1.2.5-2 [3.3-4] for the shortperiod 
and one-second period, respectively: 
2 2(1.50) 1.00 
3 3 
SDS= SMS= = 
2 2(0.78) 0.52 
3 3 
SD1= SM1= = 
The Seismic Design Category may be determined by the design spectral acceleration parameters 
combined with the Seismic Use Group. For buildings assigned to Seismic Design Category D, masonry 
shear walls must satisfy the requirements for special reinforced masonry shear walls in accordance with 
Provisions Sec. 11.3.8.2 [ACI 530 Sec. 1.13.6.4]. A summary of the seismic design parameters follows: 
Seismic Design Category (Provisions Sec. 4.2.1 [1.4]) = D 
Seismic Force Resisting System (Provisions Table 5.2.2 
[4.3-1]) = Special Reinforced 
Masonry Shear Wall 
Response Modification Factor, R (Provisions Table 5.2.2 
[4.3-1]) = 3.5 
Deflection Amplification Factor, Cd (Provisions Table 5.2.2 
[4.3-1]) = 3.5 
System Overstrength Factor, O0 (Provisions Table 5.2.2 
[4.3-1]) = 2.5 
Reliability Factor, . (Provisions Sec. 5.2.4.2 [Sec. 4.3.3]) = 1.0 
(Determination of . is discussed in Sec. 9.1.3 below [see Sec. 9.1.3.1 for changes to the reliability factor 
in the 2003 Provisions].) 
Note that the R factor for this system in the IBC and in ASCE 7 is 4.5. [5.0 in the 2003 IBC and ASCE 7- 
02] This difference would have a substantial effect on the seismic design; however, the vertical 
reinforcement in the tall 8-in. walls is controlled by wind loads so it would not change. 
9.1.2.3 Structural Design Considerations 
With respect to the load path, the roof diaphragm supports the upper 16 ft of the masonry walls (half the 
clear span plus the parapet) in the out-of-plane direction, transferring the lateral force to in-plane masonry 
shear walls.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-6 
Soil structure interaction is not considered. 
The building is of bearing wall construction. 
Other than the opening in the roof, the building is symmetric about both principal axes, and the vertical 
elements of the seismic resisting system are arrayed entirely at the perimeter. The opening is not large 
enough to be considered an irregularity (per Provisions Table 5.2.3.2[Table 4.3-2]); thus, the building is 
regular, both horizontally and vertically. Provisions Table 5.2.5.1[Table 4.4-1], permits several analytical 
procedures to be used; the equivalent lateral force (ELF) procedure (Provisions Sec. 5.4) is selected for 
used in this example. The orthogonality requirements of Provisions Sec. 5.2.5.2 Sec. 4.4.2 are potentially 
significant for the piers between the door openings at the end walls. Thus, those walls will be designed 
for 100 percent of the forces in one direction plus 30 percent of the forces in the perpendicular direction. 
There will be no inherent torsion because the building is symmetric. The effects of accidental torsion, 
and its potential amplification, need not be included because the roof diaphragm is flexible. This is the 
authors interpretation of what amounts to a conflict between Provisions Table 5.2.3.2[Table 4.3-2], Item 
1, and Provisions Sec. 5.4.4.2[Sec. 5.24.2] and Sec. 5.4.4.3[Sec. 5.2.4.3]. 
The masonry bearing walls also must be designed for forces perpendicular to their plane (Provisions Sec. 
5.2.6.2.7)[Sec. 4.6.1.3]. 
For in-plane loading, the walls will be treated as cantilevered shear walls. For out-of-plane loading, the 
walls will be treated as pinned at the bottom and simply supported at the top. The assumption of a pinned 
connection at the base is deemed appropriate because the foundation is shallow and narrow which 
permits rotation near the base of the wall. 
9.1.3 Load Combinations 
The basic load combinations (Provisions Sec. 5.2.7 [Sec. 4.2.2]) are the same as specified in ASCE 7 (and 
similar to the IBC). The seismic load effect, E, is defined by Provisions Eq. 5.2.7-1 [4.2-1] and Eq. 5.2.7- 
2 [4.2-2] as: 
E = .QE  0.2SDSD = (1.0)QE  0.2(1.00)D = QE  0.2D 
This assumes . = 1.0 as will be confirmed in the following section. 
9.1.3.1 Reliability Factor 
In accordance with Provisions Sec. 5.2.4.2[4.3.3], the reliability factor, ., applies to the in-plane load 
direction. 
For the long direction of building: 
10 
x 
wall 
max 
story w 
r V
V l 
. .. . 
=... ..... .. 
(0.5) 10 0.025 
rmaxx= ..200..= 
. .
Chapter 9, Masonry 
9-7 
2 20 2 20 3.66 
20,000 0.025 20,000 
. = - = - =- 
max x r 
. = -3.66 < 1.0 = .min, so use . = 1.0. 
For the short direction of the building: 
10 ( )(0.23) 10 (0.5)(0.23)(1) 0.115 
x 8 
wall wall 
max 
story w story 
r V V 
V l V 
=.... ....... ...=.... ....... ...= = 
Although the calculation is not shown here, note that a single 8-ft-long pier carries approximately 23 
percent (determined by considering the relative rigidities of the piers) of the in-plane load for each end 
wall. 
Also, 1.0 was used for the 10/lw term even though 10/8 ft > 1.0. According to Provisions 5.2.4.2, the 
10/lw term need not exceed 1.0 only for walls of light frame construction. This example was created based 
on a draft version of the 2000 Provisions, which limited the value of the 10/lw term to 1.0 for all shear 
walls, a requirement that was later changed for the published edition. Thus, this calculation is not strictly 
correct. Using the correct value of would result in . = 1.02 rather than the 0.77 computed below. maxx r 
This would result in a slight change in the factor on QE, 1.02 vs. 1.00, which has not been carried through 
the remainder of this example. 
(When the redundancy factor was developed by the Structural Engineers Association of California in the 
wake of the 1994 Northridge earthquake, the upper bound of 1.0 for 10/lw was simply not mentioned. The 
1997 Provisions, the UBC, and the IBC were published with no upper bound on 10/lw. However, the 
original authors of the concept published their intent with the SEAOC document in 1999 with the upper 
bound of 1.0 on 10/lw for all types of shear walls. The same change was adopted within BSSC for the 
2000 Provisions. A subsequent change to the 2000 Provisions limited the upper bound of 1.0 to apply 
only to light frame walls.) 
Therefore, 
0.12 rmaxx = 
2 20 0.77 
0.115 20,000 
. = - = 
. = 0.77 < 1.0 = .min, so use . = 1.0. 
[The redundancy requirements have been substantially changed in the 2003 Provisions. For a shear wall 
building assigned to Seismic Design Category D, . = 1.0 as long as it can be shown that failure of a shear 
wall with height-to-length-ratio greater than 1.0 would not result in more than a 33 percent reduction in 
story strength or create an extreme torsional irregularity. Therefore, the redundancy factor would have to 
be investigated only in the transverse direction where the aspect ratios of the piers between door openings 
are greater than 1.0. In the longitudinal direct, where the aspect ratio is (significantly) less than 1.0, . = 
1.0 by default.] 
9.1.3.2 Combination of Load Effects
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-8 
Load combinations for the in-plane loading direction from ASCE 7 are: 
1.2D + 1.0E + 0.5L + 0.2S 
and 
0.9D + 1.0E + 1.6H. 
L, S, H do not apply for this example so the load combinations become: 
1.2D + 1.0E 
and 
0.9D + 1.0E. 
When the effect of the earthquake determined above, 1.2D + 1.0(QE  0.2D), is inserted in each of the 
load combinations: 
1.4D + 1.0 QE 
1.0D - 1.0 QE 
and 
0.9D + 1.0(QE  0.2D) 
which results in: 
1.1D + 1.0 QE 
and 
0.7D - 1.0 QE 
Thus, the controlling cases from all of the above are: 
1.4D + 1.0 QE 
when gravity and seismic are additive and 
0.7D - 1.0 QE 
when gravity and seismic counteract. 
These load combinations are for the in-plane direction of loading. Load combinations for the out-ofplane 
direction of loading are similar except that the reliability coefficient (.) is not applicable. Thus, for 
this example (where . = 1.0), the load combinations for both the in-plane and the out-of-plane directions 
are: 
1.4D + 1.0 QE 
and
Chapter 9, Masonry 
9-9 
0.7D - 1.0 QE. 
The combination of earthquake motion (and corresponding loading) in two orthogonal directions must be 
considered (Provisions Sec. 5.2.5.2.3) [Sec. 4.4.2.3]. 
9.1.4 Seismic Forces 
9.1.4.1 Base Shear 
Base shear is computed using the parameters determined previously. The Provisions does not recognize 
the effect of long, flexible diaphragms on the fundamental period of vibration. The approximate period 
equations, which limit the computed period, are based only on the height. Since the structure is relatively 
short and stiff, short-period response will govern the design equations. According to Provisions Sec. 
5.4.1 [Sec. 5.2.1.1] and Eq. 5.4.1.1-1 [Eq. 5.2-3] (for short-period structures): 
1.0 0.286 
/ 3.5/1 
V CsW SRDSI W W W 
. . . . 
= =.. .. =.. .. = 
The seismic weight for forces in the long direction is: 
Roof = 20 psf (100)200 = 400 kips 
End walls = 103 psf (2 walls)[(30 ft)(100 ft) - 5(12 ft)(12 ft)](17.8 ft/28 ft) = 299 kips 
Side walls = 65 psf (30ft)(200ft)(2 walls) = 780 kips 
Total = 1,479 kips 
Note that the centroid of the end walls is determined to be 17.8 ft above the base, so the portion of the 
weight distributed to the roof is approximately the total weight multiplied by 17.8 ft/28 ft (weights and 
section properties of the walls are described subsequently). 
Therefore, the base shear to each of the long walls is: 
V = (0.286)(1,479 kips)/2 = 211 kips. 
The seismic weight for forces in the short direction is: 
Roof = 20 psf (100)200 = 400 kips 
Side walls = 65 psf (2 walls)(30ft)(200ft)(15ft/28ft) = 418 kips 
End walls = 103 psf (2 walls)[(30ft)(100ft)-5(12ft)(12ft)] = 470 kips 
Total = 1,288 kips 
The base shear to each of the short walls is: 
V = (0.286)(1,288 kips)/2 = 184 kips. 
9.1.4.2 Diaphragm Force 
See Sec. 10.2 for diaphragm forces and design. 
9.1.4.3 Wall Forces 
because the diaphragm is flexible with respect to the walls, shear is distributed to the walls on the basis of 
beam theory ignoring walls perpendicular to the motion (this is the "tributary" basis).
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-10 
The building is symmetric. Given the previously explained assumption that accidental torsion need not 
be applied, the force to each wall becomes half the force on the diaphragm. 
All exterior walls are bearing walls and, according to Provisions Sec. 5.2.6.2.7 [Sec. 4.6.1.3], must be 
designed for a normal (out-of-plane) force of 0.4SDSWc. The out-of-plane design is shown in Sec. 9.1.5.3 
below. 
9.1.5 Longitudinal Walls 
The total base shear is the design force. Provisions Sec. 11.7 [Sec. 11.2] is the reference for design 
strengths. The compressive strength of the masonry (fm') is 2,000 psi. Provisions Sec. 11.3.10.2 gives Em 
= 750 fm' = (750)(2 ksi) = 1,500 ksi. 
[2003 Provisions Sec. 11.2 adopts ACI 530 as a design basis for strength design masonry and provides 
some modifications to ACI 530. In general, the adoption of ACI 530 as a reference does not have a 
significant effect on this design example. Note that by adopting ACI 530 in the 2003 Provisions, Em = 
900fm per ACI 530 Sec. 1.8.2.2.1, eliminating the conflict discussed below.] 
Be careful to use values consistent with the Provisions. Different standards call for different values. To 
illustrate this point, the values of Em from different standards are shown in Table 9.1-1.
Chapter 9, Masonry 
9-11 
Table 9.1-1 Comparison of Em 
Standard Em Em for this example 
Provisions 
IBC 
ACI 530 
750 fm' 
900 fm' 
900 fm' 
1,500 ksi 
1,800 ksi 
1,800 ksi 
1.0 kip = 4.45 kN, 1.0 in. = 25.4 mm. 
For 8-inch thick CMU with vertical cells grouted at 24 in. o.c. and horizontal bond beams at 48 inch o.c., 
the weight is conservatively taken as 65 psf (recall the CMU are normal weight) and the net bedded area 
is 51.3 in.2/ft based on tabulations in NCMA-TEK 141. 
9.1.5.1 Horizontal Reinforcement 
As determined in Sec. 9.1.4.1, the design base shear tributary to each longitudinal wall is 211 kips. Based 
on Provisions Sec. 11.7.2.2 [ACI 530, Sec. 3.1.3], the design shear strength must exceed either the shear 
corresponding to the development of 1.25 times the nominal flexure strength of the wall, which is very 
unlikely in this example due to the length of wall, or 2.5 times Vu = 2.5(211) = 528 kips. 
From Provisions Eq. 11.7.3.2[ACI 530, Eq. 3-21] , the masonry component of the shear strength capacity 
for reinforced masonry is: 
m4.0 1.75 nm0.25 . 
M 
V Af P 
Vd 
= - ' + . . .. 
.. .. .... 
Conservatively treating M/Vd as equal to 1.0 for the long walls and conservatively treating P as the 
weight of the wall only without considering the roof weight contribution: 
Vm=[4.0-1.75(1.0)](51.3)(200) 2000+0.25(390)=1130 kips 
and 
fVm = 0.8(1,130) = 904 kips > 528 kips OK 
where f = 0.8 is the resistance factor for shear from Provisions Table11.5.3[ACI 530, sec. 3.1.4] . 
Horizontal reinforcement therefore is not required for shear but is required if the wall is to qualify as a 
Special Reinforced Masonry Wall. 
According to Provisions Sec. 11.3.8.3[ACI 530, Sec. 1.13.6.3] , minimum reinforcement is 
(0.0007)(7.625 in.)(8 in.) = 0.043 in.2 per course, but it may be wise to use more horizontal reinforcement 
for shrinkage in this very long wall and then use minimum reinforcement in the vertical direction (this 
concept applies even though this wall requires far more than the minimum reinforcement in the vertical 
direction due to its large height-to-thickness ratio). Two #5 bars at 48 in. on center provides 0.103 in.2 
per course. This amounts to 0.4 percent of the area of masonry plus the grout in the bond beams. The 
actual shrinkage properties of the masonry and the grout and local experience should be considered in 
deciding how much reinforcement to provide. For long walls that have no control joints, as in this 
example, providing more than minimum horizontal reinforcement is appropriate.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-12 
#7 
24" o.c. 
8" concrete 
masonry unit 
Figure 9.1-3 Trial design for 8-in.-thick CMU 
wall (1.0 in = 25.4 mm). 
9.1.5.2 Vertical Reinforcement 
Steps for verifying a trial design are noted in the sections that follow. 
9.1.5.3 Out-of Plane Flexure 
As indicated previously, the design demand for seismic out-of-plane flexure is 0.4SDSWc. For a wall 
weight of 65 psf for the 8-in.-thick CMU side walls, this demand is 0.4(1.00)(65 psf) = 26 psf. 
Calculations for out-of-plane flexure become somewhat involved and include the following: 
1. Select a trial design. 
2. Investigate to ensure ductility. 
3. Make sure the trial design is suitable for wind (or other nonseismic) lateral loadings using the IBC. 
Note that many section properties determined in accordance with the IBC are different from those 
indicated in the Provisions so section properties will have to be determined multiple times. The 
IBC portion of the calculation is not included in this example. 
[2003 Provisions and the 2003 IBC both adopt ACI 530-02 by reference, so the section properties should 
be the same for both documents.] 
4. Calculate midheight deflection due to wind by the IBC. (While the Provisions have story drift 
requirements, they do not impose a midheight deflection limit for walls). 
5. Calculate seismic demand. 
6. Determine seismic resistance and compare to demand determined in Step 5. 
Proceed with these steps as follows: 
9.1.5.3.1 Trial design 
A trial design of #7 bars at 24 in. on center is selected. See Figure 9.1-3.
Chapter 9, Masonry 
9-13 
9.1.5.3.2 Investigate to ensure ductility 
The critical strain condition corresponds to a strain in the extreme tension reinforcement (which is a 
single #7 centered in the wall in this example) equal to 1.3 times the strain at yield stress. 
Based on Provisions Sec. 11.6.2.2[ACI 530, Sec. 3.2.3.5.1] for this case: 
t = 7.63 in. 
d = t/2 = 3.81 in. 
em = 0.0025 
es = 1.3ey = 1.3(fy/Es) = 1.3(60 ksi /29,000 ksi) = 0.0027 
( m ) 1.83 in. 
m s 
c d 
e 
e e 
. . 
=. . = 
.. + .. 
a = 0.8c = 1.46 in. 
The Whitney compression stress block, a = 1.46 in. for this strain distribution, is greater than the 1.25 in. 
face shell width. Thus, the compression stress block is broken into two components: one for full 
compression against solid masonry (the face shell) and another for compression against the webs and 
grouted cells, but accounting for the open cells. These are shown as C1 and C2 in Figure 9.1-4: 
C1 = 0.80fm' (1.25 in.)b = (0.80)(2 ksi)(1.25)(24) = 48 kips (for a 24-in. length) 
C2 = 0.80 fm' (a-1.25 in.)(8 in.) = (0.80)(2 ksi)(1.46-1.25)(8) = 2.69 kips (for a 24-in. length) 
The 8-in. dimension in the C2 calculation is for combined width of grouted cell and adjacent mortared 
webs over a 24-in. length of wall. The actual width of one cell plus the two adjacent webs will vary with 
various block manufacturers, and may be larger or smaller than 8 in. The 8-in. value has the benefit of 
simplicity and is correct when considering solidly grouted walls.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-14 
1.25" 
3.81" 
1.25" 
t = 7.63" 
P = P + P 
d = 3.81" 
1.83" 1.98" 
= 0.0025 
N.A. 
1.3 = 0.0027 
1.25" 
1.46" 
N.A. 
T 
C 
C 
0.48" 
1.21" 
0.8f ' 
0.21" 
#7 at 24" o.c. 
1 
m 
2 
y 
m 
f w 
a 
c 
Figure 9.1-4 Investigation of out-of-plane ductility for the 8-in.-thick CMU side walls 
(1.0 in = 25.4 mm). 
T is based on 1.25 Fy (Provisions Sec. 11.6.2.2)[ACI 530, Sec. 3.2.3.5.1]: 
T = 1.25FyAs = (1.25)(60 ksi)(0.60 in.2) = 45 kips (for a 24-in. length) 
Use unfactored P (Provisions Sec. 11.6.2.2)[ACI 530, Sec. 3.2.3.5.1]: 
P = (Pf + Pw) = (20 psf (10 ft.) + 65 psf (16 ft.)) = 1.24 klf = 2.48 kips (for a 24-in. length) 
Check C1 + C2 > T + P:
Chapter 9, Masonry 
9-15 
T + P = 47.5 kips 
C1 + C2 = 50.7 kips > 47.5 kips. OK 
The compression capacity is greater than the tension capacity; therefore, the ductile failure mode criterion 
is satisfied. 
[The ductility (maximum reinforcement) requirements in ACI 530 are similar to those in the 2000 
Provisions. However, the 2003 Provisions also modify some of the ACI 530 requirements, including 
critical strain in extreme tensile reinforcement (1.5 times) and axial force to consider when performing the 
ductility check (factored loads).] 
9.1.5.3.3 Check for wind load using the IBC 
Load factors and section properties are not the same in the IBC and the Provisions (The wind design 
check is beyond the scope of this seismic example so it is not presented here.) Both strength and 
deflection need to be ascertained in accordance with IBC. 
Note that, for comparison, selected properties for the Provisions (and IBC) ductility check, IBC wind 
strength check, and Provisions seismic strength check are tabulated below. Keeping track of which 
version of a given parameter is used for each of the calculations can get confusing; be careful to apply the 
correct property for each analysis. 
Table 9.1-2 Comparison of Variables (explanations in the following text) 
Parameter Provisions Ductility 
Calculation 
Provisions Strength 
Calculation 
IBC Wind 
Calculation 
P 1.24 klf 0.87 & 1.74 klf 1.12 klf 
Em NA 1,500,000 psi 1,800,000 psi 
fr NA 80 psi 112 psi 
w NA 26 psf 19 psf (service) 
es 0.0027 NA NA 
d 3.82 in. 3.82 in. 3.82 in. 
c 1.83 in. 1.25 in. 1.25 in. 
a 1.46 in. 1.00 in. 1.00 in. 
Cres = C1+C2 50.1 kips 52.1 kips 56.4 kips 
n = Es/Em NA 19.33 16.11 
Ig NA 355 in.4 355 in.4 
Sg NA 93.2 in.3 93.2 in.3 
Ase NA 0.32 in.2/ft 0.32 in.2/ft 
Icr NA 48.4 in.4/ft 48.4 in.4/ft 
Mcr = frS NA 7.46 in.-kips 10.44 in.-kips 
dallow NA NA 2.32 in. 
NA = not applicable, 1.0 kip = 4.45 kN, 1.0 ft = 0.3048 m, 1.0 in = 25.4 mm, 1.0 ksi = 6.98 MPa, 
1.0 in.-kip = 0.113 kN-m.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-16 
9.1.5.3.4 Calculate midheight deflection due to wind by the IBC 
The actual calculation is not presented here. For this example the midheight deflection was calculated 
using IBC Eq. 21-41[ACI 530, Eq. 3-31] with Icr = 47.3 in.4 per ft. Using IBC Eq. 21.41[ACI 530, Eq. 3- 
31], the calculated deflection is 2.32 in., which is less than 2.35 in. = 0.007h (IBC Eq. 21-39[ACI 530, 
Eq. 3-29]). 
9.1.5.3.5 Calculate seismic demand 
For this case, the two load factors for dead load apply: 0.7D and 1.4D. Conventional wisdom holds that 
the lower dead load will result in lower moment-resisting capacity of the wall so the 0.7D load factor 
would be expected to govern. However, the lower dead load also results in lower P-delta so both cases 
should be checked. (As it turns out, the higher factor of 1.4D governs). 
Check moment capacity for 0.7D: 
Pu = 0.7(Pf + Pw). 
For this example, the iterative procedure for addressing P-delta from Amrhein will be used, not 
Provisions Eq. 11.5.4.3[ACI 530, Commentary Sec. 3.1.5.3] which is intended for in-plane deflections: 
Roof load, Pf = 0.7(0.2 klf) = 0.14 klf 
Eccentricity, e = 7.32 in. (distance from wall centerline to roof reaction centerline) 
Modulus of elasticity (Provisions Eq. 11.3.10.2 [ACI 530, 1.8.2.2 ]), Em = 750 fm' = 1,500,000 psi 
[Note that by adopting ACI 530 in the 2003 Provisions, Em = 900fm per ACI 530 Sec. 1.8.2.2.1.] 
Modular ratio, s 19.3 
m 
n E
E 
= = 
The modulus of rupture (fr) is found in Provisions Table 11.3.10.5.1[ACI 530, Sec. 3.1.7.2.1]. The values 
given in the table are for either hollow CMU or fully grouted CMU. Values for partially grouted CMU 
are not given; Footnote a indicates that interpolation between these values must be performed. As 
illustrated in Figure 9.1-6, the interpolated value for this example is 80 psi: 
(fr - 50 psi)/(103 in.2 - 60 in.2) = (136 psi - 50 psi)/(183 in.2 - 60 in.2) 
fr = 80 psi 
Ig = 355 in.4/ft 
Sg = 93.2 in.3/ft 
Mcr = frSg = 7460 in-lb/ft
Chapter 9, Masonry 
9-17 
16" 8" 
b = 24" 
1.25" 
N.A. N.A. 
1.25" 
d = 3.81" 
nAs 
b 
c 
b = 8.32" inferred from NCMA tabulations 
b = 8" used for convenience 
Figure 9.1-7 Cracked moment of inertia (Icr) for 8-in.-thick CMU side walls (1.0 in = 25.4 mm). 
CASE 1 
All cells open 
A = 60 in. 
f = 50 psi 
CASE 2 
(1) Cell grouted 
A = 103 in. 
f = 80 psi * 
CASE 3 
Fully grouted 
A = 183 in. 
f = 136 psi 
* By interpolation 
2
r r r 
2 2 
24" 24" 24" 
Figure 9.1-6 Basis for interpolation of modulus of rupture, fr (1.0 in = 25.4 mm, 1.0 psi = 6.89 kPa). 
Refer to Figure 9.1-7 for determining Icr. The neutral axis shown on the figure is not the conventional 
neutral axis by linear analysis; instead it is the plastic centroid, which is simpler to locate, especially when 
the neutral axis position results in a T beam cross-section. (For this wall, the neutral axis does not 
produce a T section, but for the other wall in this building, a T section does result.) Cracked moments of 
inertia computed by this procedure are less than those computed by linear analysis but generally not so 
much less that the difference is significant. (This is the method used for computing the cracked section 
moment of inertia for slender walls in the standard for concrete structures, ACI 318.) Axial load does 
enter the computation of the plastic neutral axis and the effective area of reinforcement. Thus: 
P = 1.24 klf 
T = ((0.60 in.2)/(2 ft.))(60 ksi) = 18.0 klf 
C = T + P = 19.24 klf 
a = C/(0.8 f'mb) = (19.24 klf)/(0.8(2.0 ksi)(12 in./ft.)) = 1.002 in. 
c = a/0.8 = 1.253 in. 
Icr = nAse(d-c)2 + bc3/3 = 19.33(0.30 in.2 + (1.24 klf)/60 ksi)(3.81 in. - 1.25 in.)2 + 
(12 in./ft)(1.25 in.)3/3 = 4.84 in.4/ft
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-18 
Note that Icr could be recomputed for P = 0.7D and P = 1.4D but that refinement is not pursued in this 
example. 
The standard technique is to compute the secondary moment in an iterative fashion as shown below: 
Axial load 
Pu = 0.7(Pf + Pw) = 0.7(0.2 klf + 1.04 klf) = 0.868 klf 
First iteration 
0.868 (0.60)(60) 0.614 in.2/ 2 ft. = 0.312 in.2/ ft 
60 
u s y 
se 
y 
P Af 
A 
f 
+ + 
= = = 
2 
1 / 8 ( ) u u 0 0 w M =wh +Pe+P+P . 
2 
1 
(26 psf/12)(336 in.) 7.32 in. 
(140 plf) (140plf + 728 plf)(0) 
8 2 u M= + .. ..+ 
. . 
Mu1 = 31,088 in.-lb/lf > Mcr = 7460 
( )
2 2 
1 
5(7460)(336) 5(31,088 7460)(336) 0.165 3.827 3.99 in. 
s 48(1,500,000) 355 48(1,500,000)(48.4) 
- 
. = + = + = 
Second iteration 
2 30,576 512 (140 728)(3.99) 34,551 in.-lb u M = + + + = 
2 
2 
0.165 5(34,551 7460)(336) 0.165 4.388 4.55 in. 
s 48(1,500,000)(48.4) 
- 
. = + = + = 
Third iteration 
Mu3 = 30,576 + 512 + (140 + 728)(4.55) = 35,040 in.-lb/lf 
( ) 2 
3 
5 35,040 7460 (336) 
0.165 0.165 4.467 4.63 in. 
s 48(1,500,000)(48.4) 
- 
. = + = + = 
Convergence check 
4.63 4.55 1.8% 5% 
4.55 
- 
= < 
Mu = 35,040 in.-lb (for the 0.7D load case) 
Using the same procedure, find Mu for the 1.4D load case. The results are summarized below: 
First iteration 
P = 7360 plf 
Mu1 = 31,601 in.-lb/ft
Chapter 9, Masonry 
9-19 
.u
1 = 4.08 in. 
Second iteration 
Mu2 = 38,684 in.-lb/ft 
.u
2 = 5.22 in. 
Third iteration 
Mu3 = 40,667 in.-lb/ft 
.3 
= 5.54 in. 
Fourth iteration 
Mu4 = 41,225 in.-lb/ft 
.u
4 = 5.63 in. 
Check convergence 
5.63 5.54 1.7% 5% 
5.54 
- 
= < 
Mu = 41,225 in.-lb (for the 1.4D load case) 
9.1.5.3.6 Determine flexural strength of wall 
Refer to Figure 9.1-8. As in the case for the ductility check, a strain diagram is drawn. Unlike the 
ductility check, the strain in the steel is not predetermined. Instead, as in conventional strength design of 
reinforced concrete, a rectangular stress block is computed first and then the flexural capacity is checked. 
T = Asfy = (0.30 in.2/ft.)60 ksi = 18.0 klf 
The results for the two axial load cases are tabulated below. 
Load Case 0.7D + E 1.4D + E 
Factored P, klf 0.87 1.74 
T + P = C, klf 18.87 19.74 
a = C / (0.8f'mb), in. 0.981 1.028 
MN = C (d - a/2), in.-kip/ft. 62.6 65.1 
fMN = 0.85MN, in.-kip/ft. 53.2 55.3 
MU, in.-kip/ft. 35.0 41.2 
Acceptance OK OK
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-20 
1.25" 
3.81" 
1.25" 
t = 7.63" 
P 
d = 3.81" 
c 
= 0.0025 
N.A. 
> = 0.00207 
a 
1.25" 
N.A. T = A F + P 
C 
0.8f ' 
#7 at 24" o.c. 
d-a/2 
u 
s 
m 
m 
t 
a/2 
y 
s y 
= a/0.80 
s 
t s t s 
Figure 9.1-8 Out-of-plane strength for 8-in.-thick CMU walls (1.0 in = 25.4 mm). 
Note that wind actually controls the stiffness and strength out-of-plane and that this is only a tentative 
acceptance for seismic. The Provisions requires a check of the combined orthogonal loads in accordance
Chapter 9, Masonry 
9-21 
with Provisions Sec. 5.2.5.2, Item a [Sec. 4.4.2.3]. However, as discussed below, a combined orthogonal 
load check was deemed unnecessary for this example. 
9.1.5.4 In-Plane Flexure 
In-plane calculations for flexure in masonry walls include two items per the Provisions: 
1. Ductility check and 
2. Strength check. 
It is recognized that this wall is very strong and stiff in the in-plane direction. In fact, most engineers 
would not even consider these checks necessary in ordinary design. The ductility check is illustrated here 
for two reasons: to show a method of implementing the requirement and to point out an unexpected 
result. (In the authors opinion, the Provisions should reconsider the application of the ductility check 
where the M/Vdv ratio is substantially less than 1.0.) 
9.1.5.4.1 Ductility check 
Provisions Sec. 11.6.2.2 [ACI 530, 3.2.3.5.1] requires that the critical strain condition correspond to a 
strain in the extreme tension reinforcement equal to 5 times the strain associated with Fy. This calculation 
uses unfactored gravity loads. (See Figure 9.1-9.) 
[The ductility (maximum reinforcement) requirements in ACI 530 are similar to those in the 2000 
Provisions. However, the 2003 Provisions also modify some of the ACI 530 requirements, including 
critical strain in extreme tensile reinforcement (4 times yield) and axial force to consider when performing 
the ductility check (factored loads).] 
P = Pw + Pf = (0.065 ksf (30 ft.) + 0.02 ksf (10 ft.))(200 ft.) = 430 kips 
P is at the base of the wall rather than at the midheight. 
0.0025 200 ft 38.94 ft 
0.0025 0.0103 
m 
m s 
c d 
e 
e e 
=... + ... =... + ... = 
a = 0.8c = 31.15 ft = 373.8 in. 
Cm = 0.8fm'abavg = 2,560 kips 
Where bavg is taken from the average area used earlier, 51.3 in.2/ft.; see Figure 9.1-9 for locations of 
tension steel and compression steel (the rebar in the compression zone will act as compression steel). 
From this it can be seen that: 
( )( ) 
(1.25 ) 40.27 (0.60) 453 kips 
Ts1 fy 2 2 ft o.c. 
. . 
= .. .. = 
. . 
(1.25 ) 120.79 (0.60) 2,718 kips 
Ts2= fy .. 2 .. = 
. . 
6.73 (0.60) 121 kips 
Cs1=fy..2 ft. o.c... = 
. .
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-22 
= 5E 
= 0.0103 
c = 38.94' 161.06' 
N.A. 
61.06' P = P + P 
31.15' 7.79' 
0.8f ' 
32.21' 
N.A. 
21.5' 
35.58' 
C 
C 
72.53 ksi 
f = 60 ksi 
26.84' 
100.66' 
T 
T 
1.25 F 
= 75 ksi 
5 F = 
300 ksi 
23.37' 
C 
y 
y 
s2 
s1 
40.27' 120.79' y 
s1 
s2 
m 
f w 
m 
s y 
m = 
0.0025 
6.73' 
a 
Figure 9.1-9 In-plane ductility check for side walls (1.0 in = 25.4 mm, 1.0 ksi = 6.89 MPa). 
( ) 32.21 (0.60) 290 kips 
Cs2 fy (2)(2) 
. . 
= . . = 
. . 
Note that some authorities would not consider the compression resistance of reinforcing steel that is not 
enclosed within ties. The Provisions clearly allows inclusion of compression in the reinforcement. 
SC > ST
Chapter 9, Masonry 
9-23 
#7 at 24" o.c. 
10'-4" 
All cells grouted 
Figure 9.1-10 Grout cells solid within 10 ft of each end of side walls (1.0 in = 25.4 mm, 1.0 ft = 0.3048 m). 
Cm + Cs1 + Cs2> P + Ts1 + Ts2 
2,560 + 121 + 290 = 2,971 < 3,601 = 430 + 453 + 2,718 
Therefore, there is not enough compression capacity to ensure ductile failure. 
In order to ensure ductile failure with #7 bars at 24 in. on center, one of the following revisions must be 
made: either add (3,601 kips - 2,971 kips) = 630 kips to Cm or reduce T by reducing As. Since this 
amount of reinforcement is needed for out-of-plane flexure, As cannot be reduced. 
Try filling all cells for 10 ft - 0 in. from each end of the wall. As shown in Figure 9.1-10, this results in 
10 additional grouted cells. 
Area of one grouted cell: (8 in.)(5.13 in.) = 41 in.2 
Volume of grout for one cell: (6 in.)(5.13 in.)(30 ft.)/(144 in.2/ft.2) = 6.41 ft.3 
Weight of grout for one cell: (0.140 kcf)(6.41) = 0.90 kips/cell 
Additional P: (10 additional cells)(0.9) = 9.0 kips 
Additional Cm: 0.8 fm' (41 in.2)(10 cells) = 656 kips 
Additional Cm - additional P: 656 kips - 9 kips = 647 kips 
Net additional Cm: 647 kips > 630 kips OK 
or, as expressed in terms of the above equation: 
SC > ST 
2,971 kips + 656 kips = 3,627 kips > 3,610 kips = 3,601 kips + 9 kips OK 
Since C > T, the ductile criterion is satisfied. 
This particular check is somewhat controversial. In the opinion of the authors, flexural yield is feasible 
for walls with M/Vd in excess of 1.0; this criterion limits the compressive strain in the masonry, which 
leads to good performance in strong ground shaking. For walls with M/Vd substantially less than 1.0, the 
wall will fail in shear before a flexural yield is possible. Therefore, the criterion does not affect 
performance. Well distributed and well developed reinforcement to control the shear cracks is the most 
important ductility attribute for such walls. 
9.1.5.4.2 Strength check 
The wall is so long with respect to its height that in-plane strength for flexure is acceptable by inspection.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-24 
9.1.5.5 Combined Loads 
Combined loads are not calculated here because the in-plane strength is obviously very high. Out-ofplane 
resistance governs the flexural design. 
9.1.5.6 Shear in Longitudinal Walls (Side Walls) 
Compute out-of-plane shear at base of wall in accordance with Provisions Sec. 5.2.6.2.7[Sec. 4.6.1.3]: 
Fp = 0.4SDSWc = (0.4)(1.00)(65 psf)(28 ft/2) = 364 plf. 
Information from the flexural design from Sec. 9.1.5.3 is needed to determine the required shear strength 
based upon development of the flexural capacity. The ratio of fMN to MU is the largest for the load case 
0.7D + E. The load that would develop the flexural capacity is approximated by ratio (a second P-delta 
analysis does not seem justified for this check): 
w w 
M
M
N
U 
' 
/ ./.
. 
=  =  = . 
f f 
26 
532 085 
350 
psf 465 psf 
1.25 times this results in a load for shear design of 58 psf. Thus VU = (58 psf)(28 ft. /2) = 818 plf. The 
capacity of computed per Provisions Eq.11.7.3.2[ACI 530, Eq. 3.2.1] : 
m4.0 1.75 nm0.25 
V MAf P 
Vd 
=... - ... ...... ' + 
M/Vd need not be taken larger than 1.0. An is taken as bwd = 8.32(3.81) = 31.7 in.2 per cell from Figure 
9.17. Because this shear exists at both the bottom and the top of the wall, conservatively neglect the 
effect of P: 
[4.0 1.75(1.0)](51.3in.2 / 2 ft.) 2,000 0 1.595 klf m V = - + = 
fVm = (0.8)(1.595) = 1.28 klf > 0.81 klf 
As indicated in Sec. 9.1.4.1 and Sec. 9.1.5.1, the in-plane demand at the base of the wall, Vu = 2.5(211 
kips) = 528 kips, and the shear capacity, fVm is larger than 904 kips. 
For the purpose of understanding likely behavior of the building somewhat better, Vn is estimated more 
accurately for these long walls: 
M/Vd = h/l = 28/200 = 0.14 
P = 0.7D = 0.7(430) = 301 kip 
Vm = [4.0 - 1.75(0.14)][200(51.3) + 2(10)91.5-51.3)](0.045) + 0.25(301) = 1870 + 75 = 1945 kip 
Vs = 0.5(Av/s)fyd = 0.5(0.62/4.0)(60)(200) = 930 kip 
Vn = 1945 + 930 = 2875 kip 
Maximum Vn = 6%f'mA = 6(0.045 ksi)(9234 in.2) = 2493 < 2875 kip 
fVn = 0.8(2493) = 1994 kip 
VE = 211kip 
Vn/VE = 11.8 >> R used in design 
In other words, it is unlikely that the long masonry walls will yield in either in-plane shear or flexure at 
the design seismic ground motion. The walls will likely yield in out-of-plane response and the roof 
diaphragm may also yield. The roof diaphragm for this building is illustrated in Sec. 10.2.
Chapter 9, Masonry 
9-25 
The combined loads for shear (orthogonal loading, per Provisions Sec. 5.2.5.2.2, Item a)[Sec. 4.4.2.3] are 
shown in Table 9.1-3. 
Table 9.1-3 Combined Loads for Shear in Side Wall 
Out-of-Plane In-Plane Total 
Case 1 
Case 2 
1.00(810/1,280)+ 
0.30(810/1,280)+ 
0.30(528/1994)= 
1.00(528/1994)= 
0.71 < 1.00 OK 
0.45 < 1.00 OK 
Values are in kips; 1.0 kip = 4.45 kN. 
9.1.6 Transverse Walls 
The transverse walls will be designed in a manner similar to the longitudinal walls. Complicating the 
design of the transverse walls are the door openings, which leave a series of masonry piers between the 
doors. 
9.1.6.1 Horizontal Reinforcement 
The minimum reinforcement, per Provisions Sec. 11.3.8.3[ACI 530, Sec.1.13.6.3] , is (0.0007)(11.625 
in.)(8 in.) = 0.065 in.2 per course. The maximum spacing of horizontal reinforcement is 48 in., for which 
the minimum reinforcement is 0.39 in.2. Two #4 in bond beams at 48 in. on center would satisfy the 
requirement. The large amount of vertical reinforcement would combine to satisfy the minimum total 
reinforcement requirement. However, given the 100-ft length of the wall, a larger amount is desired for 
control of restrained shrinkage as discussed in Sec. 9.1.5.1. Two #5 at 48 in. on center will be used. 
9.1.6.2 Vertical Reinforcement 
The area for each bay subject to out-of-plane wind is 20 ft wide by 30 ft high because wind load applied 
to the doors is transferred to the masonry piers. However, the area per bay subject to both in-plane and 
out-of-plane seismic is reduced by the area of the doors. This is because the doors are relatively light 
compared to the masonry. See Figures 9.1-12 and 9.1-13. 
9.1.6.3 Out-of-Plane Flexure 
Out-of-plane flexure will be considered in a manner similar to that illustrated in Sec. 9.1.5.3 . The design 
of this wall must account for the effect of door openings between a row of piers. The steps are the same 
as identified previously and are summarized here for convenience: 
1. Select a trial design, 
2. Investigate to ensure ductility, 
3. Make sure the trial design is suitable for wind (or other non-seismic) lateral loadings using IBC, 
4. Calculate midheight deflection due to wind by IBC, 
5. Calculate the seismic demand, and 
6. Determine the seismic resistance and compare to the demand determined in Step 5. 
9.1.6.3.1 Trial design 
A trial design of 12-in.-thick CMU reinforced with two #6 bars at 24 in. on center is selected. The selfweight 
of the wall, accounting for horizontal bond beams at 4ft on center, is conservatively taken as 103 
psf. Adjacent to each door jamb, the vertical reinforcement will be placed into two cells. See Figure 9.1- 
11.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-26 
2'-0" 2'-0" 2'-0" 
8'-0" 
(2) #6 
6.88" 
11.63" 
Figure 9.1-11 Trial design for piers on end walls (1.0 in = 25.4 
mm, 1.0 ft = 0.3048 m). 
28'-0" 
20'-0" 8'-0" 
17.8' 20'-0" 
In-plane 
loads 
P 
V (short wall) 
V + V 
(long wall) 
Area/ bay subject 
to seismic (because 
masonry walls are 
much heavier 
than doors) 
Area/ bay subject 
to wind (because 
doors transfer wind 
loads to masonry) 
Pf = 8 kips f Pf 
w
f w 
2'-0' 
Pf 
Out-of-plane loads 
applied to bay 
Figure 9.1-12 In-plane loads on end walls (1.0 ft = 0.3048 m). 
6' 
17.8' 
21' 
12' 16' 2' 
R = 6.82 kips /bay out-of-plane 
4.88 kips /bay in-plane 
w = (0.4) (103) (8') = 330 plf/bay out-of-plane 
(0.286) (103) (8') = 236 plf/bay in-plane 
R = 11.97 kips/bay out-of-plane 
8.56 kips/bay in-plane 
H 
H 
H 
w = (0.4) (103 psf) (20') = 824 plf/bay out-of-plane 
(0.286) (103 psf) (20') = 589 plf/bay in-plane 
bot 
bot 
top 
2 
1 
r 
top 
Figure 9.1-13 Out-of-plane load diagram and resultant of lateral loads (1.0 ft = 0.3048 m, 
1.0 lb = 4.45 N, 1.0 kip = 4.45 kN). 
Next, determine the design loads. The centroid for seismic loads, accounting for the door openings, is 
determined to be 17.8 ft above the base. See Figures 9.1-12 and 9.1-13.
Chapter 9, Masonry 
9-27 
9.1.6.3.2 Investigate to ensure ductility 
The critical strain condition is corresponds to a strain in the extreme tension reinforcement (which is a 
pair of #6 bars in the end cell in this example) equal to 1.3 times the strain at yield stress. See Figures 
9.1-11 and 9.1-14. 
For this case: 
t = 11.63 in. 
d = 11.63 - 2.38 = 9.25 in. 
em = 0.0025 (Provisions Sec. 11.6.2.1.b)[ACI 530, Sec. 3.2.2] 
es = 1.3 ey = 1.3 (fy/Es) = 1.3 (60 ksi /29,000 ksi) = 0.0027 (Provisions Sec. 11.6.2.2)[ACI 530, Sec. 
3.2.3.5.1] 
4.45 in. 
( ) 
m 
m s 
c d 
e 
e e 
. . 
=.. + .. = 
a = 0.8c = 3.56 in. (Provisions Sec. 11.6.2.2)
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-28 
d = 9.25" 
c 
4.45" 
1.50" 2.06" 0.89" 4.80" 
T 
0.8f ' 
1.6 ksi 
1.92" 
2.98" 
3.70" 
a = 0.8c = 3.56" 
1.3 = 0.0027 
C 
C 
= 0.0025 
m 
P = (Pf + Pw ) 
m 
2 
1 
y 
11.63" 
1.50" 1.50" 
(6) #6 (6) #6 
N.A. 
Cres 
Figure 9.1-14 Investigation of out-of-plane ductility for end wall (1.0 in = 25.4 
mm, 1.0 ksi = 6.89 MPa). 
Note that the Whitney compression stress block, a = 3.56 in. deep, is greater than the 1.50-in. face shell 
thickness. Thus, the compression stress block is broken into two components: one for full compression 
against solid masonry (the face shell) and another for compression against the webs and grouted cells but 
accounting for the open cells. These are shown as C1 and C2 on Figure 9.1-15. The values are computed 
using Provisions Sec. 11.6.2.1e:[ACI 530, 3.2.2.e]; 
C1 = 0.80fm' (1.50 in.)b = (0.80)(2 ksi)(1.50)(96) = 230 kips (for full length of pier) 
C2 = 0.80fm' (a - 1.50 in.)(6(8 in.)) = (0.80)(2 ksi)(3.56 - 1.50)(48) = 158 kips 
The 48 in. dimension in the C2 calculation is the combined width of grouted cell and adjacent mortared 
webs over the 96-in. length of the pier. 
T is based on 1.25Fy (Provisions Sec. 11.6.2.2)[ACI 530, Sec. 3.2.3.5.1]:
Chapter 9, Masonry 
9-29 
2'-0" 
2" 
b = 8.31" 
d = 9.25" 
c = 1.66" 
N.A. N.A. 
1.25" 6.06" 1" 6.06" 1.25" 
(2) #6 at 24" 
8.63" 
1.5" 1.5" 
11.63" 
w 
Figure 9.1-15 Cracked moment of inertia (Icr) for end walls (1.0 in = 25.4 mm, 1.0 ft = 0.3048 m). 
Dimension c depends on calculations shown for Figure 9.1-16. 
T = 1.25FyAs = (1.25)(60 ksi)(6  0.44 in.2) = 198 kips/pier 
P = (Pf + Pw) = 8.0 k + (0.103 ksf)(18 ft.)(20 ft.) = 45.1 kips/pier 
P is computed at the head of the doors: 
C1 + C2 > P + T 
388 kip > 243 kips 
Since the compression capacity is greater than the tension capacity, the ductility criterion is satisfied. 
[The ductility (maximum reinforcement) requirements in ACI 530 are similar to those in the 2000 
Provisions. However, the 2003 Provisions also modify some of the ACI 530 requirements, including 
critical strain in extreme tensile reinforcement (1.5 times) and axial force to consider when performing the 
ductility check (factored loads).] 
9.1.6.3.3 Check for wind loading using IBC 
Note that load factors and section properties are different in the IBC and the Provisions. Note also that 
wind per bay is over the full 20 ft wide by 30 ft high bay as discussed above. (The calculations are not 
presented here.) 
9.1.6.3.4 Calculate midheight deflection due to wind by IBC 
Although the calculations are not presented here, note that in Figure 9.1-15 the neutral axis position and 
partial grouting results in a T beam cross section for the cracked moment of inertia. Use of the plastic 
neutral axis is a simplification for computation of the cracked moment of inertia. For this example, 
midheight out-of-plane deflection is 1.27 in. < 2.35 in. = 0.007h, which is acceptable. 
9.1.6.3.5 Calculate Seismic Demand 
For this example, the load combination with 0.7D has been used and, for this calculation, forces and 
moments over a single pier (width = 96 in.) are used. This does not violate the b > 6t rule (ACI 530 
Sec. 7.3.3)[ACI 530, Sec. 3.2.4.3.3] because the pier is reinforced at 24 in. o.c. The use of the full width 
of the pier instead of a 24 in. width is simply for calculation convenience.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-30 
For this example, a P-delta analysis using RISA-2D was run. This resulted in: 
Maximum moment, Mu = 66.22 ft-kips/bay = 66.22/20 ft = 3.31 klf (does not govern) 
Moment at top of pier, Mu = 62.12 ft-kips/pier = 62.12 / 8 ft = 7.77 klf (governs) 
Shear at bottom of pier, Vu = 6.72 kips/pier 
Reaction at roof, Vu = 12.07 kips/bay 
Axial force at base, Ru = 54.97 kips/pier 
The shears do not agree with the reactions shown in Figure 9.1-13; because the results in Figure 9.1-13 do 
not include the P-delta consideration. 
9.1.6.3.6 Determine moment resistance at the top of the pier 
See Figure 9.1-16. 
As = 6-#6 = 2.64 in.2 
d = 9.25 in. 
T = 2.64(60) = 158.4 kip 
C = T + P = 203.5 kip 
a = C / (0.8f'mb) = 203.5 / [0.8(2)96] = 1.32 in. 
Because a is less than the face shell thickness (1.50 in.), compute as for a rectangular beam. Moments are 
computed about the centerline of the wall. 
MN = C (t/2 - a/2) + P (0) + T (d - t/2) 
= 203.5(5.81 - 1.32/2) + 158.4(9.25-1.32/2) = 1593 in.-kip = 132.7 ft.-kip 
fMN = 0.85(132.7) = 112.8 ft.-kip 
Because moment capacity at the top of the pier, fMn = 112.8 ft-kips, exceeds the maximum moment 
demand at top of pier, Mu = 62.1 ft-kips, the condition is acceptable but note that this is only tentative 
acceptance. 
The Provisions requires a check of the combined loads in accordance with Provisions 5.2.5.2, Item a 
[Sec. 4.4.2.3]. See Sec. 9.1.6.5 for the combined loads check.
Chapter 9, Masonry 
9-31 
d = 9.25" 
a = 4.49" 3.44" 
T 
0.8f ' 
= 1.6 ksi 
5.15" 
m 
Pu = (Pf + Pw ) D 
11.63" 
f 
1.50" 1.50" 
t tf 
1.32" 
C 
Figure 9.1-16 Out-of-plane seismic strength of pier on end wall (1.0 in = 
25.4 mm, 1.0 ksi = 6.89 MPa). 
9.1.6.4 In-Plane Flexure 
There are several possible methods to compute the shears and moments in the individual piers of the end 
wall. For this example, the end wall was modeled using RISA-2D. The horizontal beam was modeled at 
the top of the opening, rather than at its midheight. The in-plane lateral loads (from Figure 9.1-12) were 
applied at the 12-ft elevation and combined with joint moments representing transfer of the horizontal 
forces from their point of action down to the 12-ft elevation. Vertical load due to roof beams and the selfweight 
of the end wall were included. The input loads are shown on Figure 9.1-17. For this example: 
w = (18 ft.)(103 psf) + (20 ft.)(20 psf) = 2.254 klf 
H = (184 kip)/5 = 36.8 kip 
M = 0.286((400 + 418)(28 - 12) + 470(17.8 - 12)) = 452 ft-kip
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-32 
H 
M 
M/2 
H/2 
w 
H/2 H 
M 
H 
M 
H 
M 
M/2 
96'-0" 
12'-0" 
Figure 9.1-17 Input loads for in-plane end wall analysis (1.0 ft = 0.3048 m). 
El. 112'-0" = T.O. Pier 
8' 
12' 
V = 43.6 kip 
R = 55.0 kip 
bot 
bot
M b o t = 0 
V t o p = 43.6 kip 
M t o p = 523 ft-kip 
P t o p = 45.1 kip 
Figure 9.1-18 In-plane design condition for 8-ft-wide pier 
(1.0 ft = 0.3048 m). 
The input forces at the end wall are distributed over all the piers to simulate actual conditions. The RISA- 
2D frame analysis accounts for the relative stiffnesses of the 4-ft-and 8-ft-wide piers. The final 
distribution of forces, shears, and moments for an interior pier is shown on Figure 9.1-18. 
As a trial design for in-plane pier design, use two #6 bars at 24 in. on center supplemented by adding two 
#6 bars in the cells adjacent to the door jambs (see Figure 9.1-19). 
Chapter 9, Masonry 
9-33 
8'-0' = 96" 
11.63" 
92" 
m = 0.0025
14" 6" 18" 42" 66" 74" 
N.A. 
s = 5 y 
= 0.0103 
0.0092 
0.0058 
0.0025 
0.0008 
0.0019 
c = 18" 
P = (Pf + P ) 
a = 0.8c 
14.4"
10.8" 
Cm 
0.8f 'm = 1.6 ksi 0.8f 'm = 1.6 ksi 
Cs1 
Cs2 
6" 
14" 18" 
Ts4
42" 
Ts3 Ts2 Ts1 
66"
74" 
1.25Fy 
= 75 ksi 
5 y 
w 
4" 
Figure 9.1-19 In-plane ductility check for 8-ft-wide pier (1.0 in = 25.4 mm, 1.0 ksi = 6.89 MPa).
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-34 
The design values for in-plane design at the top of the pier are: 
Unfactored 0.7D + 1.0E 1.4D + 1.0E 
Axia P = 45.1 kips Pu = 31.6 kips Pu = 63.2 kips 
Shea V = 43.6 kips Vu = 43.6 kips Vu = 43.6 kips 
Mom M = 523 ft-kips Mu = 523 ft-kips Mu = 523 ft-kips 
The ductility check is illustrated in Figure 9.1-19: 
em = 0.0025 
es = 5ey = (5)(60/29,000) = 0.0103 
d = 92 in. 
From the strain diagram, the strains at the rebar locations are: 
e66 = 0.0092 
e42 = 0.0058 
e18 = 0.0025 
e6 = 0.0008 
e14 = 0.0019 
To check ductility, use unfactored loads: 
P = Pf + Pw = (0.020 ksf)(20 ft)(20 ft) + (0.103 ksf)(18 ft)(20 ft) 
P = 8 kips + 37.1 kips = 45.1 kips 
a = 0.8c = 14.4 in. 
Cm = (0.8fm' )ab = 1.6 ksi)(14.4 in.)(11.63 in.) = 268.0 kips 
Ts1 = Ts2 = Ts3 = Ts4 = (1.25Fy)(As) = (1.25)(60 ksi)(2  0.44 in.2) = 66 kips 
Cs1 = FyAs(e14/ey) = (60 ksi)(2  0.44 in.2)(0.0019/0.00207) = 48.5 kips 
Cs2 = FyAs(e6/ey) = (60 ksi)(2  0.44 in.2)(0.0008/0.00207) = 20.4 kips 
SC > S T + P 
Cm + Cs1 + Cs2 > Ts1 + Ts2 + Ts3 + Ts4 + P 
268 + 48.5 + 20.4 > 66 + +66 + 66 + 66 + 45.1 
336.9 kips > 309.1 kips 
Since compression capacity exceeds tension capacity, ductile failure is ensured. Note that 1.25Fy is used 
for tension calculations per Provisions Sec. 11.6.2.2 [ACI 530, Sec. 3.2.3.5-1] . 
[The ductility (maximum reinforcement) requirements in ACI 530 are similar to those in the 2000 
Provisions. However, the 2003 Provisions also modify some of the ACI 530 requirements, including 
critical strain in extreme tensile reinforcement (4 times yield) and axial force to consider when performing 
the ductility check (factored loads).] 
For the strength check, see Figure 9.1-20.
Chapter 9, Masonry 
9-35 
11.63" 
m = 0.0025 
N.A. 
= 0.0017 
Cm1 Cm2 Cm3 
Ts1 * Ts2 
m = 0.0025 
Ts4 Ts3 Ts2 Ts1 
y = F
E 
= 0.00207 
y 
y = 0.00207 
Balanced 
Case 
P = 0 Case 
N.A. 
0.8f'm 
0.8f'm 
y = 0.00207 
Cs1
Cs2 = 0.0019 
* 
96" 
48" 44" 4" 
a = 11.3" 
c = 14.2" 
16" 16" 8.3" 
40.3"
50.3" 
7.7" 2.3" 
48" 44" 
Center 
Line 
M 
P 
Figure 9.1-20 In-plane seismic strength of pier (1.0 in = 25.4 mm). Strain diagram superimposed on 
strength diagram for both cases. Note that low force in reinforcement is neglected in calculations. 
To ascertain the strength of the pier, a fPn - fMn curve will be developed. Only the portion below the 
balance point will be examined as that portion is sufficient for the purposes of this example. Ductile 
failures occur only at points on the curve that are below the balance point so this is consistent with the
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-36 
overall approach). 
For the P = 0 case, assume all bars in tension reach their yield stress and neglect compression steel (a 
conservative assumption): 
Ts1 = Ts2 = Ts3 = Ts4 = (2)(0.44 in.2)(60 ksi) = 52.8 kips 
Cm = S Ts = (4)(52.8) = 211.2 kips 
Cm = 0.8fmab = (0.8)(2 ksi)a(11.63 in.) = 18.6a 
Thus, a = 11.3 in. and c = a/f = 11.3 / 0.8 = 14.2 in. 
SMcl = 0: 
Mn = 42.35 Cm + 44Ts1 + 36Ts2 + 12Ts3 - 12Ts4 = 13,168 in.-kips 
fMn = (0.85)(13,168) = 11,193 in.-kips = 933 ft-kips 
For the balanced case: 
d = 92 in. 
e = 0.0025 
ey = 60/29,000 = 0.00207 
m 50.3 in. 
m y 
c d 
e 
e e 
. . 
=... + ... = 
a = 0.8c = 40.3 in. 
Compression values are determined from the Whitney compression block adjusted for fully grouted cells 
or nongrouted cells: 
Cm1 = (1.6 ksi)(16 in.)(11.63 in.) = 297.8 kips 
Cm2 = (1.6 ksi)(16 in.)(2  1.50 in.) = 76.8 kips 
Cm3 = (1.6 ksi)(8.3 in.)(11.63 in.) = 154.4 kips 
Cs1 = (0.88 in.2)(60 ksi) = 52.8 kips 
Cs2 = (0.88 in.2)(60 ksi)(0.0019 / 0.00207) = 48.5 kips 
Ts2 = (0.88 in.2)(60 ksi) = 52.8 kips 
Ts2 = (0.88 in.2)(60 ksi)(0.0017 / 0/00207) = 43.4 kips 
S Fy = 0: 
Pn = SC - ST = 297.8 + 76.8 + 154.4 + 52.8 + 48.5 -52.8 - 43.4 = 534 kips 
fPn = (0.85)(534) = 454 kips 
S Mcl = 0: 
Mn = 40Cm1 + 24Cm2 + 11.85Cm3 + 44Cs1 + 36Cs2 + 44Ts1 + 36Ts2 = 23,540 in.-kips 
fMn = (0.85)(23,540) = 20,009 in. - kips = 1,667 ft-kips 
The two cases are plotted in Figure 9.1-21 to develop the fPn - fMn curve on the pier. The demand (Pu , 
Mu) also is plotted. As can be seen, the pier design is acceptable because the demand is within the fPn - 
fMn curve. (See the Birmingham 1 example in Sec. 9.2 for additional discussion of fPn - fMn curves.) 
By linear interpolation, fMn at the minimum axial load is 968 kip.
Chapter 9, Masonry 
9-37 
9.1.6.5 Combined Loads 
Combined loads for in-plane and out-of-plane moments in piers at end walls, per Provisions Sec. 
5.2.5.2.2, Item a, are shown in Table 9.1-4. 
Table 9.1-4 Combined Loads for Flexure in End Pier 
0.7D 
Out-of-Plane In-Plane Total 
Case 1 
Case 2 
1.0(62.12/112.8) + 
0.3(62.12/112.8) + 
0.3(523/986) = 
1.0(523/986) = 
0.71 < 1.00 OK 
0.70 < 1.00 OK 
Values are in kips; 1.0 kip = 4.45 kN. 
500 
ft-kips 
200 kips 
(P max = 66 kips) 
(P min = 33 kips) 
fP 
fM 
M = 986 ft-kips 
P = 0 Case 
(933 ft-kips, 0 kips) 
Simplified fP - fM curve 
100 kips 
300 kips 
400 kips 
500 kips 
600 kips 
Balance 
(1667 ft-kips, 454 kips) 
1,000 
ft-kips 
1,500 
ft-kips 
2,000 
ft-kips 
M = 523 ft-kips 
n 
n 
fPn 
fMn 
u
u 
u 
u 
u u 
Figure 9.1-21 In-plane fP11 - fM11 diagram for pier (1.0 kip = 4.45 kN, 1.0 ft-kip = 1.36 kN-m).
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-38 
30' 
12' 
12' 
Pf 
Vroof + long wall 
(upper portion) 
Pw 
Vshort wall 
Vbase 
P base 
Vbase 
Pbase 
P 
V 
M 
P pier 
Mbase 
Figure 9.1-22 In-plane shear on end wall and pier (1.0 ft = 
0.3048 m). 
9.1.6.6 Shear at Transverse Walls (End Walls) 
The shear at the base of the pier is 43.6 kips/bay. At the head of the opening where the moment demand 
is highest, the in-plane shear is slightly less (based on the weight of the pier). There, V = 43.6 kips - 
0.286(8 ft)(12 ft)(0.103 ksf) = 40.8 kips. (This refinement in shear is not shown in Figure 9.1-18 
although the difference in axial load at the two locations is shown.) The capacity for shear must exceed 
2.5 times the demand or the shear associated with 125 percent of the flexural capacity. Using the results 
in Table 9.1-4, the 125 percent implies a factor on shear by analysis of: 
125 
1 
125 
1 
07 
1 
085 
. . 210 
. . 
. 
1 
Demand to capacity ratio 
.
. .
.
. .
.
. .
.
. .
= 
.. .
.. .
.. .
.. .
= 
f 
Therefore, the required shear capacities at the head and base of the pier are 91.6 kips and 85.7 kips, 
respectively. 
The in-plane shear capacity is computed as follows where the net area, An, of the pier is the area of face 
shells plus the area of grouted cells and adjacent webs:
Chapter 9, Masonry 
9-39 
Vm4.0 1.75 VMdAnfm 0.25P 
=.. - .. .... ' + 
. . .. 
An = (96 in.  1.50 in. 2) + (6 cells  8 in.  8.63 in.) = 702 in.2 
V 
A
s 
f d S 
V 
y V = 
.. .
.. . 
= 
.
. .
.
. . 
05 
05 
062 
48 
60 
2 
. 
. 
. 
( 
in. 
in. 
ksi)(96 in.) 
= 37.2 kips / bay 
At the head of the opening: 
Vm = [4.0 - 1.75(1.0)](702 in.2)(0.0447 ksi) + (0.25)(0.7)(45.1 kips) = 78.5 kips/bay 
fVN = (0.8)(78.5 + 37.2) = 92.6 kips/bay 
At the base: 
Vm = [4.0 - 1.75(0)](702 in.2)(0.0447 ksi) + (0.25)(0.7)(55.0 kips) = 135.2 kips/bay 
fVN = (0.8)(135.2 + 37.2) = 137.9 kips/bay 
As discussed previously, M/Vd need not exceed 1.0 in the above equation. 
For out-of-plane shear, see Figure 9.1-13. Shear at the top of wall is 12.07 kips/bay and shear at the base 
of the pier is 6.72 kips/bay. From the values in the figure, the shear at the head of the opening is 
computed as 6.72 kips - (12 ft)(0.33 kip/ft) = 2.76 kips. The same multiplier of 2.10 for development of 
125 percent of flexural capacity will be applied to out-of-plane shear resulting in 25.3 kips at the top of 
the wall, 5.80 kips at the head of the opening, and 14.11 kips at the base. 
Out-of-plane shear capacity is computed using the same equation. Sbwd is taken as the net area An. Note 
that M/Vd is zero at the support because the moment is assumed to be zero; however, a few inches into the 
span, M/Vd will exceed 1.0 so the limiting value of 1.0 is used here. This is typically the case when 
considering out-of-plane loads on a wall. 
For computing shear capacity at the top of the wall: 
An = bwd =((8 in./2 ft.)  20 ft)(9.25 in.) = 740 in.2 
Vm = [4.0 - 1.75(1)](740 in.2)(0.0447 ksi) + (0.25)(8.0) = 76.9 kips/bay 
fVm = (0.8)(76.9) = 61.5 kips/bay 
For computing shear capacity in the pier: 
An =(8 in./cell)(6 cells)(9.25 in.) = 444 in.2 
Vm = [4.0 - 1.75(1)](444 in.2)(0.0447 ksi) + (0.25)(41.67) = 55.4 kips/bay 
fVm = (0.8)(55.4) = 44.3 kips/bay 
The combined loads for shear at the end pier (per Provisions 5.2.5.2.2, Item a [Sec. 4.4-23]) are shown in
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-40 
Table 9.1-5. 
Table 9.1-5 Combined Loads for Shear in End Wall 
In-Plane Out-of-Plane Total 
Case 1 
Pier base 1.0(91.6/137.9) + 0.3(14.11/44.3) = 0.76 < 1.00 OK 
Case 2 
Pier base 0.3(91.6/137.9) + 1.0(14.11/44.3) = 0.52< 1.00 OK 
Case 1 
Pier head 1.0(85.7/92.6) + 0.3(5.80/44.3) = 0.96 > 1.00 OK 
Case 2 
Pier head 0.3(85.7/92.6) + 1.0(5.80/44.3) = 0.41 < 1.00 OK 
Values are in kips; 1.0 kip = 4.45 kN. 
9.1.7 Bond Beam 
Reinforcement for the bond beam located at the elevation of the roof diaphragm can be used for the 
diaphragm chord. The uniform lateral load for the design of the chord is the lateral load from the long 
wall plus the lateral load from the roof and is equal to 0.87 klf. The maximum tension in rebar is equal 
the maximum moment divided by the diaphragm depth: 
M/d = 4,350 ft-kips/100 ft = 43.5 kips 
The seismic load factor is 1.0. The required reinforcement is: 
Areqd = T/fFy = 43.5/(0.85)(60) = 0.85 in.2 
This will be satisfied by two #6 bars, As = (2  0.44 in.2) = 0.88 in.2 
In Sec. 10.2, the diaphragm chord is designed as a wood member utilizing the wood ledger member. 
Using either the wood ledger or the bond beam is considered acceptable. 
9.1.8 In-Plane Deflection 
Deflection of the end wall (short wall) has two components as illustrated in Figure 9.1-23.
Chapter 9, Masonry 
9-41 
17.8' 
V (short wall) 
V + V (long wall) 
s
f 
12' 
4' 12' 8' 
18' 
16' 
28' 
5.8' 
. 1 
. 2 
. tot 
Figure 9.1-23 In-plane deflection of end wall (1.0 ft = 0.3048 m). 
As obtained from the RISA 2D analysis of the piers, .1 = 0.047 in.: 
2 
VL 
AG 
a 
. = S 
where a is the form factor equal to 6/5 and 
G = Em/2(1 + ) = 1500 ksi / 2(1 + 0.15) = 652 ksi 
A = An = Area of face shells + area of grouted cells 
= (100 ft  12 in./ft  2  1.50 in.2) +(50)(8 in.)(8.63 in.) = 7,050 in.2 
Therefore: 
= 0.0013 + 0.0059 = 0.007 in. 
6(67.15)(5.8 12)6(116.9)(16 12) 
25(7,050)(652)5(7,050)(652) 
. .  . .  
. = . . + . . 
. . . . 
and,
.total = Cd(0.047 + 0.007) = 3.5(0.054 in.) = 0.19 in. < 3.36 in. 
(3.36 = 0.01hn = 0.01hsx) (Provisions Sec. 11.5.4) 
Note that the drift limits for masonry structures are smaller than for other types of structure. It is possible 
to interpret Provisions Table 5.2.8 [Table 4.5-1] to give a limit of 0.007hn for this structure but that limit 
also is easily satisfied. The real displacement in this structure is in the roof diaphragm; see Sec. 10.2.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-42 
6'-0" 
6'-8" 
Prestressed 
hollow core 
slabs 
40'-0" 24'-0" 24'-0" 24'-0" 40'-0" 
4'-0" 
14'-0" 
24'-0" 24'-0" 24'-0" 
152'-0" 
72'-0" 
8" concrete 
masonry wall 
Figure 9.2-1 Typical floor plan (1.0 in = 25.4 mm, 1.0 ft = 0.3048 m). 
9.2 FIVE-STORY MASONRY RESIDENTIAL BUILDINGS IN BIRMINGHAM, 
ALABAMA; NEW YORK, NEW YORK; AND LOS ANGELES, CALIFORNIA 
9.2.1 Building Description 
In plan, this five-story residential building has bearing walls at 24 ft on center (see Figures 9.2-1 and 9.2- 
2). All structural walls are of 8-in.-thick concrete masonry units (CMU). The floor is of 8-in.-thick 
hollow core precast, prestressed concrete planks. To demonstrate the incremental seismic requirements 
for masonry structures, the building is partially designed for four locations: two sites in Birmingham, 
Alabama; a site in New York, New York; and a site in Los Angeles, California. The two sites in 
Birmingham have been selected to illustrate the influence of different soil profiles at the same location. 
The building is designed for Site Classes C and E in Birmingham. The building falls in Seismic Design 
Categories B and D in these locations, respectively. For Site Class D soils, the building falls in Seismic 
Design Categories C and D for New York and Los Angeles, respectively. 
[Note that the method for assigning seismic design category for short period buildings has been revised in 
the 2003 Provisions. If the fundamental period, Ta, is less than 0.8Ts, the period used to determine drift is 
less than Ts, and the base shear is computed using 2003 Provisions Eq 5.2-2, then seismic design category 
is assigned using just 2003 Provisions Table 1.4-1 (rather than the greater of 2003 Provisions Tables 1.4- 
1 and 1.4-2). This change results in the Birmingham Site Class E building being assigned to Seismic 
Design Category C instead of D. The changes to this example based on the revised seismic design 
category are not noted in the remainder of the example. The New York building provides an example of 
what the Seismic Design Category C requirements would be for the Birmingham Site Class E building.]
Chapter 9, Masonry 
9-43 
152'-0" 
36'-0" 80'-0" 36'-0" 
5 at 8'-8" = 43'-4" 2'-0" 
45'-4" 
Figure 9.2-2 Building elevation (1.0 in = 25.4 mm, 1.0 ft = 0.3048 m). 
For the New York and both Birmingham sites, it is assumed that shear friction reinforcement in the joints 
of the diaphragm planks is sufficient to resist seismic forces, so no topping is used. For the Los Angeles 
site, a cast-in-place 2 -in.-thick reinforced lightweight concrete topping is applied to all floors. The 
structure is free of irregularities both in plan and elevation. The Provisions, by reference to ACI 318, 
requires reinforced cast-in-place toppings as diaphragms in Seismic Design Category D and higher. Thus, 
the Birmingham example in Site Class E would require a topping, although that is not included in this 
example. 
Provisions Chapter 9 has an appendix (intended for trial use and feedback) for the design of untopped 
precast units as diaphragms. The design of an untopped diaphragm for Seismic Design Categories A, B, 
and C is not explicitly addressed in ACI 318. The designs of both untopped and topped diaphragms for 
these buildings are described in Chapter 7 of this volume using ACI 318 for the topped diaphragm in the 
Los Angeles building and using the appendix to Provisions Chapter 9 for untopped diaphragms in the 
New York building. It is assumed here that the diaphragm for the Birmingham 2 example would be 
similar to the New York example, and the extra weight of the Birmingham 2 topping is not included in the 
illustration here. 
No foundations are designed in this example. However, for the purpose of determining the site class 
coefficient (Provisions Sec. 4.1.2.1 [Sec. 3.5]), a stiff soil profile with standard penetration test results of 
15 < N < 50 is assumed for Los Angeles and New York sites resulting in a Site Class D for these two 
locations. For Birmingham, however, one site has soft rock with N > 50 and the other has soft clay with 
N < 15, which results in Site Classes C and E, respectively. The foundation systems are assumed to be 
able to carry the superstructure loads including the overturning moments. 
The masonry walls in two perpendicular directions act as bearing and shear walls with different levels of 
axial loads. The geometry of the building in plan and elevation results in nearly equal lateral resistance in 
both directions. The walls are constructed of CMU and are typically minimally reinforced in all 
locations. The walls are assumed to act as columns in their planes. Figure 9.2-3 illustrates the wall 
layout.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-44 
A
B
B
A 
A
B
B
A 
D 
C C D
D 
D C C 
E 
F 
G 
G 
F 
E 
Wall length 
A 36'-0" 
B 34'-0' 
C 32'-8" 
D 32'-8" 
E 8'-0" 
F 8'-0" 
G 8'-0" 
Figure 9.2-3 Plan of walls (1.0 ft = 0.3048 m). 
The floors serve as horizontal diaphragms distributing the seismic forces to the walls and are assumed to 
be stiff enough to be considered rigid. There is little information about the stiffness of untopped precast 
diaphragms. The design procedure in the appendix to Provisions Chapter 9 results in a diaphragm 
intended to remain below the elastic limit until the walls reach an upper bound estimate of strength, 
therefore it appears that the assumption is reasonable. 
Material properties are as follows: 
The compressive strength of masonry, f!m, is taken as 2,000 psi and the steel reinforcement has a yield 
limit of 60 ksi. 
The design snow load (on an exposed flat roof) is taken as 20 psf for New York; design for snow does not 
control the roof design in the other locations. 
This example covers the following aspects of a seismic design: 
1. Determining the equivalent lateral forces, 
2. Design of selected masonry shear walls for their in-plane loads, and 
3. Computation of drifts. 
See Chapter 7 of this volume for the design and detailing of untopped and topped precast diaphragms. 
9.2.2 Design Requirements 
9.2.2.1 Provisions Parameters 
The basic parameters affecting the design and detailing of the buildings are shown in Table 9.2-1.
Chapter 9, Masonry 
9-45 
[The 2003 Provisions have adopted the 2002 USGS probabilistic seismic hazard maps, and the maps have 
been added to the body of the 2003 Provisions as figures in Chapter 3 (instead of the previously used 
separate map package).] 
9.2.2.2 Structural Design Considerations 
The floors act as horizontal diaphragms and the walls parallel to the motion act as shear walls for all four 
buildings 
The system is categorized as a bearing wall system (Provisions Sec. 5.2.2[Sec. 4.3]). For Seismic Design 
Category D, the bearing wall system has a height limit of 160 ft and must comply with the requirements 
for special reinforced masonry shear walls (Provisions Sec. 11.11.5[Sec. 11.2.1.5]). Note that the 
structural system is one of uncoupled shear walls. Crossing beams over the interior doorways (their 
design is not included in this example) will need to continue to support the gravity loads from the deck 
slabs above during the earthquake, but are not designed to provide coupling between the shear walls. 
The building is symmetric and appears to be regular both in plan and elevation. It will be shown, 
however, that the building is actually torsionally irregular. Provisions Table 5.2.5 [Table 4.4-1] permits 
use of the equivalent lateral force (ELF) procedure in accordance with Provisions Sec. 5.4 [Sec. 5.2] for 
Birmingham 1 and New York City (Seismic Design Categories B and C). By the same table, the 
Category D buildings must use a dynamic analysis for design. For this particular building arrangement, 
the modal response spectrum analysis does not identify any particular effect of the torsional irregularity, 
as will be illustrated. 
Table 9.2-1 Design Parameters 
Design Parameter Value for 
Birmingham 1 
Value for 
Birmingham 2 
Value for 
New York 
Value for 
Los Angeles 
Ss (Map 1) [Figure 
3.3-1] 
0.3 0.3 0.4 1.5 
S1 (Map 2) [Figure 
3.3-2] 
0.12 0.12 0.09 0.6 
Site Class C E D D 
Fa 1.2 2.34 1.48 1 
Fv 1.68 3.44 2.4 1.5 
SMS = FaSs 0.36 0.7 0.59 1.5 
SM1 = FvS1 0.2 0.41 0.22 0.9 
SDS = 2/3 SMS 0.24 0.47 0.39 1 
SD1 = 2/3 SM1 0.13 0.28 0.14 0.6 
Seismic Design 
Category 
B D C D 
Masonry Wall Type Ordinary 
Reinforced 
Special 
Reinforced 
Intermediate 
Reinforced 
Special Reinforced 
Provisions Design Coefficients (Table 5.2.2 [4.3-1]) 
R 2.0 3.5 2.5 3.5
FEMA 451, NEHRP Recommended Provisions: Design Examples 
Design Parameter Value for 
Birmingham 1 
Value for 
Birmingham 2 
Value for 
New York 
Value for 
Los Angeles 
9-46 
O0 2.5 2.5 2.5 2.5 
Cd 1.75 3.5 2.25 3.5 
IBC Design Coefficients (presented for comparison with Provisions coefficients) 
R 2.5 5.0 3.5 5.0 
O0 2.5 2.5 2.5 2.5 
Cd 1.75 3.5 2.25 3.5 
The orthogonal effect (Provisions Sec. 5.2.5.2, Item a [Sec. 4.4.2]) applies to structures assigned to 
Seismic Design Categories C and D (all of the example buildings except for Birmingham 1). However, 
the arrangement of this building is not particularly susceptible to orthogonal effects. This is because the 
stresses developed under out-of-plane loading for short-height walls (story clear height is 8 ft) are low 
and, their contribution to orthogonal effects is minimal. 
The walls are all solid and there are no significant discontinuities, as defined by Provisions Sec. 5.2.6.2.3 
[Sec. 4.3.2.3], in the vertical elements of the seismic-force-resisting system. 
Ignoring the short walls at stairs and elevators, there are eight shear walls in each direction, therefore, the 
system appears to have adequate redundancy (Provisions Sec. 5.2.6.2.4 [Sec. 4.3.3]). The reliability 
factor, however, will be computed. [See Sec. 9.2.3.1 for changes to the reliability factor.] 
Tie and continuity requirements (Provisions Sec. 5.2.6.1.2 [Sec. 4.6]) must be addressed when detailing 
connections between floors and walls (see Chapter 7 of this volume). 
Nonstructural elements (Provisions Chapter 14 [Chapter 6]) are not considered in this example. 
Collector elements are required in the diaphragm for longitudinal response (Provisions Sec. 5.2.6.2.5 
[Sec. 4.6]). Rebar in the longitudinal direction, spliced into bond beams, will be used for this purpose 
(see Chapter 7 of this volume). 
Diaphragms must be designed for the required forces (Provisions Sec. 5.2.6.2.6 [Sec. 4.6]). 
The bearing walls must be designed for the required force perpendicular to their plane (Provisions Sec. 
5.2.6.2.7 [Sec. 4.6.1.3]). 
Each wall is a vertical cantilever; there are no coupling beams. The walls are classified as masonry 
cantilever shear wall structures in Provisions Table 5.2.8 [Table 4.5-1], which limits interstory drift to 
0.01 times the story height. Provisions Sec.11.5.4.1.1 also limits drift to 0.01 times the wall height for 
such a structure. 
[The deflection limits have been removed from Chapter 11 of the 2003 Provisions because they were 
redundant with the general deflection limits. Based on ACI 530 Sec. 1.13.3.2, the maximum drift for all 
masonry structures is 0.007 times the story height. Thus, there appears to be a conflict between ACI 530 
and 2003 Provisions Table 4.5-1.]
Chapter 9, Masonry 
9-47 
Vertical accelerations must be considered for the prestressed slabs in Seismic Design Category D 
(Provisions Sec. 5.2.6.4.3 [Sec. 4.6.3.1]); refer to Chapter 7 of this volume. The evaluation of such 
components involves the earthquake effect determined using Provisions Eq. 5.2.7-1 [4.2-1] and 5.2.7-2 
[4.2-1]. The important load is the vertical effect (-0.2SDSD), which reduces the effect of dead loads. 
Because the system is prestressed, application of this load might lead to tension where there would 
otherwise be no reinforcement. The reinforcement within the topping will control this effect. Refer to 
Sec. 7.1 of this volume for the design of precast, prestressed slabs and topping. 
Design, detailing, and structural component effects are presented in the chapters of the Provisions that are 
relevant to the materials used. 
9.2.3 Load Combinations 
The basic load combinations (Provisions Sec. 5.2.7 [Sec. 4.2.2]) are the same as those in ASCE 7 (and are 
similar to those in the IBC). The seismic load effect, E, is defined by Provisions Eq. 5.2.7-1 [4.2-1] and 
5.2.7-2 [4.2-2] as: 
E = .QE  0.2SDSD 
9.2.3.1 Reliability Factor 
Note that . is a multiplier on design force effects and applies only to the in-plane direction of the shear 
walls. For structures in Seismic Design Categories A, B and C, . = 1.0 (Provisions Sec. 5.2.4.1 [Sec. 
4.3.3.1]). For structures in Seismic Design Category D, . is determined per Provisions Sec. 5.2.4.2 [Sec. 
4.3.3.2]. 
For the transverse direction, ignoring accidental torsion: 
10 1 10 0.038 
x 8 33 
wall 
max 
story w 
r V
V l 
=.... ....... ....... ...... ...= 
and, 
2 20 2 20 3.03 
0.038 10,944 maxx x r A 
. = - = - =- 
Since the computed . < 1.0 use . = 1.0 for the transverse direction. Accidental torsion does not change 
enough to change this conclusion. 
maxx r 
Based on similar calculations for the longitudinal direction, . is determined to be 1.0. 
[The redundancy requirements have been substantially changed in the 2003 Provisions. For structures 
assigned to Seismic Design Categories B and C, . = 1.0 in all cases. For a shear wall building assigned to 
Seismic Design Category D, . = 1.0 as long as it can be shown that failure of a shear wall with height-tolength-
ratio greater than 1.0 would not result in more than a 33 percent reduction in story strength or 
create an extreme torsional irregularity. The intent is that the aspect ratio is based on story height, not 
total height. Therefore, the redundancy factor would not have to be investigated (. = 1.0) for the 
structure(s) assigned to Seismic Design Category D.]
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-48 
9.2.3.2 Combination of Load Effects 
The seismic load effect, E, determined for each of the buildings is: 
Birmingham 1 E = (1.0)QE  (0.2)(0.24)D = QE  0.05D 
Birmingham 2 E = (1.0)QE  (0.2)(0.47)D = QE  0.09D 
New York E = (1.0)QE  (0.2)(0.39)D = QE  0.08D 
Los Angeles E = (1.0)QE  (0.2)(1.00)D = QE  0.20D 
The applicable load combinations from ASCE 7 are: 
1.2D + 1.0E + 0.5L + 0.2S 
when the effects of gravity and seismic loads are additive and 
0.9D + 1.0E + 1.6H 
when the effects of gravity and seismic loads are counteractive. (H is the effect of lateral pressures of soil 
and water in soil.) 
