ERBS design procedure
Terms excluding ①～⑨follow the “Structuraltechnical standard
manual of buildings”,
BCJ, 2001, in Japaneseunit stressdesignLoad combination:
Calculate the permanent and temporary force

design② Frame analysis
1st step

Allowable① Set the beam shape

2nd step

Permanent loads
and external loads

Start③ Check the frame structure stress④ Check the stress concentration of floor slab

story ≦ 3height ≦ 13m
eaves height ≦ 9m
span ≦ 6m
total floor area ≦ 500 m2⑤ Check the RBS section
height > 31m
Decision for the specified buildings

DecisionDecision⑥Check the interstory
drift angle
drift angle ≦ 1/200

Check the story stiffness ratio，and horizontal torsion
Story stiffness ratio ≧ 6/10
Eccentricity ratio ≦ 15/100⑥Check the interstory
drift angle
drift angle ≦ 1/200

NoYes○Allowable stresscheck by C 0 ≧ 0.3○Prevent the damageof the brace end○Increase brace stress with β⑦Prevent the damage of the brace
end and section connections⑧Prevent local buckling, etc.
○Prevent damage of column base

Route １

Route ２End

Design flow of ERBS

-1-⑨Check horizontal
load-carrying capacity
Qu ≧ QunQun  D s Fes QudRoute ３

① Determine beam shape
ERBS shape and size is determined using Table 1~2 and Figure 1~3 by applying experimental
and analytical results, and FEMA-350 (Reference 1).
Table 1 Shape and Size of the Beamtypesizea1  max50,5tf a  0.5～0.75 b f

b  0.65～0.85 db

R  4c  c  b  b  8c

ERBS-H typeb f 1  0.85b f 0c  0.2b f  c  0.25b f

b  0.65～0.85 db

R  4c  c  b  b  8c

figure 1H type(*1）

a1  max50,5tf a  0.5～0.75 b f

f  2b f / 3 , f1  0.2 f
t f  3  trib  t fdb  600

ERBS-BH typefigurec  0.2b f  c  0.25b f
f  1.2b f , f1  0.1 ffigure 2(*2）

BH typeb f 1  0.85b f 0db  900(Reference1)“FEMA-350,Recommended

Seismic

DesignCriteria

forNew

Steel

Moment-Frame Buildings, June 2000”(*1）Align the beam flange bottom and horizontal beam end wing, when trib  t f using theback up bar.
(*2）Set the joint at the end of f , or on the inner side toward the beam center.
(*3）Boxing in the H-type inner wing is not always required.
Definition of terms
Reduced section：reduced bottom section where beam flange width is b f 1

d b ：beam depthb f 0 ：center flange widthb f ：inner flange widthc ：depth of flange reductionb f 1 ：reduced flange widthb fe ：flange width at face

α：point at flange width is b f

a ：length between face and αb ：length of reduction

f ：length of the inner wingf 1 ：parallel part length of the inner wingt f ：flange thickness

Z p ：plastic section modulus at reduced section（including web）

M f ：end beam moment

Z b ：plastic section modulus at beam end, Z b  t f・b fe・d b  t fr ：radius of reduction

A f 1 ：flange cross sectional area at reduced section, A f 1  b f 1・t f

e ：space size at beam flange side-2-Table 2
Parameters & Details
Beam symmetry
Width-thickness ratio
of beamLateral stiffening
the beamPrequalification Data for ERBS Connections

Limitations
Wing and reduction shape should be symmetric.
FA rank. Evaluate it using the center flange width b f 0 , even with theofMinimum
span- to-depth ratio
Centre flange width

reduced section.
“Lateral stiffening ensuring horizontal load carrying capacity”.
Evaluate using A f 1 and i y at the reduction length. Gusset platesof stiffening member should not be welded, within the range a+b
from the beam end.
[span／depth]  5Centre flange width： b f 0  300

Flange thickness t f

40mm maximum

Eccentric connection tocolumn
Floor rigidity

Flange width at face b fe is within the column width.

