Seismic Design of

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Nov 25, 2013 (3 years and 4 months ago)

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Seismic Design of
Reinforced Concrete Structures



CASE STUDY #1
Three Storey Office Building
Vancouver, B.C.
Canada

SOFTEK Services Ltd.
#275 – 13500 Maycrest Way
Richmond, BC, Canada V6V 2N8
Tel: (604) 273-7737
Web: www.s-frame.com

Copyright Notice
This document is copyright © 2007 by Softek Services Ltd. All rights reserved. No part of this publication may
be reproduced, transmitted, transcribed, stored in a retrieval system, or translated into any human or computer
language, in any form or by any means, electronic, mechanical, optical, chemical, manual, or otherwise, without
the prior written permission of Softek Services Ltd.
Disclaimer
Softek Services Ltd. cannot be held responsible for the material presented in this document. This document is
intended for the use of professional personnel competent to evaluate the significance and limitations of its
content and recommendations, and who will accept the responsibility for its application. Softek Services Ltd.
disclaims any and all responsibility for the application of the contents presented in this document and for the
accuracy of any of the material contained in this document including computer software referenced herein.
CASE STUDY #1

In this project, we will design various shear walls in a three storey office building located in a
high risk seismic zone using SOFTEK’s products: S-FRAME
®
and S-CONCRETE
TM
. Key
results generated by S-FRAME
®
and S-CONCRETE
TM
will also be verified using hand
calculations.

Building Description



Design Data (NBCC 2005)


Location: Commercial Building in Vancouver, BC, Canada
(Granville & 41
st
Ave), Site Soil = Class C

Loads: Retail on Ground Floor, Live Load = 4.8 kPa
Offices on Upper Levels, Live Load = 2.4 kPa
Partition Allowance = 1.0 kPa
Front Curtain Wall Weight = 1.0 kPa
Ground Snow, S
s
= 1.9 kPa and S
r
= 0.3 kPa

Seismic: S
a
(0.2) = 0.95, S
a
(0.5) = 0.65, S
a
(1.0) = 0.34, S
a
(2.0) = 0.17, PGA = 0.47
Importance Factor I
E
= 1.0
Force Modification Factors R
d
= 2.0, R
0
= 1.4 (moderately ductile SFRS)

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2
Materials:
''
c y c c
f 25 MPa, f 400 MPa, E 4500 f 22,500 MPa= = = =


3 3
c
W 2400 kg/m 23.5 kN/m= =

c
c
E
22,500
G 9
2(1 ) 2(1 0.22)
= = =
+ ν +
221 MPa


Snow Load for Roof Level, NBCC 2005 Clause 4.1.6.2

( ) ( )
s s b w s a r
S I S C C C C S 1 1.9 0.8 1.0 1.0 1.0 0.3 1.82 kPa
⎡ ⎤ ⎡ ⎤
= + = ⋅ ⋅ ⋅ ⋅ ⋅ + =
⎣ ⎦ ⎣ ⎦


Building Seismic Weight Estimation


Building Seismic Weight Estimation at Roof Level
Item
Description
Weight (kN)
1
Snow Load = 0.25 x 1.82 kN/m
2
x 12m x 24m (no hole)
131
2
Slab = 23.5 kN/m
3
x 0.2m x 12m x 24m (no hole)
1354
3
Front Curtain Wall = 1.0 kN/m
2
x 1.5m x 12m
18
4
Columns = 23.5 kN/m
3
x (3 x 0.3m x 0.3m + 0.2m x 0.4m) x 1.5m
12
5
Walls = 23.5 kN/m
3
x [1.5mx(0.2mx64m + 0.25mx2.4mx3) –
2x0.2mx1.2mx0.75m]
506

Sub-Total W
4
=
2,021

Building Seismic Weight Estimation at 3
rd
Floor
Item
Description
Weight (kN)