Load effect H does not apply for this design, and the snow load effect, S, exceeds the minimum roof live 
load only at the building in New York. However, even for New York, the snow load effect is only used 
for combinations of gravity loading. Consideration of snow loads is not required in the effective seismic 
weight, W, of the structure when the design snow load does not exceed 30 psf (Provisions Sec. 5.3 [Sec. 
5.2.1]). 
The basic load combinations are combined with E as determined above, and the load combinations 
representing the extreme cases are: 
Birmingham 1 1.25D + QE +0.5L 
0.85D - QE 
Birmingham 2 1.29D + QE +0.5L 
0.81D - QE 
New York 1.28D + QE +0.5L +0.2S 
0.82D - QE 
Los Angeles 1.40D + QE +0.5L 
0.70D - QE 
These combinations are for the in-plane direction. Load combinations for the out-of-plane direction are 
similar except that the reliability coefficient (1.0 in all cases for in-plane loading) is not applicable. 
It is worth noting that there is an inconsistency in the treatment of snow loads combined with seismic 
loads. IBC Sec. 1605.3 clearly deletes the snow term from the ASD combinations where the design snow 
load does not exceed 30 psf. There is no similar provision for the strength load combinations in the IBC 
for reference standard, ASCE 7. 
[The strength design load combinations in the 2003 IBC do have a similar exemption for snow loads, but 
ASCE 7-02 load combinations do not.] 
9.2.4 Seismic Design for Birmingham 1
Chapter 9, Masonry 
9-49 
9.2.4.1 Birmingham 1 Weights 
Use 67 psf for 8-in.-thick, normal weight hollow core plank plus the nonmasonry partitions. This site is 
assigned to Seismic Design Category B, and the walls will be designed as ordinary reinforced masonry 
shear walls (Provisions Sec. 11.11.3 [Sec. 4.2.1.3]), which do not require prescriptive seismic 
reinforcement. However, both ACI 530 and IBC 2106.1.1.2 stipulate that ordinary reinforced masonry 
shear walls have a minimum of vertical #4 bars at 120 in. on center. [By reference to ACI 530, the 2003 
Provisions (and 2003 IBC) do have prescriptive seismic reinforcement requirements for ordinary 
reinforced masonry shear walls. Refer to ACI 530 Sec. 1.13.2.2.3.] Given the length of the walls, 
vertical reinforcement of #4 bars at 8 ft on center works well for detailing reasons and will be used here. 
For this example, 45 psf will be assumed for the 8-in.-thick lightweight CMU walls. The 45 psf value 
includes grouted cells and bond beams in the course just below the floor planks. 
Story weight, wi, is computed as follows: 
For the roof: 
Roof slab (plus roofing) = (67 psf) (152 ft)(72 ft) = 733 kips 
Walls = (45 psf)(589 ft)(8.67 ft/2) + (45 psf)(4)(36 ft)(2 ft) = 128 kips 
Total = 861 kips 
Note that there is a 2-ft-high masonry parapet on four walls and the total length of masonry wall, 
including the short walls, is 589 ft. 
For a typical floor: 
Slab (plus partitions) = 733 kips 
Walls = (45 psf)(589 ft)(8.67 ft) = 230 kips 
Total = 963 kips 
Total effective seismic weight, W = 861 + (4)(963) = 4,713 kips 
This total excludes the lower half of the first story walls, which do not contribute to seismic loads that are 
imposed on CMU shear walls. 
9.2.4.2 Birmingham 1 Base Shear Calculation 
The seismic response coefficient, Cs, is computed using Provisions Sec. 5.4.1.1 [Sec. 5.2.1.1]. 
Per Provisions Eq. 5.4.1.1-1 [Eq. 5.2-2]: 
0.24 0.12 
/ 21 
DS 
s 
C S
R I = = = 
The value of Cs need not be greater than Provisions Eq. 5.4.1.1-2 [Eq. 5.2-3]: 
( ) 
0.13 0.192 
( / ) 0.338 21 
D1 
s 
C S 
T R I 
= = = 
T is the fundamental period of the building approximated per Provisions Eq. 5.4.2.1-1[Eq. 5.2-6] as: 
= x=(0.02)(43.330.75 )=0.338 sec 
a r n T Ch
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-50 
where Cr = 0.02 and x = 0.75 are from Provisions Table 5.4.2.1 [Table 5.2-2]. 
The value for Cs is taken as 0.12 (the lesser of the two computed values). This value is still larger than 
the minimum specified in Provisions Eq. 5.4.1.1-3: 
Cs = 0.044ISDS = (0.044)(1.0)(0.24) = 0 0.0106 
[This minimum Cs value has been removed in the 2003 Provisions. In its place is a minimum Cs value 
for long-period structures, which is not applicable to this example.] 
The total seismic base shear is then calculated using Provisions Eq. 5.4.1 [Eq. 5.2-1]as: 
V = CsW = (0.12)(4,713) = 566 kips 
9.2.4.3 Birmingham 1 Vertical Distribution of Seismic Forces 
Provisions Sec. 5.4.4 [Sec. 5.2.3] stipulates the procedure for determining the portion of the total seismic 
load assigned to each floor level. The story force, Fx, is calculated using Provisions Eq. 5.4.3-1 [Eq. 5.2- 
10] and 5.4.3.-2 [Eq. 5.2-11], respectively, as: 
Fx = CvxV 
and 
1 
k 
x x 
vx n 
k 
i i 
i 
C w h 
wh 
= 
= 
S 
For T = 0.338 sec < 0.5 sec, k = 1.0. 
The seismic design shear in any story is determined from Provisions Eq. 5.4.4[Eq. 5.2-12]: 
Vx= Fi 
i=x 
nS 
The story overturning moment is computed from Provisions Eq. 5.4.5[Eq. 5.2-14]: 
( ) 
n 
x i i x 
i x 
M F h h 
= 
=S - 
The application of these equations for this building is shown in Table 9.2-2.
Chapter 9, Masonry 
9-51 
Contribution to weight 
concentrated at all stories. 
Dynamic response to 
round motion results 
in lateral load at all stories. 
Moments are 
from S Vh 
Weight of entire building 
above ground floor 
helps to resist moments. 
Contribution to weight 
concentrated at roof. 
Only upper half of walls 
out of plane contribute, 
but upper half of all walls 
used for convenience. 
Dynamic response to 
ground motion results 
in lateral load applied 
at roof. 
Moment at fifth floor 
M = V h 
P of roof slab plus 
entire height of wall 
helps to resist M. 
M5 P5 
Proof 
h
Vroof 
5 roof 
Figure 9.2-4 Location of moments due to story shears. 
Table 9.2-2 Birmingham 1 Seismic Forces and Moments by Level 
Level 
(x) 
wx 
(kips) 
hx 
(ft) 
wxhx
k 
(ft-kips) 
Cvx Fx 
(kips) 
Vx 
(kips) 
Mx 
(ft-kips) 
543213 
861 
963 
963 
963 
963 
4,715 
43.34 
34.67 
26.00 
17.33 
8.67 
37,310 
33,384 
25,038 
16,692 
8,346 
120,770 
0.3089 
0.2764 
0.2073 
0.1382 
0.0691 
1.0000 
175 
156 
117 
78 
39 
566 
1.8e+14 1,515 
4,385 
8,272 
12,836 
17,739 
1.0 kips = 4.45 kN, 1.0 ft = 0.3048 m. 
A note regarding locations of V and M: the vertical weight at the roof (5th level), which includes the 
upper half of the wall above the 5th floor (4th level), produces the shear V applied at the 5th level. That 
shear in turn produces the moment applied at the top of the 4th level. Resisting this moment is the rebar in 
the wall combined with the wall weight above the 4th level. Note that the story overturning moment is 
applied to the level below the level thatreceives the story shear. This is illustrated in Figure 9.2-4.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-52 
9.2.4.4 Birmingham 1 Horizontal Distribution of Forces 
The wall lengths are shown in Figure 9.2-3. The initial grouting pattern is basically the same for walls A, 
B, and C. Because of a low relative stiffness, the effects Walls D, E, and F are ignored in this analysis. 
Walls A, B, and C are so nearly the same length that their stiffnesses will be assumed to be the same for 
this example. 
Torsion is considered according to Provisions Sec. 5.4.4[Sec. 5.2.4]. For a symmetric plan, as in this 
example, the only torsion to be considered is the accidental torsion, Mta, caused by an assumed 
eccentricity of the mass each way from its actual location by a distance equal to 5 percent of the 
dimension of the structure perpendicular to the direction of the applied loads. 
Dynamic amplification of the torsion need not be considered for Seismic Design Category B per 
Provisions Sec. 5.4.4.3 [Sec. 5.2.4.3]. 
For this example, the building will be analyzed in the transverse direction only. The evaluation of Wall D 
is selected for this example. The rigid diaphragm distributes the lateral forces into walls in both 
directions. Two components of force must be considered: direct shear and shear induced by torsion. 
The direct shear force carried by Wall D is one-eighth of the total story shear (eight equal walls). The 
torsional moment per Provisions Sec. 5.4.4.2 [Sec. 5.2.4.2] is: 
Mta = 0.05bVx =(0.05)(152 ft)Vx = 7.6Vx 
The torsional force per wall, Vt, is: 
2 
t 
t 
V M Kd 
Kd 
= S 
where K is the stiffness (rigidity) of each wall. 
Because all the walls in this example are assumed to be equally stiff: 
t t 2 
V M d
d 
. . 
= . . 
..S .. 
where d is the distance from each wall to the center of twisting. 
3d2 = 4(36)2 + 4(12)2 + 4(36)2 + 4(12)2 = 11,520 
The maximum torsional shear force in Wall D, therefore is: 
Vt = 7.6 V(36/11,520) = 0.0238V 
Total shear in Wall D is: 
=0.125 +0.0238 =0.149 totV V V V 
The total story shear and overturning moment may now be distributed to Wall D and the wall proportions 
checked. The wall capacity will be checked before considering deflections.
Chapter 9, Masonry 
9-53 
9.2.4.5 Birmingham 1 Transverse Wall (Wall D) 
The strength or limit state design concept is used in the Provisions. This method was introduced in the 
2002 edition of ACI 530, the basic reference standard for masonry design. Because strength design was 
not in prior editions of ACI 530, strength design of masonry as defined in the Provisions is illustrated 
here. 
[The 2003 Provisions adopts by reference the ACI 530-02 provisions for strength design in masonry, and 
the previous strength design section has been removed. This adoption does not result in significant 
technical changes, and the references to the corresponding sections in ACI 530 are noted in the following 
sections.] 
9.2.4.5.1 Birmingham 1 Shear Strength 
Provisions Sec. 11.7.2 [ACI 530, Sec. 3.1.3] states that the ultimate shear loads must be compared to the 
design shear strength per Provisions Eq. 11.7.2.1: 
Vu # fVn 
The strength reduction factor, f, is 0.8 (Provisions Table 11.5.3, ACI 530 [See 3.1.4.3]). The design 
shear strength, fVn, must exceed the shear corresponding to the development of 1.25 times the nominal 
flexural strength of the member but need not exceed 2.5 times Vu (Provisions Sec. 11.7.2.2 [ACI 530, Sec. 
3.1.3]). The nominal shear strength, Vn, is (Provisions Eq. 11.7.3.1-1 [ACI 530, Eq. 3-18]): 
Vn = Vm + Vs 
The shear strength provided by masonry is (Provisions Eq. 11.7.3.2 [ACI 530, Eq. 3-21]): 
m4.0 1.75 nm0.25 
V MAf P 
Vd 
=... - ... ...... ' + 
For grouted cells at 8 ft on center: 
An = (2  1.25 in.  32.67 ft x 12 in.) + (8  5.13 in.2  5 cells) = 1,185 in.2 
The shear strength provided by reinforcement is given by Provisions Eq. 11.7.3.3 [ACI 530, Sec. 
3.2.4.1.2.2] as: 
= 0.5.. .. 
. . 
v 
s y v 
V AFd 
s 
The wall will have a bond beam with two #4 bars at each story to bear the precast floor planks and wire 
joint reinforcement at alternating courses. Common joint reinforcement with 9 gauge wires at each face 
shell will be used; each wire has a cross-sectional area of 0.017 in.2 With six courses of joint 
reinforcement and two #4 bars, the total area per story is 0.60 in.2 or 0.07 in.2/ft. 
Vs = 0.5(0.07 in.2/ft.)(60 ksi)(32.67 ft.) = 68.3 kips 
The maximum nominal shear strength of the member (Wall D in this case) for M/Vdv > 1.00 (the 
Provisions has a typographical error for the inequality sign) is given by Provisions Eq. 11.7.3.1-3 [ACI 
530, Eq. 3-22]:
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-54 
(max) = 4 ' n mn V fA 
The coefficient 4 becomes 6 for M/Vdv < 0.25. Interpolation between yields the following: 
V 
M 
Vd 
f A N 
V 
m n (max)= - 
.
. .
.
. .
(6.67 2.67 ) ' 
The shear strength of Wall D, based on the equations listed above, is summarized in Table 9.2-3. Note 
that Vx and Mx in this table are values from Table 9.2-2 multiplied by 0.149 (which represents the portion 
of direct and torsional shear assigned to Wall D). P is the dead load of the roof or floor times the 
tributary area for Wall D. (Note that there is a small load from the floor plank parallel to the wall.) 
Table 9.2-3 Shear Strength Calculations for Birmingham 1 Wall D 
Story Vx 
(kips) 
Mx 
(ft-kips) 
Mx/Vxd 2.5 Vx 
(kips) 
P 
(kips) 
fVm 
(kips) 
fVs 
(kips) 
fVn 
(kips) 
fVn max 
(kips) 
5 26 225 0.265 65.0 41 158.1 54.6 212.7 252.7 
4 49.3 652 0.405 123.3 89 157.3 54.6 211.9 236.9 
3 66.7 1230 0.564 166.8 137 155.1 54.6 209.7 218.8 
2 78.4 1910 0.746 196.0 184 151.0 54.6 205.6 198.3 
1 84.2 2640 0.960 210.5 232 144.8 54.6 199.4 174.1 
1.0 kip = 4.45 kN, 1.0 ft-kip = 1.36 kN-m. 
VU exceeds both fVn and fVn max at the first story. It would be feasible to add grouted cells in the first 
story to remedy the deficiency. However, it will be shown following the flexural design that the shear to 
develop 1.25 times the flexural capacity is 1.94(84.2 kips) = 163 kips, which is OK. 
9.2.4.5.2 Birmingham 1 Axial and Flexural Strength 
All the walls in this example are bearing shear walls since they support vertical loads as well as lateral 
forces. In-plane calculations include: 
1. Strength check and 
2. Ductility check 
9.2.4.5.2.1 Strength check 
The wall demands, using the load combinations determined previously, are presented in Table 9.2-4 for 
Wall D. In the table, Load Combination 1 is 1.25D + QE + 0.5L and Load Combination 2 is 0.85D + QE. 
Table 9.2-4 Demands for Birmingham 1 Wall D 
Load Combination 1 Load Combination 2 
Level PD 
(kips) 
PL 
(kips) 
Pu 
(kips) 
Mu 
(ft-kips) 
Pu 
(kips) 
Mu 
(ft-kips) 
54321 4.2e+12 8172534 5.111518e+13 2.2565e+17 3.576116e+12 2.256521e+17 
1.0 kip = 4.45 kN, 1.0 ft-kip = 1.36 kN-m. 
Strength at the bottom story (where P, V, and M are the greatest) will be examined. (For a real design, all 
levels should be examined). The strength design will consider Load Combination 2 from Table 9.2-4 to 
be the governing case because it has the same lateral load as Load Combination 1 but with lower values 
of axial force.
Chapter 9, Masonry 
9-55 
For the base of the shear walls: 
Pumin = 197 kips plus factored weight of lower  of 1st story wall = 197 + (0.85)(6.4) = 202 kips 
= 307 + (1.25)(6.4) = 315 kips Pumax 
Mu = 2,640 ft-kips 
Try one #4 bars in each end cell and a #4 bar at 8 ft on center for the interior cells. A fPn - fMn curve, 
representing the wall strength envelope, will be developed and used to evaluate Pu and Mu determined 
above. Three cases will be analyzed and their results will be used in plotting the fPn - fMn curve. 
In accordance with Provisions Sec. 11.6.2.1 [ACI 530, Sec. 3.2.2], the strength of the section is reached 
as the compressive strains in masonry reach their maximum usable value of 0.0025 for CMU. The force 
equilibrium in the section is attained by assuming an equivalent rectangular stress block of 0.8f!m over an 
effective depth of 0.8c, where c is the distance of the neutral axis from the fibers of maximum 
compressive strain. Stress in all steel bars is taken into account. The strains in the bars are proportional 
to their distance from the neutral axis. For strains above yield, the stress is independent of strain and is 
taken as equal to the specified yield strength Fy. See to Figure 9.2-5 for strains and stresses for all three 
cases selected. 
Case 1 (P = 0) 
Assume all tension bars yield (which can be verified later): 
Ts1 = (0.20 in.2)(60 ksi) = 12.0 kips 
Ts2 = (0.20 in.2)(60 ksi) = 12.0 kips each 
Because the neutral axis is close to the compression end of the wall, compression steel, Cs1, will be 
neglected (it would make little difference anyway) for Case 1: 
SFy = 0: 
Cm = ST 
Cm =(4)(12.0) = 48.0 kips 
The compression block will be entirely within the first grouted cell: 
Cm = 0.8 fmab 
48.0 = (0.8)(2.0 ksi)a(7.625 in) 
a = 3.9 in. = 0.33 ft 
c = a/0.8 = 0.33/0.8 = 0.41 ft 
Thus, the neutral axis is determined to be 0.41 ft from the compression end on the wall, which is within 
the first grouted cell: 
SMcl = 0: (The math will be a little easier if moments are taken about the wall centerline.) 
Mn = (16.33-0.33/2 ft)Cm + (16.00 ft) Ts1 + (0.00 ft)STs2 + (0.00 ft)Pn 
Mn = (16.17)(48.0) + (16.00)(12) + 0 + 0 = 968 ft-kips 
fMn = (0.85)(968) = 823 ft-kips
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-56 
y = 0.00207 
Ts1 
N.A. 
= 0.00207 
y 
Balanced 
Case 
P = 0 Case 
a = 14.15' 
c = 17.69' 
Center 
Line 
Ts1 
STs2 
Cm shell 
Cm cells 
a = 0.33' 
c = 0.41' 
m= 0.0025 
0.8f'm 
Cm tot 
32.66' 
16.33' 16.00' 0.33' 
#4 at 8'-0" o.c. 
#4 at 
end 
m= 0.0025 
0.8f'm 
Cs1 
Ts2 Ts2 Ts2 
*Ts2 
N.A. 
*C s2 
N.A. 
y = 0.00207 
Ts1 
ST 
m= 0.0025 
0.8f'm 
Cm cell 
Ts2 Ts2 
Intermediate 
Case 
Cm shell 
y = 0.00207 
Cs1 
Ts 2 * 
a = 6.40' 
c = 8.00' 
8.00' 
M 
P 
(typical) 
fE
y = 
16.33' 16.00' 
Figure 9.2-5 Strength of Birmingham 1 Wall D (1.0 ft = 0.3048 m). Strain diagram superimposed on strength 
diagram for the three cases. The low force in the reinforcement is neglected in the calculations. 
Chapter 9, Masonry 
9-57 
To summarize, Case 1: 
fPn = 0 kips 
fMn = 823 ft-kips 
Case 2 (Intermediate case between P = 0 and Pbal) 
Let c = 8.00 ft.(this is an arbitrary selection). Thus, the neutral axis is defined at 8 ft from the 
compression end of the wall: 
a = 0.8c = (0.8)(8.00) = 6.40 ft 
Cm shells = 0.8fm(2 shells)(1.25 in. / shell)(6.40 ft. (12 in./ft) = 307.2 kips 
Cm cells = 0.8 fm(41 in.2) = 65.6 kips 
Cm tot = Cm shells + Cm cells = 307.2 + 65.6 = 373 kips 
Cs1 = (0.20 in.2)(60 ksi) = 12 kips (Compression steel is included in this case) 
Ts1 = (0.20 in.2)(60 ksi) = 12 kips 
Ts2 = (0.20 in.2)(60 ksi) = 12 kips each 
Some authorities would not consider the compression resistance of reinforcing steel that is not enclosed 
within ties. The Provisions clearly allows inclusion of compression in the reinforcement. 
SFy = 0: 
Cm tot + Cs1 = Pn + Ts1 + STs2 
373 + 12 = Pn + (3)(12.0) 
Pn = 349 kips 
fPn = (0.85)(349) = 297 kips 
SMcl = 0: 
Mn = (13.13 ft)Cm shell + (16.00 ft)(Cm cell + Cs1) + (16.00 ft)Ts1 + (8.00 ft)Ts2 
Mn = (13.13)(307.2) + (16.00)(65.6 + 12) + (16.00)(12.0) + (8.00 ft)(12.0) = 5,563 ft-kips 
fMn = (0.85)(5,563) = 4,729 ft-kips 
To summarize Case 2: 
fPn = 297 kips 
fMn = 4,729 ft-kips 
Case 3 (Balanced case) 
In this case, Ts1 just reaches its yield stress: 
0.0025 (32.33 ft) = 17.69 ft 
(0.0025 0.00207) 
c 
. . 
=.. + .. 
a = 0.8c = (0.8)(17.69) = 14.15 ft 
Cm shells = 0.8fm(2 shells)(1.25 in. / shell)(14.15 ft.) (12 in./ft) = 679.2 kips 
Cm cells =0.8fm(2 cells)(41 in.2/cell) = 131.2 kips 
Cm tot = Cm shells + Cm cells = 810.4 kips 
Cs1 = (0.20 in.2)(60 ksi) = 12.0 kips
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-58
Ts1 = (0.20 in.2)(60 ksi) = 12.0 kips 
Cs2 and Ts2 are neglected because they are small, constituting less than 2 percent of the total Pn. 
SFy = 0: 
Pn = SC - ST 
Pn = Cm tot + Cs1 - Ts1 = 810.4 + 12.0 - 12.0 = 810.4 kips 
fPn = (0.85)(810.4) = 689 kips 
SMcl = 0: 
Mn = 9.26 Cm shells + ((16 + 8)/2) Cm cells + 16 Cs1 + 8 Ts2 + 16 Ts1 
Mn = (9.26)(679.2) + (12.0)(131.2) + (16.00)(12.0) + (ignore small Ts2) + (16.0)(12.0) = 8,248 kips 
fMn = (0.85)(8,248) = 7,011 ft-kips 
To summarize Case 3: 
fPn = 689 kips 
fMn = 7,011 ft-kips 
Using the results from the three cases above, the fPn - fMn curve shown in Figure 9.2-6 is plotted. 
Although the portion of the fPn - fMn curve above the balanced failure point could be determined, it is 
not necessary here. Thus, only the portion of the curve below the balance point will be examined. This is 
the region of high moment capacity. 
Similar to reinforced concrete beam-columns, in-plane compression failure of the cantilevered shear wall 
will occur if Pu > Pbal, and tension failure will occur if Pu < Pbal. A ductile failure mode is essential to the 
design, so the portion of the curve above the balance point is not useable. 
As can be seen, the points for Pu min , Mu and Pu max , are within the fPn - fMn envelope; thus, the strength 
design is acceptable with the minimum reinforcement. Figure 9.2-6 shows two schemes for determining 
the design flexural resistance for a given axial load. One interpolates along the straight line between pure 
bending and the balanced load. The second makes use of intermediate points for interpolation. This 
particular example illustrates that there can be a significant difference in the interpolated moment capacity 
between the two schemes for axial loads midway between the balanced load and pure bending. 
For the purpose of shear design, the value of fMN at the design axial load is necessary. Interpolating 
between the intermediate point and the P = 0 point for P = 202 kips yields fMN = 3,480 ft-kip. Thus, the 
factor on shear to represent development of 125 percent of flexural capacity is: 
1.25 / 1.25 3480/ 0.85 1.94 
2640 
N
U 
M
M 
f f
= =
Chapter 9, Masonry 
9-59 
9.2.4.5.2.2 Ductility check 
Provisions Sec.11.6.2.2 [ACI 530, Sec. 3.2.3.5] requires that the critical strain condition correspond to a 
strain in the extreme tension reinforcement equal to 5 times the strain associated with Fy. Note that this 
calculation uses unfactored gravity axial loads (Provisions Sec.11.6.2.2 [ACI 530, Sec. 3.2.3.5]). Refer to 
Figure 9.2-5 and the following calculations which illustrate this using loads at the bottom story (highest 
axial loads). Calculations for other stories are not presented in this example. 
5,000 ft-kips 10,000 ft-kips 
500 kips 
1,000 kips 
P max = 
315 kips 
P min = 
202 kips 
fP 
fM 
M u = 2640 ft-kips 
Intermediate 
(4729 ft-kips, 297 kips) 
P = 0 
(823 ft-kips, 0 kips) 
Actual fP - fM curve 
Balance 
(7011 ft-kips, 689 kips) 
Simplified fP - fM curve 
n 
n 
fPn 
fMn 
u
u 
n 
n 
n 
n 
Figure 9.2-6 fP11 - fM11 diagram for Birmingham 1 Wall D (1.0 kip = 4.45 kN, 1.0 ft-kip = 1.36 kN-m).
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-60 
m = 0.0025 
N.A. 
s = 5 y 
= 0.0103 
6.32' 
P 
Cm 
0.8f 'm = 1.6 ksi 
Cs1 
5.99' 6.51' 
Ts3 Ts2 Ts1 
Ts4 
(5) reinforced cells at 8'-0" o.c. 
1.25f y 
= 75 ksi 
5f y = 300 ksi 
26.02' 
32'-8" 
3.79' 
0.33' 
5.06' 1.26' 
2.02' 
10.02' 
18.02' 
26.02' 
0.33' 
f y = 60 ksi 
72.86 ksi 
23.29 ksi 
= 5 ( 60 
29,000 ) 
c 
a 
Figure 9.2-7 Ductility check for Birmingham 1 Wall D (1.0 ft = 0.3048 m, 1.0 ksi = 6.89 MPa). 
For Level 1 (bottom story), the unfactored axial loads are: 
P = 232 kips + weight of half of first story wall = 232 + 6.4 = 238.4 kips 
Refer to Figure 9.2-7: 
Cm = 0.8 fm(ab + Acell) = (1.6 ksi)[(5.06 ft. x 12 in./ft.)(2.5 in.) + 41 in.2] = 308.5 kips (same as above) 
Cs1 = FyAs = (60 ksi)(0.20 in.2) = 12.0 kips 
Ts1 = Ts2 = Ts3 = (1.25  60 ksi)(0.20 in.2) = 15 kips
Chapter 9, Masonry 
9-61 
Ts4 = (23.29 ksi)(0.20 in.2) = 4.6 kips 
3 C > 3P + T 
Cm + Cs1 > P + Ts1 + Ts2 + Ts3 + Ts4 
308.5 + 12.0 > 238.4 + 15 + 15 + 15 + 4.6 
320.5 kips > 288 kips OK 
There is more compression capacity than required so ductile failure condition governs. 
[The ductility (maximum reinforcement) requirements in ACI 530 are similar to those in the 2000 
Provisions. However, the 2003 Provisions also modify some of the ACI 530 requirements, including 
critical strain in extreme tensile reinforcement (4 times yield) and axial force to consider when performing 
the ductility check (factored loads).] 
9.2.4.6 Birmingham 1 Deflections 
The calculations for deflection involve many variables and assumptions, and it must be recognized that 
any calculation of deflection is approximate at best. 
Deflections are to be calculated and compared with the prescribed limits set forth by Provisions Table 
5.2.8. Deformation requirements for masonry structures are given in Provisions Sec. 11.5.4 [Table 4.5-1]. 
The following procedure will be used for calculating deflections: 
1. For each story, compare Mx (from Table 9.2-3) to Mcr = S(fr + Pu min / A) to determine if wall will 
crack. 
2. If Mcr < Mx, then use cracked moment of inertia and Provisions Eq. 11.5.4.3. 
3. If Mcr > Mx, then use In = Ig for moment of inertia of wall. 
4. Compute deflection for each level. 
Other approximations can be used such as the cubic interpolation formula given in Provisions 11.5.4.3, 
but that equation was derived for reinforced concrete members acting as single span beams, not 
cantilevers. In the authors opinion, all these approximations pale in comparison to the approximation of 
nonlinear deformation using Cd. 
For the Birmingham 1 building: 
be = effective masonry wall width 
be = [(2  1.25 in.)(32.67 ft  12) + (5 cells)(41 in.2/cell)]/(32.67 ft  12) = 3.02 in. 
S = be l2/6 = (3.02)(32.67  12)2/6 = 77,434 in.3 
fr = 0.250 ksi 
A = be l = (3.02 in.)(32.67 ft  12) = 1,185 in.2 
Pu is calculated using 1.00D (see Table 9.2-4). 1.00D is considered to be a reasonable value for axial load 
for this admittedly approximate analysis. If greater conservatism is desired, Pu could be calculated using 
0.85D. 
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-62 
V5 
V4 
V3 
V2 
V1 
d 5 
d 4 
d 3 
d 2 
d1 
l 
be 
Figure 9.2-8 Shear wall deflections. 
The results are shown in Table 9.2-5. 
Table 9.2-5 Birmingham 1 Cracked Wall Determination 
Level Pumin 
(kips) 
Mcr 
(ft-kips) 
Mu 
(ft-kips) 
Status 
54321 
41 
89 
137 
185 
232 
1836 
2098 
2359 
2621 
2877 
225 
652 
1230 
1910 
2640 
uncracked 
uncracked 
uncracked 
uncracked 
uncracked 
1.0 kip = 4.45 kN, 1.0 ft-kip = 1.36 kN-m. 
For uncracked walls: 
In = Ig = bl3/12 = (3.02 in.)(32.67  12)3 /12 = 1.52 x 107 in.4 
The calculation of d will consider flexural and shear deflections. For the final determination of 
deflection, a RISA-2D analysis was made. The result is summarized Table 9.2-6 below. Figure 9.2-8 
illustrates the deflected shape of the wall.
Chapter 9, Masonry 
9-63 
Table 9.2-6 Deflections, Birmingham 1 
Level F 
(kips) 
Ieff 
(in.4) 
dflexural 
(in.) 
dshear 
(in.) 
dtotal 
(in.) 
Cd dtotal 
(in.) 
. 
(in.) 
54321 26.0 
23.2 
17.4 
11.7 
5.8 
1.52  107 
1.52  107 
1.52  107 
1.52  107 
1.52  107 
0.108 
0.078 
0.049 
0.024 
0.007 
0.054 
0.049 
0.041 
0.028 
0.015 
0.162 
0.128 
0.090 
0.052 
0.021 
0.284 
0.223 
0.157 
0.091 
0.037 
0.061 
0.066 
0.066 
0.054 
0.037 
1.0 kip = 4.45 kN, 1.0 in. = 25.4 mm. 
The maximum story drift occurs at Level 4 (Provisions Table 5.2.8 [Table 4.5-1]): 
[The specific procedures for computing deflection of shear walls have been removed from the 2003 
Provisions. ACI 530 does not contain the corresponding provisions in the text, however, the commentary 
contains a discussion and equations that are similar to the procedures in the 2000 Provisions. However, 
as indicated previously, there is a potential conflict between the drift limits in 2003 Provisions Table 4.5- 
1 and ACI 530 Sec. 1.13.3.2.] 
. = 0.066 in. < 1.04 in. = 0.01hn OK 
9.2.4.7 Birmingham 1 Out-of-Plane Forces 
Provisions Sec 5.2.6.2.7 [Sec. 4-6.1.3] requires that the bearing walls be designed for out-of-plane loads 
determined as follows: 
w = 0.40SDSWc $ 0.1Wc 
w = (0.40)(0.24)(45 psf) = 4.3 psf < 4.5 psf = 0.1Wc 
The calculated seismic load, w = 4.5 psf, is much less than wind pressure for exterior walls and is also 
less than the 5 psf required by IBC Sec. 1607.13 for interior walls. Thus, seismic loads do not govern the 
design of any of the walls for loading in the out-of-plane direction. 
9.2.4.8 Birmingham 1 Orthogonal Effects 
Orthogonal effects do not have to be considered for Seismic Design Category B (Provisions Sec. 5.2.5.2.1 
[Sec. 4.4.2.1]). 
This completes the design of Transverse Wall D. 
9.2.4.9 Summary of Design for Birmingham 1 Wall D 
8 in. CMU 
f!m = 2,000 psi 
Reinforcement: 
One vertical #4 bar at wall end cells 
Vertical #4 bars at 8 ft on center at intermediate cells throughout 
Bond beam with two - #4 bars at each story just below the floor and roof slabs
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-64 
Horizontal joint reinforcement at 16 inches 
Grout at cells with reinforcement and at bond beams. 
9.2.5 Seismic Design for New York City 
This example focuses on differences from the design for the Birmingham 1 site. 
9.2.5.1 New York City Weights 
As before, use 67 psf for 8-in.-thick normal weight hollow core plank plus the nonmasonry partitions. 
This site is assigned to Seismic Design Category C, and the walls will be designed as intermediate 
reinforced masonry shear walls (Provisions Sec. 11.11.4 [Sec. 11.2.1.4] and Sec. 11.3.7 [Sec. 11.2.1.4]), 
which requires prescriptive seismic reinforcement (Provisions Sec. 11.3.7.3 [ACI 530, Sec. 1.13.2.2.4]). 
Intermediate reinforced masonry shear walls have a minimum of #4 bars at 4 ft on center. For this 
example, 48 psf will be assumed for the 8-in. CMU walls. The 48 psf value includes grouted cells and 
bond beams in the course just below the floor planks. In Seismic Design Category C, more of the 
regularity requirement must be checked. It will be shown that this symmetric building with a seemingly 
well distributed lateral force system is torsionally irregular by the Provisions. 
Story weight, wi: 
Roof 
Roof slab (plus roofing) = (67 psf) (152 ft)(72 ft) = 733 kips 
Walls = (48 psf)(589 ft)(8.67 ft/2) + (48 psf)(4)(36 ft)(2 ft) = 136 kips 
Total = 869 kips 
There is a 2-ft high masonry parapet on four walls and the total length of masonry wall is 589 ft. 
Typical floor 
Slab (plus partitions) = 733 kips 
Walls = (48 psf)(589 ft)(8.67 ft) = 245 kips 
Total = 978 kips 
Total effective seismic weight, W = 869 + (4)(978) = 4781 kips 
This total excludes the lower half of the first story walls, which do not contribute to seismic loads that are 
imposed on CMU shear walls. 
9.2.5.2 New York City Base Shear Calculation 
The seismic response coefficient, Cs, is computed from Provisions Sec. 5.4.1.1 [Sec. 5.2-1.1]: 
0.39 0.156 
/ 2.51 
DS 
s 
C S
R I = = = 
The value of Cs need not be greater than:
Chapter 9, Masonry 
9-65 
( ) 
0.14 0.166 
( / ) 0.338 2.5 1 
D1 
s 
C S 
T R I = = = 
where T is the same as found in Sec. 9.2.4.2. 
The value for Cs is taken as 0.156 (the lesser of the two computed values). This value is still larger than 
the minimum specified in Provisions Eq. 5.3.2.1-3. Using Provisions Eq. 5.4.1.1-3: 
Cs = 0.044SD1I = (0.044)(0.14)(1) = 0.00616 
[This minimum Cs value has been removed in the 2003 Provisions. In its place is a minimum Cs value 
for long-period structures, which is not applicable to this example.] 
The total seismic base shear is then calculated using Provisions Eq. 5.4.1 [Eq. 5.2-1]: 
V = CsW = (0.156)(4,781) = 746 kips 
9.2.5.3 New York City Vertical Distribution of Seismic Forces 
The vertical distribution of seismic forces is determined in accordance with Provisions Sec. 5.4.4 [Sec. 
5.2.3], which was described in Sec. 9.2.4.3. Note that for Provisions Eq. 5.4.3-2 [Eq. 5.2-11], k = 1.0 
since T = 0.338 sec (similar to the Birmingham 1 building). 
The application of the Provisions equations for this building is shown in Table 9.2-7: 
Table 9.2-7 New York City Seismic Forces and Moments by Level 
Level 
(x) 
wx 
(kips) 
hx 
(ft) 
wxhx
k 
(ft-kips) 
Cvx Fx 
(kips) 
VX 
(kips) 
Mx 
(ft-kips) 
543213 
869 
978 
978 
978 
978 
4,781 
43.34 
34.67 
26.00 
17.33 
8.67 
37,657 
33,904 
25,428 
16,949 
8,476 
122,414 
0.3076 
0.2770 
0.2077 
0.1385 
0.0692 
1.000 
229 
207 
155 
103 
52 
746 
2.3e+14 1,985 
5,765 
10,889 
16,907 
23,370 
1.0 kip = 4.45 kN, 1.0 ft = 0.3048 m, 1.0 ft-kip = 1.36 kN-m. 
9.2.5.4 New York City Horizontal Distribution of Forces 
The initial distribution is the same as Birmingham 1. See Sec. 9.2.4.4 and Figure 9.2-3 for wall 
designations. 
Total shear in Wall Type D: 
0.125 0.0238 0.149 totV= V+ V= V 
Provisions Sec.5.4.4.3 [Sec. 4.3.2.2] requires a check of torsional irregularity using the ratio of maximum 
displacement at the end of the structure, including accidental torsion, to the average displacement of the 
two ends of the building. For this simple and symmetric structure, the actual displacements do not have 
to be computed to find the ratio. Relying on symmetry and the assumption of rigid diaphragm behavior
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-66 
used to distribute the forces, the ratio of the maximum displacement of Wall D to the average 
displacement of the floor will be the same as the ratio of the wall shears with and without accidental 
torsion: 
0.149 1.190 
0.125 
max 
ave
F V 
F V 
= = 
This can be extrapolated to the end of the rigid diaphragm therefore: 
1 0.190 152 / 2 1.402 
36 
max 
ave 
d
d 
= + .. ..= 
. . 
Provisions Table 5.2.3.2 [Table 4.3-2] defines a building as having a Torsional Irregularity if this ratio 
exceeds 1.2 and as having an Extreme Torsional Irregularity if this ratio exceeds 1.4. Thus, an 
important result of the Seismic Design Category C classification is that the total torsion must be amplified 
by the factor: 
2 2 1.402 1.365 
1.2 1.2 
max 
x 
ave 
A 
d 
d 
=.. .. =.. .. = 
. . . . 
Therefore, the portion of the base shear for design of Wall D is now: 
0.125 1.365(0.0238 ) 0.158 DV= V+ V= V 
which is a 5.8 percent increase from the fraction before considering torsional irregularity. 
The total story shear and overturning moment may now be distributed to Wall D and the wall proportions 
checked. The wall capacity will be checked before considering deflections. 
9.2.5.5 New York City Transverse Wall D 
The strength or limit state design concept is used in the Provisions. 
9.2.5.5.1 New York City Shear Strength 
Similar to the design for Birmingham 1, the shear wall design is governed by: 
Vu=fVn 
Vn = Vm + Vs 
V fAdepending on M/Vd n mn max =4 to 6 ' 
m= 4-1.75( )nm0.25 
M 
V Af P 
Vd 
' + . . 
.. .. 
0.5 v 
s y v 
V Afd 
s 
= .. .. 
. . 
where
Chapter 9, Masonry 
9-67 
An = (2  1.25 in.  32.67 ft  12 in.) + (41 in.2  9 cells) = 1,349 in.2 
The shear strength of each Wall D, based on the aforementioned formulas and the strength reduction 
factor of f = 0.8 for shear from Provisions Table 11.5.3 [ACI 530, Sec. 3.1.4.3], is summarized in Table 
9.2-8. Note that Vx and Mx in this table are values from Table 9.2-7 multiplied by 0.158 (representing the 
portion of direct and indirect shear assigned to Wall D), and P is the dead load of the roof or floor times 
the tributary area for Wall D.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-68 
Table 9.2-8 New York City Shear Strength Calculation for Wall D 
Story Vx 
(kips) 
Mx 
(ft-kips) 
Mx/Vxd 2.5 Vx 
(kips) 
P 
(kips) 
fVm 
(kips) 
fVs 
(kips) 
fVn 
(kips) 
fVn max 
(kips) 
5 36.1 313 0.265 90.3 42 179.0 54.6 233.6 287.6 
4 68.7 908 0.405 171.8 90 176.9 54.6 231.5 269.7 
3 93.1 1715 0.564 232.8 139 173.2 54.6 227.8 249.2 
2 109.3 2663 0.746 273.3 188 167.7 54.6 222.3 225.8 
1 117.5 3680 0.959 293.8 236 159.3 54.6 213.9 198.4 
1.0 kip = 4.45 kN, 1.0 ft-kip = 1.36 kN-m. 
Vu exceeds fVn at the lower three stories. As will be shown at the conclusion of the design for flexure, the 
factor to achieve 125 percent of the nominal flexural capacity is 1.58. This results in Vu being less than 
fVn at all stories. If that were not the case, it would be necessary to grout more cells to increase An or to 
increase fm. 
9.2.5.5.2 New York City Axial and Flexural Strength 
The walls in this example are all load-bearing shear walls because they support vertical loads as well as 
lateral forces. In-plane calculations include: 
1. Strength check and 
2. Ductility check. 
9.2.5.5.2.1 Strength check 
Wall demands, using load combinations determined previously, are presented in Table 9.2-9 for Wall D. 
In the table, Load Combination 1 is 1.28D + QE + 0.5L and Load Combination 2 is 0.82D + QE. 
Table 9.2-9 Demands for New York City Wall D 
Load Combination 1 Load Combination 2 
Level PD 
(kips) 
PL 
(kips) 
Pu 
(kips) 
Mu 
(ft-kips) 
Pu 
(kips) 
Mu 
(ft-kips) 
54321 
42 
90 
139 
188 
236 
08 
17 
25 
34 
54 
119 
186 
253 
319 
313 
908 
1715 
2663 
3680 
34 
74 
114 
154 
194 
313 
908 
1715 
2663 
3680 
1.0 kip = 4.45 kN, 1.0 ft-kip = 1.36 kN-m. 
As in Sec. 9.2.4.5.2, strength at the bottom story (where P, V, and M are the greatest) will be examined. 
The strength design will consider Load Combination 2 from Table 9.2-9 to be the governing case because 
it has the same lateral load as Load Combination 1 but with lower values of axial force. Refer to Fig. 9.2- 
9 for notation and dimensions.
Chapter 9, Masonry 
9-69 
y = 0.00207 
Ts1 
N.A. 
y = 0.00207 
Balanced 
Case 
P = 0 Case 
a = 14.15' 
c = 17.69' 
Center 
Line 
Ts1 
S Ts2 
Cm shell 
Cm cells 
a = 0.66' 
c = 0.82' 
Cs1 
y = 0.00207 
m= 0.0025 
0.8f'm 
Cm tot 
32.67' 
16.33' 16.00' 0.33' 
#4 at 4.00' 
on center 
(1) #4 
y = 0.00207 
m= 0.0025 
0.8f'm 
Cs1 
M 
P 
N.A. 
Figure 9.2-9 Strength of New York City and Birmingham 2 Wall D. Strength diagrams are superimposed over the 
strain diagrams for the two cases (intermediate case is not shown) (1.0 ft = 0.3048 m). 
Examine the strength of Wall D at Level 1: 
= 0.82 D = 0.82 (236 + factored weight of lower half of first story wall) Pumin 
= 0.82(236 + 6.4) = 199 kips 
= 1.28 D + 0.5 L 319 = 1.28(236 + 6.4) + 0.5(34) = 327 kips Pumax 
Mu = 3,680 ft-kips 
Because intermediate reinforced masonry shear walls are used (Seismic Design Category C), vertical 
reinforcement at is required at 4 ft on center in accordance with Provisions Sec. 11.3.7.3 [ACI 530, Sec. 
1.13.2.2.4]. Therefore, try one #4 bar in each end cell and #4 bars at 4 ft on center at all intermediate 
cells.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-70 
The calculation procedure is similar to that for the Birmingham 1 building presented in Sec. 9.2.4.5.2. 
The results of the calculations (not shown) for the New York building are summarized below. 
P = 0 case 
fPn = 0 
fMn = 1,475 ft-kips 
Intermediate case 
c = 8.0 ft 
fPn = 330 kips 
fMn = 5,600 ft-kips 
Balanced case 
fPn = 807 
fMn = 8,214 ft-kips 
With the intermediate case, it is simple to use the three points to make two straight lines on the interaction 
diagram. Use the simplified fPn - fMn curve shown in Figure 9.2-10. The straight line from pure 
bending to the balanced point is conservative and can easily be used where the design is not as close to 
the criterion. It is the nature of lightly reinforced and lightly loaded masonry walls that the intermediate 
point is frequently useful. 
Use one #4 bar in each end cell and one #4 bar at 4 ft on center throughout the remainder of the wall. 
As shown in the design for Birmingham 1,for the purpose of shear design, the value of fMN at the design 
axial load is necessary. Interpolating between the intermediate point and the P = 0 point for P = 199 kips 
yields fMN = 3,960 ft-kip. Thus, the factor on shear to represent development of 125 percent of flexural 
capacity is: 
125 125
3960 085 
3680 
. 158 
/ 
. 
/ . 
. 
fM f 
M
N
U 
= =
Chapter 9, Masonry 
9-71 
Figure 9.2-10 fP11 - fM11 Diagram for New York City and Birmingham 2 Wall D (1.0 kip = 4.45 kN, 1.0 ft-kip = 
1.36 kN-m). 
9.2.5.5.2.2 Ductility check 
Refer to Sec. 9.2.4.5.2, Item 2, for explanation [see Sec. 9.2.4.5.2 for discussion of revisions to the 
ductility requirements in the 2003 Provisions.]. For Level 1 (bottom story), the unfactored loads are: 
P = 236 + weight of lower  of first story wall = 236 + 6.4 = 242.4 kips 
M = 3,483 ft-kips 
Cm = 0.8 fm' [(a)(b) + Acells] 
where b = face shells = (2  1.25 in.) and Acell = 41 in.2 
5,000 ft-kips 10,000 ft-kips 
500 kips 
1,000 kips 
P max = 342 kips 
P min = 206 kips 
fP 
fM 
Birm #2, M = 3283 ft-kips 
P = 0 
(1475 ft-kips, 0 kips) 
Balance 
(8214 ft-kips, 807 kips) 
P max = 
327 kips 
P min = 
199 kips 
NYC, M = 3680 ft-kips 
3 point fP - fM curve 
Intermediate 
(5600 ft-kips, 330 kips) 
Simplified fP - fM curve 
n 
n 
fMn 
fPn 
u
u 
u 
u 
u 
u 
n 
n 
n 
n
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-72
Cm = (1.6 ksi)[(5.03 ft  12)(2.5 in.) + (2)(41)] = 372.6 kips 
Cs1 = FyAs = (60 ksi)(0.20 in.2) = 12 kips 
Cs2 = (22.6 ksi)(0.20 in.2) = 4.5 kips 
Ts1 = Ts2 = Ts3 = Ts4 = Ts5 = (75 ksi)(0.20 in.2 ) = 15 kips 
Ts6 = (69.6 ksi)(0.20 in.2) = 13.9 kips 
Ts7= (23.5 ksi)(0.20 sq. in.) = 4.7 kips 
3 C > 3 P + T 
Cm + Cs1 + Cs2 > P + Ts1 + Ts2 + Ts3 + Ts4 + Ts5 + Ts6 + Ts7 
372.6 + 12.0 + 4.5 > 242.5 + 5(15) + 13.9 + 4.7 
389 kips > 336 kips OK
Chapter 9, Masonry 
9-73 
m = 0.0025 
N.A. 
s = 5 Ey 
= 0.0103 
6.29' 
P 
Cm 
0.8f m' = 1.6 ksi 
Cs1 
5.96' 6.51' 
Ts5 Ts3 Ts1 
Ts7 
(9) #4 at 4'-0" o.c. 
1.25f y 
= 75 ksi 
5f y = 300 ksi 
26.05' 
32'-8" 
3.78' 
0.33' 
5.03' 1.26' 
26.05' 
0.33' 
f y = 60 ksi 
72.5 ksi 
23.5 ksi 
= 5 ( 60 
29,000 ) 
c 
Ts6 Ts4 Ts2 
Cs2 
22.6 ksi 
69.6 ksi 
Figure 9.2-11 Ductility check for New York City and Birmingham 2 Wall D (1.0 ft = 0.3048 m, 1.0 ksi 
= 6.89 MPa). 
9.2.5.6 New York Deflections 
Refer to 9.2.4.6 for more explanation [see Sec. 9.2.4.6 for discussion of revisions to the deflection 
computations and requirements in the 2003 Provisions, as well as the potentially conflicting drift limits]. 
For the New York City building, the determination of whether the walls will be cracked is: 
be = effective masonry wall width 
be = [(2  1.25 in.)(32.67 ft  12) + (9 cells)(41 in.2/cell)]/(32.67 ft  12) = 3.44 in. 
A = be l = (3.44 in.)(32.67  12) = 1,349 in.2
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-74 
S = be l2/6 = (3.44)(32.67  12)2 /6 = 88,100 in.3 
fr = 0.250 ksi 
Pu is calculated using 1.00D (see Table 9.2-8 for values and refer to Sec. 9.2.4.6 for discussion). Table 
9.2-10 a summarizes of these calculations. 
Table 9.2-10 New York City Cracked Wall Determination 
Level Pu 
(kips) 
Mcr 
(ft-kips) 
Mx 
(ft-kips) 
Status 
54321 
42 
90 
139 
188 
236 
2064 
2325 
2592 
2860 
3120 
313 
908 
1715 
2663 
3680 
uncracked 
uncracked 
uncracked 
uncracked 
cracked 
1.0 kip = 4.45 kN, 1.0 ft-kip = 1.36 kN-m. 
For the uncracked walls: 
In = Ig = bl3/12 = (3.44 in.)(32.67  12)3/12 = 1.73  107 in.4 
For the cracked wall, observe that the intermediate point on the interaction diagram is relatively close to 
the design point. Therefore, as a different type of approximation, compute a cracked moment of inertia 
using the depth to the neutral axis of 8.0 ft: 
Icr = bec3/3 + 3nAsd2 
Icr = (3.44 in.)(8.0 ft  12)3/3 + 19.3(0.2)(4.332 + 8.332 + 12.332 + 16.332 + 20.332 + 24.332)144 = 
= 1.01 x 106 + 0.84 x 106 = 1.85 x 106 in.4 
Per Provisions Eq. 11.5.4.3: 
3 3 
cr 1 cr 
eff n cr n 
a a 
I IM I M I 
M M 
. . . . .. 
= . .+ .-. ..= 
. . .. . ... 
Ieff = 1.13 x 107 in.4 
Provisions 11.5.4.3 would imply that Ieff would be used for the full height. Another reasonable option is 
to use Icr at the first story and Ig above that. The calculation of d should consider shear deflections in 
addition to the flexural deflections. For this example Ieff will be used over the full height for the final 
determination of deflection (a RISA 2D analysis was made). The result is summarized in Table 9.2-11.
Chapter 9, Masonry 
9-75 
Table 9.2-11 New York City Deflections 
Level F 
(kips) 
Ieff 
(in.4) 
dflexural 
(in.) 
dshear 
(in.) 
dtotal 
(in.) 
Cd dtotal 
(in.) 
. 
(in.) 
54321 34.1 
30.9 
23.1 
15.3 
7.7 
1.13  107 
1.13  107 
1.13  107 
1.13  107 
1.13  106 
0.256 
0.189 
0.124 
0.065 
0.020 
0.080 
0.075 
0.064 
0.050 
0.033 
0.336 
0.264 
0.188 
0.115 
0.053 
0.757 
0.593 
0.422 
0.259 
0.118 
0.163 
0.171 
0.163 
0.141 
0.118 
1.0 kip = 4.45 kN, 1.0 in. = 25.4 mm 
The maximum story drift occurs at Level 4: 
.4 = 0.171 in. < 1.04 in. = 0.01 h4 (Provisions Table 5.2.8 [Table 4.5-1]) OK 
The total displacement at the top of the wall is 
. = 0.757 in. < 5.2 in. = 0.01 hn (Provisions 11.5.4.1.1) OK 
9.2.5.7 New York City Out-of-Plane Forces 
Provisions Sec 5.2.6.2.7 [Sec. 4.4.2.2] requires that the bearing walls be designed for out-of-plane loads 
determined as 
w = 0.40 SDS Wc $ 0.1Wc 
With SDS = 0.39 , w = 0.156Wc > 0.1Wc , so w = (0.156)(48 psf) = 7.5 psf, which is much less than wind 
pressure for exterior walls. Even though Wall D is not an exterior wall, the lateral pressure is sufficiently 
low that it is considered acceptable by inspection, without further calculation. Seismic loads do not 
govern the design of Wall D for loading in the out-of-plane direction. 
9.2.5.8 New York City Orthogonal Effects 
According to Provisions Sec. 5.2.5.2.2, orthogonal interaction effects have to be considered for Seismic 
Design Category C when the equivalent lateral force (ELF) procedure is used (as it is here). However, the 
out-of-plane component of only 30 percent of 7.5 psf on the wall will not produce a significant effect 
when combined with the in-plane direction of loads, so no further calculation will be made. 
This completes the design of the transverse Wall D for the New York building. 
9.2.5.9 Summary of New York City Wall D Design 
8 in. CMU 
f!m = 2,000 psi
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-76
Reinforcement: 
Vertical #4 bars at 4 ft on center throughout the wall 
Bond beam with two #4 at each story just below the floor or roof slabs 
Horizontal joint reinforcement at alternate courses 
9.2.6 Birmingham 2 Seismic Design 
The emphasis here is on differences from the previous two locations for the same building. Per 
Provisions Table 5.2.5.1 [Table 4.4-1], the torsional irregularity requires that the design of a Seismic 
Design Category D building be based on a dynamic analysis. Although not explicitly stated, the 
implication is that the analytical model should be three-dimensional in order to capture the torsional 
response. This example will compare both the equivalent lateral force procedure and the modal response 
spectrum analysis procedure and will demonstrate that, as long as the torsional effects are accounted for, 
the static analysis could be considered adequate for design. 
9.2.6.1 Birmingham 2 Weights 
The floor weight for this examples will use the same 67 psf for 8-in.-thick, normal weight hollow core 
plank plus roofing and the nonmasonry partitions as used in the prior examples (see Sec. 9.2.1). This 
site is assigned to Seismic Design Category D, and the walls will be designed as special reinforced 
masonry shear walls (Provisions Sec. 11.11.5 and Sec. 11.3.8[ACI 530, Sec. 1.13.2.2.5), which requires 
prescriptive seismic reinforcement (Provisions Sec. 11.3.7.3). Special reinforced masonry shear walls 
have a maximum spacing of rebar at 4 ft on center both horizontally and vertically. Also, the total area of 
horizontal and vertical reinforcement must exceed 0.0020 times the gross area of the wall, and neither 
direction may have a ratio of less than 0.0007. The vertical #4 bars at 48 in. used for the New York City 
design yields a ratio of 0.00055, so it must be increased. Two viable options are #5 bars at 48 in. 
(yielding 0.00085) and #4 bars at alternating spaces of 32 in. and 40 in. (12 bars in the wall), which yields 
0.0080. The latter is chosen in order to avoid unnecessarily increasing the shear demand. Therefore, the 
horizontal reinforcement must be (0.0020 - 0.0008)(7.625 in.)(12 in./ft.) = 0.11 in.2/ft. or 0.95 in.2 per 
story. Two #5 bars in bond beams at 48 in. on center will be adequate. For this example, 56 psf weight 
for the 8-in.-thick CMU walls will be assumed. The 56 psf value includes grouted cells and bond beams. 
Story weight, wi: 
Roof: 
Roof slab (plus roofing) = (67 psf) (152 ft)(72 ft) = 733 kips 
Walls = (56 psf)(589 ft)(8.67 ft/2) + (56 psf)(4)(36 ft)(2 ft) = 159 kips 
Total = 892 kips 
There is a 2-ft-high masonry parapet on four walls and the total length of masonry wall is 589 ft. 
Typical floor: 
Slab (plus partitions) = 733 kips 
Walls = (56 psf)(589 ft)(8.67 ft) = 286 kips 
Total = 1,019 kips 
Total effective seismic weight, W = 892 + (4)(1,019) = 4,968 kips
Chapter 9, Masonry 
9-77 
This total excludes the lower half of the first story walls which do not contribute to seismic loads that are 
imposed on CMU shear walls. 
9.2.6.2 Birmingham 2 Base Shear Calculation 
The ELF analysis proceeds as described for the other locations. The seismic response coefficient, Cs, is 
computed using Provisions Eq. 5.4.1.1-1 [Eq. 5.2-2] and 5.4.1.1-2 [Eq.5.2-3]: 
(Controls) 
0.47 0.134 
/ 3.51 
DS 
s 
C S
R I = = = 
( ) 
0.28 0.237 
( / ) 0.338 3.51 
D1 
s 
C S 
T R I = = = 
This is somewhat less than the 746 kips computed for the New York City design due to the larger R 
factor. 
The fundamental period of the building, based on Provisions Eq. 5.4.2.1-1 [Eq.5.2-6], is 0.338 sec as 
computed previously (the approximate period, based on building system and building height, will be the 
same for all locations). The value for Cs is taken as 0.134 (the lesser of the two values). This value is still 
larger than the minimum specified in Provisions Eq. 5.3.2.1-3 which is: 
Cs = 0.044SD1I = (0.044)(0.28)(1) = 0.012 
[This minimum Cs value has been removed in the 2003 Provisions. In its place is a minimum Cs value for 
long-period structures, which is not applicable to this example.] 
The total seismic base shear is then calculated using Provisions Eq. 5.4.1 [Eq.5.2-1] as: 
V = CsW = (0.134)(4,968) = 666 kips 
A three-dimensional (3D) model was created in SAP 2000 for the modal response spectrum analysis. The 
masonry walls were modeled as shell bending elements and the floors were modeled as an assembly of 
beams and shell membrane elements. The beams have very little mass and a large flexural moment of 
inertia to avoid consideration of models of vertical vibration of the floors. The flexural stiffness of the 
beams was released at the bearing walls in order to avoid a wall slab frame that would inadvertently 
increase the torsional resistance. The mass of the floors was captured by the shell membrane elements. 
Table 9.2-12 shows data on the modes of vibration used in the analysis. 
Provisions Sec. 4.1.2.6 [Sec. 3.3.4] was used to create the response spectrum for the modal analysis. The 
key points that define the spectrum are: 
TS = SD1/SDS = 0.28/0.47 = 0.60 sec 
T0 = 0.2 TS = 0.12 sec 
at T = 0, Sa = 0.4 SDS/R = 0.0537 g 
from T = T0 to TS, Sa = SDS/R = 0.1343 g 
for T > TS, Sa = SD1/(RT) = 0.080/T 
The computed fundamental period is less than the approximate period. The transverse direction base 
shear from the SRSS combination of the modes is 457.6 kips, which is considerably less than that 
obtained using the ELF method.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-78 
Provisions Sec. 5.5.7 [Sec. 5.3.7] requires that the modal base shear be compared with the ELF base shear 
computed using a period somewhat larger than the approximate fundamental period (CuTa). Per Sec. 
9.2.4.2, Ta = 0.338 sec. and per Provisions Table 5.4.2 [Table 5.2-1] Cu = 1.4. Thus, CuTa = 0.48 sec., 
which is less that SD1/SDS. Therefore, the ELF base shear for comparison is 666 kips as just computed. 
Because 85 percent of 666 kips = 566 kips, Provisions Sec. 5.5.7 [Sec. 5.3.7] dictates that all the results 
of the modal analysis be factored by: 
0.85 566 1.24 
458 
ELF 
Modal 
V 
V 
= = 
Both analyses will be carried forward as discussed in the subsequent sections. 
Table 9.2-12 Birmington 2 Periods, Mass Participation Factors, and Modal Base Shears in the 
Transverse Direction for Modes Used in Analysis 
Mode Period, Individual mode (percent) Cumulative sum (percent) Trans. 
number (seconds) Long. Trans. Vert. Long. Trans. Vert. base shear 
1 0.2467 0.00 0.00 0.00 0.00 0.00 0.00 0.0 
2 0.1919 0.00 70.18 0.00 0.00 70.18 0.00 451.1 
3 0.1915 70.55 0.00 0.00 70.55 70.18 0.00 0.0 
4 0.0579 0.00 18.20 0.00 70.55 88.39 0.00 73.9 
5 0.0574 17.86 0.00 0.00 88.41 88.39 0.00 0.0 
6 0.0535 0.00 4.09 0.00 88.41 92.48 0.00 16.1 
7 0.0532 4.17 0.00 0.00 92.58 92.48 0.00 0.0 
8 0.0413 0.00 0.01 0.00 92.58 92.48 0.00 0.0 
9 0.0332 1.50 0.24 0.00 94.08 92.72 0.00 0.8 
10 0.0329 0.30 2.07 0.00 94.38 94.79 0.00 7.1 
11 0.0310 1.28 0.22 0.00 95.66 95.01 0.00 0.8 
12 0.0295 0.22 1.13 0.00 95.89 96.14 0.00 3.8 
13 0.0253 1.97 0.53 0.00 97.86 96.67 0.00 1.7 
14 0.0244 0.53 1.85 0.00 98.39 98.52 0.00 5.9 
15 0.0190 1.05 0.36 0.00 99.44 98.89 0.00 1.1 
16 0.0179 0.33 0.94 0.00 99.77 99.82 0.00 2.8 
17 0.0128 0.19 0.07 0.00 99.95 99.90 0.00 0.2 
18 0.0105 0.03 0.10 0.00 99.99 99.99 0.00 0.3 
1 kip = 4.45 kN. 
9.2.6.3 Birmingham 2 Vertical Distribution of Seismic Forces 
The dynamic analysis will be revisited for the horizontal distribution of forces in the next section but as 
demonstrated there, the ELF procedure is considered adequate to account for the torsional behavior in this 
example. The dynamic analysis can certainly be used to deduce the vertical distribution of forces. This 
analysis was constructed to study amplification of accidental torsion. It would be necessary to integrate 
the shell forces to find specific story forces, and it is not necessary to complete the design. Therefore, the 
vertical distribution of seismic forces for the ELF analysis is determined in accordance with Provisions 
Sec. 5.4.4 [Sec. 5.2.3], which was described in Sec. 9.2.4.3. For Provisions Eq. 5.4.3-2 [Sec. 5.2-11], k = 
1.0 since T = 0.338 sec (similar to the Birmingham 1 and New York City buildings). It should be noted 
that the response spectrum analysis may result in moments that are less than those calculated using the 
ELF method; however, because of its relative simplicity, the ELF is used in this example. 
Application of the Provisions equations for this building is shown in Table 9.2-13:
Chapter 9, Masonry 
9-79 
Table 9.2-13 Birmingham 2 Seismic Forces and Moments by Level 
Level 
(x) 
wx 
(kips) 
hx 
(ft) 
wxhx 
(ft-kips) 
Cvx Fx 
(kips) 
Vx 
(kips) 
Mx 
(ft-kips) 
543213 
892 
1,019 
1,019 
1,019 
1,019 
4,968 
43.34 
34.67 
26.00 
17.33 
8.67 
38,659 
35,329 
26,494 
17,659 
8,835 
126,976 
0.3045 
0.2782 
0.2086 
0.1391 
0.0695 
1.000 
203 
185 
139 
93 
46 
666 
203 
388 
527 
620 
666 
1,760 
5,124 
9,693 
15,068 
20,843 
1.0 kip = 4.45 kN, 1.0 ft = 0.3048 m, 1.0 ft-kip = 1.36 kN-m. 
9.2.6.4 Birmingham 2 Horizontal Distribution of Forces 
For the ELF analysis, this is the same as that for New York City location; see Sec. 9.2.5.4. 
Total shear in wall type D: 
0.125 Vtot= V+1.365(0.0238)V=0.158V=104.9kips 
The dynamic analysis shows that the fundamental mode is a pure torsional mode. The fact that the 
fundamental mode is torsional does confirm, to an extent, that the structure is torsionally sensitive. This 
modal analysis does not show any significant effect of the torsion, however. The pure symmetry of this 
structure is somewhat idealistic. Real structures usually have some real eccentricity between mass and 
stiffness, and dynamic analysis then yields coupled modes, which contribute to computed forces. 
The Provisions does not require that the accidental eccentricity be analyzed dynamically. For illustration, 
however, this was done by adjusting the mass of the floor elements to generate an eccentricity of 5 
percent of the 152-ft length of the building. Table 9.2-14 shows the results of such an analysis. 
(Accidental torsion could also be considered using a linear combination of the dynamic results and a 
statically applied moment equal to the accidental torsional moment.) 
The transverse direction base shear from the SRSS combination of the modes is 403.8 kips, significantly 
less than the 457.6 kips for the symmetric model. The amplification factor for this base shear is 566/404 
= 1.4. This smaller base shear from modal analysis of a model with an artificially introduced eccentricity 
is normal. For two primary reasons. First, the mass participates in more modes. The participation in the 
largest mode is generally less, and the combined result is dominated by the largest single mode. Second, 
the period for the fundamental mode generally increases, which will reduce the spectral response except 
for structures with short periods (such as this one). 
The base shear in Wall D was computed by adding the in-plane reactions. For the symmetric model the 
result was 57 kips, which is 12.5 percent of the total of 458 kips, as would be expected. Amplifying this 
by the 1.24 factor yields 71 kips The application of a static horizontal torsion equal to the 5 percent 
eccentricity times a base shear of 566 kips (the floor) adds 13 kips, for a total of 84 kips. If the static 
horizontal torsion is amplified by 1.365, as found in the analysis for the New York location, the total 
becomes 89 kips, which is less than the 99 kips and 105 kips computed in the ELF analysis without and 
with, respectively, the amplification of accidental torsion. The Wall D base shear from the eccentric 
model was 66 kips; with the amplification of base shear = 1.4, this becomes 92 kips. Note that this value 
is less than the direct shear from the symmetric model plus the amplified static torsion. The obvious 
conclusion is that more careful consideration of torsional instability than actually required by the
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-80 
Provisions does not indicate any more penalty than already given by the procedures for the ELF in the 
Provisions. Therefore the remainder of the example designs for this building are completed using the 
ELF. 
Table 9.2-14 Birmingham periods, Mass Participation Factors, and Modal Base 
Shears in the Transverse Direction for Modes Used in Analysis 
Mode Period Individual mode (percent) Cumulative sum (percent) Trans. 
Number (sec) Long. Trans. Vert. Long. Trans. Vert. Base Shear 
1 0.2507 0.0 8.8 0.0 0.0 8.8 0.0 56.3 
2 0.1915 70.5 0.0 0.1 70.5 8.8 0.1 0.0 
3 0.1867 0.0 61.4 0.0 70.5 70.2 0.1 394.9 
4 0.0698 0.0 2.9 0.0 70.5 73.1 0.1 12.7 
5 0.0613 1.1 0.0 23.0 71.6 73.1 23.1 0.0 
6 0.0575 19.2 0.0 0.0 90.9 73.1 23.2 0.0 
7 0.0570 0.0 13.7 0.0 90.9 86.8 23.2 55.5 
8 0.0533 0.0 5.6 0.0 90.9 92.4 23.2 22.0 
9 0.0480 1.2 0.0 12.8 92.0 92.4 35.9 0.0 
10 0.0380 1.4 0.0 0.0 93.5 92.4 35.9 0.0 
11 0.0374 0.0 0.4 0.0 93.5 92.8 35.9 1.3 
12 0.0327 1.7 0.0 0.2 95.2 92.8 36.1 0.0 
13 0.0322 0.0 3.1 0.0 95.2 95.9 36.1 10.4 
14 0.0263 2.8 0.0 0.1 98.0 95.9 36.2 0.0 
15 0.0243 0.0 3.0 0.0 98.0 98.8 36.2 9.5 
16 0.0201 1.6 0.0 0.1 99.6 98.8 36.3 0.0 
17 0.0164 0.0 1.1 0.0 99.6 100.0 36.3 3.4 
18 0.0141 0.4 0.0 0.1 100.0 100.0 36.3 0 
The total story shear and overturning moment (from the ELF analysis) may now be distributed to Wall D 
and the wall proportions checked. The wall capacity will be checked before considering deflections. 
The extreme torsional irregularity has an additional consequence for Seismic Design Category D: 
Provisions 5.6.2.4.2 [Sec. 4.6.3.2] requires that the design forces for connections between diaphragms, 
collectors, and vertical elements (walls) be increased by 25 percent above the diaphragm forces given in 
Provisions 5.4.1 [Sec. 4.6.3.4]. For this example, the diaphragm of precast elements is designed using the 
different requirements of the appendix to Provisions Chapter 9 (see Chapter 7 of this volume). 
9.2.6.5 Birmingham 2 Transverse Wall (Wall D) 
The design demands are slightly smaller than for the New York City design, yet there is more 
reinforcement, both vertical and horizontal in the walls. This illustration will focus on those items where 
the additional reinforcement has special significance. 
9.2.6.5.1 Birmingham 2 Shear Strength 
Refer to Sec. 9.2.5.5.1 for most quantities. The additional horizontal reinforcement raises Vs and the 
additional grouted cells raises An and, therefore both Vm and Vn max. 
Av/s = (4)(0.31 in.2)/(8.67 ft.) = 0.1431 in.2/ft 
Vs = 0.5(0.1431)(60 ksi)(32.67 ft) = 140.2 kips 
An = (2  1.25 in.  32.67 ft x 12 in.) + (41 in.2  12 cells) = 1,472 in.2
Chapter 9, Masonry 
9-81 
The shear strength of Wall D is summarized in Table 9.2-15 below. (Note that Vx and Mx in this table are 
values from Table 9.2-13 multiplied by 0.158, the portion of direct and torsional shear assigned to the 
wall). Clearly, the dynamic analysis would make it possible to design this wall for smaller forces, but the 
minimum configuration suffices. The 1.96 multiplier on Vx to determine Vu is explained in the subsequent 
section on flexural design. 
Table 9.2-15 Shear Strength Calculations for Wall D, Birmingham 2 
Level 
(x) 
Vx 
(kips) 
Mx 
(ft-kips) 
Mx/Vxd 1.98Vx 
(kips) 
P 
(kips) 
fVm 
(kips) 
fVs 
(kips) 
fVn 
(kips) 
fVn max 
(kips) 
5 32.0 277 0.265 63.4 42 194.6 112.2 306.8 313.9 
4 61.1 907 0.454 121 90 186.8 112.2 299 287.3 
3 83.0 1527 0.563 164.3 139 186.6 112.2 298.8 272.0 
2 97.7 2373 0.743 193.4 188 179.7 112.2 291.9 246.7 
1 104.9 3283 0.958 207.7 236 169.6 112.2 281.8 216.6 
1.0 kip = 4.45 kN, 1.0 ft-kip = 1.36 kN-m. 
Note that Vn max is less than Vn at all levels except the top story. The capacity is greater than the demand 
at all stories, therefore, the design is satisfactory for shear. 
9.2.6.5.2 Birmingham 2 Axial and Flexural Strength 
Once again, the similarities to the design for the New York City location will be exploited. Normally, the 
in-plane calculations include: 
1. Strength check 
2. Ductility check 
9.2.6.5.2.1 Strength check 
The wall demands, using the load combinations determined previously, are presented in Table 9.2-16 for 
Wall D. In the table, Load Combination 1 is 1.29D + QE + 0.5L and Load Combination 2 is 0.81D + QE. 
Table 9.2-16 Birmingham 2 Demands for Wall D 
Load Combination 1 Load Combination 2 
Level PD 
(kips) 
PL 
(kips) 
Pu 
(kips) 
Mu 
(ft-kips) 
Pu 
(kips) 
Mu 
(ft-kips) 
54321 
43 
94 
145 
196 
247 
0 
8 
17 
25 
34 
55 
125 
196 
265 
336 
277 
807 
1527 
2373 
3283 
36 
76 
117 
159 
200 
277 
807 
1527 
2373 
3283 
1.0 kip = 4.45 kN, 1.0 ft-kip = 1.36 kN-m 
Strength at the bottom story (where P, V, and M are the greatest) are less than required for the New York 
City design. The demands are plotted on Figure 9.2-10, showing that the design for New York City has 
sufficient axial and flexural capacity for this Birmingham 2 location. For this design, the interaction 
capacity line will be shifted to the right, due to the presence of additional reinforcing bars. The only 
calculation here will be an estimate of the factor to develop the flexural capacity at the design axial load. 
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-82 
The flexural capacity for lightly load walls is approximately proportional to the sum of axial load plus the 
yield of the reinforcing steel: 
Birmingham #2 capacity 
NewYorkCapacity 
kips 12 0.20 in. ksi 
kips 9 0.20 in. ksi 
2 
2 = 
+   
+   
= = 
200 60 
199 60 
344 
307 
1.12 
Therefore the factor by which the walls shears must be multiplied to represent 125 percent of flexural 
capacity, given that the factor was 1.58 for the New York design is: 
158 112 177 
746 
666 
.. =. =1.98 
New York base shear 
Birmingham #2 base shear 
9.2.6.5.2.2 Ductility check 
The Provisions requirements for ductility are described in Sec. 9.2.4.5.2 and 9.2.5.5.2. Since the wall 
reinforcement and loads are so similar to those for the New York City building, the computations are not 
repeated here. 
[Refer to Sec. 9.2.4.5.2 for discussion of revisions to the ductility requirements in the 2003 Provisions.] 
9.2.6.6 Birmingham 2 Deflections 
The calculations for deflection would be very similar to that for the New York City location. Ironically, 
that procedure will indicate that the wall is not cracked at the design load. The Cd factor is larger, 3.5 vs. 
2.25. However, the calculation is not repeated here; refer to Sec. 9.2.4.6 and Sec. 9.2.5.6. 
[Refer to Sec. 9.2.4.6 for discussion of revisions to the deflection computations and requirements in the 
2003 Provisions, as well as the potentially conflicting drift limits.] 
9.2.6.7 Birmingham 2 Out-of-Plane Forces 
Provisions Sec. 5.2.6.2.7 [Sec. 4.6.1.3] requires that the bearing walls be designed for out-of-plane loads 
determined: 
w = 0.40 SDS Wc $ 0.1Wc 
w = (0.40)(0.47)(56 psf) = 10.5 psf $ 0.1Wc 
The calculated seismic load, w = 10.5 psf, is less than wind pressure for exterior walls. Even though Wall 
D is not an exterior wall, the lateral pressure is sufficiently low that it is considered acceptable by 
inspection without further calculation. Seismic loads do not govern the design of Wall D for loading in 
the out-of-plane direction. 
9.2.6.8 Birmingham 2 Orthogonal Effects 
According to Provisions Sec. 5.2.5.2.2 [Sec. 4.4.2.3], orthogonal interaction effects have to be considered 
for Seismic Design Category D when the ELF procedure is used (as it is here). However, the out-of-plane 
component of only 30 percent of 10.5 psf on the wall will not produce a significant effect when combined 
with the in-plane direction of loads so no further calculation will be made. 
This completes the design of the Transverse Wall D.
Chapter 9, Masonry 
9-83 
9.2.6.9 Birmingham 2 Summary of Wall Design for Wall D 
8-in. CMU 
f'm = 2,000 psi 
Reinforcement: 
12 vertical #4 bars per wall (spaces alternate at 32 and 40 in. on center) 
Two bond beams with 2 - #5 at each story, at bearing for the planks, and at 4 ft above each floor. 
Horizontal joint reinforcement at alternate courses is recommended, but not required. 
9.2.7 Seismic Design for Los Angeles 
Once again, the differences from the designs for the other locations will be emphasized. As explained for 
the Birmingham 2 building, the Provisions would require a dynamic analysis for design of this building. 
For the reasons explained in Sec. 9.2.6.4, this design is illustrated using the ELF procedure. 
9.2.7.1 Los Angeles Weights 
Use 91 psf for 8-in.-thick, normal weight hollow core plank, 2.5 in. lightweight concrete topping (115 
pcf), plus the nonmasonry partitions. This building is Seismic Design Category D, and the walls will be 
designed as special reinforced masonry shear walls (Provisions Sec. 11.11.5 and Sec. 11.3.8 
[Sec.11.2.1.5]), which requires prescriptive seismic reinforcement (Provisions Sec. 11.3.8.3 [ACI 530, 
Sec. 1.13.2.2.5]). Special reinforced masonry shear walls have a minimum spacing of vertical 
reinforcement of 4 ft on center. For this example, 60 psf weight for the 8-in. CMU walls will be assumed. 
The 60 psf value includes grouted cells and bond beams in the course just below the floor planks and in 
the course 4 ft above the floors. A typical wall section is shown in Figure 9.2-12.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-84 
8 " p recast 
concrete p lank 
8 " con crete 
m asonry w all 
See F igure 7 .1 -8 
fo r g ro u ted 
sp a c e d e ta il. 
V ertic a l b ar 
at 4 '-0" o .c. 
B o n d b eam 
w / (2) # 5 (ty p ) 
9 g a . horizo n tal 
jo in t re in fo rc em en t 
at 1 '-4" o .c. 
8" 
4'-0" 
7'-91 8" 
2" 
4'-0" 
21
2" 21
2" 
L ig h tw eig h t 
concrete 
to p p ing 
Figure 9.2-12 Typical wall section for the Los Angeles location (1.0 in. = 25.4 
mm, 1.0 ft = 0.3048 m) 
Story weight ,wi: 
Roof weight: 
Roof slab (plus roofing) = (91 psf) (152 ft)(72 ft) = 996 kips 
Walls = (60 psf)(589 ft)(8.67 ft/2) + (60 psf)(4)(36 ft)(2 ft) = 170 kips 
Total = 1,166 kips 
There is a 2-ft-high masonry parapet on four walls and the total length of masonry wall is 589 ft. 
Typical floor: 
Slab (plus partitions) = 996 kips 
Walls = (60 psf)(589 ft)(8.67 ft) = 306 kips 
Total = 1,302 kips 
Total effective seismic weight, W = 1,166 + (4)(1,302) = 6,374 kips 
This total excludes the lower half of the first story walls, which do not contribute to seismic loads that are 
not imposed on the CMU shear walls.
Chapter 9, Masonry 
9-85 
9.2.7.2 Los Angeles Base Shear Calculation 
The seismic response coefficient, Cs, is computed using Provisions Eq. 5.4.1.1-1 [Eq. 5.2-2] and 5.4.1.1-2 
[Eq. 5.2-3]: 
Controls 
1.00 0.286 
/ 3.51 
DS 
s 
C S
R I = = = 
( ) 
0.60 0.507 
( / ) 0.338 3.5 1 
D1 
s 
C S 
T R I = = = 
where T is the fundamental period of the building, which is 0.338 sec as computed previously (the 
approximate period, based on building system and building height, will be the same for all locations). 
The value for Cs is taken as 0.286 (the lesser of these two). This value is still larger than the minimum 
specified in Provisions Eq. 5.3.2.1-3 which is: 
Cs = 0.044SD1I = (0.044)(0.60)(1) = 0.026 
[This minimum Cs value has been removed in the 2003 Provisions. In its place is a minimum Cs value 
for long-period structures, which is not applicable to this example.] 
The total seismic base shear is then calculated Provisions Eq. 5.4.1 [Eq.5.2-1]: 
V = CsW = (0.286)(6,374) = 1,823 kips 
9.2.7.3 Los Angeles Vertical Distribution of Seismic Forces 
The vertical distribution of seismic forces is determined in accordance with Provisions Sec. 5.4.4 [Sec. 
5.2.3], which as described in Sec. 9.2.4.3. Note that for Provisions Eq. 5.4.3-2 [Eq. 5.2-11], k = 1.0 since 
T = 0.338 sec (similar to the previous example buildings). 
The application of the Provisions equations for this building is shown in Table 9.2-17: 
Table 9.2-17 Los Angeles Seismic Forces and Moments by Level 
Level 
(x) 
wx 
(kips) 
hx 
(ft) 
wxhx
k 
(ft-kips) 
Cvx Fx 
(kips) 
Vx 
(kips) 
Mx 
(ft-kips) 
543213 
1,166 
1,302 
1,302 
1,302 
1,302 
6,374 
43.34 
34.67 
26.00 
17.33 
8.67 
50,534 
45,140 
33,852 
22,564 
11,288 
163,378 
0.309 
0.276 
0.207 
0.138 
0.069 
1.000 
564 
504 
378 
252 
126 
1,824 
564 
1,608 
1,446 
1,698 
1,824 
4,890 
14,150 
26,686 
41,409 
57,222 
1.0 kip = 4.45 kN, 1.0 ft = 0.3048 m, 1.0 ft-kip = 1.36 kN-m 
9.2.7.4 Los Angeles Horizontal Distribution of Forces 
This is the same as for the Birmingham 2 design; see Sec. 9.2.6.4. 
Total shear in Wall Type D: 
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-86 
Vtot = 0.125V + 1.365(0.0238)V = 0.158V 
The total story shear and overturning moment may now be distributed to each wall and the wall 
proportions checked. The wall capacity will be checked before considering deflections. 
9.2.7.5 Los Angeles Transverse Wall D 
The strength or limit state design concept is used in the Provisions. 
9.2.7.5.1 Los Angeles Shear Strength 
The equations are the same as for the prior locations for this example building. Looking forward to the 
design for flexural and axial load, the amplification factor on the shear is computed as: 
125 125
9156 085 
9012 
. . 149 
/ . 
. 
M
M
n
u
= = (which is less than the 2.5 upper bound) 
Therefore, the demand shear is 1.49 times the value from analysis. (This design continues to illustrate the 
ELF analysis and; as explained for the Birmingham 2 design, smaller demands could be derived from the 
dynamic analysis.) All other parameters are similar to those for Birmingham 2 except that: 
An = (2  1.25 in.  32.67 ft x 12 in.) + (41 in.2  15 cells) = 1,595 in.2 
The shear strength of each Wall D, based on the aforementioned formulas and data, are summarized in 
Table 9.2-18. 