Welded studs
of composite slab
Supplementary
information about theshapeConcrete slab or a metal deck and concrete slab or horizontal bracing
should be used.
Welded studs should not be placed in the area of the beam flange
between the column face and 6 inches beyond the extreme end of theERBS(a+b+15cm).
・Point α provides length a and b.
・When flange width of the face is increased by more than b f , extend
it nearly in the tangential direction from the point α. Flange widthat face b fe is more than b f .
・Beam end size related to the ultrasonic inspection and welding isshown in Figure 3.

Figure 1 ERBS-H shape and size

-3-

Figure 2 ERBS-BH shape and size

Figure3 Beam end size for various methods-4-The beam shape design flow of H and BH types is shown in figure 4.
Structural planning：assume member section of the frame
Set the beam section d b , b f 0 , t f
Assume the hinge location a  b / 2 (note 1)
Set depth of flange reduction（ c ）0.2b f  c  0.25b fb f  1.5b f 0b f 1  b f  2c ≦ 0.85b f 0
Set the reduction length （ b ）Adjust the flange reducing angle （ b / 2c ） (note 2)
Nob  0.65～0.85 d bYESNoa  0.5～0.75b f
YES

Check the limitations(table 2)/beam end size（figure 3)/
shape and size（table １ and figure 1~２）
Figure 4 Beam shape design flow
(Note 1) Initial value of the hinge location referenced according to the experiment is shown
below.
Sectiondb / b f 0 = 3

d b / b f 0 =2.5(experiment)
d b / b f 0 =2
(Note 2)

Hinge location for referencea  b / 2 / d b  0.65
a  b / 2 / d b  0.68
a  b / 2 / d b  0.80

Adjust the flange reducing angle,

estimated using analysis and experiment.

b / 2c ≒2.9-5-

② Frame analysis
Frame analysis should be conducted according to the models accurately expressing beam
reduction shape characteristics. However, it should be implemented using the following
frame analysis method having a rigid zone, when the one described above is not used.
・Model the reduced section and rigid zone using the frame analysis program.
・Set the rigid zone length a + b / 2 from the column face. Set the reduced section length b/2.
・Set the joint location a + b from the column face dividing it into the reduced section andcenter section.

Joint

・The beam stiffness does not consider the composite action with concrete slab.

Rigid zone

Center section

Reduced section

Elevation of analysis model

Beam plana+b/2 b/2lb

Figure 5 Frame analysis with a rigid zone
③ Check stress on frame structure
・The allowable unit stress of buckling of the bending member shall be evaluated using b f 1
and t f in the reduced section. The interval between the support points of the compression
flange l b must be the length between the column face (or column center) and first support
point from the column face where the reduced section is located.
・Set the location for the design of the section at the rigid zone end.
・The allowable unit stress of buckling of the compression member of the beams shall be
evaluated using the radius of gyration of the reduced section, where the reduced section islocated.
④ Check stress concentration of the floor slabWelded studs should not be placed in the beam flange area between the column face and 6
inches beyond the extreme end of the ERBS (a + b + 15 cm). The shearing force transfer ofthe floor slab should be designed taking into account the fewer studs effects, where an extra
shear force occurs at the brace. The allowable strength of studs meets the requirement of AIJ,
1985, Design Recommendations for Composite Constructions.

-6-⑤ Check the RBS the section
Verify that the bending and shear strength satisfy the RBS design procedure. The beam end
connection bears the moment of the flange and the shear force of the web.
ⅰ)Design for the beam end moment
・Evaluation of the bending strength of connections
Bending moment working at the end connection of the beam is calculated as follows.
In this procedure, the beam end moment M f is determined using the load condition andmechanism, as shown in figure 6~7. When the permanent load condition is different, thecorresponding shearing force V p should be calculated.

M f  M pr  V p x = C pr R y Z p F y  V p x(1)

M pr  C pr Ry Z p Fy(2)M f ：the moment demands at the column face.
M pr ：the probable peak plastic hinge moment
C pr ：the peak connection strength coefficient, including strain hardening, local restraint,additional reinforcement, and other connection conditions. C pr  1.15 for RBS.

R y ：A coefficient applicable to the beam or girder material, Ry  1.1
Z p ：effective plastic modulus of the section at the location of the plastic hinge (including
the web).