Area of Holes = 4.8m x 2.4m + 2.4m x 2.4m = 17.28 m
2

Net Area = 12m x 24m – 17.28 m
2
= 270.7 m
2


1
Partition Load = 0.5 kN/m
2
x 270.7 m
2
(max 0.5 kPa)
135
2
Slab = 23.5 kN/m
3
x 0.2m x 270.7 m
2

1272
3
Front Curtain Wall = 1.0 kN/m
2
x 3m x 12m
36
4
Columns = 23.5 kN/m
3
x (3 x 0.3m x 0.3m + 0.2m x 0.4m) x 3m
25
5
Walls = 23.5 kN/m
3
x [3mx(0.2mx64m + 0.25mx2.4mx3) - 2x0.2x1.2mx1.5m]
1012

Sub-Total W
3
=
2,480

Building Seismic Weight Estimation at 2
nd
Floor
Item
Description
Weight (kN)

Area of Holes = 4.8m x 2.4m + 2.4m x 2.4m = 17.28 m
2

Net Area = 12m x 24m – 17.28 m
2
= 270.7 m
2


1
Partition Load = 0.5 kN/m
2
x 270.7 m
2
(max 0.5 kPa)
135
2
Slab = 23.5 kN/m
3
x 0.2m x 270.7 m
2

1272
3
Front Curtain Wall = 1.0 kN/m
2
x 3.25m x 12m
39
4
Columns = 23.5 kN/m
3
x (3 x 0.3m x 0.3m + 0.2m x 0.4m) x 3.25m
27
5
Walls = 23.5 kN/m
3
x [3.25mx(0.2mx64m + 0.25mx2.4mx3) – 2x0.2x1.2mx1.5m]
1098

Sub-Total W
2
=
2,571

Total Weight = W = W
x
= W
y
= W
2
+ W
3
+ W
4
= 2021 + 2480 + 2571 = 7072 kN

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3
Building Period, NBCC 2005 Clause 4.1.8.11(3)

( )
( )
3
4
0.75
a n
T 0.05 h 0.05 9.5 0.27 sec= = =
same for each direction

According to Article 4.1.8.7(2), Equivalent Static Force Procedure as described in Article
4.1.8.11 may be used for structures that meet the following criteria: “regular structures that are
less than 60m in height and have a fundamental lateral period, T
a
, less than 2sec in each of the
orthogonal directions”. For this building, it meets this criteria.

Seismic Base Shear Calculation


4.1.8.11 Table0.1J,5.0T and 86.5
)0.2(S
)2.0(S
For
4.1.8.11 Table0.1M,0.1T and 86.5
)0.2(S
)2.0(S
For
a
a
a
va
a
a
=<<=
=<<=

∴ No reduction in overturning moment

4.1.8.4.C Table0.1F,34.0)0.1(S and Class C Site For
4.1.8.4.B Table0.1F,95.0)2.0(S and Class C Site For
va
aa
==
==


Minimum Lateral Earthquake Force
0d
E
3
2
d
0d
Ev
0d
Eva
RR
WI)2.0(S
V,5.1R For
RR
WIM)0.2(S
V
4.1.8.11 Article
RR
WIM)T(S
V
≤≥≥
=


S(T = 0.2s) = F
a
S
a
(0.2) = 1.0x0.95 = 0.95
S(T = 0.5s) = F
v
S
a
(0.5) or F
a
S
a
(0.2), whichever is smaller = 0.65
S(T = 2.0s) = F
v
S
a
(2.0) = 0.17

Using linear interpolation: S(T = 0.27s) = 0.88

kN 1600
4.1x0.2
kN 7072x0.1x95.0x
V
kN 429
4.1x0.2
7072x0.1x0.1x17.0
kN 2223
4.1x0.2
kN 7072x0.1x0.1x88.0
V
3
2
=≤
=≥==


∴Seismic Base Shear = V = 1600 kN

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4
Centre of Mass Calculation


Since most of the weight is distributed uniformly around the building, we will place the axes at
the centre of the slab (see figure below). The only weights that will influence the location of the
centre of mass will be Walls #2, #4, #5, #6 and the holes in the slab for the 2
nd
and 3
rd
floor. We
will assume there are no holes at the roof level.