Table 9.2-18 Los Angeles Shear Strength Calculations for Wall D 
Story 
Vx 
(kips) 
Mx 
(ft-kips) 
Mx/Vxd 1.49Vx 
(kips) 
P 
(kips) 
fVm 
(kips) 
fVs 
(kips) 
fVn 
(kips) 
fVn max 
(kips) 
5 88.8 770 0.265 132.3 42 210.1 112.2 322.3 340 
4 168.2 2229 0.406 250.6 90 205.7 112.2 317.9 318.7 
3 227.7 4203 0.565 339.3 139 199.6 112.2 311.8 294.5 
2 267.4 6522 0.747 398.4 188 191.3 112.2 303.8 266.8 
1 287.2 9012 0.960 427.9 236 179.5 112.2 291.7 234.3 
1.0 kip = 4.45 kN, 1.0 ft-kip = 1.36 kN-m 
Just as for the Birmingham 2 design, the maximum on Vn controls over the sum of Vm and Vs at all stories 
except the top. Unlike the prior design, the shear capacity is inadequate in the lower three stories. The 
solution is to add grout. At the first story, solid grouting is necessary: 
An = (7.625 in.)(32.67 ft.)(12 in./ft.) = 2989 in.2 
fVn max = 0.8(4.11)(0.0447 ksi)(2989 in.2) = 439 kips > 428 kips OK 
At the third story, six additional cells are necessary, and at the second story, approximately two out of 
three cells must be grouted. The additional weight adds somewhat to the demand but only about 2 
percent. If the entire building were grouted solid (which would be common practice in the hypothetical 
location), the weight would increase enough that the shear strength criterion might be violated.
Chapter 9, Masonry 
9-87 
9.2.7.5.2 Los Angeles Axial and Flexural Strength 
The basics of the flexural design have been demonstrated for the previous locations. The demand is much 
higher at this location, however, which introduces issues about the amount and distribution of 
reinforcement in excess of the minimum requirements. Therefore, the strength and ductility checks will 
both be examined. 
9.2.7.5.2.1 Strength check 
Load combinations, using factored loads, are presented in Table 9.2-19 for Wall D. In the table, Load 
Combination 1 is 1.4D + QE + 0.5L, and Load Combination 2 is 0.7D + QE. 
Table 9.2-19 Los Angeles Load Combinations for Wall D 
Load Combination 1 Load Combination 2 
Level 
(x) 
PD 
(kips) 
PL 
(kips) 
Pu 
(kips) 
Mu 
(ft-kips) 
Pu 
(kips) 
Mu 
(ft-kips) 
54321 
63 
126 
189 
251 
314 
0 
8 
17 
25 
34 
88 
180 
273 
364 
456 
770 
2229 
4203 
6522 
9012 
44 
88 
132 
176 
220 
770 
2229 
4203 
6522 
9012 
1.0 kip = 4.45 kN, 1.0 ft-kip = 1.36 kN-m 
Strength at the bottom story (where P, V, and M are the greatest) is examined. This example considers 
Load Combination 2 from Table 9.2.19 to be the governing case, because it has the same lateral load as 
Load Combination 1 but lower values of axial force. 
Refer to Figure 9.2-13 for notation and dimensions.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-88 
Figure 9.2-13 Los Angeles: Strength of wall D (1.0 ft = 0.3048 m). 
Strength diagrams superimposed on strain diagrams for the two cases. 
Examine the strength of Wall D at Level 1: 
Pumin = 220 kips 
= 456 kips max u P 
Mu = 9,012 ft-kips 
Because special reinforced masonry shear walls are used (Seismic Design Category D), vertical 
reinforcement at 4 ft. on center and horizontal bond beams at 4 ft on center are prescribed (Provisions 
Sec. 11.3.7.3 [ACI 530, Sec. 1.13.2.2.5]). (Note that the wall is 43.33 ft high, not 8 ft high, for purposes 
of determining the maximum spacing of vertical and horizontal reinforcement.) 
y = 0.00207 
Ts1 
N.A. 
y = 0.00207 
P = 0 Case 
a = 14.15' 
c = 17.69' 
Center 
Line 
S Ts2 
Cm cell 1 Cm shell 
a = 1.86' 
c = 2.33' 
y = 0.00207 
m= 0.0025 
0.8f' 
C m + C 
32.67' 
16.33' 16.00' 0.33' 
y = 0.00207 
m= 0.0025 
0.8f' 
Cs1 
M 
P 
(7) #5 at 4.00' on center 
(4) #8 at 0.67' on center 
Cs2 Cs2 Cs2 
Ts1 
Ts2 Ts2 Ts2 
= 0.00179 
Cm cell 2 
Ts1 
N.A. 
Balanced 
Case 
s 
m
m
Chapter 9, Masonry 
9-89 
For this bending moment, the minimum vertical reinforcement will not suffice. For reinforcement 
uniformly distributed, a first approximation could be taken from a simple model using an effective 
internal moment arm of 80 percent of the overall length of the wall:
2 2 
ksi
0.8 / 2 9012 0.8 220 32.67 / 2 7.8 in 0.24 in /ft. 
60( )(0.8 / 2) 60 0.8 32.67 / 2 s 
A M Pl
l 
- -   
= = = = 
  
The minimum vertical steel is 0.0007 times the gross area, which is 0.064 in.2/ft. At the maximum 
spacing of 4 ft, a #5 bar is slightly above the minimum. Experimental evidence indicates that uniformly 
distributed reinforcement will deliver good performance. This could be accomplished with a #9 bar at 48 
in. or a #8 bar at 40 in. This design will work well in a wall that is solidly grouted; however, for walls 
that are grouted only at cells containing reinforcement, it will be found that this wall fails the ductility 
check (which can be remedied by placing several extra grouted cells near each end of the wall as was 
shown in Sec. 9.1.5.4). The flexural design was completed before the shear design (described in the 
previous section) discovered the need for solid grout in the first story. The remainder of this flexural 
design check is carried out without consideration of the added grout. (It is unlikely that the interaction 
line will be affected near the design points, but the balanced point will definitely change.) 
It has long been common engineering practice to concentrate flexural reinforcement near the ends of the 
wall. (This a normal result of walls that intersect to form flanges with reinforcement in both web and 
flange.) For this design, if one uses the minimum #5 bar at 48 inches, then the extra steel at the ends of 
the walls is approximately: 
(7.8 7 0.31) / 2 2.8 in2 s end A = -  = 
Try #8 bars in each of the first four end cells and #5 bars at 4 ft on center at all intermediate cells. 
The calculation procedure is similar to that presented in Sec. 9.2.4.5.2. The strain and stress diagrams are 
shown in Figure 9.2-13 for the Birmingham 1 building and the results are as follows: 
P = 0 case 
fPn = 0 
fMn = 6,636 ft-kips 
Intermediate case, setting c = 4.0 ft 
fPn = 223 kips 
fMn = 9,190 ft-kips 
Balanced case 
fPn = 1049 kips 
fMn = 14,436 ft-kips 
The simplified fPn - fMn curve is shown in Figure 9.2-14 and indicates the design with #8 bars in the first 
four end cells and #5 bars at 4 ft on center throughout the remainder of the wall is satisfactory.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-90 
9.2.7.5.2.2 Ductility check 
Provisions Sec. 11.6.2.2 [ACI 530, Sec. 3.2.3.5] has been illustrated in the prior designs. Recall that this 
calculation uses unfactored gravity axial loads (Provisions Sec. 11.6.2.2 [ACI 530, Sec. 3.2.3.5]). Refer 
to Figure 9.2-15 and the following calculations which illustrate this using loads at the bottom story 
(highest axial loads). The extra grout required for shear is also ignored here. More grout gives higher 
compression capacity, which is conservative. 
5,000 ft-kips 10,000 ft-kips 
500 kips 
1,000 kips 
P max = 456 kips 
P min = 220 kips 
fP 
fM 
P = 0 
(6636 ft-kips, 0 kips) 
Balance 
(15,320 ft-kips, 989 kips) 
M = 9012 ft-kips 
15,000 ft-kips 
Simplified fP - fM curve 
u 
3 point fP - fM curve 
Intermediate 
(9190 ft-kips, 223 kips) 
n 
n 
fPn 
fMn 
u 
u 
n 
n 
n 
n 
Figure 9.2-14 fP11 - fM11 diagram for Los Angeles Wall D (1.0 kip = 4.45 kN, 1.0 kip-ft = 1.36 kN-m). 
Chapter 9, Masonry 
9-91 
m = 0.0025 
N.A. 
s = 5 y 
= 0.0103 
6.29' 
P 
Cm 
0.8f 'm = 1.6 ksi 
Cs1 
5.96' 6.51' 
Ts4 Ts1 
(7) #5 at 4'-0" o.c. 
1.25f y 
= 75 ksi 
5f y = 300 ksi 
26.05' 
32'-8" 
3.78' 
0.33' 
5.03' 1.26' 
26.05' 
0.33' 
f y = 60 ksi 
72.86 ksi 
23.5 ksi 
= 5 ( 60 
29,000 ) 
c 
S T s2 
Cs2 
22.6 ksi 
(4) #8 at 
0.67' o.c. 
1.08' 5.21' 
(4) #8 at 
0.67' o.c. 
53.2 ksi 
Ts3 
69.6 ksi 
45.6 ksi 
Figure 9.2-15 Ductility check for Los Angeles Wall D (1.0 ft = 0.3048 m, 1.0 ksi = 6.89 MPa) 
For Level 1 (bottom story), the unfactored loads are: 
P = 314 kips 
Cm = 0.8fm[(a)(b) + Acells] 
where b = flange width = (2  1.25 = 2.5 in.) and Acells = 41 in.2 
Cm = (1.6 ksi)[(5.03 ft  12)(2.5 in.) + (5 cells)(41)] = 569.4 kips 
Cs1 = 0.79(2 x 60 + 53.2 + 45.6) = 172.9 kips 
Cs2 = (22.6 ksi)(0.31 in.2) = 7.0 kips
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-92 
3C = 749 kips 
3Ts1 = (4  0.79 in.2)(75 ksi) = 237 kips 
3Ts2 = (4  0.31 in.2)(75 ksi ) = 93.0 kips 
Ts3 = (0.31 in.2)(69.6 ksi) = 21.6 kips 
Ts4 = (0.31 in.2)(23.5 ksi) = 7.3 kips 
3T = 359 kips 
3 C > 3 P + T 
749 kips > 673 kips 
If a solution with fully distributed reinforcement were used, the tension from reinforcement would 
increase while the compression from grout at the end of the wall, as well as compression of steel at the 
compression would also decrease. The criterion would not be satisfied. Adding grout would be required. 
[Refer to Sec. 9.2.4.5.2 for discussion of revisions to the ductility requirements in the 2003 Provisions.] 
9.2.7.6 Los Angeles Deflections 
Recall the assertion that the calculations for deflection involve many variables and assumptions and that 
any calculation of deflection is approximate at best. The requirements and procedures for computing 
deflection are provided in Sec. 9.2.4.6. [Refer to Sec. 9.2.4.6 for discussion of revisions to the deflection 
computations and requirements in the 2003 Provisions, as well as the potentially conflicting drift limits.] 
For the Los Angeles building, the determination of whether the walls will be cracked is as follows: 
be = effective masonry wall width 
be = [(2  1.25 in.)(32.67 ft  12) + (15 cells)(41 in.2/cell)]/32.67 ft  12) = 4.07 in. 
A = be l = (4.07 in.)(32.67  12) = 1595 in.2 
S = be l2/6 = (7.07)(32.67  12)2/6 = 104,207 in.3 
fr = 0.250 ksi 
Pu is calculated using 1.00D (See Table 9.2-18 for values, and refer to Sec. 9.2.4.6 for discussion). Table 
9.2-20 provides a summary of these calculations. (The extra grout required for shear strength is also not 
considered here; the revision would slightly reduce the computed deflections by raising the cracking 
moment.) 
Table 9.2-20 Los Angeles Cracked Wall Determination 
Level Pumin 
(kips) 
Mcr 
(ft-kips) 
Mx 
(ft-kips) 
Status 
54321 
63 
126 
189 
251 
314 
2514 
2857 
3200 
3538 
3880 
770 
2229 
4203 
6522 
9012 
uncracked 
uncracked 
cracked 
cracked 
cracked 
1.0 kip = 4.45 kN, 1.0 ft-kip = 1.36 kN-m. 
For the uncracked walls (Levels 4 and 5): 
In = Ig = bel3/12 = (4.07 in.)(32.67  12)3/12 = 2.04  107 in.4
Chapter 9, Masonry 
9-93 
For the cracked walls, the transformed cross section will computed by classic methods. Assuming the 
neutral axis to be about 10 ft in from the compression face gives five #5 bars in tension. The tension 
reinforcement totals: 
As = 4(0.79) + 5(0.31) = 3.16 + 1.55 = 4.71 in.2 
The axial compression stiffens the wall. The effect is approximated with an equivalent area of tension 
reinforcement equal to half the compression. Thus, the total reinforcement becomes: 
Ase = 4.71 + 0.5(314)/60 = 4.71 + 2.62 = 7.33 in.2 
The centroid of this equivalent reinforcement is 29.5 ft from the compression face. Following the classic 
method for transformed cracked cross sections and with n = 19.3: 
. = 7.33/(4.04 x 29.5 x 12) = 0.0051 
.n = 0.0051(19.3) = 0.099 
k = sqrt(.n2 + 2.n) - .n = 0.36 
kd = c = 10.5 feet (which is close enough to the assumed 10 feet) 
Icr = bc3/3 + 3nAsd2 = 4.04(29.5 x 12)3 + 19.3(7.33)(29.5 - 10.5)2(144) = 1.01 x 107 in.4 
The Provisions encourages the use of the cubic interpolation formula illustrated for the previous 
locations. For the values here, this yields Ieff = 1.09 x 107 in.4, which is about half the gross moment of 
inertia (which in itself is not a bad approximation for a cracked and well reinforced cross section). For 
this example, the deflection computation will instead use the cracked moment of inertia in the lower three 
stories and the gross moment of inertia in the upper two stories. The results from a RISA 2D analysis are 
shown in Table 9.2-21, and are about 5 percent higher than use of Ieff over the full height. 
Table 9.2-21 Los Angeles Deflections 
Level F 
(kips) 
Ieff 
(in.4) 
dflexural 
(in.) 
dshear 
(in.) 
dtotal 
(in.) 
Cd dtotal 
(in.) 
. 
(in.) 
54321 
84.0 
75.1 
56.3 
37.6 
18.8 
2.04  107 
2.04  107 
1.01  107 
1.01  107 
1.01  107 
0.491 
0.370 
0.237 
0.118 
0.033 
0.127 
0.118 
0.096 
0.068 
0.035 
0.618 
0.488 
0.333 
0.186 
0.068 
2.163 
1.708 
1.166 
0.651 
0.238 
0.455 
0.543 
0.515 
0.413 
0.238 
1.0 kip = 4.45 kN, 1.0 in. = 25.4 mm. 
The maximum drift occurs at Level 4 per Provisions Table 5.2.8 is: 
. = 0.543 in. < 1.04 in. = 0.01 hn (Provisions Table 5.2.8 [Table 4.5-1]) OK 
9.2.7.7 Los Angeles Out-of-Plane Forces 
Provisions Sec 5.2.6.2.7 [Sec. 4.6.1.3] requires that the bearing walls be designed for out-of-plane loads 
determined as follows: 
w = 0.40 SDS Wc $ 0.1Wc 
w = (0.40)(1.00)(60 psf) = 24psf $ 0.1Wc
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-94 
The out-of-plane bending moment, using the strength design method for masonry, for a pressure, w = 24 
psf and considering the P-delta effect, is computed to be 2,232 in.-lb/ft. This compares to a computed 
strength of the wall of 14,378 in.-lb/ft, considering only the #5 bars at 4 ft on center. Thus, the wall is 
loaded to about 16 percent of its capacity in flexure in the out-of-plane direction. (See Sec. 9.1 for a more 
detailed discussion of strength design of masonry walls, including the P-delta effect.) 
9.2.7.8 Los Angeles Orthogonal Effects 
According to Provisions Sec. 5.2.5.2.2 [Sec. 4.4.2.3], orthogonal interaction effects have to be considered 
for Seismic Design Category D when the ELF procedure is used (as it is here). 
The out-of-plane effect is 16 percent of capacity, as discussed in Sec. 9.2.7.7 above. When considering 
the 0.3 combination factor, the out-of-plane action adds about 5 percent overall to the interaction effect. 
For the lowest story of the wall, this could conceivably require a slight increase in capacity for in-plane 
actions. In the authors opinion, this is on the fringe of requiring real consideration (in contrast to the end 
walls of Example 9.1). 
This completes the design of the transverse Wall D. 
9.2.7.9 Los Angeles Summary of Wall Design for Wall D 
8-in. CMU 
f!m = 2,000 psi 
Reinforcement: 
Four vertical #8 bars, one bar in each cell for the four end cells 
Vertical #5 bars at 4 ft on center at intermediate cells 
Two bond beams with two #5 bars at each story, at floor bearing and at 4 ft above each floor 
Horizontal joint reinforcement at alternate courses recommended, but not required 
Grout: 
All cells with reinforcement and bond beams, plus solid grout at first story, at two out of three 
cells in the second story, and at six extra cells in the third story 
Table 9.2-22 compares the reinforcement and grout for Wall D designed for each of the four locations. 
Table 9.2-22 Variation in Reinforcement and Grout by Location 
Birmingham 1 New York City Birmingham 2 Los Angeles 
Vertical bars 5 - #4 9 - #4 12 - #4 8 - #8 + 7 - #5 
Horizontal bars 10 - -#4 + jt. reinf 10 - #4 + jt. reinf 20 - #5 20 - #5 
Grout (cu. ft.) 91 122 189 295 
1 cu. ft. = 0.0283 m3.
Chapter 9, Masonry 
9-95 
40'-0" 24'-0" 24'-0" 24'-0" 40'-0" 
4'-0" 
72'-0" 
24'-0" 24'-0" 24'-0' 
152'-0" 
A
B
B
A 
Joists 
at 2'-6" 
A
B
B
A 
C C 
C C 
D D 
D D 
Bulbs at ends 
of walls 
101
2" grouted 
brick walls at 
lower 6 stories. 
8" brick in 
upper 6 stories. 
Figure 9.3-1 Floor plan (1.0 ft = 0.3048 m, 1.0 in. = 25.4 mm) 
9.3 TWELVE-STORY RESIDENTIAL BUILDING IN 
LOS ANGELES, CALIFORNIA 
9.3.1 Building Description 
This 12-story residential building has a plan form similar to that of the five-story masonry building 
described in Sec. 9.2. The floor plan and building elevation are illustrated in Figures 9.3-1 and 9.3-2, 
respectively. The floors are composed of 14-in.-deep open web steel joists spaced at 30 in. that support a 
3-in. concrete slab on steel form deck. A fire-rated ceiling is included at the bottom chord of the joists. 
Partitions, including the shaft openings, are gypsum board on metal studs, and the exterior nonstructural 
curtain walls are glass and aluminum. 
All structural walls are of grouted brick. For purposes of illustration, two styles of wall are included. The 
lower six stories have 10-1/2in.-thick walls consisting of two wythes of 4-in. (nominal) brick and a 3-1/4- 
in. grout space. The upper six stories have 8-in. (nominal) brick, hollow unit style, with the vertical 
reinforcing in the cells and the horizontal reinforcing in bond beams. (In actual construction, however, a 
single style wall might be used throughout: either a two-wythe grouted wall or a through-the-wall unit of 
an appropriate thickness). The walls are subject to high overturning moments and have a reinforced 
masonry column at each end. The column concentrates the flexural reinforcement and increases resistance 
to overturning. (Similar concentration and strength could be obtained with transverse masonry walls 
serving as flanges for the shear walls had the architectural arrangement been conducive to this approach.) 
Although there is experimental evidence of improved performance of walls with all vertical reinforcement 
uniformly distributed, concentration at the ends is common in engineering practice and the flexural 
demands are such for this tall masonry building that the concentration of masonry and reinforcement at 
the ends is simply much more economical. 
The compressive strength of masonry, f!m, used in this design is 2,500 psi for Levels 1 through 6 and 
3,000 psi for Levels 7 through 12.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-96 
152'-0" 
36'-0" 80'-0" 36'-0" 
12 at 10'-0"=120'-0" 2'-0" 
122'-0" 
Figure 9.3-2 Elevation (1.0 ft = 0.3048 m, 1.0 in. = 25.4 mm) 
This example illustrates the following aspects of the seismic design of the structure: 
1. Development of equivalent lateral forces 
2. Reinforced masonry shear wall design 
3. Check for building deflection and story drift 
4. Check of diaphragm strength. 
9.3.2 Design Requirements 
9.3.2.1 Provisions Design Parameters 
Table 9.3-1 shows the design parameters for building design.
Chapter 9, Masonry 
9-97 
Table 9.3-1 Design Parameters 
Design Parameter Value 
Ss (Map 1 [Figure 3.3-3]) 1.5 
S1 (Map 2 [Figure 3.3-4]) 0.6 
Site Class C 
Fa 1 
Fv 1.3 
SMS = FaSs 1.5 
SM1 = FvS1 0.78 
SDS = 2/3 SMS 1 
SD1 = 2/3 SM1 0.52 
Seismic Design Category D 
Masonry Wall Type Special Reinforced 
R 3.5 
O0 2.5 
Cd 3.5 
[The 2003 Provisions have adopted the 2002 USGS probabilistic seismic hazard maps, and the maps have 
been added to the body of the 2003 Provisions as figures in Chapter 3 (instead of the previously used 
separate map package).] 
9.3.2.2 Structural Design Requirements 
The load path consists of the floors acting as horizontal diaphragms and the walls parallel to the motion 
acting as shear walls. 
Soil-structure interaction is not considered. 
The building is a bearing wall system (Provisions Table 5.2.2 [4.3-1]) . 
Deformational compatibility must be assured (Provisions Sec. 5.2.2.4.3 [Sec. 4.5.3]). The structural 
system is one of non-coupled shear walls. Crossing beams over the halls (their design is not included in 
this example) will need to continue to support the gravity loads from the floors and roof during an 
earthquake but will not provide coupling between the shear walls. 
The building is symmetric in plan but has the same torsional irregularity described in Sec. 9.2.5.4. The 
vertical configuration is regular except for the change in wall type between the sixth and seventh stories, 
which produces a significant discontinuity in stiffness and strength, both for shear and flexure (Provisions 
Sec. 5.2.3.3 [Sec. 4.3.2.3] and 5.2.6.2.3 [Sec. 4.6.1.6]). There is no weak story because the strength does 
not increase as one goes upward. The stiffness discontinuity will be shown to qualify as regular. 
Provisions Table 5.2.5.1 [Table 4.4-1] would not permit the use of the ELF procedure of Provisions Sec. 
5.4 [Sec. 5.2]; instead a dynamic analysis of some type is required. As will be illustrated, this particular 
building does not really benefit from this requirement.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-98 
The design and detailing must comply with the requirements of Provisions Sec. 5.2.6 [Sec. 4.6]. 
The walls must resist forces normal to their plane (Provisions Sec. 5.2.6.2.7 [Sec. 4.6.1.3]). These forces 
will be used when addressing the orthogonal effects (Provisions Sec.5.2.5.2.2 [Sec. 4.4.2.3]). 
With eight walls in each direction, the system is expected to be redundant. 
Tie and continuity requirements for anchorage of masonry walls must be considered when detailing the 
connections between floors and walls (Provisions Sec. 5.2.6.1.2 [Sec. 4.6.2.1] and 5.2.6.1.3). 
Openings in walls and diaphragms need to be reinforced (Provisions Sec. 5.2.6.2.2 [Sec. 4.6.1.4]). 
Diaphragms need to be designed to comply with Provisions Sec. 5.2.6.2.6 [Sec. 4.6.3.4]. 
The story drift limit is 0.01hsx (Provisions Sec. 5.2.8 [Sec. 4.5.1]) and the overall drift limit is 0.01hsn 
(Provisions Sec. 11.5.4.1). For this structure the difference between these two is significant, as will be 
shown. 
[The deflection limits have been removed from Chapter 11 of the 2003 Provisions because they were 
redundant with the general deflection limits. Based on ACI 530 Sec. 1.13.3.2, the maximum drift for all 
masonry structures is 0.007 times the story height. Thus, there appears to be a conflict between ACI 530 
and 2003 Provisions Table 4.5-1.] 
9.3.2.3 Load Combinations 
The basic load combinations (Provisions Sec. 5.2.7 [Sec. 4.2.2]) are the same as those in ASCE 7 except 
that the seismic load effect, E, is defined by Provisions Eq. 5.2.7-1 [Eq. 4.2-1] and 5.2.7-2 [Eq. 4.2-2] as: 
E = .QE  0.2SDSD 
Based on the configuration of the shear walls and the results presented in Sec. 9.2, the reliability factor, ., 
is treated as equal to 1.0 for both directions of loading. Refer to Sec. 9.2.3.1 for additional information. 
[The redundancy requirements have been substantially changed in the 2003 Provisions. For a shear wall 
building assigned to Seismic Design Category D, . = 1.0 as long as it can be shown that failure of a shear 
wall with height-to-length-ratio greater than 1.0 would not result in more than a 33 percent reduction in 
story strength or create an extreme torsional irregularity. The intent is that the aspect ratio is based on 
story height, not total height. Therefore, the redundancy factor would not have to be investigated (. = 
1.0) for this building.] 
The discussion on load combinations for the Los Angeles site in Sec.9.2 is equally applicable to this 
example. Refer to Sec. 9.2.3.2 for determination of load combinations. 
The load combinations representing the extreme cases are: 
1.4D + QE + 0.5L 
0.7D + QE 
9.3.3 Seismic Force Analysis 
The analysis is performed using the ELF procedure of Provisions Sec. 5.4 [Sec. 5.2] and checked with a 
modal response spectrum (MRS) analysis in conformance with Provisions Sec. 5.5 [Sec. 5.3]. This
Chapter 9, Masonry 
9-99 
example illustrates an analysis for earthquake motions acting in the transverse direction only. Earthquake 
motions in all directions will need to be addressed for an actual project. 
9.3.3.1 Building Weights 
For the ELF analysis, the masses are considered to be concentrated at each floor level whereas, for the 
MRS analysis, it is distributed on both wall and floor elements. Note that the term level corresponds to 
the slab above each story. Thus Level 1 is the second floor; Level 12 is the roof. 
Lower Levels: 
Slab, joists, partitions, ceiling, mechanical/electrical (M/E), curtain wall at 53 psf 
(0.053 ksf)(152 ft)(72 ft) = 580 kips/story 
Walls: 10.5 in. at 114 psf (brick at 73 psf + grout at 41 psf) 
(0.114 ksf)(10 ft)[(8)(29 ft) + (4)(30 ft) + (4)(32 ft)] = 547 kips/story 
Bulbs at ends of walls: (24 in.  24 in. bulb) 
Brick: 2.01 ft2 /bulb; Grout: 1.99 ft2/bulb 
[(2.01 ft2)(0.120 kcf) + (1.99 ft2)(0.150 kcf)] (10 ft)(32 bulbs) = 173 kips/story 
Upper Levels: 
Slab, joists, partitions, ceiling, M/E, curtain wall at 53 psf 
(0.053 ksf)(152 ft)(72 ft) = 580 kips/story 
Walls: 8 in. Partially grouted brick at 48 psf 
(0.048 ksf)(10 ft)[(8)(29.67 ft) + (4)(30.67 ft) + (4)(32.67 ft)] = 236 kips/story 
Bulbs at ends of walls (grouted 20 in.  20 in. brick bulb) 
(0.315 klf/bulb)(10 ft)(32 bulbs) = 101 kips/story 
Roof: 
Slab, roofing, joists,, ceiling, M&E, curtain wall at 53 psf : 
(0.053 ksf)(152 ft)(72 ft) = 580 kips 
Walls 
(238 kips/story + 101 kips/story)/2 = 170 kips 
Parapet 
(4 parapets)(2 ft)[(0.048 kips/lf)(33 ft) +(3.15 kips/bulb)(2 bulbs)/(10 ft)] = 18 kips 
Preliminary design indicates a 10-1/2-in. wall with bulbs for the six lower stories and an 8-in. wall with 
bulbs for the six upper stories. Therefore, effective seismic weight, W, is computed as follows: 
Levels 1-5 (5)(580 + 547 + 173) = (5)(1,300 kips/level) = 6,500 kips 
Level 6 580 + (547 + 236)/2 + (173 + 101)/2 = 1,109 kips = 1,109 kips 
Levels 7-11 (5)(580 + 236 + 101) = (5)(917 kips/level) = 4,585 kips 
Level 12 (roof) (580 + 170 + 18) = 768 kips = 768 kips 
Total = 12,962 kips
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-100 
The weight of the lower half of walls for the first story are not included with the walls for Level 1 because 
the walls do not contribute to the seismic loads. 
9.3.3.2 Base Shear 
The seismic coefficient, Cs, for the ELF analysis is computed as: 
1.00 0.286 
s / 3.5/1.0 
C SDS 
R I = = = 
The value of Cs need not be greater than: 
0.52 
0.198 
( / ) (0.75)(3.5 /1) 
= = = D1 
s 
S 
C 
T R I 
The value of the fundamental period, T, was determined from a dynamic analysis of the building modeled 
as a cantilevered shear wall. RISA 2D was used for this analysis, with cracked sections taken into 
account. From this analysis, a period of T = 0.75 sec was determined. See Sec. 9.3.4. This value is also 
obtained from the 3D dynamic analysis (described subsequently) for the first translational mode in the 
transverse direction when using a reduced modulus of elasticity to account for cracking in the masonry 
(approximately 60 percent of the nominal value for E). Provisions Sec. 5.4.2 [Sec. 5.2-2] requires that the 
fundamental period, T, established in a properly substantiated analysis be no larger than the approximate 
period, Ta , multiplied by Cu, determined from Provisions Table 5.4.2 [Table 5.2-1]. The approximate 
period of the building, Ta, is calculated based as: 
= 3/ 4=(0.02)(120)0.75=0.725 sec a r n T Ch 
where Cr = 0.02 from Provisions Table 5.4.2.1 [Table 5.2-2], and hn = 120 ft 
TaCu = (0.725)(1.4) = 1.015 sec > 0.75 sec = T 
(Note that T = 0.75 sec will be verified later when deflections are examined). 
The value for Cs is taken to be 0.198 (the minimum of the two values computed above). This value is 
still larger than the minimum specified: 
=0.044 =(0.044)(1.0)(0.60)=0.0264 Cs ISD1 
[This minimum Cs value has been removed in the 2003 Provisions. In its place is a minimum Cs value 
for long-period structures, which is not applicable to this example.] 
The total seismic base shear is then calculated by Provisions Eq. 5.4.1 [Eq. 5.2-1]: 
V = CsW = (0.198)(12,962 kips) = 2,568 kips 
A 3-D model was created in SAP 2000 for the MRS analysis. Just as for the five-story building described 
in Sec. 9.2, the masonry walls were modeled as shell bending elements and the floors were modeled as an 
assembly of beams and shell membrane elements. See Sec. 9.2.6.2 for further description. The difference 
in fm between upper and lower stories was not modeled; the value of Em used was 1,100 ksi, which is 59 
percent of the value from Provisions Eq. 11.3.10.2 for the lower stories. [Note that by adopting ACI 530 
in the 2003 Provisions, Em = 900fm per ACI 530 Sec. 1.8.2.2.1.] As mentioned, this value was selected as
Chapter 9, Masonry 
9-101 
an approximation of the effects of flexural cracking. Unlike the five-story building, the difference in 
length between the longitudinal and transverse walls was modeled. However, to simplify construction of 
the model, wall types A and B are the same length. Because this example illustrates design in the 
transverse direction, this liberty has little effect. Table 9.3-2 shows data on the modes of vibration used in 
the analysis.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-102 
Table 9.3-2 Periods, mass participation ratios, and modal base shears in the transverse 
direction for modes used in analysis 
Mode Period Individual mode (percent) Cumulative sum (percent) Trans. 
number (seconds) Long. Trans. Vert. Long. Trans. Vert. base shear 
1 0.9471 0.00 0.00 0.00 0.00 0.00 0.00 0.0 
2 0.7469 0.00 59.12 0.00 0.00 59.12 0.00 1528.0 
3 0.6941 59.16 0.00 0.00 59.16 59.12 0.00 0.0 
4 0.2247 0.00 0.00 0.00 59.16 59.12 0.00 0.0 
5 0.1763 0.00 24.38 0.00 59.16 83.50 0.00 896.2 
6 0.1669 24.57 0.00 0.00 83.73 83.50 0.00 0.0 
7 0.1070 0.00 0.01 0.00 83.73 83.51 0.00 0.5 
8 0.1059 0.00 0.00 0.28 83.74 83.51 0.28 0.0 
9 0.1050 0.00 0.00 29.48 83.74 83.51 29.76 0.0 
10 0.0953 0.00 0.00 0.00 83.74 83.51 29.76 0.0 
11 0.0900 0.00 0.00 1.51 83.74 83.51 31.27 0.0 
12 0.0858 0.00 0.03 0.01 83.74 83.54 31.28 1.1 
13 0.0832 0.00 7.25 0.00 83.74 90.79 31.28 234.4 
14 0.0795 7.11 0.00 0.00 90.85 90.79 31.28 0.0 
15 0.0778 0.04 0.00 0.19 90.88 90.79 31.48 0.0 
16 0.0545 0.00 4.47 0.00 90.88 95.26 31.48 117.5 
17 0.0526 4.44 0.00 0.00 95.32 95.26 31.48 0.0 
18 0.0413 0.01 1.24 0.00 95.33 96.51 31.48 29.1 
19 0.0392 1.66 0.05 0.00 96.99 96.55 31.48 1.1 
20 0.0358 0.07 0.87 0.00 97.06 97.43 31.48 19.5 
21 0.0288 1.59 0.33 0.01 98.66 97.76 31.49 7.0 
22 0.0278 0.33 1.40 0.00 98.99 99.16 31.49 28.9 
23 0.0191 0.76 0.23 0.01 99.75 99.39 31.50 4.3 
24 0.0186 0.23 0.60 0.00 99.98 99.98 31.50 11.1 
1 kip = 4.45kN. 
The combined modal base shear is 1,791 kips 
The fundamental mode captures no translation of mass; it is a pure torsional response. This is a 
confirmation of the intent of the torsional irregularity provision. The first translational mode has a period 
of 0.75 sec, confirming the earlier statements. Also note that the base shear is only about 70 percent of 
the ELF base shear (2,568 kips) even though the fundamental period is the same. The ELF analysis 
assumes that all the mass participates in the fundamental mode whereas the dynamic analysis does not. 
The absolute sum of modal base shears is higher than the ELF but the statistical sum is not. Provisions 
Sec. 5.5.7 requires that the modal base shear be compared with 85 percent of the ELF base shear. The 
comparison value is 0.85(2,568 kips), which is 2,183 kips. Because this is greater than the value from the 
modal analysis, the modal analysis results would have to be factored upwards by the ratio 2,183/1,791 = 
1.22. The period used for this comparison cannot exceed CuTa, which is 1.015 sec as described 
previously. Note that the period used is from Mode 2, because Mode 1 is a purely torsional mode. The 
1.22 factor is very close to the factor for the five story building computed in Sec. 9.2.6.2; an additional 
comparison will follow. 
9.3.3.3 Vertical Distribution of Seismic Forces 
Carrying forward with the ELF analysis, Provisions Sec. 5.4.3 [Sec. 5.2.3] provides the procedure for 
determining the portion of the total seismic loads assigned to each floor level. The story force, Fx, is 
calculated as:
Chapter 9, Masonry 
9-103 
Fx=CvxV 
and 
=1 
= 
S 
k 
x x 
vx n 
k 
i i 
i 
C w h 
wh 
For T = 0.75 sec, which is between 0.5 sec and 2.5 sec, the value of k is determined to be 1.125 based on 
interpolation (Provisions Sec. 5.4.3 [Sec. 5.2.3]). 
The seismic design shear in any story shall be determined from: 
n 
x i 
i x 
V F 
= 
= S 
The story overturning moment is computed from: 
( ) 
n 
x i i x 
i x 
M F h h 
= 
=S - 
Table 9.3-3 shows the application of these equations for this building. 
Table 9.3-3 Seismic Forces and Moments by Level 
Level 
(x) 
wx 
(kips) 
hx 
(kips) 
1.125 
wxhx 
(ft-kips) 
Cvx Fx 
(kips) 
Vx 
(kips) 
Mx 
(kips) 
12 
11 
10 
9 
8 
7 
6 
5 
4 
3 
2 
1 
768 
917 
917 
917 
917 
917 
1,109 
1,300 
1,300 
1,300 
1,300 
1,300 
120 
110 
100 
90 
80 
70 
60 
50 
40 
30 
20 
10 
167,700 
181,500 
163,100 
144,800 
126,900 
109,200 
111,000 
106,000 
82,500 
59,700 
37,800 
17,300 
1,307,400 
0.128 
0.139 
0.125 
0.111 
0.097 
0.084 
0.085 
0.081 
0.063 
0.046 
0.029 
0.013 
1.00 
329 
357 
320 
284 
249 
214 
218 
208 
162 
117 
74 
34 
2,568 
329 
686 
1,006 
1,291 
1,540 
1,754 
1,972 
2,181 
2,342 
2,460 
2.534 
2,568 
3,300 
10,200 
20,200 
33,100 
48,500 
66,000 
85,800 
107,600 
131,000 
155,600 
181,000 
206,600 
1.0 kip = 4.45 kN, 1.0 ft = 0.3048 m, 1.0 ft-kip = 1.36 kN-m. 
The dynamic modal analysis does give a direct output for the gross overturning moment, of about 66 
percent of the moment from the ELF analysis. Because the model is built with shell elements, there is no 
direct value for the variation of moment with height.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-104 
9.3.3.4 Horizontal Distribution 
For the ELF analysis, the approach is essentially the same as used for the five-story masonry building 
described in Sec. 9.2.4.4: 
Direct shear: All transverse walls have the same properties, except axial load. Axial load affects cracking 
but, each wall considered has the same stiffness. Therefore, each will resist an equivalent amount in 
direct shear: 
V = V/8 = 0.125Vx 
Torsion: The center of mass corresponds with the center of resistance; therefore, the only torsion is due 
to the 5 percent accidental eccentricity in accordance with Provisions Sec. 5.4.4.2 [Sec. 5.2.4.2]: 
Mta = 0.05 bV = (0.05)(152 ft)V = 7.6V 
The longitudinal walls are slightly longer than the transverse walls. Unlike the example in Sec. 9.2, this 
difference will be illustrated here in a simple fashion. The diaphragm is assumed to be rigid. When the 
walls are not identical, a measure of the actual stiffness is necessary; for masonry walls, this involves both 
flexural and shear deformations. The conventional technique is an application of the following equation 
for deformation of a simple cantilever wall without bulbs or flanges at the ends: 
3 6 
3 5 wall 
m m 
Vh Vh 
E I G A 
. = + 
Considering Gm = 0.4 Em, A = Lt, and I = L3t/12, this can be simplified to : 
3 
4 3 wall 
V h h 
Et L L 
. = .... .. + .. .... 
... . . ... 
Rigidity, K, is inversely proportional to deflection. Considering E and t as equal for all walls: 
3
1 
4 3 
K 
h h 
L L 
= 
.. .. + .. .. 
. . . . 
Figure 9.3-3 identifies the walls. 
For a multistory building, the quantity h is not easy to pin down. For this example, the authors suggest 
the following approach: use h= 10 ft (one story) to evaluate the shear in the wall at the base and also use h 
= 80 ft (two thirds of total height) to evaluate the moments in the walls. Table 9.3-4 shows some of the 
intermediate steps for these two assumptions. 
(d, as used here, is the distance of the wall to the centroid of the building, not the length of the 
wall, as used elsewhere) 
Chapter 9, Masonry 
9-105 
Table 9.3-4 Relative Rigidities 
Wall Length 
(ft) 
Arm, d 
(ft) 
for shear, h = 10 ft for moment, h = 80 ft 
h/d K Kd2 (ft2) h/d K Kd2 (ft2) 
ABCD 
36 
34 
33 
33 
36 
12 
12 
36 
0.278 
0.294 
0.303 
0.303 
1.088 
1.017 
0.980 
0.980 
1,410 
146 
141 
1,270 
2,967 
2.22 
2.35 
2.42 
2.42 
0.01978 
0.01690 
0.01556 
0.01556 
25.63 
2.43 
2.24 
20.17 
50.47 
1.0 ft = 03.048 m 
The total torsional rigidity is four times the amount in Table 9.3-4, since there are four walls of each type. 
When considering torsion, Wall D is the critical member (shortest length, greatest d). 
For shear due to accidental torsion: 
2 
(0.980)(36) 
7.6 0.0226 or 7.6 
4(2,967) 
0.01556(36) 0.0211 
4(50.47) t 
MKd 
V V V V 
Kd 
= = = V 
S 
... ... ... ...=
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-106 
When considering the approximations involved, the remainder of the ELF example will simply use 
0.0226V for Vt. Because a 3D analytical model exists, a simplistic load case with a static horizontal 
torsion at each level was defined. The couple varied directly with height, so the variation of mass with 
height was ignored. Examining the base reactions for Wall D yields a torsional shear equal to 0.0221V 
and an overturning moment corresponding to 0.0191V. Therefore, the hand computations illustrated are 
somewhat conservative. 
Total shear for Wall D is equal to the direct shear plus shear due to accidental torsion, which is computed 
as: 
0.125V + 0.0226V = 0.148V 
2'-0" 1'-8" 
1'-8" 
2'-0" 
d
d 
Wall at Stories 7 to 12 
Wall at Stories 1 to 6 
8" 
101
2" 
Wall length 
Wall d 
A 36'-0" 
B 34'-0" 
C 33'-0" 
D 33'-0" 
Figure 9.3-3 Wall dimensions (1.0 in. = 25.4 mm, 1.0 ft = 0.0348 m).
Chapter 9, Masonry 
9-107 
The resulting shears and overturning moments for Wall D are shown in Table 9.3-5. 
Table 9.3-5 Shear for Wall D 
Level Story Shear 
(kips) 
Wall Shear 
(kips) 
Story Moment 
(ft-kips) 
Wall Moment 
(ft-kips) 
12 
11 
10 
9 
8 
7 
6 
5 
4 
3 
2 
1 
329 
686 
1,006 
1,291 
1,540 
1,754 
1,972 
2,181 
2,342 
2,460 
2,534 
2,568 
49 
102 
149 
191 
228 
260 
292 
323 
347 
364 
375 
380 
3,300 
10,200 
20,200 
33,100 
48,500 
66,000 
85,800 
107,600 
131,000 
155,600 
181,000 
206,600 
500 
1,500 
3,000 
4,900 
7,200 
9,800 
12,700 
15,900 
19,400 
23,000 
26,800 
30,60 
1.0 kip = 4.45 kN, 1.0 ft-kip = 1.36 kN-m. 
When considering accidental torsion, a check for torsional irregularity must be made. First consider the 
case used for design: a direct shear of 0.125V and a torsional shear of 0.0226V. The ratio of extreme 
displacement to average displacement can be found from these values and the dimensions, considering 
symmetry: 
Average displacement is proportional to 0.125V 
Torsional displacement at Wall D is proportional to 0.0226V 
Torsional displacement at the corner is proportional to (0.0226V)((152 ft./2)/36 ft. = 0.0447V 
Ratio of corner to average displacement = (0.125 + 0.0447)/0.125 = 1.38 
If the lower value of torsional shear, 0.0191V, found from the 3D computer analysis for the static torsion 
is used, the ratio becomes 1.32. In either case, the result is a torsional irregularity (ratio exceeds 1.2) but 
not an extreme torsional irregularity (ratio does not exceed 1.4). The reason for the difference from the 
five-story building, in which the ratio exceeded 1.4, is that the longer walls in the longitudinal directions. 
For the ELF analysis, Provisions Sec. 5.4.4.3 [Sec. 5.2.4.3] requires the accidental torsion to be 
amplified: 
2 
3.0 
1.2 x 
A Max displacement 
Ave displacement 
. . 
=. .= 
. . 
If one uses the ratio of 1.32 based on the 3D computer analysis, the amplifier is 1.21 and the torsional 
shear becomes 1.32(0.0191V) = 0.0231V. This is close enough to the unamplified 0.0226V that the ELF 
analysis will simply proceed with a torsional shear of 0.0226V. 
As described in Sec. 9.2.6.4 for the five-story building, the 3D analytical model was altered to offset the 
center of mass from the center of rigidity. The modal periods, mass participation ratios, and base shears 
are given in Table 9.3-6. The total base shear is 1620 kips, down from the 1,791 kips found without the 
eccentricity. The 1,620 kips still slightly exceeds the minimum for design of 1,613 kips described earlier. 
Thus, MRS analysis can be used directly in the load combinations and can be considered to include the
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-108 
amplified accidental torsion. 
Table 9.3-6 Periods, Mass Participation Ratios, and Modal Base Shears in the Transverse 
Direction for Modes Used in Analysis of Building with Deliberate Eccentricity 
Mode Period Individual mode (percent) Cumulative sum (percent) Trans. 
number (seconds) Long. Trans. Vert. Long. Trans. Vert. base shear 
1 0.965 0.0 8.5 0.0 0.0 8.5 0.0 169.4 
2 0.723 0.0 50.6 0.0 0.0 59.1 0.0 1352.7 
3 0.694 59.2 0.0 0.0 59.2 59.1 0.0 0.0 
4 0.229 0.0 3.3 0.0 59.2 62.5 0.0 122.7 
5 0.171 0.0 21.0 0.0 59.2 83.5 0.0 772.7 
6 0.167 24.6 0.0 0.0 83.7 83.5 0.0 0.0 
7 0.120 0.0 0.0 20.3 83.7 83.5 20.3 0.0 
8 0.108 0.0 1.0 0.0 83.7 84.4 20.3 35.3 
9 0.105 0.0 0.0 0.0 83.7 84.4 20.3 0.1 
10 0.097 0.0 0.0 0.0 83.7 84.4 20.3 0.1 
11 0.090 0.0 0.0 11.4 83.8 84.4 31.7 0.0 
12 0.081 0.0 6.2 0.0 83.8 90.7 31.7 198.6 
13 0.079 7.1 0.0 0.1 90.9 90.7 31.8 0.0 
14 0.074 0.0 0.0 3.1 90.9 90.7 34.8 0.1 
15 0.072 0.0 0.6 0.1 90.9 91.3 34.9 17.5 
16 0.061 0.0 0.5 0.1 90.9 91.7 34.9 12.8 
17 0.053 4.3 0.0 0.0 95.2 91.7 34.9 0.0 
18 0.052 0.0 4.1 0.0 95.2 95.8 34.9 104.8 
19 0.043 0.7 0.0 0.0 95.9 95.8 34.9 0.0 
20 0.037 1.5 0.0 0.0 97.4 95.8 35.0 0.1 
21 0.035 0.0 2.5 0.0 97.4 98.3 35.0 54.7 
22 0.027 1.8 0.0 0.0 99.2 98.3 35.0 0.0 
23 0.022 0.0 1.7 0.0 99.2 100.0 35.0 32.8 
24 0.018 0.8 0.0 0.0 100.0 100.0 35.0 0 
1 kip = 4.45 kN 
The combined modal base shear is 1620 kips. 
Mode 1 now includes a translational component, and the comparison to an ELF base shear would be 
performed with its period. For T = 0.965 sec, the ELF base shear becomes 1,996 kips, and the 
comparison value is 0.85 (1996) = 1,696 kips. This is 78 percent of the value for the symmetric model 
and illustrates one of the problems in handling accidental torsion in a consistent fashion. 
Without factoring the modal results up to achieve a base shear of 1,696 kips (a factor of 1.047), the 
reactions indicate that the base shear for wall D is 266.5 kips, or 0.1645 times the total base shear. If one 
takes the direct shear as one-eighth (0.125V), that leaves 0.0395V for the dynamically amplified shear due 
to accidental torsion, which could be interpreted to be an amplification of 1.79 over the shear of 0.0221V 
found for the static torsion. Thus, it is clear that the amplification value of 1.21 from the equation given 
for the ELF analysis underestimates the dynamic amplification of accidental torsion. The bottom line is 
that the Wall D shear of 266.5 kips from the dynamic analysis is significantly less than the shear of 380 
kips found in the ELF analysis without amplification of accidental torsion. The example will proceed 
based upon the shear of 380 kips. 
9.3.3.5 Transverse Wall (Wall D)
Chapter 9, Masonry 
9-109 
The strength or limit state design concept is used in the Provisions. 
[The 2003 Provisions adopts by reference the ACI 530-02 provisions for strength design in masonry, and 
the previous strength design section has been removed. This adoption does not result in significant 
technical changes, and the references to the corresponding sections in ACI 530 are noted in the following 
sections.] 
9.3.3.5.1 Axial and Flexural Strength General 
The walls in this example are all bearing shear walls since they support vertical loads as well as lateral 
forces. 
The demands for the representative design example, Wall D, are presented in this section. The design of 
the lower and upper portions of Wall D is presented in the next two sections. For both locations, in-plane 
calculations include: 
1. Strength check and 
2. Ductility check. 
The axial and flexural demands for Wall D, using the load combinations identified in Sec. 9.3.2.3, are 
presented in Table 9.3-7. In the table, Load Combination 1 represents 1.4D + 1.0 QE + 0.5L, and Load 
Combination 2 represents 0.7D + 1.0 QE. 
Table 9.3-7 Load Combinations for Wall D 
Load Combination 1 Load Combination 2 
Level PD 
(kips) 
PL 
(kips) 
Pu 
(kips) 
Mu 
(ft-kips) 
Pu 
(kips) 
Mu 
(ft-kips) 
12 
11 
10 
9 
8 
7 
6 
5 
4 
3 
2 
1 
37 
80 
124 
168 
212 
255 
308 
370 
432 
494 
556 
618 
0 
8 
17 
25 
34 
43 
50 
59 
67 
76 
84 
92 
51 
117 
182 
247 
313 
379 
457 
548 
639 
730 
821 
912 
500 
1,500 
3,000 
4,900 
7,200 
9,800 
12,700 
15,900 
19,400 
23,000 
26,800 
30,600 
26 
56 
87 
117 
148 
179 
216 
259 
303 
346 
389 
433 
500 
1,500 
3,000 
4,900 
7,200 
9,800 
12,700 
15,900 
19,400 
23,000 
26,800 
30,60 
1.0 kip = 4.45 kN, 1.0 ft-kip = 1.36 kN-m. 
Strength at the lowest story (where P, V, and M are the greatest) for both the lower wall (Level 1) and the 
upper wall (Level 7) constructions will be examined. The design for both locations is based on the values 
for Load Combination 2 in Table 9.3-7.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-110 
2'-0" 
2'-0" 
1012" 
358" 358" 
(16) #9 
with #3 ties 
at 1'-4" o.c. 
Figure 9.3-4 Bulb reinforcement at lower levels 
(1.0 in. = 25.4 mm, 1.0 ft = 0.3048 m). 
9.3.3.5.2 Axial and Flexural Strength Lower Levels 
Examine the strength of Wall D at Level 1: 
9.3.3.5.2.1 Strength Check (Level 1) 
Pumin = 433 kips + factored weight of half of 1st story wall = 433 + (0.7)(21.9) = 448 kips 
Mu = 30,600 ft-kips 
For this Seismic Design Category D building, the special reinforced masonry shear walls must have 
vertical and horizontal reinforcement spaced at no more than 4 ft on center. The minimum in either 
direction is 0.0007(10.5 in.) = 0.0074 in.2/in. (vertical #5 at 42 in. on center). That will be used as the 
vertical reinforcement (although some of the subsequent calculations of flexural resistance are based upon 
a spacing of 48 in. on center); the shear strength demands for horizontal reinforcement will be greater and 
will satisfy the total amount of 0.0020(10.5 in.) = 0.021 in.2/in.(horizontal #5 at 22 in. on center will 
suffice). 
Try 16 #9 bars in each bulb. Refer to Figure 9.3-4 for the placement of the reinforcement in the bulb. In 
some of the strength calculations, the #5 bars in the wall will be neglected as a conservative 
simplification.
Chapter 9, Masonry 
9-111 
Cm bulb 
N.A. 
P = 0 Case 
Intermediate 
Case 
Balanced 
Case 
m = 0.0035 
m = 0.0035 
m = 0.0035 
y = 0.00207 
y = 0.00207 
Ts 
y = 0.00207 
y = 0.00207 
Ts
Ts 
y = 0.00207 
Cs 
Cm wall 
Cs 
N.A. 
Cm bulb Cm wall 
Cm bulb 
N.A. 
2.00' 2.00' 
33.00' 
29.00' 
M 
P 
15.50' 
a = 0.8c 
= 1.37' 
c = 1.71' 
2.00' 
c = 3.00' 
a = 0.8c 
= 2.40' 
a = 0.8c = 16.09' 
c = 20.11' 
S Ts2 
S Ts2 
Figure 9.3-5 Strength of Wall D, Level 1 (1.0 ft = 0.3048 m) 
For evaluating the capacity of the wall, a fPn - fMn curve will be developed to represent the wall strength 
envelope. The demands (Pu and Mu determined above) will then be compared to this curve. Several cases 
will be analyzed and their results used in plotting the fPn - fMn curve. Refer to Figure 9.3-5 for notation 
and dimensions. 
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-112 
Case 1 (P = 0) 
The neutral axis will be within the compression bulb, so assume that only the bars closest to the 
compression face are effective in compression. (Also recall that the Provisions clearly endorses the use of 
compression reinforcement in strength computations.) 
Ts1 = (16 bars)(1.00 in.2)(60 ksi) = 960 kips 
Ts2 = (7 bars)(0.31 in.2)(60 ksi) = 130 kips 
Cs = (5 bars)(1.00 in.2)(60 ksi) = 300 kips 
SC = ST + P 
Cm + Cs = Ts1 + Ts2 + P 
Cm = 960 + 130 +0 - 300 = 790 kips 
Cm = 790 kips = ffm(24 in.)a = (0.8)(2.5 ksi)(24 in.)a 
a = 16.46 in. = 1.37 ft 
c = a/0.8 = 1.37/0.8 = 1.71 ft = 20.7 in. 
Check strain in compression steel 
es = 0.0035(20.7 in. - 6 in.)/(20.7 in.) = 0.0025 > yield; assumption OK 
SMcl = 0 
Mn = (790 kips)(16.5 ft- 1.37 ft/2) + (300 + 960 kips)(15.5 ft) + (130 kips)(0 ft.) = 32,170 ft-kips 
fMn = (0.85)(32,170) = 27,340 ft-kips 
Case 2 (Intermediate case between P = 0 and balanced case): 
Select an intermediate value of c. Let c = 3.0 ft, and determine Pn and Mn for this case. 
a = 0.8c = 2.4 ft 
Cm bulb = (0.8)(2.5 ksi)(24 in.)2 = 1152 kips 
Cm wall = (0.8)(2.5 ksi)(10.5 in.)(0.4 ft.  12) = 101 kips 
Cs = (16 bars)(1.00 in.2)(60 ksi) = 960 kips (approximate; not all bars reach full yield) 
SC = (1152 + 101 + 960) = 2213 kips 
ST = 960 +130 kips = 1090 kips 
SFy = 0 
Pn = GC - GT = 2213 - 1090 = 1123 kips 
fPn = (0.85)(1123) = 955 kips 
SMcl = 0 
Mu = (1152 +960 kips)(15.5 ft) + (101 kips)(14.1 ft) + (960 kips)(15.5 ft) = 49,040 ft-kips 
fMn = (0.85)(49,040) = 41,680 ft-kips 
Case 3 (Balanced case): 
0.0035 (32.00 ft) 20.11 ft 
0.0035 0.00207 
=... + ... = 
c 
a = (0.8)c = 16.09 ft 
Ignore the distributed #5 bars for this case.
Chapter 9, Masonry 
9-113 
Cm bulb = 1152 kips 
Cm wall = (0.8)(2.5 ksi)(10.5 in.)(14.09 ft.  12) = 3,550 kips 
Cs = 960 kips 
SC = (1152 + 3550 + 960) = 5,662 kips 
Ts = 960 kips 
SFy = 0 
Pn = GC - GT = 5662 - 960 = 4,702 kips 
fPn = (0.85)(4,702) = 3,997 kips 
SMcl = 0 
Mu = (1152 +960 kips)(15.5 ft) + (3550 kips)(7.46 ft) + (960 kips)(15.5 ft) = 74,100 ft-kips 
fMn = (0.85)(74,100) = 62,980 ft-kips 
The actual design strength, fMn, at the level of minimum axial load can be found by interpolation to be 
33,840 ft.-kip.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-114 
9.3.3.5.2.2 Ductility check (Level 1) 
Provisions Sec. 11.6.2.2 [ACI 530, Sec. 3.2.3.5] requires that the critical strain condition correspond to a 
strain in the extreme tension reinforcement equal to 5 times the strain associated with Fy. Note that this 
calculation uses unfactored gravity axial loads (Provisions 11.6.2.2 [ACI 530, Sec. 3.2.3.5]). See Figure 
9.3-7 and the following calculations. 
1,000 kips P max = 912 kips 
P min = 433 kips 
fP 
fM 
M = 30,600 ft-kips 
Balance 
(62,980 ft-kips, 
3,997 kips) 
Intermediate 
(41,680 ft kips, 955 kips) 
20,000 
ft-kips 
30,000 
ft-kips 
40,000 
ft-kips 
50,000 
ft-kips 
60,000 
ft-kips 
2,000 kips 
3,000 kips 
4,000 kips 
(27,340 ft kips, 0 kips) 
n 
n 
fPn 
fMn 
u 
u
u 
Figure 9.3-6 fP11 - fM11 Diagram for Level 1 (1.0 kip = 4.45 kN, 1.0 ft-kip = 1.36 kN-m). 
Chapter 9, Masonry 
9-115 
2.00' 
33.00' 
2.00' 
24.88' 
P 
8.12' 
s = 5 
= 0.0103 
1.25 f = 
Ts1 
4.49' 
6.49' 
0.8f 'm = 
2.00' 
Cs1 
m = 0.0035 
c 
2.0 ksi 
Cm bulb Cm wall 
N.A. 
y 
102 ksi 
300 ksi 
5f = 
23.88' 1.00' 
1.00' 
N.A. 
75 ksi 
a 
2.00' 
f y = 60 ksi 
y 
y 
1.63' 
S Ts2 
Figure 9.3-7 Ductility check for Wall D, Level 1 (1.0 ft = 0.3048 m, 1.0 ksi = 6.89 MPa). 
(0.0003.50003.50103) (32.00 ft) = 8.12 ft 
. . 
. . 
.. .. 
= 
c + 
a = 0.8 c = 6.49 ft 
For Level 1, the unfactored loads are: 
P = 618 kips 
M = 30,600 ft-kips 
=0.8 ' =1,152 kips 
Cmbulb fmAbulb 
=0.8 ' (10.5 in.)(4.49 ft  12)=1131 kips 
Cmwall fm
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-116 
The distributed wall rebar that is near the neutral axis is divided between tension and compression, and 
therefore it will not have much effect on the result of this check, so it will be neglected. 
Cs1 = (60 ksi)(16  1.00 in.2) = 960 kips 
Ts1 = (16  1.00 in.2)(75 ksi) = 1,200 kips 
Ts2 = (4  0.31 in.2)(75 ksi) = 93 kips 
P = 618 kips (unfactored dead load) 
3C > 3P + 3T 
1152 +1131 + 960 > 618 + 1200 + 93 
3,243 kips > 1,818 kips 
There is more compression capacity than tension capacity, so a ductile failure condition governs. 
[The ductility (maximum reinforcement) requirements in ACI 530 are similar to those in the 2000 
Provisions. However, the 2003 Provisions also modify some of the ACI 530 requirements, including 
critical strain in extreme tensile reinforcement (4 times yield) and axial force to consider when performing 
the ductility check (factored loads).] 
9.3.3.5.3 Axial and Flexural Strength Upper Levels 
Examine the strength of Wall D at Level 7. 
Pumin = 179 kips + factored weight of  of 7th story wall = 179 + (0.7)(11.4) =190 kips 
Mu = 9,800 ft-kips 
This is a point, however, where some of the reinforcement in the lower wall will be terminated. Although 
not required by the Provisions, most design standards require the longitudinal reinforcement to be 
extended a distance d beyond the point where it could theoretically be terminated. (The ASD chapter of 
ACI 530 has such a requirement.) Therefore, the reinforcement at level 6 (7th floor) should be capable of 
resisting the moment d below. d is approximately three stories for this wall, therefore, take Mu = 19,400 
ft-kip (and P = 303 kip) from Level 3. 
Try eight #9 in each bulb and vertical #5 bars at 4 ft on center in the wall. Refer to Figure 9.3-8 for the 
placement of the bulb reinforcement.
Chapter 9, Masonry 
9-117 
8" 
1'-8" 
1'-8" 
Figure 9.3-8 Bulb reinforcement at upper levels 
(1.0 ft = 0.3048 m, 1.0 in. = 25.4 mm). 
For evaluating the capacity of the wall, a fPn - fMn curve will be developed to represent the wall strength 
envelope for Level 7. The demands (Pu and Mu determined above) will then be compared to this curve. 
Several cases will be analyzed and their results used in plotting the fPn - fMn curve. Refer to Figure. 9.3- 
9 for notation and dimensions. 
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-118 
T 
N.A. 
C 
m = 0.0035 
m bulb Cm wall 
s 
Cs 
y = 0.00207 
Balance 
Case 
P = 0 Case 
N.A. 
Cm bulb 
m = 0.0035 
Ts 
y = 0.00207 
y = 0.00207 
1.67' 1.67' 
33.00' 
29.67' 
b e = 2.96" 
a = 1.16' 
c = 1.45' 
1.67' 14.50' 
16.17' 
c = 20.21' 
15.67' 
M 
P 
S Ts2 
+ Cs 
Figure 9.3-9 Strength of Wall D at Level 7 (1.0 ft = 0.3048 m) 
Case 1 (P = 0) 
Tension forces: 
Ts1 = (8 bars)(1.00 in.2)(60 ksi) = 480 kips 
Ts2 = (7 bars)(0.31 in.2)(60 ksi) = 130 kips 
Equilibrium: 
SC = ST + P
Chapter 9, Masonry 
9-119 
SC = 480 + 130 + 0 = 610 kips 
Assume bars closest to compression face yield: 
SC = Cs + Cm 
Cs = (3 bars)(1.00 in.2)(60 ksi) = 180 kips 
Cm = 610 - 180 = 430 kips 
Locate equivalent stress block and neutral axis: 
430 = ffm(20 in.)a = (0.8)(3 ksi)(20 in.)a 
a = 8.96 in. = 0.75 ft 
c = a/0.8 = 0.75/0.8 = 0.93 ft = 11.2 in. 
Verify strain in compression steel: 
At the outside layer, e = (0.0035)(7.2 in. / 11.2 in.) = 0.0023 > yield, 
At the central layer, e = (0.0035)(1.2 in. / 11.2 in.) = 0.0004 => fs = 11 ksi 
Resultant moment: 
SMcl = 0: 
Mn = (430 kips)(16.5 ft- 0.75 ft/2) + (180 kips)(16.5 - 0.33 ft) + (480 kips)(16.5 - 0.83 ft) 
+ (130 kips)(0 ft.) = 17,370 ft-kips 
fMn = (0.85)(17,280) = 14,760 ft-kips 
Case 2 (Intermediate) 
Assume the neutral axis at the face of the bulb, c = 1.67 ft 
a = 0.8c = 1.33 ft. = 16 in. 
Cm = (2.4 ksi)(16 in.)(20 in.) = 768 kip 
For the compression steel, it is necessary to compute the strains: 
At the outside layer, e = (0.0035)(16 in. / 20 in.) = 0.0028 > yield 
At the central layer, e = (0.0035)(10 in. / 20 in.) = 0.00175 => fs = 50 ksi 
At the inside layer, e = (0.0035)(4 in. / 20 in.) = 0.0007 => fs = 20 ksi 
Cs = (3.0 x 60 ksi + 2.0 x 50 ksi + 3.0 x 20 ksi) = 340 kips 
Ts1 = (8 bars)(1.00 in.2)(60 ksi) = 480 kips 
Ts2 = (7 bars)(0.31 in.2)(60 ksi) = 130 kips 
Pn = 768 + 340 - 480 -130 = 498 kips 
fPn = (0.85)(498) = 423 kips 
Mn = (768 kips)(16.5 ft- 1.33 ft/2) + (340 + 480 kips)(16.5 - 1.67/2 ft) + (130 kips)(0 ft) 
= 25,010 ft-kips 
fMn = (0.85)(25,010) = 21,260 ft-kips 
At P = 303 kips, fMn = 19,990 ft.-kips by interpolation (exceeds 19,400 ft.-kips, OK)
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-120 
Case 3 (Balanced Case): 
0.0035 (32.17 ft) = 20.21 ft 
0.0035 0.00207 
. . 
. . 
.. .. 
= 
+ c
a = (0.8)c = 16.17 ft 
be = effective width of wall 
be = [(2)(1.3125 in.)(29.67 ft x 12) + (8 cells)(15 in.2/cell)] = 2.96 in./ft 
Cm bulb = 960 kips 
Cm wall = (0.8)(3 ksi)(2.96 in.)(14.17 ft x 12) = 101 kips 
Cs = 480 kips 
SC = (960 + 101 + 480) = 1,541 kips 
Ts = 480 kips, ignoring the distributed #5 bars 
SFy = 0 
Pn = GC - GT = 1,541 - 480 = 1,061 kips 
fPn = (0.85)(1,061) = 902 kips 
SMcl = 0 
Mu = (480 +960 kips)(15.67 ft) + (101 kips)(7.42 ft) + (480 kips)(15.67 ft) = 30,830 ft-kips 
fMn = (0.85)(30,830) = 26,210 ft-kips
Chapter 9, Masonry 
9-121 
The ductility check is performed similar to that for the wall at Level 1. See Figure 9.3-11 and the 
following calculations. 
20,000 ft-kips 30,000 ft-kips 
500 kips 
1,000 kips 
379 kips 
179 kips 
fP 
fM 
M = 9,800 ft-kips 
Balance 
(26,210 ft-kips, 902 kips) 
Simplified fP - fM curve 
10,000 ft-kips 
P = 0 Case 
(14,760 ft-kips, 0 kips) 
Intermediate 
(21,260 ft-kips, 423 kips) 
Extension of demand 
a distance "d" 
(19,400 ft-kips, 303 kips) 
n 
n 
fMn 
fPn 
u u 
u 
Figure 9.3-10 fP11 - fM11 Diagram for Level 7 (1.0 kip = 4.45 kN, 1.0 ft-kip = 1.36 kN-m). 
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-122 
1.67' 
33.00' 
1.67' 
24.63' 
P 
8.37' 
s = 5 
= 0.0103 
1.67' 
25.2 ksi 
1.25 fy = 
Ts1 
5.03' 
6.70' 
8.37' 
0.8f 'm = 
1.67' 
Cs1 
m= 0.0035 
c 
2.4 ksi 
Cm wall 
Cm bulb 
N.A. 
y 
102 ksi 
300 ksi 
5fy = 
24.63' 
0.84' 
N.A. 
0.84' 
75 ksi 
21.9 ksi 
fy = 60 ksi 
Cs2 
Ts2 
Ts3 
Cs3 
a 
0.67' 
1.67' 
c 
Figure 9.3-11 Ductility check for Wall D, Level 7 (1.0 ft = 0.3048 m, 1.0 ksi = 6.89 MPa) 
For Level 7, the unfactored loads are: 
P = 255 kips 
M = 11,600 ft-kips
Chapter 9, Masonry 
9-123 
Cmb = 0.8fmAb = 0.8(3.0 ksi)(400 in.2) = 960 kips 
Cmw = 0.8fmAw = 0.8(3.0 ksi)[2(1.3125 in.)(5.03 ft.)(12 in/ft) + (1 cell)(25.6 in.2/cell)] 
= 442 kips 
Cs1 = (60 ksi)(8 1.00 in.2) = 480 kips 
Cs2 = (60 ksi)(0.31 in.2) = 19 kips 
Ts1 = (8  1.00 in.2)(75 ksi) = 600 kips 
Ts2 = (4  0.31 in.2)(75 ksi ) = 116 kips 
Ts3 = (0.31 in.2)(25.2 ksi) = 7 kips 
P = 255 kips (unfactored dead load) 
GC > P + GT 
960 + 442 + 480 + 19 > 255 + 600 + 116 + 7 ? 
1901 kips > 978 kips OK 
There is more compression capacity than tension capacity, so a ductile failure condition governs. 
9.3.3.5.4 Shear Strength 
The first step is to determine the net area, An, for Wall D. The definition of An in the Provisions, however, 
does not explicitly address bulbs or flanges at the ends of walls. Following an analogy with reinforced 
concrete design, the area is taken as the thickness of the web of the wall times the overall length. In 
partially grouted walls this is not extended to the point of ignoring the grouted cores because the 
implication is that grouted cores are intended to be included. (However, if the spacing of bond beams 
greatly exceeds the spacing of grouted cores, even that assumption might be questionable.) 
For Levels 1 through 6: 
An = (10.5 in.(33 ft  12))= 4,158 in.2 
For Levels 7 through 12 (using 8 x 8 x 12 clay brick units): 
An = (2)(1.3125 in.)(12 in.)(33 ft) + (7 + 2  3 cells)(25.6 in.2 /cell with adjacent webs) 
= 1,372 in.2 
Shear strength is determined as described in Sec. 9.2 using Provisions Eq. 11.7.2.1 [ACI 520, Sec. 3.13] 
and Provisions Eq. 11.7.3.1-1 [ACI 530, Eq. 3-18], respectively: 
Vu # fVn 
Vn = Vm + Vs 
For Levels 1 through 6 using Provisions Eq. 11.7.3.1-3 where x > 1.0 : 
x 
M
V d 
(max)=4 ' = (4)( 2,500)(4,806) = 961 kips (P3) n mn V fA
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-124 
For Levels 7 through 12 where varies from 0.30 to 1.14: x 
x 
M
V d 
Vn (max) varies from 5.87 m'n to 4 m'n f A f A 
Therefore, Vn (max) varies from 5.87 3,000(1836) = 590 kips to 4 3,000(1836) = 402 kips 
depending on the value of . The masonry shear strength is computed as: x 
x 
M
V d 
m= 4 - 1.75 nm0.25 
V MAf P 
Vd 
... ... ...... ' + 
The shear strength of Wall D, based on the aforementioned formulas and the strength reduction factor of 
f = 0.8 for shear from Provisions Table 11.5.3 [ACI 530, Sec. 3.1.4.3] , is summarized in Table 9.3-8. Vx 
and Mx in this table are values from Table 9.3-2 multiplied by 0.148 (the portion of direct and torsional 
shear assigned to Wall D). P is the dead load of the roof or floor multiplied by the tributary area for Wall 
D, and d is the wall length, not height (d = 32.67 ft for Wall D). 
The demand shear, Vu, is found by amplifying the loads to a level that produces a moment of 125 percent 
of the nominal flexural strength at the base of the wall. Given the basic flexural demand of 30,600 ft-kips, 
a design resistance of 33,840 ft-kips, f = 0.85, and the 1.25 factor, the overall amplification of design 
load is 1.63. 
Table 9.3-8 Shear Strength for Wall D 
Level Vx/wall 
(kips) 
Mx/wall 
(ft-kips) 
Mx/Vxd 1.63Vx fVn max 
(kips) 
fVm 
(kips) 
P 
(kips) 
fVm 
(kips) 
Reqd fVs 
(kips) 
12 49 500 0.309 80 351.2 OK 37 215 - 
11 102 1500 0.446 166 329.3 OK 80 210 - 
10 149 3000 0.610 243 303.0 OK 124 201 42 
9 191 4900 0.777 311 276.2 NG 168 192 119 
8 228 7200 0.957 372 247.4 NG 212 182 190 
7 260 9800 1.142 424 240.5 NG 255 186 238 
6 292 12700 1.318 476 665.3 OK 308 436 40 
5 323 15900 1.492 526 665.3 OK 370 448 78 
4 347 19400 1.694 566 665.3 OK 432 461 106 
3 364 23000 1.915 593 665.3 OK 494 473 120 
2 375 26800 2.166 611 665.3 OK 556 485 126 
1 380 30600 2.440 619 665.3 OK 618 498 121 
1.0 kip = 4.45 kN, 1.0 ft-kip = 1.36 kN-m. 
Note that 1.63 Vx exceeds Vn max at Levels 7, 8, and 9. The next most economical solution appears to be to 
add grout to increase An and, therefore, both Vm and Vn max. Check Level 7 using solid grout: 
An = (7.5 in.)( 33 ft)(12 in./ft) = 2970 in.2 
(max) 4 (4)( 2,500)(2,970) = 650 kips n mn V = f'A= 
fVn max = 0.8(650) = 520 kips > 424 kips OK
Chapter 9, Masonry 
9-125 
fVm =(0.80)..(4.0-1.75(1.0))2,970 3000 /1000+0.25(255)..=344 kips 
fVs = Vu - fVm = 424 - 344 = 80 kips 
Check minimum reinforcement for capacity. With vertical #5 at 48 in., a reinforcement ratio of 0.00086 
is provided. Thus the horizontal reinforcement must exceed (0.0020 - 0.00086)(7.5 in.)(12 in.) = 0.1025 
in.2/ft. With story heights of 10 ft., bond beams at 40 in. on center are convenient, which would require 
0.34 in.2 Therefore, for 2 - #4 at 40 in. on center: 
fVs = 0.80(0.5)(0.4 / 40) (60 ksi)(33 ft.)(12 in./ft.) = 95 kips > 80 kips OK 
The largest demand for fVx in the lower levels is 126 kips at level 2. As explained in the design of the 
lower level walls for flexural and axial loads (Sec. 9.3.3.5.1), horizontal #5 at 22 in. are required to satisfy 
minimum reinforcement. Given the story height, check for horizontal #5 at 20 in.: 
fVs = 0.8(0.5)(0.31 / 20) (60 ksi)(33 ft.)(12 in./ft.) = 147 kips > 126 kips OK 
In summary, for shear it is necessary to grout the hollow units at story 7 solid, and to add some grout at 
stories 8 and 9. Horizontal reinforcement is 2 - #4 in bond beams at 40 in. on center in the upper stories 
and one #5 at 20 in. on center in the grouted cavity of the lower stories. 
9.3.4 Deflections 
The calculations for deflection involve many variables and assumptions, and it must be recognized that 
any calculation of deflection is approximate at best. 
Deflections are to be calculated and compared with the prescribed limits set forth by Provisions Table 
5.2.8 [Table 4.5-1]. Deformation requirements for masonry structures are discussed in Provisions Sec. 
11.5.4. 
The following procedure will be used for calculating deflections: 
1. Determine if the wall at each story will crack by comparing Mx (see Table 9.3-6) to Mcr where 
= ( + ) 
cr r umin M S f P A 
2. If Mcr < Mx, then use cracked moment of inertia and Provisions Eq. 11.5.4.3. 
3. If Mcr > Mx, then use In = Ig for moment of inertia of wall. 
4. Compute deflection for each level. 
5. dmax = 3 story drift 
[The specific procedures for computing deflection of shear walls have been removed from the 2003 
Provisions. ACI 530 does not contain the corresponding provisions in the text, however, the commentary 
contains a discussion and equations that are similar to the procedures in the 2000 Provisions. Based on 
ACI 530 Sec. 1.13.3.2, the maximum drift for all masonry structures is 0.007 times the story height. 
Thus, there appears to be a conflict between ACI 530 and 2003 Provisions Table 4.5-1.] 
For the upper levels (the additional grout required for shear strength is not considered here): 
be = effective masonry wall width 
be = [(2  1.3125 in.)(356) + (7 cells)(15 in.2/cell)]/(356) = 2.92 in. 
A = Awall + 2Abulb = (2.92 in.)(356 in.) + (2)(400 in.2) = 1,840 in.2
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-126 
[Note that by adopting ACI 530 in the 2003 Provisions, Em = 900fm per ACI 530 Sec. 1.8.2.2.1] 
Per Provisions Eq. 11.3.10.2 [ACI 530, 1.8.2.2]: 
E = 750 fm = 2,250 ksi (n = 12.89) 
Ig = Iwall + I bulb 
3 2 
6 4 (2.96)(356) 376 
(2 bulbs)(20 20) 39.4 10 in. 
12 2 
= +  . .=  
. . 
. . g I 
S = Ig/c = 39.4  106 /(198) = 199,000 in.3 
fr = 0.250 ksi 
= 1.00D (see Table 9.3-6.) Pumin 
For the lower levels: 
A = A wall + 2Abulb = (10.5 in.)(348 in.) + (2)(576 in.2) = 4,806 in.2 
E = 750 fm = 1,875 ksi (n = 15.47) 
I = Iwall + I bulb 
3 2 
6 4 (10.5)(348) 372 
(2 bulbs)(24 24) 76.7 10 in. 
12 2 
= +  . .=  
. . 
. . g I 
S = Ig / c = 76.7  106 /(198) = 387,000 in.3 
fr = 0.250 ksi 
= 1.00D (see Table 9.3-6.) Pumin 
Table 9.3-9 provides a summary of these calculations. 
Table 9.3-9 Cracked Wall Determination 
Level Pumin 
(kips) 
Mcr 
(ft-kips) 
Mx 
(ft-kips) 
Status 
12 
11 
10 
987654321 
37 
80 
124 
168 
212 
255 
308 
370 
432 
494 
556 
618 
7,620 
8,820 
8,950 
9,620 
10,300 
11,000 
15,400 
16,000 
16,700 
17,300 
18,000 
18,600 
500 
1,500 
3,000 
4,900 
7,200 
9,800 
12,700 
15,900 
19,400 
23,000 
26,800 
30,600 
uncracked 
uncracked 
uncracked 
uncracked 
uncracked 
uncracked 
uncracked 
uncracked 
cracked 
cracked 
cracked 
cracked 
1.0 kip = 4.45 kN, 1.0 ft-kip = 1.36 kN-m. 
For the uncracked walls at the upper levels:
Chapter 9, Masonry 
9-127 
In = Ig = 39.4  106 in.4 
For the uncracked walls at the lower levels: 
In = Ig = 76.7  106 in.4 
For the cracked walls at the lower levels, the determination of Icr will be for the load combination of 1.0D 
+ 0.5L. The 0.5L represents an average condition of live load. Making reference to Figure 9.3-6, it can 
be observed that at this level of Pu, the point on the fPn - fMn curve is near the intermediate point 
previously determined. This is where c = 3.0 ft. (The actual c dimension will be very close to 3.0 ft). 
For this case, and referring to Figure 9.3-5, the cracked moment of inertia is: 
I cr = I bulb + I wall + I nAs 
= [244 + (24  24)(24)2 ] + [10.5  123/3 ] + [(15.47  16 in.2)(29 ft  12)2] 
= 30.3  106 in4. 
Note that 98.9 percent of the value comes from one term: the reinforcement in the tension bulb. If the 
distributed #5 bars is added to this computation, the value becomes 31.5  106 in.4 
With the other masonry examples, the interpolation between gross and cracked section properties was 
used. The application of that is less clear here, where the properties step at midheight, so two analyses are 
performed. First, each story is considered to be cracked or uncracked. Second is the authors 
interpretation of the effective moment of inertia equation as: 
For all the cracked walls (Provisions Eq. 11.5.4.3 [ACI 530, Commentary Sec. 3.1.5.3]): 
3 3 
= + 1- = 
. . . . .. 
. . . . .. 
. . .. . ... 
cr cr 
eff n cr n 
a a 
M M 
I I I I 
M M 
( ) ( ) 3 3 
618, 600 6 18,600 64 
1 
30,600 30,600 
=76.7 10 +30.6 10 - 41.0 10 in. 
 .. ..  .. .. ....=  
. . . . .. 
eff I 
The entire 12-story Wall D will be treated as a stepped, vertical masonry cantilever shear wall. For the 
lower step (Levels 1-6), Ieff = 41.0  106 in.4 Even though the upper walls are uncracked, Ieff of the upper 
step (Levels 7-12), will be In reduced in the same proportion as the lower levels: 
(upper levels) 
6 
6 6 
6 
(39.4 10 ) 41.0 10 = 21.1 10 
76.7 10 
.  . 
=  ..  ..  eff I 
Both the deflections and the fundamental period can now be found. Two RISA 2D analyses were run, 
and the deflections shown in Table 9.3-10 were obtained. The deflection from the RISA 2D analysis at 
each level is multiplied by Cd (= 3.5) to determine the inelastic deflection at each level. From these, the 
story drift, ., at each level can be found. 
The periods shown in the table validate the period of T = 0.75 sec previously used to determine the base 
shear in Sec. 9.3.3.2. 
Table 9.3-10 Deflections for ELF Analysis (inches) 
Using gross and cracked properties, Using effective moment of inertia
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-128 
story by story (T = 0.798 sec) (T = 0.755 sec) 
Level Elastic Total Drift Ratio Elastic Total Drift Ratio 
12 3.40 11.90 3.14 10.99 
11 3.04 10.65 1.25 2.77 9.71 1.28 
10 2.68 9.38 1.27 1.01 2.40 8.41 1.30 1.01 
9 2.32 8.10 1.27 1.00 2.04 7.13 1.28 0.99 
8 1.95 6.84 1.26 0.99 1.68 5.87 1.26 0.98 
7 1.60 5.60 1.24 0.98 1.34 4.68 1.19 0.95 
6 1.26 4.41 1.19 0.96 1.02 3.58 1.10 0.92 
5 0.95 3.34 1.07 0.90 0.76 2.65 0.93 0.85 
4 0.66 2.31 1.03 0.96 0.52 1.82 0.83 0.90 
3 0.40 1.40 0.91 0.88 0.32 1.11 0.71 0.85 
2 0.20 0.68 0.71 0.78 0.16 0.55 0.56 0.79 
1 0.06 0.20 0.48 0.67 0.05 0.17 0.38 0.68 
0 0 0.00 0.20 0.42 0 0.00 0.17 0.44 
1.0 kip = 4.45 kN, 1.0 in. = 25.4 mm 
The two methods give comparable results. The maximum building deflection is compared to the 
maximum deflection permitted by Provisions Sec. 11.5.4.1.1 as follows: 
Cddmax = 11.90 in. < 14.4 in. = 0.1hn OK 
The maximum story drift occurs at Story 11 and is compared to the maximum story drift permitted by 
Provisions Table 5.2.8 [Table 4.5-1] as follows: 
. = 1.30 in. > 1.20 in. = 0.01hsx NG 
Although this indicates a failure to satisfy the Provisions, in the authors opinion the drift is satisfactory 
for two reasons. First the MRS analysis shows smaller drifts (Table 9.3-11) that are within the criteria. 
