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Timestamp: 2019-04-23 16:46:57+00:00

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elevation of the building on line A is shown in Figure 4-4. A CMU wall section is shown in Figure 4-5, and a plan view of an 8'-0&quot; CMU wall/pier is shown in Figure 4-6. The design example illustrates the strength design approach to CMU wall design for both in-plane and out-of-plane seismic forces.
This example will illustrate the following parts of the design process.
1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 10 Design base shear coefficient. Base shear in the transverse direction. Shear in wall on line A. Design 8'-0&quot; shear wall on line A for out-of-plane seismic forces. Design 8'-0&quot; shear wall on line A for in-plane seismic forces. Design 8'-0&quot; shear wall on line A for axial and in-plane bending forces. Deflection of shear wall on line A. Requirements for shear wall boundary elements. Wall-roof out-of-plane anchorage for lines 1 and 3. Chord design.
Base shear in transverse direction.
For longitudinal wall weight (out-of-plane walls), note that the upper half of the wall weight is tributary to the roof diaphragm. This example neglects openings in the top half of the walls.
Shear wall on line A.
Design 8'-0&quot; shear wall on line A for out-of-plane seismic forces.
depending on the direction of seismic ground motion. In this Part, the first of these two analyses will be performed. The analysis will be done using the &quot;slender wall&quot; design provisions of §2108.2.4. The analysis incorporates static plus P deflections caused by combined gravity loads and out-of-plane seismic forces and calculates an axial plus bending capacity for the wall under the defined loading.
Wall load on 8-foot wall (at wall mid-height):  16 ft  Pwall DL = (75 psf )(8 ft ) + 3 ft  = 6,600 lb  2  6,600 lb w wall DL = = 825 plf 8 ft Dead load from wall lintels:  20 ft  PL intel D = (75 psf )(9 ft )  = 6,750 lb  2  l = (96 in. - 44 in.) 2 = 26 in. w L int elD = 6,750 lb = 3,115 plf 26 in. 12 in.
Thus, use the average value of F p = (1 2 )(27.8 psf + 52.5 psf ) = 40.2 psf Calculation of wall moments due to out-of-plane forces is done using the standard beam formula for a propped cantilever. See Figure 4-7 for wall out-of-plane loading diagram and Figure 4-8 for tributary widths of wall used to determine the loading diagram.
W1 = (10 ft + 8 ft + 10 ft )(40.2 psf ) = 1,125 plf W2 = 8 ft (40.2 psf ) = 322 plf Using simple beam theory to calculate moment M oop for out-of-plane forces, the location of maximum moment is at h = 9.8 feet: M oop = 15,530 lb - ft = 186,360 lb - in. Comparison of seismic out-of-plane forces with wind (approximately 25 psf) indicate that seismic forces control the design.
The wall section shown in Figure 4-6 will be designed. The controlling load combinations for masonry are: 1.2 D + 1.6 Lr 1.1(1.2 D + 1.0 E ) = 1.32 D + 1.1( E h + E v ) 1.1E v = 1.1(0.5)C a ID = 0.55(0.53)(1.0) D = 0.30 D Note: Exception 2 of §1612.2.1 requires that a 1.1 factor be applied to the load combinations for strength design of masonry elements including seismic forces. The SEAOC Seismology Committee has recommended that this factor be deleted. However; this example shows use of the factor because it is a present requirement of the code, thus: PD + RLL = 1.2 (27 ,750 lb ) + 1.6 (6840 lb ) = 44 ,244 lb Pu = PD + L + E = PD + 1.1E v = 1.32(27 ,750 lb ) + (0.30 )(27 ,750 lb ) = 44 ,955 lb The controlling load case by examination is Equation (12-5) for gravity plus seismic out-of-plane forces.
' Slender wall design of masonry walls with an axial load of 0.04 f m or less are designed under the requirements of §2108.2.4.4.
Calculate M cr using the value for f r from §2108.2.4.6, Equation (8-31):  96 in.(7.625 in.)2 M cr = S g f r =   6    (4.0 )(2,500 )1 2 = 186,050 lb - in.
= 0.11 in. + 0.28 in. = 0.38 in.
= 0.11 in. + 0.37 in. = 0.48 in.