For the roof level:
kNm399305943x8.4mx4.2mx25.0mx5.1xm/kN5.23m6.3mx7.3mx2.0mx5.1xm/kN5.23WiUi
33
=+=+=

m2.0
2021
399
Wi
WiUi
U ===




kNm41033080)m4.862.1(mx4.2mx25.0mx5.1xm/kN5.23
m)05.3(mx7.3mx2.0mx5.1xm/kN5.23WiVi
3
3
−=−−=−−−+
−=

m2.0
2021
410
Wi
WiVi
V −=

==






For practical purposes, the centre of mass equals to the centre of the slab (
0VU ==
). The
influence of Wall #2, #4, #5, #6 and the holes in the slab is minimal because the weight of the
slab and the other walls (#1, #3, and #7) dominate the centre of mass for this building.

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5
Centre of Rigidity Calculation


This hand calculation is based on the assumption that walls are “strong” in bending in one
direction and “very weak” or “negligible” in bending in the other direction. Lateral stiffness (k)
of each wall element must also be estimated – assuming some form of flexural behaviour.
Special attention is given to Wall #2 and Wall #6. The two walls will likely be reinforced in
such a manner that they will deflect and behave as a single unit – an “L-Shaped” wall. This will
be reflected in the computations below. For the purpose of this calculation, the following
reference axes will be used.


Walls #1 and #3 (use h = 3m for calculations)
A
g
= 24m x 0.2m = 4.8 m
2

I
g
= 0.2m x (24m)
3
/ 12 = 230.4 m
4

If we apply a 100kN force at the top, the lateral deflection will be approximately:
mm 0083.00081.000017.0
10x8.4x9221
3000x000,100x2.1
10x4.230x500,22x3
3000x000,100
AG
Vh2.1
IE3
Vh
612
3
gcgc
3
=+=+=+=∆

(assuming a pin-fixed end condition)
direction) weakin stiffness (no 0kandkN/mm 087,12
mm 0.0083
kN 100
k
xy
===


Walls #4 and #5 (use h = 3m for calculations)
A
g
= 2.4m x 0.25m = 0.6 m
2

I
g
= 0.25m x (2.4m)
3
/ 12 = 0.288 m
4

If we apply a 100kN force at the top, the lateral deflection will be approximately:
mm 204.00651.01389.0
10x6.0x9221
3000x000,100x2.1
10x288.0x500,22x3
3000x000,100
AG
Vh2.1
IE3
Vh
612
3
gcgc
3
=+=+=+=∆

(assuming a pin-fixed end condition)
direction) weakin stiffness (no 0kandkN/mm 490
mm 0.204
kN 100
k
yx
===

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6

Wall #2 and Wall #6 as an L-Shape (use h = 3m for calculations)

S-CONCRETE L-Shape Results

If we apply a 100kN force at the top in the X-direction, the
lateral deflection will be approximately:
23
g
49
gy
mm10x1290A,mm10x6.678I ==

mm089.00303.00589.0
10x29.1x9221
3000x000,100x2.1
10x6.678x500,22x3
3000x000,100
AG
Vh2.1
IE3
Vh
x
69
3
gcgyc
3
x
=+=∆
+=+=∆

mm/kN1120
mm0.089
kN 100
k
x
==


If we apply a 100kN force at the top in the Y-direction, the
lateral deflection will be approximately:
23
g
49
gx
mm10x1290A,mm10x9.1785I ==

mm0527.00303.00224.0
10x29.1x9221
3000x000,100x2.1
10x9.1785x500,22x3
3000x000,100
AG
Vh2.1
IE3
Vh
y
69
3
gcgyc
3
y
=+=∆
+=+=∆

mm/kN1898
mm0.0527
kN 100
k
y
==


Wall #7 (use h = 3m for calculations)
This was modelled in S-FRAME using quadrilateral elements and a rigid diaphragm on the roof
(see figure below). A force of 100kN was applied at the roof and a lateral deflection of
0.0213mm was obtained.
direction) weakin stiffness (no 0kandkN/mm 4700
mm 0.0213
kN 100F
k
yx
===

=



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7

Element
Xi
(m)
Yi
(m)
Kxi
(kN/m)
(x 10
3
)
Kyi
(kN/m)
(x 10
3
)
Xi Kyi
(kN x 10
3
)
Yi Kxi
(kN x 10
3
)
1
0
---
---
12,087
0
---
2 & 6
10.2
9.8
1,120
1,898
19,360
10,976
3
12
---
---
12,087
145,044
---
4
N/A
3.6
490
---
---
1,764
5
N/A
6.0
490
---
---
2,940
7
N/A
24
4,700
---
---
112,800