On a more fundamental level, however, the authors believe the basic check for drift of a masonry wall is 
performed according to Provisions Sec. 11.5.4.1.1, which applies only to the total displacement at the top 
of the wall, and that the story drift for any particular story is more properly related to the values for nonmasonry 
buildings. That limit is 0.020 hsx, or 2.4 in. per story. For a building with a torsional 
irregularity, Provisions Sec. 5.4.6.1 [Sec. 4.5.1] requires that the story drift be checked at the plan 
location with the largest drift, which would be a corner for this building. That limit is satisfied for this 
building, both by ELF and MRS analyses.
Chapter 9, Masonry 
9-129 
Table 9.3-11 Displacements from Modal Analysis, inches 
At corner of floor plate with 
maximum displacements. Story 
drift would be pertinent, 
although not at 0.010 
At wall with maximum in-plane 
displacement. Roof limit for 
masonry would be pertinent. 
Level Elastic Total Approx. 
Drift 
Elastic Total Approx. 
Drift 
12 2.48 8.67 1.91 6.70 
11 2.20 7.69 0.98 1.69 5.90 0.79 
10 1.91 6.70 0.99 1.48 5.17 0.74 
9 1.63 5.70 1.00 1.26 4.40 0.77 
8 1.35 4.72 0.98 1.04 3.64 0.76 
7 1.08 3.77 0.95 0.83 2.91 0.73 
6 0.83 2.90 0.88 0.64 2.23 0.68 
5 0.62 2.16 0.74 0.48 1.67 0.57 
4 0.43 1.50 0.66 0.33 1.16 0.51 
3 0.27 0.93 0.57 0.21 0.72 0.44 
2 0.14 0.48 0.46 0.11 0.37 0.35 
1 0.05 0.16 0.32 0.04 0.12 0.25 
0 0.00 0.00 0.16 0.00 0.00 0.12 
1.0 kip = 4.45 kN, 1.0 in. = 25.4 mm. 
The drifts in Table 9.3-11 are not the true modal drifts. The values are computed from the modal sum 
maximum displacements, rather than being a modal sum of drifts in each mode. The values in the table 
are less than the true value. 
Both tables also confirm that the change in stiffness at midheight does not produce a stiffness irregularity. 
Provisions Sec. 5.2.3.3, Exception 1 [Sec. 4.3.2.3, Exception 1], clarifies that if the drift in a story never 
exceeds 130 percent of the drift in the story above, then there is no vertical stiffness irregularity. Note 
that the inverse does not apply; even though the drift in Story 2 is more than double that in Story 1, it 
does not constitute a stiffness irregularity. 
9.3.5 Out-of-Plane Forces 
Provisions Sec 5.2.6.2.7 [Sec. 4.6.1.3] states that the bearing walls shall be designed for out-of-plane 
loads equal to: 
w = 0.40 SDS Wc $ 0.1Wc 
w = (0.40)(1.00)(114 psf) = 45.6 psf $ 0.1Wc 
Therefore, w = 45.6 psf. Out-of-plane bending, using the strength design method for masonry, for a load 
of 45.6 psf acting on a 10 ft story height is approximated as 456 ft.-lb. per linear ft of wall. This 
compares to a computed strength of the wall of 1,600 in.-lb per linear foot of wall, considering only the 
#5 bars at 4 ft on center. Thus the wall is loaded to 28.5 percent of its capacity in flexure in the out-ofplane 
direction. The upper wall has the same reinforcement, about 42 percent of the load and about 71 
percent of the thickness. Therefore, it will be loaded to a smaller fraction of its capacity. (Refer to 
Example 9.1 for a more detailed discussion of strength design of masonry walls, including the P-delta 
effect.)
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-130 
9.3.6 Orthogonal Effects 
In accordance with Provisions Sec. 5.2.5.2.2 [Sec. 4.4.2.3], orthogonal interaction effects must be 
considered for buildings in Seismic Design Category D when the ELF procedure is used. Any out-ofplane 
effect on the heavily reinforced bulbs is negligible compared to the in-plane effect, so orthogonal 
effects on the bulbs need not be considered further. Considering only the #5 bars and the load 
combination of 100 percent of in-plane load plus 30 percent of the out-of-plane load, yields a result that 
0.3(0.285), or 8.6 percent of the capacity of the #5 bars is not available for in-plane resistance. Given that 
the #5 bars contribute about 12 percent to the tension resistance (130 kips, vs 960 kips for the bulb 
reinforcement), the overall effect is a change of about 1 percent in in-plane resistance, which is negligible. 
This completes the design of the transverse Wall D. 
9.3.7 Wall Anchorage 
The anchorage for the bearing walls must be designed for the force, Fp, determined in accordance with 
Provisions Sec. 5.2.6.2.7 [Sec. 4.6.1.3] as: 
Fp = 0.4SDSWc = (0.4)(1.00)(10 ft)(114 psf) = 456 lb/ft 
Minimum force = 0.10Wc = (0.10)(10 ft)(114 psf) = 114 lb/ft 
Provisions Sec. 5.2.6.3.2 [Sec. 4.6.2.1] references Provisions Sec. 6.1.3 [Sec. 6.2.2] for anchorage of 
walls where diaphragms are not flexible. For the lower wall: 
0.4 (1 2 / ) 0.4(1.0)(1.0)(10ft. 114 psf )(1 2(0.5)) 364 lb./ft. 
/ 2.5/1.0 
p DS p 
p 
p p 
a S W 
F zh 
R I 
 
= + = + = 
Therefore, design for 456 lb/ft. For a 2 ft-6 in. joist spacing, the anchorage force at each joist is Fp = 
1,140 lb. 
Refer to Figure 9.3-12 for the connection detail. A 3/16-in. fillet, weld 2 in. long on each side of the joist 
seat to its bearing plate will be more than sufficient. Two1/2 in.-diameter headed anchor studs on the 
bottom of the bearing plate also will be more than sufficient to transfer 4,560 lb into the wall. 
9.3.8 Diaphragm Strength 
See Example 7.1 for a more detailed discussion on the design of horizontal diaphragms. 
To compute the story force associated with the diaphragm on each level, use Provisions Eq. 5.2.6.4-4 [Eq. 
4.6-2]: 
=
= 
= 
S
S
n 
i 
i x 
px n px 
i 
i x 
F 
F w
w
Chapter 9, Masonry 
9-131 
The results are shown in Table 9.3-12. Note that wi is approximately the same as wpx for this case, the 
only difference being the weight of the walls perpendicular to the force direction, so the wi values were 
used for both. 
Table 9.3-12 Diaphragm Seismic Forces 
Level wi 
(kips) 
Fi 
(kips) 
Fpx 
(kips) 
12 
11 
10
987654321 
768 
917 
917 
917 
917 
917 
1109 
1300 
1300 
1300 
1300 
1300 
329 
357 
320 
284 
249 
214 
218 
208 
162 
117 
74 
34 
329 
373 
355 
336 
318 
301 
338 
365 
336 
309 
282 
258 
1.0 kip = 4.45 kN 
The maximum story force is 373 kips. Therefore, use 373 kips/152 ft = 2.45 kips/ft in the transverse 
direction. The shear in the diaphragm is shown in Figure 9.3-13b. The reaction, R, at each wall pair is 
373/4 = 93.25 kips. The diaphragm force at each wall pair is 93.25 kips/(2  33 ft) = 1.41 kips/ft. 
Joist to 
plate 
Slab reinforcement at 
perimeter of building 
is continuous through 
the wall. Use 
#4 x 4'-0" at 16" o.c. 
dowels elsewhere 
(center on wall). 
3
16 2" 
Plate 3
8"x 6" x 0'-10" 
at each joist (2'-6" o.c.) 
with (2) 1
2" dia. x 6" H.A.S. 
Figure 9.3-12 Floor anchorage to wall (1.0 in. = 25.4 mm, 1.0 ft = 0.3048 m).
FEMA 451, NEHRP Recommended Provisions: Design Examples 
9-132 
The maximum diaphragm shear stress is v = V/td = 1410 plf/(2.5 in.)(12 in.) = 47 psi. This compares to 
an allowable shear of 
fvc=(0.85)(2)fc'=(0.85)(2) 3,000=93 psi 
for 3,000 psi concrete. Thus, no shear reinforcing is necessary. Provide . = 0.0018 as minimum 
reinforcement, so As = 0.054 in.2/ft. Use WWF 6  6-2.9/2.9, which has As = 0.058 in.2/ft. 
The moment in the diaphragm is shown in Figure 9.3-13c. The maximum moment is 2,460 ft-kips. 
Perimeter reinforcement in the diaphragm is determined from: 
T = M/d = (2,460 ft-kips)/(72 ft) = 34.1 kips 
As = T/fFy = 34.1 kips / (0.85)(60 ksi) = 0.67 in.2 
Boundary elements of diaphragms may also serve as collectors. The collector force is not usually the 
same as the chord force. Provisions Sec. 5.2.6.4.1 [Sec. 4.6.2.2] requires that collector forces be 
amplified by O0. Collector elements are required in this diaphragm for the longitudinal direction. A 
similar design problem is illustrated in Chapter 7 of this volume. Where reinforcing steel withing a 
topping slab is used for chords or collectors, ACI 318, Sec. 21.9.8 (2002 edition) imposes special spacing 
and cover requirements. Given the thin slab in this building, the chord reinforcement will have to be 
limited to bars with couplers at the splices or a thickened edge will be required.
Chapter 9, Masonry 
9-133 
Figure 9.3-13 Shears and moments for diaphragm 
(1.0 kip/ft = 14.6 kN/m, 1.0 kip = 4.45 kN, 1.0 kip-ft = 1.36kN-m) 
Fp = 2.45 kips/ft 
2,460 ft-kips 
93.25 kips 
a) Load 
b) Shear 
c) Moment 
Figure 9.3-13.
10-1 
10 
WOOD DESIGN 
Peter W. Somers, P.E. and Michael Valley, P.E. 
This chapter examines the design of a variety of wood building elements. Section10.1 features a BSSC 
trial design prepared in the early 1980s as a starting point. Section 10.2 completes the roof diaphragm 
design for the building featured in Section 9.1. In both cases, only those portions of the designs necessary 
to illustrate specific points are included. 
Typically, the weak links in wood systems are the connections, but the desired ductility must be 
developed by means of these connections. Wood members have some ductility in compression 
(particularly perpendicular to grain) but little in tension. Nailed plywood shear panels develop 
considerable ductility through yielding of nails and crushing of wood adjacent to nails. Because wood 
structures are composed of many elements that must act as a whole, the connections must be considered 
carefully to ensure that the load path is complete. Tying the structure together , which is as much an art 
as a science, is essential to good earthquake-resistant construction. 
Wood elements often are used in low-rise masonry and concrete buildings. The same basic principles 
apply to the design of wood elements, but certain aspects of the design (for example, wall-to-diaphragm 
anchorage) are more critical in mixed systems than in all-wood construction. 
Wood structural panel sheathing is referred to as plywood in this chapter. As referenced in the 2000 
NEHRP Recommended Provisions, wood structural panel sheathing includes plywood and other products, 
such as oriented-strand board (OSB), that conform to the materials standards of Chapter 12. According to 
Provisions Chapter 12, panel materials other than wood structural panel sheathing do not have a 
recognized capacity for seismic-force resistance in engineered construction. 
In addition to the 2000 NEHRP Recommended Provisions and Commentary (hereafter, the Provisions and 
Commentary), the documents edited below are either referenced directly, or are useful design aids for 
wood construction. 
AF&PA Manual American Forest & Paper Association 1996. Manual for Engineered Wood 
Construction (LRFD), including supplements, guidelines, and ASCE 16-95, 
AF&PA. 
[AF&PA Wind American Forest and Paper Association. 2001. Special Design Provisions for 
& Seismic Wind and Seismic, ADS/LRFD Supplement. AF&PA.] 
FEMA 451, NEHRP Recommended Provisions: Design Examples 
10-2 
ANSI/AITC A190.1 American National Standards Institute/American Institute of Timber 
Construction. 1992. American National Standard for Wood Products: 
Structural Glued-Laminated Timber, A190.1. AITC. 
APA PDS American Plywood Association. 1998. Plywood Design Specification, APA. 
APA 138 American Plywood Association. 1998. Plywood Diaphragms, APA Research 
Report 138. APA. 
ASCE 7 American Society of Civil Engineers. 1998 [2002]. Minimum Design Loads for 
Buildings and Other Structures. ASCE. 
ASCE 16 American Society of Civil Engineers. 1995. Standard for Load and Resistance 
Factor Design (LRFD) for Engineered Wood Construction. ASCE. 
UBC Std 23-2 International Conference of Building Officials. 1997. UBC Standard 23-2 
Construction and Industrial Plywood, Uniform Building Code. ICBO. 
Roark Roark, Raymond. 1985. Formulas for Stress and Strain, 4th Ed. McGraw-Hill. 
USGS CD-ROM United States Geological Survey. 1996. Seismic Design Parameters, CD-ROM. 
USGS. 
WWPA Rules Western Wood Products Association. 1991. Western Lumber Grading Rules. 
WWPA. 
Although this volume of design examples is based on the 2000 Provisions, it has been annotated to reflect 
changes made to the 2003 Provisions. Annotations within brackets,[ ], indicated both organizational 
changes (as a result of a reformat of all the chapters of the 2003 Provisions) and substantive technical 
changes to the 2003 Provisions and its primary reference documents. While the general concepts of the 
changes are described, the design examples and calculations have not been revised to reflect the changes 
to the 2003 Provisions. 
The most significant change to the wood chapter in the 2003 Provisions is the incorporation by reference 
of the AF&PA, ADS/LRFD Supplement, Special Design Provisions for Wind and Seismic for design of 
the engineered wood construction. A significant portion of the 2003 Provisions Chapter 12, including the 
diaphragm and shear wall tables, has been replaced by a reference to this document. This updated 
chapter, however, does not result in significant technical changes , as the Supplement, (referred to herein 
as AF&PA Wind&Seismic) is in substantial agreement with the 2000 Provisions. There are, however, 
some changes to the provisions for perforated shear walls, which are covered in Section 10.1 
Some general technical changes in the 2003 Provisions that relate to the calculations and/or design in this 
chapter include updated seismic hazard maps, changes to the Seismic Design Category classification for 
short period structures, revisions to the redundancy requirements, revisions to the wall anchorage design 
requirement for flexible diaphragms, and a new Simplified Design Procedure that could be applicable 
to the examples in this chapter. 
Where the affect the design examples in this chapter, other significant changes to the 2003 Provisions and 
primary reference documents are noted. However, some minor changes to the 2003 Provisions and the 
reference documents may not be noted.
Chapter 10, Wood Design 
10-3 
10.1 THREE-STORY WOOD APARTMENT BUILDING; SEATTLE, WASHINGTON 
This example features a wood frame building with plywood diaphragms and shear walls. It is based on a 
BSSC trial design by Bruce C. Olsen, Structural Engineer, Seattle, Washington. 
10.1.1 BUILDING DESCRIPTION 
This three-story, wood frame apartment building has plywood floor diaphragms and shear walls. The 
building has a double-loaded central corridor. Figure 10.1-1 shows a typical floor plan, and Figure 10.1-2 
shows a longitudinal section and elevation. The building is located in a neighborhood a few miles north 
of downtown Seattle. The site coordinates for determining the seismic design parameters are 47.69 N, 
122.32 W. 
The shear walls in the longitudinal direction are located on the exterior faces of the building and along the 
corridor. (In previous versions of this volume of design examples, the corridor walls were gypsum 
wallboard sheathed shear walls; however, gypsum wallboard sheathing, is no longer recognized for 
engineered design of shear walls per Provisions Sec. 12.3.5. Therefore, plywood shear walls are provided 
at the corridors.) The entire solid (non-glazed) area of exterior walls plywood sheathing, but only a 
portion of the corridor walls will require sheathing. For the purposes of this example, assume that each 
corridor wall will have a net of 55 ft of plywood (the reason for this is explained later). In the transverse 
direction, the end walls and one line of interior shear walls provide lateral resistance. (In previous 
versions of this example, only the end walls were shear walls. The interior walls now are required for 
control of diaphragm deflections given the increased seismic ground motion design parameters for the 
Seattle area.) 
The floor and roof systems consist of wood joists supported on bearing walls at the perimeter of the 
building, the corridor lines, plus one post-and-beam line running through each bank of apartments. 
Exterior walls are framed with 26 studs for the full height of the building to accommodate insulation. 
Interior bearing walls require 26 or 34 studs on the corridor line up to the second floor and 24 studs 
above the second floor. Apartment party walls are not load bearing; however, they are double walls and 
are constructed of staggered, 24 studs at 16 in. on center. Surfaced, dry (seasoned) lumber, is used for 
all framing to minimize shrinkage. Floor framing members are assumed to be composed of Douglas 
Fir-Larch material, and wall framing is Hem-fir No. 2, as graded by the WWPA. The material and 
grading of other framing members associated with the lateral design is as indicated in the example. The 
lightweight concrete floor fill is for sound isolation, and is interrupted by the party walls, corridor walls, 
and bearing walls. 
The building is founded on interior footing pads, continuous strip footings, and grade walls (Figure 
10.1-3). The depth of the footings, and the height of the grade walls, are sufficient to provide crawl space 
clearance beneath the first floor. 
The building is typical of apartment construction throughout the western United States, and has the 
weight necessary to balance potential overturning forces in the transverse direction. If the ground floor 
were a slab-on-grade, however, the resulting shallower grade wall might well require special attention, 
due to the possibility of overturning on some of the shear wall units.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
10-4 
Typical apartment 
partitions 
Post & 
beam 
lines 
148'-0" 
28'-0" 8'-0" 
56'-0" 
9'-0" 9'-0" 
25'-0" 6'-0" 25'-0" 
11
2" Lightweight concrete 
over plywood deck on 
joists at 16" o.c. 
Figure 10.1-1 Typical floor plan (1.0 ft = 0.3048 m). 
148'-0" 
15'-0" 26'-0" 30'-0" 26'-0" 30'-0" 13'-0" 
Sheathed 
wall 
Glazed 
wall 
4' 9' 9' 9' 
Doors 
Corridor 
walls Stairs 
8'-0" 
Figure 10.1-2 Longitudinal section and elevation (1.0 ft = 0.3048 m).
Chapter 10, Wood Design 
10-5 
Pad 
footings 
Continuous footing 
and grade wall 
Figure 10.1-3 Foundation plan. 
10.1.1.1 Scope 
In this example, the structure is designed and detailed for forces acting in the transverse and longitudinal 
directions. However, greater attention is paid to the transverse direction, because of diaphragm 
flexibility. The example includes the following 
1. Development of equivalent static loads, including torsional effects on plywood diaphragms, 
2. Design and detailing of transverse plywood walls for shear and overturning moment, 
3. Design and detailing of plywood floor and roof diaphragms, 
4. Design and detailing of wall and diaphragm chord members, 
5. Detailed deflection and P-delta calculations, and 
6. Design and detailing of longitudinal plywood walls using the requirements for perforated shear walls. 
[Note that the new Simplified Design Procedure contained in 2003 Provisions Simplified Alternate 
Chapter 4 as referenced by 2003 Provisions Sec. 4.1.1 is likely to be applicable to this example, subject to 
the limitations specified in 2003 Provisions Sec. Alt. 4.1.1.] 
10.1.2 BASIC REQUIREMENTS 
10.1.2.1 Provisions Parameters 
Seismic Use Group (Provisions Sec. 1.3 [1.2]) = I 
Occupancy Importance Factor, I (Provisions Sec. 1.4 [1.3]) = 1.0 
Site Coordinates = 47.69 N, 122.32 W 
Short Period Response, SS (USGS CD-ROM) = 1.34 
One Second Period Response, S1 (USGS CD-ROM) = 0.46 
Site Class (Provisions Sec. 4.1.2.1 [3.5]) = D 
Seismic Design Category (Provisions Sec. 4.2 [1.4]) = D 
Seismic-Force-Resisting System (Provisions Table 5.2.2 [4.3-1]) = Wood panel shear wall 
Response Modification Coefficient, R (Provisions Table 5.2.2 [4.3-1) = 6.5 
System Overstrength Factor, O0 (Provisions Table 5.2.2 [4.3-1) = 3 
Deflection Amplification Factor, Cd (Provisions Table 5.2.2 [4.3-1) = 4
FEMA 451, NEHRP Recommended Provisions: Design Examples 
10-6 
10.1.2.2 Structural Design Criteria 
10.1.2.2.1 Ground Motion (Provisions Sec. 4.1.2 [3.3]) 
Based on the site location, the spectral response acceleration values can be obtained from either the 
seismic hazard maps accompanying the Provisions or from the USGS CD-ROM. For site coordinates 
47.69 N, 122.32 W, the USGS CD-ROM returns short period response, SS = 1.34 and one second period 
response, S1 = 0.46. 
[The 2003 Provisions have adopted the 2002 USGS probabilistic seismic hazard maps, and the maps have 
been added to the body of the 2003 Provisions as figures in Chapter 3 (instead of the previously used 
separate map package). A CD-ROM containing the site response parameters based on the 2002 maps is 
also available.] 
The spectral response factors are then adjusted for the site class (Provisions Sec. 4.1.2.4 [3.5]). For this 
example, it is assumed that a site class recommendation was not part of the soils investigation, which 
would not be uncommon for this type of construction. When soil properties are not known, Provisions 
Sec. 4.1.2.1 [3.5] defaults to Site Class D, provided that soft soils (Site Class E or F) are not expected to 
be present at the site (a reasonable assumption for soils sufficient to support a multistory building on 
shallow spread footings). The adjusted spectral response acceleration parameters are computed, 
according to Provisions Eq. 4.1.2.4-1 and 4.1.2.4-2 [3.3-1 and 3.3-4], for the short period and one second 
period, respectively, as: 
SMS = FaSS = 1.0(1.34) = 1.34 
SM1 = FvS1 = 1.54(0.46) = 0.71 
where Fa and Fv are site coefficients defined in Provisions Tables 4.1.2.4a and 4.1.2.4b [3.3-1 and 3.3-2], 
respectively. Note that Straight line interpolation was used for Fv. 
Finally, the design spectral response acceleration parameters (Provisions Sec. 4.1.2.5 [3.3.3]) are 
determined in accordance with Provisions Eq. 4.1.2.5-1 and 4.1.2.5-2 [3.3-3 and 3.3-4], for the short 
period and one second period, respectively, as: 
2 2(1.34) 0.89 
SDS=3SMS=3 = 
2 2(0.71) 0.47 
SD1=3SM1=3 = 
10.1.2.2.3 Seismic Design Category (Provisions Sec. 4.2 [1.4]) 
Based on the Seismic Use Group and the design spectral response acceleration parameters, the Seismic 
Design Category is assigned to the building based on Provisions Tables 4.2.1a and 4.2.1b [1.4-1 and 1.4- 
2]. For this example, the building is assigned Seismic Design Category D. 
[Note that the method for assigning seismic design category for short period buildings has been revised in 
the 2003 Provisions. If the fundamental period, Ta, is less than 0.8Ts, the period used to determine drift is 
less than Ts, and the base shear is computed using 2003 Provisions Eq 5.2-2, then seismic design category 
is assigned using just 2003 Provisions Table 1.4-1 (rather than the greater of 2003 Provisions Tables 1.4- 
1 and 1.4-2). The change does not affect this example.]
Chapter 10, Wood Design 
10-7 
148'-0" 
56'-0" 
25'-0" 25'-0" 
A 
15'-0" 30'-0" 30'-0" 
15'-0" 30'-0" 30'-0" 
A: Another solution would be to use the full 
end for the shear wall, not just 25 feet. 
Solid 
interior 
wall 
Solid end 
wall 
55 feet of net shear wall 
length, each side of 
corridor, distributed over 
length of building. 
Perforated 
56'-0" 84'-0" exterior wall 
Figure 10.1-4 Load path and shear walls (1.0 ft = 0.3048 m). 
10.1.2.2.4 Load Path (Provisions Sec. 5.2.1 [14.2-1]) 
See Figure 10.1-4. For both directions, the load path for seismic loading consists of plywood floor and 
roof diaphragms and plywood shear walls. Because the lightweight concrete floor topping is 
discontinuous at each partition and wall, it is not considered to be a structural diaphragm. 
10.1.2.2.5 Basic Seismic-Force-Resisting Systems (Provisions Sec. 5.2.2 [4.3]) 
Building Class (Provisions Table 5.2.2 [4.3-1]): Bearing Wall System 
Seismic-Force-Resisting System (Provisions Table 5.2.2 [4.3-1]): Light frame walls with shear panels 
with R = 6.5, O0 = 3, and Cd = 4 for both directions 
10.1.2.2.6 Structure Configuration (Provisions Sec. 5.2.3 [5.3.2]) 
Diaphragm Flexibility (Provisions Sec. 5.2.3.1 [4.3.2.1]): Rigid (wood panel diaphragm in light frame 
structure with structural panels for shear resistance). Provisions Sec. 12.4.1.1 [4.3.2.1]defines a structural 
panel diaphragm as flexible, if the maximum diaphragm deformation is more than two times the average 
story drift. Due to the central shear wall, this is not expected to be the case in this building. 
Plan Irregularity (Provisions Sec. 5.2.3.2 [4.3.2.2]): Since the shear walls are not balanced for loading in 
the transverse direction (see Figure 10.1-4), there will be some torsional response of the system. The 
potential for torsional response combined with the rigid diaphragm requires that the building be checked 
for a torsional irregularity (Provisions Table 5.2.3.2 [4.3-2]). This check will be performed following the 
initial determination of seismic forces, and the final seismic forces will be modified if required. 
Vertical Irregularity (Provisions Sec. 5.2.3.3 [4.3.2.3]): Regular
FEMA 451, NEHRP Recommended Provisions: Design Examples 
10-8 
10.12.2.7 Redundancy (Provisions Sec. 5.2.4 [4.3.3]) 
For Seismic Design Category D, the reliability factor, ., is computed per Provisions Eq. 5.2.4.2 [4.3.3.2]. 
Since the computation requires more detailed information than is known prior to the design, assume . = 
1.0 for the initial analysis and verify later. If, in the engineer's judgement, the initial design appears to 
possess relatively few lateral elements, the designer may wish to use an initial . greater than 1.0 (but no 
greater than 1.5). 
[The redundancy requirements have been substantially changed in the 2003 Provisions. For a shear wall 
building assigned to Seismic Design Category D, . = 1.0 as long as it can be shown that failure of a shear 
wall with height-to-length-ratio greater than 1.0 would not result in more than a 33 percent reduction in 
story strength or create an extreme torsional irregularity. The intent is that the aspect ratio is based on 
story height, not total height. Therefore, the redundancy factor would have to be investigated only in the 
longitudinal direction where the aspect ratios of the perforated shear walls would be interpreted as having 
aspect ratios greater than 1.0 at individual piers. In the longitudinal direct, where the aspect ratio is less 
than 1.0, . = 1.0 by default.] 
10.1.2.2.8 Analysis Procedure (Provisions Sec. 5.2.5 [4.4.1]) 
Design in accordance with the equivalent lateral force (ELF) procedure (Provisions Sec. 5.4 [5.2]): No 
special requirements. In accordance with Provisions Sec. 5.2.5.2.3 [4.4.2], the structural analysis must 
consider the most critical load effect due to application of seismic forces in any direction for structures 
assigned to Seismic Design Category D. For the ELF procedure, this requirement is commonly satisfied 
by applying 100 percent of the seismic force in one direction, and 30 percent of the seismic force in the 
perpendicular direction; as specified in Provisions Sec. 5.2.5.2.2, Item a [4.4.2.2, Item 1]. For a lightframed 
shear wall building, with shear walls in two orthogonal directions, the only element affected by 
this directional combination would be the design of the shear wall end post and tie-down, located where 
the ends of two perpendicular walls intersect. In this example, the requirement only affects the shear wall 
intersections at the upper left and lower left corners of Figure 10.1-4. The directional requirement is 
satisfied using a two-dimensional analysis in the design of the remainder of the shear wall and diaphragm 
elements. 
10.1.2.2.9 Design and Detailing Requirements (Provisions Sec. 5.2.6 [4.6]) 
See Provisions Chapters 7 and 12 for special foundation and wood requirements, respectively. As 
discussed in greater detail below, Provisions Sec. 12.2.1, now utilizes Load and Resistance Factor Design 
(LRFD) for the design of engineered wood structures. Therefore, the design capacities are consistent with 
the strength design demands of Provisions Chapter 5 [4 and 5]. 
10.1.2.2.10 Combination of Load Effects (Provisions Sec. 5.2.7 [4.2.2]) 
The basic design load combinations are as stipulated in ASCE 7 and modified by the Provisions 
Eq. 5.2.7-1 and 5.2.7-2 [4.2-1 and 4.2-1]. Seismic load effects according to the Provisions are: 
E = .QE + 0.2SDSD 
and 
E = .QE - 0.2SDSD 
when seismic and gravity are additive and counteractive, respectively.
Chapter 10, Wood Design 
10-9 
For SDS = 0.89 and assuming . = 1.0 (both discussed previously), the design load combinations are as 
stipulated in ASCE 7: 
1.2D + 1.0E + 0.5L + 0.2S = 1.38D + 1.0QE + 0.5L + 0.2S 
and 
0.9D - 1.0E = 0.72D - 1.0QE 
10.1.2.2.11 Deflection and Drift Limits (Provisions Sec. 5.2.8 [4.5.1]) 
Assuming that interior and exterior finishes have not been designed to accommodate story drifts, then the 
allowable story drift is (Provisions Table 2.5.8 [4.5-1]): 
.a = 0.020 hsx 
where interstory drift is computed from story drift as (Provisions Eq. 5.4.6.1 [5.2-15]): 
( 1) 
1 
d xe x e 
x x x 
C 
I 
d d 
d d - 
- 
. = - = .. - .. 
where Cd is the deflection amplification factor, I is the occupancy importance factor, and dxe is the total 
elastic deflection at Level x. 
10.1.2.3 Basic Gravity Loads 
Roof: 
Live/Snow load (in Seattle, snow load governs over roof live = 25 psf 
load; in other areas this may not be the case) 
Dead load (including roofing, sheathing, joists, = 15 psf 
insulation, and gypsum ceiling) 
Floor: 
Live load = 40 psf 
Dead load (1 in. lightweight concrete, sheathing, joists, and = 20 psf 
gypsum ceiling. At first floor, omit ceiling but add insulation.) 
Interior partitions, corridor walls (8 ft high at 11 psf) = 7 psf distributed floor 
load 
Exterior frame walls (wood siding, plywood sheathing, = 15 psf of wall surface 
26 studs, batt insulation, and 5/8-in. gypsum drywall) 
Exterior double glazed window wall = 9 psf of wall surface 
Party walls (double-stud sound barrier) = 15 psf of wall surface 
Stairways = 20 psf of horiz. 
projection
FEMA 451, NEHRP Recommended Provisions: Design Examples 
10-10 
Perimeter footing (10 in. by 1 ft-4 in.) and grade beam = 562 plf 
(10 in. by 3 ft-2 in.) 
Corridor footing (10 in by 1 ft-4 in.) and grade wall = 292 plf 
(8 in. by 1 ft-3 in.); 18 in. minimum crawl space under first floor 
Applicable seismic weights at each level = 182.8 kips 
Wroof = area (roof dead load + interior partitions + party walls) 
+ end walls + longitudinal walls 
W3 = W2 = area (floor dead load + interior partitions + party walls) = 284.2 kips 
+ end walls + longitudinal walls 
Effective total building weight, W = 751 kips 
For modeling the structure, the first floor is assumed to be the seismic base, because the short crawl space 
with concrete foundation walls is quite stiff compared to the superstructure. 
10.1.3 SEISMIC FORCE ANALYSIS 
The analysis is performed manually following a step-by-step procedure for determining the base shear 
(Provisions Sec. 5.4.1 [5.2.1]), and the distribution of vertical (Provisions Sec. 5.4.3 [5.2.3]) and 
horizontal (Provisions Sec. 5.4.4 [5.2.4]) shear forces. Since there is no basic irregularity in the building 
mass, the horizontal distribution of forces to the individual shear walls is easily determined. These forces 
need only be increased to account for accidental torsion (see subsequent discussion). 
No consideration is given to soil-structure interaction since there is no relevant soil information available; 
a common situation for a building of this size and type. The soil investigation ordinarily performed for 
this type of structure is important, but is not generally focused on this issue. Indeed, the cost of an 
investigation sufficiently detailed to permit soil-structure interaction effects to be considered, would 
probably exceed the benefits to be derived. 
10.1.3.1 Period Determination and Calculation of Seismic Coefficient (Provisions Sec. 5.4.1 [5.2.1]) 
Using the values for SD1, SDS, R, and I from Sec. 10.1.2.1, the base shear is computed per Provisions Sec. 
5.4.1 [5.2.1]. The building period is based on Provisions Eq. 5.4.2.1-1 [5.2-6]: 
x 0.237 
Ta=Crhn= 
where Cr = 0.020, hn = 27 ft, and x = 0.75. 
According to Provisions Eq. 5.4.1.1-1 [5.2-2]: 
0.89 0.137 
/ 6.5/1.0 
DS 
s 
C S
R I 
= = = 
but need not exceed Provisions Eq. 5.4.1.1-2 [5.2-3]: 
0.47 0.305 
( / ) (0.237)(6.5 /1.0) 
D1 
s 
C S 
T R I 
= = =
Chapter 10, Wood Design 
10-11 
Ground floor 
Roof 
56'-0" 
9'-0" 9'-0" 9'-0" 
2 floor 
3 floor F3rd 
Froof 
F2nd 
nd 
rd 
Ground floor 
Roof 
56'-0" 
9'-0" 9'-0" 9'-0" 
2 floor 
3 floor F3rd 
Froof 
F2nd 
nd 
rd 
Figure 10.1-5 Vertical shear distribution (1.0 ft = 0.3048 m). 
Although it would probably never govern for this type of structure, also check minimum value according 
to Provisions Eq. 5.4.1.1-3: 
Cs = 0.044ISDS = (0.044)(1.0)(0.89) = 0.039 
[This minimum Cs value has been removed in the 2003 Provisions. In its place is a minimum Cs value for 
long-period structures, which is not applicable to this example.] 
The calculation of actual T as based on the true dynamic characteristics of the structure would not affect 
Cs; thus, there is no need to compute the actual period because the Provisions does not allow a calculated 
period that exceeds CuTa where Cu = 1.4 (see Provisions Sec. 5.4.2 [5.2.2]). Computing the base shear 
coefficient, per Provisions Eq. 5.4.1.1-1 [5.2-2], using this maximum period, would give Cs = 0.218, 
which still exceeds 0.137. Therefore, short period response governs the seismic design of the structure, 
which is common for low-rise buildings. 
10.1.3.2 Base Shear Determination 
According to Provisions Eq. 5.4.1 [5.2-1]: 
V = CsW = 0.137(751) = 103 kips (both directions) 
where effective total weight is W = 751 kips as computed previously. 
10.1.3.3 Vertical Distribution of Forces 
Forces are distributed as shown in Figure 10.1-5, where the story forces are calculated according to 
Provisions Eq. 5.4.3-1 and 5.4.3-2 [5.2-10 and 5.2-11]: 
1 
k 
x x 
x vx n k 
i i 
i 
F C V w h V 
w h 
= 
. . 
. . 
= =. . 
..S .. 
. .
FEMA 451, NEHRP Recommended Provisions: Design Examples 
10-12 
For T < 0.5, k = 1.0 and k 182.2(27) 284.2(18) 284.2(9) 12,610 
Swihi = + + = 
Froof = [182.8(27)/12,608](103.2) = 40.4 kips 
F3rd = [284.2(18)/12,608](103.2) = 41.9 kips 
F2nd = [284.2(9)/12,608](103.2) = 20.9 kips 
S = 103 kips 
It is convenient and common practice, to perform this calculation along with the overturning moment 
calculation. Such a tabulation is given in Table 10.1-1. 
Table 10.1-1 Seismic Coefficients, Forces, and Moments 
Level 
x 
Wx 
(kips) 
hx 
(ft) 
wxh k
x 
(k = 1) 
Cvx Fx 
(kips) 
Vx 
(kips) 
Mx 
(ft-kips) 
Roof 182.8 27 4,936 0.391 40.4 
40.4 364 
3 284.2 18 5,115 0.406 41.9 
82.3 1,104 
2 284.2 9 2,557 0.203 20.9 
103.2 2,033 
S 751.2 12,608 
1.0 ft = 0.3048 m, 1.0 kip = 4.45 kN, 1.0 ft-kip= 1.36 kN-m. 
10.1.3.4 Horizontal Distribution of Shear Forces to Walls 
Since the diaphragms are defined as rigid by the Provisions (Sec 5.2.3.1 [4.3.2.1] and 12.4.1.1), the 
horizontal distribution of forces must account for relative rigidity of the shear walls, and horizontal 
torsion must be included. As discussed below, for buildings with a Type 1a or 1b torsional irregularity 
(per Provisions Sec. 5.2.3.2 [4.3.2.2]) the torsional amplification factor (Provisions Sec. 5.4.4.3 [5.2.4.3]) 
must be calculated. 
It has been common practice for engineers to consider wood diaphragms as flexible, regardless of the 
relative stiffness between the walls and the diaphragms. Under the flexible diaphragm assumption, loads 
are distributed to shear walls based on tributary area, without taking diaphragm continuity, or relative 
wall rigidity into account. Recognizing that diaphragm stiffness should not be ignored (even for wood 
structural panel sheathing), the Provisions provides limits on when the flexible diaphragm assumption 
may be used and when it may not be used. 
The calculation of horizontal force distribution for rigid diaphragms, can be significantly more laborious 
than the relatively simple tributary area method. Therefore, this example illustrates some simplifications 
that can be used as relatively good approximations (as confirmed by using more detailed calculations). 
The design engineer is encouraged to verify all simplifying assumptions that are used to approximate 
rigid diaphragm force distribution. 
For this example, forces are distributed as described below.
Chapter 10, Wood Design 
10-13 
10.1.3.4.1 Longitudinal Direction 
Based on the rigid diaphragm assumption, force is distributed based on the relative rigidity of the 
longitudinal walls, and the transverse walls are included for resisting torsion. By inspection, however, the 
center of mass coincides with the center of rigidity and , thus, the torsional demand is just the accidental 
torsional moment resulting from a 5 percent eccentricity of force from the center of mass (Provisions Sec. 
5.4.4.2 [5.2.4.2]). 
In this direction, there are four lines of resistance and the total torsional moment is relatively small. 
Although the walls are unequal in length, the horizontal distribution of the forces can be simplified by 
making two reasonable assumptions. First, for plywood shear walls, it is common to assume that stiffness 
is proportional to net in-plane length of sheathing (assuming sheathing thickness, nailing, and chord 
elements are roughly similar). Second, for this example, assume that all of the torsional moment is 
resisted by the end walls in the transverse directions. This is a reasonable assumption because the walls 
have a greater net length than the longitudinal exterior shear walls and are located much farther from the 
center of rigidity (and thus contribute more significantly to the rotational resistance). 
Therefore, direct shear is distributed in proportion to wall length and torsional shear is neglected. Each 
exterior wall has 45 ft net length and each corridor wall has 55 ft net length for a total of 200 ft of net 
shear wall. The approximate load to each exterior wall is (45/200)Fx = 0.225Fx, and the load to each 
corridor wall is (55/200)Fx = 0.275Fx. (The force distribution was also computed using a complete 
rigidity model, including accidental torsion, with all transverse and longitudinal wall segments. The 
resulting distribution is 0.231Fx to the exterior walls and 0.276 Fx to the corridor walls, for a difference of 
2.7 percent and 0.4 percent, respectively). 
10.1.3.4.2 Transverse Direction 
Again, based on the rigid diaphragm assumption, force is to be distributed based on relative rigidity of the 
transverse walls, and the longitudinal walls are included for resisting torsion because the center of mass 
does not coincide with the center of rigidity. The torsional demand must be computed. Assuming that all 
six transverse wall segments have the same rigidity, the distance from the center of rigidity (CR) to the 
center of mass (CM) can be computed as: 
148/ 2 4 60 144 4.67ft 
3 
CM CR 
+ + 
- = - = 
The accidental torsional moment resulting from a 5 percent eccentricity of force from the center of mass 
(Provisions Sec. 5.4.4.2 [5.2.4.2]) must also be considered. 
As in the longitudinal direction, the force distribution in the transverse direction can be computed with 
reasonable accuracy by utilizing a simplified model. This simplification is made possible largely because 
the transverse wall segments are all of the same length and, thus, the same rigidity (assuming nailing and 
chord members are also the same for all wall segments). First, although the two wall segments at each 
line of resistance are offset in plan (Figure 10.1-4), assume that the wall segments do align, and are 
located at the centroid of the two segments. Second, assume that the longitudinal walls do not resist the 
torsional moment. Third, since interior transverse wall segments are located close to the center of rigidity, 
assume that they do not contribute to the resistance of the torsional moment. 
To determine the forces on each wall, split the seismic force into two parts: that due to direct shear and 
that due to the torsional moment. Because their rigidities are the same, the direct shear is resisted by all 
six wall segments equally and is computed as 0.167Fx. The torsional moment is resisted by the four end 
wall segments. Including accidental torsion, the total torsional moment is computed as:
FEMA 451, NEHRP Recommended Provisions: Design Examples 
10-14 
Total Torsional Moment, Mt + Mta = Fx  (4.67 ft + 0.05  148 ft) = 12.07Fx 
Torsional Shear to End Walls = 12.07Fx/140 ft = 0.086Fx 
Therefore, the simplified assumption yields a total design force of 0.167Fx + 0.086Fx/2= 0.210Fx, at each 
of the end wall segments on the right side of Figure 10.4-1, 0.167Fx - 0.086Fx/2= 0.124Fx at each of the 
end wall segments on the left side of the figure, and 0.167Fx at each interior wall segment. 
Next, determine the relationship between the maximum and average story drifts, to determine if a 
torsional irregularity exists, as defined in Provisions Table 5.2.3.2 [4.3-2]. Assuming that all of the shear 
walls will have the same plywood and nailing, the deflection of each wall will be proportional to the force 
in that wall. Therefore, the maximum drift for any level, will be proportional to 0.210Fx, and the average 
drift proportional to (0.210Fx + 0.124Fx)/2 = 0.167Fx. The ratio of maximum to average deflection is 
0.210Fx/0.167Fx = 1.26. Since the ratio is greater than 1.2, the structure is considered to have a torsional 
irregularity Type 1a, in accordance with Provisions Table 5.2.3.2 [4.3-2]. Therefore, the horizontal force 
distribution must be computed again using the torsional amplification factor, Ax, from Provisions Sec. 
5.4.4.3 [5.2.4.3]. (Diaphragm connections and collectors, not considered in this example, must satisfy 
Provisions Sec. 5.2.6.4.2. [4.6.3.2]) 
In accordance with Provisions Eq 5.4.4.3-1 [5.2-13]: 
( ) 
2 2 
0.210 1.10 
1.2 1.2 0.167 
max x 
x 
avg x 
A F
F 
d 
d 
. . . . 
=. . =. .= 
.. .. .. .. 
Therefore, recompute the torsional moment and forces as: 
Total torsional moment, Ax(Mt + Mta) = 1.10[Fx  (4.67 ft + 0.05  148 ft)] = 13.28Fx 
Torsional shear to End walls = 13.28Fx/140 ft = 0.095Fx 
The revised simplified assumption yields a maximum total design force of 0.167Fx + 0.095Fx/2= 0.214Fx 
at the end walls (right side of Figure 10.4-1), which will be designed in the following sections. 
(The force distribution was also computed using a complete rigidity model including all transverse and 
longitudinal wall segments. Relative rigidities were based on length of wall and the torsional moment 
was resisted based on polar moment of inertia. The resulting design forces to the shear wall segments 
were 0.208Fx, 0.130Fx, and 0.168Fx for the right end, left end, and interior segments, respectively. The 
structure is still torsionally irregular (dmax / dave = 1.23), but the irregularity is less substantial and the 
resulting Ax = 1.05. This more rigorous analysis, has shown that the simplified approach provides 
reasonable results in this case. Since the simplified method is more likely to be used in design practice, 
the values from that approach will be used for the remainder of this example.) 
10.1.3.5 Verification of Redundancy Factor 
Once the horizontal force distribution is determined, the assumed redundancy factor must be verified. 
Provisions Eq. 5.2.4.2, is used to compute the redundancy factor, ., for each story as: 
2 20 
x 
x 
rmax Ax 
. = -
Chapter 10, Wood Design 
10-15 
where Ax is the floor area in square feet and rmaxx is the ratio of the design story shear resisted by the 
single element carrying the most shear force in the story to the total story shear. As defined for shear 
walls in Provisions Sec. 5.2.4.2, shall be taken as the shear in the most heavily loaded wall, rmaxx 
multiplied by 10/lw, where lw is the wall length in feet. 
The most heavily loaded wall is the end wall, farthest away from the interior shear wall. Since the force 
distribution is the same for all three levels, the redundancy factor need only be determined for one level, 
in this case, at the first floor. The redundancy factor is computed as: 
Floor area = (140)(56) = 7,840 ft2 
Max load to wall = 0.214V 
Wall length = 25 ft 
= 0.214V(10/25)/V = 0.0856 rmaxx 
Therefore: 
2 20 0.64 
0.0856 7,840 . x = - = - 
Because the calculated redundancy factor is less than the minimum permitted value of 1.0, the initial 
assumption of . = 1.0 is correct, and the design can proceed using the previously computed lateral forces. 
[See Sec. 10.1.2.2 for discussion of the changes to the redundancy requirements in the 2003 Provisions.] 
10.1.3.6 Diaphragm Design Forces 
As specified in Provisions Sec. 5.2.6.4.4 [4.6.3.4], floor and roof diaphragms must be designed to resist a 
force, Fpx, in accordance with Provisions Eq. 5.2.6.4.4 [4.6-2]: 
n 
i 
i x 
px n px 
i 
i x 
F 
F w 
w 
=
=
S 
= 
S 
plus any force due to offset walls (not applicable for this example). The diaphragm force as computed 
above need not exceed 0.4SDSIwpx, but shall not be less than 0.2SDSIwpx. This latter value often governs at 
the lower levels of the building, as it does here. The maximum required diaphragm demand does not 
govern in this example. 
The weight tributary to the diaphragm, wpx, need not include the weight of walls parallel to the force. For 
this example, however, since the shear walls in both directions are relatively light compared to the total 
tributary diaphragm weight, the diaphragm force is computed based on the total story weight, for 
convenience. 
Transverse direction 
Roof Level 
SFi = 40.4 
Swi = 182.8 
FP,roof = (40.4/182.8)(182.8) = 40.4 kips (controls for roof) 
0.2SDSIwpx = (0.2)(0.89)(1.0)(182.8) = 32.7 kips
FEMA 451, NEHRP Recommended Provisions: Design Examples 
10-16 
Third Floor 
SFi = 40.4 + 41.9 = 82.3 
Swi = 182.8 + 284.2 = 447.0 
FP,3rd =(82.3/447.0)(284.2) = 50.1 kips 
0.2SDSIwpx = (0.2)(0.89)(1.0)(284.2) = 50.8 kips (controls for 3rd floor) 
Second Floor 
SFi = 40.4 + 41.9 + 20.9 = 103,231 
Swi = 182.8 + 284.2 + 284.2 = 751.2 
FP,2nd = (103.2/751.2)(284.2) = 39.1 kips 
0.2SDSIwpx = (0.2)(0.89)(1.0)(284.2) = 50.8 kips (controls for 2nd floor) 
Diaphragm forces in the longitudinal direction are computed in a similar manner. Since the weight of the 
exterior walls is more significant in the longitudinal direction, the designer may wish to subtract this 
weight from the story force in order to compute the diaphragm demands. 
10.1.4 Basic Proportioning 
Designing a plywood diaphragm and plywood shear wall building, principally involves the determination 
of sheathing thicknesses and nailing patterns to accommodate the applied loads. In doing so, some 
successive iteration of steps may be needed to satisfy the deflection limits. 
Nailing patterns in diaphragms and shear walls have been established on the basis of tabulated 
requirements included in the Provisions. It is important to consider the framing requirements for a given 
nailing pattern and capacity as indicated in the notes following the tables. In addition to strength 
requirements, Provisions Sec. 12.4.1.2 [12.4.1.1] places aspect ratio limits on plywood diaphragms 
(length/width shall not exceed 4/1 for blocked diaphragms), and Provisions Sec. 12.4.2.3 [12.4.2.3] places 
similar limits on shear walls (height/width shall not exceed 2/1 for full design capacities). However, it 
should be taken into consideration that compliance with these aspect ratios does not guarantee that the 
drift limits will be satisfied. 
Therefore, diaphragms and shear walls have been analyzed for deflection as well as for shear capacity. A 
procedure for computing diaphragm and shear wall deflections is provided in Commentary Sec. 12.4.1. 
This procedure is illustrated below. The Commentary does not indicate how to compute the nail 
deformation (nail slip) factor, but there is a procedure contained in the commentary of ASCE16. 
In the calculation of diaphragm deflections, the chord slip factor can result in large additions to the total 
deflection. This can be overcome by using neat holes for bolts, and proper shimming at butt joints. 
However, careful attention to detailing and field inspection are essential, to ensure that they are provided. 
[AF&PA Wind & Seismic also contains procedures for computing diaphragm and shear wall deflections. 
The equations are slightly different from the more commonly used equations that appear in the 
Commentary and AF&PA LRFD Manual. In AF&PA Wind & Seismic, the shear and nail slip terms are 
combined using an apparent shear stiffness parameter. However, the apparent shear stiffness values are 
only provided for OSB. Therefore, the deflection equations in the Commentary or AF&PA LRFD 
Manual must be used in this example which has plywood diaphragms and shear walls. The apparent 
shear stiffness values for plywood will likely be available in future editions of AF&PA Wind & Seismic.] 
10.1.4.1 Strength of Members and Connections
Chapter 10, Wood Design 
10-17 
The 2000 Provisions has adopted Load and Resistance Factor Design (LRFD) for engineered wood 
structures. The Provisions includes the ASCE 16 standard by reference and uses it as the primary design 
procedure for engineered wood construction. Strength design of members and connections is based on 
the requirements of ASCE 16. The AF&PA Manual and supplements contain reference resistance values 
for use in design. For convenience, the Provisions contain design tables for diaphragms and shear walls 
that are identical to those contained in the AF&PA Structural-Use Panels Supplement. However, the 
modification of the tabulated design resistance for shear walls and diaphragms with framing other than 
Douglas Fir-Larch or Southern Pine is different between the two documents. This example illustrates the 
modification procedure contained in the Provisions tables, which is new to the 2000 edition. 
Throughout this example, the resistance of members and connections subjected to seismic forces, acting 
alone, or in combination with other prescribed loads, is determined in accordance with ASCE 16 and the 
AF&PA Manual; with the exception of shear walls and diaphragms for which design resistance values are 
taken from the Provisions. The LRFD standard incorporates the notation D, T, Z, etc. to represent 
adjusted resistance values, which are then modified by a time effect factor, ., and a capacity reduction 
factor, f, to compute a design resistance, which is defined as factored resistance in the Provisions and 
ASCE 16. Additional discussion on the use of LFRD is included in Commentary Sec. 12.2 and 12.3, and 
in the ASCE 16 commentary. It is important to note that ACSE 16 and the AF&PA Manual use the term, 
resistance, to refer to the design capacities of members and connections while strength refers to 
material property values. 
It is worth noting that the AF&PA Manual contains a Pre-engineered Metal Connections Guideline for 
converting allowable stress design values for cataloged metal connection hardware (for example, tiedown 
anchors) into ultimate capacities for use with strength design. This procedure is utilized in this 
example. 
[The primary reference for design of wood diaphragms and shear walls in the 2003 Provisions is AF&PA 
Wind & Seismic. Much of the remaining text in the 2003 Provisions results from differences between 
AF&PA Wind & Seismic and Chapter 12 of the 2000 Provisions as well as areas not addressed by 
AF&PA Wind & Seismic. Because the AF&PA Wind & Seismic tabulated design values for diaphragms 
and shear walls do not completely replace the tables in the 2000 Provisions, portions of the tables remain 
in the 2003 Provisions. Therefore, some diaphragm and shear wall design values are in the 2003 
Provisions and some are in AF&PA Wind & Seismic. The design values in the tables are different 
between the two documents. The values in the 2003 Provisions represent factored shear resistance 
(.fD), while the values in AF&PA Wind & Seismic represent nominal shear resistance that must then be 
multiplied by a resistance factor, f, (0.65) and a time effect factor ,., (1.0 for seismic loads). Therefore, 
while the referenced tables may be different, the factored resistance values based on the 2003 Provisions 
should be the same as those in examples based on the 2000 Provisions. The calculations that follow are 
annotated to indicate from which table the design values are taken.] 
10.1.4.2 Transverse Shear Wall Nailing 
The design will focus on the more highly loaded end walls; interior walls are assumed to be similar. 
10.1.4.2.1 Load to Any One of Four 25-ft End Walls 
Froof = 0.214(40.4) = 8.65 kips 
F3rd = 0.214(41.9) = 8.97 kips 
F2nd = 0.214(20.9) = 4.47 kips 
S = 22.09 kips
FEMA 451, NEHRP Recommended Provisions: Design Examples 
10-18 
9' 9' 9' 
25'-0" 
27'-0" 
22.09 kips 
v = 0.883 kip/ft 
v = 0.704 kip/ft 
v = 0.346 kip/ft 
Fr o o f = 8.65 kips 
F3 rd = 8.97 kips 
F2 n d = 4.47 kips 
9' 9' 9' 
25'-0" 
27'-0" 
22.09 kips 
v = 0.883 kip/ft 
v = 0.704 kip/ft 
v = 0.346 kip/ft 
Fr o o f = 8.65 kips 
F3 rd = 8.97 kips 
F2 n d = 4.47 kips 
Figure 10.1-6 Transverse section: end wall (1.0 ft = 0.3048 m, 1.0 
kip = 4.45 kN, 1.0 kip/ft = 14.6 kN/m). 
10.1.4.2.2 Roof to Third Floor Design 
V = 8.65 kips 
v = 8.65/25 = 0.346 klf 
Try a -in. (15/32) plywood rated sheathing (not Structural I) on blocked 2-in. Douglas fir-Larch 
members at 16 in. on center with 10d common nails at six-in. on center at panel edges and 12 in. on center 
at intermediate framing members. According to Note b of Provisions Table 12.4.3-2a [12.4-3a], the 
design shear resistance must be adjusted for Hem-fir wall framing. The specific gravity adjustment factor 
equals 1-(0.5-SG) where SG is the specific gravity of the framing lumber. From Table 12A of the 
AF&PA Structural Connections Supplement, the SG = 0.43 for Hem-fir. Therefore, the adjustment factor 
is 1-(0.5-0.43) = 0.93. 
From Provisions Table 12.4.3-2a [AF&PA Wind/Seismic Table 4.3A], .fD = 0.40 klf 
Adjusted shear resistance = 0.93(0.40) = 0.37 klf > 0.346 klf OK 
8d nails could be used at this level but, because 10d nails are required below, 10d nails are used here for 
consistency. 
Deflection of plywood shear panels depends, in part, on the slip at the nails which, in turn, depends on the 
load per nail. See Sec. 10.1.4.3 for a detailed discussion of the appropriate slip for use with the 
Provisions and see Table 10.1-2 for fastener slip equations used here. 
Load per nail = 0.346(6/12)(1000) = 173 lb 
Nail slip en = 1.2(173/769)3.276 = 0.00904 in. 
In the above equation, 1.2 = factor for plywood other than Structural I, and 769 and 3.276 are constants 
explained in Sec. 10.1.4.3. 
[As indicated previously, AF&PA Wind & Seismic does not provide apparent shear stiffness values for 
plywood sheathing. Therefore, the deflection equations in the Commentary or AF&PA LRFD Manual 
must be used in this case.] 
10.1.4.2.3 Third Floor to Second Floor 
V = 8.65 + 8.97 = 17.62 kips
Chapter 10, Wood Design 
10-19 
v = 17.62/25 = 0.704 klf 
Try -in. (15/32) plywood rated sheathing (not Structural I) on blocked 2-in. Douglas fir-Larch members 
at 16 in. on center with 10d nails at 3-in. on center at panel edges, and at 12 in. on center at intermediate 
framing members. 
From Provisions Table 12.4.3-2a [AF&PA Wind&Seismic Table 4.3A], .fD = 0.78 klf 
Adjusted shear resistance = 0.93(0.78) = 0.73 klf > 0.704 klf OK 
Table Note e [AF&PA Wind&Seismic Sec. 4.3.7.1] requires 3-in. framing at adjoining panel edges. 
Load per nail = 0.704(3/12)(1000) = 176 lb 
Nail slip en = 1.2(176/769)3.276 = 0.00960 in. 
10.1.4.2 Second Floor to First Floor 
V = 17.62 + 4.47 = 22.09 kips 
v = 22.09/25 = 0.883 klf 
Try a -in. (15/32) plywood rated sheathing (not Structural I) on blocked 2-in. Douglas fir-Larch 
members at 16 in. on center with 10d common nails at 2-in. on center at panel edges and 12 in. on center 
at intermediate framing members. 
From Provisions Table 12.4.3-2a [AF&PA Wind&Seismic Table 4.3A], .fD = 1.00 klf 
Adjusted shear resistance = 0.93(1.00) = 0.93 klf > 0.883 klf OK 
Table Notes d and e [AF&PA Wind&Seismic Sec. 4.3.7.1]require 3-in. framing at adjoining panel edges. 
Load per nail = 0.883(2/12)(1000) = 147 lb 
Nail slip en = 1.2(147/769)3.276 = 0.00534 in. 
10.1.4.3 General Note on the Calculation of Deflections for Plywood Shear Panels 
The commonly used expressions for deflection of diaphragms and shear walls, which are contained in 
Commentary Sec. 12.4.1, and the ASCE 16 commentary standard (see also UBC Std 23-2 or APA 138), 
include a term for nail slip. The ASCE 16 commentary includes a procedure for estimating nail slip that 
is based on LRFD design values. 
The values for en used in this example (and in Sec. 10.2) are calculated according to Table 10.1-2, which 
is taken from Table C9.5-1 of ASCE 16.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
10-20 
Table 10.1-2 Fastener Slip Equations 
Fastener 
Minimum 
Penetration 
(in.) 
For Maximum 
Loads Up to 
(lb.) 
Approximate Slip, en 
* 
Green/Dry Dry/Dry 
6d common nail 1-1/4 180 (Vn/434)2.314 (Vn/456)3.144 
8d common nail 1-7/16 220 (Vn/857)1.869 (Vn/616)3.018 
10d common nail 1-5/8 260 (Vn/977)1.894 (Vn/769)3.276 
14-ga staple 1 to 2 140 (Vn/902)1.464 (Vn/596)1.999 
14-ga staple 2 170 (Vn/674)1.873 (Vn/361)2.887 
*Fabricated green/tested dry (seasoned); fabricated dry/tested dry. Vn = fastener load in pounds. Values 
based on Structural I plywood fastened to Group II lumber. Increase slip by 20 percent when plywood is not 
Structural I. 
1.0 in. = 25.4 mm, 1.0 lb = 4.45 N. 
This example is based on the use of surfaced dry lumber so the dry/dry values are used. The 
appearance of the equations for both shear wall and diaphragm deflection, implies greater accuracy than is 
justified. The reader should keep in mind that the deflections calculated are only rough estimates. 
10.1.4.4 Transverse Shear Wall Deflection 
From Commentary Sec. 12.4, modified as described below, shear wall deflection is computed as: 
8 3 0.75 n a 
vh vh he h d 
wEA Gt w 
d = + + + 
The above equation produces displacements in inches and the individual variables must be entered in the 
force or length units as described below: 
v = V/w where V is the total shear on the wall and v is in units of pounds/foot 
8vh3/wEA = bending deflection, as derived from the formula db = Vh3/3EI, where V is the total shear 
in pounds, acting on the wall and I = Aw2/2 (in.4) 
vh/Gt = shear deflection, as derived from the formula dv = Vh/GA', where A' = wt (in.2) 
0.75 hen = nail slip, in inches. Note that with h being given in feet, the coefficient 0.75 carries units 
of 1/ft. 
(h/w)da = deflection due to anchorage slip, in inches. Note that for use in the deflection equation 
contained in the Commentary, the term da represents the vertical deflection due to anchorage details. 
In the deflection equation contained in the commentary in the AF&PA Manual, there is no h/w factor, 
so it should be assumed that the term da represents the horizontal deflection at the top of the wall due 
to anchorage details. 
For this example: 
E = 1,600,000 psi 
G = 75,000 psi 
A (area) = 2(2.5)5.5 = 27.5 in.2 for assumed double 36 end posts
Chapter 10, Wood Design 
10-21 
Roof 
2 floor 
3 floor 
nd 
rd 
Roof 
2 floor 
3 floor 
nd 
rd 
Figure 10.1-7 Force 
distribution for flexural 
deflections. 
w (shear wall length) = 25 ft 
h (story height) = 9 ft 
t (effective thickness) = 0.298 in. for -in. unsanded plywood (not Structural I) 
en = nail deformation factor from prior calculations, inches. 
This equation is designed for a one-story panel and some modifications are in order for a multistory 
panel. The components due to shear distortion and nail slip are easily separable (see Table 10.1-3). 
Table 10.1-3 Wall Deflection (per story) Due to Shear and Nail Slip 
Story v 
(plf) 
vh/Gt 
(in.) 
en 
(in.) 
0.75enh 
(in.) 
Roof 346 0.139 0.00904 0.061 
3 704 0.284 0.00960 0.065 
2 883 0.356 0.00534 0.036 
1.0 in. = 25.4 mm, 1.0 plf = 14.6 N/m. 
Likewise, the component due to anchorage slip is easily separable; it is a rigid body rotation. If a 1/8-in. 
upward slip is assumed (on the tension side only), the deflection per story is (9/25)(1/8) = 0.045 in. (Table 
10.1-4). 
The component due to bending is more difficult to separate. For this example, a grossly simplified, 
distributed triangular load on a cantilever beam is used (Figure 10.1-7). 
The total load V is taken as 21.3 kips, the sum of the story forces on the wall. The equation for the deflection, 
taken from Roark's Formulas for Stress and Strain, is: 
5 4 4 5 
2 2 
2 (11 15 5 ) 
5 x 
V h hx hx x 
EAb h 
d= - + - 
where x is the distance (ft) from the top of the building to the story in question, h is the total height (ft), b is 
the wall width (ft), A is the chord cross sectional area (in.2), E is the modulus of elasticity (psi), and the 
resulting displacement, d, is in inches. 
This somewhat underestimates the deflections, but it is close enough for design. The results are shown in 
Table 10.1-4. 
The total wall deflections, shown in Table 10.1-5, are combined with the diaphragm deflections (see Sec. 
10.1.4.7, 10.1.4.8, and 10.1.4.9). Drift limits are checked after diaphragm deflections are computed.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
10-22 
Table 10.1-4 Wall Deflection (per story) Due to Bending and Anchorage Slip 
Level Effective 
8vh3/wEA (in.) 
(h/w)da 
(in.) 
Roof 0.031 0.045 
3 0.027 0.045 
2 0.012 0.045 
1.0 in. = 25.4 mm. 
Table 10.1-5 Total Elastic Deflection and Drift of End Wall 
Level Shear 
(in.) 
Bending 
(in.) 
Nail Slip 
(in.) 
Anchor 
Slip (in.) 
Drift .e 
(in.) 
Total de 
(in.) 
Roof 0.139 0.031 0.061 0.045 0.277 1.145 
3 0.284 0.027 0.065 0.045 0.420 0.869 
2 0.356 0.012 0.036 0.045 0.448 0.448 
1.0 in. = 25.4 mm. 
10.1.4.5 Transverse Shear Wall Anchorage 
Provisions Sec. 12.4.2.4 [12.4.2.4.1] requires tie-down (hold-down) anchorage at the ends of shear walls 
where net uplift is induced. Net uplift is computed as the combination of the seismic overturning moment 
and the dead load counter-balancing moment using the load combination indicated in Provisions Eq. 
5.2.7-2 [4.2-2]. 
Traditionally, tie-down devices were designed to resist this net overturning demand. However, 
Provisions Sec. 12.4.2.4 [12.4.2.4.1] requires that the nominal strength of tie-down devices exceeds the 
expected maximum vertical force that the shear panels can deliver to the tension-side end post. 
Specifically, the nominal strength (f = 1.0) of the tie-down device must be equal to, or greater than the 
net uplift forces resulting from O0/1.3 times the factored shear resistance of the shear panels, where O0 is 
the system overstrength factor. These uplift forces are cumulative over the height of the building. Also, 
dead load forces are not used in these calculations to offset the uplift forces for the design of the tie-down 
anchorage. 
[The requirements for uplift anchorage are slightly different in the 2003 Provisions. The tie down force is 
based on the nominal strength of the shear wall rather than the O0/1.3 times the factored resistance as 
specified in the 2000 Provisions. This change is primarily intended to provide consistent terminology and 
should not result in a significant impact on the design of the tie down.] 
An additional requirement of Provisions Sec. 12.4.2.4 [12.4.2.4.1], is that end posts must be sized such 
that failure across the net section does not control the capacity of the system. That is, the tensile strength 
of the net section must be greater than the tie-down strength. Note that the tie-downs designed in the 
following section are not located at wall intersections where the directional combination requirements of 
Provisions Sec. 5.2.5.2.3 [4.4.2.3] would apply (see also the Analysis Procedure in Sec. 10.1.2.2). 
10.1.4.5.1 Tie-down Anchors at Third Floor 
For the typical 25-ft end wall segment, the overturning moment at the third floor is: 
M0 = 9(8.65) = 77.8 ft-kip = QE
Chapter 10, Wood Design 
10-23 
The counter-balancing moment, 0.72 QD = 0.72D(25 ft/2). The width of the floor contributing to this, is 
taken as half the span of the exterior window header equal to 6.5 feet (see Figures 10.1-1 and 10.1-2). 
For convenience, the same length is used for the longitudinal walls, the weight of which (interior and 
exterior glazed wall) is assumed to be 9 psf. 
End wall self weight = 9 ft (25ft )15 psf/1000 = 3.4 kips 
Tributary floor = 6.5 ft (25 ft) (15) psf/1000 = 2.4 kips 
Tributary longitudinal walls = 9 ft (6.5 ft) 9 psf (2)/1000 = 1.1 kips 
S = 6.9 kips 
0.72QD = 0.72(6.9)12.5 = 62.1 ft-kip 
M0 (net) = 77.8 - 62.1 = 15.7 ft-kip 
Therefore, uplift anchorage is required. Using the procedure described above, nominal tie-down strength 
must be equal to or exceed the tension force, T, computed as: 
T = 0.37 klf (3/1.3)(8ft) = 6.83 kips 
where 
0.37 klf = factored shear wall resistance at the third floor 
3/1.3 = O0/1.3 
8 ft = net third floor wall height (9-ft story height minus approximately 1 ft of framing). 
Use a double tie-down device to eliminate the eccentricity associated with a single tie-down. Also, use 
the same size end post over the full height of the wall to simplify the connections and alignments. As will 
be computed below, a 66 (Douglas fir-Larch) post is required at the first floor. At the third floor, try 
two sets of double tie-down anchors connected through the floor with a 5/8-in. threaded rod to the end 
posts with two 5/8-in. bolts similar to Figure 10.1-8. 
According to Provisions Sec. 12.4.2.4 [12.4.2.4.1], the nominal tie-down strength is defined as the 
maximum test load the device can resist under cyclic testing without connection failure by either metal 
or wood failure. [Note that this definition for nominal tie down strength has been removed in the 2003 
Provisions. This change is primarily intended to provide consistent terminology and should not result in 
a significant impact on the design of the tie down.] For a single tie-down device as described above, the 
cataloged allowable uplift capacity is 2.76 kips and the cataloged average ultimate load is 12.15 kips. 
However, according to the documentation in this particular suppliers product catalog, the ultimate values 
are based on static testing and the type of failure is not indicated. Therefore, for this example, the 
cataloged ultimate value is not considered to satisfy the requirements of the Provisions. As an alternate 
approach, this example will utilize the methodology contained in the AF&PA Manual for converting 
cataloged allowable stress values to strength values. 
Based on the procedure in the AF&PA Manual, Pre-Engineered Metal Connectors Guide, the strength 
conversion factor for a connector for which the catalog provides an allowable stress value with a 1.33 
factor for seismic is 2.88/1.33. 
Since the typical product catalogs provide design capacities only for single tie-downs, the design of 
double tie-downs requires two checks. First, consider twice the capacity of one tie-down, and second, the 
capacity of the bolts in double shear. f = 1.0 for both these calculations. 
For the double tie-down, the nominal strength is computed as:
FEMA 451, NEHRP Recommended Provisions: Design Examples 
10-24 
2.fZ = 2(1.0)(1.0)(2.76)(2.88/1.33) = 12.0 kips > 6.83 kips OK 
For the two bolts through the end post in double shear, the AF&PA Manual gives: 
2.fZ = 2(1.0)(1.0)(4.90) = 9.80 kips > 6.83 kips OK 
The factored capacity of the tie-downs must also be checked for the design loads, which in this case will 
not govern the design. 
10.1.4.5.2 Tie-down Anchors at Second Floor 
Since tie-downs are required at the third floor, it would be common practice to provide tie-downs at the 
second and first floors, whether or not calculations indicate that they are required. Nevertheless, the 
overturning calculations are performed for illustrative purposes. The overturning moment at the second 
floor is: 
M0 = 18(8.65) + 9(8.97) = 236 ft-kip. 
The counter-balancing moment, 0.72 QD = 0.72(DL)12.5. 
End wall self weight = 18 ft (25 ft) 15 psf/1000 = 6.75 kips 
Tributary floor = 6.5 ft (25 ft) (15 + 27) psf/1000 = 6.83 kips 
Tributary longitudinal walls = 18 ft (6.5 ft) 9 psf (2)/1000 = 2.10 kips 
S = 15.7 kips 
0.72QD = 0.72(15.7)12.5 = 141 ft-kips 
M0 (net) = 236-141 = 95 ft-kips 
As expected, uplift anchorage is required. The design uplift force is computed using an adjusted factored 
shear resistance of 0.73 klf at the second floor, and a net length of wall height equal to 8 ft. Note that 8 ft. 
is appropriate for this calculation given the detailing for this structure. As shown in Figure 10.1-10, the 
plywood sheathing is not detailed as continuous across the floor framing, which results in a net sheathing 
height of about 8ft. If the sheathing were detailed across the floor framing, then 9 ft would be the 
appropriate wall height for use in computing tie-down demands. Combined with the uplift force at the 
third floor, the total design uplift force at the second floor is: 
T = 6.83 kips + 0.73 klf (3/1.3)(8ft) = 20.3 kips 
Use two sets of double tie-down anchors to connect the 66 end posts. Using the same procedure as for 
the third floor, tie-downs with a seven-eighths-in. threaded rod and three seven-eighths-in. bolts are 
computed to be adequate. See Figure 10.1-8.
Chapter 10, Wood Design 
10-25 
Threaded rod connected 
to tie-down anchors 
with thru bolts 7 
diameters from 
end of stud 
Post and 
tie down 
Floor framing 
Solid blocking 
at post 
Corner 
of wall 
Figure 10.1-8 Shear wall tie down at suspended floor framing. 
10.1.4.5.3 Tie-down Anchors at First Floor 
The overturning moment at the first floor is: 
M0 = 27(8.65) + 18(8.97) + 9(4.47) = 435 ft-kip = QE. 
The counter-balancing moment, 0.72 QD = 0.72D(25 ft/2). 
End wall self weight = 27 ft (25ft )15 psf/1000 = 10.1 kips 
Tributary floor = 6.5 ft (25 ft) (15 + 27 + 27) psf/1000 = 11.2 kips 
Tributary longitudinal walls = 27 ft (6.5 ft) 9 psf (2)/1000 = 3.2 kips 
S = 24.5 kips 
0.72QD = 0.72(24.5)12.5 = 221 ft-kip 
M0 (net) = 435 - 221 = 214 ft-kip 
As expected, uplift anchorage is required. The design uplift force is computed using an adjusted factored 
shear resistance of 0.93 klf at the first floor, and a net wall height of 8ft. Combined with the uplift force 
at the floors above, the total design uplift force at the first floor is: 
T = 20.3 kips + 0.93 klf (3/1.3)(8ft.) = 37.5 kips. 
Use a double tie-down anchor that extends through the floor with an anchor bolt into the foundation. 
Tie-downs with a 7/8-in. threaded rod, and four 7/8-in. bolts are adequate.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
10-26 
The strength of the end post, based on failure across the net section, must also be checked (Provisions 
Sec. 12.4.2.4 [12.4.2.4.1]). For convenience, the same size post has been used over the height of the 
building, so the critical section is at the first floor. A reasonable approach to preclude net tension failure 
from being a limit state would be to provide an end post, whose factored resistance exceeds the nominal 
strength of the tie-down device. (Note that using the factored resistance rather than nominal strength of 
the end post provides an added margin of safety that is not explicitly required by the Provisions.) The 
nominal strength of the first floor double tie-down is 42.9 kips, as computed using the procedure 
described above. Therefore, the tension capacity at the net section must be greater than 42.9 kips. 
Try a 66 Douglas Fir-Larch No. 1 end post. In some locations, the shear wall end post also provides 
bearing for the window header, so this size is reasonable. Accounting for 1-in. bolt holes, the net area of 
the post is 24.75 in.2 According to the AF&PA Manual, Structural Lumber Supplement: 
.fT = (1.0)(0.8)(2.23ksi)(24.75in2) = 44.2 kips > 42.9 kips OK 
For the maximum compressive load at the end post, combine maximum gravity load, plus the seismic 
overturning load. In the governing condition, the end post supports the header over the glazed portion of 
the exterior wall (end wall at right side of Figure 10.1-1). Assume that the end post at the exterior side of 
the wall supports all the gravity load from the header, and resists one-half of the seismic overturning load. 
Compute gravity loads based on a 6.5-foot tributary length of the header: 
Tributary DL = ((27 ft)(9 psf) + (8 ft)(15 + 27 + 27)psf)(6.5ft)/1000 = 5.17 kips 
Tributary LL = 8 ft (7 ft) (40 + 40) psf/1000 = 4.48 kips 
Tributary SL = 8 ft (6.5 ft) (25 psf)/1000 = 1.30 kips 
The overturning force is based on the seismic demand (not wall capacity as used for tension anchorage) 
assuming a moment arm of 23 ft: 
Overturning, QE = 435/23 ft = 18.9 kips 
Per load combination associated with Provisions Eq. 5.2.7.1-1 [4.2-1]: 
Maximum compression = 1.38(5.17) + 1.0(18.9/2) + 0.5(4.48) + 0.2(1.30) = 19.1 kips. 
Due to the relatively short clear height of the post, the governing condition is bearing perpendicular to the 
grain on the bottom plate. Check bearing of the 66 end post on a 36 Douglas fir-Larch No. 2 plate, per 
the AF&PA Manual, Structural Lumber Supplement: 
.fP= (1.0)(0.8)(1.30ksi)(30.25 in.2) = 31.5 kips > 19.1 kips. OK 
The 66 end post is slightly larger than the double 36 studs assumed above for the shear wall deflection 
calculations. Therefore, the computed shear wall deflection is slightly conservative, but the effect is 
minimal. 
10.1.4.5.4 Check Overturning at the Soil Interface 
A summary of the overturning forces is shown in Figure 10.1-9. To compute the overturning at the soil 
interface, the overturning moment must be increased for the 4-ft foundation height: 
M0 = 435 + 22.09(4.0) = 523 ft-kip
Chapter 10, Wood Design 
10-27 
However, it then may be reduced in accordance with Provisions Sec. 5.3.6: 
M0 = 0.75(523) = 392 ft-kip 
To determine the total resistance, combine the weight above with the dead load of the first floor and 
foundation. 
Load from first floor = 25 ft (6.5 ft) (27 - 4 + 1)psf /1000 = 3.9 kips 
where 4 psf is the weight reduction due to the absence of a ceiling, and 1 psf is the weight of insulation. 
The length of the longitudinal foundation wall included, is a conservative approximation of the amount 
carried by minimum nominal reinforcement in the foundation. 
Foundation weight = (562 plf (13 ft + 25 ft) + 292 plf (13 ft))/1000 = 25.1 kips 
First floor = 3.9 kips 
Structure above = 24.5 kips 
S = 53.5 kips 
Therefore, 0.72D - 1.0QE = 0.72(53.5)12.5 ft - 1.0(392) = 89.5 ft-kips, which is greater than zero, so the 
wall will not overturn. 
10.1.4.5.5 Anchor Bolts for Shear 
At the first floor: 
v = 0.883 klf. 
A common anchorage for non-engineered construction is a -in. bolt at 4ft-0 in. For a -in. bolt in a 
Douglas fir-Larch 36 plate, in single shear, parallel to the grain: 
.fZ/4ft = (1.0)(0.65)(2.38 kips)/4 = 0.39 klf < 0.883 klf NG 
Try a larger bolt and tighter spacing. For this example, use a 5/8-in. bolt at 32 in. on center: 
.fZ/2.67ft = (1.0)(0.65)(3.72 kips)/2.67 = 0.91 klf > 0.883 klf OK 
Provisions Sec. 12.4.2.4 [12.4.2.4.2]requires plate washers at all shear wall anchor bolts. A summary of 
the transverse shear wall elements is shown in Figure 10.1-9.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
10-28 
10.1.4.6 Remarks on Shear Wall Connection Details 
In normal platform frame construction, details must be developed that will transfer the lateral loads 
through the floor system and, at the same time, accommodate normal material sizes and the cross-grain 
shrinkage in the floor system. The connections for wall overturning in Sec. 10.1.4.5 are an example of 
one of the necessary force transfers. The transfer of diaphragm shear to supporting shear walls is another 
important transfer as is the transfer from a shear wall on one level to the level below. 
The floor-to-floor height is nine-ft with about one-ft occupied by the floor framing. Using standard 8-ftlong 
plywood sheets for the shear walls, a gap occurs over the depth of the floor framing. It is common to 
use the floor framing to transfer the lateral shear force. Figures 10.1-10 and 10.1-11 depict this 
accomplished by nailing the plywood to the bottom plate of the shear wall, which is nailed through the 
floor plywood to the double 212 chord in the floor system. 
9' 9' 9' 
Provide tie-down 
anchor at each end 
bolted to end post 
at each level. 
F 3 r d = 8.97 kips 
25'-0" 
F r o o f = 8.65 kips 
F 2 n d = 4.47 kips 
Provide 5
8" anchor 
bolts at 2'-8" o.c. 
WD 
Figure 10.1-9 Transverse wall: overturning 
(1.0 ft = 0.3048 m, 1.0 in. = 25.4 mm, 1.0 kip = 4.45 kN).
Chapter 10, Wood Design 
10-29 
Wall and floor nailing 
as in bearing wall 
detail (Figure 10.1-10) 
Framing clip or 
alternate toe nailing 
as in bearing wall 
detail (Figure 10.1-10) 
8d at 6" o.c. 
into blocking 
2x12 joists 
Sheet metal strap 
connecting blocks 
across bottom of 
joists 
2x12 blocking between joists at 4'-0" o.c. extending 
inward from wall to third joist and toe-nailed to 
each joist with 2-8d and to wall plate with 2-16d. 
Connect across joist with sheet metal strap and 
2-10d each side. Remainder of blocking at 
plywood joists to be 2x3 lumber. 
Figure 10.1-11 Nonbearing wall (1.0 in. = 25.4 mm). 
Joists at 16" o.c. 
toe-nailed to wall 
plate with 3-8d 
2x12 blocking between 
joists toe-nailed to 
wall plate with 2-16d 
and to 2-2x12 chord 
with 2-16d 
Provide sheet metal framing 
clips (under blocking) to 
connect 2-2x12 
chord to wall top plate 
Where shear forces are 
small, sheet metal 
framing clips may be 
replaced by toe-nailing 
the 2-2x12 chord 
to the wall top plate 
Floor edge 
nailing 
Wall edge 
nailing 
Bottom plate nailing to 
develop capacity of shear wall 
2-2x12 chord 
for diaphragm 
Figure 10.1-10 Bearing wall (1.0 in = 25.4 mm).
FEMA 451, NEHRP Recommended Provisions: Design Examples 
10-30 
Bottom plate nailing 
2x12 joist toe-nailed 
to sill with 3-16d 
2x12 blocking between 
joists. Connect to sill 
with sheet metal 
framing clips. 
3x6 sill, dense, treated 
Wall edge nailing 
Flashing, sheathing, 
and finish siding 
not shown 
5
8" dia. anchor 
bolt at 2'-8" o.c. 
w/ plate washer 
Figure 10.1-12 Foundation wall detail (1.0 in. = 25.4 mm). 
The top plate of the lower shear wall also is connected to the double 212 by means of sheet metal 
framing clips to the double 212 to transfer the force back out to the lower plywood. (Where the forces 
are small, using toe nails between the double 212 and the top plate may be used for this connection.) 
This technique leaves the floor framing free for cross-grain shrinkage. Although some designers in the 
past may have used a short tier of plywood nailed to the plates of the stud walls to accomplish the 
transfer, Provisions Sec. 12.4.2.6 prohibits this type of detailing. 
