Contents Singapore CP65 99 WALL 002

User Manual: Singapore CP65-99 WALL-002

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Software Verification
PROGRAM NAME: ETABS
REVISION NO.:

0

EXAMPLE Singapore CP65-99 Wall-002
P-M INTERACTION CHECK FOR A WALL
EXAMPLE DESCRIPTION
The Demand/Capacity ratio for a given axial load and moment are tested in this
example. A reinforced concrete wall is subjected to a factored axial load Pu =
8368 kN and moments Muy = 11967 kN-m. This wall is reinforced as noted
below. The design capacity ratio is checked by hand calculations and the results
are compared with ETABS program results.
GEOMETRY, PROPERTIES AND LOADING

EXAMPLE Singapore CP65-99 Wall-002 - 1

Software Verification
PROGRAM NAME: ETABS
REVISION NO.:

Material Properties
E=
ν=

25000 MPa
0.2

0

Design Properties

Section Properties
tb = 200 mm
H = 2500 mm
d=
2400 mm
s=
460 mm
As1= As5 = 4-35M+2-20M (4600 mm^2)
As2, As3, As4, As5 = 2-20M (600 mm^2)

f ′c = 30 MPa
fy = 460 MPa

TECHNICAL FEATURES OF ETABS TESTED
 Concrete wall flexural demand/capacity ratio
RESULTS COMPARISON
Independent results are hand calculated and compared with ETABS design
check.

Output Parameter

ETABS

Independent

Percent
Difference

Wall Demand/Capacity Ratio

1.001

1.00

0.10%

COMPUTER FILE: SINGAPORE CP65-99 WALL-002
CONCLUSION
The ETABS results show an acceptable comparison with the independent results.

EXAMPLE Singapore CP65-99 Wall-002 - 2

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PROGRAM NAME: ETABS
REVISION NO.:

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HAND CALCULATION
Wall Strength Determined as follows:
1) A value of e = 1430 mm was determined using e = M u / Pu where M u and Pu were
taken from the ETABS test model PMM interaction diagram for pier P1. Values for
M u and Pu were taken near the balanced condition and large enough to produce a
flexural D/C ratio very close to or equal to one. The depth to the neutral axis, c, was
determined by iteration using an excel spreadsheet so that equations 1 and 2 below
were equal.
2) From the equation of equilibrium:
Pn = Cc + Cs − T
where

Cc =Ccw +Ccf , where Ccw and Ccf are the area of the concrete web and flange in
compression
0.67
=
Ccw
fcu • 200 • ( a − 200 )
γm
0.67
Ccf =
f cu ( 200•2500 )
γm
As′1 
0.67  As′2 
0.67  As′3 
0.67 
fc′  +
fc′  +
fc′ 
 fs1 −
 fs 2 −
 fs 3 −
γs 
γm
γm
γm
 γs 
 γs 

A
A
A
T = s 4 fs 4 + s 5 fs 5 + s 6 fs 6
γs
γs
γs

Cs =

=
Pn1

A′ 
0.67
0.67
0.67 
fcu • 200 • ( a - 200 ) +
fcu ( 200 • 2500 ) + s1  fs1 −
fc′  +
γm
γm
γs 
γm

(Eqn. 1)
As′2 
As 5
As 6
0.67  As′3 
0.67  As 4
fc′  +
fc′  −
fs 4 +
fs 5 +
fs 6
 fs 2 −
 fs 3 −
γs 
γm
γm
γs
γs
 γs 
 γs

3) Taking moments about As6:


a −tf


− t f  + Cs1 ( d − d ′ ) + Cs 2 ( 4 s ) + 
1 Ccf ( d − d ′ ) + Ccw  d −
2
Pn 2 = 



e′ 

Cs 3 ( 3s ) − Ts 4 ( 2 s ) − Ts 5 ( s )


(Eqn. 2)

EXAMPLE Singapore CP65-99 Wall-002 - 3

Software Verification
PROGRAM NAME: ETABS
REVISION NO.:

0

As1 
Asn 
Asn 
0.67 
0.67 
0.67 
f c′=
f c′=
f c′ 
 f s1 −
 ; Csn
 f sn −
 ; Tsn
 f sn −
γs 
γm
γs 
γm
γs 
γm



and the bar strains and stresses are determined below.
=
Cs1
where

The plastic centroid is at the center of the section and d ′′ = 1150 mm
e′ =e + d ′′ =1430 + 1150 =2580 mm.

4) Using c = 1160 mm (from iteration),
a=
β1c =
0.9 •1160 =
1044 mm
5) Assuming the extreme fiber strain equals 0.0035 and c= 1160 mm, the steel stresses
and strains can be calculated. When the bar strain exceeds the yield strain then,
fs = f y :
 c − d′ 
= 0.00320; f s =
ε s E ≤ Fy ;
ε s1 =

 0.0035
 c 
c−s−d 
= 0.00181
εs 2 =

 0.0035
c


 c − 2s − d ′ 
εs3 =

 0.0035 = 0.00042
c


 d − c − 2s 
= 0.00097
=
εs 4 
 εs6
 d −c 
 d −c−s 
= 0.00235
=
εs5 
 εs6
 d −c 
 d −c 
= 0.00374
εs6 =

 0.0035
 c 

f s1 = 460 MPa
f s 2 = 362.0 MPa
f s 3 = 84.4 MPa
f s 4 = 193.2 MPa
f s 5 = 460.00 MPa
f s 6 = 460.00 MPa

Substituting in Eqn. 1 and 2 and iterating the value of the neutral axis depth until the
two equations are equal give
Pn1 = 8368 kN
Pn2 = 8368 kN
M=
P=
8368(1430) /1000 = 11,967 kN-m
n
ne

EXAMPLE Singapore CP65-99 Wall-002 - 4



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