Fy ：the specified design strength
V p ：the shear force at the plastic hinge.
Vp M pr  M pr  PL' / 2  wL'2 / 2(3)L'x ：the length between the plastic hinge and column face, x  a  b / 2p ：the concentrated load on the beam centerw ：the distributed load on the beam.

column柱

Mpr

VAhinge
ヒンジw
VPx

P

Mpr

L’L’=7.3mx

column柱Figure 6 Sample calculation of shear at a plastic hinge taking into account the gravity loads-7-Mf

Mpr
VPx

Figure 7 Calculation of demands at column face
・Check the bending strength of the connection
If the following equation is satisfied, the design is acceptable.

M f  R y Z b Fy

(4)Z b ：plastic section modulus at beam end， Z b  t f・b fe・d b  t fⅱ）Design for the shear of the connection
Calculate the shear at the column face V f , according to the equation:
Vf  2

VfMf
L  dc QL

t w d b  2t f  2 S r(5) f s ≦1.0
where,

Q L ：shear due to permanent load.
f s ：allowable shear stress under temporary forces
S r ：height of weld access hole

For the rest, panel zone and continuity plates should be calculated according to “DesignStandard for Steel Structures”, “AIJ, 2001, Recommendation for Design of Connections in
Steel Structures,” etc.
⑥ Check the interstory drift angle
When the interstory drift angle is evaluated using the “frame analysis method with rigidzone” in frame analysis, the calculated displacement is smaller than the actual one withrespect to the less flexural length of the beam. Therefore, it should be checked minutely by
the following the simple calculation of safety side estimation. （In order to calculate thedisplacement accurately, it should be evaluated using another non-uniform beam section
model, etc. The calculated result of frame displacement with a non-uniform beam section
element is shown in reference A.）・Model the beam using the uniform section element. The section is the center of the steel
beam section. The beam stiffness does not take into account the composite action withconcrete slab in the same way as the frame analysis.
・Check that the interstory drift angle is within the allowable range of the calculateddisplacement.
This frame displacement having a uniform section element provides safety estimation,
because the displacement is greater than that of the actual accurate RBS section analysis.
The beam deflection caused by permanent loads is safe, because the displacement is greaterthan the actual one.

-8-

⑦ Prevent the damage of the brace end and section connection
Check the “connection to ensure the horizontal

M *pload carrying capacity” at the beam end as
follows.
・The moment demands of the connection

M *p  M f

The bending moment acting at the beamhinge

end of the horizontal load-carrying capacity
is the moment demanded at the columnface( M f ).
Check that the connection does not breakby multiplying the safety factor with theAssumed plastic area
(more than l /10 or 2d)where、 l :length of beam，d:beam depth

moment. While calculating this moment M f , the permanent load and C pr is considered asa supplementary factor.
The moment demands of connection M *p is given by the following expression

M *p  M f   C pr R y Z p F y  V p x ・Check the ultimate bending moment of the connection
Check that the connection does not break, assuming that the moment demands ofconnection is M *p . However, it is always satisfied using the following calculation, because
the RBS section check is executed using the allowable stress.
Multiplying the safety factor α with the RBS equation, R y Z b F y   M f  M *p（400N quality carbon steel：α＝1.3）

Z b σU ＝ 400 Z b ＞  R y Z b F y

 = 1.3 × 1.1 Z b × 235 = 336 Z b ＞ M *pOK

（490N quality carbon steel：α ＝ 1.2）

Z b σU ＝ 490 Z b ＞  R y Z b Fy

 = 1.2 × 1.1 Z b × 345 = 455 Z b ＞ M *p

OK⑧ Prevent local buckling, etc.
To prevent lateral buckling, the procedure for checking the beam is as follows:・Applying the case “secure the lateral stiffening supports at equal distance along the fulllength of the beam”
The lateral stiffening supports the distance at the beam end within the range for l / n e . The
lateral stiffening supports number n e with a slenderness ratio along the minor axis of thereduced beam section  ye by satisfying the following equation.

 ye ≦170  20ne (400 Ｎ quality carbon steel) ye ≦130  20ne (490 Ｎ quality carbon steel)
The lateral stiffening interval at the center of the beam is within the range for l / n c . n c isthe lateral stiffening number with a slenderness ratio along the minor axis of the beam
center section  yc by satisfying the following equation.