The floor plywood is nailed directly to the framing at the edge of the floor, before the plate for the upper 
wall is placed. Also, the floor diaphragm is connected directly to framing that spans over the openings 
between shear walls. The axial strength, and the connections of the double 212 chords, allow them to 
function as collectors to move the force from the full length of the diaphragm to the discrete shear walls. 
(According to Provisions Sec. 5.2.6.4.1 [4.6.2.2], the design of collector elements in wood shear wall 
buildings in Seismic Design Category D need not consider increased seismic demands due to 
overstrength.) 
The floor joist is toe nailed to the wall below for forces normal to the wall. Likewise, full-depth blocking 
is provided adjacent to walls that are parallel to the floor joists, as shown in Figure 10.1-11. (Elsewhere 
the blocking for the floor diaphragm only need be small pieces, flat 24s, for example.) The connections 
at the foundation are similar (see Figure 10.1-12). 
The particular combinations of nails and bent steel framing clips shown in Figures 10.1-10, 10.1-11, and 
10.1-12, to accomplish the necessary force transfers, are not the only possible solutions. A great amount
Chapter 10, Wood Design 
10-31 
of leeway exists for individual preference, as long as the load path has no gaps. Common carpentry 
practices often will provide most of the necessary transfers but, just as often, a critical few will be missed. 
As a result, careful attention to detailing and inspection is an absolute necessity. 
10.1.4.7 Roof Diaphragm Design 
While it has been common practice to design plywood diaphragms as simply supported beams spanning 
between shear walls, the diaphragm design for this example must consider the continuity associated with 
rigid diaphragms. The design will be based on the maximum shears and moments that occur over the 
entire diaphragm. From Sec. 10.1.3.5, the diaphragm design force at the roof is, FP,roof = 40.4 kips. 
As discussed previously, the design force computed in this example includes the internal force due to the 
weight of the walls parallel to the motion. Particularly for one-story buildings, it is common practice to 
remove that portion of the design force. It is conservative to include it, as is done here. 
10.1.4.7.1 Diaphragm Nailing 
The maximum diaphragm shear occurs at the end walls. From Sec. 10.1.3.4, each 25-ft end wall segment 
resists 21 percent of the total story (diaphragm, in this case) force. Distributing the diaphragm force at the 
same rate, the diaphragm shear over the entire diaphragm width at the end walls is: 
V = (0.214)(2)(40.4) = 17.3 kips 
v = 17.3/56 ft. = 0.308 klf 
Try -in. (15/32) plywood rated sheathing (not Structural I) on blocked 2-in. Douglas fir-Larch members 
at 16 in. on center, with 8d nails at 6 in. on center at panel edges and 12 in. on center at intermediate 
framing members. 
From Provisions Table 12.4.3-1a [AF&PA Wind&Seismic Table 4.2A], .fD = 0.35 klf > 0.308 klfOK 
The determination of nail slip for diaphragms is included below. 
10.1.4.7.2 Chord and Splice Connection 
Diaphragm continuity is an important factor in the design of the chords. The design must consider the 
tension/compression forces, due to positive moment at the middle of the span, as well as negative moment 
at the interior shear wall. It is reasonable (and conservative) to design the chord for the positive moment 
assuming a simply supported beam and for the negative moment accounting for continuity. The positive 
moment is wl2/8, where w is the unit diaphragm force, and l is the length of the governing diaphragm 
span. For a continuous beam of two unequal spans, under a uniform load, the maximum negative moment 
is: 
3 3 
1 2 
8(1 2) 
M wl wl 
l l 
- + 
= 
+ 
where w is the unit diaphragm force, and l1 and l2 are the lengths of the two diaphragm spans. For w = 
40.4 kips /140 ft = 0.289 klf, the maximum positive moment is: 
0.289(84)2 / 8 = 255 ft-kip 
and the maximum negative moment is:
FEMA 451, NEHRP Recommended Provisions: Design Examples 
10-32 
Stagger joints in 
opposite faces. Use 
16'-0" minimum 
lengths. 
Drive shim in joint 
2-2x12 chord - Douglas 
Fir-Larch, No. 1 
7" 7" 
4" dia. split ring. 
3 
4" dia. machine 
bolt. 3 
4" dia. hole. 1'-2" 
Figure 10.1-13 Diaphragm chord splice (1.0 ft = 0.3048 m, 1.0 in. = 25.4 mm). 
0.289(84)30.289(56)3198 ft-kip 
8(84 56) 
+ 
= 
+ 
The positive moment controls, and the design chord force is 255/56 = 4.55 kips. Try a double 212 
Douglas fir-Larch No. 2 chord. Due to staggered splices, compute the tension capacity based on a single 
212, with a net area of An = 15.37 in.2 (accounting for 1-in. bolt holes). Per the AF&PA Manual, 
Structural Lumber Supplement: 
.fT = (1.0)(0.8)(1.55psi)(15.37in2) = 19.1 kips > 4.55 kips OK 
For chord splices, use 4 in. diameter split-ring connectors with 3/4-in. bolts. For split rings is single 
shear, the capacity of one connector is: 
.fT = (1.0)(0.65)(17.1) = 11.1 kips > 4.55 kips. OK 
This type of chord splice connection, shown in Figure 10.1-13, is generally used only for heavily loaded 
chords and is shown here for illustrative purposes. A typical chord splice connection for less heavily 
loaded chords can be accomplished more easily by using 16d nails to splice the staggered chord members. 
10.1.4.7.3 Diaphragm Deflection 
The procedure for computing diaphragm deflections, contained in Commentary Sec. 12.4 (similar to the 
commentary of ASCE 16), is intended for the single span, flexible diaphragm model that has been used 
in common practice. The actual deflections of multiple span rigid diaphragms may, in general, be 
similar to those of single-span diaphragms because shear deflection and nail slip (both based on shear 
demand) tend to dominate the behavior. As multiple span deflection computations tend to be 
cumbersome, it is suggested that the design engineer compute diaphragm deflections based on the single
Chapter 10, Wood Design 
10-33 
span model. This will result with a reasonable (and conservative), estimation of overall displacements. If 
these displacements satisfy the drift criteria, then the design is assumed to be adequate; if not, then a more 
rigorous computation of displacements could be performed. 
It is the authors opinion that for diaphragm displacement computations, the diaphragm loading should be 
according to Provisions Eq. 5.2.6.4.4 [5.2-11]; the minimum design demand (0.2SDSIwpx) need not be 
considered. Although the Provisions does not provide a specific requirement either way, this 
interpretation is consistent with Provisions Sec. 5.4.6.1 [5.2.6.1], which indicates that the minimum base 
shear equation (Provisions Eq. 5.4.1.1-3) need not be used for calculation of story drift. 
[See Sec. 10.1.3.1 for a discussion of diaphragm deflection computation using the 2003 Provisions and 
AF&PA Wind & Seismic reference.] 
From Commentary Sec. 12.4, the diaphragm deflection is computed as: 
. 
5 3 0.188 ( ) 
8 4 2
c 
n 
vL vL Le X 
wEA Gt w 
d 
S . 
= + + + 
The equation produces the midspan diaphragm displacement in inches, and the individual variables must 
be entered in the force or length units as described below. For the single span approximation, consider 
the longer, 84 ft long span so that: 
(84/140) 
2 
Fpx 
v 
w 
= 
where Fpx is the story diaphragm force at Level x, w is the diaphragm width, and v is in pounds/ft. For 
this calculation, the unit shear will be the same at both ends of the diaphragm. Therefore, it 
underestimates the actual unit shear at the end wall but overestimates the actual unit shear at the interior 
wall. These inaccuracies are assumed to be roughly offsetting. The individual terms of the above 
equation represent the following: 
5vL3/8wEA = bending deflection, as derived from the formula db = 5vL4/384EI, where v is the 
diaphragm unit force in pounds per ft and I = Aw2/2 (in2) 
vL/4Gt = shear deflection as derived from the formula dv = vL2/8GA, where v is the diaphragm unit 
force in pounds per ft and A = wt (in2) 
0.188Len = nail slip deflection in inches 
S(.cX)/2w = deflection due to chord slip in inches 
For this example: 
v = 40.4(84/140)/[(2(56))(1000)] = 216 plf (ignoring torsion) 
L = 84 ft and w = 56 ft, diaphragm length and width 
A = 2(1.5) 11.25 = 33.75 in.2 (double 212) 
t = 0.298 in. for -in. unsanded plywood 
E = 1,600,000 psi 
G =75,000 psi 
Nail slip is computed using the same procedure as for shear walls (Sec. 10.1.4.3), with the load per nail 
based on the diaphragm shear from this section. The diaphragm nailing is 8d at 6 in. on center.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
10-34 
Load per nail = 216(6/12) = 108 lb 
Nail slip en = 1.2(108/616)3.018 = 0.0063 in. 
In the above equation, 1.2 is the factor for plywood other than Structural I and 616 and 3.018 are 
coefficients for seasoned lumber. 
Although a tight chord splice has been specified, the chord slip is computed for illustrative purposes. 
Assume chord slip occurs only at the side tension since the compression splices are tightly shimmed as 
shown in Figure 10.1-13. For this example: 
.c = chord splice slip (1/16 in. used for this example) 
X = distance from chord splice to nearest support 
Assuming splices at 20 ft on center, along the 84-ft diaphragm length (ignore diaphragm continuity for 
this term), the sum of the chord splice slip is: 
S(.cX) = (1/16)(20 + 40 + 24 + 4) = 5.5 
Thus: 
5(216)(843 ) 216(84) 0.188(84)(0.0063) 5.5 
8(1,600,000)(33.75)(56) 4(75,000)(0.298) 2(56) 
d= + + + 
= 0.027 + 0.203 + 0.100 + 0.055 = 0.385 in. 
Wall and floor drifts are added, and checked in Sec. 10.1.4.9. 
10.1.4.8 Second and Third Floor Diaphragm Design 
The design of the second and third floor diaphragms follows the same procedure as the roof diaphragm. 
From Sec. 10.1.3.6, the diaphragm design force for both floors is Fp,3rd = Fp,2nd = 50.8 kips. 
10.1.4.8.1 Diaphragm Nailing 
The maximum diaphragm shear occurs at the end walls. From Sec. 10.1.3.4, each end wall segment 
resists 21 percent of the total story (diaphragm, in this case) force. Distributing the diaphragm design 
force at the same rate, the diaphragm shear at the end walls is: 
V = (0.214)(2)(50.8) = 21.7 kips 
v = 21.7/56 ft = 0.388 klf 
Try -in. (15/32) plywood rated sheathing (not Structural I) on blocked 2-in. Douglas fir-Larch members 
at 16 in. on center, with 8d nails at 4 in. on center at boundaries and continuous panel edges, at 6 in. on 
center at other panel edges, and 12 in. on center at intermediate framing members. 
From Provisions Table 12.4.3-1a [AF&PA Wind&Seismic Table 4.2A], .fD' = 0.47 klf > 0.388 klfOK 
10.1.4.8.2 Chord and Splice Connection 
Computed as described above for the roof diaphragm, the maximum positive moment is 320 kips and the 
design chord force is 5.71 kips.
Chapter 10, Wood Design 
10-35 
By inspection, a double 212 chord spliced with 4-in. diameter split ring connectors, as at to the roof 
level, is adequate. A typical chord splice connection is shown in Figure 10.1-13. 
10.1.4.8.3 Diaphragm Deflection 
Using the same procedure as before: 
L = 84 ft and w = 56 ft, diaphragm length and width 
A = 2(1.5) 11.25 = 33.75 in.2 (double 212) 
t = 0.298 in. for -in. unsanded plywood 
E = 1,600,000 psi 
G = 75,000 psi 
As discussed previously, diaphragm deflection computations need not consider the minimum diaphragm 
design forces. 
Therefore, at the third floor: 
v = 50.1(84/140)/[(2(56))(1000)] = 268 plf 
Load per nail = 268(4/12) = 89 lb 
Nail slip en = 1.2(89/616)3.018 = 0.0035 in. 
5(268)(843 ) 268(84) 0.188(84)(0.0035) 5.5 
8(1,600,000)(33.75)(56) 4(75,000)(0.298) 2(56) 
d= + + + 
= 0.033 + 0.252 + 0.056 + 0.055 = 0.396 in. 
At the second floor: 
v = 39.1(84/140)/(2(56))(1000) = 209 plf 
Load per nail = 209(4/12) = 70 lb 
Nail slip en = 1.2(70/616)3.018 = 0.0017 in. 
5(209)(843 ) 209(84) 0.188(84)(0.0017) 5.5 
8(1,600,000)(33.75)(56) 4(75,000)(0.298) 2(56) 
d= + + + 
= 0.026 + 0.197 + 0.026 + 0.055 = 0.304 in. 
10.1.4.9 Transverse Deflections, Drift, and P-delta Effects 
Transverse deflections for walls and diaphragms were calculated above. The diaphragm deflections for 
the second and third floors are based on the seismic force analysis and not the minimum diaphragm 
design forces (Provisions Sec. 5.2.6.4.4 [5.2.3]). 
To determine the maximum story deflections and drifts (see Section 10.1.2.2), the midspan diaphragm 
deflection is combined with the wall deflection. This is summarized in Table 10.1-6, which shows the 
drift below the level considered. 
Table 10.1-6 Total Deflection and Drift 
Level Wall (in.) Diaphragm (in.) Total de (in.) d = Cdde/I (in.) .,drift (in.)
FEMA 451, NEHRP Recommended Provisions: Design Examples 
10-36 
Roof 1.14 0.38 1.52 6.08 1.00 
3 0.87 0.40 1.27 5.08 2.08 
2 0.45 0.30 0.75 3.00 3.00 
1.0 in. = 25.4 mm. 
The maximum permissible drift is 0.020 hs = 2.16 in. Therefore, the drift limitations are satisfied at the 
third floor and roof. The drift in the first story is about 38 percent too large so the design must be revised. 
One beneficial aspect of the design that has been ignored in these calculations is the lightweight concrete 
floor fill. Although it is discontinuous at the stud walls, it will certainly stiffen the diaphragm. No 
studies that would support a quantitative estimate of the effect have been found. Because the shear 
deformation of the -in. plywood diaphragm represents a significant portion of the total drift at the first 
and second levels, any significant increase in shear stiffness that might be provided by the concrete would 
further reduce the expected drifts. In this case, about 60 percent of the drift is contributed by the shear 
walls. The diaphragm stiffness would need to be tripled to meet the drift criteria, so the shear walls will 
be revised. 
If Structural I plywood is used for the shear walls in this direction, the drift criteria are satisfied at all 
levels. In Sec. 10.1.4.4, G becomes 90,000 ksi, and t (effective thickness) becomes 0.535 in. The 
deflections due to shear and nail slip (the 1.2 factor no longer applies) are reduced. The resulting total 
elastic deflection of the shear walls at Levels 2, 3, and Roof are, 0.242 in., 0.488 in., and 0.669 in., 
respectively. Table 10.1-6b shows the revised results, which satisfy the drift criteria. 
Table 10.1-6b Total Deflection and Drift (Structural I Plywood Shear Walls) 
Level Wall (in.) Diaphragm (in.) Total de (in.) d = Cdde/I (in.) .,drift (in.) 
Roof 0.67 0.38 1.05 4.20 0.64 
3 0.49 0.40 0.89 3.56 1.40 
2 0.24 0.30 0.54 2.16 2.16 
1.0 in. = 25.4 mm. 
Because the tie-down calculations (Sec. 10.1.4.5) depend on the tabulated capacities of the shear wall 
panels and Structural I panels have slightly higher capacities, the connection designs must be verified. 
Additional calculations (not shown here) confirm that the hardware that was previously selected still 
works. It is also worth noting that although many designers fail to perform deflection calculations for 
wood construction, the drift criteria control the selection of sheathing grade in this example. 
The P-delta provision also must be examined. Following Provisions Sec. 5.4.6.2 [5.2.6.2] and assuming 
the total mass deflects two-thirds of the maximum diaphragm deflection, the P-delta coefficient is as 
shown in Table 10.1-7. 
[Note that the equation to determine the stability coefficient has been changed in the 2003 Provisions. 
The importance factor, I, has been added to 2003 Provisions Eq. 5.2-16. However, this does not affect 
this example because I = 1.0.] 
Table 10.1-7 P-delta Stability Coefficient 
Level PD 
(kips) 
PL 
(kips) 
SP 
(kips) 
. 
(in.) 
V 
(kips) . = P./VhCd
Chapter 10, Wood Design 
10-37 
Roof 183 204 387 0.67 40.4 0.015 
3 284 130 801 1.27 82.3 0.029 
2 284 130 1,215 1.76 103.2 0.048 
1.0 in. = 25.4 mm, 1.0 kip = 4.45 kN. 
For example, to compute the effective P-delta drift at the roof: 
droof = Cd[dwall + 2/3(ddiaphragm)] = 4[0.67 + 2/3(0.38)] = 3.693 in. 
d3 = 4[0.49 + 2/3(0.40)] = 3.027 in. 
d = 3.693 - 3.027 = 0.67 in. 
The story dead load is the same as shown in Table 10.1-1, and the story live load is based on 25 psf snow 
load for the roof and 16 psf reduced live load (0.440 psf) acting over the entire area of Levels 2 and 3. 
For . < 0.10, no deflection amplification due to P-delta effects is necessary. 
10.1.4.10 Longitudinal Direction 
Only one exterior shear wall section will be designed here. The design of the corridor shear walls would 
be similar to the transverse walls. (This example has assumed a greater length of corridor wall than 
would likely be required to resist the design forces. This increased length is intended to reduce the 
demand to the exterior walls, assuming rigid diaphragm distribution, to a level below the maximum 
permitted in-plane shear for the perforated shear wall design procedure as discussed below.) For loads in 
the longitudinal direction, diaphragm stresses and deflections are negligible. 
The design of the exterior wall will utilize the guidelines for perforated shear walls (Provisions Sec. 
12.4.3 [AF&PA Wind&Seismic Sec. 4.3]), which are new to the 2000 Provisions. [The provisions for 
perforated shear walls are contained in AF&PA Wind & Seismic and therefore have been removed from 
the 2003 Provisions. The design provisions are spread throughout AF&PA Wind & Seismic Sec. 4.3.3, 
but are not substantially different for the provisions contained in the 2000 Provisions except for the 
revisions to the tie down requirements as noted below.] The procedure for perforated shear walls applies 
to walls with openings that have not been specifically designed and detailed for forces around the 
openings. Essentially, a perforated wall is treated in its entirety, rather than as a series of discrete wall 
piers. The use of this design procedure is limited by several conditions (Provisions Sec. 12.4.3.2 
[AF&PA Wind&Seismic Sec. 4.3.5.2]), the most relevant to this example is that the factored design shear 
resistance shall not exceed 0.64 klf. This requirement essentially limits the demand on perforated shear 
walls such that the required factored design shear resistance is less than 0.64 klf. If the configuration 
required higher design values, then walls must be added in order to reduce the demand. 
The main aspects of the perforated shear wall design procedure are as follows. The design shear capacity 
of the shear wall is the sum of the capacities of each segment (all segments shall have the same sheathing 
and nailing) reduced by an adjustment factor that accounts for the geometry of the openings. Uplift 
anchorage (tie-down) is required only at the ends of the wall (not at the ends of all wall segments), but all 
wall segments must resist a specified tension force (using anchor bolts at the foundation and with 
strapping or other means at upper floors). Requirements for shear anchorage and collectors (drag struts) 
across the openings are also specified. It should be taken into account that the design capacity of a 
perforated shear wall, is less than a standard segmented wall with all segments restrained against 
overturning. However, the procedure is useful in eliminating interior hold downs for specific conditions 
and, thus, is illustrated in this example.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
10-38 
The portion of the story force resisted by each exterior wall was computed previously as 0.225Fx. The 
exterior shear walls are composed of three separate perforated shear wall segments (two at 30 ft long and 
one at 15 ft long, all with the same relative length of full height sheathing), as shown in Figure 10.1-2. 
This section will focus on the design of a 30-ft section. Assuming that load is distributed to the wall 
sections based on relative length of shear panel, then the total story force to the 30-ft section is 
(30/75)0.225Fx = 0.090Fx per floor. The load per floor is 
Froof = 0.090(40.4) = 3.64 kips 
F3rd = 0.090(41.9) = 3.77 kips 
F2nd = 0.090(20.9) = 1.88 kips 
S = 9.29 kips 
10.1.4.10.1 Perforated Shear Wall Resistance 
The design shear capacity (Provisions Sec. 12.4.3.3 [AF&PA Wind&Seismic Table 4.3.3.4]) is computed 
as the factored shear resistance for the sum of the wall segments, multiplied by an adjustment factor that 
accounts for the percentage of full height (solid) sheathing and the ratio of the maximum opening to the 
story height. At each level, the design shear capacity, Vwall, is: 
Vwall = (vC0)SLi 
where 
v = unadjusted factored shear resistance (Provisions Table 12.4.3-2a [Table 12.4.3a or AF&PA 
Wind&Seismic Table 4.3A). 
C0 = shear capacity adjustment factor (Provisions Table 12.4.3-1 [AF&PA Wind&Seismic Table 
4.3.3.4]) = 0.83 
The percent of full height sheathing is (4 + 10 + 4)/30 = 0.60, and the maximum opening height ratio is 4 
ft /8 ft = 0.5. Per Provisions Table 12.4.3-1 [AF&PA Wind&Seismic Table 4.3.3.4], C0 = 0.83. 
SLi = sum of widths of perforated shear wall segments = 4 + 10 + 4 = 18 ft
Chapter 10, Wood Design 
10-39 
Froof 
4'-0" 6'-0" 10'-0" 6'-0" 4'-0" 
F3rd 
F2nd 
2'-6" 2'-6" 4'-0" 
Floor 
framing 
Perforated 
shear wall 
h = 8'-0" 
Provide 
tie-down 
for uplift 
at ends 
Perforated 
shear wall 
segment 
Figure 10.1-14 Perforated shear wall at exterior (1.0 ft = 0.3048 m, 1.0 in. = 25.4 mm) 
The wall geometry (and thus the adjustment factor and total length of wall segments) is the same at all 
three levels, as shown in Figure 10.1-14. Perforated shear wall plywood and nailing are determined 
below. 
Roof to third floor: 
V = 3.64 kips 
Required v = 3.64/0.83/18 = 0.244 klf 
Try -in. (15/32) Structural I plywood rated sheathing on blocked 2-in. Douglas fir-Larch members at 16 
in. on center with 10d common nails at 6 in. on center at panel edges and 12 in. on center at intermediate 
framing members. (Structural I plywood would not be required to satisfy the strength requirements. 
However, to minimize the possibility for construction errors, the same grade of sheathing is used on walls 
in both directions.) 
From Provisions Table 12.4.3-2a [AF&PA Wind&Seismic Table 4.3A], .fD = 0.44 klf > 0.244 klf OK 
Third floor to second floor: 
V = 3.64 + 3.77 = 7.41 kips 
Required v = 7.41/0.83/18 = 0.496 klf
FEMA 451, NEHRP Recommended Provisions: Design Examples 
10-40 
Try -in. (15/32) Structural I plywood rated sheathing on blocked 2 in. Douglas fir-larch members at 16 
in. on center with 10d common nails at four in. on center at panel edges and 12 in. on center at 
intermediate framing members. 
From Provisions Table 12.4.3-2a [AF&PA Wind&Seismic Table 4.3A]: 
.fD = 0.66 klf (0.64 klf max) > 0.496 klf OK 
As discussed above, Provisions Sec. 12.4.3.2, Item b [12.4.3], limits the factored shear resistance in 
Provisions Table 12.4.3-2a [AF&PA Wind&Seismic Table 4.3A] to 0.64 klf, which still exceeds the 
demand at this level, so the limitation is satisfied. 
Second floor to first floor: 
V = 7.41 + 1.88 = 9.29 kips 
Required v = 9.29/0.83/18 = 0.622 klf 
By inspection, the same plywood and nailing from above will work. 
10.1.4.10.2 Perforated Shear Wall Uplift Anchorage 
According to Provisions Sec. 12.4.3.4.1 [AF&PA Wind&Seismic Table 4.3A], the requirements for uplift 
anchorage must be evaluated at the ends of the wall only. Uplift at each wall segment is treated 
separately as described later. Uplift forces, based on the strength of the shear panels (per Provisions Sec. 
12.4.2.4 [12.4.2.4.1]), are to be computed as discussed in Sec. 10.1.4.5. For this example, calculations 
involving seismic overturning and counter-balancing moments are assumed not to be applicable for 
perforated shear walls, as they are not expected to act as rigid bodies in resisting global overturning. 
The tie-down design force is determined as O0/1.3 times the factored shear resistance of the shear panels, 
as discussed previously. For this example, the tie-down will be designed at the first floor only; the other 
floors would be computed similarly, and tie-down devices, as shown in Figure 10.1-8, would be used. 
[The requirements for uplift anchorage are slightly different in the 2003 Provisions. The tie down force is 
based on the nominal strength of the shear wall rather than the O0/1.3 times the factored resistance as 
specified in the 2000 Provisions. This change is primarily intended to provide consistent terminology and 
should not result in a significant impact on the design of the tie down.] 
The uplift forces are computed as: 
Roof: T = 0.44 klf (3.0/1.3) (8 ft) = 8.12 kips 
Third floor: T = 0.66 klf (3.0/1.3) (8 ft) = 12.18 kips 
Second floor: T = 0.66 klf (3.0/1.3) (8 ft) = 12.18 kips 
S = 32.48 kips 
Since the chord member supports the window header as well, use a 66 Douglas fir-Larch No. 1 similar 
to the transverse walls. Try a double tie-down device with a 7/8-in. anchor bolt and three 7/8-in. stud 
bolts. Using the method described above for computing the strength of a double tie-down, the nominal 
design strength is 34.3 kips, which is greater than the demand of 32.48 kips. 
The design of the tie-downs at the second and third floors is similar. 
10.1.4.10.3 Perforated Shear Wall Compression Chords
Chapter 10, Wood Design 
10-41 
Provisions Sec. 12.4.3.4.4 [AF&PA Wind&Seismic Sec. 4.3.6.1], requires each end of a perforated shear 
wall to have a chord member designed for the following compression force from each story: 
0S i 
C Vh 
C L 
= 
where 
V = design shear force in the shear wall (not wall capacity as used for uplift) 
h = shear wall height (per floor) 
C0 = shear capacity adjustment factor 
SLi = sum of widths of perforated shear wall segments 
For h = 8 ft, C0 = 0.83 and SLi = 18 ft, the compression force is computed as: 
Third floor: C = 3.64(8)/0.83/18 = 1.95 kips 
Second floor: C = (3.64 + 3.77)(8)/0.83/18 = 3.96 kips 
First floor: C = (3.64 + 3.77 + 1.88)(8)/0.83/18 = 4.98 kips 
S = 10.89 kips 
Again, just the chord at the first floor will be designed here; the design at the upper floors would be 
similar. Although not explicitly required by Provisions Sec. 12.4.3.4.4 [AF&PA Wind&Seismic Sec. 
4.3.6.1], it is rational to combine the chord compression with gravity loading (using the load combination 
1.38D + 1.0QE + 0.5L + 0.2S in accordance with Provisions Eq. 5.2.7.1-1 [4.2-1]), in order to design the 
chord member. The end post of the longitudinal shear wall supports the same tributary weight at the end 
post of the transverse shear walls. Using the weights computed previously in Sec. 10.1.4.5, the design 
compression force is: 
1.38(5.17) + 1.0(10.89) + 0.5(4.48) + 0.2(1.30) = 20.5 kips. 
The bearing capacity on the bottom plate was computed previously as 31.5 kips, which is greater than 
20.5 kips. Where end posts are loaded in both directions, orthogonal effects must be considered in 
accordance with Provisions Sec. 5.2.5.2 [4.4.2.3]. 
10.1.4.10.4 Anchorage at Shear Wall Segments 
The anchorage at the base of a shear wall segment (bottom plate to floor framing or foundation wall), is 
designed per Provisions Sec. 12.4.3.4.2 [AF&PA Wind&Seismic Sec. 4.3.6.4]. While this anchorage 
need only be provided at the full height sheathing, it is usually extended over the entire length of the 
perforated shear wall to simplify the detailing and reduce the possibility of construction errors. 
0 i 
v V 
C L 
= 
S 
where 
V = design shear force in the shear wall 
C0 = shear capacity adjustment factor 
SLi = sum of widths of perforated shear wall segments
FEMA 451, NEHRP Recommended Provisions: Design Examples 
10-42 
This equation is the same as was previously used to compute unit shear demand on the wall segments. 
Therefore, the in-plane anchorage will be designed to meet the following unit, in-plane shear forces: 
Third floor: v = 0.244 klf 
Second floor: v = 0.496 klf 
First floor: v = 0.622 klf. 
In addition to resisting the in-plane shear force, Provisions Sec. 12.4.3.4.3 [AF&PA Wind&Seismic Sec. 
4.3.6.4] requires that the shear wall bottom plates be designed to resist a uniform uplift force, t, equal to 
the unit in-plane shear force. Per Provisions Sec. 12.4.3.4.5 [AF&PA Wind&Seismic Sec. 4.3.6.4], this 
uplift force must be provided with a complete load path to the foundation. That is, the uplift force at each 
level must be combined with the uplift forces at the levels above (similar to the way overturning moments 
are accumulated down the building). 
At the foundation level, the unit in-plane shear force, v, and the unit uplift force, t, are combined for the 
design of the bottom plate anchorage to the foundation wall. The design unit forces are: 
Shear: v = 0.622 klf 
Tension: t = 0.244 + 0.496 + 0.622 = 1.36 klf 
Assuming that stresses on the wood bottom plate govern the design of the anchor bolts, the anchorage is 
designed for shear (single shear, wood-to-concrete connection) and tension (plate washer bearing on 
bottom plate). The interaction between shear and tension need not be considered in the wood design for 
this configuration of loading. As for the transverse shear walls, try a 5/8-in. bolt at 32 in. on center with a 
3-in. square plate washer (Provisions Sec. 12.4.2.4 [12.4.2.4.2] requires 1/433 in. plate washer for 
5/8-in. anchor bolts). As computed previously, the shear capacity is 0.91 klf so the bolts are adequate. 
For anchor bolts at 32 in. on center, the tension demand per bolt is 1.36 klf (32/12) = 3.63 kips. Bearing 
capacity of the plate washer (using a Douglas fir No. 2 bottom plate) is computed per the AF&PA 
Manual, Structural Connections Supplement, as: 
.fP = (1.0)(0.8)(1.30 ksi)(9 in.2) = 9.36 kips > 3.63 kips OK 
The anchor bolts themselves must be designed for combined shear, and tension in accordance with 
Provisions Sec. 9.2 [11.2]. 
In addition to designing the anchor bolts for uplift, a positive load path must be provided to transfer the 
uplift forces into the bottom plate. One method for providing this load path continuity, is to use metal 
straps nailed to the studs and lapped around the bottom plate, as shown in Figure 10.1-15. Attaching the 
studs directly to the foundation wall (using embedded metal straps) for uplift and using the anchor bolts 
for shear only is an alternative approach.
Chapter 10, Wood Design 
10-43 
1
4x3x3 steel plate 
washer at anchor bolt 
Metal strap used 
to resist uplift, 
lapped under 
sill plate. 
5
8" dia. anchor 
bolt at 2'-8" o.c. 
(Used to resist 
shear and tension) 
Nail to 
stud 
Figure 10.1-15 Perforated shear wall detail at foundation (1.0 ft = 0.3048 m, 1.0 in. = 25.4 mm). 
Nail to stud 
(typical) 
Nails to resist shear 
Sheet metal framing 
clips (Under blocking) 
used to resist shear. 
Metal strap used 
to resist uplift 
Figure 10.1-16 Perforated shear wall detail at floor framing. 
At the upper floors, the load transfer for in-plane shear is accomplished by using nailing or framing clips 
between the bottom plates, rim joists, and top plates in a manner similar to that for standard shear walls. 
The uniform uplift force can be resisted either by using the nails in withdrawal (for small uplift demand) 
or by providing vertical metal strapping between studs above and below the level considered. This type 
of connection is shown in Figure 10.1-16. For this type of connection (and the one shown in Figure 10.1- 
15) to be effective, shrinkage of the floor framing must be minimized using dry or manufactured lumber.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
10-44 
~ 
~ 
Typ. articulated 
glue-lam beams 
Concrete 
masonry walls 
Typ. roof joists 
at 24" o.c. 
Steel column on 
spread footing 
Plywood roof 
sheathing 
Open 
5 bays at 40'-0" = 200'-0" 
5 bays at 20'-0" = 100'-0" 
Figure 10.2-1 Building plan (1.0 ft = 0.3048 m, 1.0 in. = 25.4 mm). 
For example, consider the second floor. The required uniform uplift force, t = 0.244 + 0.496 = 0.740 klf. 
Place straps at every other stud, so the required strap force is 0.740 (32/12) = 1.97 kips. Provide an 18- 
gauge strap with 1210d nails at each end. 
10.2 WAREHOUSE WITH MASONRY WALLS AND WOOD ROOF, LOS ANGELES, 
CALIFORNIA 
This example features the design of the wood roof diaphragm, and wall-to-diaphragm anchorage for the 
one-story masonry building, described in Sec. 9.1 of this volume of design examples. Refer to that 
example for more detailed building information and the design of the masonry walls. 
10.2.1 Building Description 
This is a very simple rectangular warehouse, 100 ft by 200 ft in plan (Figure 10.2-1), with a roof height of 
28 ft. The wood roof structure slopes slightly, but it is nominally flat. The long walls (side walls) are 8 
in. thick and solid, and the shorter end walls are 12 in. thick and penetrated by several large openings. 
Based on gravity loading requirements, the roof structure consists of wood joists, supported by 8-in. 
wide, by 24-in. deep, glued-laminated beams, on steel columns. The joists span 20 ft and the beams span 
40 ft, as an articulated system. Typical roof framing is assumed to be Douglas fir-Larch No.1 as graded 
by the WWPA. The glued-laminated beams meet the requirements of combination 24F-V4 per 
ANSI/AITC A190.1. 
The plywood roof deck acts as a diaphragm to carry lateral loads to the exterior walls. There are no 
interior walls for seismic resistance. The roof contains a large opening that interrupts the diaphragm 
continuity. 
The diaphragm contains continuous cross ties in both principal directions. The details of these cross ties 
and the masonry wall-to-diaphragm anchorage are substantially different from those shown in previous 
versions of this example. This is primarily due to significant revisions to the Provisions requirements for 
anchorage of masonry (and concrete) walls to flexible wood diaphragms.
Chapter 10, Wood Design 
10-45 
The following aspects of the structural design are considered in this example: 
1. Development of diaphragm forces based on the equivalent lateral force procedure used for the 
masonry wall design ( Sec. 9.1) 
2. Design and detailing of a plywood roof diaphragm with a significant opening 
3. Computation of drift and P-delta effects 
4. Anchorage of diaphragm and roof joists to masonry walls and 
5. Design of cross ties and subdiaphragms 
[Note that as noted in Sec. 9.1, the new Simplified Design Procedure contained in 2003 Provisions 
Simplified Alternate Chapter 4 as referenced by 2003 Provisions Sec. 4.1.1 is likely to be applicable to 
this example, subject to the limitations specified in 2003 Provisions Sec. Alt. 4.1.1.] 
10.2.2 Basic Requirements 
10.2.2.1 Provisions Parameters 
SS (Provisions Maps [Figure 3.3.3]) = 1.50 
S1 (Provisions Maps [Figure 3.3.4]) = 0.60 
Site Class (Provisions Sec. 4.1.2.1 [3.5]) = C 
Seismic Use Group (Provisions Sec. 1.3 [1.2]) = I 
Seismic Design Category (Provisions Sec. 4.2 [1.4]) = D 
Seismic Force Resisting System (Provisions Table 5.2.2 [4.3-1]) = Special reinforced 
masonry shear wall 
Response Modification Factor, R (Provisions Table 5.2.2 [4.3-1]) = 3.5 
System Overstrength Factor, O0 (Provisions Table 5.2.2 [4.3-1]) = 2.5 
Deflection Amplification Factor, Cd (Provisions Table 5.2.2 4.3-1]) = 3.5 
[The 2003 Provisions have adopted the 2002 USGS probabilistic seismic hazard maps, and the maps have 
been added to the body of the 2003 Provisions as figures in Chapter 3 (instead of the previously used 
separate map package).] 
10.2.2.2 Structural Design Criteria 
A complete discussion on the criteria for ground motion, seismic design category, load path, structural 
configuration, redundancy, analysis procedure, and shear wall design, is included in Sec. 9.1 of this 
volume of design examples. 
10.2.2.2.1 Design and Detailing Requirements (Provisions Sec. 5.2.6 [4.6]) 
See Provisions Chapter 12, for wood design requirements. As discussed in greater detail in Sec. 10.1, 
Provisions Sec. 12.2.1 utilizes load and resistance factor design (LRFD) for the design of engineered 
wood structures. The design capacities are therefore, consistent with the strength design demands of 
Provisions Chapter 5. 
The large opening in the diaphragm must be fitted with edge reinforcement (Provisions Sec. 5.2.6.2.2 
[4.6.1.4]). However, the diaphragm does not require any collector elements that would have to be 
designed for the special load combinations (Provisions Sec. 5.2.6.4.1 [4.6.2.2]). 
The requirements for anchoring of masonry walls to flexible diaphragms (Provisions Sec. 5.2.6.3.2 
[4.6.2.1]) are of great significance in this example.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
10-46 
10.2.2.2.2 Combination of Load Effects (Provisions Sec. 5.2.7 [4.2.2]) 
The basic design load combinations for the lateral design, as stipulated in ASCE 7, and modified by the 
Provisions Eq. 5.2.7-1 and 5.2.7-2 [4.2-1 and 4.2-2], were computed in Sec. 9.1 of this volume of design 
examples as: 
1.4D + 1.0QE 
and 
0.7D - 1.0QE. 
The roof live load, Lr, is not combined with seismic loads, and the design snow load is zero for this Los 
Angeles location. 
10.2.2.2.3 Deflection and Drift Limits (Provisions Sec. 5.2.8 [4.5.1]) 
In-plane deflection and drift limits for the masonry shear walls are considered in Sec. 9.1. 
As illustrated below, the diaphragm deflection is much greater than the shear wall deflection. According 
to Provisions Sec. 5.2.6.2.6 [4.5.2], in-plane diaphragm deflection shall not exceed the permissible 
deflection of the attached elements. Because the walls are essentially pinned at the base, and simply 
supported at the roof, they are capable of accommodating large deflections at the roof diaphragm. 
For illustrative purposes, story drift is determined and compared to the requirements of Provisions Table 
5.2.8 [4.5-1]. However, according to this table, there is essentially no drift limit for a single story 
structure as long as the architectural elements can accommodate the drift (assumed to be likely in a 
warehouse structure with no interior partitions). As a further check on the deflection, P-delta effects 
(Provisions Sec. 5.4.6.2 [5.2.6.2]) are evaluated. 
10.2.3 Seismic Force Analysis 
Building weights and base shears are as computed in Sec. 9.1 of this volume of design examples. (The 
building weights used in this example are based on a preliminary version of Example 9.1 and, thus, minor 
numerical differences may exist between the two examples). Provisions Sec. 5.2.6.4.4 [4.6.3.4]specifies 
that floor and roof diaphragms be designed to resist a force, Fpx, in accordance with Provisions Eq. 
5.2.6.4.4 [4.6-2]as follows: 
n 
i 
i x 
px n px 
i 
i x 
F 
F w
w 
=
=
S 
= 
S 
plus any force due to offset walls (not applicable for this example). For one-story buildings, the first term 
of this equation will be equal to the seismic response coefficient, Cs, which is 0.286. The effective 
diaphragm weight, wpx, is equal to the weight of the roof, plus the tributary weight of the walls 
perpendicular to the direction of the motion. The tributary weights are: 
Roof = 20(100)(200) = 400 kips 
Side walls = 2(65)(28/2+2)(200) = 416 kips 
End walls = 2(103)(28/2+2)(100) = 330 kips
Chapter 10, Wood Design 
10-47 
The diaphragm design force is computed as: 
Transverse Fp,roof = 0.286(400 + 416) = 233 kips 
Longitudinal Fp,roof = 0.286(400 + 330) = 209 kips 
These forces exceed the minimum diaphragm design forces given in Provisions Sec. 5.2.6.4.4 [4.6.3.4], 
because Cs exceeds the minimum factor of 0.2SDS. 
10.2.4 Basic Proportioning of Diaphragm Elements 
The design of plywood diaphragms primarily involves the determination of sheathing sizes and nailing 
patterns to accommodate the applied loads. Large openings in the diaphragm and wall anchorage 
requirements, however, can place special requirements on the diaphragm capacity. Diaphragm deflection 
is also a consideration. 
Nailing patterns for diaphragms are established on the basis of tabulated requirements included in the 
Provisions. It is important to consider the framing requirements for a given nailing pattern and capacity 
as indicated in the notes following the tables. In addition to strength requirements, Provisions Sec. 
12.4.1.2 places aspect ratio limits on plywood diaphragms (length-to-width shall not exceed 4/1 for 
blocked diaphragms). However, it should be taken into consideration that compliance with this aspect 
ratio does not guarantee that drift limits will be satisfied. 
While there is no specific limitation on deflection for this example, the diaphragm has been analyzed for 
deflection as well as for shear capacity. A procedure for computing diaphragm deflections is illustrated in 
detail, in Sec. 10.1.4.7. 
In the calculation of diaphragm deflections, the chord splice slip factor can result in large additions to the 
total deflection. This chord splice slip, however, is often negligible where the diaphragm is continuously 
anchored to a bond beam in a masonry wall. Therefore, chord splice slip is assumed to be zero in this 
example. 
10.2.4.1 Strength of Members and Connections 
The 2000 Provisions have adopted Load and Resistance Factor Design (LRFD) for engineered wood 
structures. The Provisions includes the ASCE 16 standard by reference and uses it as the primary design 
procedure for engineered wood construction. Strength design of members and connections is based on 
the requirements of ASCE 16. The AF&PA Manual and supplements contain reference resistance values 
for use in design. For convenience, the Provisions contains design tables for diaphragms that are 
identical to those contained in the AF&PA Structural-Use Panels Supplement. Refer to Sec. 10.1.4.1 for 
a more complete discussion of the design criteria. 
[The primary reference for design of wood diaphragms in the 2003 Provisions is AF&PA Wind & 
Seismic. Much of the remaining text in the 2003 Provisions results from differences between AF&PA 
Wind & Seismic and Chapter 12 of the 2000 Provisions as well as areas not addressed by AF&PA Wind 
& Seismic. Because the AF&PA Wind & Seismic tabulated design values for diaphragms do not 
completely replace the tables in the 2000 Provisions, portions of the tables remain in the 2003 Provisions. 
Therefore, some diaphragm design values are in the 2003 Provisions and some are in AF&PA Wind & 
Seismic. The design values in the tables are different between the two documents. The values in the 
2003 Provisions represent factored shear resistance (.fD), while the values in AF&PA Wind & Seismic 
represent nominal shear resistance that must then be multiplied by a resistance factor, f, (0.65) and a time 
effect factor ,., (1.0 for seismic loads). Therefore, while the referenced tables may be different, the 
factored resistance values based on the 2003 Provisions should be the same as those in examples based on
FEMA 451, NEHRP Recommended Provisions: Design Examples 
10-48 
the 2000 Provisions. The calculations that follow are annotated to indicate from which table the design 
values are taken.] 
10.2.4.2 Roof Diaphragm Design for Transverse Direction 
10.2.4.2.1 Plywood and Nailing 
The diaphragm design force, Fp,roof = 233 kips. Accounting for accidental torsion (Provisions Sec. 5.4.4.2 
[5.2.4.2]), the maximum end shear = 0.55Fp,roof = 128 kips. This corresponds to a unit shear force v = 
(128/100) = 1.28 klf. Although there is not a specific requirement in the Provisions, it is the authors 
opinion that accidental torsion should be considered, even for flexible diaphragms, to account for the 
possibility of a non-uniform mass distribution in the building. 
Because the diaphragm shear demand is relatively high, the plywood sheathing and roof framing at the 
ends of the building, must be increased in size over the standard roof construction for this type of 
building. Assuming 3-in. nominal framing, try blocked 3/4-in. (23/32) Structural I plywood rated 
sheathing with two lines of 10d common nails at 2-1/2 in. on center at diaphragm boundaries, continuous 
panel edges, and two lines at 3 in. on center at other panel edges. 
From Provisions Table 12.4.3-1a [12.4-1a], .fD = 1.60 klf > 1.28 klf OK 
Because the diaphragm shear decreases towards the midspan of the diaphragm, the diaphragm capacity 
may be reduced towards the center of the building. Framing size and plywood thickness are likely to 
have a more significant impact on cost than nail spacing, determine a reasonable location to transition to 2 
in. nominal roof joists and  in. (15/32) plywood. A reasonable configuration for the interior of the 
building utilizes  -in. (15/32) Structural I plywood rated sheathing with a single line of 10d at 2  in. on 
center nailing at diaphragm boundaries, continuous panels edges, and 4 in. on center nailing at other panel 
edges. Using 24 flat blocking at continuous panel edges, the requirements found in Notes f, and g of 
Provisions Table 12.4.3-1a [AF&PA Wind&Seismic Sec. 4.2.7.1] are met. Determine the distance, X, 
from the end wall where the transition can be made as: 
.fD = 0.83 klf (Provisions Table 12.4.3-1a [AF&PA Wind&Seismic Table 4.2A]), 
Shear Capacity = 0.83(100) = 83.0 kips 
Uniform Diaphragm Demand = 233/200 = 1.165 klf 
X = (128 - 83)/1.165 = 38.6 ft, say 40 ft from the diaphragm edge 
[Here is an example where both the Provisions tables and the AF&PA Wind & Seismic tables are 
required to complete the design. The design value for this plywood thickness and nailing pattern is 
contained in AF&PA Wind & Seismic, but the design value for the higher-capacity diaphragm at the ends 
is contained in the Provisions.] 
In a building of this size, it may be beneficial to further reduce the diaphragm nailing towards the middle 
of the roof. However, due to the requirements for subdiaphragms, (see below) and diaphragm capacity, in 
the longitudinal direction and for simplicity of design, no additional nailing pattern is used. 
Table 10.2-1 contains a summary of the diaphragm framing and nailing requirements (All nails are 10d 
common). See Figure 10.2-2 for designation of framing and nailing zones, and Figure 10.2-3 for typical 
plywood layout. 
Table 10.2-1 Roof Diaphragm Framing and Nailing Requirements 
Zone* Framing Plywood Nail Spacing (in.) Capacity 
(kip/ft)
Chapter 10, Wood Design 
10-49 
40'-0" 100'-0" 20'-0" 40'-0" 
100'-0" 
Zone a Zone b Zone a 
Extended zone a due to 
diaphragm opening. 
Figure 10.2-2 Diaphragm framing and nailing layout (1.0 ft = 0.3048 m). 
Boundaries 
and Cont. 
Panel Edges 
Other Panel 
Edges 
Intermediate 
Framing 
Members 
a 312  in. 2 (2 lines) 3 (2 lines) 12 (1 line) 1.60 
b 212  in. 2 (1 line) 4 (1 line) 12 (1 line) 0.83 
1.0 in. = 25.4 mm, 1.0 kip/ft = 14.6 kN/m. 
* Refer to Figure 10.2-2 for zone designation.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
10-50 
Longitudinal 
Transverse 
2x12 or 3x12 joists at 24" o.c. 
2x4 flat blocking at edges 
of plywood panels 
4'x8' plywood with 
grain to joists and 
end joints staggered 
Figure 10.2-3 Plywood layout (1.0 ft = 0.3048 m, 1.0 in. = 25.4 mm). 
10.2.4.2.2 Chord Design 
Although the bond beam at the masonry wall could be used as a diaphragm chord, this example illustrates 
the design of the wood ledger member as a chord. Chord forces are computed using a simply supported 
beam analogy, where the design force is the maximum moment divided by the diaphragm depth. 
Diaphragm moment, M = wL2/8 = Fp,roofL/8 = 233(200/8) = 5,825 ft-kips 
Chord force, T = C = 5,825/(100 - 16/12) = 59.0 kips 
Try a select structural Douglas fir-larch 412, for the chord. Assuming two 1/16 in. bolt holes (for 1-in. 
bolts) at splice locations, the net chord area is 31.9 in.2 Tension strength (parallel to wood grain), per the 
AF&PA Manual, Structural Lumber Supplement: 
.fT (1.0)(0.8)(2.70)(31.9) = 68.9 kips > 59.0 kips OK 
Design the splice for the maximum chord force of 59.0 kips. Try bolts with steel side plates using 1 in. 
A307 bolts, with a 3 in. length in the main member. The capacity, according to the AF&PA Manual, 
Structural Connections Supplement, is: 
.fZ = (1.0)(0.65)(16.29) = 10.6 kips per bolt. 
Number of bolts required = 59.0/10.6 = 5.6 
Use two rows of three bolts. The reduction (group action factor) for multiple bolts is negligible. Net area 
of the 412 chord with two rows of 1-1/16 in. holes is 31.9 in.2 as assumed above. Therefore, use six 1 
in. A307 bolts on each side of the chord splice (Figure 10.2-4). Although it is shown for illustration, this 
type of chord splice may not be the preferred splice against a masonry wall since the bolts, and side plate, 
would have to be recessed into the wall.
Chapter 10, Wood Design 
10-51 
7" 4" 
6-1" A-307 bolts on 
each side of chord. 
Splice in tight hole. 
4x12 Select 
Structural 
Doug. fir chord 
Drive shim in joint 
2 plates 
1
4"x7"x2'-9" 
31
2" 
4" 
31 
41 2" 
31 4" 
2" 
Figure 10.2-4 Chord splice detail (1.0 ft = 0.3048 m, 1.0 in. = 25.4 mm). 
10.2.4.2.3 Diaphragm Deflection and P-delta Check 
The procedure for computing diaphragm deflection is described in Sec. 10.1.4.7. 
[AF&PA Wind & Seismic also contains procedures for computing diaphragm deflections. The equations 
are slightly different from the more commonly used equations that appear in the Commentary and 
AF&PA LRFD Manual. In AF&PA Wind & Seismic, the shear and nail slip terms are combined using an 
apparent shear stiffness parameter. However, the apparent shear stiffness values are only provided for 
OSB. Therefore, the deflection equations in the Commentary or AF&PA LRFD Manual must be used in 
this example which has plywood diaphragms. The apparent shear stiffness values for plywood will likely 
be available in future editions of AF&PA Wind & Seismic.] 
As stated in Commentary Sec. 12.4, the diaphragm deflection is computed as: 
5 3 0.188 ( ) 
8 4 2 
c 
n 
vL vL Le X 
wEA Gt w 
d 
S . 
= + + + 
The equation produces the midspan diaphragm displacement in inches, and the individual variables must 
be entered in the force or length units as described below. A small increase in diaphragm deflection due 
to the large opening is neglected. An adjustment factor, F, for non-uniform nailing is applied to the third 
term of the above equation for this example. 
Bending deflection = 5vL3/8wEA = 0.580 in. 
Shear deflection = vL/4Gt = 1.210 in. 
Effective nails slip deflection = 0.188 LenF = 0.265 in. 
Deflection due to chord slip at splices = S(.cX)/2w = 0.000 in. 
(The chord slip deflection is assumed to be zero because the chord is connected to the continuous bond 
beam at the top of the masonry wall.) 
The variables above and associated units used for computations are: 
v = (233/2)/100 = 1,165 plf (shear per foot at boundary, ignoring torsion)
FEMA 451, NEHRP Recommended Provisions: Design Examples 
10-52 
40'-0" 60'-0" 
End Center 
0.146 kips 
0.233 kips 
0.088 kips 
1 
2 
Figure 10.2-5 Adjustment for nonuniform nail spacing (1.0 ft = 0.3048 m, 1.0 kip = 
4.45 kN) 
L = 200 ft and w = 100 ft (diaphragm length and depth) 
A = effective area of 412 chord and two-#6 rebars assumed to be in the bond beam 
= 39.38 in.2 + 2(0.44)(29,000,000/1,900,000) = 52.81 in.2 
t = 0.535 in. (effective, for  in. Structural I plywood, unsanded; neglect  in. plywood at edge) 
E = 1,900,000 psi (for Douglas fir-larch select structural chord) 
G = 90,000 psi (Structural I plywood) 
en = nail slip for 10d nail at end of diaphragm (use one and one-half-in. nail spacing, two lines at 3 in., 
not the 1-1/4 in. perimeter spacing) 
Design nail load = 1,165/(12/1.5) = 146 lb/nail 
Using the procedure in Sec. 10.1.4.3 and the value for dry/dry lumber from Table 10.1-2, en = 
(146/769)3.276 = 0.00433 in. 
F = adjustment for non-uniform nail spacing 
The 0.188 coefficient assumes a uniform nail spacing, which implies an average load per nail of one-half the 
maximum. One common practice for calculating deformation is simply to use the load per nail that would 
result from the larger spacing should that spacing be used throughout. However, this gives a large estimate 
for deformation due to nail slip. Figure 10.2-5 shows the graphic basis for computing a more accurate nail 
slip term. The basic amount is taken as that for the smaller nail spacing (1-1/2 in.), which gives 146 lbs per 
nail. 
Using a larger nail spacing at the interior increases the deformation. The necessary increase may be 
determined as the ratio of the areas of the triangles in Figure 10.2-5, which represents the load per nail 
along the length of the diaphragm. The ratio is for Triangle 2 representing zone b compared to Triangle 1 
representing zone a, where zones a and b are as shown Figure 10.2-2. 
Ratio = (233 - 88)60 / 146(100) = 0.63 
thus, F = 1 + 0.63 = 1.63. 
Total for diaphragm: 
d = 0.580 + 1.210 + 0.265 + 0.00 = 2.055 in. 
End wall deflection = 0.037 in. (see Sec. 9.1 of this volume of design examples)
Chapter 10, Wood Design 
10-53 
Therefore, the total elastic deflection dxe = 2.055 + 0.037 = 2.092 in. 
Total deflection, dx = Cd dxe/I = 3.5(2.092)/1.0 = 7.32 in. = . 
P-delta effects are computed according to Provisions Sec. 5.4.6.2 [5.2.6.2]using the stability coefficient 
computed per Provisions Eq. 5.4.6.2-1 [5.2-16]: 
x 
x sx d 
P 
V h C 
. 
. 
= 
Because the midspan diaphragm deflection is substantially greater than the deflection at the top of the 
masonry end walls, it would be overly conservative to consider the entire design load at the maximum 
deflection. Therefore, the stability coefficient is computed by splitting the P-delta product into two terms 
one for the diaphragm and one for the end walls. 
For the diaphragm, consider the weight of the roof and side walls at the maximum displacement. (This 
overestimates the P-delta effect. The computation could consider the average displacement of the total 
weight, which would lead to a reduced effective delta. Also, the roof live load need not be included.) 
P = 400 + 416 = 816 kips 
.= 7.32 in. 
V = 233 kips (diaphragm force) 
For the end walls, consider the weight of the end walls at the wall displacement: 
P = 330 kips 
. = (3.5)(0.037) = 0.13 in. 
V = 264 kips (additional base shear for wall design) 
For story height, h = 28 feet, the stability coefficient is: 
816(7.32) 330(0.13) (28)(12)(3.5) 0.022 
d 233 264 
P P hC 
V V 
. 
=.. .+ ... =.. + .. = 
. . . . 
For . < 0.10, no deflection amplification due to P-delta effects is necessary. 
[Note that the equation to determine the stability coefficient has been changed in the 2003 Provisions. 
The importance factor, I, has been added to 2003 Provisions Eq. 5.2-16. However, this does not affect 
this example because I = 1.0.] 
Drift index =./hsx = 7.32/[28(12)] = 0.022. 
This is slightly less than the limiting drift ratio of 0.025 applied for most low-rise buildings in Seismic 
Use Group I (Provisions Sec. 5.2.8 [4.5.1] and Table 5.2.8 [4.5-1]). However, for one-story buildings, 
Table 5.2.8, Note b [4.5-1, Note c], and Provisions Sec. 5.2.6.2.6 [4.5.2], permit unlimited drift provided 
that the structural elements and finishes can accommodate the drift. The limit for masonry cantilever 
shear wall structures (0.007) should only be applied to the in-plane movement of the end walls (0.13/h = 
0.0004 << 0.007). The construction of the out-of-plane walls allows them to accommodate very large 
drifts. It is further expected that the building does not contain interior elements that are sensitive to drift.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
10-54 
Open 
20' 
Glued-laminated 
beams 
20' 20' 40' to 
diaphragm 
boundary 
Open 
20' 
Glued-laminated 
beams 
20' 20' 
Open 
20' 
Glued-laminated 
beams 
20' 20' 40' to 
diaphragm 
boundary 
Figure 10.2-6 Diaphragm at roof opening (1.0 ft = 0.3048 m). 
Given the above conditions, and the fact that P-delta effects are not significant for this structure, the 
computed diaphragm deflections appear acceptable. 
10.2.4.2.4 Detail at Opening 
Consider diaphragm strength at the roof opening (Figure 10.2-6), as required by Provisions Sec. 5.2.6.2.2 
[4.6.1.4]. 
Check diaphragm nailing for 
required shear area (shear in diaphragm at edge of opening): 
Shear = 128 - [40(1.165)] = 81.4 kips 
v = 81.4/(100 - 20) = 1.02 klf 
Because the opening is centered in the width of the diaphragm, half the force to the diaphragm must be 
distributed on each side of the opening. 
Diaphragm capacity in this area = 0.830 klf as computed previously (see Table 10.2-1 and Figure 10.2-2). 
Because the diaphragm demand at the reduced section exceeds the capacity, the extent of the Zone a 
nailing and framing should be increased. For simplicity, extend the Zone a nailing to the interior edge of 
the opening (60 ft from the end wall). The diaphragm strength is now adequate for the reduced overall 
width at the opening. 
10.2.4.2.5 Framing around Opening 
The opening is located 40 ft from one end of the building and is centered in the other direction (Figure 
10.2-6). This does not create any panels with very high aspect ratios. 
In order to develop the chord forces, continuity will be required across the glued-laminated beams in one 
direction and across the roof joists in the other direction. 
10.2.4.2.6 Chord Forces at Opening 
To determine the chord forces on the edge joists, model the diaphragm opening as a Vierendeel truss and 
assume the inflection points will be at the midpoint of the elements (Figure 10.2-7). Compute forces at 
the opening using a uniformly distributed diaphragm demand of 233/200 = 1.165 klf. 
For Element 1 (shown in Figure 10.2-7):
Chapter 10, Wood Design 
10-55 
3 
1 2 
Element 1 
109 kips 
1.165 kip/ft 
w1 
M1 
V1A 
M1 
V1 B 
40'-0" 
20'-0" 40'-0" 
Figure 10.2-7 Chord forces and Element 1 free-body diagram (1.0 ft = 0.3048 
m, 1.0 kip = 4.45 kN, 1.0 kip/ft = 14.6 kN/m). 
w1 = 1.165/2 = 0.582 kips/ft 
V1B = 0.5[128-(40)(1.165)] = 40.7 kips 
V1A = 40.7 - 20(0.582) = 29.1 kips 
M1 due to Vierendeel action = ()[40.7(10) + 29.1(10)] = 349 ft-kips 
Chord force due to M1 = 349/40 = 8.72 kips. This is only 41 psi on the glued-laminated beam on the edge 
of the opening. This member is adequate by inspection. On the other side of this diaphragm element, the 
chord force is much less than the maximum global chord force (59.0 kips), so the ledger and ledger splice 
are adequate. 
For Element 3, analyze Element 2 (shown in Figure 10.2-8): 
w2 = 1.165(40/100) = 0.466 kips/ft 
V3 = 128(40/100) = 51.2 kips 
V1B = 40.7 kips 
M1 = 349 ft-kips. 
T1B is the chord force due to moment on the total diaphragm: 
M = 128(40) - 1.165(402/2) = 4,188 ft-kips 
T1B = 4,188/100 = 41.9 kips 
SM0: M3 = M1 + 40V3 - 40T1B - w2402/2 = 348 ft-kips 
Therefore, the chord force on the roof joist = 348/40 = 8.7 kips
FEMA 451, NEHRP Recommended Provisions: Design Examples 
10-56 
M1 
M 3 = M 
V3 
2 
O w 
2 
V1B
T1B 
2 
Figure 10.2-8 Free-body 
diagram for Element 2. 
Alternatively, the chord design should consider the wall anchorage force interrupted by the opening. As 
described in Sec. 10.2.4.3, the edge members on each side of the opening are used as continuous crossties, 
with maximum cross-tie force of 25.0 kips. Therefore, the cross-tie will adequately serve as a chord 
at the opening. 
10.2.4.3 Roof Diaphragm Design for Longitudinal Direction 
Force = 209 kips 
Maximum end shear = 0.55(209) = 115 kips 
Diaphragm unit shear v = 115/200 = 0.58 klf 
For this direction, the plywood layout is Case 3 in Provisions Table 12.4.3-1a [AF&PA Wind&Seismic 
Table 4.2A]. Using  in. Structural I plywood rated sheathing, blocked, with 10d common nails at 2  
in. on center at diaphragm boundaries and continuous panel edges parallel to the load (ignoring the 
capacity of the extra nails in the outer zones): 
.fD = 0.83 plf > 0.58 plf, Provisions Table 12.4.3-1a [AF&PA Wind&Seismic Table 4.2A] OK 
Therefore, use the same nailing designed for the transverse direction. Compared with the transverse 
direction, the diaphragm deflection and P-delta effects will be satisfactory. 
10.2.4.4 Masonry Wall Anchorage to Roof Diaphragm 
As stipulated in Provisions Sec. 5.2.6.3.2 [4.6.2.1], masonry walls must be anchored to flexible 
diaphragms to resist out-of-plane forces computed per Provisions Eq. 5.2.6.3.2 [4.6-1]as: 
FP = 1.2SDSIWp = 1.2(1.0)(1.0)Wp = 1.2 Wp 
Side walls, FP = 1.2(65psf)(2+28/2)/1000 = 1.25 klf 
End walls, FP = 1.2(103psf)(2 + 28/2)/1000 = 1.98 klf 
[In the 2003 Provisions the anchorage force for masonry walls connected to a flexible diaphragm has 
been reduced to 0.8SDSIWp.] 
10.2.4.3.1 Anchoring Joists Perpendicular to Walls (Side Walls)
Chapter 10, Wood Design 
10-57 
Because the roof joists are spaced at 2 ft on center, provide a connection at each joist that will develop 
2(1.25) = 2.50 kip/joist. 
A common connection for this application is metal tension tie or hold-down device that is anchored to the 
masonry wall with an embedded bolt and is either nailed or bolted to the roof joist. Other types of 
anchors include metal straps that are embedded in the wall and nailed to the top of the joist. The ledger is 
not used for this force transfer because the eccentricity between the anchor bolt and the plywood creates 
tension perpendicular to the grain in the ledger (cross-grain bending), which is prohibited. Also, using 
the edge nails to resist tension perpendicular to the edge of the plywood, is not permitted. 
Try a tension tie with a 3/4 in. headed anchor bolt, embedded in the bond beam and 18 10d nails into the 
side of the joist (Figure 10.2-10). Modifying the allowable values using the procedure in Sec. 10.1.4.5 
results in a design strength of: 
.fZ = (1.0)(0.65)(4.73) = 3.07 kips per anchor > 2.50 kips OK 
The joists anchored to the masonry wall must also be adequately connected to the diaphragm sheathing. 
Determine the adequacy of the typical nailing for intermediate framing members. The nail spacing is 12 
in. and the joist length is 20 ft, so there are 20 nails per joist. From the AF&PA Manual, Structural 
Connections Supplement, the strength of a single 10d common nail is: 
.fZ = (1.0)(0.65)(0.298) = 0.194 kips per nail 
20(0.194) = 3.88 kips > 2.50 kips OK 
The embedded bolt also serves as the ledger connection, for both gravity loading, and in-plane shear 
transfer at the diaphragm. Therefore, the strength of the anchorage to masonry, and the strength of the 
bolt in the wood ledger, must be checked. 
For the anchorage to masonry, check the combined tension and shear, resulting from the out-of-plane 
seismic loading (2.50 kips/bolt), and the vertical gravity loading. Assuming 20 psf dead load (roof live 
load is not included), and a 10-foot tributary roof width, the vertical load per bolt = (20 psf)(10 ft)(2 
ft)/1000 = 0.40 kip. Using the load combination described previously, the design horizontal tension and 
vertical shear on the bolt are: 
ba = 1.0QE = 2.50 kips 
bv = 1.4D = 1.4(0.40) = 0.56 kip 
The anchor bolts in masonry, are designed according to Provisions Sec. 11.3.12. Using 3/4 in. headed 
anchor bolts, both axial strength, Ba, and shear strength, Bv, will be governed by masonry breakout. Per 
Provisions Eq. 11.3.12.3-1, and 11.3.12.1-1, respectively, the design strengths are: 
Ba= 4f Ap fm' 
Bv=1750f 4fm' Ab 
where f = 0.5, fm = 2,000 psi, Ab = tensile area of bolt = 0.44 in.2, and Ap = projected area on the masonry 
surface of a right circular cone = 113 in.2 (assuming 6 in. effective embedment). Therefore, 
Ba = 10.1 kips and Bv = 4.8 kips. Shear and tension are combined per Provisions Eq. 11.3.12.4 as: 
2.50 0.56 0.36 1.0 OK 
10.1 4.8 
a v 
a v 
b b 
B B 
+ = + = <
FEMA 451, NEHRP Recommended Provisions: Design Examples 
10-58 
Bond beam at top 
Vert reinf. to top 
Bond beam 
with 2-#7 cont. 
for chord 
4x12 ledger 
2x12 or 3x12 joist 
with joist hanger 
Structural sheathing 
(See plan for thickness 
and nailing) 
Tension tie (e.g. Simpson 
LTT131) at each joist 
3
4" dia. bolt at tension 
tie (24" o.c.) 
Figure 10.2-9 Anchorage of masonry wall perpendicular to joists (1.0 in. = 25.4 mm). 
[The 2003 Provisions refers to ACI 530, Sec. 3.1.6 for strength design of anchorage to masonry, as 
modified by 2003 Provisions Sec. 11.2. In general, the methodology for anchorage design using ACI 530 
should be comparable to the 2000 Provisions, though some of the equations and reduction factors may be 
different. In addition, the 2003 Provisions require that anchors are either controlled by a ductile failure 
mode or are designed for 2.5 times the anchorage force.] 
Figure 10.2-9 summarizes the details of the connection. In-plane seismic shear transfer (combined with 
gravity) and orthogonal effects are considered in a subsequent section. 
According to Provisions Sec. 5.2.6.3.2 [4.6.2.1, diaphragms must have continuous cross-ties to distribute 
the anchorage forces into the diaphragms. Although the Provisions do not specify a maximum spacing, 
20 ft is common practice for this type of construction and Seismic Design Category. 
For cross-ties at 20 ft on center, the wall anchorage force per cross-tie is: 
(1.25 klf)(20 ft) = 25.0 kips 
Try a 312 (Douglas fir-Larch No. 1) as a cross-tie. Assuming one row of 15/16 in. bolt holes, the net 
area of the section is 25.8 in.2 Tension strength (parallel to wood grain) per the AF&PA Structural 
Lumber Supplement, is: 
.fT = (1.0)(0.8)(1.82)(25.8) = 37.5 kips > 25.0 kips OK 
At the splices, try a double tie-down device with four 7/8 in. bolts in double shear through the 312 
(Figure 10.2-10). Product catalogs provide design capacities for single tie-downs only; the design of 
double hold downs requires two checks. First, consider twice the capacity of one tie-down and, second, 
consider the capacity of the bolts in double shear. 
For the double tie-down, use the procedure in Sec. 10.1.4.5 to modify the allowable values: 
2.fZ = 2(1.0)(0.65)(20.66) = 26.9 kips > 25.0 kips OK
Chapter 10, Wood Design 
10-59 
3x12 with 
joist hanger 
Glue-lam beam 
3x12 with 
joist hanger 
7
8" dia. threaded 
rod at hold downs 
Plywood with boundary 
nailing along 3x12 
Tie-down on both sides 
of 3x12 with 4-7
8" dia. thru 
bolts. (Each side of beam) 
Figure 10.2-10 Chord tie at roof opening (1.0 in. = 25.4 mm). 
The reduction (group action) factor for multiple bolts, Cg = 0.97. 
For the four bolts, the AF&PA Manual, Structural Connections Supplement, gives: 
4.fCgZ = 4(1.0)(0.65)(0.97)(10.17) = 25.6 kips > 25.0 kips 
OK 
In order to transfer the wall anchorage forces into the cross-ties, the subdiaphragms between these ties 
must be checked per Provisions Sec. 5.2.6.3.2 [4.6.2.1]. There are several ways to perform these 
subdiaphragm calculations. One method is illustrated in Figure 10.2-11. The subdiaphragm spans 
between cross-ties and utilizes the glued-laminated beam and ledger as its chords. The 1-to-1 aspect 
ratio meets the requirement of 2 to 1 for subdiaphragms per Provisions Sec. 5.2.6.3.2 [4.2.6.1]. 
For the typical subdiaphragm (Figure 10.2-11): 
Fp = 1.25 klf 
v = (1.25)(20/2)/20 = 0.625 klf. 
The subdiaphragm demand is less than the minimum diaphragm capacity (0.83 klf along the center of the 
side walls). In order to develop the subdiaphragm strength, and boundary nailing must be provided along 
the cross-tie beams.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
10-60 
Fp 
V V 
Subdiaphragm 
F p = 1.25 kips/ft 
Subdiaphragm 
20'-0" 20'-0" 
20'-0" 
Glued - 
laminated 
beam 
Spliced 
continuous 
cross ties 
Masonry 
wall 
Typical joist, 
anchored to 
masonry wall 
~ 
~ 
Figure 10.2-11 Cross tie plan layout and subdiaphragm free-body 
diagram for side walls (1.0 ft = 0.3048 m, 1.0 kip/ft = 14.6 kN/m). 
10.2.4.3.2 Anchorage at Joists Parallel to Walls (End Walls) 
Where the joists are parallel to the walls, tied elements must transfer the forces into the main body of the 
diaphragm, which can be accomplished by using either metal strapping and blocking or metal rods and 
blocking. This example uses threaded rods that are inserted through the joists and coupled to the anchor 
bolt (Figure 10.2-12). Blocking is added on both sides of the rod to transfer the force into the plywood 
sheathing. The tension force in the rod causes a compression force on the blocking through the nut, and 
bearing plate at the innermost joist. 
Chapter 10, Wood Design 
10-61 
Bond beam 
with chord 
reinf. 
4x12 ledger 
3x12 joist 
at 24" o.c. 
3
4" dia. threaded rod 
at 4'-0" o.c. extend 
through joists 
3
4" Structural I 
plywood 
3
4" dia. bolt 
at 2'-0" o.c. 
Coupler 
12'-0" 
10d nails at 6" 
o.c. at blocking 
Plate washer 
and nut at 
last joist 
2x12 blocking on 
each side of 
threaded rod 
Bond beam 
at top 
Vertical 
reinforcement 
Figure 10.2-12 Anchorage of masonry wall parallel to joists (1.0 ft = 0.3048 m, 1.0 in. = 25.4 mm) 
The anchorage force at the end walls is 1.98 klf. Space the connections at 4 ft on center so that the wall 
need not be designed for flexure (Provisions Sec. 5.2.6.1.2 [4.6.1.2]). Thus, the anchorage force is 7.92 
kips per anchor. 
Try a 3/4 in. headed anchor bolt, embedded into the masonry. In this case, gravity loading on the ledger 
is negligible and can be ignored, and the anchor can be designed for tension only. (In-plane shear transfer 
and orthogonal effects are considered later.) 
As computed for 3/4 in. headed anchor bolts (with 6 in. embedment), the design axial strength Ba = 10.1 
kips > 7.92 kips. Therefore, the bolt is acceptable. 
Using couplers rated for 125 percent of the strength of the rod material, the threaded rods are then 
coupled to the anchor bolts and extend six joist spaces (12 ft) into the roof framing. (The 12 ft are 
required for the subdiaphragm force transfer as discussed below.) 
Nailing the blocking to the plywood sheathing is determined, using the AF&PA Structural Connections 
Supplement. As computed previously, the capacity of a single 10d common nail, .fZ = 0.194 kips. 
Thus, 7.92/0.194 = 41 nails are required. This corresponds to a nail spacing of about 7 in. for two 12-ft 
rows of blocking. Space nails at 6 in. for convenience. 
Use the glued-laminated beams (at 20 ft on center) to provide continuous cross-ties, and check the 
subdiaphragms between the beams to provide adequate load transfer to the beams per Provisions Sec. 
5.2.6.3.2 [4.6.2.1]. 
Design tension force on beam = (1.98 klf)(20 ft) = 39.6 kips 
The stress on the beam is ft = 39.600/[8.75(24)] = 189 psi, which is small. The beam is adequate for 
combined moment due to gravity loading and axial tension. 
At the beam splices, try one-in. bolts with steel side plates. Per the AF&PA Manual, Structural 
Connections Supplement:
FEMA 451, NEHRP Recommended Provisions: Design Examples 
10-62 
Fp 
V
V 
Subdiaphragm 
Fp = 1.98 kips/ft 
Subdiaphragm 
20'-0" 
Glued - laminated 
beam, spliced as 
continuous cross tie 
Masonry 
wall 
Typical wall, 
anchor at 
4'-0" o.c. 
12'-0" 
~
~ 
Roof joist for 
subdiaphragm 
chord 
~ 
Figure 10.2-13 Cross tie plan layout and subdiaphragm free-body diagram for end 
walls (1.0 ft = 0.3048 m, 1.0 kip/ft = 14.6 kN/m). 
lfZ = (1.0)(0.65)(18.35) = 11.93 kips per bolt. 
The number of bolts required = 39.6/11.93 = 3.3. 
Use four bolts in a single row at midheight of the beam, with 1/4-in. by 4-in. steel side plates. The 
reduction (group action factor) for multiple bolts is negligible. Though not included in this example, the 
steel side plates should be checked for tension capacity on the gross and net sections. There are 
pre-engineered hinged connectors for glued-laminated beams that could provide sufficient tension 
capacity for the splices. 
In order to transfer the wall anchorage forces into the cross-ties, the subdiaphragms between these ties 
must be checked per Provisions Sec. 5.2.6.3.2 [4.6.2.1]. The procedure is similar to that used for the side 
walls as described previously. The end wall condition is illustrated in Figure 10.2-13. The subdiaphragm 
spans between beams and utilizes a roof joist as its chord. In order to adequately engage the 
subdiaphragm, the wall anchorage ties must extend back to this chord. The maximum aspect ratio for 
subdiaphragms is 2 to 1 per Provisions Sec. 5.2.6.3.2 [4.6.2.1]. Therefore, the minimum depth is 20/2.5 
= 8 ft. 
For the typical subdiaphragm (Figure 10.2-13):
Chapter 10, Wood Design 
10-63 
Fp = 1.98 klf 
v = (1.98)(20/2)/8 = 2.48 klf 
As computed previously (see Table 10.2-1 and Figure 10.2.2), the diaphragm strength in this area is 1.60 
klf < 2.48 klf. Therefore, increase the subdiaphragm depth to 12 ft (six joist spaces): 
v = (1.98)(20/2)/12 = 1.65 klf .1.60 klf OK 
In order to develop the subdiaphragm strength, boundary nailing must be provided along the cross-tie 
beams. There are methods of refining this analysis using multiple subdiaphragms so that all of the tension 
anchors need not extend 12 ft into the building. 
10.2.4.3.3 Transfer of Shear Wall Forces 
The in-plane diaphragm shear must be transferred to the masonry wall by the ledger, parallel to the wood 
grain. The connection must have sufficient capacity for the diaphragm demands as: 
Side walls = 0.575 klf 
End walls = 1.282 klf 
For each case, the capacity of the bolted wood ledger and the capacity of the anchor bolts embedded into 
masonry must be checked. Because the wall connections provide a load path for both in-plane shear 
transfer and out-of-plane wall forces, the bolts must be checked for orthogonal load effects in accordance 
with Provisions Sec. 5.2.5.2.3 [4.4.2.3]. That is, the combined demand must be checked for 100 percent 
of the lateral load effect in one direction (e.g., shear) and 30 percent of the lateral load effect in the other 
direction (e.g., tension). 
At the side walls, the wood ledger with 3/4-in. bolts (Figure 10.2-9) must be designed for gravity loading 
(0.56 kip per bolt as computed above) as well as seismic shear transfer. The seismic load per bolt (at 2 ft 
on center) is 0.575(2) = 1.15 kips. 
Combining gravity shear and seismic shear, produces a resultant force of 1.38 kips at an angle of 26 
degrees from the axis of the wood grain. Using the formulas for bolts at an angle to the grain per the 
AF&PA Structural Connections Supplement gives 
lfZ = (1.0)(0.65)(3.47) = 2.26 kips > 1.38. OK 
This bolt spacing satisfies the load combination for gravity loading only. 
For the check of the embedded anchor bolts, the factored demand on a single bolt is 1.15 kips in 
horizontal shear (in-plane shear transfer), 2.50 kips in tension (out-of-plane wall anchorage), 0.56 kip in 
vertical shear (gravity). Orthogonal effects are checked, using the following two equations: 
(0.3)(2.5) 1.152 0.562 0.34 
19.1 4.8 
+ 
+ = 
and 
(controls) < 1.0 OK 
(2.5) [0.3(1.15)]2 0.562 0.38 
10.1 4.8 
+ 
+ =
FEMA 451, NEHRP Recommended Provisions: Design Examples 
10-64 
At the end walls, the ledger with 3/4-in. bolts (Figure 10.2-12) need only be checked for in-plane seismic 
shear because gravity loading is negligible. For bolts spaced at 4 ft on center, the demand per bolt is 
1.282(4) = 5.13 kips parallel to the grain of the wood. Per the AF&PA Structural Manual, Connections 
Supplement: 
Z = (1.0)(0.65)(5.37) = 3.49 kips < 5.13 kips NG 
Therefore, add 3/4-in. headed bolts evenly spaced between the tension ties such that the bolt spacing is 2 
ft on center and the demand per bolt is 1.282(2) = 2.56 kips. These added bolts are used for in-plane 
shear only and do not have coupled tension tie rods. 
For the check of the embedded bolts, the factored demand on a single bolt is 2.56 kips in horizontal shear 
(in-plane shear transfer), 7.92 kips in tension (out-of-plane wall anchorage), 0 kip in vertical shear 
(gravity is negligible). Orthogonal effects are checked using the following two equations: 
(0.3)(7.92) 2.56 0.77 
10.1 4.8 
+ = 
and 
(controls) < 1.0 OK 
7.92 0.3(2.56) 0.94 
10.1 4.8 
+ = 
Therefore, the wall connections satisfy the requirements for combined gravity, and seismic loading, 
including orthogonal effects.
11-1 
11 
SEISMICALLY ISOLATED STRUCTURES 
Charles A. Kircher, P.E., Ph.D. 
Chapter 13 of the 2000 NEHRP Recommended Provisions addresses the design of buildings that 
incorporate a seismic isolation system. The Provisions provides essentially a stand alone set of design 
and analysis criteria for an isolation system. Chapter 13 defines load, design, and testing requirements 
specific to the isolation system and interfaces with the appropriate materials chapters for design of the 
structure above the isolation system and of the foundation and structural elements below. 
A discussion of background, basic concepts, and analysis methods is followed by an example that 
illustrates the application of the Provisions to the structural design of a building with an isolation system. 
In this example, the building is a three-story emergency operations center (EOC) with a steel 
concentrically braced frame above the isolation system. Although the facility is hypothetical, it is of 
comparable size and configuration to actual base-isolated EOCs, and is generally representative of 
base-isolated buildings. 
The EOC is located in San Francisco and has an isolation system that utilizes elastomeric bearings, a type 
of bearing commonly used for seismic isolation of buildings. The example comprehensively describes 
the EOCs configuration, defines appropriate criteria and design parameters, and develops a preliminary 
design using the equivalent lateral force (ELF) procedure of Chapter 13. It also includes a check of the 
preliminary design using dynamic analysis as required by the Provisions and specifies isolation system 
design and testing criteria. 
Located in a region of very high seismicity, the building is subject to particularly strong ground motions. 
Large seismic demands pose a challenge for the design of base-isolated structures in terms of the capacity 
of the isolation system and the configuration of the structure above the isolation system. The isolation 
system must accommodate large lateral displacements (e.g., in excess of 2 ft). The structure above the 
isolation system should be configured to produce the smallest practical overturning loads (and uplift 
displacements) on the isolators. The example addresses these issues and illustrates that isolation systems 
can be designed to meet the requirements of the Provisions, even in regions of very high seismicity. 
Designing an isolated structure in a region of lower seismicity would follow the same approach. The 
isolation system displacement, overturning forces, and so forth would all be reduced, and therefore, easier 
to accommodate using available isolation system devices. 
The isolation system for the building in the example is composed of high-damping rubber (HDR) 
elastomeric bearings. HDR bearings are constructed with alternating layers of rubber and steel plates all 
sheathed in rubber. The first base-isolated building in the United States employed this type of isolation 
system. Other types of isolation systems used to base isolate buildings employed lead-core elastomeric 
bearings (LR) and sliding isolators, such as the friction pendulum system (FPS). In regions of very high 
seismicity, viscous dampers have been used to supplement isolation system damping (and reduce 
displacement demand). Using HDR bearings in this example should not be taken as an endorsement of 
this particular type of isolator to the exclusion of others. The concepts of the Provisions apply to all types
FEMA 451, NEHRP Recommended Provisions: Design Examples 
11-2 
of isolations systems, and other types of isolators (and possible supplementary dampers) could have been 
used equally well in the example. 
In addition to the 2000 NEHRP Recommended Provisions and Commentary (hereafter, the Provisions and 
Commentary), the following documents are either referenced directly or are useful aids for the analysis 
and design of seismically isolated structures. 
ATC 1996 Applied Technology Council. 1996. Seismic Evaluation and Retrofit of 
Buildings, ATC40. 
Constantinou Constantinou, M. C., P. Tsopelas, A. Kasalanati, and E. D. Wolff. 1999. 