Totals
6,800
26,072
164,404
128,480

m 30.6
072,26
404,164
Kyi
KyiXi
X Rigidity of Centre to Distance
cr
====




Small difference between centre of mass and centre of rigidity (e
x
= 0.3m)

m 9.18
800,6
480,128
Kxi
KxiYi
YRigidity of Centre to Distance
cr
====




Significant difference between centre of mass and centre of rigidity (e
y
= 6.89m)


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8
Distribution of Base Shear, NBCC Clause 4.1.8.11(6)


F
t
= 0 because T
a
< 0.7sec
kN1600Vwhere
hW
hW
)FV(F
n
1
ii
xx
tx
=−=



Level
Height
h
x
(m)
Storey Weight
W
x
(kN)
W
x
h
x
(kNm)
Lateral Force
F
x
(kN)
Storey Shear
V
x
(kN)
Roof
9.5
2,021
19,199.5
693
693
3
6.5
2,480
16,120
582
1275
2
3.5
2,571
8,998.5
325
1600


∑ = 7,072
∑ = 44,318
∑ = 1600


Design Eccentricities, NBCC Clause 4.1.8.11(10)


T
x
= F
x
(e
x
± 0.10 D
nx
) = F
x
(0.3 ± 0.1 x 12) = F
x
(0.3 ± 1.2) kNm
T
x
= F
x
(e
y
± 0.10 D
ny
) = F
x
(6.9 ± 0.1 x 24) = F
x
(6.9 ± 2.4) kNm

Hand Calculations Versus 3D Modelling in S-FRAME
®

For this project and for one load case, hand calculations will be used to compute the distribution
of lateral forces and torsional moments to all the lateral force resisting elements in this building
for the upper most level only. A 3D model using S-FRAME
®
will also be created and the results
will be compared to hand computed values. The 3D model in S-FRAME
®
can also give us an
estimate of the torsional sensitivity, B, for this building, NBCC Clause 4.1.8.11(9). Numerous
other load cases and load combinations will also be generated using S-FRAME
®
.

3D Beam Model in S-FRAME
®


In this 3D model, the walls are modelled as beam elements with a computed moment of inertia
for strong axis bending and zero for weak axis bending. This approach should produce similar
results as hand calculations because the same assumption is applied – weak or non-existent in
one direction and strong in the other direction. Shear areas are also provided to give more
accurate deflections for evaluation purposes. “Rigid” members are provided to model the
behavior at the ends of each wall. “Rigid” means a relatively high moment of inertia.

According to Clause 4.1.8.3 of NBCC 2005, structural modelling shall be representative of the
magnitude and spatial distribution of the mass of the building and of the stiffness of all elements
of the SFRS. The model shall account for the effect of cracked sections in reinforced concrete
and sway effects arising from the interaction of gravity loads with the displaced configuration of
the structure (P-Delta). S-FRAME
®
can perform geometric non-linear analysis (P-Delta).

Furthermore, according to Clause 21.2.5.2.1 of CSA-A23.3-04, for the purpose of determining
forces in and deflections of the structure, reduced section properties shall be used. The effective
property to be used as a fraction of the gross section property shall be as specified below:

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9
Coupling Beams A
ve
= 0.15 A
g
; I
e
= 0.4 I
g
without diagonal reinforcement

Column I
e
=
α
c
I
g
;
0.1
Af
P
6.05.0
g
'
c
s
c
≤+=α

Wall A
xe
=
α
w
A
g
; I
e
=
α
w
I
g
;
0.1
Af
P
6.0
g
'
c
s
w
≤+=α


For walls, P
s
shall be determined at the base of the wall. Preliminary calculations indicate an α
w

value in the range of 0.62 and α
c
value in the range of 0.65 which will be confirmed later.