= 1.76W p = 1.76 (75 psf ) = 132.5 psf M u (132.5 psf )(3 ft )2 8 = 596 lb - ft = 7,155 lb - in.  408,439 lb - in.  Wall section is okay at parapet.
Design 8'-0&quot; shear wall on line A for in-plane seismic forces.
The in-plane shear strength of the wall must be determined and compared to demand. The strength of the wall is determined as follows. Vertical reinforcement is #[email protected] inches o.c. Try #[email protected] inches o.c. horizontally. Note that concrete masonry cells are spaced at 8-inch centers, thus reinforcement arrangements must have spacings in increments of 8 inches (such as 8 inches, 16 inches, 24 inches, 32 inches, 40 inches, and 48 inches). Typical reinforcement spacings are 16 inches and 24 inches for horizontal and vertical reinforcement.
Design 8'-0&quot; shear wall on line A for combined axial and in-plane bending actions.
The reader is referred to an excellent book for the strength design of masonry Design of Reinforced Masonry Structures, by Brandow, Hart, Verdee, published by Concrete Masonry Association of California and Nevada, Sacramento, CA, Second Edition, 1997. This book describes the calculation of masonry wall/pier strength design in detail. The axial load vs. bending moment capacity (P-M) diagram for the wall must be calculated. For this, the designer must understand the controlling strain levels that define yielding and ultimate strength. At yield moment, the steel strain is the yielding strain (0.00207 in./in. strain) and the masonry strain must be below 0.002 in./in. (for under-reinforced sections). At ultimate strength, the masonry has reached maximum permissible strain (0.003 in./in.) and the steel strain is considered to have gone beyond yield strain level (see§2108.2.1.2 for a list of design assumptions). See Figure 4-10 for concrete masonry stress-strain behavior. A representation of these strain states is shown in Figures 4-11 and 4-12 (the pier width is defined as h ).
Pb = 0.85 (2,500)(7.625 in.)(0.85)(92 in.)(0.003 0.00507 ) = 750 kips Pb = 0.65(750 kips ) = 487 kips A P-M diagram can thus be developed. The P-M diagrams were calculated and plotted using a spreadsheet program. By observation, the design values Pu and M u (Pu = 43 k, M u = 167 k - ft ) are within the nominal strength limits of Pn , M n values shown in Figure 4-13. Plots for Pn vs. M n can be seen in Figure 4-13 and for Pn vs. M n in Figure 4-14.
Figure 4-13. The Pn-Mn nominal strength curve with masonry strain at 0.003 in./in.
Figure 4-14. The Pn-Mn design strength curve with masonry strain at 0.003 in./in.
Check for type of wall failure by calculating wall moment at shear Vn :  82.7 k   (10') Vn (10')  0.60  Mu = = 689 k - ft = 2 2 Pu = 43.7 k By looking at the Pn - M n curve, this Pu - M u load is just outside the Pn , M n curve. The shear wall failure will likely be a bending failure. However, the designer might still consider a  = 0.60 for shear design to be conservative.
Deflection of shear wall on line A.
m = 0.7 R s = 0.7 (4.5)(0.021 in.) = 0.066 in. Thus, deflections are less than 0.025h = 3.0 in.
Requirements for shear wall boundary elements.
This P-M point is not within the P-M curve using a limiting masonry strain of 0.0015 in./in. (see Figure 4-15). From an analysis it can be determined that the maximum c distance to the neutral axis is approximately 22 inches. For this example, boundary ties are required. Note that narrow shear wall performance is greatly increased with the use of boundary ties. The code requires boundary elements to have a minimum dimension of 3 × wall thickness, which is 24 inches due to yield moments. After yield moment capacity is exceeded, the c distance is reduced. Thus, if boundary element ties are provided at each end of the wall/pier extending 24 inches inward, the regions experiencing strain greater than 0.0015 in./in. are confined. Space boundary ties at 8-inch centers. The purpose of masonry boundary ties is not to confine the masonry for compression, but to support the reinforcement in compression to prevent buckling. Tests have been performed to show that masonry walls can achieve 0.006 in./in. compressive strains when boundary ties are present.
Figure 4-15. P-M curve for boundary element requirements; masonry strain is limited to 0.0015 in./in.