Property Modification Factors for Seismic Isolation Bearings, Technical Report 
MCEER-99-0012. State University of New York. 
CSI Computers and Structures, Inc. (CSI). 1999. ETABS Linear and Nonlinear 
Static and Dynamic Analysis and Design of Building Systems. 
FEMA 273 Federal Emergency Management Agency. 1997. NEHRP Guidelines for the 
Seismic Rehabilitation of Buildings, FEMA 273. 
FEMA 222A Federal Emergency Management Agency. 1995. NEHRP Recommended 
Provisions for Seismic Regulations for New Buildings, FEMA 222A. 
91 UBC International Conference of Building Officials. 1991. Uniform Building Code. 
94 UBC International Conference of Building Officials. 1994. Uniform Building Code. 
Kircher Kircher, C. A., G. C. Hart, and K. M. Romstad. 1989. "Development of Design 
Requirements for Seismically Isolated Structures" in Seismic Engineering and 
Practice, Proceedings of the ASCE Structures Congress, American Society of 
Civil Engineers, May 1989. 
SEAOC 1999 Seismology Committee, Structural Engineers Association of California. 1999. 
Recommended Lateral Force Requirements and Commentary, 7th Ed. 
SEAOC 1990 Seismology Committee, Structural Engineers Association of California. 1990. 
Recommended Lateral Force Requirements and Commentary, 5th Ed. 
SEAONC Isolation Structural Engineers Association of Northern California. 1986. Tentative 
Seismic Isolation Design Requirements. 
Although the guide is based on the 2000 Provisions , it has been annotated to reflect changes made to the 
2003 Provisions. Annotations within brackets, [ ], indicate both organizational changes (as a result of a 
reformat of all of the chapters of the 2003 Provisions) and substantiative technical changes to the 2003 
Provisions and its primary reference documents. While the general changes to the document are 
described , the deign examples and calculations have not been revised to reflect the changes to the 2003 
Provisions. 
In the 2003 edition of the Provisions, Chapter 13 has been restructured so that it is better integrated into 
the Provisions as a whole and is less of a stand alone set of requirements. Where they affect the design 
examples in this chapter, other significant changes to the 2003 Provisions and primary reference 
documents may be noted.
Chapter 11, Seismically Isolated Structures 
11-3
FEMA 451, NEHRP Recommended Provisions: Design Examples 
11-4 
11.1 BACKGROUND AND BASIC CONCEPTS 
Seismic isolation, commonly referred to as base isolation, is a design concept that presumes a structure 
can be substantially decoupled from potentially damaging earthquake ground motions. By decoupling the 
structure from ground shaking, isolation reduces the level of response in the structure that would 
otherwise occur in a conventional, fixed-base building. Conversely, base-isolated buildings may be 
designed with a reduced level of earthquake load to produce the same degree of seismic protection. That 
decoupling is achieved when the isolation scheme makes the fundamental period of the isolated structure 
several times greater than the period of the structure above the isolation system. 
The potential advantages of seismic isolation and the advancements in isolation system products led to the 
design and construction of a number of isolated buildings and bridges in the early 1980s. This activity, in 
turn, identified a need to supplement existing seismic codes with design requirements developed 
specifically for such structures. These requirements assure the public that isolated buildings are safe and 
provide engineers with a basis for preparing designs and building officials with minimum standards for 
regulating construction. 
Initial efforts developing design requirements for base-isolated buildings began with ad hoc groups of the 
Structural Engineers Association of California (SEAOC), whose Seismology Committee has a long 
history of contributing to codes. The northern section of SEAOC was the first to develop guidelines for 
the use of elastomeric bearings in hospitals. These guidelines were adopted in the late 1980s by the 
California Office of Statewide Health Planning and Development (OSHPD) and were used to regulate the 
first base-isolated hospital in California. At about the same time, the northern section of SEAOC 
published SEAONC Isolation, first set of general requirements to govern the design of base-isolated 
buildings. Most of the basic concepts for the design of seismically isolated structures found in the 
Provisions can be traced back to the initial work by the northern section of SEAOC. 
By the end of the 1980s, the Seismology Committee of SEAOC recognized the need to have a more 
broadly based document and formed a statewide committee to develop design requirements for isolated 
structures Kircher. The "isolation" recommendations became an appendix to the 1990 SEAOC Blue 
Book. The isolation appendix was adopted with minor changes as a new appendix in the 1991 Uniform 
Building Code and has been updated every three years, although it remains largely the same as the 
original 91 UBC appendix. (SEAOC 1990 and 1999 are editions of SEAOCs Recommended Lateral 
Force Requirements and Commentary, which is also known as the Blue Book.) 
In the mid-1990s, the Provisions Update Committee of the Building Seismic Safety Council incorporated 
the isolation appendix of the 94 UBC into the 1994 Provisions (FEMA 222A). Differences between the 
Uniform Building Code (UBC) and the Provisions were intentionally minimized and subsequent editions 
of the UBC and the Provisions are nearly identical. Additional background may be found in the 
commentary to the 1999 SEAOC Blue Book. 
The Provisions for designing the isolation system of a new building were used as the starting point for the 
isolation system requirements of the NEHRP Guidelines for Seismic Rehabilitation of Buildings (FEMA 
273). FEMA 273 follows the philosophy that the isolation system for a rehabilitated building should be 
comparable to that for a new building (for comparable ground shaking criteria, etc.). The superstructure, 
however, could be quite different, and FEMA 273 provides more suitable design requirements for 
rehabilitating existing buildings using an isolation system. 
11.1.1 Types of Isolation Systems
Chapter 11, Seismically Isolated Structures 
11-5 
The Provisions requirements are intentionally broad, accommodating all types of acceptable isolation 
systems. To be acceptable, the Provisions requires the isolation system to: 
1. Remain stable for maximum earthquake displacements, 
2. Provide increasing resistance with increasing displacement, 
3. Have limited degradation under repeated cycles of earthquake load, and 
4. Have well-established and repeatable engineering properties (effective stiffness and damping). 
The Provisions recognizes that the engineering properties of an isolation system, such as effective 
stiffness and damping, can change during repeated cycles of earthquake response (or otherwise have a 
range of values). Such changes or variability of design parameters are acceptable provided that the design 
is based on analyses that conservatively bound (limit) the range of possible values of design parameters. 
The first seismic isolation systems used in buildings in the United States were composed of elastomeric 
bearings that had either a high-damping rubber compound or a lead core to provide damping to isolated 
modes of vibration. Other types of isolation systems now include sliding systems, such as the friction 
pendulum system (FPS), or some combination of elastomeric and sliding isolators. Some applications at 
sites with very strong ground shaking use supplementary fluid-viscous dampers in parallel with either 
sliding or elastomeric isolators to control displacement. While generally applicable to all types of 
systems, certain requirements of the Provisions (in particular, prototype testing criteria) were developed 
primarily for isolation systems with elastomeric bearings. 
Isolation systems typically provide only horizontal isolation and are rigid or semi-rigid in the vertical 
direction. A rare exception to this rule is the full isolation (horizontal and vertical) of a building in 
southern California isolated by large helical coil springs and viscous dampers. While the basic concepts 
of the Provisions can be extended to full isolation systems, the requirements are only for horizontal 
isolation systems. The design of a full isolation system requires special analyses that explicitly include 
vertical ground shaking and the potential for rocking response. 
Seismic isolation is commonly referred to as base isolation because the most common location of the 
isolation system is at or near the base of the structure. The Provisions does not restrict the plane of 
isolation to the base of the structure but does require the foundation and other structural elements below 
the isolation system to be designed for unreduced (RI = 1.0) earthquake forces. 
11.1.2 Definition of Elements of an Isolated Structure 
The design requirements of the Provisions distinguish between structural elements that are either 
components of the isolation system or part of the structure below the isolation system (e.g., foundation) 
and elements of the structure above the isolation system. The isolation system is defined by the 
Provisions as: 
The collection of structural elements that includes all individual isolator units, all structural elements that 
transfer force between elements of the isolation system, and all connections to other structural elements. 
The isolation system also includes the wind-restraint system, energy-dissipation devices, and/or the 
displacement restraint system if such systems and devices are used to meet the design requirements of 
Chapter 13. 
Figure 11.1-1 illustrates this definition and shows that the isolation system consists not only of the 
isolator units but also of the entire collection of structural elements required for the system to function 
properly. The isolation system typically includes segments of columns and connecting girders just above 
the isolator units because such elements resist moments (due to isolation system displacement) and their 
yielding or failure could adversely affect the stability of isolator units.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
11-6 
Structure Below The 
Isolation System 
Structural Elements That Transfer 
Force Between Isolator Units 
Isolator 
Unit 
Isolator 
Unit 
Structure Above The 
Isolation System 
Isolation 
Interface 
Figure 11.1-1 Isolation system terminology. 
The isolation interface is an imaginary boundary between the upper portion of the structure, which is 
isolated, and the lower portion of the structure, which is assumed to move rigidly with the ground. 
Typically, the isolation interface is a horizontal plane, but it may be staggered in elevation in certain 
applications. The isolation interface is important for design of nonstructural components, including 
components of electrical and mechanical systems that cross the interface and must accommodate large 
relative displacements. 
The wind-restraint system is typically an integral part of isolator units. Elastomeric isolator units are very 
stiff at very low strains and usually satisfy drift criteria for wind loads, and the static (breakaway) friction 
force of sliding isolator units is usually greater than the wind force. 
11.1.3 Design Approach 
The design of isolated structures using the Provisions (like the UBC and SEAOCs Blue Book) has two 
objectives: achieving life safety in a major earthquake and limiting damage due to ground shaking. To 
meet the first performance objective, the isolation system must be stable and capable of sustaining forces 
and displacements associated with the maximum considered earthquake and the structure above the 
isolation system must remain essentially elastic when subjected to the design earthquake. Limited 
ductility demand is considered necessary for proper functioning of the isolation system. If significant 
inelastic response was permitted in the structure above the isolation system, unacceptably large drifts 
could result due to the nature of long-period vibration. Limiting ductility demand on the superstructure 
has the additional benefit of meeting the second performance objective of damage control. 
The Provisions addresses the performance objectives by requiring:
Chapter 11, Seismically Isolated Structures 
11-7 
1. Design of the superstructure for forces associated with the design earthquake, reduced by only a 
fraction of the factor permitted for design of conventional, fixed-base buildings (i.e., RI = 3/8 R #2.0). 
2. Design of the isolation system and elements of the structure below the isolation system (e.g., 
foundation) for unreduced design earthquake forces. 
3. Design and prototype testing of isolator units for forces (including effects of overturning) and 
displacements associated with the maximum considered earthquake. 
4. Provision of sufficient separation between the isolated structure and surrounding retaining walls and 
other fixed obstructions to allow unrestricted movement during the maximum considered earthquake. 
11.1.4 Effective Stiffness and Effective Damping 
The Provisions utilizes the concepts of effective stiffness and damping to define key parameters of 
inherently nonlinear, inelastic isolation systems in terms of amplitude-dependent linear properties. 
Effective stiffness is the secant stiffness of the isolation system at the amplitude of interest. Effective 
damping is the amount of equivalent viscous damping described by the hysteresis loop at the amplitude of 
interest. Figure 11.1-2 shows the application of these concepts to both hysteretic isolator units (e.g., 
friction or yielding devices) and viscous isolator units and shows the Provisions equations used to 
determine effective stiffness and damping from tests of prototypes. Ideally, the effective damping of 
velocity-dependent devices (including viscous isolator units) should be based on the area of hysteresis 
loops measured during cyclic testing of the isolation system at full-scale earthquake velocities. Tests of 
prototypes are usually performed at lower velocities (due to test facility limitations), resulting in 
hysteresis loops with less area, which produce lower (conservative) estimates of effective damping. 
11.2 CRITERIA SELECTION 
As specified in the Provisions the design of isolated structures must be based on the results of 
the equivalent lateral force (ELF) procedure, response spectrum analysis, or (nonlinear) time 
history analysis. Because isolation systems are typically nonlinear, linear methods (ELF 
procedure and response spectrum analysis) use effective stiffness and damping properties to 
model nonlinear isolation system components. 
The ELF procedure is intended primarily to prescribe minimum design criteria and may be used 
for design of a very limited class of isolated structures (without confirmatory dynamic analyses). 
The simple equations of the ELF procedure are useful tools for preliminary design and provide a 
means of expeditious review and checking of more complex calculations. The Provisions also 
uses these equations to establish lower-bound limits on results of dynamic analysis that may be 
used for design. Table 11.2-1 summarizes site conditions and structure configuration criteria 
that influence the selection of an acceptable method of analysis for designing of isolated 
structures. Where none of the conditions in Table 11.2-1 applies, all three methods are 
permitted.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
11-8 
Force 
Hysteretic Isolator Viscous Isolator 
Force 
Disp. Disp. 
Eloop Eloop 
eff 
F F 
k 
. . 
+ - 
+ - 
+ 
= 
+ 
( )2 
2 loop 
eff 
eff 
E 
k 
 
p .+ .- 
. . 
= .. .. 
.. + .. 
. + . + 
. - . - 
F+ F+ 
F- F- 
Figure 11.1-2 Effective stiffness and effective damping. 
Table 11.2-1 Acceptable Methods of Analysis* 
Site condition or 
Structure Configuration Criteria 
ELF 
Procedure 
Response 
Spectrum 
Analysis 
Time 
History 
Analysis 
Site Conditions 
Near-source (S1 > 0.6) NP P P 
Soft soil (Site Class E or F) NP NP P 
Superstructure Configuration 
Flexible or irregular superstructure 
(height > 4 stories, height > 65 ft, or 
TM > 3.0 sec., or TD #3T) 
NP P P 
Nonlinear superstructure (requiring explicit 
modeling of nonlinear elements; 
Provisions Sec. 13.2.5.3.1) [13.4.1.2] 
NP NP P 
Isolation System Configuration 
Highly nonlinear isolation system or 
system that otherwise does not meet the 
criteria of Provisions Sec. 13.2.5.2, Item 7 
[13.2.4.1, Item 7] 
NP NP P 
* P indicates permitted and NP indicates not permitted by the Provisions.
Chapter 11, Seismically Isolated Structures 
11-9 
Seismic criteria are based on the same site and seismic coefficients as conventional, fixed-base structures 
(e.g., mapped value of S1 as defined in Provisions Chapter 4 [3]). Additionally, site-specific design 
criteria are required for isolated structures located on soft soil (Site Class E of F) or near an active source 
such that S1 is greater than 0.6, or when nonlinear time history analysis is used for design. 
11.3 EQUIVALENT LATERAL FORCE PROCEDURE 
The equivalent lateral force (ELF) procedure is a displacement-based method that uses simple equations 
to determine isolated structure response. The equations are based on ground shaking defined by 1 second 
spectral acceleration and the assumption that the shape of the design response spectrum at long periods is 
inversely proportional to period as shown in Provisions Figure 4.1.2.6 [3.3-15]. [In the 2003 edition of 
the Provisions , there is also a 1/T2 portion of the spectrum at periods greater than TL. However, in most 
parts of the Unites States TL is longer than the period of typical isolated structures.] Although the ELF 
procedure is considered a linear method of analysis, the equations incorporate amplitude-dependent 
values of effective stiffness and damping to implicitly account for the nonlinear properties of the isolation 
system. The equations are consistent with the nonlinear static procedure of FEMA 273 assuming the 
superstructure is rigid and lateral displacements to occur primarily in the isolation system. 
11.3.1 Isolation System Displacement 
The isolation system displacement for the design earthquake is determined by using Provisions Eq. 
13.3.3.1 [13.3-1]: 
4 2 
D1 D 
D 
D 
D g S T 
p B 
=.. .. 
. . 
where the damping factor ,BD, is based on effective damping, D, using Provisions Table 13.3.3.1 [13.3- 
1]. This equation describes the peak (spectral) displacement of a single-degree-of-freedom (SDOF) 
system with period, TD, and damping, D, for the design earthquake spectrum defined by the seismic 
coefficient, SD1. SD1 corresponds to 5 percent damped spectral response at a period of 1 second. BD, 
converts 5 percent damped response to the level of damping of the isolation system. BD is 1.0 when 
effective damping, D, is 5 percent of critical. Figure 11.3-1 illustrates the underlying concepts of 
Provisions Eq. 13.3.3.1 [13.3-1] and the amplitude-dependent equations of the Provisions for effective 
period, TD, and effective damping, D.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
11-10 
Mass .. Design earthquake 
spectral acceleration 
(W/g .. SD1/T) 
Mass .. Spectral 
acceleration 
Spectral 
displacement 
Maximum 
stiffness 
curve 
Minimum 
stiffness 
curve 
Response reduction, BD 
(effective damping, 
D > 5% of critical) 
kDmax 
kDmin 
4 2 D1 D 
g S T 
p 
. . 
. . 
. . 
D 2 
Dmin 
T W 
k g 
= p 
2 
1 
2 
D 
D 
Dmax D 
E 
k D 
 
p
. . 
= .. .. 
. . 
S 
SED DD 
Vb 
Figure 11.3-1 Isolation system capacity and earthquake demand. 
The equations for maximum displacement, DM, and design displacement, DD, reflect differences due to the 
corresponding levels of ground shaking. The maximum displacement is associated with the maximum 
considered earthquake (characterized by SM1) whereas the design displacement corresponds to the design 
earthquake (characterized by SD1). In general, the effective period and the damping factor (TM and BM, 
respectively) used to calculate the maximum displacement are different from those used to calculate the 
design displacement (TD and BD) because the effective period tends to shift and effective damping may 
change with the increase in the level of ground shaking. 
As shown in Figure 11.3-1, the calculation of effective period, TD, is based on the minimum effective 
stiffness of the isolation system, kDmin, as determined by prototype testing of individual isolator units. 
Similarly, the calculation of effective damping is based on the minimum loop area, ED, as determined by 
prototype testing. Use of minimum effective stiffness and damping produces larger estimates of effective 
period and peak displacement of the isolation system. 
The design displacement, DD, and maximum displacement, DM, represent peak earthquake displacements 
at the center of mass of the building without the additional displacement, that can occur at other locations 
due to actual or accidental mass eccentricity. Equations for determining total displacement, including the 
effects of mass eccentricity as an increase in the displacement at the center of mass, are based on the plan 
dimensions of the building and the underlying assumption that building mass and isolation stiffness have 
a similar distribution in plan. The increase in displacement at corners for 5 percent mass eccentricity is 
about 15 percent if the building is square in plan, and as much as 30 percent if the building is long in plan. 
Figure 11.3-2 illustrates design displacement, DD, and maximum displacement, DM, at the center of mass 
of the building and total maximum displacement, DTM, at the corners of an isolated building.
Chapter 11, Seismically Isolated Structures 
11-11 
Total Maximum Displacement 
(maximum considered earthquake 
corner of building) 
Plan View of Building 
Maximum Displacement 
(maximum considered earthquake 
center of building) 
Design Displacement 
(design earthquake center 
of building) 
DTM 
DM
DD 
Figure 11.3-2 Design, maximum, and total maximum displacement. 
11.3.2 Design Forces 
Forces required by the Provisions for design of isolated structures are different for design of the 
superstructure and design of the isolation system and other elements of the structure below the isolation 
system (e.g., foundation). In both cases, however, use of the maximum effective stiffness of the isolation 
system is required to determine a conservative value of design force. 
In order to provide appropriate overstrength, peak design earthquake response (without reduction) is used 
directly for design of the isolation system and the structure below. Design for unreduced design 
earthquake forces is considered sufficient to avoid inelastic response or failure of connections and other 
elements for ground shaking as strong as that associated with the maximum considered earthquake (i.e., 
shaking as much as 1.5 times that of the design earthquake). The design earthquake base shear, Vb, is 
given by Provisions Eq. 13.3.4.1 [13.3-7]: 
Vb = kDmaxDD, 
where kDmax is the maximum effective stiffness of the isolation system at the design displacement, DD. 
Because the design displacement is conservatively based on minimum effective stiffness, Provisions Eq. 
13.3.4.1 implicitly induces an additional conservatism of a worst case combination mixing maximum and 
minimum effective stiffness in the same equation. Rigorous modeling of the isolation system for dynamic
FEMA 451, NEHRP Recommended Provisions: Design Examples 
11-12 
analyses precludes mixing of maximum and minimum stiffness in the same analysis (although separate 
analyses are typically required to determine bounding values of both displacement and force). 
Design earthquake response is reduced by a modest factor for design of the superstructure above the 
isolation interface, as given by Provisions Eq. 13.3.4.2 [13.3-8]: 
b Dmax D 
s 
I I 
V V k D 
R R 
= = 
The reduction factor, RI, is defined as three-eighths of the R factor for the seismic-force-resisting system 
of the superstructure, as specified in Provisions Table 5.2.2 [4.3-1], with an upper-bound value of 2.0. A 
relatively small RI factor is intended to keep the superstructure essentially elastic for the design 
earthquake (i.e., keep earthquake forces at or below the true strength of the seismic-force-resisting 
system). The Provisions also impose three limits on design forces that require the value of Vs to be at 
least as large as each of: 
1. The shear force required for design of a conventional, fixed-base structure of period TD. 
2. The shear force required for wind design, and/or 
3. A factor of 1.5 times the shear force required for activation of the isolation system. 
These limits seldom govern design but reflect principles of good design. In particular, the third limit is 
included in the Provisions to ensure that isolation system displaces significantly before lateral forces 
reach the strength of the seismic-force-resisting system. 
For designs using the ELF procedure, the lateral forces, Fx, must be distributed to each story over the 
height of the structure, assuming an inverted triangular pattern of lateral load (Provisions Eq.13.3.5 [13.3- 
9]): 
1 
s x x 
x n 
i i 
i
F V w h 
w h 
= 
= 
S 
Because the lateral displacement of the isolated structure is dominated by isolation system displacement, 
the actual pattern of lateral force in the isolated mode of response is distributed almost uniformly over 
height. The Provisions require an inverted triangular pattern of lateral load to capture possible 
higher-mode effects that might be missed by not modeling superstructure flexibility. Rigorous modeling 
of superstructure flexibility for dynamic analysis would directly incorporate higher-mode effects in the 
results. 
Example plots of the design displacement, DD, total maximum displacement, DTM, and design forces for 
the isolation system, Vb, and the superstructure, Vs (RI = 2), are shown in Figure 11.3-3 as functions of the 
effective period of the isolation system. The figure also shows the design base shear required for a 
conventional building, V (R/I = 5). The example plots are for a building assigned to Seismic Design 
Category D with a one-second spectral acceleration parameter, SD1, equal to 0.6, representing a stiff soil 
site (Site Class D) located in a region of high seismicity but not close to an active fault. In this example, 
the isolation system is assumed to have 20 percent effective damping (at all amplitudes of interest) and 
building geometry is assumed to require 25 percent additional displacement (at corners/edges) due to the 
requisite 5 percent accidental eccentricity.
Chapter 11, Seismically Isolated Structures 
11-13 
0.00 
0.05 
0.10 
0.15 
0.20 
0.25 
0.30 
0.35 
0.40 
1 1.5 2 2.5 3 3.5 4 
Effective period of isolation system (seconds) 
Design base shear (fraction of W) 
0
5
10 
15 
20 
25 
30 
35 
40 
Isolation system displacement (in.) 
SD1=0.6 BD=1.5 DTMDM=1.25 
Vb 
Vs(RI =2) 
V(R I =5) 
DTM 
DD 
Figure 11.3-3 Isolation system displacement and shear force as function of period (1.0 in. = 25.4 mm). 
The plots in Figure 11.3-3 illustrate the fundamental trade off between displacement and force as a 
function of isolation system displacement. As the period is increased, design forces decrease and design 
displacements increase linearly. Plots like those shown in Figure 11.3-3 can be constructed during 
conceptual design once site seismicity and soil conditions are known (or are assumed) to investigate trial 
values of effective stiffness and damping of the isolation system. In this particular example, an isolation 
system with an effective period falling between 2.5 and 3.0 seconds would not require more than 22 in. of 
total maximum displacement capacity (assuming TM # 3.0 seconds). Design force on the superstructure 
would be less than about eight percent of the building weight (assuming TD $ 2.5 seconds and RI $ 2.0). 
11.4 DYNAMIC LATERAL RESPONSE PROCEDURE 
While the ELF procedure equations are useful tools for preliminary design of the isolations system, the 
Provisions requires a dynamic analysis for most isolated structures. Even where not strictly required by 
the Provisions, the use of dynamic analysis (usually time history analysis) to verify the design is 
common. 
11.4.1 Minimum Design Criteria 
The Provisions encourages the use of dynamic analysis but recognize that along with the benefits of more 
complex models and analyses also comes an increased chance of design error. To avoid possible under 
design, the Provisions establishes lower-bound limits on results of dynamic analysis used for design. 
The limits distinguish between response spectrum analysis (a linear, dynamic method) and time history 
analysis (a nonlinear, dynamic method). In all cases, the lower-bound limit on dynamic analysis is 
established as a percentage of the corresponding design parameter calculated using the ELF procedure 
equations. Table 11.4-1 summarizes the percentages that define lower-bound limits on dynamic analysis.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
11-14 
Table 11.4-1 Summary of Minimum Design Criteria for Dynamic Analysis 
Design Parameter Response 
Spectrum Analysis 
Time History 
Analysis 
Total design displacement, DTD 90% DTD 90% DTD 
Total maximum displacement, DTM 80% DTM 80% DTM 
Design force on isolation system, Vb 90% Vb 90% Vb 
Design force on irregular superstructure, Vs 100% Vs 80% Vs 
Design force on regular superstructure, Vs 80% Vs 60% Vs 
The Provisions permits more liberal drift limits when the design of the superstructure is based on dynamic 
analysis. The ELF procedure drift limits of 0.010hsx are increased to 0.015hsx for response spectrum 
analysis and to 0.020hsx for time history analysis (where hsx is the story height at level x). Usually a stiff 
system (e.g., braced frames) is selected for the superstructure and drift demand is typically less than about 
0.005hsx. Provisions Sec. 13.4.7.4 [13.4.4] requires an explicit check of superstructure stability at the 
maximum considered earthquake displacement if the design earthquake story drift ratio exceeds 0.010/RI. 
11.4.2 Modeling Requirements 
As for the ELF procedure, the Provisions requires the isolation system to be modeled for dynamic 
analysis using stiffness and damping properties that are based on tests of prototype isolator units. 
Additionally, dynamic analysis models are required to account for: 
1. Spatial distribution of individual isolator units, 
2. Effects of actual and accidental mass eccentricity, 
3. Overturning forces and uplift of individual isolator units, and 
4. Variability of isolation system properties (due to rate of loading, etc.). 
The Provisions requires explicit nonlinear modeling of elements if time history analysis is used to justify 
design loads less than those permitted for ELF or response spectrum analysis. This option is seldom 
exercised and the superstructure is typically modeled using linear elements and conventional methods. 
Special modeling concerns for isolated structures include two important and related issues: the uplift of 
isolator units, and the P-delta effects on the isolated structure. Isolator units tend to have little or no 
ability to resist tension forces and can uplift when earthquake overturning (upward) loads exceed factored 
gravity (downward) loads. Local uplift of individual elements is permitted (Provisions Sec. 13.6.2.7 
[13.2.5.7]) provided the resulting deflections do not cause overstress or instability. To calculate uplift 
effects, gap elements may be used in nonlinear models or tension may be released manually in linear 
models. 
The effects of P-delta loads on the isolation system and adjacent elements of the structure can be quite 
significant. The compression load, P, can be large due to earthquake overturning (and factored gravity 
loads) at the same time that large displacements occur in the isolation system. Computer analysis 
programs (most of which are based on small-deflection theory) may not correctly calculate P-delta 
moments at the isolator level in the structure above or in the foundation below. Figure 11.4-1 illustrates 
moments due to P-delta effects (and horizontal shear loads) for an elastomeric bearing isolation system 
and a sliding isolation system. For the elastomeric system, the P-delta moment is split one-half up and 
one-half down. For the sliding system, the full P-delta moment is applied to the foundation below (due to 
the orientation of the sliding surface). A reverse (upside down) orientation would apply the full P-delta 
moment on the structure above.
Chapter 11, Seismically Isolated Structures 
11-15 
C 
A
B D 
. 
H2 
H1 
H4 
H3 
Elastomeric Isolator Sliding Isolator 
V 
P 
V 
P 
P 
V 
V 
P 
. 
Figure 11.4-1 Moments due to horizontal shear and P-delta effects. 
1
2 
3 
4 
2
2 
A
B
C
D 
M P VH 
M P VH 
M VH 
M P VH 
. 
= + 
. 
= + 
=
= . + 
For time history analysis, nonlinear force-deflection characteristics of isolator units are explicitly modeled 
(rather than using effective stiffness and damping). Force-deflection properties of isolator units are 
typically approximated by a bilinear, hysteretic curve whose properties can be accommodated by 
commercially available nonlinear structural analysis programs. Such bilinear hysteretic curves should 
have approximately the same effective stiffness and damping at amplitudes of interest as the true 
force-deflection characteristics of isolator units (as determined by prototype testing). 
Figure 11.4-2 shows a bilinear idealization of the response of a typical nonlinear isolator unit. Figure 
11.4-2 also includes simple equations defining the yield point (Dy, Fy) and end point (D, F) of a bilinear 
approximation that has the same effective stiffness and damping as the true curve (at a displacement, D).
FEMA 451, NEHRP Recommended Provisions: Design Examples 
11-16 
Force 
Displacement 
Eloop 
F 
F 
D 
D 
4( ) 
eff
y y loop 
F k 
D
DF F D E 
 
-  
0.32 (using in inches) 
0.637 
eff 
y y 
eff 
T DW D 
F
DF F D 
F D 
 
= 
. - . 
= . . 
. . 
At the displacement of interest: 
Dy 
Fy 
Figure 11.4-2 Bilinear idealization of isolator unit behavior. 
11.4.3 Response Spectrum Analysis 
Response spectrum analysis requires that isolator units be modeled using amplitude-dependent values of 
effective stiffness and damping that are the same as those for the ELF procedure. The effective damping 
of the isolated modes of response is limited to 30 percent of critical. Higher modes of response are 
usually assumed to have five percent dampinga value of damping appropriate for the superstructure, 
which remains essentially elastic. As previously noted, maximum and minimum values of effective 
stiffness are typically used to individually capture maximum displacement of the isolation system and 
maximum forces in the superstructure. Horizontal loads are applied in the two orthogonal directions, and 
peak response of the isolation system and other structural elements is determined using the 100 percent 
plus 30 percent combination method. 
11.4.4 Time History Analysis 
Time history analysis with explicit modeling of nonlinear isolator units is commonly used for the 
evaluation of isolated structures. Where at least seven pairs of time history components are employed, the 
values used in design for each response parameter of interest may be the average of the corresponding 
analysis maxima. Where fewer pairs are used (with three pairs of time history components being the 
minimum number permitted), the maximum value of each parameter of interest must be used for design. 
The time history method is not a particularly useful design tool due to the complexity of results, the 
number of analyses required (e.g., to account for different locations of eccentric mass), the need to 
combine different types of response at each point in time, etc. It should be noted that while Provisions 
Chapter 5 does not require consideration of accidental torsion for either the linear or nonlinear response 
history procedures, Chapter 13 does require explicit consideration of accidental torsion, regardless of the 
analysis method employed. Time history analysis is most useful when used to verify a design by
Chapter 11, Seismically Isolated Structures 
11-17 
Figure 11.5-1 Three-dimensional model of the structural system. 
checking a few key design parameters, such as: isolation displacement, overturning loads and uplift, and 
story shear force. 
11.5 EMERGENCY OPERATIONS CENTER USING ELASTOMERIC BEARINGS, 
SAN FRANCISCO, CALIFORNIA 
This example features the seismic isolation of a hypothetical emergency operations center (EOC), located 
in the center of San Francisco, California, an area of very high seismicity. Using high-damping rubber 
bearings, other types of isolators could be designed to have comparable response properties. Isolation is 
an appropriate design strategy for EOCs and other buildings where the goal is to limit earthquake damage 
and protect facility function. The example illustrates the following design topics: 
1. Determination of seismic design parameters, 
2. Preliminary design of superstructure and isolation systems (using the ELF procedure), 
3. Dynamic analysis of seismically isolated structures, and 
4. Specification of isolation system design and testing criteria. 
While the example includes development of the entire structural system, the primary focus is on the 
design and analysis of the isolation system. Examples in other chapters may be referred to for more 
in-depth descriptions of the provisions governing detailed design of the superstructure (i.e., the structure 
above the isolation system) and the foundation. 
11.5.1 System Description 
This EOC is a three-story, steel-braced frame structure with a large, centrally located mechanical 
penthouse. Story heights of 15 ft at all floors accommodate computer access floors and other 
architectural and mechanical systems. The roof and penthouse roof decks are designed for significant live 
load to accommodate a helicopter-landing pad and meet other functional requirements of the EOC. Figure 
11.5-1 shows the three-dimensional model of the structural system.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
11-18 
30'-0" 30'-0" 30'-0" 30'-0" 
30'-0" 30'-0" 30'-0" 30'-0" 30'-0" 30'-0" 
Figure 11.5-2 Typical floor framing plan (1.0 ft = 0.3048 m). 
30'-0" 30'-0" 30'-0" 30'-0" 
30'-0" 30'-0" 30'-0" 30'-0" 30'-0" 30'-0" 
Figure 11.5-3 Penthouse roof framing plan (1.0 ft = 0.3048 m). 
The structure (which is regular in configuration) has plan dimensions of 120 ft. by 180 ft. at all floors 
except for the penthouse, which is approximately 60 ft by 120 ft in plan. Columns are spaced at 30 ft in 
both directions. This EOCs relatively large column spacing (bay size) is used to reduce the number of 
isolator units for design economy and to increase gravity loads on isolator units for improved earthquake 
performance. Figures 11.5-2 and 11.5-3, are framing plans for the typical floor levels (1, 2, 3, and roof) 
and the penthouse roof.
Chapter 11, Seismically Isolated Structures 
11-19 
Penthouse Roof 
Roof 
Third Floor 
Second Floor 
First Floor 
Base 
Figure 11.5-4 Longitudinal bracing elevation (Column Lines B and D). 
Penthouse Roof 
Roof 
Third Floor 
Second Floor 
First Floor 
Base 
a) b) 
Figure 11.5-5 Transverse bracing elevations: (a) on Column Lines 2 and 6 and ( b) on Column Line 4. 
The vertical load carrying system consists of concrete fill on steel deck floors, roofs supported by steel 
beams at 10 ft on center, and steel girders at column lines. Isolator units support the columns below the 
first floor. The foundation is a heavy mat (although the spread footings or piles could be used depending 
on the soil type, depth to the water table, and other site conditions). 
The lateral system consists of a symmetrical pattern of concentrically braced frames. These frames are 
located on Column Lines B and D in the longitudinal direction, and on Column Lines 2, 4 and 6 in the 
transverse direction. Figures 11.5-4 and 11.5-5, respectively, show the longitudinal and transverse 
elevations. Braces are specifically configured to reduce the concentration of earthquake overturning, and 
uplift loads on isolator units by: 
1. Increasing the continuous length of (number of) braced bays at lower stories, 
2. Locating braces at interior (rather than perimeter) column lines (more hold-down weight), and 
3. Avoiding common end columns for transverse and longitudinal bays with braces. 
The isolation system is composed of 35 identical elastomeric isolator units, located below columns. The 
first floor is just above grade, and the isolator units are approximately 3 ft below grade to provide
FEMA 451, NEHRP Recommended Provisions: Design Examples 
11-20 
clearance below the first floor, for construction and maintenance personnel. A short retaining wall 
borders the perimeter of the facility and provides 30 in. of moat clearance for lateral displacement of 
the isolated structure. Access to the EOC is provided at the entrances by segments of the first floor slab, 
which cantilever over the moat. 
Girders at the first-floor column lines are much heavier than the girders at other floor levels, and have 
moment-resisting connections to columns. These girders stabilize the isolator units by resisting moments 
due to vertical (P-delta effect) and horizontal (shear) loads. Column extensions from the first floor to the 
top plates of the isolator units are stiffened in both horizontal directions, to resist these moments and to 
serve as stabilizing haunches for the beam-column moment connections. 
11.5.2 Basic Requirements 
11.5.2.1 Specifications 
General: 1997 Uniform Building Code (UBC) 
Seismic: NEHRP Recommended Provisions 
11.5.2.2 Materials 
Concrete: 
Strength fc ' = 3 ksi 
Weight (normal) . =150 pcf 
Steel: 
Columns Fy = 50 ksi 
Primary first-floor girders (at column lines) Fy = 50 ksi 
Other girders and floor beams Fy = 36 ksi 
Braces Fy = 46 ksi 
Steel deck: 3-in.-deep, 20-gauge deck 
Seismic isolator units (high-damping rubber): 
Maximum long-term-load (1.2D + 1.6L) face pressure, sLT 1,400 psi 
Maximum short-term-load (1.5D + 1.0L + QMCE) face pressure, sST 2,800 psi 
Minimum bearing diameter (excluding protective cover) 1.25DTM 
Minimum rubber shear strain capacity (isolator unit), .max 300 percent 
Minimum effective horizontal shear modulus, Gmin 65 - 110 psi 
Third cycle at . = 150 percent (after scragging/recovery) 
Maximum effective horizontal shear modulus, Gmax 1.3  Gmin 
First cycle at . = 150 percent (after scragging/recovery) 
Minimum effective damping at 150 percent rubber shear strain, eff 15 percent 
11.5.2.3 Gravity Loads 
Dead loads: 
Main structural elements (slab, deck, and framing) self weight
Chapter 11, Seismically Isolated Structures 
11-21 
Miscellaneous structural elements (and slab allowance) 10 psf 
Architectural facades (all exterior walls) 750 plf 
Roof parapets 150 plf 
Partitions (all enclosed areas) 20 psf 
Suspended MEP/ceiling systems and supported flooring 15 psf 
Mechanical equipment (penthouse floor) 50 psf 
Roofing 10 psf 
Reducible live loads: 
Floors (1-3) 100 psf 
Roof decks and penthouse floor 50 psf 
Live load reduction: 
The 1997 UBC permits area-based live load reduction, of not more than 40 percent for elements with 
live loads from a single story (e.g., girders), and not more than 60 percent for elements with live loads 
from multiple stories (e.g., axial component of live load on columns at lower levels and isolator 
units). 
EOC weight (dead load) and live load: 
Penthouse roof WPR = 965 kips 
Roof (penthouse) WR = 3,500 kips 
Third floor W3 = 3,400 kips 
Second floor W2 = 3,425 kips 
First floor W1 = 3,425 kips 
Total EOC weight W = 14,715 kips 
(See Chapter 1 for a discussion of live load contributions to the seismic weight.) 
Live load (L) above isolation system L = 7,954 kips 
Reduced live load (RL) above isolation system RL = 3,977 kips 
Table 11.5-1 Gravity Loads on Isolator Units* 
Dead/live loads (kips) 
Column 
line 1 2 3 4 
ABC 
182/73 
336/166 
295/149 
349/175 
570/329 
520/307 
303/153 
606/346 
621/356 
345/173 
539/309 
639/358 
1.0 kip = 4.45 kN. 
* Loads at Column Lines 5, 6 and 7 (not shown) are the same as those at Column Lines 3, 2, and 1, 
respectively; loads at Column Lines D and E (not shown) are the same as those at Column Lines B 
and A, respectively.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
11-22 
11.5.2.4 Provisions Parameters 
11.5.2.4.1 Performance Criteria (Provisions Sec. 1.3 [1.2 and 1.3]) 
Designated Emergency Operation Center Seismic Use Group III 
Occupancy Importance Factor I = 1.5 (conventional) 
Occupancy Importance Factor (Provisions Chapter 13) I = 1.0 (isolated) 
Chapter 13 does not require use of the occupancy importance factor to determine the design loads on 
the structural system of an isolated building (that is, I = 1.0), but the component importance factor is 
required by Chapter 6, to determine seismic forces on components (Ip = 1.5 for some facilities). 
11.5.2.4.2 Ground Motion (Provisions Chapter 4 [3]) 
Site soil type (assumed) Site Class D 
Maximum considered earthquake (MCE) spectral acceleration at SS = 1.50 
short periods (Provisions Map 7) 
[The 2003 Provisions have adopted the 2002 USGS probabilistic seismic hazard maps, and the maps have 
been added to the body of the 2003 Provisions as figures in Chapter 3 (instead of the previously used 
separate map package.] 
Site coefficient (Provisions Table 4.1.2.4b [3.3-1]) Fa = 1.0 
MCE spectral acceleration adjusted for site class (FaSS) SMS = 1.50 
Design earthquake (DE) spectral acceleration (2/3SMS) SDS = 1.0 
MCE spectral acceleration at a period of 1 second (Provisions Map 8) S1 = 0.9 
On Map 8, the contour line of S1 = 0.9 runs approximately through the center of the San Francisco 
peninsula, with other San Francisco peninsula contour lines ranging from 0.6 (greatest distance from 
San Andreas Fault), to about 1.6 (directly on top of the San Andreas Fault). 
Site coefficient (Provisions Table 4.1.2.4b [3.3-2]) Fv = 1.5 
MCE spectral acceleration adjusted for site class (FvS1) SM1 = 1.35 
Design earthquake (DE) spectral acceleration (2/3SM1) SD1 = 0.9 
Seismic Design Category (Provisions Table 4.2.1b [1.4-2]) Seismic Design Category F 
[In the 2003 edition of the Provisions, the ground motion trigger for Seismic Design Categories E and F 
have been changed to S1 $0.60. No change would result for this example.]
Chapter 11, Seismically Isolated Structures 
11-23 
MCE 
80% 
of MCE 
DE 
80% 
of DE 
0.0 
0.2 
0.4 
0.6 
0.8 
1.0 
1.2 
1.4 
1.6
0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 
Period (seconds) 
Spectral Acceleration (g) 
Figure 11.5-6 Example design spectra (5 percent damping). 
11.5.2.4.3 Design Spectra (Provisions Sec. 13.4.4.1) 
Figure 11.5-6 plots design earthquake and maximum considered earthquake response spectra as 
constructed in accordance with the procedure of Provisions Sec. 13.4.4.1 [3.4] using the spectrum shape 
defined by Provisions Figure 4.1.2.6 [3.3-15]. [In the 2003 edition of Provisions, the shape of the design 
spectrum changes at period beyond TL, but no change would result for this example.] Provisions Sec. 
13.4.4.1[13.2.3.1] requires site-specific design spectra to be calculated for sites with S1 greater than 0.6 
(e.g., sites near active sources). Site-specific design spectra may be taken as less than 100 percent but not 
less than 80 percent of the default design spectra of Provisions Figure 4.1.2.6 [3.3-15]. 
For this example, site-specific spectra for the design earthquake and the maximum considered earthquake, 
were assumed to be 80 percent (at long periods) and 100 percent (at short periods) of the respective 
spectra shown in Figure 11.5-6. The 80 percent factor reflects a reduction in demand that could be 
achieved through a detailed geotechnical investigation of site soil conditions. In general, site-specific 
spectra for regions of high seismicity, with well defined fault systems (like those of the San Francisco 
Bay Area), would be expected to be similar to the default design spectra of Provisions Figure 4.1.2.6 [3.3- 
15]. 
11.5.2.4.4 Design Time Histories (Provisions Sec. 13.4.4.2 [13.2.3.2]) 
For time history analysis, Provisions Sec. 13.4.4.7 [13.4.2.3] requires no less than three pairs of 
horizontal ground motion time history components to be selected from actual earthquake records, and 
scaled to match either the design earthquake (DE) or the maximum considered earthquake (MCE) 
spectrum. The selection and scaling of appropriate time histories is usually done by an earth scientist, or 
a geotechnical engineer experienced in the seismology of the region; and takes the earthquake
FEMA 451, NEHRP Recommended Provisions: Design Examples 
11-24 
magnitudes, fault distances, and source mechanisms that influence hazards at the building site into 
consideration. 
For this example, records from the El Centro Array Station No. 6, the Newhall Fire Station, and the 
Sylmar Hospital are selected from those recommended by ATC-40 [ATC, 1996], for analysis of buildings 
at stiff soil sites with ground shaking of 0.2g or greater. These records, and the scaling factors required to 
match the design spectra are summarized in Table 11.5-2. These three records represent strong ground 
shaking recorded relatively close to fault rupture; and contain long-period pulses in the direction of 
strongest shaking, which can cause large displacement of the isolated structures. MCE scaling factors for 
the El Centro No. 6 and Sylmar records are 1.0, indicating that these records are used without 
modification for MCE time history analysis of the EOC. 
Provisions Sec. 13.4.4.2 [13.2.3.2] provides criteria for scaling earthquake records to match a target 
spectrum over the period range of interest, defined as 0.5TD to 1.25TM. In this example, TD and TM are 
both assumed to be 2.5 seconds, so the period range of interest is from 1.25 seconds to 3.125 seconds. 
For each period in this range, the average of the square-root-of-the-sum-of-the-squares (SRSS) 
combination of each pair of horizontal components of scaled ground motion, may not fall below the target 
spectrum by more 10 percent. The target spectrum is defined as 1.3 times the design spectrum of interest 
(either the DE or the MCE spectrum). 
Figure 11.5-7 compares the spectra of scaled time history components with the spectrum for the design 
earthquake. Rather than comparing the average of the SRSS spectra with 1.3 times the design spectrum, 
as indicated in the Provisions, Figure 11.5-7 shows the (unmodified) design spectrum and the average of 
the SRSS spectra divided by 1.3. The effect is the same, but the method employed eliminates one level of 
obscurity and permits direct comparison of the calculated spectra without additional manipulation. A 
comparison of maximum, considered earthquake spectra would look similar (with values that were 1.5 
times larger). The fit is very good at long periods - the period range of interest for isolated structures. At 
2.5 seconds, the spectrum of the scaled time histories has the same value as the target spectrum. Between 
1.25 seconds and 3.125 seconds, the spectrum of the scaled time histories is never less than the target 
spectrum by more than 10 percent. At short periods, the spectrum of scaled time history motions are 
somewhat greater than the target spectrum - a common result of scaling real time histories to match the 
long-period portion of the design spectrum. 
Figure 11.5-7 also includes plots of upper-bound (maximum demand) and lower-bound (minimum 
demand) envelopes of the spectra of the six individual components of scaled time histories. The 
maximum demand envelope illustrates that the strongest direction (component) of shaking of at least one 
of the three scaled time histories is typically about twice the site-specific spectrum in the period range of 
the isolated structure - consistent with strong ground shaking recorded near sources in the fault normal 
direction. 
Table 11.5-2 Earthquake Time History Records and Scaling Factors 
Record 
No. 
Earthquake Source and Recording Station Scaling factor 
Year Earthquake Station (owner) DE MCE 
1 1979 Imperial Valley, CA El Centro Array Station 6 (USGS) 0.67 1.0 
2 1994 Northridge, CA Newhall Fire Station (CDMG) 1.0 1.5 
3 1994 Northridge, CA Sylmar Hospital (CDMG) 0.67 1.0
Chapter 11, Seismically Isolated Structures 
11-25 
0.0 
0.5 
1.0 
1.5 
2.0 
2.5 
0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 
Period (seconds) 
Spectral Acceleration (g) 
Target spectrum 
Average of SRSS spectra divided by 1.3 
Maximum and minimum demands 
Period range of interest for 
scaling time history records 
Figure 11.5-7 Comparison of design earthquake spectra. 
11.5.2.5 Structural Design Criteria 
11.5.2.5.1 Design Basis (Provisions Sec. 5.2 and 13.2 [4.3 and 13.2]) 
Seismic-force-resisting system. Special steel concentrically 
braced frames (height < 100 ft) 
Response modification factor, R (Provisions Table 5.2.2 [4.3-1] 6 (conventional) 
Response modification factor for design of the superstructure, RI 2 (isolated) 
( Provisions Sec. 13.3.4.2 [13.3.3.2], 3/8R # 2) 
Plan irregularity (of superstructure) - (Provisions Table 5.2.3.2) None 
Vertical irregularity (of superstructure) - from Table 5.2.3.3 None 
Lateral response procedure (Provisions Sec. 13.2.5.2 [13.2.4.1], S1 > 0.6) Dynamic analysis 
Redundancy/reliability factor - (Provisions Sec. 5.2.4.2 [4.3.3]) . > 1.0 (conventional) 
. = 1.0 (isolated) 
Provisions Sec. 5.2.4.2 [4.3.3] requires the use of a calculated . value, which would be greater than 
1.0 for a conventional structure with a brace configuration similar to the superstructure of the 
base-isolated EOC. However, in the author's opinion, the use of RI equal to 2.0 (rather than R equal
FEMA 451, NEHRP Recommended Provisions: Design Examples 
11-26 
to 6) as required by Provisions Sec. 13.3.4.2 precludes the need to further increase superstructure 
design forces for redundancy/reliability. Future editions of the Provisions should address this issue. 
[The redundancy factor is changed substantially in the 2003 Provisions. However, the rationale set forth 
above by the author still holds; use of . = 1.0 is reasonable since RI = 2.0.] 
11.5.2.5.2 Horizontal Earthquake Loads and Effects (Provisions Chapters 5 and 13) 
Design earthquake (acting in either the X or Y direction) DE (site specific) 
Maximum considered earthquake (acting in either the X or Y direction) MCE (site specific) 
Mass eccentricity - actual plus accidental 0.05b = 6 ft (z to X axis) 
0.05d = 9 ft (z to Y axis) 
The superstructure is symmetric about both principal axes, however, the placement of the braced 
frames results in a ratio of maximum corner displacement to average displacement of 1.25, exceeding 
the threshold of 1.2 per the definition of the Provisions. If the building were not on isolators, the 
accidental torsional moment would need to be increased from 5 percent to 5.4 percent of the building 
dimension. The input to the superstructure is controlled by the isolation system, and it is the authors 
opinion that the amplification of accidental torsion is not necessary for such otherwise regular 
structures. Future editions of the Provisions should address this issue. Also refer to the discussion of 
analytical modeling of accidental eccentricities in Guide Chapter 1. 
Superstructure design (reduced DE response) QE = QDE/2 = DE/2.0 
Isolation system and foundation design (unreduced DE response) QE = QDE = DE/1.0 
Check of isolation system stability (unreduced MCE response) QE = QMCE = MCE/1.0 
11.5.2.5.3 Combination of Horizontal Earthquake Load Effects (Provisions Sec. 5.25.2.3 [4.4.2.3] and 
13.4.6.3 [13.4.2.2]) 
Response due earthquake loading in the X and Y directions QE = Max (1.0QEX+0.3QEY, 
0.3QEX+1.0QEY) 
In general, the horizontal earthquake load effect, QE, on the response parameter of interest is 
influenced by only one direction of horizontal earthquake load, and QE = QEX or QE = QEY. 
Exceptions include vertical load on isolator units due to earthquake overturning forces. 
11.5.2.5.4 Combination of Horizontal and Vertical Earthquake Load Effects (Provisions Sec. 5.2.7 
[4.2.2.1]) 
Design earthquake (.QE  0.2SDSD) E = QE  0.2D 
Maximum considered earthquake (.QE  0.2SMSD) E = QE  0.3D 
11.5.2.5.5 Superstructure Design Load Combinations (UBC Sec. 1612.2, using RI = 2) 
Gravity loads (dead load and reduced live load) 1.2D + 1.6L 
Gravity and earthquake loads (1.2D + 0.5L + 1.0E) 1.4D + 0.5L + QDE/2
Chapter 11, Seismically Isolated Structures 
11-27 
Gravity and earthquake loads (0.9D - 1.0E) 0.7D - QDE/2 
11.5.2.5.6 Isolation System and Foundation Design Load Combinations (UBC Sec. 1612.2) 
Gravity loads (for example, long term load on isolator units) 1.2D + 1.6L 
Gravity and earthquake loads (1.2D + 0.5L + 1.0E) 1.4D + 0.5L + QDE 
Gravity and earthquake loads (0.9D - 1.0E) 0.7D - QDE 
11.5.2.5.7 Isolation System Stability Load Combinations (Provisions Sec. 13.6.2.6 [13.2.5.6]) 
Maximum short term load on isolator units (1.2D + 1.0L + |E|) 1.5D + 1.0L + QMCE 
Minimum short term load on isolator units (0.8D - |E|) 0.8D - QMCE 
Note that in the above combinations, the vertical earthquake load (0.2SMSD) component of |E| is 
included in the maximum (downward) load combination, but excluded from the minimum (uplift) 
load combination. It is the author's opinion that vertical earthquake ground shaking is of a dynamic 
nature, changing direction too rapidly to affect appreciably uplift of isolator units and need not be 
used with the load combinations of Provisions Sec. 13.6.2.6 [13.2.5.6] for determining minimum 
(uplift) vertical loads on isolator units due to the MCE. Future editions of the Provisions should 
explicitly address this issue. 
11.5.3 Seismic Force Analysis 
11.5.3.1 Basic Approach to Modeling 
To expedite calculation of loads on isolator units, and other elements of the seismic-force-resisting 
system, a 3-dimensional model of the EOC is developed and analyzed using the ETABS computer 
program (CSI, 1999). While there are a number of programs to choose from, ETABS (version 7.0) is 
selected for this example since it permits the automated release of tension in isolator units subject to uplift 
and has built-in elements for modeling other nonlinear properties of isolator units. Arguably, all of the 
analyses performed by the ETABS program could be done by hand, or by spreadsheet calculation (except 
for confirmatory time history analyses). 
The ETABS model is used to perform the following types of analyses and calculations: 
1. Dead load weight and live loads for the building--calculate maximum long-term load on isolator units 
(Guide Table 11.5-1) 
2. ELF procedure for gravity and reduced design earthquake loads--design superstructure (ignoring 
uplift of isolator units) 
3. Nonlinear static analysis with ELF loads for gravity and unreduced design earthquake loads--design 
isolation system and foundation (considering uplift of isolator units) 
4. Nonlinear static analysis with ELF loads - gravity and unreduced design earthquake loads--calculate 
maximum short-term load (downward force) on isolator units (Guide Table 11.5-5) and calculate 
minimum short-term load (downward force) of isolator units (Guide Table 11.5-6)
FEMA 451, NEHRP Recommended Provisions: Design Examples 
11-28 
5. Nonlinear static analysis with ELF loads for gravity and unreduced MCE loads--calculate maximum 
short-term load (downward force) on isolator units (Guide Table 11.5-7) and calculate minimum 
short-term load (uplift displacement) of isolator units (Guide Table 11.5-8) 
6. Nonlinear time history analysis - gravity and scaled DE or MCE time histories--verify DE 
displacement of isolator units and design story shear (Guide Table 11.5-10), verify MCE 
displacement of isolator units (Guide Table 11.5-11), verify MCE short-term load (downward force) 
on isolator units (Guide Table 11.5-12), and verify MCE short-term load (uplift displacement) of 
isolator units (Guide Table 11.5-13). 
The Provisions requires a response spectrum analysis for the EOC (see Guide Table 11.2-1). In general, 
the response spectrum method of dynamic analysis is considered sufficient for facilities that are located at 
a stiff soil site, which have an isolation system meeting the criteria of Provisions Sec. 13.2.5.2.7, Item 7 
[13.2.4.1, Item 7]. However, nonlinear static analysis is used for the design of the EOC, in lieu of 
response spectrum analysis, to permit explicit modeling of uplift of isolator units. For similar reasons, 
nonlinear time history analysis is used to verify design parameters. 
11.5.3.2 Detailed Modeling Considerations 
Although a complete description of the ETABS model is not possible, key assumptions and methods used 
to model elements of the isolation system and superstructure are described below. 
11.5.3.2.1 Mass Eccentricity 
Provisions Sec. 13.4.5.2 [13.4.1.1] requires consideration of mass eccentricity. Because the building in 
the example is doubly symmetric, there is no actual eccentricity of building mass (but such would be 
modeled if the building were not symmetric). Modeling of accidental mass eccentricity would require 
several analyses, each with the building mass located at different eccentric locations (for example, four 
quadrant locations in plan). This is problematic, particularly for dynamic analysis using multiple time 
history inputs. In this example, only a single (actual) location of mass eccentricity is considered, and 
calculated demands are increased moderately for the design of the seismic-force-resisting system, and 
isolation system (for example, peak displacements calculated by dynamic analysis are increased by 10 
percent for design of the isolation system). 
11.5.3.2.2 P-delta Effects 
P-delta moments in the foundation and the first floor girders just above isolator units due to the large 
lateral displacement of the superstructure are explicitly modeled. The model distributes half of the 
P-delta moment to the structure above, and half of the P-delta moment to the foundation below the 
isolator units. ETABS (version 7.0) permits explicit modeling of the P-delta moment, but most computer 
programs (including older versions of ETABS) do not. Therefore, the designer must separately calculate 
these moments and add them to other forces for the design of affected elements. The P-delta moments are 
quite significant, particularly at isolator units that resist large earthquake overturning loads along lines of 
lateral bracing. 
11.5.3.2.5 Isolator Unit Uplift 
Provisions Sec. 13.6.2.7 [13.2.5.7] permits local uplift of isolator units provided the resulting deflections 
do not cause overstress of isolator units or other structural elements. Uplift of some isolator units is likely 
(for unreduced earthquake loads) due to the high seismic demand associated with the site. Accordingly, 
isolator units are modeled with gap elements that permit uplift when the tension load exceeds the tensile 
capacity of an isolator unit. Although most elastomeric bearings can resist some tension stress before
Chapter 11, Seismically Isolated Structures 
11-29 
yielding (typically about 150 psi), isolator units are assumed to yield as soon as they are loaded in 
tension, producing larger estimates of uplift displacement and overturning loads on isolator units that are 
in compression. 
The assumption that isolator units have no tension capacity is not conservative, however, for design of the 
connections of isolator units to the structure above and the foundation below. The design of anchor bolts 
and other connection elements, must include the effects of tension in isolator units (typically based on a 
maximum stress of 150 psi). 
11.5.3.2.4 Bounding Values of Bilinear Stiffness of Isolator Units 
The design of elements of the seismic-force-resisting system is usually based on a linear, elastic model of 
the superstructure. When such models are used, Provisions Sec. 13.4.5.3.2 [13.4.1.2] requires that the 
stiffness properties of nonlinear isolation system components be based on the maximum effective stiffness 
of the isolation system (since this assumption produces larger earthquake forces in the superstructure). In 
contrast, the Provisions require that calculation of isolation system displacements be based on the 
minimum effective stiffness of the isolation system (since this assumption produces larger isolation 
system displacements). 
The concept of bounding values, as discussed above, applies to all of the available analysis methods. For 
the ELF procedure, the Provisions equations are based directly on maximum effective stiffness, kDmax and 
kMmax, are used for calculating design forces; and on minimum effective stiffness, kDmin and kDmax, are used 
for calculating design displacements. Where (nonlinear) time history analysis is used, isolators are 
explicitly modeled as bilinear hysteretic elements with upper or lower-bound stiffness curves. 
Upper-bound stiffness curves are used to verify the forces used for the design of the superstructure and 
lower-bound stiffness curves are used to verify design displacements of the isolation system. 
11.5.4 Preliminary Design Based on the ELF Procedure 
11.5.4.1 Calculation of Design Values 
11.5.4.1.1 Design Displacements 
Preliminary design begins with the engineer's selection of the effective period (and damping) of the 
isolated structure, and the calculation of the design displacement, DD. In this example, the effective 
period of the EOC facility (at the design earthquake displacement) is TD = 2.5 seconds; and the design 
displacement is calculated as follows, using Provisions Eq. 13.3.3.1: 
2 
(9.8) 0.9(2.5) 16.3 in. 
4 1.35 
D1 D 
D 
D 
D g S T 
p B 
=.. .. = = 
. . 
The 1.35 value of the damping coefficient, BD, is given in Provisions Table 13.3.3.1 [13.3-1] assuming 15 
percent effective damping at 16.3 in. of isolation system displacement. Effective periods of 2 to 3 
seconds and effective damping values of 10 to 15 percent are typical of high-damping rubber (and other 
types of) bearings. 
Stability of the isolation system must be checked for the maximum displacement, DM, which is calculated 
using Provisions Eq. 13.3.3.3 as follows: 
( ) 2 
9.8 1.35(2.5) 24.5 in. 
4 1.35 
M1 M 
M 
M 
Dg S T 
p B 
=.. .. = = 
. .
FEMA 451, NEHRP Recommended Provisions: Design Examples 
11-30 
For preliminary design, the effective period and effective damping at maximum displacement are assumed 
to be the same as the values at the design displacement. While both the effective period and damping 
values may reduce slightly at larger rubber strains, the ratio of the two parameters tend to be relatively 
consistent. 
The total displacement of specific isolator units (considering the effects of torsion) is calculated based on 
the plan dimensions of the building, the total torsion (due to actual, plus accidental eccentricity), and the 
distance from the center of resistance of the building to the isolator unit of interest. Using Provisions Eq. 
13.3.3.5-1 and 13.3.3.5-2 [13.3-5 and 13.3-6], the total design displacement, DTD, and the total maximum 
displacement, DTM, of isolator units located on Column Lines 1 and 7 are calculated for the critical 
(transverse) direction of earthquake load as follows: 
2 2 2 2 
1 12 16.3 1 90 12(0.05)(180) 16.3(1.21) 19.7 in. 
120 180 TD D 
D D y e 
b d 
= ...+ ...+ ......= ...+ ... + ......= = 
2 2 2 2 
1 12 24.5 1 90 12(0.05)(180) 24.5(1.21) 29.6 in. 
120 180 TM M 
D D y e 
b d 
= ...+ ...+ ......= ...+ ... + ......= = 
The equations above assume that mass is distributed in plan in proportion to isolation system stiffness and 
shifted by 5percent, providing no special resistance to rotation of the building on the isolation system. In 
fact, building mass is considerably greater toward the center of the building, as shown by the schedule of 
gravity loads in Table 11.5-1. The stiffness of the isolation system is uniform in plan (since all isolators 
are of the same size) providing significant resistance to dynamic earthquake rotation of the building. 
While the Provisions permit a reduction in the total displacements calculated using the ELF procedure 
(with proper substantiation of resistance to torsion), in this example the 21 percent increase is considered 
to be conservative for use in preliminary design and for establishing lower-bound limits on dynamic 
analysis results. 
11.5.4.1.2 Minimum and Maximum Effective Stiffness 
Provisions Eq. 13.3.3.2 [13.3-2] expresses the effective period at the design displacement in terms of 
building weight (dead load) and the minimum effective stiffness of the isolation system, kDmin. 
Rearranging terms and solving for minimum effective stiffness: 
2 
2 2 
4 114,715240 kips/in. 
9.8 2.5 Dmin 
D 
k W 
g T 
=.. p .. =.. .. = 
. . . . 
This stiffness is about 6.9 kips/in. for each of 35 identical isolator units. The effective stiffness can vary 
substantially from one isolator unit to another and from one cycle of prototype test to another. Typically, 
an isolator units effective stiffness is defined by a range of values for judging acceptability of prototype 
(and production) bearings. The minimum value of the stiffness range, kDmin, is used to calculate isolation 
system design displacements; the maximum value of the stiffness, kDmax, is used to define design forces. 
The variation in effective stiffness depends on the specific type of isolator, elastomeric compound, 
loading history, etc., but must, in all cases, be broad enough to comply with the Provisions Sec. 13.9.5.1 
[13.6.4.1]requirements that define maximum and minimum values of effective stiffness based on testing 
of isolator unit prototypes. Over the three required cycles of test at DD, the maximum value of effective 
stiffness (for example, at the first cycle) should not be more than about 30 percent greater than the 
minimum value of effective stiffness (for example, at the third cycle) to comply with Provisions Sec.
Chapter 11, Seismically Isolated Structures 
11-31 
13.9.4 [13.6.3]. On this basis, the maximum effective stiffness, kDmax, of the isolation system in this 
example, is limited to 312 kips/in. (that is, 1.3  240 kips/in.). 
The range of effective stiffness defined by kDmin and kDmax (as based on cyclic tests of prototype isolator 
units) does not necessarily bound all the possible variations in the effective stiffness of elastomeric 
bearings. Other possible sources of variation include: stiffness reduction, due to post-fabrication 
"scragging" of bearings (by the manufacturer), and partial recovery of this stiffness over time. 
Temperature and aging effects of the rubber material, and other changes in properties that can also occur 
over the design life of isolator units. (Elastomeric bearings are typically scragged immediately 
following molding and curing to loosen up the rubber molecules by application of vertical load.) 
Provisions Sec. 13.6.2.1 [13.2.5.1] requires that such variations in isolator unit properties be considered in 
design, but does not provide specific criteria. A report - Property Modification Factors for Seismic 
Isolation Bearings (Constantinou, 1999) - provides guidance for establishing a range of effective stiffness 
(and effective damping) properties that captures all sources of variation over the design life of the isolator 
units. The full range of effective stiffness has a corresponding range of effective periods (with different 
levels of spectral demand). The longest effective period (corresponding to the minimum effective 
stiffness) of the range would be used to define isolation system design displacements; and the shortest 
effective period (corresponding to the maximum effective stiffness) of the range would be used to define 
the design forces on the superstructure. 
11.5.4.1.3 Lateral Design Forces 
The lateral force required for the design of the isolation system, foundation, and other structural elements 
below the isolation system, is given by Provisions Eq. 13.3.4.1 [13.3-7]: 
Vb = kDmax DD = 312(16.3) = 5,100 kips 
The lateral force required for checking stability and ultimate capacity of elements of the isolation system, 
may be calculated as follows: 
VMCE = kDmax DM = 312(24.5) = 7,650 kips 
The (unreduced) base shear of the design earthquake is about 35 percent of the weight of the EOC, and 
the (unreduced) base shear of the MCE is just over 50 percent of the weight. In order to design the 
structure above the isolation system, the design earthquake base shear is reduced by the RI factor in 
Provisions Eq. 13.3.4.2 [13.3-8]: 
312(16.3) 2,550 kips 
2.0 
Dmax D 
s 
1 
Vk D 
R 
= = = 
This force is about 17 percent of the dead load weight of the EOC, which is somewhat less than, but 
comparable to, the force that would be required for the design of a conventional, fixed-base building of 
the same size and height, seismic-force-resisting system, and site seismic conditions. Story shear forces 
on the superstructure are distributed vertically over the height of the structure in accordance with 
Provisions Eq. 13.3.5 [13.3-9], as shown in Table 11.5-3. 
Table 11.5-3 Vertical Distribution of Reduced Design Earthquake Forces (DE/2)
FEMA 451, NEHRP Recommended Provisions: Design Examples 
11-32 
Story level, x Weight, wx 
(kips) 
Height above 
isolation 
system, hx (ft) 
Force, 
x x s 
x 
i i 
i 
F w h V 
w h 
= 
S 
(kips) 
Cumulative 
force (kips) 
Force divided 
by weight, 
x
x 
F
w 
Penthouse roof 965 64 370 370 0.38 
Roof 3,500 49 1,020 1,390 0.29 
Third floor 3,400 34 690 2,080 0.20 
Second floor 3,425 19 390 2,470 0.11 
First floor 3,425 4 80 2,550 0.023 
1.0 ft = 0.3048 m, 1.0 kip = 4.45 kN. 
Provisions Eq. 13.3.5 distributes lateral seismic design forces (DE/2) over the height of the building in an 
inverted, triangular, pattern as indicated by the ratio of Fx/wx, shown in Table 11.5-3. Because the 
superstructure is much stiffer laterally, than the isolation system, it tends to move as a rigid body in the 
first mode with a pattern of lateral seismic forces that is more uniformly distributed over the height of the 
building. The use of a triangular load pattern for design is intended to account for higher-mode response 
that may be excited due to flexibility of the superstructure. Provisions Eq. 13.3.5 [13.3-9] is also used to 
distribute forces over the height of the building for unreduced DE and MCE forces, as summarized in 
Table 11.5-4. 
Table 11.5-4 Vertical Distribution of Unreduced DE and MCE Forces 
Story level, x 
Design earthquake (DE) Maximum considered earthquake (MCE) 
Force 
(kips) 
Cumulative 
force (kips) 
Force 
divided by 
weight 
Force 
(kips) 
Cumulative 
force (kips) 
Force 
divided by 
weight 
Penthouse roof 740 740 0.76 1,110 1,110 1.14 
Roof 2,040 2,780 0.59 3,060 4,170 0.88 
Third floor 1,380 4,160 0.41 2,070 6,240 0.61 
Second floor 780 4,940 0.23 1,170 7,410 0.34 
First floor 160 5,100 0.047 240 7,650 0.071 
1.0 kip = 4.45 kN. 
11.5.4.1.4 Design Earthquake Forces for Isolator Units 
Tables 11.5-5 and 11.5-6 show the maximum and minimum downward forces for design of the isolator units. 
These forces are a result from the simultaneous application of unreduced design earthquake story forces, 
summarized in Table 11.5-4 and appropriate gravity loads to the model of the EOC. (See Guide Sec. 11.5.2.5 
for the design load combinations.) As described in Guide Sec. 11.5.2.5, loads are applied simultaneously in 
two horizontal directions. The tables report the results for both of the orientations: 100 percent in the X 
direction, plus 30 percent in the Y direction, and 30 percent in the X direction, plus 100 percent in the Y 
direction. Where the analyses indicate that certain isolator units could uplift during peak DE response (as 
indicated by zero downward force), the amount of uplift is small and would not appreciably affect the 
distribution of earthquake forces in the superstructure.
Chapter 11, Seismically Isolated Structures 
11-33 
Table 11.5-5 Maximum Downward Force (kips) for Isolator Design (1.4D + 0.5L + QDE)* 
Maximum downward force (kips) 
100%(X)  30%(Y) / 30%(X)  100%(Y) 
Column 
line 1 2 3 4 
A 347/348 661/892 522/557 652/856 
B 833/634 1,335/1,519 1,418/1,278 984/1,186 
C 537/502 939/890 1,053/1,046 1,074/1,070 
1.0 kip = 4.45 kN 
.*Forces at column lines 5, 6 and 7 (not shown) are the same as those at column lines 3, 2, and 1, respectively; loads at 
column lines D and E (not shown) are the same as those at column lines B and A, respectively. 
Table 11.5-6 Minimum Downward Force (kips) for Isolator Design (0.7D - QDE)* 
Maximum downward force (kips) 
100%(X)  30%(Y) / 30%(X)  100%(Y) 
Column 
line 1 2 3 4 
A 87/84 165/0 200/169 163/0 
B 0/123 11/0 23/69 280/59 
C 169/196 299/243 428/427 447/442 
1.0 kip = 4.45 kN 
* Forces at column lines 5, 6 and 7 (not shown) are the same as those at column lines 3, 2, and 1, respectively; loads at 
column lines D and E (not shown) are the same as those at column lines B and A, respectively. 
11.5.4.1.5 Maximum Considered Earthquake Forces and Displacements for Isolator Units 
Simultaneous application of the unreduced MCE story forces, as summarized in Table 11.5-4 and appropriate 
gravity loads to the model of the EOC, result in the maximum downward forces on isolator units shown in 
Guide Table 11.5-7, and the maximum uplift displacements shown in Table 11.5-8. The load orientations 
and MCE load combinations, are described in Guide Sec. 11.5.2.5. The tables report the results for both of 
the load orientations: 100 percent in the X direction, plus 30 percent in the Y direction, and 30 percent in the 
X direction, plus 100 percent in the Y direction. Since the nonlinear model assumes that the isolators have 
no tension capacity, the values given in Guide Table 11.5-8 are upper bounds on uplift displacements.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
11-34 
Table 11.5-7 Maximum Downward Force (kips) on Isolator Units (1.5D + 1.0L + QMCE)* 
Maximum downward force (kips) 
100%(X)  30%(Y) / 30%(X)  100%(Y) 
Column 
line 1 2 3 4 
A 444/445 829/1,202 645/710 819/1,146 
B 1,112/794 1,739/2,006 1,848/1,635 1,219/1,521 
C 680/618 1,171/1,091 1,298/1,282 1,316/1,307 
1.0 kip = 4.45 kN. 
* Forces at column lines 5, 6 and 7 (not shown) are the same as those at column lines 3, 2, and 1, respectively; loads at 
column lines D and E (not shown) are the same as those at column lines B and A, respectively. 
Table 11.5-8 Maximum Uplift Displacement (in.) of Isolator Units (0.8D - QMCE)* 
Maximum uplift displacement (in.) 
100%(X)  30%(Y) / 30%(X)  100%(Y) 
Column 
line 1 2 3 4 
A No uplift 0.00/0.94 No uplift 0.00/0.50 
B 0.54/0.00 0.19/0.45 0.13/0.00 0.00/0.14 
C No uplift No uplift No uplift No uplift 
1.0 in. = 25.4 mm. 
* Displacements at column lines 5, 6 and 7 (not shown) are the same as those at column lines 3, 2, and 1, respectively; 
displacements at column lines D and E (not shown) are the same as those at column lines B and A, respectively. 
11.5.4.1.6 Limits on Dynamic Analysis 
The displacements and forces determined by the ELF procedure provide a basis for expeditious assessment 
of size and capacity of isolator units and the required strength of the superstructure. The results of the ELF 
procedure also establish limits on design parameters when dynamic analysis is used as the basis for design. 
Specifically, the total design displacement, DTD, and the total maximum displacement of the isolation system, 
DTM, determined by dynamic analysis cannot be less than 90 percent and 80 percent, respectively, of the 
corresponding ELF procedure values: 
DTD, dynamic $0.9DTD, ELF = 0.9(19.7) = 17.7 in. 
DTM, dynamic $0.8DTM, ELF = 0.8(29.6) = 23.7 in. 
The superstructure, if regular, can also be designed for less base shear, but not less than 80 percent of the base 
shear from the ELF procedure: 
Vs, dynamic $0.8Vs, ELF = 0.8(2,550) = 2,040 kips (= 0.14W) 
As an exception to the above, design forces less than 80 percent of the ELF results are permitted if justified 
by time history analysis (which is seldom, if ever, the case).
Chapter 11, Seismically Isolated Structures 
11-35 
Third Floor 
Base 
First Floor 
Second Floor 
Roof 
W24x84 W24x84 Penthouse Roof 
W24x104 W24x104 W24x104 W24x104 
W24x104 W24x104 W24x104 W24x104 
W24x104 W24x104 W24x104 W24x104 
W24x192 W24x131 W24x131 W24x192 
TS12x12x3 
4 TS12x12x3 
4 TS12x12x3 
4 
TS12x12x3
4 
TS12x12x3 
4 
W12x120 W12x120 
W12x120 W12x120 
W12x96 W12x65 
W12x65 W12x65 W12x53 
W12x65 W12x65 
W12x53 W12x53 
W12x65 W12x65 
W12x65 W12x65 W12x53 
TS12x12x3 
4 
TS12x12x3
4 TS12x12x3 
4 TS12x12x3 
4 
TS12x12x3 
4 
Figure 11.5-8 Elevation of framing on Column Line 2 (Column Line 6 is similar). 
11.5.4.2 Design of the Superstructure 
The lateral forces, developed in the previous section, in combination with gravity loads, provide a basis for 
the design of the superstructure, using methods similar to those used for a conventional building. In this 
example, selection of member sizes were made based on the results of ETABS model calculations. Detailed 
descriptions of the design calculations are omitted, since the focus of this section is on design aspects unique 
to isolated structures (i.e., design of the isolation system, which is described in the next section). 
Figures 11.5-8 and 11.5-9 are elevation views at Column Lines 2 and B, respectively. Figure 11.5-10 is a 
plan view of the building that shows the framing at the first floor level.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
11-36 
W24 x146 
W24 x131 
W24 x131 W24 x131 W24 x131 W24 x131 W24 x146 
W24 x192 W24 x192 W24 x192 W24 x192 W24 x192 W24 x192 
W24 x162 W24 x146 W24 x146 W24 x146 W24 x146 W24 x162 
W24 x192 W24 x192 W24 x192 W24 x192 W24 x192 W24 x192 
W24 x146 W24 x131 W24 x117 W24 x117 W24 x131 W24 x146 
W24 x104 W24 x104 W24 x131 
W18x40 
W24 x192 W24 x131 W24 x131 W24 x192 
W24 x146 W24 x117 W24 x117 W24 x146 
W24 x176 W24 x131 W24 x131 W24 x176 
W24 x146 W24 x117 W24 x117 W24 x146 
W24 x192 W24 x131 W24 x131 W24 x192 
W24 x131 W24 x104 W24 x104 W24 x131 
W18x40 W18x40 
W18x40 
W18x40 
W18x40 
W18x40 
W18x40 
W18x40 
W18x40 
W18x40 
W18x40 
W18x40 
W18x40 
W18x40 
W18x40 
W18x40 
W18x40 
W18x40 
W18x40 
W18x40 
W18x40 
W18x40 
W18x40 
W18x40 
W18x40 
W18x40 
W18x40 
W18x40 
W18x40 
W18x40 
W18x40 
W18x40 
W18x40 
W18x40 
W18x40 
W18x40 
W18x40 
W18x40 
W18x40 
W18x40 
W18x40 
W18x40 
W18x40 
W18x40 
W18x40 
W18x40 
W18x40 
Figure 11.5-10 First floor framing plan. 
Penthouse Roof 
Roof 
Third Floor 
Second Floor 
First Floor 
Base 
W24x68 W24x68 W24x68 W24x68 
W24x104 W24x131 W24x146 W24x146 W24x131 
W24x117 W24x117 W24x131 W24x131 W24x117 
W24x117 W24x131 W24x131 W24x131 W24x131 W24x117 
W24x192 W24x192 W24x192 W24x192 W24x192 W24x192 
W12x65 W12x65 W12x53 
W12x120 W12x120 W12x65 
W12x120 W12x120 W12x53 
W12x120 W12x120 W12x53 
W12x120 W12x120 W12x53 
W12x120 W12x120 W12x65 
W12x65 W12x65 W12x53 
W12x65 
W12x53 
W12x53 
W12x53 
W12x65 
TS12x12x3 
4 
TS12x12x3 
4 
TS12x12x3 
4 
TS12x12x3 
4 TS12x12x3
4 
TS12x12x3 
4 
TS12x12x3 
4 TS12x12x3 
4 
TS12x12x3 
4 
TS12x12x3 
4 
TS12x12x3 
4 
TS12x12x3 
4 TS12x12x3 
4 TS12x12x3
4 
Figure 11.5-9 Elevation of framing on Column Line B (Column Line D is similar).
Chapter 11, Seismically Isolated Structures 
11-37 
As shown in the elevations (Figures 11.5-8 and 11.5-9), fairly large (1212׾ in.) tubes are consistently used 
throughout the structure for diagonal bracing. A quick check of these braces indicates that stresses will be 
at, or below yield for design earthquake loads. The six braces at the third floor, on lines 2, 4, and 6 (critical 
floor and direction of bracing) carry a reduced design earthquake force of about 400 kips each (= 2,080 kips/6 
braces  cos(30)). The corresponding stress is about 12.5 ksi for reduced design earthquake forces, or about 
25 ksi for unreduced design earthquake forces; indicating that the structure is expected to remain elastic 
during the design earthquake. 
As shown in Figure 11.5-10, the first floor framing has heavy, W24 girders along lines of bracing (lines B, 
D, 2, 4, and 6). These girders resist P-delta moments, as well as other forces. A quick check of these girders 
indicates that only limited yielding is likely, even for the MCE loads (up to about 2 ft of MCE displacement). 
Girders on Line 2 that frame into the column at Line B (critical columns and direction of framing), resist a 
P-delta moment due to the MCE of about 1,000 kip-ft (2,000 kips/2 girders  2 ft/2). Moment in these girders 
due to MCE shear force in isolators is about 450 kip-ft (8.9 kips/in.  24 in.  4 ft/2 girders). Considering 
additional moment due to gravity loads, the plastic capacity of the first floor girders (1,680 kip-ft) would not 
be reached until isolation system displacements exceed about 2 ft. Even beyond this displacement, post-yield 
deformation would be limited (due to the limited extent and duration of MCE displacements beyond 2 ft), 
and the first floor girders would remain capable of stabilizing the isolator units. 
11.5.4.3 Design of the Isolation System 
The displacements and forces calculated in Guide Sec. 11.5.4.1 provide a basis to either: 
1. Develop a detailed design of the isolator units or 
2. Include appropriate design properties in performance-based specifications. 
Developing a detailed design of an elastomeric bearing, requires a familiarity with rubber bearing technology, 
that is usually beyond the expertise of most structural designers; and often varies based on the materials used 
by different manufacturers. This example, like most recent isolation projects, will define design properties 
for isolator units that are appropriate for incorporation into a performance specification (and can be bid by 
more than one bearing manufacturer). Even though the specifications will place the responsibility for meeting 
performance standards with the supplier, the designer must still be knowledgeable of available products and 
potential suppliers, to ensure success of the design. 
11.5.4.3.1 Size of Isolator Units 
The design properties of the seismic isolator units are established based on the calculations of ELF demand, 
recognizing that dynamic analysis is required to verify these properties (and will likely justify slightly more 
lenient properties). The key parameters influencing size are: 
1. Peak displacement of isolator units, 
2. Average long-term (gravity) load on all isolator units, 
3. Maximum long-term (gravity) load on individual isolator units, and 
4. Maximum short-term load on individual isolator units (gravity plus MCE loads) including maximum 
uplift displacement. 
These parameters are summarized in Guide Table 11.5-9. Loads or displacements on individual isolator units 
are taken as the maximum load or displacement on all isolator units (since the design is based on only one 
size of isolator unit). Reduced live load is used for determining long-term loads on isolators.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
11-38 
Table 11.5-9 Summary of Key Design Parameters [isolator unit location] 
Key Design Parameter ELF Procedure 
or Gravity Analysis 
Dynamic Analysis 
Limit 
DE displacement at corner of building (DTD) 19.7 in. [A1] > 17.7 in. [A1] 
MCE displacement at corner of building (DTM) 29.6 in. [A1] > 23.7 in. [A1] 
Average long-term load (1.0D + 0.5L) 477 kips [ALL] N/A 
Maximum long-term load (1.2D + 1.6L) 1,053 kips [C4] N/A 
DE max short-term load (1.4D + 0.5L + QDE) 1,519 kips [B2] N/A 
DE minimum short-term load (0.7D - QDE) 150 kips uplift [A2] N/A 
MCE max short-term load (1.5D + 1.0L + QMCE) 2,006 kips [B2] N/A 
MCE minimum short-term load (0.8D - QMCE) 0.94 in. uplift [A2] N/A 
1.0 in. = 25.4 mm, 1.0 kip = 4.45 kN. 