Element
Gross Properties
Effective Section Properties
Wall #1 & #3
414
3
g
mm 10x304.2
12
000,24x200
I ==

2
g
mm 000,800,4000,24x200A ==

2
v
mm 000,000,4000,24x200x
6
5
A ==

I
e
= 0.62 x 2.304 x 10
14
= 1.428 x 10
14
mm
4

A
e
= 0.62 x 4,800,000 = 2,976,000 mm
2

A
ev
= 0.62 x 4,000,000 = 2,480,000 mm
2

Walls #2 & #6
49
gy
mm 10x6.678I =

49
gx
mm 10x9.1785I =

23
g
mm 10x1290A =

233
v
mm 10x107510x1290x
6
5
A ==

I
ey
= 0.62 x 678.6 x 10
9
= 4.21 x 10
11
mm
4

I
ex
= 0.62 x 1785.9 x 10
9
= 11.07 x 10
11
mm
4

A
e
= 0.62 x 1290 x 10
3
= 799,800 mm
2

A
ev
= 0.62 x 1075 x 10
3
= 666,500 mm
2

Walls #4 & #5
411
3
g
mm 10x88.2
12
2400x250
I ==

2
g
mm 000,6002400x250A ==

2
v
mm 000,5002400x250x
6
5
A ==

I
e
= 0.62 x 2.88 x 10
11
= 1.786 x 10
11
mm
4

A
e
= 0.62 x 600,000 = 372,000 mm
2

A
ev
= 0.62 x 500,000 = 310,000 mm
2

Column
300x300
48
4
gygx
mm 10x75.6
12
300
II ===

2
g
mm 000,90300x300A ==


I
e
= 0.65 x 6.75 x 10
8
= 4.387 x 10
8
mm
4

A
e
= A
g
= 90,000 mm
2

Column
200x400
49
3
gx
mm 10x066.1
12
400x200
I ==

48
3
gy
mm 10x67.2
12
200x400
I ==

2
g
mm 000,80400x200A ==


I
ex
= 0.65 x 1.066 x 10
9
= 6.929 x 10
8
mm
4

I
ey
= 0.65 x 2.67 x 10
8
= 1.735 x 10
8
mm
4
A
e
= A
g
= 80,000 mm
2



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10
Wall #7 is modelled as a “frame” with “rigid members” that connect the end faces of each “pier”
to the ends of “coupling beams”. The “coupling beams” represent the segments at the openings.
This is illustrated below in elevation.

Element
Gross Properties
Effective Section Properties
Wall #7a
411
3
g
mm 10x304.2
12
2400x200
I ==

2
g
mm 000,4802400x200A ==

2
v
mm 000,4002400x200x
6
5
A ==

I
e
= 0.62 x 2.304 x 10
11
= 1.428 x 10
11
mm
4

A
e
= 0.62 x 480,000 = 297,600 mm
2

A
ev
= 0.62 x 400,000 = 248,000 mm
2

Wall #7b
412
3
g
mm 10x8432.1
12
4800x200
I ==

2
g
mm 000,9604800x200A ==

2
v
mm 000,8004800x200x
6
5
A ==

I
e
= 0.62 x 1.8432 x 10
12
= 1.1428x 10
12
mm
4

A
e
= 0.62 x 960,000 = 595,200 mm
2

A
ev
= 0.62 x 800,000 = 496,000 mm
2

Coupling Beam B1
49
3
g
mm 10x717.5
12
700x200
I ==

2
g
mm 000,140700x200A ==

I
e
= 0.4 x 5.717 x 10
9
= 2.287x 10
9
mm
4

A
e
= 0.15 x 140,000 = 21,000 mm
2

Coupling Beam B2
410
3
g
mm 10x625.5
12
1500x200
I ==

2
g
mm 000,3001500x200A ==

I
e
= 0.4 x 5.625 x 10
10
= 2.25x 10
10
mm
4

A
e
= 0.15 x 300,000 = 45,000 mm
2

Coupling Beam B3
410
3
g
mm 10x333.13
12
2000x200
I ==

2
g
mm 000,4002000x200A ==

I
e
= 0.4 x 13.333 x 10
10
= 5.333x 10
10
mm
4

A
e
= 0.15 x 400,000 = 60,000 mm
2

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11
S-FRAME Model using Beam Type Members Only and Rigid Diaphragms




The S-FRAME 3D model of the office building shown here consists only of “beam” type
members with rigid diaphragms specified for each floor level. Only the 2
nd
floor diaphragm is
displayed above.