The P-M curve shown in Figure 4-15 is derived by setting masonry strain at the compression edge at 0.0015 in./in. and by increasing the steel tension strain at the opposite wall reinforcement bars. Moments are calculated about the center of the wall pier and axial forces are calculated about the cross-section. P-M points located at the outside of the denoted P-M boundary element curve will have masonry strains exceeding the allowable, and thus will require boundary element reinforcement or devices. It can be seen that boundary reinforcement is required for the point (Pu = 45 k, M u = 619 k ) . Boundary element confinement ties may consist of #3 or #4 closed reinforcement in 10-inch and 12-inch CMU walls. At 8-inch CMU walls pre-fabricated products such as the &quot;masonry comb&quot; are the best choice for boundary reinforcement because these walls are too narrow for reinforcement ties (even #3 and #4 bars). The boundary reinforcement should extend around three vertical #4 bars at the ends of the wall.
Wall-roof out-of-plane anchorage for lines 1 and 3.
f p = 1.06w p , where w p is the panel weight of 75 psf (see Figure 4-16) loading.
Section 1633.2.8.1 requires a minimum wall-roof anchorage of q roof = 420 plf q roof = 897 plf  420 plf  use q roof = 897 plf The design anchorage reaction at different anchor spacings is thus: at 4'-0&quot; centers, q roof = 3,588 lb at 6'-0&quot; centers, q roof = 5,382 lb at 8'-0&quot; centers, q roof = 7,175 lb Therefore, choose wall-roof anchors that will develop the required force at the chosen spacing. The roof diaphragm must also be designed to resist the required force with the use of subdiaphragms (or other means). The subject of diaphragm design is discussed in Design Example 5. For this example, a double holdown connection spaced at 8'-0&quot; centers will be used (see Figure 4-19). This type of connection must be secured into a solid roof framing member capable of developing the anchorage force. First check anchor capacity in concrete block of Tables 21-E-1 and 21-E-2 of Chapter 21. Alternately, the strength provisions of §2108.1.5.2 can be used. The required tension, T, for bolt embedment is T = E 1.4 = 7,175 lbs 1.4 = 5,125 lb . For ¾-inch diameter bolts embedded 6 inches, T = 2,830 lb per Table 21-E-1 and 3,180 lb per Table 21-E-2. These values are for use with allowable stress design (ASD).
By choosing a pair of pre-fabricated holdown brackets with adequate capacity for a double shear connection into a 2½-inch glued-laminated framing member, the brackets are good for 2 × 3,685 lb = 7 ,307 lb (ASD) &gt; 7,175 lb × 1.4 steel element factor/1.4 ASD factor = 7,175 lb. Thus, the brackets are okay.
a  M n = As f y  d -  2  .314 in.   1    = 11,689 lb - ft  7,176 lb - ft M n = 0.8 (4 ) .20 in. 2 (60,000 psi ) 3.81 in. -  2   12 in.
Per §1633.2.8.1, item 5, the wall-roof connections must be made with 2½-inch minimum net width roof framing members (2½-inch GLB members or similar) and developed into the roof diaphragm with diaphragm nailing and subdiaphragm design.
Anchor bolt embedment and edge distances are controlled by §2106.2.14.1 and §2106.2.14.2. Section 2106.2.14.1 requires that the shell of the masonry unit wall next to the wood ledger have a hole cored or drilled that allows for 1-inch grout all around the anchor bolt. Thus, for a 7/8-inch diameter anchor bolt, the core hole is 2-7/8-inch in diameter at the inside face masonry unit wall. Section 2106.2.14.2 requires that the anchor bolt end must have 1½ inches clearance to the outside face of masonry. The face shell thickness for 8-inch masonry is 1¼ inches, thus the anchor bolt end distance to the inside face of the exterior shell is 7-5/8&quot;-1¼&quot;-6&quot; = 3/8&quot;. It is recommended that the minimum clear dimension is ¼-inch if fine grout is used and ½-inch if coarse pea gravel grout is used (Figure 4-18).
Analysis of transverse roof diaphragm chords is determined by calculation of the diaphragm simple span moment wl 2 8 divided by the diaphragm depth.
Thus 2-#5 chord bars As = 0.62 in.2 are adequate to resist the chord forces. Place chord bars close to the roof diaphragm level. Since roof framing often is sloped to drainage, the chord placement is a matter of judgment.

References: §2108
 §1612
 §2108
 §2108
 §2108
 §1633
 §2106
 §2106