As rule of thumb, elastomeric isolators should have a diameter, excluding the protective layer of cover, of 
no less than 1.25 times maximum earthquake displacement demand. In this case, the full displacement 
determined by the ELF procedure would require an isolator diameter of: 
iISO $ 1.25(29.6) = 37 in. (0.95 m) 
If justified by dynamic analysis, then 80 percent of the ELF procedure displacement would require an isolator 
diameter of: 
iISO $ 1.25(23.7) = 30 in. (0.75 m) 
For this example, a single size of isolator unit is selected with a nominal diameter of no less than 35.4 in. 
(0.90 m). Although the maximum vertical loads vary enough to suggest smaller diameter of isolator units 
at certain locations (such as at building corners), all of the isolator units must be large enough to sustain MCE 
displacements; which are largest at building corners, due to torsion. An isolator unit with a diameter of 35.4 
in., has a corresponding bearing area of about Ab = 950 square in. The maximum long-term face pressure is 
about 1,100 psi (i.e., 1,053 kips/950 in.2), which is less than the limit for most elastomeric bearing 
compounds. Average long-term face pressure is about 500 psi (i.e., 477 kips/950 in.2) indicating the 
reasonably good distribution of loads among all isolator units. 
The minimum effective stiffness is 6.9 kips/in. per isolator unit at the design displacement (i.e., about 16 in.). 
The height of the isolator unit is primarily a function of the height of the rubber, hr, required to achieve this 
stiffness, given the bearing area, Ab, and the effective stiffness of the rubber compound. The EOC design 
accommodates rubber compounds with minimum effective shear modulus (at 150 percent shear strain) 
ranging from G150% = 65 psi to 110 psi. Numerous elastomeric bearing manufacturers have rubber compounds 
with a shear modulus that falls within this range. Since rubber compounds (and in particular, high-damping 
rubber compounds) are nonlinear, the effective stiffness used for design must be associated with a shear strain 
that is close to the strain level for the design earthquake (e.g., 150 percent shear strain). For a minimum 
effective shear modulus of 65 psi, the total height of the rubber, hr, would be:
Chapter 11, Seismically Isolated Structures 
11-39 
W12x120 
W24x192 
Figure 11.5-11 Typical detail of the isolation system at columns (for clarity, 
some elements not shown). 
2 2 
150% 65 lb/in. 950 in. 8.9 in. 9 in. 
6,900 lb/in. 
b 
r 
eff 
hG A 
k 
 
= = = . 
The overall height of the isolator unit, H, including steel shim and flange plates, would be about 15 in. For 
compounds with an effective minimum shear modulus of 110 psi, the rubber height would be proportionally 
taller (about 15 in.), and the total height of the isolator unit would be about 24 in. 
11.5.4.3.2 Typical Isolation System Detail 
For the EOC design, the isolation system has a similar detail at each column, as shown in Figure 11.5-11. 
The column has an extra large base plate that bears directly on the top of the isolator unit. The column base 
plate is circular, with a diameter comparable to that of the top plate of the isolator unit. Heavy, first floor 
girders frame into, and are moment connected to the columns (moment connections are required at this floor 
only). The columns and base plates are strengthened by plates that run in both horizontal directions, from 
the bottom flange of the girder to the base. Girders are stiffened above the seat plates, and at temporary 
jacking locations. The top plate of the isolator unit is bolted to the column base plate, and the bottom plate 
of the isolator unit is bolted to the foundation. 
The foundation connection accommodates the removal and replacement of isolator units, as required by 
Provisions Sec. 13.6.2.8 [13.2.5.8]. The bottom plate of the isolator unit bears on a steel plate, that has a 
shear lug at the center grouted to the reinforced concrete foundation. Anchor bolts pass through holes in this 
plate and connect to threaded couplers that are attached to deeply embedded rods. 
With the exception of the portion of the column above the first floor slab, each element shown in Figure 
11.5-11 is an integral part of the isolation system (or foundation), and is designed for the gravity and
FEMA 451, NEHRP Recommended Provisions: Design Examples 
11-40 
unreduced design earthquake loads. In particular, the first floor girder, the connection of the girder to the 
column, and the connection of the column to the base plate, are designed for gravity loads and forces caused 
by horizontal shear and P-delta effects due to the unreduced design earthquake load (as shown earlier in 
Figure 11.4-1). 
11.5.4.3.3 Design of Connections of Isolator Units 
Connection of the top plate of the isolator unit, to the column base plate; and the connection of the bottom 
plate to the foundation, are designed for load combinations that include maximum downward forces (1.4D 
+ 0.5L + QDE), and minimum downward (uplift) forces (0.7D - QDE). The reactions to bolts at the top and 
bottom plates of isolator units (ignoring shear friction and shear capacity of the lug at the base) are 
approximately equal, and include shear, axial load, and moment, in the most critical direction of response for 
individual bolts. Moments include the effects of P-delta and horizontal shear across the isolator unit, as 
described by the equations shown in Figure 11.4-1: 
2 2 
M P VH 
. 
= + 
In this case, H1 = H2 = H/2, where H is the height of the isolator unit (assumed to be 24 in., the maximum 
permissible height of isolator units). For maximum downward acting loads, maximum tension on any one 
of N (12) uniformly spaced bolts located in a circular pattern of diameter, Db, must be equal to 43 in. and is 
estimated as follows: 
b 4 b14 12( ) 
b b b b b b 
F M PA M PA MP P VHP 
S A N A D A N N D N D 
. . . . . . . . + . 
=.. - .. =..  - .. = .. - ..= .. - .. 
1 2(1,519 kips)(19.7 in.) 175 kips(24 in.) 1,519 kips 1,587 1,519 6 kips 
12 43 in. 12 
=.. + - ..= - . 
. . 
In this calculation, the vertical load, P = 1,519 kips, is the maximum force occurring at Column B2, the 
deflection is based on DTD = 19.7 in., and the shear force is calculated as V = 8.9 kips/in.  19.7 in. The 
calculation indicates only a modest amount of tension force in anchor bolts, due to the maximum downward 
loads on the isolator units. However, the underlying assumption of the plane sections remaining plane is not 
conservative if top and bottom flange plates yield during a large lateral displacement of the isolator units. 
Rather than fabricate bearings with overly thick flange plates, manufacturers usually recommend anchor bolts 
that can carry substantial amounts of shear and tension, which can arise due to local bending of flanges. 
Tension forces in anchor bolts can occur due to local uplift of isolator units. The ETABS model did not 
calculate uplift forces, since it was assumed that the isolator units would yield in tension. An estimate of the 
uplift load may be based on a maximum tension yield stress of about 150 psi, which produces a tension force, 
Ft, of no more than 150 kips for isolator units with a bearing area of 950 in.2 Applying uplift and shear loads 
to the isolator unit, the maximum tension force in each bolt is estimated at: 
b 1 2( ) 
t 
b b 
FM P A P VHP 
S A N N D 
. . . . + . 
=. + . = . + . 
. . . . 
1 2(150 kips(19.7 in.) +175 kips(24 in.) 150 kips 333 150 40 kips 
12 43 in. 12 
=.. + ..=+. 
. . 
The maximum shear load per bolt is:
Chapter 11, Seismically Isolated Structures 
11-41 
175 15 kips 
v 12 
FV 
N 
= = . 
Twelve 1-in. diameter anchor bolts of A325 material are used for these connections. Alternatively, eight 
1-in. diameter bolts are permitted for isolator units manufactured with eight, rather than twelve, bolt holes. 
Bolts are tightened as required for a slip-critical connection (to avoid slip in the unlikely event of uplift). 
Design of stiffeners and other components of the isolation system detail are not included in this example, but 
would follow conventional procedures for design loads. Design of the isolator unit including top and bottom 
flange plates is the responsibility of the manufacturer. Guide Sec. 11.5.6 includes example performance 
requirements of specifications that govern isolator design (and testing) by the manufacturer. 
11.5.5 Design Verification Using Nonlinear Time History Analysis 
The design is verified (and in some cases isolation system design properties are improved) using time history 
analysis of models that explicitly incorporate isolation system nonlinearity, including lateral force-deflection 
properties and uplift of isolator units, which are subject to net tension loads. Using three sets of horizontal 
earthquake components the EOC is analyzed separately for design earthquake (DE) and MCE ground 
shaking,. All analyses are repeated for two models of the EOC, one with upper-bound (UB) stiffness 
properties and the other with lower-bound (LB) stiffness properties of isolator units. 
11.5.5.1 Ground Motion 
Each pair of horizontal earthquake components is applied to the model in three different orientations relative 
to the principal axes of the building. Time histories are first applied to produce the maximum response along 
the X axis of the EOC model. The analyses are then repeated with the time histories, they are rotated 90 to 
produce the maximum response along the Y axis of the EOC model. Additionally, the time histories are 
rotated 45 to produce the maximum response along an X-Y line (to check if this orientation would produce 
greater response in certain elements than the two principal axis orientations). A total of 18 analyses (2 models 
 3 component orientations  3 sets of earthquake components) are performed separately for both DE and 
the MCE loads. Results are based on the maximum response of the parameter of interest calculated by each 
DE or MCE analysis, respectively. Parameters of interest include: 
Design earthquake 
1. Peak isolation system displacement 
2. Maximum story shear forces (envelope over building height) 
Maximum considered earthquake 
1. Peak isolation system displacement 
2. Maximum downward load on any isolator unit (1.5D + 1.0L + QMCE) 
3. Maximum uplift displacement of any isolator unit (0.8L - QMCE). 
11.5.5.2 Bilinear Stiffness Modeling of Isolator Units 
Nonlinear force-deflection properties of the isolator units are modeled using a bilinear curve; whose hysteretic 
behavior is a parallelogram, which is supplemented by a small amount of viscous damping. A bilinear curve 
is commonly used to model the nonlinear properties of elastomeric bearings; although other approaches are 
sometimes used, including a trilinear curve that captures stiffening effects of some rubber compounds at very 
high strains.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
11-42 
0 
100 
200 
300 
400 
0 5 10 15 20 25 30 
Displacement (in.) 
Force (kip) 
Upper-bound stiffness 
Lower-bound stiffness 
Property Lower-bound Upper-bound 
Effective s tiffnes s (kip/in.) 7 11 
Initial s tiffnes s (kip/in.) 13.0 20.5 
Yield force (kip) 52 82 
Pos t-/pre-yield ratio 0.408 0.408 
Supplementary damping 3% 3% 
Figure 11.5-12 Stiffness and damping properties of EOC isolator units (1.0 in. = 25.4 mm, 1.0 kip = 
4.45 kN). 
Using engineering judgement, the initial stiffness, the yield force, the ratio of post-yield to pre-yield stiffness 
for the bilinear curves, and the amount of supplementary viscous damping, v, are selected. The effective 
stiffness and effective damping values are used for preliminary design. The lower-bound bilinear stiffness 
curve is based on kDmin = 6.9 kips/in. (rounded to 7 kips/in.). The upper-bound bilinear stiffness curve is based 
on an effective stiffness of kDmax = 8.9 kips/in. times 1.2 (which the result is rounded to 11 kips/in.). The 1.2 
factor is taken into account for the effects of aging over the design life of the isolators. Both upper-bound 
and lower-bound stiffness curves are based on effective damping (combined hysteretic and supplementary 
viscous) of D  15 percent. Figure 11.5-12 illustrates upper-bound and lower-bound bilinear stiffness curves, 
and summarizes the values of the parameters that define these curves. 
For isolators with known properties, parameters defining bilinear stiffness may be based on test data provided 
by the manufacturer. For example, the bilinear properties shown in Figure 11.5-12, are compared with 
effective stiffness and effective damping test data that are representative of a HK090H6 high-damping rubber 
bearing manufactured by Bridgestone Engineered Products Company, Inc. Bridgestone is one of several 
manufacturers of high-damping rubber bearings and provides catalog data on design properties of standard 
isolators. The HK090H6 bearing has a rubber height of about 10 in., a diameter of 0.9 m, without the cover, 
and a rubber compound with a relatively low shear modulus (less than 100 psi at high strains). 
Plots of the effective stiffness and damping of the HK090H6 bearing, are shown in Figure 11.5-13 (solid 
symbols), and are compared with plots of the calculated effective stiffness and damping (using the equations 
from Figure 11.4-2 and the bilinear stiffness curves shown in Figure 11.5-12). The effective stiffness curve 
of the HK090H6 bearing, is based on data from the third cycle of testing; and therefore, best represents the 
minimum effective (or lower-bound) stiffness. The plots indicate the common trend in effective stiffness of 
high-damping elastomeric bearings significant softening up to 100 percent shear strain, fairly stable stiffness 
from 100 percent to 300 percent shear strain, and significant stiffening beyond 300 percent shear strain. The 
trend in effective damping is a steady decrease in damping with amplitude beyond about 150 percent shear 
strain.
Chapter 11, Seismically Isolated Structures 
11-43 
0
5 
10 
15 
20 
25 
30 
0 5 10 15 20 25 30 
Displacement (in.) 
Stiffness (kip/in.) or 
Damping (% of critical) 
Effective damping - Test Data 
Effective damping 
Upper-bound effective stiffness 
Effective stiffness - Test Data 
Lower-bound effective stiffness 
Figure 11.5-13 Comparison of modeled isolator properties to test data (1.0 in. = 25.4 mm, 1.0 
kip/in. = 0.175 kN/mm). 
The effective stiffness and damping of the bilinear stiffness curves capture the trends of the HK090H6 bearing 
reasonably well, as shown in Figure 11.5-13. Lower-bound effective stiffness fits the HK090H6 data well 
for response amplitudes of interest (i.e., the displacements in the range of about 15 in. through 25 in.). 
Likewise, the effective damping of the bilinear stiffness curves, is conservatively less than the effective 
damping based on test data for the same displacement range. These comparisons confirm that the bilinear 
stiffness properties of isolator units used for nonlinear analysis, are valid characterizations of the 
force-deflection properties of the Bridgestone HK090H6 bearing, also presumably, the bearings of other 
manufacturers that have comparable values of effective stiffness and damping. 
11.5.5.3 Summary of Results for Time History Analyses 
Tables 11.5-10 and 11.5-11, compare the results of the design earthquake and maximum considered 
earthquake time history analyses, respectively, with the corresponding values calculated using the ELF 
procedure. The peak response values noted for the time history analyses are, the maxima of X-axis and 
Y-axis directions of response. Because torsional effects were neglected in the time history analyses for this 
example, the displacement at the corner of the EOC was assumed to be 1.1 times the center displacement; this 
assumption may not be conservative. The reported shears occur below the indicated level.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
11-44 
Table 11.5-10 Design Earthquake Response Parameters - Results of ELF Procedure and Nonlinear 
Time History Analysis 
Response parameter ELF 
procedure 
Time history analysis 
Peak response Model Record 
Isolation system displacement (in.) 
Displacement at center of EOC, DD 16.3 in. 14.3 in. LB 
stiffness Sylmar 
Displacement at corner of EOC, DTD 19.7 in. 15.7 in.* 
Superstructure forces  story shear (kips) 
Penthouse roof 740 kips 400 kips 
UB stiffness Newhall 
Roof (penthouse) 2,780 kips 1,815 kips 
Third floor 4,160 kips 3,023 kips 
Second floor 4,940 kips 4,180 kips 
First floor 5,100 kips 5,438 kips 
1.0 in. = 25.4 mm, 1.0 kip = 4.45 kN. 
* Displacement includes an arbitrary 10 percent increase for possible torsional response. 
Table 11.5-11 Maximum Considered Earthquake Response Parameters  Results of ELF Procedure 
and Nonlinear Time History Analysis 
Response parameter 
ELF 
procedure 
Time history analysis 
Peak response Model Record 
Displacement at center of EOC, DM 24.5 in. 26.4 in. LB 
stiffness 
El Centro 
Displacement at corner of EOC, DTM 29.7 in. 29.0 in.1 No. 6 
1.0 in. = 25.4 mm. 
* Displacement includes an arbitrary 10 percent increase for possible torsional response. 
The results of the design earthquake time history analyses, verify that the results of the ELF procedure are 
generally conservative. A design displacement of 16 in., and total design displacement of 20 in., are 
conservative bounds on calculated displacements, even if significant torsion should occur. The story shears 
calculated using the ELF procedure, are larger than those from the dynamic analyses at all superstructure 
elevations. The higher the elevation, the larger the margin between the story shears, which is an indication 
that the inverted triangular pattern (the results from Provisions Eq. 13.3.5 [13.3-9]), produces conservative 
results. At the second story (below the third floor), the critical level for the brace design the story shear from 
the dynamic analyses (3,023 kips) is only about three-quarters of the story shear from the ELF procedure 
(4,160 kips). The results of the MCE time history analyses show that the ELF procedure can underestimate 
the maximum displacement. Accordingly, a maximum displacement of 27 in., and total maximum 
displacement of 30 in., are used for the design specifications. 
Tables 11.5-12 and 11.5-13, respectively, report the maximum downward forces on isolator units and the 
maximum uplift displacements determined from the maximum considered earthquake time history analyses.
Chapter 11, Seismically Isolated Structures 
11-45 
These tables report two values for each isolator location representing both the X-axis and Y-axis orientations 
of the strongest direction of shaking of the time history record. 
Table 11.5-12 Maximum Downward Force (kips) on Isolator Units (1.5D + 1.0L + QMCE)* 
Upper-bound stiffness model  Newhall record  X-axis/Y-axis orientations 
Column 
line 1 2 3 4 
A 417/411 930/1,076 651/683 889/1,027 
B 1,048/881 1,612/1,766 1,696/1,588 1,308/1,429 
C 656/628 1,145/1,127 1,301/1,300 1,319/1,318 
1.0 kip = 4.45 kN. 
* Forces on Column Lines 5, 6 and 7 (not shown) are enveloped by those on Column Lines 3, 2, and 1, respectively; forces 
on Column Lines D and E (not shown) are enveloped by those on Column Lines B and A, respectively. 
Table 11.5-13 Maximum Uplift Displacement (in.) Of Isolator Units (0.8D - QMCE)1 
Upper-bound stiffness model  Newhall record  X-axis/Y-axis orientations 
Column 
line 1 2 3 4 
A No uplift 0.00/0.39 No uplift 0.00/0.17 
B 0.26/0.00 0.19/0.12 No uplift No uplift 
C No uplift No uplift No uplift No uplift 
1.0 in. = 25.4 mm. 
* Displacement on Column Lines 5, 6 and 7 (not shown) enveloped by those on Column Lines 3, 2, and 1, respectively; 
displacement on Column Lines D and E (not shown) are enveloped by those on Column Lines B and A, respectively. 
Table 11.5-12 indicates a maximum downward force of 1,766 kips (at column B2), which is somewhat less 
than the value predicted using the ELF procedure (2,006 kips in Table 11.5-7). Table 11.5-13 includes the 
results from the controlling analysis, and indicates a maximum uplift displacement of 0.39 in. (at column A2), 
which is significantly less than the value predicted using the ELF procedure (0.94 in. as noted in Table 
11.5-8). Based on these results, the design specification (Guide Sec. 11.5.2.10) uses values of 2,000 kips 
maximum downward force and  in. of maximum uplift displacement. 
11.5.6 Design and Testing Criteria for Isolator Units 
Detailed design of the isolator units for the EOC facility, is the responsibility of the bearing manufacturer 
subject to the design and testing (performance)of the criteria included in the construction documents 
(drawings and/or specifications). Performance criteria typically includes a basic description, and 
requirements for isolator unit sizes, the design life and durability, the environmental loads and fire-resistance 
criteria, Quality Assurance and Quality Control (including QC testing of production units), the design forces 
and displacements, and prototype testing requirements. This section summarizes key data and performance 
criteria for the EOC, including the criteria for prototype testing of isolator units, as required by Provisions 
Sec. 13.9.2 [13.6.1].
FEMA 451, NEHRP Recommended Provisions: Design Examples 
11-46 
d b 
D b 
H 
tf tf 
D 
D + 2t 
D
o
f 
o o 
Figure 11.5-14 Isolator dimensions. 
11.5.6.1 Dimensions and Configuration 
Steel shim plate diameter, Do $ 35.4 in. 
Rubber cover thickness, to $ 0.5 in. 
Rubber layer thickness, tr # 0.375 in. 
Gross rubber diameter, Do + 2to # 38 in. 
Bolt pattern diameter, Db = 43 in. 
Bolt hole diameter, db = 1.5 in. 
Steel flange plate diameter, Df = 48 in. 
Steel flange plate thickness, tf $ 1.5 in. 
Total height, H # 24 in. 
11.5.6.2 Prototype Stiffness and Damping Criteria 
Minimum effective stiffness (third cycle of test, typical vertical load) kDmin, kMmin $ 7.0 kips/in. 
Maximum effective stiffness (first cycle test, typical vertical load) kDmax, kMmax # 9.0 kips/in. 
The kDmin, kDmax, kMmin, and kMmax, are properties of the isolation system as a whole (calculated from the 
properties of individual isolator units using Provisions Eq. 13.9.5.1-1 through 13.9.5.1-4 [13.6-3 through 
13.6-6]). Individual isolator units may have stiffness properties that fall outside the limits (by, perhaps, 
10 percent), provided the average stiffness of all isolator units complies with the limits.
Chapter 11, Seismically Isolated Structures 
11-47 
Effective damping 
(minimum of three cycles of test, typical vertical load) D $ 15 percent 
(minimum of three cycles of test, typical vertical load) M $ 12 percent 
Vertical Stiffness 
(average of three cycles of test, typical vertical load kv $ 5,000 kips/in. 
 50 percent) 
11.5.6.3 Design Forces (Vertical Loads) 
Maximum long-term load (individual isolator) 1.2D + 1.6L = 1,200 kips 
Typical load - cyclic load tests (average of all isolators) 1.0D + 0.5L = 500 kips 
Upper-bound load - cyclic load tests (all isolators) 1.2D + 0.5L + |E| = 750 kips 
Lower-bound load - cyclic load tests (all isolators) 0.8D - |E| = 250 kips 
Maximum short-term load (individual isolator) 1.2D + 1.0L + |E| = 2,000 kips 
Minimum short-term load (individual isolator) 0.8D - |E| = tension force due 
to  in. of uplift 
11.5.6.4 Design Displacements 
Design earthquake displacement DD = 16 in. 
Total design earthquake displacement DTD = 20 in. 
Maximum considered earthquake displacement DM = 27 in. 
Total maximum considered earthquake displacement DTM = 30 in. 
11.5.6.5 Prototype Testing Criteria 
Table 11.5-14 summarizes the prototype test criteria found in Provisions Sec. 13.9.2 [13.6.1] as well as the 
corresponding loads on isolator units of the EOC.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
11-48 
Table 11.5-14 Prototype Test Requirements 
Provisions criteria EOC criteria 
No. of cycles Vertical load Lateral load Vertical load Lateral load 
Vertical stiffness test 
3 cycles Typical  50% None 500  250 kips None 
Cyclic load tests to check wind effects (Provisions Sec. 13.9.2.3 [13.6.1.2]) 
20 cycles Typical Design force 500 kips  20 kips 
Cyclic load tests to establish effective stiffness and damping (Provisions Sec. 13.9.2.3 [13.6.1.2]) 
3 cycles Typical 0.25DD 500 kips  4 in. 
3 cycles Upper-bound* 0.25DD 750 kips  4 in. 
3 cycles Lower-bound* 0.25DD 250 kips  4 in. 
3 cycles Typical 0.5DD 500 kips  8 in. 
3 cycles Upper-bound* 0.5DD 750 kips  8 in. 
3 cycles Lower-bound* 0.5DD 250 kips  8 in. 
3 cycles Typical 1.0DD 500 kips  16 in. 
3 cycles Upper-bound* 1.0DD 750 kips  16 in. 
3 cycles Lower-bound* 1.0DD 250 kips  16 in. 
3 cycles Typical 1.0DM 500 kips  27 in. 
3 cycles Upper-bound* 1.0DM 750 kips  27 in. 
3 cycles Lower-bound* 1.0DM 250 kips  27 in. 
3 cycles Typical 1.0DTM 500 kips  30 in. 
Cyclic load tests to check durability (Provisions Sec. 13.9.2.3 [13.6.1.2]) 
30SD1/SDSBD(= 20)** Typical load 1.0DTD 500 kips  20 in. 
Static load test of isolator stability (Provisions Sec. 13.9.2.6 [13.6.1.5]) 
N/A Maximum 1.0DTM 2,000 kips 30 in. 
N/A Minimum 1.0DTM -in. of uplift 30 in. 
1.0 in. = 25.4 mm, 1.0 kip = 4.45 kN. 
* Tests with upper-bound and lower-bound vertical loads are required by Provisions Sec. 13.9.2.3 [13.6.1.2] for isolator units 
that are vertical-load-carrying elements. 
** The Provisions contains a typographical error where presenting this expression. The errata to the Provisions correct the 
error. 

12-1 
12 
NONBUILDING STRUCTURE DESIGN 
Harold O. Sprague Jr., P.E. 
Chapter 14 of the 2000 NEHRP Recommended Provisions and Commentary (hereafter, the Provisions and 
Commentary) is devoted to nonbuilding structures. Nonbuilding structures comprise a myriad of 
structures constructed of all types of materials with markedly different dynamic characteristics and a wide 
range of performance requirements. 
Nonbuilding structures are a general category of structure distinct from buildings. Key features that 
differentiate nonbuilding structures from buildings include human occupancy, function, dynamic 
response, and risk to society. Human occupancy, which is incidental to most nonbuilding structures, is 
the primary purpose of most buildings. The primary purpose and function of nonbuilding structures can 
be incidental to society or the purpose and function can be critical for society. 
In the past, many nonbuilding structures were designed for seismic resistance using building code 
provisions developed specifically for buildings. These code provisions were not adequate to address the 
performance requirements and expectations that are unique to nonbuilding structures. For example 
consider secondary containment for a vertical vessel containing hazardous materials. Nonlinear 
performance and collapse prevention, which are performance expectations for buildings, are inappropriate 
for a secondary containment structure, which must not leak. 
Traditionally, the seismic design of nonbuilding structures depended on the various trade organizations 
and standards development organizations that were disconnected from the building codes. The Provisions 
have always been based upon strength design and multiple maps for seismic ground motion definition, 
whereas most of the industry standards were based on allowable stress design and a single zone map. The 
advent of the 1997 Provisions exacerbated the problems of the disconnect for nonbuilding structures with 
direct use of seismic spectral ordinates, and with the change to a longer recurrence interval for the seismic 
ground motion. It became clear that a more coordinated effort was required to develop appropriate 
seismic design provisions for nonbuilding structures. 
This chapter develops examples specifically to help clarify Chapter 14 of the Provisions. The solutions 
developed are not intended to be comprehensive but instead focus on interpretation of Provisions Chapter 
14 (Nonbuilding Structure Design Requirements). Complete solutions to the examples cited are beyond 
the scope of this chapter. 
Although this volume of design examples is based on the 2000 Provisions, it has been annotated to reflect 
changes made to the 2003 Provisions. Annotations within brackets, [ ], indicate both organizational 
changes (as a result of a reformat of all of the chapters of the 2003 Provisions) and substantive technical 
changes to the 2003 Provisions and its primary reference documents. While the general concepts of the 
changes are described, the design examples and calculations have not been revised to reflect the changes 
to the 2003 Provisions.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
12-2 
Several noteworthy changes were made to the nonbuilding structures requirements of the 2003 
Provisions. These include clearer definition of the scopes of Chapters 6 and 14, expanded, direct 
definition of structural systems (along with design parameters and detailing requirements) in Chapter 14, 
and a few specific changes for particular nonbuilding structural systems. 
In addition to changes Provisions Chapter 14, the basic earthquake hazard maps were updated, the 
redundancy factor calculation was completely revised, and the minimum base shear equation for areas 
without near-source effects was eliminated. 
Where they affect the design examples in this chapter, significant changes to the 2003 Provisions and 
primary reference documents are noted. However, some minor changes to the 2003 Provisions and the 
reference documents may not be noted. 
In addition to the Provisions and Commentary, the following publications are referenced in this chapter: 
United States Geological Survey, 1996. Seismic Design Parameters (CD-ROM) USGS. 
[The 2003 Provisions have adopted the 2002 USGS probabilistic seismic hazard maps, and the 
maps have been added to the body of the 2003 Provisions as figures in Chapter 3 (instead of the 
previously used separate map package). The CD-ROM also has been updated.] 
American Water Works Association. 1996. Welded Steel Tanks for Water Storage. AWWA. 
American Petroleum Institute (API), Welded steel tanks for oil storage. API 650, 10th Edition, 
November 1998. 
12.1 NONBUILDING STRUCTURES VERSUS NONSTRUCTURAL COMPONENTS 
Many industrial structures are classified as either nonbuilding structures or nonstructural components. 
This distinction is necessary to determine how the practicing engineer designs the structure. The intent of 
the Provisions is to provide a clear and consistent design methodology for engineers to follow regardless 
of whether the structure is a nonbuilding structure or a nonstructural component. Central to the 
methodology is how to determine which classification is appropriate. 
The design methodology contained in Provisions Chapter 6, Architectural, Mechanical, and Electrical 
Components Design Requirements, focuses on nonstructural component design. As such, the 
amplification by the supporting structure of the earthquake-induced accelerations is critical to the design 
of the component and its supports and attachments. The design methodology contained in Provisions 
Chapter 14 focuses on the direct effects of earthquake ground motion on the nonbuilding structure. 
Table 12-1 Applicability of the Chapters of the Provisions 
Supporting 
Structure 
Supported Item 
Nonstructural Component Nonbuilding Structure 
Building 
Chapter 5 [4 and 5]for supporting 
structure 
Chapter 6 for supported item 
Chapter 5 [4 and 5]for supporting 
structure 
Chapter 14 for supported item 
Nonbuilding Chapter 14 for supporting structure 
Chapter 6 for supported item 
Chapter 14 for both supporting 
structure and supported item
Chapter 12, Nonbuilding Structure Design 
12-3 
30' 30' 30' 30' 
25' 
80' 
Inlet filter 
Figure 12-1 Combustion turbine building (1.0 ft = 0.3048 m). 
The example shown in Figure 12-1 is a combustion turbine, electric-power-generating facility with four 
bays. Each bay contains a combustion turbine and supports an inlet filter on the roof. The uniform 
seismic dead load of the supporting roof structure is 30 psf. Each filter weighs 34 kips. 
The following two examples illustrate the difference between nonbuilding structures that are treated as 
nonstructural components, using Provisions Chapter 6, and those which are designed in accordance with 
Provisions Chapter 14. There is a subtle difference between the two chapters: 
6.1: . . .if the combined weight of the supported components and nonbuilding structures with flexible dynamic 
characteristics exceeds 25 percent of the weight of the structure, the structure shall be designed considering 
interaction effects between the structure and the supported items. 
14.4: If the weight of a nonbuilding structure is 25 percent or more of the combined weight of the nonbuilding 
structure and the supporting structure, the design seismic forces of the nonbuilding structure shall be 
determined based on the combined nonbuilding structure and supporting structural system.. . .  
The difference is the plural components and the singular nonbuilding structure, and that difference is 
explored in this example. 
[The text has been cleaned up considerably in the 2003 edition but some inconsistencies persist. Sec. 
14.1.5 indicates the scopes of Chapters 6 and 14. Both chapters consider the weight of an individual 
supported component or nonbuilding structure in comparison to the total seismic weight. Where the 
weight of such an individual item does not exceed 25 percent of the seismic weight, forces are determined 
in accordance with Chapter 6. Where a nonbuilding structures weight exceeds 25 percent of the seismic 
weight, Sec. 14.1.5 requires a combined system analysis and the rigidity or flexibility of the supported 
nonbuilding structure is used in determining the R factor. In contrast, Sec. 6.1.1 requires consideration of 
interaction effects only where the weight exceeds 25 percent of the seismic weight and the supported item 
has flexible dynamic characteristics.] 
12.1.1 Nonbuilding Structure
FEMA 451, NEHRP Recommended Provisions: Design Examples 
12-4 
For the purpose of illustration assume that the four filter units are connected in a fashion that couples their 
dynamic response. Therefore, the plural components used in Provisions Sec. 6.1 is apparently the most 
meaningful provision. 
[The text no longer contains a plural, but conceptually the frame could be considered a single item in this 
instance (just as the separate items within a single roof-top unit would be lumped together).] 
12.1.1.1 Calculation of Seismic Weights 
All four inlet filters = WIF = 4(34 kips) = 136 kips 
Support structure = WSS = 4 (30 ft)(80 ft)(30 psf) = 288 kips 
The combined weight of the nonbuilding structure (inlet filters) and the supporting structural system is 
Wcombined = 136 kips + 288 kips = 424 kips 
12.1.1.2 Selection of Design Method 
The ratio of the supported weight to the total weight is: 
136 0.321 25% 
424 
IF 
Combined 
W 
W 
= = > 
Because the weight of the inlet filters is 25 percent or more of the combined weight of the nonbuilding 
structure and the supporting structure (Provisions Sec. 14.4 [14.1.5]), the inlet filters are classified as 
nonbuilding structures and the seismic design forces must be determined from analysis of the combined 
seismic-resistant structural systems. This would require modeling the filters, the structural components of 
the filters, and the structural components of the combustion turbine supporting structure to determine 
accurately the seismic forces on the structural elements as opposed to modeling the filters as lumped 
masses. [See the discussion added to Sec. 12.1.] 
12.1.2 NONSTRUCTURAL COMPONENT 
For the purpose of illustration assume that the inlet filters are independent structures, although each is 
supported on the same basic structure. In this instance, one filter is the nonbuilding structure. The 
question is whether it is heavy enough to significantly change the response of the combined system. 
12.1.2.1 Calculation of Seismic Weights 
One inlet filter = WIF = 34 kips 
Support structure = WSS = 4 (30 ft)(80 ft)(30 psf) = 288 kips 
The combined weight of the nonbuilding structures (all four inlet filters) and the supporting structural 
system is 
Wcombined = 4 (34 kips) + 288 kips = 424 kips 
12.1.2.2 Selection of Design Method 
The ratio of the supported weight to the total weight is:
Chapter 12, Nonbuilding Structure Design 
12-5 
W 
W 
IF 
Combined 
= 34 = < 
424 
0.08 25% 
Because the weight of an inlet filter is less than 25 percent of the combined weight of the nonbuilding 
structures and the supporting structure (Provisions Sec. 14.4 [14.1.5]), the inlet filters are classified as 
nonstructural components and the seismic design forces must be determined in accordance with 
Provisions Chapter 6. In this example, the filters could be modeled as lumped masses. The filters and the 
filter supports could then be designed as nonstructural components. 
12.2 PIPE RACK, OXFORD, MISSISSIPPI 
This example illustrates the calculation of design base shears and maximum inelastic displacements for a 
pipe rack using the equivalent lateral force (ELF) procedure. 
12.2.1 Description 
A two-tier, 12-bay pipe rack in a petrochemical facility has concentrically braced frames in the 
longitudinal direction and ordinary moment frames in the transverse direction. The pipe rack supports 
four runs of 12-in.-diameter pipe carrying naphtha on the top tier and four runs of 8-in.-diameter pipe 
carrying water for fire suppression on the bottom tier. The minimum seismic dead load for piping is 35 
psf on each tier to allow for future piping loads. The seismic dead load for the steel support structure is 
10 psf on each tier. 
Pipe supports connect the pipe to the structural steel frame and are designed to support the gravity load 
and resist the seismic and wind forces perpendicular to the pipe. The typical pipe support allows the pipe 
to move in the longitudinal direction of the pipe to avoid restraining thermal movement. The pipe support 
near the center of the run is designed to resist longitudinal and transverse pipe movement as well as 
provide gravity support; such supports are generally referred to as fixed supports. 
Pipes themselves must be designed to resist gravity, wind, seismic, and thermally induced forces, 
spanning from support to support. 
If the pipe run is continuous for hundreds of feet, thermal/seismic loops are provided to avoid a 
cumulative thermal growth effect. The longitudinal runs of pipe are broken up into sections by providing 
thermal/seismic loops at spaced intervals. In Figure 12-2, it is assumed thermal/seismic loops are 
provided at each end of the pipe run.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
12-6 
20'-0" 
6 bays @ 20'-0" 
= 120'-0" 
PLAN 
ELEVATION SECTION 
10'-0" 8'-0" 
15'-0" 3'-0" 
20'-0" 
5 bays @ 20'-0" 
= 100'-0" 
Expansion loop 
breaks the 
continuity 
Horizontal bracing 
at braced bay only 
Figure 12-2 Pipe rack (1.0 ft = 0.3048 m). 
12.2.2 Provisions Parameters 
12.2.2.1 Ground Motion 
The spectral response acceleration coefficients at the site are 
SDS = 0.40 
SD1 = 0.18. 
[The 2003 Provisions have adopted the 2002 USGS probabilistic seismic hazard maps, and the maps have 
been added to the body of the 2003 Provisions as figures in Chapter 3 (instead of the previously used 
separate map package).] 
12.2.2.2 Seismic Use Group and Importance Factor 
The upper piping carries a hazardous material (naphtha) and the lower piping is required for fire 
suppression. The naphtha piping is included in Provisions Sec. 1.3.1, Item 11 [Sec. 1.2.1, Item 11], 
therefore, the pipe rack is assigned to Seismic Use Group III. 
According to Provisions Sec. 14.5.1.2 [14.2.1], the importance factor, I, is 1.5 based on Seismic Use 
Group, Hazard, and Function. If these three measures yield different importance factors, the largest factor 
applies. 
12.2.2.3 Seismic Design Category 
For this structure assigned to Seismic Use Group III with SDS = 0.40 and SD1 = 0.18, the Seismic Design 
Category is D according to Provisions Sec. 4.2.1 [1.4].
Chapter 12, Nonbuilding Structure Design 
12-7 
12.2.3 Design in the Transverse Direction 
[Chapter 14 has been revised so that it no longer refers to Table 4.3-1. Instead values for design 
coefficients and detailing requirements are provided with the chapter.] 
12.2.3.1 Design Coefficients 
Using Provisions Table 14.5.1.1 [14.4-2] (which refers to Provisions Table 5.2.2 [4.3-1]), the parameters 
for this ordinary steel moment frame are 
R = 4 
O0 = 3 
Cd = 3 
[In the 2003 Provisions, R factor options are presented that correspond to required levels of detailing. R = 
3.5, O = 3; Cd = 3.] 
Ordinary steel moment frames are retained for use in nonbuilding structures such as pipe racks because 
they allow greater flexibility for accommodating process piping and are easier to design and construct 
than special steel moment frames. 
12.2.3.2 Seismic Response Coefficient 
Using Provisions Eq. 5.4.1.1-1 [5.2-2]: 
C S 
s R I 
= DS = 04 = 
4 15 
. 015 
. 
. 
From analysis, T = 0.42 sec. For nonbuilding structures, the fundamental period is generally 
approximated for the first iteration and must be verified with final calculations. For many nonbuilding 
structures the maximum period limit contained in the first paragraph of Provisions Sec. 5.4.2 [5.2.2] is not 
appropriate. As a result, the examples in this chapter neglect that limit. Future editions of the Provisions 
will clarify that this limit does not apply to nonbuilding structures. [In the 2003 Provisions, Sec. 14.2.9 
makes clear that the approximate period equations do not apply to nonbuilding structures.] 
Using Provisions Eq. 5.4.1.1-2 [5.2-3], CS does not need to exceed 
( ) ( ) 
0.18 0.161 
0.42 4 1.5 
D1 
s 
C S 
T R I 
= = = 
Using Provisions Eq. 5.4.1.1-3, Cs shall not be less than 
Cs = 0.044ISDS = 0.044(1.5)(0.4) = 0.0264 
[This minimum Cs value has been removed in the 2003 Provisions. In its place is a minimum Cs value for 
long-period structures, which is not applicable to this example.] 
Provisions Eq. 5.4.1.1-1 [5.2-2] controls; Cs = 0.15.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
12-8 
12.2.3.3 Seismic Weight 
W = 2(20 ft)(20 ft)(35 psf + 10 psf) = 36 kips 
12.2.3.4 Base Shear (Provisions Sec. 5.3.2 [5.2.1]) 
V = CsW = 0.15(36 kips) = 5.4 kips 
12.2.3.5 Drift 
Although not shown here, drift of the pipe rack in the transverse direction was calculated by elastic 
analysis using the design forces calculated above. The calculated lateral drift, dxe = 0.328 in. 
Using Provisions Eq. 14.3.2.1 [5.2-15], 
3.5(0.328 in.) 
0.765 in. 
1.5 
d xe 
x 
C 
I
d 
d = = = 
The lateral drift must be checked with regard to acceptable limits. The acceptable limits for nonbuilding 
structures are not found in codes. Rather, the limits are what is acceptable for the performance of the 
piping. In general, piping can safely accommodate the amount of lateral drift calculated in this example. 
P-delta effects must also be considered and checked as required in Provisions Sec. 5.4.6.2 [5.2.6.2]. 
12.2.3.6 Redundancy Factor 
Some nonbuilding structures are designed with parameters from Provisions Table 5.2.2 [4.3-1]; if they are 
termed nonbuilding structures similar to buildings. For such structures the redundancy factor applies, 
if the structure is in Seismic Design Category D, E, or F. Pipe racks, being fairly simple moment frames 
or braced frames, are in the category similar to buildings. Because this structure is assigned to Seismic 
Design Category D, Provisions Sec. 5.2.4.2 [4.3.3.2] applies. The redundancy factor is calculated as 
. = 2 - 20 
rmaxx Ax 
where r is the fraction of the seismic force at a given level resisted by one component of the vertical maxx 
seismic-force-resisting system at that level, and Ax is defined as the area of the diaphragm immediately 
above the story in question. Some interpretation is necessary for the pipe rack. Considering the 
transverse direction, the seismic-force-resisting system is an ordinary moment resisting frame with only 
two columns in a single frame. The frames repeat in an identical pattern. The diaphragm is the pipes 
themselves, which are not rigid enough to make one consider the 240 ft length between expansion joints 
as a diaphragm. Therefore, for the computation of . in the transverse direction, each 20-by-20 ft bay will 
be considered independently. 
The maximum of the sum of the shears in the two columns equals the story shear, so the ratio rmax is 1.0. 
The diaphragm area is simply the bay area: 
Ax = 20 ft  20 ft = 400 ft2, 
therefore, 
. = 2- = 
20 
10 
10 
. 
. 
400
Chapter 12, Nonbuilding Structure Design 
12-9 
[The redundancy requirements have been changed substantially in the 2003 Provisions.] 
12.2.3.7 Determining E 
E is defined to include the effects of horizontal and vertical ground motions as follows: 
E = .QE  0.2 SDS D 
where QE is the effect of the horizontal earthquake ground motions, which is determined primarily by the 
base shear just computed, and D is the effect of dead load. By putting a simple multiplier on the effect of 
dead load, the last term is an approximation of the effect of vertical ground motion. For the moment 
frame, the joint moment is influenced by both terms. E with the + on the second term when combined 
with dead and live loads will generally produce the largest negative moment at the joints, while E with the 
-on the second term when combined with the minimum dead load (0.9D) will produce the largest 
positive joint moments. 
The Provisions also requires the consideration of an overstrength factor, O0, on the effect of horizontal 
motions in defining E for components susceptible to brittle failure. 
E = . O0 QE  0.2 SDS 
The pipe rack does not appear to have components that require such consideration. 
12.2.4 Design in the Longitudinal Direction 
[In the 2003 Provisions, Chapter 14 no longer refers to Table 4.3-1. Instead, Tables 14.2-2 and 14.2-3 
have design coefficient values and corresponding detailing requirements for each system.] 
12.2.4.1 Design Coefficients 
Using Provisions Table 14.5.1.1 [14.2-2] (which refers to Provisions Table 5.2.2 [4.3-1]), the parameters 
for this ordinary steel concentrically braced frame are: 
R = 4 
O0 = 2 
Cd = 4 
[The 2003 Provisions allow selection of appropriate design coefficients and corresponding detailing for 
several systems. In the case of this example, R would equal 5, but the calculations that follow are not 
updated.] 
Where Provisions Table 5.2.2 [4.3-1] is used to determine the values for design coefficients, the detailing 
reference sections noted in the table also apply. A concentric braced frame has an assigned R of 5, but an 
R of 4 is used to comply with Provisions Sec. 5.2.2.2.1 [4.3.1.2.1]. 
[In the 2003 Provisions, Chapter 14 no longer refers to Table 4.3-1. Instead, Tables 14.2-2 and 14.2-3 
have design coefficient values and corresponding detailing requirements for each system. Chapter 14 
contains no requirements corresponding to that found in Sec. 4.3.1.2.1 (related to R factors for systems in 
orthogonal directions).] 
12.2.4.2 Seismic Response Coefficient 
Using Provisions Eq. 5.4.1.1-1 [5.2-2]:
FEMA 451, NEHRP Recommended Provisions: Design Examples 
12-10
C S 
s R I 
= DS = 04 = 
4 15 
. 015 
. 
. 
From analysis, T = 0.24 seconds. The fundamental period for nonbuilding structures, is generally 
approximated for the first iteration and must be verified with final calculations. For many nonbuilding 
structures the maximum period limit contained in the first paragraph of Provisions Sec. 5.4.2 [5.2.2] is not 
appropriate. As a result, the examples in this chapter neglect that limit. Future editions of the Provisions 
are expected to clarify that this limit does not apply to nonbuilding structures. [In the 2003 Provisions, 
Sec. 14.2.9 makes clear that the approximate period equations do not apply to nonbuilding structures.] 
Using Provisions Eq. 5.4.1.1-2 [5.2-3], CS does not need to exceed: 
( ) ( ) C S 
s T R I 
= D1 = 018 = 
024 4 15 
. 0281 
. . 
. 
Provisions Sec. 14.5.1 [14.2.8] provides equations for minimum values of Cs that replace corresponding 
equations in Provisions Sec. 5.4.1.1 [5.2.1.1]. However, according to Item 2 of Sec. 14.5.1 [14.2.8, 
replacement of Chapter 5 equations for minima occurs only for nonbuilding systems that have an R value 
provided in Table 14.5.1.1 [14.4-2]. In the present example the R values are taken from Table 5.2.2 so 
the minima defined in Sec. 5.4.1.1 apply. [In the 2003 Provisions this is no longer the case as reference 
to Table 4.3-1 has been eliminated. Since the example structure would satisfy exception 1 of Sec. 14.2.8 
and the minimum base shear equation in Chapter 5 was removed, no additional minimum base shear must 
be considered.] 
Using Provisions Eq. 5.4.1.1-3, Cs shall not be less than: 
Cs = 0.044ISDS = 0.044(1.5)(0.4) = 0.0264 
Provisions Eq. 5.4.1.1-1 [5.2-2] controls; Cs = 0.12. 
12.2.4.3 Seismic Weight 
W = 2(240 ft)(20 ft)(35 psf + 10 psf) = 432 kips 
12.2.4.4 Base Shear 
Using Provisions Eq. 5.3.2 [5.2-1]: 
V = CsW = 0.15(432 kips) = 64.8 kips 
12.2.4.5 Redundancy Factor 
For the longitudinal direction, the diaphragm is the horizontal bracing in the bay with the braced frames. 
However, given the basis for the redundancy factor, it appears that a more appropriate definition of Ax 
would be the area contributing to horizontal forces in the diagonal braces. Thus Ax = 20(240) = 4800 ft2. 
The ratio rx is 0.25; each of the four braces has the same stiffness, and each is capable of tension and 
compression. Therefore: 
. = 2 - 20 = < 
025 
085 10 10 
. 
. ., . 
4800 
use
Chapter 12, Nonbuilding Structure Design 
12-11 
8'-0" 
3'-0" 3'-0" 3'-0" 3'-0" 
3'-0" 
8'-0" 8'-0" 8'-0" 8'-0" 
N 
W E 
S 
Figure 12-3 Steel storage rack (1.0 ft = 0.3048 m). 
[The redundancy requirements have been changed substantially in the 2003 Provisions.] 
12.3 STEEL STORAGE RACK, OXFORD, MISSISSIPPI 
This example uses the equivalent lateral force (ELF) procedure to calculate the seismic base shear in the 
east-west direction for a steel storage rack. 
12.3.1 Description 
A four-tier, five-bay steel storage rack is located in a retail discount warehouse. There are concentrically 
braced frames in the north-south and east-west directions. The general public has direct access to the 
aisles and merchandise is stored on the upper racks. The rack is supported on a slab on grade. The design 
operating load for the rack contents is 125 psf on each tier. The weight of the steel support structure is 
assumed to be 5 psf on each tier. 
12.3.2 Provisions Parameters 
12.3.2.1 Ground Motion
FEMA 451, NEHRP Recommended Provisions: Design Examples 
12-12 
The spectral response acceleration coefficients at the site are as follows: 
SDS = 0.40 
SD1 = 0.18 
[The 2003 Provisions have adopted the 2002 USGS probabilistic seismic hazard maps, and the maps have 
been added to the body of the 2003 Provisions as figures in Chapter 3 (instead of the previously used 
separate map package).] 
12.3.2.2 Seismic Use Group and Importance Factor 
Use Provisions Sec. 1.3 [1.2]. The storage rack is in a retail facility. Therefore the storage rack is 
assigned to Seismic Use Group I. According to Provisions Sec. 14.6.3.1 and 6.1.5 [14.3.5.2], I = Ip = 1.5 
because the rack is in an area open to the general public. 
12.3.2.3 Seismic Design Category 
Use Provisions Tables 4.2.1a and 4.2.1b [1.4-1 and 1.4-2]. Given Seismic Use Group I, SDS = 0.40, and 
SD1 = 0.18, the Seismic Design Category is C. 
12.3.2.4 Design Coefficients 
According to Provisions Table 14.5.1.1 [14.2-3], the design coefficients for this steel storage rack are 
R = 4 
O0 = 2 
Cd = 3 
12.3.3 Design of the System 
12.3.3.1 Seismic Response Coefficient 
Provisions Sec. 14.6.3 [14.3.5]allows designers some latitude in selecting the seismic design 
methodology. Designers may use the Rack Manufacturers Institute specification if they modify the 
equations to incorporate the seismic spectral ordinates contained in the Provisions; or they may use an R 
of 4 and use Provisions Chapter 5 according to the exception in Provisions Sec. 14.6.3.1. The exception 
is used in this example. [In the 2003 Provisions these requirements have been restructured so that the 
primary method is use of Chapter 5 with the design coefficients of Chapter 14; racks designed using the 
RMI method of Sec. 14.3.5.6 are deemed to comply.] 
Using Provisions Eq. 5.4.1.1-1 [5.2-3]: 
0.4 0.15 
4 1.5 
DS 
s 
C S
R I 
= = = 
From analysis, T = 0.24 seconds. For this particular example the short period spectral value controls the 
design. The period, for taller racks, however, may be significant and will be a function of the operating 
weight. Using Provisions Eq. 5.4.1.1-2 [5.2-3], CS does not need to exceed 
( ) ( ) 
0.18 0.281 
0.24 4 1.5 
D1 
s 
C S 
T R I 
= = =
Chapter 12, Nonbuilding Structure Design 
12-13 
Provisions Sec. 14.5.1 [14.2.8]provides equations for minimum values of Cs that replace corresponding 
equations in Sec. 5.4.1.1 [5.2.1.1]. The equations in Sec. 14.5.1 [14.2.8] are more conservative than those 
in Sec. 5.4.1.1 [5.2.1.1] because nonbuilding structures generally lack redundancy and are not as highly 
damped as building structures. These equations generally govern the design of systems with long periods. 
According to Item 2 of Sec. 14.5.1 [14.2.8], replacement of the Chapter 5 equations for minima occurs 
only for nonbuilding systems that have an R value provided in Table 14.5.1.1 [14.2-2]. In the present 
example the R value is taken from Table 14.5.1.1 [14.2-2]and the Seismic Design Category is C so Eq. 
14.5.1-1 [14.2-2] applies. Using that equation, Cs shall not be less than the following: 
Cs = 0.14SDSI = 0.14(0.4)(1.5) = 0.084 
Provisions Eq. 5.4.1.1-1[5.2-2] controls; Cs = 0.15. 
12.3.3.2 Condition a (each rack loaded) 
12.3.3.2.1 Seismic Weight 
In accordance with Provisions Sec. 14.6.3.2 [14.3.5.3], Item a: 
Wa = 4(5)(8 ft)(3 ft)[0.67(125 psf)+5 psf] = 42.6 kips 
12.3.3.2.2 Design Forces and Moments 
Using Provisions Eq. 5.4.1 [5.2-1], the design base shear for condition a is calculated 
Va = CsW = 0.15(42.6 kips) = 6.39 kips 
In order to calculate the design forces, shears, and overturning moments at each level, seismic forces must 
be distributed vertically in accordance with Provisions Sec. 14.6.3.3 [14.3.5.4]. The calculations are 
shown in Table 12.3-1.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
12-14 
Table 12.3-1 Seismic Forces, Shears, and Overturning Moments 
Level 
x 
Wx 
(kips) 
hx 
(ft) 
wxh k
x 
(k = 1) 
Cvx Fx 
(kips) 
Vx 
(kips) 
Mx 
(ft-kips) 
5 10.65 12 127.80 0.40 2.56 
2.56 7.68 
4 10.65 9 95.85 0.30 1.92 
4.48 21.1 
3 10.65 6 63.90 0.20 1.28 
5.76 38.4 
2 10.65 3 31.95 0.10 0.63 
6.39 57.6 
S 42.6 319.5 
1.0 ft = 0.3048 m, 1.0 kip = 4.45 kN, 1.0 ft-kip = 1.36 kN-m. 
12.3.3.2.3 Resisting Moment at the Base 
MOT, resisting = Wa (1.5 ft) = 42.6(1.5 ft) = 63.9 ft-kips 
12.3.3.3 Condition b (only top rack loaded) 
12.3.3.3.1 Seismic Weight 
In accordance with Provisions Sec. 14.6.3.2 [14.3.5.3], Item b: 
Wb = 1(5)(8 ft)(3 ft)(125 psf) + 4(5)(8 ft)(3 ft)(5 psf) = 17.4 kips 
12.3.3.3.2 Base Shear 
Using Provisions Eq. 5.4.1 [5.2-1], the design base shear for condition b is calculated as follows: 
Vb = CsW = 0.15(17.4 kips) = 2.61 kips 
12.3.3.3.3 Overturning Moment at the Base 
Although the forces could be distributed as shown above for condition a, a simpler, conservative 
approach for condition b is to assume that a seismic force equal to the entire base shear is applied at the 
top level. Using that simplifying assumption, 
MOT = Vb (12 ft) = 2.61 kip (12 ft) = 31.3 ft-kips 
12.3.3.3.4 Resisting Moment at the Base 
MOT, resisting = Wb (1.5 ft) = 17.4(1.5 ft) = 26.1 ft-kips 
12.3.3.4 Controlling Conditions 
Condition a controls shear demands at all but the top level. 
Although the overturning moment is larger under condition a , the resisting moment is larger than the 
overturning moment. Under condition b the resistance to overturning is less than the applied
Chapter 12, Nonbuilding Structure Design 
12-15 
overturning moment. Therefore, the rack anchors must be designed to resist the uplift induced by the 
base shear for condition b. 
12.3.3.5 Torsion 
It should be noted that the distribution of east-west seismic shear will induce torsion in the rack system 
because the east-west brace is only on the back of the storage rack. The torsion should be resisted by the 
north-south braces at each end of the bay where the east-west braces are placed. If the torsion were to be 
distributed to each end of the storage rack, the engineer would be required to calculate the transfer of 
torsional forces in diaphragm action in the shelving, which may be impractical. 
12.4 ELECTRIC GENERATING POWER PLANT, MERNA, WYOMING 
This example highlights some of the differences between the design of nonbuilding structures and the 
design of building structures. The boiler building in this example illustrates a solution using the 
equivalent lateral force (ELF) procedure. Due to mass irregularities, the boiler building would probably 
also require a modal analysis. For brevity, the modal analysis is not illustrated. 
12.4.1 Description 
Large boilers in coal-fired electric power plants are generally suspended from support steel near the roof 
level. Additional lateral supports (called buck stays) are provided near the bottom of the boiler. The 
buck stays resist lateral forces but allow the boiler to move vertically. Lateral seismic forces are resisted 
at the roof and at the buck stay level. Close coordination with the boiler manufacturer is required in order 
to determine the proper distribution of seismic forces. 
In this example, a boiler building for a 950 mW coal-fired electric power generating plant is braced 
laterally with ordinary concentrically braced frames in both the north-south and the east-west directions. 
The facility is part of a grid and is not for emergency back up of a Seismic Use Group III facility. 
The dead load of the structure, equipment, and piping, WDL, is 16,700 kips. 
The weight of the boiler in service, WBoiler, is 31,600 kips. 
The natural period of the structure (determined from analysis) is as follows: 
North-South, TNS = 1.90 seconds 
East-West, TEW = 2.60 seconds
FEMA 451, NEHRP Recommended Provisions: Design Examples 
12-16 
N S
E 
W 
Buck stays 
(Typical) 
BOILER 
Plate girder to 
support boiler 
(Typical) 
240'-0" 
185'-0" 
230'-0" 
Section A-A 
A A
Chapter 12, Nonbuilding Structure Design 
12-17 
12.4.2 Provisions Parameters 
Seismic Use Group (Provisions Sec. 1.3 [1.2]) = II 
(for continuous operation, but not for emergency 
back up of a Seismic Use Group III facility) 
Occupancy Importance Factor, I (Provisions Sec. 1.4 [14.2.1]) = 1.25 
Site Coordinates = 42.800 N, 110.500 W 
Short Period Response, SS (Seismic Design Parameters) = 0.966 
One Second Period Response, S1 (Seismic Design Parameters) = 0.278 
Site Class (Provisions Sec. 4.1.2.1 [3.5]) = D (default) 
Acceleration-based site coefficient, Fa (Provisions Table 4.1.2.4a 
[3.3-1]) = 1.11 
Velocity-based site coefficient, Fv (Provisions Table 4.1.2.4b 
[3.3-2]) = 1.84 
Design spectral acceleration response parameters 
SDS = (2/3)SMS = (2/3)FaSS = (2/3)(1.11)(0.966) = 0.715 
SD1 = (2/3)SM1 = (2/3)FvS1 = (2/3)(1.84)(0.278) = 0.341 
Seismic Design Category (Provisions Sec. 4.2 [1.4]) = D 
Seismic-Force-Resisting System (Provisions Table 14.5.1.1 
[14.2-2]) = Steel concentrically braced 
frame (Ordinary) 
Response Modification Coefficient, R (Provisions Table 5.2.2) = 5 
System Overstrength Factor, O0 (Provisions Table 5.2.2) = 2 
Deflection Amplification Factor, Cd (Provisions Table 5.2.2) = 4 
Height limit (Provisions Table 14.5.1.1) = None 
Note: If the structure were classified as a building, its height would be limited to 35 ft for a Seismic 
Design Category D ordinary steel concentrically braced frame, according to the Provisions Table 5.2.2. 
The structure is, however, defined as a nonbuilding structure according to Provisions Sec. 14.6.3.4. 
Provisions Table 14.5.1.1 does not restrict the height of a nonbuilding structure using an ordinary steel 
concentrically braced frame. 
[Changes in the 2003 Provisions would affect this example significantly. Table 14.2-2 would be used to 
determine design coefficients and corresponding levels of detailing. For structures of this height using an 
ordinary concentrically braced frame system, R = 1.5, O0 
= 1, and Cd = 1.5. Alternatively, a special 
concentrically braced frame system could be employed.]
FEMA 451, NEHRP Recommended Provisions: Design Examples 
12-18 
12.4.3 Design in the North-South Direction 
12.4.3.1 Seismic Response Coefficient 
Using Provisions Eq. 5.4.1.1-1[5.2-2]: 
0.715 0.179 
5 1.25 
DS 
s 
C S
R I 
= = = 
From analysis, T = 1.90 seconds. Using Provisions Eq. 5.4.1.1-2 [5.2-3], CS does not need to exceed 
( ) ( ) 
0.341 0.045 
1.90 5 1.25 
D1 
s 
C S 
T R I 
= = = 
but using Provisions Eq. 5.4.1.1-3, Cs shall not be less than: 
Cs = 0.044ISDS = 0.044(1.25)(0.715) = 0.0393 
[Under the 2003 Provisions no additional minimum base shear must be considered since the example 
structure would satisfy exception 1 of Sec. 14.2.8 and the minimum base shear equation in Chapter 5 was 
removed.] 
Provisions Eq. 5.4.1.1-2 [5.2-3] controls; Cs = 0.045. 
12.4.3.2 Seismic Weight 
Calculate the total seismic weight, W, as: 
W = WDL + WBoiler = 16,700 kips + 31,600 kips = 48,300 kips 
12.4.3.3 Base Shear 
Using Provisions Eq. 5.4.1 [5.2-1]: 
V = CsW = 0.045(48,300 kips) = 2170 kips 
12.4.3.4 Redundancy Factor 
Refer to Sec. 12.2.3.6 for an explanation of the application of this factor to nonbuilding structures similar 
to buildings. The seismic force resisting system is an ordinary concentric braced frame with five columns 
in a single line of framing. The number of bays of bracing diminishes near the top, and the overall plan 
area is large. For the purposes of this example, it will be assumed that the structure lacks redundancy and 
. = 1.5. 
[The redundancy requirements have been substantially changed in the 2003 Provisions. If it is assumed 
that the structure would fail the redundancy criteria, . = 1.3.] 
12.4.3.5 Determining E 
See Sec. 12.2.3.7.
Chapter 12, Nonbuilding Structure Design 
12-19 
12.4.4 Design in the East-West Direction 
12.4.4.1 Seismic Response Coefficient 
Using Provisions Eq. 5.4.1.1-1 [5.2-2]: 
0.715 0.179 
5 1.25 
DS 
s 
C S
R I 
= = = 
From analysis, T = 2.60 seconds. Using Provisions Eq. 5.4.1.1-2 [5.2-3], CS does not need to exceed: 
( ) ( ) 
0.341 0.0328 
2.60 5 1.25 
D1 
s 
C S 
T R I 
= = = 
Using Provisions Eq. 5.4.1.1-3, Cs shall not be less than: 
Cs = 0.044ISDS = 0.044(1.25)(0.715) = 0.0393 
[Under the 2003 Provisions no additional minimum base shear must be considered since the example 
structure would satisfy exception 1 of Sec. 14.2.8 and the minimum base shear equation in Chapter 5 was 
removed.] 
Provisions Eq. 5.4.1.1-3 controls; Cs = 0.0393. [Under the 2003 Provisions, Eq. 5.2-3 would control the 
base shear coefficient for this example.] 
12.4.4.2 Seismic Weight 
Calculate the total seismic weight, W, as 
W = WDL + WBoiler = 16,700 kips + 31,600 kips = 48,300 kips 
12.4.4.3 Base Shear 
Using Provisions Eq. 5.4.1 [5.2-1]: 
V = CsW = 0.0393(48,300 kips) = 1900 kips 
12.5 PIER/WHARF DESIGN, LONG BEACH, CALIFORNIA 
This example illustrates the calculation of the seismic base shear in the east-west direction for the pier 
using the ELF procedure. 
12.5.1 Description 
A private shipping company is developing a pier in Long Beach, California, to service container vessels. 
In the north-south direction, the pier is tied directly to an abutment structure supported on grade. In the 
east-west direction, the pier resists seismic forces using moment frames. 
The design live load for container storage is 1000 psf.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
12-20 
15'-0" 
PLAN N 
ELEVATION 
3'-0" 3'-0" 
S 
W
E 
30'-0" 20'-0" 10'-0" 
Mean Sea Level 
Mud 
Dense 
sand 
10'-0" 10'-0" 10'-0" 10'-0" 3'-0" 
Figure 12-5 Pier plan and elevation (1.0 ft = 0.3048 m). 
12.5.2 Provisions Parameters 
Seismic Use Group (Provisions Sec. 1.3 [1.2]) = I 
(The pier serves container vessels that carry no 
hazardous materials.) 
Importance Factor, I (Provisions Sec. 14.5.1.2 [14.2.1]) = 1.0 
Short Period Response, SS = 1.75 
One Second Period Response, S1 = 0.60 
Site Class (Provisions Sec. 4.1.2.1 [3.5]) = D (dense sand) 
Acceleration-based Site Coefficient, Fa (Provisions Table 4.1.2.4a 
[3.3-1]) = 1.0
Chapter 12, Nonbuilding Structure Design 
12-21 
Velocity-based Site Coefficient, Fv (Provisions Table 4.1.2.4b 
[3.3-2]) = 1.5 
Design spectral acceleration response parameters 
SDS = (2/3)SMS = (2/3)FaSS = (2/3)(1.0)(1.75) = 1.167 
SD1 = (2/3)SM1 = (2/3)FvS1 = (2/3)(1.5)(0.60) = 0.60 
Seismic Design Category (Provisions Sec. 4.2) = D 
Seismic-Force-Resisting System (Provisions Table 14.5.1.1 
[14.2-2]) = Intermediate concrete 
moment frame 
Response Modification Coefficient, R (Provisions Table 5.2.2) = 5 
(The International Building Code and the 2002 edition of 
ASCE 7 would require an R value of 3.) 
System Overstrength Factor, O0 (Provisions Table 5.2.2) = 3 
Deflection Amplification Factor, Cd (Provisions Table 5.2.2) = 4 
Height limit (Provisions Table 14.5.1.1) = 50 ft 
If the structure was classified as a building, an intermediate reinforced concrete moment frame 
would not be permitted in Seismic Design Category D. 
[Changes in the 2003 Provisions would affect this example significantly. Table 14.2-2 would be used to 
determine design coefficients and corresponding levels of detailing. For structures of this height using an 
intermediate concrete moment frame system, R = 3, O0 
= 2, and Cd = 2.5.] 
12.5.3 Design of the System 
12.5.3.1 Seismic Response Coefficient 
Using Provisions Eq. 5.4.1.1-1 [5.2-2]: 
1.167 0.233 
5 1.0 
DS 
s 
C S
R I 
= = = 
From analysis, T = 0.596 seconds. Using Provisions Eq. 5.4.1.1-2 [5.2-3], CS does not need to exceed: 
( ) ( ) 
0.60 0.201 
0.596 5 1.0 
D1 
s 
C S 
T R I 
= = = 
Using Provisions Eq. 5.4.1.1-3 , Cs shall not be less than: 
Cs = 0.044ISDS = 0.044(1.0)(1.167) = 0.0513
FEMA 451, NEHRP Recommended Provisions: Design Examples 
12-22 
[Under the 2003 Provisions no additional minimum base shear must be considered since the example 
structure would satisfy exception 1 of Sec. 14.2.8 and the minimum base shear equation in Chapter 5 was 
removed.] 
Provisions Eq. 5.4.1.1-2 [5.2-3] controls; Cs = 0.201. 
12.5.3.2 Seismic Weight 
In accordance with Provisions Sec. 5.3 [5.2.1] and 14.6.6 [14.2.6], calculate the dead load due to the deck, 
beams, and support piers, as follows: 
WDeck = 1.0 ft(43 ft)(21 ft)(0.150 kip/ft3) = 135.5 kips 
WBeam = 4(2 ft)(2 ft)(21 ft)(0.150 kip/ft3) = 50.4 kips 
WPier = 8[p(1.25 ft)2][(10 ft - 3 ft) + (20 ft)/2](0.150 kip/ft3) = 100.1 kips 
WDL = WDeck + WBeams + WPiers = 135.5 + 50.4 + 100.1 = 286.0 kips 
Calculate 25 percent of the storage live load 
W1/4 LL = 0.25(1000 psf)(43 ft)(21 ft) = 225.8 kips 
Calculate the weight of the displaced water (Provisions Sec. 14.6.6 [14.3.3.1]) 
WDisp. water = 8[p(1.25 ft)2](20 ft)(64 pcf) = 50.27 kips 
Therefore, the total seismic weight is 
W = WDL + W1/4LL + WDisp. water = 286.0 + 225.8 + 50.27 = 562.1 kips 
12.5.3.3 Base Shear 
Using Provisions Eq. 5.3.2 [5.2-1]: 
V = CsW = 0.201(562.1 kips) = 113.0 kips 
12.5.3.4 Redundancy Factor 
This structure is small in area and has a large number of piles. Following the method described in Sec. 
12.2.3.6, yields . = 1.0. 
12.6 TANKS AND VESSELS, EVERETT, WASHINGTON 
The seismic response of tanks and vessels can be significantly different from that of buildings. For a 
structure composed of interconnected solid elements, it is not difficult to recognize how ground motions 
accelerate the structure and cause inert forces within the structure. Tanks and vessels, when empty, 
respond in a similar manner. 
When there is liquid in the tank, the response is much more complicated. As earthquake ground motions 
accelerate the tank shell, the shell applies lateral forces to the liquid. The liquid, which responds to those 
lateral forces. The liquid response may be amplified significantly if the period content of the earthquake 
ground motion is similar to the natural sloshing period of the liquid.
Chapter 12, Nonbuilding Structure Design 
12-23 
H = 10'-0" 5'-0" 
H R = 15'-0" 
D = 20'-0" 
Freeboard 
d s 
Figure 12-6 Storage tank section (1.0 ft = 0.3048 m). 
Earthquake-induced impulsive fluid forces are those calculated assuming that the liquid is a solid mass. 
The convective fluid forces are those that result from sloshing in the tank. It is important to account for 
the convective forces on columns and appurtenances inside the tank, because they are affected by 
sloshing in the same way that waves affect a pier in the ocean. 
The freeboard considerations are critical. Often times, the roof acts as a structural diaphragm. If a tank 
does not have sufficient freeboard, the sloshing wave can rip the roof from the wall of the tank. This 
could result in the failure of the wall and loss of the liquid within. 
The nature of seismic design for liquid containing tanks and vessels is complicated. The fluid mass that is 
effective for impulsive and convective seismic forces is discussed in the literature referenced in the 
NEHRP Provisions and Commentary. 
12.6.1 Flat-Bottom Water Storage Tank 
12.6.1.1 Description 
This example illustrates the calculation of the design base shear using the equivalent lateral force (ELF) 
procedure for a steel water storage tank used to store potable water for a process within a chemical plant 
(Figure 12-6). 
The tank is located away from personnel working within the facility. 
The weight of the tank shell, roof, and equipment is 15,400 lb. 
12.6.1.2 Provisions Parameters 
Seismic Use Group (Provisions Sec. 1.3 [1.2]) = I 
Importance Factor, I (Provisions Sec. 14.5.1.2 [14.2.1]) = 1.0 
Site Coordinates = 48.000 N, 122.250 W 
Short Period Response, SS = 1.236
FEMA 451, NEHRP Recommended Provisions: Design Examples 
12-24 
One Second Period Response, S1 = 0.406 
Site Class (Provisions Sec. 4.1.2.1 [3.5]) = C (per geotech) 
Acceleration-based Site Coefficient, Fa (Provisions Table 4.1.2.4a 
[3.3-1]) = 1.0 
Velocity-based Site Coefficient, Fv (Provisions Table 4.1.2.4b 
[3.3-2]) = 1.39 
Design spectral acceleration response parameters 
SDS = (2/3)SMS = (2/3)FaSS = (2/3)(1.0)(1.236) = 0.824 
SD1 = (2/3)SM1 = (2/3)FvS1 = (2/3)(1.39)(0.406) = 0.376 
Seismic-Force-Resisting System (Provisions Table 14.5.1.1 
[14.2-3]) = Flat-bottom, groundsupported, 
anchored, bolted 
steel tank 
Response Modification Coefficient, R (Provisions Table 14.5.1.1 
[14.2-3]) = 3 
System Overstrength Factor, O0 (Provisions Table 5.2.2 
[14.2-3]) = 2 
Deflection Amplification Factor, Cd (Provisions Table 5.2.2 
[14.2-3]) = 2 
[The 2003 Provisions have adopted the 2002 USGS probabilistic seismic hazard maps, and the maps have 
been added to the body of the 2003 Provisions as figures in Chapter 3 (instead of the previously used 
separate map package). The CD-ROM also has been updated.] 
12.6.1.3 Calculations for Impulsive Response 
12.6.1.3.1 Natural Period for the First Mode of Vibration 
Based on analysis, the period for impulsive response of the tank and its contents is Ti = 0.14 sec. 
12.6.1.3.2 Spectral Acceleration 
Based on Provisions Figure 14.7.3.6-1 [14.4-1]: 
0.376 0.456 seconds 
0.824 
D1 
s 
DS 
T S
S 
= = = 
Using Provisions Sec. 14.7.3.6.1 [114.4.7.5.1] with Ti < Ts,: 
Sai = SDS = 0.824 
12.6.1.3.3 Seismic (Impulsive) Weight
Chapter 12, Nonbuilding Structure Design 
12-25 
Wtank = 15.4 kips 
W1water = p(10 ft)2(10 ft)(0.0624 kip/ft3) (W1/WT)= 196.0 (0.75) kips = 147 kips 
The ratio W1/WT (= 0.75) was determined from AWWA D100 (it depends on the ratio of height 
to diameter) 
Wi = Wtank + W1water = 15.4 + 147 = 162.4 kips 
12.6.1.3.4 Base Shear 
According to Provisions Sec. 14.7.3.6.1 [14.4.7.5.1]: 
( ) 
V S W 
i R 
= ai i = = 
0824 162 4 
3 
44 6 
. . 
. 
kips 
kips 
12.6.1.4 Calculations for Convective Response Natural Period for the First Mode of Sloshing 
12.6.1.4.1 Natural Period for the First Mode of Sloshing 
Using Provisions Section 14.7.3.6.1 [14.4.7.5.1]: 
( )2 
ft 
s 
2 2 20 ft 2.58 s 
3.68 tanh 3.68 3.68 32.174 tanh 3.68(10 ft) 
10 ft 
c 
T D 
g H 
D 
=p =p = 
. . . . 
. . . . 
. . . . 
12.6.1.4.2 Spectral Acceleration 
Using Provisions Sec. 14.7.3.6.1 [14.4.7.5.1] with Tc < 4 seconds: 
1.5 1.5(0.376) 0.219 
2.58 
D1 
ac 
c 
S S 
T 
= = = 
12.6.1.4.3 Seismic (Convective) Weight 
Wc = Wwater (W2/WT) = 196 (0.30) = 58.8 kips 
The ratio W2/WT (= 0.30) was determined from AWWA D100. 
12.6.1.4.4 Base Shear 
According to Provisions Sec. 14.7.3.6.1 [14.4.7.5.1]: 
( ) 
V S W
c R 
= ac c = = 
0 219 588 
3 
4 29 
. . 
. 
kips 
kips
FEMA 451, NEHRP Recommended Provisions: Design Examples 
12-26 
12.6.1.5 Design Base Shear 
Although Item b of Provisions Sec. 14.7.3.2 [14.4.7.1] indicates that impulsive and convective 
components may, in general, be combined using the SRSS method, Provisions Sec. 14.7.3.6.1 [14.4.7.5.1] 
requires that the direct sum be used for ground-supported storage tanks for liquids. Using Provisions Eq. 
14.7.3.6.1 [14.4-1]: 
V = Vi + Vc = 44.6 + 4.29 = 48.9 kips 
[In the 2003 Provisions, use of the SRSS method is also permitted for ground-supported storage tanks for 
liquids.] 
12.6.2 FLAT-BOTTOM GASOLINE TANK 
12.6.2.1 Description 
This example illustrates the calculation of the base shear and the required freeboard using the ELF 
procedure for a petro-chemical storage tank in a refinery tank farm near a populated city neighborhood. 
An impoundment dike is not provided to control liquid spills. 
The tank is a flat-bottom, ground-supported, anchored, bolted steel tank constructed in accordance with 
API 650. The weight of the tank shell, roof, and equipment is 15,400 lb. 
12.6.2.2 Provisions Parameters 
Seismic Use Group (Provisions Sec. 1.3 [1.2]) = III 
(The tank is used for storage of hazardous material.) 
Importance Factor, I (Provisions Sec. 14.5.1.2 [14.2.1]) = 1.5 
Site Coordinates = 48.000 N, 122.250 W 
Short Period Response, SS = 1.236 
One Second Period Response, S1 = 0.406 
Site Class (Provisions Sec. 4.1.2.1 [3.5]) = C (per geotech) 
Acceleration-based Site Coefficient, Fa (Provisions Table 4.1.2.4a 
[3.3-1]) = 1.0 
Velocity-based Site Coefficient, Fv (Provisions Table 4.1.2.4b 
[3.3-2]) = 1.39 
Design spectral acceleration response parameters 
SDS = (2/3)SMS = (2/3)FaSS = (2/3)(1.0)(1.236) = 0.824 
SD1 = (2/3)SM1 = (2/3)FvS1 = (2/3)(1.39)(0.406) = 0.376
Chapter 12, Nonbuilding Structure Design 
12-27 
Seismic-Force-Resisting System (Provisions Table 14.5.1.1 
[14.2-3]) = Flat-bottom, groundsupported, 
anchored, bolted 
steel tank 
Response Modification Coefficient, R (Provisions Table 14.5.1.1 
[14.2-3]) = 3 
System Overstrength Factor, O0 (Provisions Table 5.2.2 
[14.2-3]) = 2 
Deflection Amplification Factor, Cd (Provisions Table 5.2.2 
[14.2-3]) = 2 
[The 2003 Provisions have adopted the 2002 USGS probabilistic seismic hazard maps, and the maps have 
been added to the body of the 2003 Provisions as figures in Chapter 3 (instead of the previously used 
separate map package). The CD-ROM also has been updated.] 
12.6.2.3 Calculations for Impulsive Response 
12.6.2.3.1 Natural Period for the First Mode of Vibration 
Based on analysis, the period for impulsive response of the tank and its contents is Ti = 0.14 sec. 
12.6.2.3.2 Spectral Acceleration 
Based on Provisions Figure 14.7.3.6-1 [14.4-1]: 
0.376 0.456 seconds 
0.824 
D1 
s 
DS 
T S
S 
= = = 
Using Provisions Sec. 14.7.3.6.1 [ 14.4.7.5.1] with Ti < Ts,: 
Sai = SDS = 0.824 
12.6.2.3.3 Seismic (Impulsive) Weight 
Wtank = 15.4 kips 
WGas = p(10 ft)2(10 ft)(0.046 kip/ft3)(W1/WT) = 144.5 kips (0.75) = 108.4 kips 
Note: The ratio W1/WT was determined from AWWA D100, but API 650 should be used. 
Wi = Wtank + WGas = 15.4 + 108.4 = 123.8 kips 
12.6.2.3.4 Base Shear 
According to Provisions Sec. 14.7.3.6.1 [14.4.7.5.1]: 
( )( ) 
V 
S IW 
i R 
= ai i= = 
0824 15 1238 
3 
510 
. . . 
. 
kips 
kips
FEMA 451, NEHRP Recommended Provisions: Design Examples 
12-28 
12.6.2.4 Calculations for Convective Response 
12.6.2.4.1 Natural Period for the First Mode of Sloshing 
The dimensions are the same as those used for the water tank in Sec. 12.6.1; therefore, Tc = 2.58 sec. 
12.6.2.4.2 Spectral Acceleration 
Likewise, Sac = 0.219. 
12.6.2.4.3 Seismic (Convective) Weight 
Wc = WLNG (W2/WT) = 144.5 ( 0.30) = 43.4 kips 
The ratio W2/WT was determined from AWWA D100. 
12.6.2.4.4 Base shear 
According to Provisions Sec. 14.7.3.6.1 [14.4.7.5.1]: 
( )( ) 
V 
S IW 
c R 
= ac c= = 
0824 15 435 
3 
17 9 
. . . 
. 
kips 
kips 
12.6.2.5 Design Base Shear 
Using Provisions Eq. 14.7.3.6.1 [14.4-1]: 
V = Vi + Vc = 51.0 + 17.9 = 68.9 kips 
12.6.2.6 Minimum Freeboard 
Provisions Table 14.7.3.6.1.2 [14.4-2] indicates that a minimum freeboard equal to ds is required for this 
tank. Using Provisions Eq. 14.7.3.6.1.2 [14.4-9]: 
ds = 0.5DISac = 0.5(20 ft)(1.5)(0.219) = 3.29 ft 
The 5 ft freeboard provided is adequate. 
12.7 EMERGENCY ELECTRIC POWER SUBSTATION STRUCTURE, ASHPORT, 
TENNESSEE 
The main section addressing electrical transmission, substation, and distribution structures is in the 
appendix to Chapter 14 of the Provisions. The information is in an appendix so that designers can take 
time to evaluate and comment on the seismic design procedures before they are included in the main text 
of the Provisions. 
[In the 2003 Provisions Sections A14.2.1 and A14.2.2 were removed because the appropriate industry 
standards had been updated to include seismic design criteria and earthquake ground motions consistent 
with the Provisions. Therefore, all references to the Provisions in Sec. 12.7 of this chapter are obsolete.]
Chapter 12, Nonbuilding Structure Design 
12-29 
20'-0" 15'-0" 
7'-0" 
15'-0" 
15'-0" 
cg 
PLAN 
ELEVATION 
Figure 12-7 Platform for elevated transformer (1.0 ft = 0.3048 m). 
12.7.1 Description 
This example illustrates the calculation of the base shear using the ELF procedure for a braced frame that 
supports a large transformer (Figure 12-7). The substation is intended to provide emergency electric 
power to the emergency control center for the fire and police departments of a community. There is only 
one center designed for this purpose. 
The weight of the transformer equipment is 17,300 lb. 
The weight of the support structure is 12,400 lb. 
The period of the structure is T = 0.240 sec. 
Although the ratio of the supported structure over the total weight is greater than 25 percent, experience 
indicates that the transformer will behave as a lumped rigid mass.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
12-30 
12.7.2 Provisions Parameters 
12.7.2.1 Ground Motion 
The design response spectral accelerations are defined as 
SDS = 1.86 
SD1 = 0.79 
12.7.2.2 Seismic Use Group and Importance Factor 
The structure is for emergency electric power for a Seismic Use Group III facility. Therefore, the 
platform is assigned to Seismic Use Group III, as required by Provisions Sec. 1.3 [1.2]. Using Provisions 
Table 14.5.1.2 [14.2-1], the Importance Factor, I, is equal to 1.5. 