Special attention is given to Walls #2 and #6. Walls #2 and #6 is modelled as one column which
will be subjected to biaxial bending. The properties of this “column” is given the section
properties of the L-Shape (i.e. I
x
and I
y
). Note that to minimize the amount of torsion that will be
attracted to each wall, the torsional constants, J, for each wall were assigned negligible values.
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12
S-FRAME Model – Center of Rigidity Evaluation


To assess the accuracy of “C of R” calculation, we will apply 1000 kN force at each level in the
X-direction at the computed “C of R”. In theory, loading the building at the “C of R” will
generate deflections without rotation – “pure translation”. The results are displayed below (X-
deflections in mm).

As you can see above, the building is rotating in a clockwise direction. This most likely means
that we have underestimated the stiffness of the L-Shape (Walls #2 & #6). Using a trial-and-
error approach in S-FRAME, we discovered the “true” center of rigidity near e
y
= 5.5m for this
building (as indicated below).


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13
Torsional Sensitivity Evaluation


The next step is to use this 3D model in S-FRAME to give us an estimate of the torsional
sensitivity, B, for this building. Here, the primary concern would be loading in the X-direction
creating a twist in the building. According to NBCC Clause 4.1.8.11(9), the equivalent static
forces, F
x
, shall be applied at distances of ±0.10D
ny
= ±2.4m from the center of mass at each
floor level. The critical load case for this evaluation would be applying the forces at a distance
of +2.4m away from the “C of M”. This is implemented in S-FRAME by applying the
equivalent static forces, F
x
, at the center of mass at each level plus a torsional moment of F
x
x
2.4m in the appropriate direction.


Torsional Sensitivity:
avg
max
x
B
δ
δ
=


Level
Corner Deflections Total
(mm)
# of
Corners
max
δ

(mm)
avg
δ

(mm)
B
x
Roof
2x2.85 + 2x2.13 = 9.96
4
2.85
2.49
1.14
3
rd
2x1.83 + 2x1.27 = 6.20
4
1.83
1.55
1.18
2
nd
2x0.794+2x0.475 = 2.54
4
0.794
0.634
1.25

Base on the results above, B = 1.25 for this building. According to NBCC Clause 4.1.8.11(10),
for a building with B

1.7, torsional effects can be accounted for by applying equivalent static
forces, F
x
, to the building located at ±0.10D
nx
and ±0.10D
ny
from the “C of M” for each principle
direction.

Technically, we should also evaluate the torsional sensitivity for loading in the y-direction (N-S
direction). Since large walls (Wall #1 and #3) dominate the rigidity in the y-direction, it is
unlikely that the torsional sensitivity parameter, B, for loading in this direction will be greater
than that computed above.
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Distribution of Lateral Force to Walls


For hand calculations in tabular form, we will consider only one load case (E-W direction) and
applied to the roof only. The results of these hand calculations will then be compared to the
results generated by S-FRAME for the walls in the top floor. To complete the design of this
building, other load cases will be generated in S-FRAME including loading in the N-S direction,
dead loads, and factored load combinations.

V
x
= F
x
= 693 kN, V
y
= 0 kN
T = V
x
(e
y
+ 0.10D
ny
) = 693 x (6.9 + 0.10 x 24) = 6445 kNm

Note: T = V
x
(e
y
- 0.10D
ny
) = 693 x (6.9 - 0.10 x 24) = 3119 kNm done in S-FRAME only




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Distribution of E-W Lateral Force and Torsional Moment to Walls


V
x
= F
x
= 693 kN, V
y
= 0 kN
T = -V
x
(e
y
+ 0.10D
ny
) = -693 x (6.9 + 0.10 x 24) = -6445 kNm