 yc ≦170  20nc (400 Ｎ quality carbon steel)-9- yc ≦130  20nc (490 Ｎ quality carbon steel),where l ：beam length ye ：slenderness ratio along the minor axis of the reduced beam section（＝ l / i ye ） yc ：slenderness ratio along the minor axis of the beam center section（＝ l / i yc ）i ye ：radius of gyration along the minor axis of the reduced beam section, i ye  I y1 A1i yc ：radius of gyration along the minor axis of the beam center section, i yc  I y A
I y1 ， I y ：moment of inertia of the reduced beam section and beam center section along
the minor axis.
A1 ， A ：area of the reduced beam section and the beam center section.
・Applying the case “secure the lateral stiffening member mainly near the beam end”
Secure the lateral stiffening supports at the distance calculated using the following
equation in the area where the bending moment exceeds the yield moment. The moment
distribution used to calculate the lateral stiffening supports is also estimated, assuming thatthe moment at the column face is M f . Furthermore, the yield moment should be calculatedusing the reduced beam section at the end and the beam center section at the center of thebeam.
Multiply the safety factor α with the moment distribution for calculating the lateral
stiffening supports. The safety factor α is 1.2 for 400 N quality carbon steel, and 1.1 for 490
N quality carbon steel.
For 400 Ｎ quality carbon steel,the distance between the supports at the end：

l be・d bl≦ 250 and be ≦ 65
Af1i yethe distance between the supports at the center：l bc・d bl≦ 250 and bc ≦ 65
Af 0i yc

For 490 Ｎ quality carbon steel,the distance between the supports at the end：

l be・d bl≦ 200 and be ≦ 50
Af1i yethe distance between the supports at the center：l bc・d bl≦ 200 and bc ≦ 50
Af 0i ycA f 1 ：flange cross-sectional area at reduced section, A f 1  b f 1・t f

A f 0 ：flange cross-sectional area at the center, A f 0  b f 0・t f
l be ：the distance between the supports at the reduced beam section
l bc ：the distance between the supports at the center⑨ Check the horizontal load-carrying capacity
Assume that the plastic hinges of the beam develop at the reduced beam section while
calculating the horizontal load carrying capacity. C pr is not multiplied while calculating thehorizontal load-carrying capacity.
If it is calculated using the frame analysis program, it can be executed using the same
rigid zone model employing the frame analysis method of the 1st step design.

- 10 -＜Reference A＞ An example of the frame displacement calculation
An example of the frame displacement calculation using the non-uniform beam element isshown below.
The element and model used in the analysis is shown in Figure A-1, and the analyzeｄ cases
are shown in Table A-1.

Table A-1 Analyzed cases

Case
1 non-uniform beam
section2 uniform beam
section3 rigid zone + steelbeam

End model
ERBSbeam

Hinge model

Center model

Notes
Calculation byERBS beam
Center section
actual shape
For displacement
Center section
calculation

Rigid zone

Reduced section

Center section

For strength calculation

Figure A-1 Element and model used in the analysis- 11 -The input load and frame model is shown in Figure A-2.
“Case 2: uniform beam section” gives greater displacement than “Case 1: non-uniform beam
section” using the calculated result shown in Table A-2, and it provides an estimate of safety.

PinP/2

P/2
P=4(KN/ m2)×7×7m=206KNｗ

4.0ｍ

Columnw=5.2(KN/ m2)×7m=36.4KN/m

Beam
7.0ｍ

Earthquake load

Permanent load

Figure A-2 Input load and frame model
Table A-2 Calculated Results

Case

Horizontal
displacement inearthquake
ratio
displacement
(％)
(cm)

1. non-uniform beam section(Calculation by actual shape)
2. uniform beam section（For displacement calculation）
3. rigid zone + steel beam
(For strength calculation)

Vertical deflection inpermanent load
displacement
(cm)

ratio
(％)

2.17

100

0.28

100

2.46

113

0.31

111

1.71

79

0.2279＜Reference B＞ Flange reduction fabrication
The side of an arc along the reduction length b should be ground conforming both H and BHtypes. In H type, the point of intersection of the reduced arc and beam flange (marked ○ in thefigure) should be included in the welding area, as shown in the following figure.

- 12 -