12.7.2.3 Response Modification Coefficient 
From Provisions Table 14A.2.1, R is 3. 
12.7.3 Design of the System 
12.7.3.1 Seismic Response Coefficient 
Provisions Sec. 14A.2.2 defines Cs in a manner that is not consistent with the rest of the Provisions. This 
inconsistency will be eliminated in future editions of the Provisions. In this example, the equations are 
applied in a manner that is consistent with Chapters 5 [4 and 5] and 14  that is, R is applied in the 
calculation of Cs rather than in the calculation of V. 
Using Provisions Section 14A.2.2: 
1.86 0.93 
3 1.5 
DS 
s 
C S
R I 
= = = 
but Cs need not be larger than: 
( ) ( ) 
0.79 1.646 
0.24 3 1.5 
D1 
s 
C S 
T R I 
= = = 
Therefore, Cs = 0.93. 
12.7.3.2 Seismic Weight 
W = WTransformer + WSupport structure = 17.3 + 12.4 = 29.7 kips 
12.7.3.3 Base Shear 
Using Provisions Section 14A.2.2: V = CsW = 0.93(29.7 kips) = 27.6 kips
13-1 
13 
DESIGN FOR 
NONSTRUCTURAL COMPONENTS 
Robert Bachman, P.E. and Richard Drake, P.E. 
Chapter 6 of the 2000 NEHRP Recommended Provisions and Commentary (hereinafter, the Provisions 
and Commentary) addresses architectural, mechanical, and electrical components of buildings. Two 
examples are presented here to illustrate many of the requirements and procedures. Design and anchorage 
are illustrated for exterior precast concrete cladding, and for a roof-mounted HVAC unit. The rooftop 
unit is examined in two common installations: directly attached, and isolated with snubbers. This chapter 
also contains an explanation of the fundamental aspects of the Provisions, and an explanation of how 
piping, designed according to the ASME Power Piping code, is checked for the force and displacement 
requirements of the Provisions. 
A large variety of materials and industries are involved with nonstructural components is large, and 
numerous documents define and describe methods of design, construction, manufacture, installation, 
attachment, etc. Some of the documents address seismic issues but many do not. Provisions Sec. 6.1.1 
[6.1.2] contains a listing of approved standards for various nonstructural components. 
Although the Guide is based on the 2000 Provisions, it has been annotated to reflect changes made to the 
2003 Provisions. Annotations within brackets, [ ], indicate both organizational changes (as a result of a 
reformat of all of the chapters of the 2003 Provisions) and substantive technical changes to the 2003 
Provisions and its primary reference documents. While the general concepts of the changes are 
described, the design examples and calculations have not been revised to reflect the changes to the 2003 
Provisions. 
A few noteworthy changes were made to the nonstructural components requirements of the 2003 
Provisions. These include explicit definition of load effects (including vertical seismic forces) within the 
chapter and revised classification of nonductile anchors (based on demonstrated ductility or 
prequalification rather than embedment-length-to-diameter ratio). 
In addition to changes Provisions Chapter 6, the basic earthquake hazard maps were updated and the 
concrete design reference was updated to ACI 318-02 (with a significant resulting changes to the 
calculations for anchors in concrete). 
Where they affect the design examples in this chapter of the Guide, significant changes to the 2003 
Provisions and primary reference documents are noted. However, some minor changes to the 2003 
Provisions and the reference documents may not be noted. 
In addition to the Provisions, the following are referenced in this chapter:
FEMA 451, NEHRP Recommended Provisions: Design Examples 
13-2 
ACI 318 American Concrete Institute. 1999 [2002]. Building Code Requirements and 
Commentary for Reinforced Concrete. 
ASCE 7 American Society of Civil Engineers. 1998 [2002]. Minimum Design Loads for 
Buildings and Other Structures. 
ASHRAE APP IP American Society of Heating, Refrigeration, and Air-Conditioning Engineers 
(ASHRAE). 1999. Seismic and Wind Restraint Design, Chapter 53. 
ASME B31.1 American Society of Mechanical Engineers. Power Piping Code. 
IBC International Code Council. 2000. International Building Code. 
The symbols used in this chapter are drawn from Chapter 2 of the Provisions or reflect common 
engineering usage. The examples are presented in U.S. customary units. 
[In the 2003 Provisions, definitions and symbols specific to nonstructural components appear in Sec. 
6.1.3 and 6.1.4, respectively.] 
13.1 DEVELOPMENT AND BACKGROUND OF THE PROVISIONS FOR 
NONSTRUCTURAL COMPONENTS 
13.1.1 Approach to Nonstructural Components 
The Provisions requires that nonstructural components be checked for two fundamentally different 
demands placed upon them by the response of the structure to earthquake ground motion: resistance to 
inertial forces and accommodation of imposed displacements. Building codes have long had 
requirements for resistance to inertial forces. Most such requirements apply to the component mass an 
acceleration that vary with the basic ground motion parameter and a few broad categories of components. 
The broad categories are intended distinguish between components whose dynamic response couples with 
that of the supporting structure in such a fashion as to cause the component response accelerations to be 
amplified above the accelerations of the structure and those components that are rigid enough with respect 
to the structure so that the component response is not amplified over the structural response. In recent 
years, a coefficient based on the function of the building or of the component have been introduced as 
another multiplier for components important to life safety or essential facilities. 
The Provisions includes an equation to compute the inertial force that involves two additional concepts: 
variation of the acceleration with relative height within the structure, and reduction in design force based 
upon available ductility in the component, or its attachment. The Provisions also includes a quantitative 
measure for the deformation imposed upon nonstructural components. The inertial force demands tend to 
control the seismic design for isolated or heavy components, whereas, the imposed deformations are 
important for the seismic design for elements that are continuous through multiple levels of a structure, or 
across expansion joints between adjacent structures, such as cladding or piping. 
The remaining portions of this section describe the sequence of steps and decisions prescribed by the 
Provisions to check these two seismic demands on nonstructural components. 
13.1.2 Force Equations 
The following seismic force equations are prescribed for nonstructural components:(Provisions Eq. 
6.1.3-1 [6.2-1], 6.1.3-2 [6.2-3], and 6.1.3-3 [6.2-4]):
Chapter 13, Nonstructural Components 
13-3 
0.4 
p DS p 1 2 
p 
p 
p 
F aRS W hz 
I 
= ..+ .. 
. . 
1.6 Fpmax = SDSIpWp 
0.3 Fpmin = SDSIpWp 
where: 
Fp = horizontal equivalent static seismic design force centered at the components center of gravity 
and distributed relative to the components mass distribution. 
ap = component amplification factor (either 1.0 or 2.5) as tabulated in Provisions Table 6.2.2 [6.3-1] 
for architectural components and Provisions Table 6.3.2 [6.4-1] for mechanical and electrical 
components (Alternatively, may be computed by dynamic analysis) 
SDS = five percent damped spectral response acceleration parameter at short period as defined in 
Provisions Sec. 4.1.2 [3.3.3] 
Wp = component operating weight 
Rp = component response modification factor (varies from 1.0 to 5.0) as tabulated in Provisions 
Table 6.2.2 [6.3-1]for architectural components and Provisions Table 6.3.2 [6.4-1] for 
mechanical and electrical components 
Ip = component importance factor (either 1.0 or 1.5) as indicated in Provisions Sec. 6.1.5 [6.2-2] 
z = elevation in structure of component point of attachment relative to the base 
h = roof elevation of the structure or elevation of highest point of the seismic-force-resisting system 
of the structure relative to the base 
The seismic design force, Fp, is to be applied independently in the longitudinal, and transverse directions. 
The effects of these loads on the component are combined with the effects of static loads. Provisions Eq. 
6.1.3-2 [6.2-3 and 6.2-4] and 6.1.3-3, provide maximum and minimum limits for the seismic design force. 
For each point of attachment, a force, Fp, should be determined based on Provisions Eq. 6.1.3-1 [6.2-1]. 
The minima and maxima determined from Provisions Eq. 6.1.3-2 and 6.1.3-3 [6.2-1] must be considered 
in determining each Fp. The weight, Wp, used to determine each Fp should be based on the tributary 
weight of the component associated with the point of attachment. For designing the component, the 
attachment force, Fp, should be distributed relative to the components mass distribution over the area 
used to establish the tributary weight. With the exception of the bearing walls, which are covered by 
Provisions Sec. 5.2.6.2.7 [4.6.1.3], and out-of-plane wall anchorage to flexible diaphragms, which is 
covered by Provisions Sec. 5.2.6.3.2 [4.6.2.1], each anchorage force should be based on simple statics 
determined by using all the distributed loads applied to the complete component. Cantilever parapets that 
are part of a continuous element, should be separately checked for parapet forces. 
13.1.3 Load Combinations and Acceptance Criteria
FEMA 451, NEHRP Recommended Provisions: Design Examples 
13-4 
13.1.3.1 Seismic Load Effects 
When the effects of vertical gravity loads and horizontal earthquake loads are additive, Provisions Eq. 
5.2.7.1-1 [4.2-1] is used: 
E = .QE + 0.2SDSD 
When the effects of vertical gravity load counteract those of horizontal earthquake loads, Provisions Eq. 
5.2.7.1-2 [4.2-2] is used: 
E = .QE - 0.2SDSD 
where: 
E = effect of horizontal and vertical earthquake-induced forces 
. = redundancy factor (= 1.0 for nonstructural components) 
QE = effect of horizontal seismic forces (due to application of Fp for nonstructural components) 
D = effect of dead load 
0.2SDSD = effect of vertical seismic forces 
13.1.3.2 Strength Load Combinations 
Provisions Sec. 5.2.7 [4.2.2] requires the use of ASCE 7 factored load combinations. The combinations 
from ASCE 7 Sec. 2.3.2 that include earthquake effects are: 
U = 1.2D + 1.0E + 0.5L + 0.2S 
U = 0.9D + 1.0E + 1.6H 
13.1.4 Component Amplification Factor 
The component amplification factor, ap, found in Provisions Eq. 6.1.3-1 [6.2-1] represents the dynamic 
amplification of the component relative to the maximum acceleration of the component support point(s). 
Typically, this amplification is a function of the fundamental period of the component, Tp, and the 
fundamental period of the support structure, T. It is recognized that at the time the components are 
designed or selected, the effective fundamental period of the structure, T, is not always available. It is 
also recognized that for a majority of nonstructural components, the component fundamental period, Tp, 
can be accurately obtained only by expensive shake-table or pullback tests. As a result, the determination 
of a components fundamental period by dynamic analysis, considering T/Tp ratios, is not always 
practicable. For this reason, acceptable values of ap have been provided in the Provisions tables. 
Therefore, component amplification factors from either these tables or a dynamic analysis may be used. 
Values for ap are tabulated for each component based on the expectation that the component will behave 
in either a rigid or a flexible manner. For simplicity, a step function increase based on input motion 
amplifications is provided to help distinguish between rigid and flexible behavior. If the fundamental 
period of the component is less than 0.06 seconds, no dynamic amplification is expected and ap may be 
taken to equal 1.00. If the fundamental period of the component is greater than 0.06 seconds, dynamic 
amplification is expected, and ap is taken to equal 2.50. In addition, a rational analysis determination of
Chapter 13, Nonstructural Components 
13-5 
ap is permitted if reasonable values of both T and Tp are available. Acceptable procedures for determining 
ap are provided in Commentary Chapter 6. 
13.1.5 Seismic Coefficient at Grade 
The short period design spectral acceleration, SDS, considers the site seismicity and local soil conditions. 
The site seismicity is obtained from the design value maps (or CD-ROM), and SDS is determined in 
accordance with Provisions Sec. 4.1.2.5 [3.3.3]. The coefficient SDS is the used to design the structure. 
The Provisions approximates the effective peak ground acceleration as 0.4SDS, which is why 0.4 appears 
in Provisions Eq. 6.1.3-1 [6.2-1]. 
[The 2003 Provisions have adopted the 2002 USGS probabilistic seismic hazard maps, and the maps have 
been added to the body of the 2003 Provisions as figures in Chapter 3 (instead of the previously used 
separate map package). The CD-ROM also has been updated.] 
13.1.6 Relative Location Factor 
The relative location factor, 1 2z , scales the seismic coefficient at grade, resulting in values linearly 
h 
.. + .. 
. . 
varying from 1.0 at grade to 3.0 at roof level. This factor approximates the dynamic amplification of 
ground acceleration by the supporting structure. 
13.1.7 Component Response Modification Factor 
The component response modification factor, Rp, represents the energy absorption capability of the 
components construction and attachments. In the absence of applicable research, these factors are based 
on judgment with respect to the following benchmark values: 
1. Rp = 1.0 or 1.5, brittle or buckling failure mode is expected 
2. Rp = 2.5, some minimal level of energy dissipation capacity 
3. Rp = 3.5 or 5.0, highly ductile materials and detailing 
13.1.8 Component Importance Factor 
The component importance factor, Ip, represents the greater of the life safety importance and/or the hazard 
exposure importance of the component. The factor indirectly accounts for the functionality of the 
component or structure by requiring design for a lesser amount of inelastic behavior (or higher force 
level). It is assumed that a lesser amount of inelastic behavior will result in a component that will have a 
higher likelihood of functioning after a major earthquake. 
13.1.9 Accommodation of Seismic Relative Displacements 
The Provisions requires that seismic relative displacements, Dp, be determined in accordance with several 
equations. For two connection points on Structure A (or on the same structural system), one at Level x 
and the other at Level y, Dp is determined from Provisions Eq. 6.1.4-1 [6.2-5] as: 
Dp =d xA-d yA 
Because the computed displacements are frequently not available to the designer of nonstructural 
components, one may use the maximum permissible structural displacements per Provisions Eq. 6.1.4-2 
[6.2-6]:
FEMA 451, NEHRP Recommended Provisions: Design Examples 
13-6 
( ) max 
aA 
p 
sx 
D X Yh
. 
= - 
For two connection points on Structures A and B (or on two separate structural systems), one at Level x, 
and the other at Level y, DP and DPmax are determined from Provisions Eq. 6.1.4-3 [6.2-7] and 6.1.4-4 
[6.2-8], respectively, as: 
Dp =d xA +d yB 
max 
aA aB 
p 
sx sx 
D X Y 
h h 
. . 
= + 
where: 
Dp = seismic relative displacement that the component must be designed to accommodate 
dxA = deflection of building Level x of Structure A, determined by an elastic analysis as defined 
in Provisions Sec. 5.4.6.1 or 5.5.5 [5.2.6, 5.3.5, or 5.4.3] and multiplied by the Cd factor 
dyA = deflection of building Level y of Structure A, determined in the same fashion as dxA 
X = height of upper support attachment at Level x as measured from the base 
Y = height of lower support attachment at Level y as measured from the base 
.aA = allowable story drift for Structure A as defined in Provisions Table 5.2.8 [4.5-1] 
hsx = story height used in the definition of the allowable drift, .a, in Provisions Table 5.2.8 
[4.5-1] 
dyB = deflection of building Level y of Structure B, determined in the same fashion as dxA 
.aB = allowable story drift for Structure B as defined in Provisions Table 5.2.8 [4.5-1] 
The effects of seismic relative displacements must be considered in combination with displacements 
caused by other loads as appropriate. Specific methods for evaluating seismic relative displacement 
effects of components and associated acceptance criteria are not specified in the Provisions. However, 
the intention is to satisfy the purpose of the Provisions. Therefore, for nonessential facilities, 
nonstructural components can experience serious damage during the design level earthquake provided 
they do not constitute a serious life safety hazard. For essential facilities, nonstructural components can 
experience some damage or inelastic deformation during the design level earthquake provided they do not 
significantly impair the function of the facility. 
13.1.10 Component Anchorage Factors and Acceptance Criteria 
Design seismic forces in the connected parts, Fp, are prescribed in Provisions Sec. 6.1.3 [6.2.6]. 
When component anchorage is provided by expansion anchors or shallow anchors, a value of Rp = 1.5 is 
used. Shallow anchors are defined as those with embedment length-to-diameter ratios of less than 8. 
Anchors embedded in concrete or masonry are proportioned to carry the least of the following: 
1. The design strength of the connected part
Chapter 13, Nonstructural Components 
13-7 
2. 1.3 times the prescribed seismic design force or 
3. The maximum force that can be transferred to the connected part by the component structural system 
Determination of design seismic forces in anchors must consider installation eccentricities, prying effects, 
multiple anchor effects, and the stiffness of the connected system. 
Use of powder-driven fasteners is not permitted for seismic design tension forces in Seismic Design 
Categories D, E, and F unless approved for such loading. 
[In the 2003 Provisions reference is made to power-actuated fasteners so as to cover a broader range of 
fastener types than is implied by powder-driven.] 
The design strength of anchors in concrete is determined in accordance with, Provisions Chapter 9, which 
is basically the same as IBC Sec. 1913, and has been updated somewhat in Appendix D of ACI 318-2002. 
(These rules for anchors in concrete will probably be deleted from the next edition of the Provisions in 
favor of a reference to ACI 318.) 
[The 2003 Provisions refer to Appendix D of ACI 318-02 rather than providing specific, detailed 
requirements.] 
13.1.11 Construction Documents 
Construction documents must be prepared by a registered design professional and must include sufficient 
detail for use by the owner, building officials, contractors, and special inspectors; Provisions Table 6.1.7 
[6.2-1] includes specific requirements. 
13.2 ARCHITECTURAL CONCRETE WALL PANEL 
13.2.1 Example Description 
In this example, the architectural components are a 4.5-in.-thick precast normal weight concrete spandrel 
panel and a column cover supported by the structural steel frame of a five-story building as shown in 
Figures 13.2-1 and 13.2-2.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
13-8 
5 at 13'-6" = 67'-6" 
24'-0" 
Structural 
steel frame 
Typical window 
frame system 
Spandrel panel 
under consideration 
Column cover 
under consideration 
Figure 13.2-1 Five-story building elevation showing panel location (1.0 ft = 0.3048 m)
Chapter 13, Nonstructural Components 
13-9 
13'-6" story height 
T.O.S 
T.O.S 
24'-0" 
6'-6" panel 7'-0" column cover 
3'-0" 
Spandrel panel 
under consideration 
Typical window 
frame system 
Column cover 
under consideration 
~ 
~ 
Figure 13.2-2 Detailed building elevation (1.0 ft = 0.3048 m). 
The columns, at the third level of the five-story office building, support the spandrel panel under 
consideration. The columns between the third and fourth levels of the building support the column cover 
under consideration. The building, located near a significant active fault in Los Angeles, California, is 
assigned to Seismic Use Group I. Wind pressures normal to the building are 17 psf determined in 
accordance with ASCE 7. The spandrel panel supports glass windows weighing 10 psf. 
This example develops prescribed seismic forces for the selected spandrel panel and prescribed seismic 
displacements for the selected column cover. 
It should be noted that details of precast connections vary according to the preferences and local practices 
of the precast panel supplier. In addition, some connections may involve patented designs. As a result, 
this example will concentrate on quantifying the prescribed seismic forces and displacements. After the 
prescribed seismic forces and displacements are determined, the connections can be detailed and designed 
according to the appropriate AISC and ACI codes and PCI (Precast/Prestressed Concrete Institute) 
recommendations. 
13.2.2 Design Requirements 
13.2.2.1 Provisions Parameters and Coefficients 
ap = 1.0 for wall panels (Provisions Table 6.2.2 [6.3-1]) 
ap = 1.25 for connection fasteners (Provisions Table 6.2.2 [6.3-1])
FEMA 451, NEHRP Recommended Provisions: Design Examples 
13-10 
SDS = 1.487 (Design Values CD-ROM for the selected location and site class) 
[The 2003 Provisions have adopted the 2002 USGS probabilistic seismic hazard maps, and the maps have 
been added to the body of the 2003 Provisions as figures in Chapter 3 (instead of the previously used 
separate map package). The CD-ROM also has been updated.] 
Seismic Design Category = D (Provisions Table 4.2.1a [1.4-1] for S1 < 0.75) 
[In the footnote to 2003 Provisions Table 1.4-1, the value of S1 used to trigger assignment to Seismic 
Design Category E or F was changed from 0.75 to 0.6.] 
Spandrel Panel Wp = (150 lb/ft3)(24 ft)(6.5 ft)(0.375 ft) = 8775 lb 
Glass Wp = (10 lb/ft2)(21 ft)(7 ft) = 1470 lb (supported by spandrel panel) 
Column Cover Wp = (150 lb/ft3)(3 ft)(7 ft)(0.375 ft) = 1181 lb 
Rp = 2.5 for wall panels (Provisions Table 6.3.2 [6.3-1]) 
Rp = 1.0 for connection fasteners ((Provisions Sec. 6.1.6.1 [6.2.8.1]) 
[In the 2003 Provisions component anchorage is designed using Rp = 1.5 unless specific ductility or 
prequalification requirements are satisfied.] 
Ip = 1.0 (Provisions Sec. 6.1.5 [6.2.2]) 
40.5 ft 0.6 (at third floor) 
67.5 ft 
z
h
= = 
. = 1.0 (Provisions Sec. 6.1.3) 
[The 2003 Provisions indicate that the redundancy factor does not apply to the design of nonstructural 
components. Although the effect is similar to stating that . = 1, there is a real difference since load 
effects for such components and their supports and attachments are now defined in Chapter 6 rather than 
by reference to Chapter 4.] 
13.2.2.2 Performance Criteria 
Component failure should not cause failure of an essential architectural, mechanical, or electrical 
component (Provisions Sec. 6.1 [6.2.3]). 
Component seismic attachments must be bolted, welded, or otherwise positively fastened without 
considering the frictional resistance produced by the effects of gravity (Provisions Sec. 6.1.2 [6.2.5]). 
The effects of seismic relative displacements must be considered in combination with displacements 
caused by other loads as appropriate (Provisions Sec. 6.1.4 [6.2.7]). 
Exterior nonstructural wall panels that are attached to or enclose the structure shall be designed to resist 
the forces in accordance with Provisions Eq. 6.1.3-1 or 6.1.3-2 [6.2.6] and must be able to accommodate 
movements of the structure resulting from response to the design basis ground motion, Dp, or temperature 
changes (Provisions Sec. 6.2.4 [6.3.2]).
Chapter 13, Nonstructural Components 
13-11 
A
A1 
A
A1 
B
B 
3rd Story beam 
6'-6" 
panel 
7'-0" 
column cover / 
window frame 
24'-0" 
Panel C.G. 
Figure 13.2-3 Spandrel panel connection layout from interior (1.0 ft = 0.3048 m). 
13.2.3 Spandrel Panel 
13.2.3.1 Connection Details 
Figure 13.2-3 shows the types and locations of connections that support one spandrel panel. 
The connection system must resist the weight of the panel and supported construction including the 
eccentricity between that load and the supports as well as inertial forces generated by response to the 
seismic motions in all three dimensions. Furthermore, the connection system must not create undue 
interaction between the structural frame and the panel, such as restraint of the natural shrinkage of the 
panel or the transfer of floor live load from the beam to the panel. The panels are usually very stiff 
compared to the frame, and this requires careful release of potential constraints at connections. PCIs 
Architectural Precast Concrete (2nd Ed. 1989), provides an extended discussion of important design 
concepts for such panels. 
For this example, the basic gravity load, and vertical accelerations are resisted at Points A, which 
provides the recommended simple and statically determinant system for the main gravity weight. The 
eccentricity of vertical loads is resisted by a force couple at the two pairs of A1 and A connections. 
Horizontal loads parallel to the panel are resisted by the A connections. Horizontal loads perpendicular to 
the panel are resisted by all six connections. The A connections, therefore, restrain movement in three 
dimensions while the A1 and B connections restrain movement in only one dimension, perpendicular to 
the panel. Connection components can be designed to resolve some eccentricities by bending of the 
element; for example, the eccentricity of the horizontal in-plane force with the structural frame can be 
resisted by bending the A connection. 
The practice of resisting the horizontal in-plane force at two points varies with seismic demand and local 
industry practice. The option is to resist all of the in-plane horizontal force at one connection in order to 
avoid restraint of panel shrinkage. The choice made here depends on local experience indicating that 
precast panels of this length have been restrained at the two ends without undue shrinkage restraint 
problems.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
13-12 
The A and A1 connections are often designed to take the loads directly to the columns, particularly on 
steel moment frames where attachments to the flexural hinging regions of beams are difficult to 
accomplish. The lower B connection often require an intersecting beam to provide sufficient stiffness and 
strength to resist the loads. 
The column cover is supported both vertically and horizontally by the column, transfers no loads to the 
spandrel panel, and provides no support for the window frame. 
The window frame is supported both vertically and horizontally along the length of the spandrel panel 
and transfers no loads to the column covers. 
13.2.3.2 Prescribed Seismic Forces 
Lateral forces on the wall panels and connection fasteners include seismic loads in accordance with the 
Provisions and wind loads in accordance with ASCE 7 as indicated in the problem statement. Wind 
forces are not illustrated here. 
13.2.3.2.1 Panels 
D = Wp = 8775 lb + 1470 lb = 10245 lb (vertical gravity effect) 
(Provisions Eq. 6.1.3-1 [6.2-1]) 
( )( )( ) 
0.4 1.0 1(.487 )10245 lb( ( )) 
1 2 0.6 5362 lb 
2.5
1.0 
Fp= + = 
1.6(1.487)(1.0)(10245 lb) = 24375 lb (Provisions Eq. 6.1.3-2 [6.2-3]) Fpmax = 
0.3(1.487)(1.0)(10245 lb) = 4570 lb (Provisions Eq. 6.1.3-3 [6.2-4]) Fpmin = 
[2003 Provisions Sec. 6.2.6 now treats load effects differently. The vertical forces that must be 
considered in design are indicated directly and the redundancy factor does not apply, so the following five 
steps would be cast differently; the result is the same.] 
QE (due to application of Fp) = 5362 lb (Provisions Sec. 6.1.3) 
.QE = (1.0)(5362 lb) = 5362 lb (horizontal earthquake effect) 
0.2SDSD = (0.2)(1.487)(10245 lb) = 3047 lb (vertical earthquake effect) 
E = .QE +0.2SDSD (Provisions Eq. 5.2.7.1-1 [4.2-1]) 
E = .QE - 0.2SDSD (Provisions Eq. 5.2.7.1-2 [4.2-2]) 
13.2.3.2.2 Connection Fasteners 
The Provisions specifies a reduced RP, and an increased aP for Fasteners, which is intended to prevent 
premature failure in those elements of connections that are inherently brittle, such as embedments that 
depend on concrete breakout strength, or are simply too small to adequately dissipate energy inelastically, 
such as welds or bolts. The net effect more than triples the design seismic force. 
(Provisions Eq. 6.1.3-1 [6.2-1]) 
( )( )( ) 
( ) 0.4 1.25 1.487 10245 lb ( ( )) 
1 2 0.6 16757 lb 
1.0
1.0 
Fp= + =
Chapter 13, Nonstructural Components 
13-13 
1.6(1.487)(1.0)(10245 lb) = 24375 lb (Provisions Eq. 6.1.3-2 [6.2-3]) Fpmax = 
0.3(1.487)(1.0)(10245 lb) = 4570 lb (Provisions Eq. 6.1.3-3 [6.2-4]) Fpmin = 
[2003 Provisions Sec. 6.2.6 now treats load effects differently. The vertical forces that must be 
considered in design are indicated directly and the redundancy factor does not apply, so the following five 
steps would be cast differently; the result is the same.] 
QE (due to application of Fp) = 16757 lb (Provisions Sec. 6.1.3) 
.QE = (1.0)(16757 lb) = 16757 lb (horizontal earthquake effect) 
0.2SDSD = (0.2)(1.487)(10245 lb) = 3047 lb (vertical earthquake effect) 
E = .QE +0.2SDSD (Provisions Eq. 5.2.7.1-1 [4.2-1]) 
E = .QE - 0.2SDSD (Provisions Eq. 5.2.7.1-2 [4.2-2]) 
13.2.3.3 Proportioning and Design 
13.2.3.3.1 Panels 
The wall panels should be designed for the following loads in accordance with ACI 318. The design of 
the reinforced concrete panel is standard and is not illustrated in this example. Spandrel panel moments 
are shown in Figure 13.2-4. Reaction shears (Vu), forces (Hu), and moments (Mu) are calculated for 
applicable strength load combinations.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
13-14 
L = Column spacing 
V = Dead and/or earthquake load 
H = Wind or earthquake load 
Strong 
axis 
M
Weak 
axis 
M 
u 
ux 
uy 
Vu L 
8 
32 
H u L 
u 
Figure 13.2-4 Spandrel panel moments.
Chapter 13, Nonstructural Components 
13-15 
U = 1.4D 
Vu = 1.4(10245 lb) = 14343 lb (vertical load downward) 
(strong axis moment) 
(14343 lb)(24 ft) 
43029 ft-lb 
Mux= 8 = 
U = 1.2D + 1.0E 
1.2(10245 lb) 1.0(3047 lb) 15341 lb (vertical load downward) Vumax= + = 
.Hu =1.0(5362 lb)=5362 lb (horizontal load parallel to panel) 
.Hu =1.0(5362 lb)=5362 lb (horizontal load perpendicular to panel) 
(strong axis moment) (15341 lb)(24 ft) 
46023 ft-lb 
Muxmax= 8 = 
(weak axis moment) 
(5362 lb)(24 ft) 
4022 ft-lb 
Muy= 32 = 
U = 0.9D + 1.0E 
0.9(10245 lb) 1.0(3047 lb) 6174 lb (vertical load downward) Vumin= - = 
.Hu =1.0(5362 lb)=5362 lb (horizontal load parallel to panel) 
.Hu =1.0(5362 lb)=5362 lb (horizontal load perpendicular to panel) 
(strong axis moment) 
(6174 lb)(24 ft) 
18522 ft-lb 
Muxmin= 8 = 
(weak axis moment) 
(5362 lb)(24 ft) 
4022 ft-lb 
Muy= 32 = 
13.2.3.3.2 Connection Fasteners 
The connection fasteners should be designed for the following loads in accordance with ACI 318-2002 
(Appendix D) and the AISC specification. There are special reduction factors for anchorage in high 
seismic demand locations, and the parameters for this project would invoke those reduction factors. The 
design of the connection fasteners is not illustrated in this example. Spandrel panel connection forces are 
shown in Figure 13.2-5. Reaction shears (Vu), forces (Hu), and moments (Mu) are calculated for applicable 
strength load combinations.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
13-16 
D = Dead 
E = Earthquake 
W = Wind 
R = Reaction 
Panel C.G. 
RA 
RA1 
RA 
E, W 
D
E
1'-6" 
8" 1'-4" 
1'-4" 
Figure 13.2-5 Spandrel panel connection forces. 
U = 1.4D 
(vertical load downward at Point A and A1) 
1.4(10245 lb) 
7172 lb 
VuA= 2 = 
MuA = (7172 lb)(1.5 ft) = 10758 ft-lb (moment resisted by paired Points A and A1) 
Horizontal couple from moment at A and A1 = 10758 / 1.33 = 8071 lb 
U = 1.2D + 1.0E 
1.2(10245 lb) 1.0(3047 lb) 7671 lb (vertical load downward at Point A) 
uAmax 2 V 
+ 
= = 
1.0(16757 lb) 3 3142 lb (horizontal load perpendicular to panel at Points A and A1) 
16 uA .H = = 
HAin = (7671 lb)(1.5 ft) / (1.33 ft) + (3142 lb)(2.0 ft) / (1.33 ft) = 13366 lb (inward force at Point A) 
HA1out = (7671 lb)(1.5 ft) / (1.33 ft) + (3142 lb)(0.67) / (1.33 ft) = 10222 lb(outward force at Point A1) 
1.0(16757 lb) 8378 lb (horizontal load parallel to panel at Point A) 
2 uA .H = = 
Mu2A = (8378 lb)(1.5 ft) = 12568 ft-lb (flexural moment at Point A) 
1.0(16757 lb) 5 10473 lb (horizontal load perpendicular to panel at Points B and B1) 
8 uB .H = =
Chapter 13, Nonstructural Components 
13-17 
1'-3" 6'-0" 
1'-6" 
24'-0" 
Cover C.G. 
C E 
D E 
column cover / 
window frame 
6'-6" 7'-0" 
panel 
Figure 13.2-6 Column cover connection layout (1.0 ft = 0.3048 m). 
HB = (10743 lb)(2.0 ft) / (1.33 ft) = 15714 lb (inward or outward force at B) 
HB1 = (10473 lb)(0.67 ft) / (1.33 ft) = 5237 lb (inward or outward force at B1) 
U = 0.9D + 1.0E 
0.9(10245 lb) 1.0(3047 lb) 3086 lb (vertical load downward at Point A) 
uAmin 2 V 
- 
= = 
Horizontal forces are the same as combination 1.2 D + 1.0 E. No uplift occurs; the net reaction at A is 
downward. Maximum forces are controlled by prior combination. It is important to realize that inward 
and outward acting horizontal forces generate different demands when the connections are eccentric to the 
center of mass as it is in this example. Only the maximum reactions are computed above. 
13.2.3.4 Prescribed Seismic Displacements 
Prescribed seismic displacements are not applicable to the building panel because all connections are 
essentially at the same elevation. 
13.2.4 Column Cover 
13.2.4.1 Connection Details 
Figure 13.2-6 shows the key to the types of forces resisted at each column cover connection. 
Vertical loads, horizontal loads parallel to the panel, and horizontal loads perpendicular to the panel are 
resisted at Point C. The eccentricity of vertical loads is resisted by a force couple at Points C and D. The 
horizontal load parallel to the panel eccentricity between the panel and the support is resisted in flexure of 
the connection. The connection is designed to take the loads directly to the columns. 
Horizontal loads parallel to the panel and horizontal loads perpendicular to the panel are resisted at Point 
D. The vertical load eccentricity between the panel and the support is resisted by a force couple of Points
FEMA 451, NEHRP Recommended Provisions: Design Examples 
13-18 
C and D. The eccentricity of horizontal loads parallel to the panel is resisted by flexure at the connection. 
The connection must not restrict vertical movement of the panel due to thermal effects or seismic input. 
The connection is designed to take the loads directly to the columns. 
Horizontal loads perpendicular to the panel are resisted at Points E. The connection is designed to take 
the loads directly to the columns. 
There is no load eccentricity associated with the horizontal loads perpendicular to the panel. 
In this example, all connections are made to the sides of the column because there usually is not enough 
room between the outside face of the column and the inside face of the cover to allow a feasible 
load-carrying connection. 
13.2.4.2 Prescribed Seismic Forces 
Calculation of prescribed seismic forces for the column cover are not shown in this example. They 
should be determined in the same manner as illustrated for the spandrel panels. 
13.2.4.3 Prescribed Seismic Displacements 
The results of an elastic analysis of the building structure are not usually available in time for use in the 
design of the precast cladding system. As a result, prescribed seismic displacements are usually 
calculated based on allowable story drift requirements: 
hsx = story height = 13 ft 6 in 
X = height of upper support attachment = 47 ft 9 in 
Y = height of lower support attachment = 41 ft 9 in 
.a = 0.020hsx (Provisions Table 5.2.8 [4.5-1]) 
( ) (72 in.) 0.020 1.44 in. (Provisions Eq. 6.1.4-2 [6.2-6]) max 
a sx 
p 
sx sx
D X Y h 
h h 
. 
= - = = 
The joints at the top and bottom of the column cover must be designed to accommodate an in-plane 
relative displacement of 1.44 inches. The column cover will rotate somewhat as these displacements 
occur, depending on the nature of the connections to the column. If the supports at one level are fixed 
to the columns while the other level is designed to float, then the rotation will be that of the column at 
the point of attachment. 
13.2.5 Additional Design Considerations 
13.2.5.1 Window Frame System 
The window frame system is supported by the spandrel panels above and below. Assuming that the 
spandrel panels move rigidly in-plane with each floor level, the window frame system must accommodate 
a prescribed seismic displacement based on the full story height. 
(Provisions Eq. 6.1.4-2 [6.2-6]) ( ) (162 in.) 0.020 3.24 in. max 
a sx 
p 
sx sx
D X Y h 
h h 
. 
= - = =
Chapter 13, Nonstructural Components 
13-19 
The window frame system must be designed to accommodate an in-plane relative displacement of 3.24 in. 
between the supports. This is normally accommodated by a clearance between the glass and the frame. 
Provisions Sec. 6.2.10.1 [6.3.7], prescribes a method of checking such a clearance. It requires that the 
clearance be large enough so that the glass panel will not fall out of the frame unless the relative seismic 
displacement at the top and bottom of the panel exceeds 125 percent of the value predicted amplified by 
the building importance factor. If hp and bp are the respective height and width of individual panes and if 
the horizontal and vertical clearances are designated c1 and c2, respectively, then the following expression 
applies: 
2 
1 
1 
2 1 p 1.25 
clear p 
p 
h c 
D c D 
b c 
. . 
= .. + ..= 
. . 
For hp = 7 ft, bp = 5 ft, and Dp = 3.24 in., and setting c1 = c2, the required clearance is 0.84 in. 
13.2.5.2 Building Corners 
Some thought needs to be given to seismic behavior at external building corners. The preferred approach 
is to detail the corners with two separate panel pieces, mitered at a 45 degree angle, with high grade 
sealant between the sections. An alternative choice of detailing L-shaped corner pieces, would introduce 
more seismic mass and load eccentricity into connections on both sides of the corner column. 
13.2.5.3 Dimensional Coordination 
It is important to coordinate dimensions with the architect and structural engineer. Precast concrete 
panels must be located a sufficient distance from the building structural frame to allow room for the 
design of efficient load transfer connection pieces. However, distances must not be so large as to 
unnecessarily increase the load eccentricities between the panels and the frame. 
13.3 HVAC FAN UNIT SUPPORT 
13.3.1 Example Description 
In this example, the mechanical component is a 4-ft-high, 5-ft-wide, 8-ft-long, 3000-lb HVAC fan unit 
that is supported on the two long sides near each corner (Figure 13.3-1). The component is located at the 
roof level of a five-story office building, near a significant active fault in Los Angeles, California. The 
building is assigned to Seismic Use Group I. Two methods of attaching the component to the 4,000 psi, 
normal-weight roof slab, are considered as follows: 
1. Direct attachment to the structure with 36 ksi, carbon steel, cast-in-place anchors and 
2. Support on vibration isolation springs, that are attached to the slab with 36 ksi carbon steel 
post-installed expansion anchors.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
13-20 
Concrete Concrete 
a = 7'-0" b = 5'-6" 
4'-0" 
2'-0" 
Center-of-mass 
Elevation 
Plan 
8'-0" 
5'-0" 
Attachment location (typical). 
HVAC fan unit Direct attachment shown 
W = 3,000 lbs 
Elevation 
Figure 13.3-1 Air handling fan unit (1.0 ft = 0.3048 m, 1.0 lb = 4.45 N). 
13.3.2 Design Requirements 
13.3.2.1 Provisions Parameters and Coefficients 
ap = 1.0 for direct attachment (Provisions Table 6.3.2 [6.4-1]) 
ap = 2.5 for vibration isolated (Provisions Table 6.3.2 [6.4-1]) 
SDS = 1.487 (Design Values CD-ROM) 
[The 2003 Provisions have adopted the 2002 USGS probabilistic seismic hazard maps, and the maps have 
been added to the body of the 2003 Provisions as figures in Chapter 3 (instead of the previously used 
separate map package).] 
Seismic Design Category = D (Provisions Table 4.2.1a [14.4-1]) 
Wp = 3000 lb (given) 
Rp = 2.5 for HVAC system equipment (Provisions Table 6.3.2 [6.4-1]) 
Rp = 1.5 for expansion anchors and shallow, 
cast-in-place anchors* ((Provisions Sec. 6.1.6.1 [6.2.8.1]) 
Shallow anchors are defined by Provisions Sec. 2.1 as anchors having embedment-to-diameter 
ratios of less than 8.
Chapter 13, Nonstructural Components 
13-21 
[In the 2003 Provisions component anchorage is treated differently. Rather than making distinctions 
based on an anchor being shallow, component anchorage is designed using Rp = 1.5 unless specific 
ductility or prequalification requirements are satisfied.] 
Ip = 1.0 (Provisions Sec. 6.1.5 [6.2.2]) 
z/h = 1.0 (for roof mounted equipment) 
. = 1.0 (Provisions Sec. 6.1.3 [6.2.6]) 
[The 2003 Provisions indicate that the redundancy factor does not apply to the design of nonstructural 
components. Although the effect is similar to stating that . = 1, there is a real difference since load 
effects for such components and their supports and attachments are now defined in Chapter 6 rather than 
by reference to Chapter 4.] 
13.3.2.2 Performance Criteria 
Component failure should not cause failure of an essential architectural, mechanical, or electrical 
component (Provisions Sec. 6.1 [6.2.3]). 
Component seismic attachments must be bolted, welded, or otherwise positively fastened without 
consideration of frictional resistance produced by the effects of gravity (Provisions Sec. 6.1.2 [6.2.5]). 
Anchors embedded in concrete or masonry must be proportioned to carry the least of: (a) the design 
strength of the connected part, (b) 1.3 times the force in the connected part due to the prescribed forces, or 
(c) the maximum force that can be transferred to the connected part by the component structural system 
(Provisions Sec. 6.1.6.2 [6.2.8.2]). 
Attachments and supports transferring seismic loads must be constructed of materials suitable for the 
application and must be designed and constructed in accordance with a nationally recognized structural 
standard (Provisions Sec. 6.3.13.2.a [6.4.4, item 6]). 
Components mounted on vibration isolation systems must have a bumper restraint or snubber in each 
horizontal direction. Vertical restraints must be provided where required to resist overturning. Isolator 
housings and restraints must also be constructed of ductile materials. A viscoelastic pad, or similar 
material of appropriate thickness, must be used between the bumper and equipment item to limit the 
impact load (Provisions Sec. 6.3.13.2.e). Such components must also resist an amplified design force. 
13.3.3 Direct Attachment to Structure 
This section illustrates design for cast-in-place concrete anchors with embedment-length-to-diameters 
ratios of 8 or greater; thus, the use of Rp = 2.5 is permitted. [In 2003 Provisions Sec. 6.2.8.1, the value of 
Rp used in designing component anchorage is no longer based on the embedment depth-to-diameter ratio. 
Instead Rp = 1.5 unless specific ductility or prequalification requirements are satisfied.] 
13.3.3.1 Prescribed Seismic Forces 
See Figure 13.3-2 for freebody diagram for seismic force analysis.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
13-22 
(Provisions Eq. 6.1.3-1 [6.2-1]) ( )( )( ) 
0.4 1.0 (1.487 )3000 lb ( ( )) 
1 21 2141 lb 
2.5
1.0 
Fp= + = 
1.6(1.487)(1.0)(3000 lb) 7138 lb (Provisions Eq. 6.1.3-2 [6.2-3]) Fpmax= = 
0.3(1.487)(1.0)(3000 lb) 1338 lb (Provisions Eq. 6.1.3-3 [6.2-4]) Fpmin= = 
[2003 Provisions Sec. 6.2.6 now treats load effects differently. The vertical forces that must be 
considered in design are indicated directly and the redundancy factor does not apply, so the following six 
steps would be cast differently; the result is the same.] 
QE (due to application of Fp) = 2141 lb (Provisions Sec. 6.1.3) 
.QE = (1.0)(2141 lb) = 2141 lb (horizontal earthquake effect) 
0.2SDSD = (0.2)(1.487)(3000 lb) = 892 lb (vertical earthquake effect) 
D = Wp = 3000 lb (vertical gravity effect) 
E = .QE + 0.2SDSD (Provisions Eq. 5.2.7.1-1 [4.2-1]) 
E = .QE - 0.2SDSD (Provisions Eq. 5.2.7.1-2 [4.2-2]) 
U = 1.2D + 1.0E + 0.5L + 0.2S 
1.0(2141 lb) 
535 lb/bolt 
Vu= 4 bolts = 
no tension 
( )( ) ( )( ) ( )( ) 
( )( ) 
1.2 3000 lb 2.75 ft 1.0 2141 lb 2 ft 1.0 892 lb 2.75 ft 
288 lb/bolt 
Tu 5.5 ft 2 bolts 
- + + 
= =- 
U = 0.9D - (1.3W or 1.0E) 
R
V V
T 
D 
.QE 
0.2S D S D 
2'-0" 
5'-6" 
u 
u 
u 
u 
Figure 13.3-2 Free-body diagram for seismic force analysis 
(1.0 ft = 0.348 m).
Chapter 13, Nonstructural Components 
13-23 
Component base 
Tu 
Vu 
Figure 13.3-3 Anchor for direct 
attachment to structure. 
1.0(2141 lb) 
535 lb/bolt 
Vu= 4 bolts = 
no tension 
( )( ) ( )( ) ( )( ) 
( )( ) 
0.9 3000 lb 2.75 ft 1.0 2141 lb 2 ft 1.0 892 lb 2.75 ft 
63 lb/bolt 
Tu 5.5 ft 2 bolts 
- + + 
= =- 
13.3.3.2 Proportioning and Design 
See Figure 13.3-3 for anchor for direct attachment to structure. 
Check one -in.-diameter cast-in-place anchor embedded 2 in. into the concrete slab with no transverse 
reinforcing engaging the anchor and extending through the failure surface. Although there is no required 
tension strength on these anchors, design strengths and tension/shear interaction acceptance relationships 
are calculated to demonstrate the use of the Provisions equations. 
5.3.3.2.1 Design Tension Strength on Isolated Anchor in Slab, Away from Edge, Loaded Concentrically 
[The 2003 Provisions refer to Appendix D of ACI 318-02 rather than providing specific, detailed 
requirements. Note also that some of the resistance factors, f, are different.] 
Following Provisions Sec. 9.2, for a headed bolt or a rod with a nut at the bottom: 
(not shallow, Rp = 2.5 is permitted) 
2 in. 8.0 
0.25 in. 
L
d
= = 
[In 2003 Provisions selection of Rp is no longer based on the L/d ratio; see Sec. 6.2.8.1.] 
Tension capacity of steel, f = 0.80: 
Ns = AseFy = (0.049 in.2)(36000 psi) = 1764 lb (Provisions Sec. 9.2.5.1.2 [ACI 318-02 Eq. D-3]) 
[In Appendix D of ACI 318-02 this capacity calculation is based on fut rather than Fy, since the large 
majority of anchor materials do not exhibit a well-defined yield point.] 
Tension capacity of concrete, f = 0.70, with no eccentricity, no pullthrough, and no edge or group effect: 
(Provisions Eq. 9.2.5.2.2-1 [ACI 318-02 Eq. D-7]) 1.5 24 4000 psi(81.5 ) 4293 lb c cef N =k f'h = =
FEMA 451, NEHRP Recommended Provisions: Design Examples 
13-24 
[In Appendix D of ACI 318-02 this item is defined as Nb rather than Nc.] 
The steel controls but, with no tension demand, the point is moot. 
5.3.3.2.2 Design Shear Strength on Isolated Anchor, Away from Edge 
Shear capacity of steel, f = 0.80: 
Vs = AseFy = (0.049 in.2)(36000 psi) = 1764 lb (Provisions Eq. 9.2.6.1.2-1 [ACI 318-02 Eq. D-17) 
[In Appendix D of ACI 318-02 this capacity calculation is based on fut rather than Fy, since the large 
majority of anchor materials do not exhibit a well-defined yield point.] 
Shear capacity of concrete, far from edge, limited to pryout, f = 0.70: 
Vcp = 2Nc = 2(4293) = 8493 lb (Provisions Eq. 9.2.6.3.1 [ACI 318-02 Eq. D-28) 
The steel controls, with fVN = 0.8(1764) = 1411 lb 
Per Provisions Sec. 6.1.6.2 [6.2.8.2], anchors embedded in concrete or masonry are to be proportioned to 
carry at least 1.3 times the force in the connected part due to the prescribed forces. Thus, Vu = 1.3(535) = 
696 lb and the anchor is clearly adequate. 
5.3.3.2.3 Combined Tension and Shear 
The Provisions gives a new equation (Eq. 9.2.7.3 [ACI 318-02 Eq. D-29]) for the interaction of tension 
and shear on an anchor or a group of anchors: 
u u 1.2 , which applies when either term exceeds 0.2 
N N 
N V 
fN fV 
+ = 
5.3.3.2.3 Summary 
At each corner of the component, provide one -in.-diameter cast-in-place anchor embedded 2 in. into 
the concrete slab. Transverse reinforcement engaging the anchor and extending through the failure 
surface is not necessary. 
13.3.4 Support on Vibration Isolation Springs 
13.3.4.1 Prescribed Seismic Forces 
Design forces for vibration isolation springs are determined by an analysis of earthquake forces applied in 
a diagonal horizontal direction as shown in Figure 13.3-4. Terminology and concept are taken from 
ASHRAE APP IP. 
Angle of diagonal loading: 
tan-1 b (ASHRAE APP IP. Eq. 17) 
a 
. = .. .. 
. . 
Tension per isolator:
Chapter 13, Nonstructural Components 
13-25 
h 
1 a 2 2 b 3 
FP sin (.) 
WP 
FPV FPV 
WP
FP cos (.) 
1 2 
X 
Plan view 
X 
Y Y 
F P sin (.) 
. F P cos (.) 
FP
4 3 
F P = Seismic horizontal force 
WP = Operating weight of equipment 
F P V = Seismic vertical force 
a = Distance between vibration isolators along Y-Y 
b = Distance between vibration isolators along X-X 
h = Height of center of gravity 
= Vibration isolator location 
Figure 13.3-4 ASHRAE diagonal seismic force analysis for vibration isolation springs. 
(ASHRAE APP IP. Eq. 18) 
cos sin 
4 2 
p pv p 
u 
W F Fh 
T 
b a 
= - - .. .+ ... 
. . 
Compression per isolator: 
(ASHRAE APP IP. Eq. 19) 
cos sin 
4 2 
p pv p 
u 
W F Fh 
C 
b a 
= + + .. .+ ... 
. . 
Shear per isolator: 
(ASHRAE APP IP. Eq. 20) 
4
p 
u 
F 
V = 
Select worst case assumption: Design for post-installed expansion anchors, requiring the use of Rp = 1.5. 
(Provisions Eq. 6.1.3-1 [6.2-1]) ( )( )( ) 
( ) 0.4 2.5 1.487 3000 lb ( ( )) 
1 2 1 8922 lb 
1.5
1.0 
Fp= + = 
1.6(1.487)(1.0)(3000 lb) 7138 lb (Provisions Eq. 6.1.3-2 [6.2-3]) Fpmax= = 
0.3(1.487)(1.0)(3000 lb) 1338 lb (Provisions Eq. 6.1.3-3 [6.2-4]) Fpmin= =
FEMA 451, NEHRP Recommended Provisions: Design Examples 
13-26 
Components mounted on vibration isolation systems shall have a bumper restraint or snubber in each 
horizontal direction. Per Provisions Table 6.3.2 [6.4-1], Footnote B, the design force shall be taken as 
2Fp. 
[2003 Provisions Sec. 6.2.6 now treats load effects differently. The vertical forces that must be 
considered in design are indicated directly and the redundancy factor does not apply, so the following 
steps would be cast differently; the result is the same.] 
QE = Fp = 2(7138 lb) = 14276 lb (Provisions Sec. 6.1.3) 
.QE = (1.0)(14276 lb) = 14276 lb (horizontal earthquake effect) 
Fpv(ASHRAE) = 0.2SDSD = (0.2)(1.487)(3000 lb) = 892 lb (vertical earthquake effect) 
D = Wp = 3000 lb (vertical gravity effect) 
E = .QE + 0.2SDSD (Eq. 5.2.7.1-1 [4.2-1]) 
E = .QE - 0.2SDSD (Eq. 5.2.7.1-2 [4.2-2]) 
U = 1.2D + 1.0E + 0.5L + 0.2S 
tan17 ft 51.8o 
5.5 ft 
.= - .. ..= 
. . 
1.2(3000 lb) (892 lb) (14276 lb)(2 ft) cos(51.8 ) sin(51.8 ) 2624 lb 
4 2 7 ft 5.5 ft u T 
= - - .. + ..=- 
. . 
1.2(3000 lb)+(892 lb) (14276 lb)(2 ft) cos(51.8 ) sin(51.8 ) 4424 lb 
4 2 7 ft 5.5 ft u C 
= + .. + ..= 
. . 
14276 lb 3569 lb 
4 u V = = 
U = 0.9D + 1.0E
tan17 ft 51.8o 
5.5 ft 
.= - .. ..= 
. . 
0.9(3000 lb) (892 lb) (14276 lb)(2 ft) cos(51.8 ) sin(51.8 ) 2849 lb 
4 2 7 ft 5.5 ft u T 
= - - .. + ..=- 
. . 
0.9(3000 lb) (892 lb) (14276 lb)(2 ft) cos(51.8 ) sin(51.8 ) 4199 lb 
4 2 7 ft 5.5 ft u C = + + .. + ..= 
. . 
14276 lb 3569 lb 
4 u V = = 
13.3.4.2 Proportioning and Details
Chapter 13, Nonstructural Components 
13-27 
T 
W = 4" 
V 
"O" 
T 
V 
H = 5" 
Vibration isolator 
w/ seismic housing 
Equipment frame 
u 
u 
b 
b 
Figure 13.3-5 Anchor and snubber loads for 
support on vibration isolation springs (1.0 in. = 25.4 
mm). 
Anchor and snubber loads for support on vibration isolation springs are shown in Figure 13.3-5. 
Check vibration isolation system within housing anchored with two 1-in.-diameter post-installed 
expansion anchors embedded 9 inches into the concrete. 
The Provisions does not provide a basis for determining the design strength of post-installed expansion 
anchors. Although manufacturers provide ultimate strength shear and tension loads for their products, the 
Provisions does not provide resistance or quality values, f, to allow determination of design shear and 
tension strengths. The 2002 edition of ACI 318 contains provisions for post-installed anchors which 
depend on anchor testing per the ACI 355.2-01 testing standard. Prior to the adoption of this method, the 
best course available was to use allowable stress loads published in evaluation reports prepared by model 
building code agencies. These allowable stress loads were then multiplied by 1.4 to convert to design 
strengths. This technique will be illustrated in the following. 
[The 2003 Provisions refer to Appendix D of ACI 318-02.] 
Allowable stress combinations are included in the 2000 IBC. The 1.4 strength conversion factor is 
unnecessary when using allowable stress loads published in evaluation reports prepared by model 
building code agencies. 
13.3.4.2.1 Design Tension Strength on Isolated Anchor in Slab, Away from Edge 
Allowable stress tension values are obtained from ICBO Evaluation Services, Inc., ER-4627 for Hilti 
Kwik Bolt II concrete anchors. Similar certified allowable values are expected with anchors from other 
manufacturers. 
anchor diameter = 1 in. 
anchor depth = 9 in. 
fc' = 4000 psi
FEMA 451, NEHRP Recommended Provisions: Design Examples 
13-28 
with special inspection 
Tallow = 8800 lb 
Ps = fPc = 1.4(8800 lb) = 12320 lb 
13.3.4.2.2 Design Shear Strength on Isolated Anchor in Slab, Away from Edge 
Allowable stress shear values are obtained from ICBO Evaluation Services, Inc., ER-4627 for Hilti Kwik 
Bolt II concrete anchors. Similar certified allowable values are expected with anchors from other 
manufacturers. 
anchor diameter = 1 in. 
anchor depth = 9 in. 
fc' = 4000 psi 
Vallow = 8055 lb 
Vs = fVc = 1.4(8055 lb) = 11277 lb 
13.3.4.2.3 Combined Tension and Shear 
Per Provisions Sec. 6.1.6.2 [6.2.8.2], anchors embedded in concrete or masonry shall be proportioned to 
carry at least 1.3 times the force in the connected part due to the prescribed forces. 
In the Provisions and in the 2000 IBC, the factor of 2.0 is reduced to 1.3. This will greatly reduce the 
prescribed seismic forces. 
Interaction relationships for combined shear and tension loads are obtained from ICBO Evaluation 
Services, Inc. ER-4627 for Hilti Kwik Bolt II concrete anchors. Similar results are expected using other 
anchors. 
As stated in ICBO ES evaluation report: 
5 5 
3 3 
s s 1 
t t 
P V 
P V 
. . . . 
. . +. . = 
. . . . 
Using Provisions terminology: 
5 5 
3 3
b b 1 
c c 
T V 
fP fV 
. . . . 
. . +. . = 
. . . . 
OK 
5 5 
2 4462 lb32 1785 lb30.68 0.13 0.81 1 
11277 lb 12075 lb 
.. .. +.. .. = + = < 
. . . . 
13.3.4.2.4 Summary
Chapter 13, Nonstructural Components 
13-29 
Gap Gap 
Equipment 
frame 
Impact 
material 
Steel bushing 
bolted or welded 
to equipment frame 
Figure 13.3-6 Lateral restraint required to resist seismic forces. 
At each corner of the HVAC fan unit, provide a vibration isolation system within a housing anchored 
with two 1-in.-diameter post-installed expansion anchors embedded 9 in. into the concrete slab. Special 
inspection is required. A raised concrete pad is probably required to allow proper embedment of the 
post-installed expansion anchors. 
Other post-installed anchors, such as chemical (adhesive) or undercut post-installed anchors also could be 
investigated. These anchors may require more involved installation procedures, but they may allow the 
use of Rp = 2.5 if they have an embedment depth-to-diameter ratio of at least 8 (i.e., are not shallow 
anchors). The higher Rp value will result in much smaller prescribed seismic forces and, therefore, a 
much reduced embedment depth. 
[Again note that in 2003 Provisions selection of Rp is no longer based on the L/d ratio; see Sec. 6.2.8.1.] 
13.3.5 Additional Considerations for Support on Vibration Isolators 
Vibration isolation springs are provided for equipment to prevent vibration from being transmitted to the 
building structure. However, they provide virtually no resistance to horizontal seismic forces. In such 
cases, some type of restraint is required to resist the seismic forces. Figure 13.3-6 illustrates one concept 
where a bolt attached to the equipment base is allowed to slide a controlled distance (gap) in either 
direction along its longitudinal axis before it contacts resilient impact material. 
Design of restraints for vibration-isolated equipment varies for different applications and for different 
manufacturers. In most cases, restraint design incorporates all directional capability with an air gap, a 
soft impact material, and a ductile restraint or housing. 
Restraints should have all-directional restraint capability to resist both horizontal and vertical motion. 
Vibration isolators have little or no resistance to overturning forces. Therefore, if there is a difference in 
height between the equipment center of gravity and the support points of the springs, rocking is inevitable 
and vertical restraint is required.
FEMA 451, NEHRP Recommended Provisions: Design Examples 
13-30 
An air gap between the restraint device and the equipment prevents vibration from transmitting to the 
structure during normal operation of the equipment. Air gaps are generally no greater than  in. 
Dynamic tests indicate a significant increase in acceleration for air gaps larger than  in. 
A soft impact material often an elastomer such as bridge bearing neoprene reduces accelerations and 
impact loads by preventing steel-to-steel contact. The thickness of the elastomer can significantly reduce 
accelerations to both the equipment and the restraint device and should be specifically addressed for life 
safety applications. 
A ductile restraint or housing is critical to prevent catastrophic failure. Unfortunately, housed isolators 
made of brittle materials such as cast iron often are assumed to be capable of resisting seismic loads and 
continue to be installed in seismic zones. 
Overturning calculations for vibration- isolated equipment must consider a worst case scenario as 
illustrated in Guide Sec. 13.3.4.1. However, important variations in calculation procedures merit further 
discussion. For equipment that is usually directly attached to the structure, or mounted on housed 
vibration isolators, the weight can be used as a restoring force since the equipment will not transfer a 
tension load to the anchors until the entire equipment weight is overcome at any corner. For equipment 
installed on any other vibration isolated system (such as the separate spring and snubber arrangement 
shown in Figure 13.3-5), the weight cannot be used as a restoring force in the overturning calculations. 
As the foregoing illustrates, design of restraints for resiliently mounted equipment is a specialized topic. 
The Provisions sets out only a few of the governing criteria. Some suppliers of vibration isolators in the 
highest seismic zones are familiar with the appropriate criteria and procedures. Consultation with these 
suppliers may be beneficial. 
13.4 ANALYSIS OF PIPING SYSTEMS 
13.4.1 ASME Code Allowable Stress Approach 
Piping systems are typically designed to satisfy national standards such as ASME B31.1. Piping required 
to be designed to other ASME piping codes use similar approaches with similar definition of terms. 
13.4.1.1 Earthquake Design Requirements 
ASME B31.1 Sec. 101.5.3 requires that the effects of earthquakes, where applicable, be considered in the 
design of piping, piping supports, and restraints using data for the site as a guide in assessing the forces 
involved. However, earthquakes need not be considered as acting concurrently with wind. 
13.4.1.2 Stresses Due to Sustained Loads 
The effects of pressure, weight, and other sustained loads must meet the requirements of ASME B31.1 
Eq. 11A: 
0.75 1.0 
4 
o A 
L h 
n 
SPD iM S 
t Z 
= + = 
where: 
SL = sum of the longitudinal stresses due to pressure, weight, and other sustained loads 
P = internal design pressure, psig
Chapter 13, Nonstructural Components 
13-31 
Do = outside diameter of pipe, in. 
tn = nominal pipe wall thickness, in. 
i = stress intensification factor from ASME Piping Code Appendix D, unitless 
= 1.0 for straight pipe 
$ 1.0 for fittings and connections 
MA = resultant moment loading on cross section due to weight and other sustained loads, in.-lb 
Z = section modulus, in.3 
Sh = basic material allowable stress at maximum (hot) temperature from ASME Piping Code 
Appendix A 
For example, ASTM A53 seamless pipe and tube, Grade B: Sh = 15.0 ksi for -20 to 650 degrees F. 
13.4.1.3 Stresses Due to Occasional Loads 
The effects of pressure, weight, and other sustained loads, and occasional loads including earthquake 
must meet the requirements of ASME B31.1 Eq. 12A: 
0.75 0.75 
4 
o A B 
h 
n 
PD iM iM kS 
t Z Z 
+ + = 
where: 
MB = resultant moment loading on cross-section due to occasional loads, such as from thrust loads, 
pressure and flow transients, and earthquake. Use one-half the earthquake moment range. 
Effects of earthquake anchor displacements may be excluded if they are considered in Eq. 13A, 
in.-lb 
k = duration factor, unitless 
= 1.15 for occasional loads acting less than 10% of any 24 hour operating period 
= 1.20 for occasional loads acting less than 1% of any 24 hour operating period 
= 2.00 for rarely occurring earthquake loads resulting from both inertial forces and anchor 
movements (per ASME interpretation) 
13.4.1.4 Thermal Expansion Stress Range 
The effects of thermal expansion must meet the requirements of ASME B31.1 Eq. 13A: 
C ( ) 
E A h L 
S iM S f S S 
Z 
= = + - 
where: 
SE = sum of the longitudinal stresses due to thermal expansion, ksi
FEMA 451, NEHRP Recommended Provisions: Design Examples 
13-32
MC = range of resultant moments due to thermal expansion. Also includes the effects of earthquake 
anchor displacements if not considered in Eq. 12A, in.-lb 
SA = allowable stress range, ksi (per ASME B31.1 Eq. 1, (1.25 0.25 ) A c h S =f S + S 
f = stress range reduction factor for cyclic conditions from the ASME Piping Code Table 102.3.2. 
Sc = basic material allowable stress at minimum (cold) temperature from the ASME Piping Code 
Appendix A 
13.4.1.5 Summary 
In the ASME B31.1 allowable stress approach, the earthquakes effects only appear in the MB and MC 
terms. 
Earthquake inertial effects | MB term 
Earthquake displacement effects | MC term 
13.4.2 Allowable Stress Load Combinations 
ASME B31.1 utilizes an allowable stress approach; therefore, allowable stress force levels and allowable 
stress load combinations should be used. While the Provisions are based on strength design, the IBC 
provides the following two sets of allowable stress loads and load combinations. The IBC load 
combinations are appropriate for use for piping systems when considering earthquake effects. When 
earthquake effects are not considered, load combinations should be taken from the appropriate piping 
system design code. 
13.4.2.1 IBC Basic Allowable Stress Load Combinations 
No increases in allowable stress are permitted for the following set of load combinations: 
D (IBC Eq. 16-7) 
D + L + (Lr or S or R) (IBC Eq. 16-9) 
D + (W or 0.7 E) (IBC Eq. 16-10) 
0.6D - 0.7 E (IBC Eq. 16-12) 
13.4.2.2 IBC Alternate Basic Allowable Stress Load Combinations 
Increases in allowable stress (typically 1/3) are permitted for the following alternate set of load 
combinations that include W or E: 
D + L + (Lr or S or R) (IBC Eq. 16-13) 
D + L + S + E/1.4 (IBC Eq. 16-17) 
0.9D + E/1.4 (IBC Eq. 16-18) 
13.4.2.3 Modified IBC Allowable Stress Load Combinations
Chapter 13, Nonstructural Components 
13-33 
It is convenient to define separate earthquake load terms to represent the separate inertial and 
displacement effects. 
EI = Earthquake inertial effects | MB term 
E. = Earthquake displacement effects | MC term 
It is also convenient to use the IBC Alternate Basic Allowable Stress Load Combinations modified to use 
ASME Piping Code terminology, deleting roof load effects (Lr or S or R) and multiplying by 0.75 to 
account for the 1.33 allowable stress increase when W or E is included. Only modified IBC Eq. 16-17 
and 16-18 will be considered in the discussion that follows. 
0.75[D + L + S + (EI + E.)/1.4] (modified IBC Eq. 16-17) 
0.75[0.9D + (EI + E.)/1.4] (modified IBC Eq. 16-18) 
13.4.3 Application of the Provisions 
13.4.3.1 Overview 
Provisions Sec. 6.3.11 [6.4.2, item 4] requires that, in addition to their attachments and supports, piping 
systems assigned an Ip greater than 1.0 must themselves be designed to meet the force and displacement 
requirements of Provisions Sec. 6.1.3 and 6.1.4 [6.2.6 and 6.2.7] and the additional requirements of this 
section. 
13.4.3.1 Forces 
Provisions Sec. 6.1.3 [6.2.6] provides specific guidance regarding the equivalent static forces that must be 
considered. In computing the earthquake forces for piping systems, the inertial portion of the forces 
(noted as EI in this example) are computed using Provisions Eq. 6.1.3-1, 6.1.3-2, and 6.1.3-3 [6.2-1, 6.2- 
3, and 6.2-4] for Fp with ap = 1 and Rp = 3.5. For anchor points with different elevations, the average 
value of the Fp may be used with minimum and maximums observed. In addition, when computing the 
inertial forces, the vertical seismic effects ( 0.2SDSWp) should be considered. 
The term EI can be expressed in terms of the forces defined in the Provisions converted to an allowable 
stress basis by the 1.4 factor: 
0.2 
1.4 1.4 
p DSp 
I 
horizontal vertical 
F SW 
E 
. . . . 
=. . . . 
. . . . 
It is convenient to designate the term by the variable . 
1 0.2
1.4
DS ..  S .. 
. . 
The vertical component of EI can now be defined as Ma and applied to all load combinations that include 
EI. 
MB can now be defined as the resultant moment induced by the design force Fp/1.4 where Fp is as defined 
by Provisions Eq. 6.1.3-1, 6.1.3-2, or 6.1.3-3. 
13.4.3.3 Displacements
FEMA 451, NEHRP Recommended Provisions: Design Examples 
13-34 
Provisions Sec. 6.1.4 [6.2.7] provides specific guidance regarding the relative displacements that must be 
considered. Typically piping systems are designed considering forces and displacements using elastic 
analysis and allowable stresses for code prescribed wind and seismic equivalent static forces in 
combination with operational loads. 
However, no specific guidance is provided in the Provisions except to say that the relative displacements 
should be accommodated. The intent of the word "accommodate" was not to require that a piping system 
remain elastic. Indeed, many types of piping systems typically are very ductile and can accommodate 
large amounts of inelastic strain while still functioning quite satisfactorily. What was intended was that 
the relative displacements between anchor and constraining points that displace significantly relative to 
one another be demonstrated to be accommodated by some rational means. This accommodation can be 
made by demonstrating that the pipe has enough flexibility and/or inelastic strain capacity to 
accommodate the displacement by providing loops in the pipeline to permit the displacement or by adding 
flex lines or articulating couplings which provide free movement to accommodate the displacement. 
Sufficient flexibility may not exist where branch lines, may be forced to move with a ceiling or other 
structural system are connected to main lines. Often this accommodation is done by using engineering 
judgment, without calculations. However, if relative displacement calculations were required for a piping 
system, a flexibility analysis would be required. A flexibility analysis is one in which a pipe is modeled 
as a finite element system with commercial pipe stress analysis programs (such as Autopipe or CAESAR 
II) and the points of attachment are displaced by the prescribed relative displacements. The allowable 
stress for such a condition may be significantly greater than the normal allowable stress for the pipe. 
The internal moments resulting from support displacement may be computed by means of elastic analysis 
programs using the maximum computed relative displacements as described earlier and then adjusted. 
As with elastic inertial forces, the internal moments caused by relative displacement can be divided by Rp 
to account for reserve capacity. A reduction factor of 1.4 may be used to convert them for use with 
allowable stress equations. Therefore, MC can now be defined as the resultant moment induced by the 
design relative seismic displacement Dp/1.4Rp where Dp is defined by Provisions Eq. 6.1.4-1, 6.1.4-2, 
6.1.4-3, or 6.1.4-4 [6.2-5, 6.2-6, 6.2-7, or 6.2-8]. 
13.4.3.4 Load Combinations 
Combining ASME B31.1 Eq. 12A and 13A with modified IBC Eq. 1605.3.2.5 and 1605.3.2.6 yields the 
following: 
For modified IBC Eq. 16-17 
0.75 0.75 
4 
o A B 
h 
n 
PD iM iM kS 
t Z Z 
+.. .. = 
. . 
C ( ) 
A h L 
iM S f S S 
Z 
= + - 
and for modified IBC Eq. 16-18 
0.9 0.75 0.75 
4 
o A B 
h 
n 
PD iM iM kS 
t Z Z 
+ .. ..- = 
. . 
C ( ) 
A h L 
iM S f S S 
Z 
= + -
Chapter 13, Nonstructural Components 
13-35 
where: 
1 0.2
1.4
DS  =..  S .. 
. . 
MA = the resultant moment due to weight 
MB = the resultant moment induced by the design force Fp/1.4 where Fp is as defined by Provisions 
Eq. 6.1.3-1, 6.1.3-2, or 6.1.3-3 [6.2-1, 6.2-3, or 6.2-4]. 
MC = the resultant moment induced by the design relative seismic displacement Dp/1.4Rp where Dp is 
as defined by Provisions Eq. 6.1.4-1, 6.1.4-2, 6.1.4-3, or 6.1.4-4 [6.2-5, 6.2-6, 6.2-7, or 6.2-8]. 
SDS, Wp, and Rp are as defined in the Provisions. 
P, Do, tn, I, Z, k, Sh, SA, f, and SL are as defined in ASME B31.1.
A-1 
Appendix A 
THE BUILDING SEISMIC 
SAFETY COUNCIL 
The purpose of the Building Seismic Safety Council is to enhance the public's safety by providing a 
national forum to foster improved seismic safety provisions for use by the building community. For the 
purposes of the Council, the building community is taken to include all those involved in the planning, 
design, construction, regulation, and utilization of buildings. 
To achieve its purposes, the Council shall conduct activities and provide the leadership needed to: 
1. Promote development of seismic safety provisions suitable for use throughout the United States; 
2. Recommend, encourage, and promote adoption of appropriate seismic safety provisions in 
voluntary standards and model codes; 
3. Assess implementation progress by federal, state, and local regulatory and construction agencies; 
4. Identify opportunities for the improvement of seismic regulations and practices and encourage 
public and private organizations to effect such improvements; 
5. Promote the development of training and educational courses and materials for use by design 
professionals, builders, building regulatory officials, elected officials, industry representatives, 
other members of the building community and the public. 
6. Provide advice to governmental bodies on their programs of research, development, and 
implementation; and 
7. Periodically review and evaluate research findings, practice, and experience and make 
recommendations for incorporation into seismic design practices. 
The scope of the Council's activities encompasses seismic safety of structures with explicit consideration 
and assessment of the social, technical, administrative, political, legal, and economic implications of its 
deliberations and recommendations. 
Achievement of the Council's purpose is important to all in the public and private sectors. Council 
activities will provide an opportunity for participation by those at interest, including local, State, and 
Federal Government, voluntary organizations, business, industry, the design professions, the construction 
industry, the research community and the public. Regional and local differences in the nature and 
magnitude of potentially hazardous earthquake events require a flexible approach adaptable to the relative 
risk, resources and capabilities of each community. The Council recognizes that appropriate earthquake 
hazard reduction measures and initiatives should be adopted by existing organizations and institutions and 
incorporated into their legislation, regulations, practices, rules, codes, relief procedures and loan
A-2 
requirements, whenever possible, so that these measures and initiatives become part of established 
activities rather than being superposed as separate and additional. 
The Council is established as a voluntary advisory, facilitative council of the National Institute of 
Building Sciences, a nonprofit corporation incorporated in the District of Columbia, under the authority 
given the Institute by the Housing and Community Development Act of 1974, (Public Law 93-383), Title 
VIII, in furtherance of the objectives of the Earthquake Hazards Reduction Act of 1977 (Public Law 95- 
124) and in support of the President's National Earthquake Hazards Reduction Program, June 22, 1978. 
2005-2006 BSSC BOARD OF DIRECTION 
Chair -- Jim W. Sealy, FAIA, Architect/Consultant, Dallas, TX 
Vice Chair -- David Bonneville, Degenkolb Engineers, San Francisco, California 
Secretary -- Jim Rinner, Project Manager II, Kitchell CEM, Sacramento, California 
Ex-Officio Member -- Charles Thornton, Chairman/Principal, Thornton-Tomasetti Group, Inc., New 
York, New York 
Members 
Edwin Dean, Nishkian Dean, Portland, Oregon 
Bradford K. Douglas, Director of Engineering, American Forest and Paper Association, Washington, 
D.C. 
Cynthia J. Duncan, Director of Specifications, American Institute of Steel Construction, Chicago, Illinois 
Henry Green, Executive Director, Bureau of Construction Codes and Fire Safety, State of Michigan, 
Department of Labor and Economic Growth, Lansing, Michigan (representing the National Institute of 
Building Sciences) 
Jay W. Larson, American Iron and Steel Institute, Betlehem, Pennsylvania 
Joseph Messersmith, Coordinating Manager, Regional Code Services, Portland Cement Association, 
Rockville, Virginia (representing the Portland Cement Association) 
Ronald E. Piester, Assistant Director for Code Development, New York State, Department of State, 
Kinderhook, New York 
James Rossberg, Manager, Technical Activities for the Structural Engineering Institute, American Society 
of Civil Engineers, Reston Virginia 
W. Lee Shoemaker, Director, Engineering and Research, Metal Building Manufacturers Association, 
Cleveland, Ohio 
Howard Simpson, Simpson Gumpertz and Heger, Arlington, Massachusetts (representing National 
Council of Structural Engineers Associations) 
Shyam Sunder, Deputy Director, Building Fire Research Laboratory, National Institute of Standards and 
Technology, Gaithersburg, Maryland (representing Interagency Committee on Seismic Safety in 
Construction) 
Charles A. Spitz, Architect/Planner/Code Consultant, Wall New Jersey (representing the American 
Institute of Architects) 
Robert D. Thomas, Vice President Engineering, National Concrete Masonry Association, Herndon, 
Virginia 
BSSC Staff 
Claret M. Heider, Vice President for BSSC Programs
A-3 
Bernard F. Murphy, PE, Director, Special Projects 
Carita Tanner, Communications Manager 
BSSC MEMBER ORGANIZATIONS 
Voting Members 
AFL-CIO Building and Construction Trades Department 
American Concrete Institute 
American Consulting Engineers Council 
American Forest and Paper Association 
American Institute of Architects 
American Institute of Steel Construction 
American Iron and Steel Institute 
American Society of Civil Engineers 
ASCE, Kansas City Chapter 
American Society of Heating, Refrigeration, and Air-Conditioning Engineers 
American Society of Mechanical Engineers 
American Welding Society 
APA - The Engineered Wood Association 
Applied Technology Council 
Associated General Contractors of America 
Association of Engineering Geologists 
Association of Major City Building Officials 
Brick Industry Association 
California Geotechnical Engineers Association 
California Division of the State Architect, Office of Regulation Services 
Canadian National Committee on Earthquake Engineering 
Concrete Masonry Association of California and Nevada 
Concrete Reinforcing Steel Institute 
Earthquake Engineering Research Institute 
General Services Administration Seismic Program 
Hawaii State Earthquake Advisory Board 
Institute for Business and Home Safety 
Interagency Committee on Seismic Safety in Construction 
International Code Council 
International Masonry Institute 
Masonry Institute of America 
Metal Building Manufacturers Association 
Mid-America Earthquake Center 
National Association of Home Builders 
National Concrete Masonry Association 
National Conference of States on Building Codes and Standards 
National Council of Structural Engineers Associations 
National Elevator Industry, Inc. 
National Fire Sprinkler Association 
National Institute of Building Sciences 
National Ready Mixed Concrete Association
A-4 
Portland Cement Association 
Precast/Prestressed Concrete Institute 
Rack Manufacturers Institute 
Santa Clara University 
Seismic Safety Commission (California) 
Steel Deck Institute, Inc. 
Structural Engineers Association of California 
Structural Engineers Association of Central California 
Structural Engineers Association of Colorado 
Structural Engineers Association of Illinois 
Structural Engineers Association of Kentucky 
Structural Engineers Association of Northern California 
Structural Engineers Association of Oregon 
Structural Engineers Association of San Diego 
Structural Engineers Association of Southern California 
Structural Engineers Association of Texas 
Structural Engineers Association of Utah 
Structural Engineers Association of Washington 
The Masonry Society 
U.S. Army CERL 
Western States Clay Products Association 
Wire Reinforcement Institute 
Affiliate Members 
Bay Area Structural, Inc. 
Building Technology, Incorporated 
City of Hayward, California 
Felten Engineering Group, Inc. 
H&H Group 
HLM Design 
LaPay Consulting, Inc. 
Niehoff, Dennis, PE 
Square D Company 
Steel Joist Institute 
Vibration Mountings and Controls 
York, a Johnson Controls Company
A-5 
BSSC PUBLICATIONS 
Available free from the Federal Emergency Management Agency at 1-800-480-2520 ( by FEMA 
Publication Number). For detailed information about the BSSC and its projects, contact: BSSC, 
1090 Vermont Avenue, N.W., Suite 700, Washington, D.C. 20005 Phone 202-289-7800; Fax 202- 
289-1092; e-mail ctanner@nibs.org 
NEW BUILDINGS PUBLICATIONS 
NEHRP (National Earthquake Hazards Reduction Program) Recommended Provisions for Seismic 
Regulations for New Buildings, 2003 Edition, 2 volumes and maps, FEMA 450 (issued as a CD with only 
limited print copies available) 
NEHRP (National Earthquake Hazards Reduction Program) Recommended Provisions for Seismic 
Regulations for New Buildings, 2000 Edition, 2 volumes and maps, FEMA 368 and 369 
NEHRP Recommended Provisions: Design Examples, 2006, FEMA 451 (issued as a CD) 
A Nontechnical Explanation of the NEHRP Recommended Provisions, Revised Edition, 1995, FEMA 99 
Homebuilders= Guide to Earthquake-Resistant Design and Construction, 2006, FEMA 232 
Seismic Considerations for Steel Storage Racks Located in Areas Accessible to the Public. 2005, FEMA 
460 
Seismic Considerations for Communities at Risk, Revised Edition, 1995, FEMA 83 
Seismic Considerations: Apartment Buildings, Revised Edition, 1996, FEMA 152 
Seismic Considerations: Elementary and Secondary Schools, Revised Edition, 1990, FEMA 149 
Seismic Considerations: Health Care Facilities, Revised Edition, 1990, FEMA 150 
Seismic Considerations: Hotels and Motels, Revised Edition, 1990, FEMA 151 
Seismic Considerations: Office Buildings, Revised Edition, 1996, FEMA 153 
Societal Implications: Selected Readings, 1985, FEMA 84 
EXISTING BUILDINGS 
NEHRP Guidelines for the Seismic Rehabilitation of Buildings, 1997, FEMA 273 
NEHRP Guidelines for the Seismic Rehabilitation of Buildings: Commentary, 1997, FEMA 274 
Case Studies: An Assessment of the NEHRP Guidelines for the Seismic Rehabilitation of Buildings, 1999, 
FEMA 343 
Planning for Seismic Rehabilitation: Societal Issues, 1998, FEMA 275 
Example Applications of the NEHRP Guidelines for the Seismic Rehabilitation of Buildings, 1999, FEMA 
276
A-6 
NEHRP Handbook of Techniques for the Seismic Rehabilitation of Existing Buildings, 1992, FEMA 172 
NEHRP Handbook for the Seismic Evaluation of Existing Buildings, 1992, FEMA 178 
An Action Plan for Reducing Earthquake Hazards of Existing Buildings, 1985, FEMA 90 
MULTIHAZARD 
An Integrated Approach to Natural Hazard Risk Mitigation, 1995, FEMA 261/2-95 
LIFELINES 
Abatement of Seismic Hazards to Lifelines: An Action Plan, 1987, FEMA 142 
Abatement of Seismic Hazards to Lifelines: Proceedings of a Workshop on Development of An Action 
Plan, 6 volumes: 
Papers on Water and Sewer Lifelines, 1987, FEMA 135 
Papers on Transportation Lifelines, 1987, FEMA 136 
Papers on Communication Lifelines, 1987, FEMA 137 
Papers on Power Lifelines, 1987, FEMA 138 
Papers on Gas and Liquid Fuel Lifelines, 1987, FEMA 139 
Papers on Political, Economic, Social, Legal, and Regulatory Issues and General Workshop 
Presentations, 1987, FEMA 143 
(August 2006)