Force and Torsional Moment Applied at Roof Level

Wall
X
i
(m)
Y
i
(m)
K
xi
(kN/m)
(x 10
3
)
K
yi
(kN/m)
(x 10
3
)
x
xi
xi
V
K
K


(kN)
T
J
ky
r
xii

(kN)
V
xi
(kN)
y
yi
yi
V
K
K


(kN)
T
J
kx
r
yii


(kN)
V
yi
(kN)
#1
-6.3
---
0
12,087
0
0
0
0
-374
-374
#2/6
3.9
-9.1
1,120
1,898
114
50
164
0
36
36
#3
5.7
---
0
12,087
0
0
0
0
338
338
#4
---
-15.3
490
0
50
37
87
0
0
0
#5
---
-12.9
490
0
50
31
81
0
0
0
#7
---
5.1
4700
0
479
-118
361
0
0
0



6800
26,072
693
0
693
0
0
0

[ ]
kNm10x313.1J
kNm10x47001.5490)9.123.15(087,127.511201.9898,19.3087,123.6J
KYKXJ
9
r
32222222
r
xi
2
iyi
2
ir
=
⋅+⋅++⋅+⋅+⋅+⋅=
+=
∑∑



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Comparison of Hand Calculations Versus S-FRAME 3D Results


Wall
Hand Calculations
S-FRAME
Comments
#1
-374 kN
-347 kN
less stiff in S-FRAME
#2 & #6
36 kN / 164 kN*
28 kN / 233 kN*
more stiff in S-FRAME
#3
338 kN
319 kN
less stiff in S-FRAME
#4
87 kN
66 kN
less stiff in S-FRAME
#5
81 kN
68 kN
less stiff in S-FRAME
#7
361 kN
68 + 187 + 71 = 326 kN
less stiff in S-FRAME

* Shear in the weak direction (V
y
= 233 kN) for the L-Shape (Walls #2 & #6) is not displayed
in the above plot but can be obtained easily in a plot for “y Shear”.

Overall, hand calculated results give similar values to S-FRAME. Reasonable numbers were
obtained using simple assumptions on flexural behaviour which otherwise would be considered
rather complex in the 3D world.

The key to structural design is to develop a complete load path, determine the sectional forces
from this load path, and reinforce the members appropriately. This has been accomplished using
both hand calculations and in S-FRAME.

To complete the design of this building, other load cases and load combinations will be
generated using S-FRAME including earthquake loading E-W (-0.10D
ny
), earthquake loading N-
S (±0.10D
nx
), and dead load.

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Dead Load Estimation (at base of the wall)


Wall #1
Tributary Area ≈ 0.375 x 6m x 24m = 54.0 m
2

Slab = 0.2m x 54.0 m
2
x 23.5 kN/m
3
= 253.8 kN at each level
Partitions = 54.0 m
2
x 1 kN/m
2
= 54.0 kN at each level (except roof)
Self Weight = 9.5m x 0.2m x 24m x 23.5 kN/m
3
= 1,071.6 kN
Total at base = 3 x 253.8 + 2 x 54.0 + 1071.6 = 1,941 kN
Walls #2 & #6

Tributary Area ≈ 5m x 5m = 25 m
2

Slab = 0.2m x 25 m
2
x 23.5 kN/m
3
= 117.5 kN at each level
Partitions = 25 m
2
x 1 kN/m
2
= 25 kN at each level (except roof)
Self Weight = 9.5m x 0.25m x 2.4m x 23.5 kN/m
3
= 134 kN
Total at base = 3 x 117.5 + 2 x 25 + 134 = 537 kN
Wall #7b

Tributary Area ≈ 0.8 x 3m x 4.8m = 11.5 m
2

Slab = 0.2m x 11.5 m
2
x 23.5 kN/m
3
= 54.0 kN at each level
Partitions = 11.5 m
2
x 1 kN/m
2
= 11.5 kN at each level (except roof)
Self Weight = 9.5m x 0.2m x 4.8m x 23.5 kN/m
3
= 214 kN
Total at base = 3 x 54.0 + 2 x 11.5 + 214 = 399 kN



Hand calculations for dead load at the base of each wall are similar to the results generated by S-
FRAME. Since S-FRAME is relatively more accurate than the hand computed values, we will
use S-FRAME results to evaluate the effective section properties as outlined in Clause 21.2.5.2.1
of CSA-A23.3-04.
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Wall #1 – Effective Section Properties


P
s
= 1919 kN, f
c
' = 25 MPa, A
g
= 200 x 24,000 = 4,800,000 mm
2

62.0
000,800,4x25
000,919,1
6.0
Af
P
6.0
g
'
c
s
w
≈+=+=α

A
xe
= 0.62 x A
g
= 2,976,000 mm
2
(effective cross-sectional area)
A
ve
= 0.62 x 5/6 x A
g
= 2,480,000 mm
2
(effective shear area)
I
e
= 0.62 x I
g
= 0.62 x 2.304x10
14
= 1.428x10
14
mm
4


Walls #2 & #6 – Effective Section Properties


P
s
= 490 kN, f
c
' = 25 MPa, A
g
= 1,290,000 mm
2

62.0
000,290,1x25
000,490
6.0
Af
P
6.0
g
'
c
s
w
≈+=+=α

A
xe
= 0.62 x A
g
= 799,800 mm
2
(effective cross-sectional area)
A
ve
= 0.62 x 5/6 x A
g
= 666,500 mm
2
(effective shear area)
I
ey
= 0.62 x I
gy
= 0.62 x 678.6 x10
9
= 4.21x10
11
mm
4

I
ex
= 0.62 x I
gx
= 0.62 x 1785.9 x10
9
= 11.1x10
11
mm
4


Wall #7b – Effective Section Properties


P
s
= 440 kN, f
c
' = 25 MPa, A
g
= 200 x 4,800 = 960,000 mm
2

62.0
000,960x25
000,440
6.0
Af
P
6.0
g
'
c
s
w
≈+=+=α

A
xe
= 0.62 x A
g
= 595,200 mm
2
(effective cross-sectional area)
A
ve
= 0.62 x 5/6 x A
g
= 496,000 mm
2
(effective shear area)
I
e
= 0.62 x I
g
= 0.62 x 1.84x10
12
= 1.14x10
12
mm
4



For practical purposes, all the walls in this building appear to have an effective moment of inertia
of 0.62 x I
g
and effective cross-sectional area of 0.62 x A
g
. This was used in the S-FRAME
model to compute the factored lateral deflections (∆
f
) and the factored sectional forces (N
f
, V
f
,
and M
f
) used for analysis and design of these walls.

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19
Design Load Combinations


S-FRAME will be used to generate the load cases and load combinations for design purposes.
The following load combinations will be created for the design of Wall #1, #2 & #6, and #7b
which is based on 1.0 x Earthquake + 1.0 x Dead load factors.

Load Combination #1: 1.0 x E-W (+0.10Dny) + 1.0 x D
Load Combination #2: 1.0 x E-W (-0.10Dny) + 1.0 x D
Load Combination #3: 1.0 x N-S (+0.10Dnx) + 1.0 x D
Load Combination #4: 1.0 x N-S (-0.10Dnx) + 1.0 x D

Lateral Load in Opposite Direction (primarily used to design the L-Shape)
Load Combination #5: -1.0 x E-W (+0.10Dny) + 1.0 x D
Load Combination #6: -1.0 x E-W (-0.10Dny) + 1.0 x D
Load Combination #7: -1.0 x N-S (+0.10Dnx) + 1.0 x D
Load Combination #8: -1.0 x N-S (-0.10Dnx) + 1.0 x D

Companion loads associated with Live and Snow Loads may easily be added to the above load
combinations but, in this case, they will not likely govern the design of this building. The
primary purpose here is to illustrate the use and application of S-FRAME and S-CONCRETE in
the analysis and design of this office building.


Design Sectional Forces at base for Wall #7b (generated by S-FRAME)


Load Combination #1: N
f
= -424 kN, V
f
= 413 kN, M
f
= 2118 kNm, ∆
f
= 1.5 mm
Load Combination #2: N
f
= -437 kN, V
f
= 472 kN, M
f
= 2341 kNm, ∆
f
= 1.6 mm
Load Combination #5: N
f
= -443 kN, V
f
= 463 kN, M
f
= 2478 kNm, ∆
f
= 1.8 mm
Load Combination #6: N
f
= -425 kN, V
f
= 522 kN, M
f
= 2700 kNm, ∆
f
= 1.9 mm

Note: The shear forces displayed here must be “magnified” for design purposes. According
to Clause 21.7.3.4.1 of CSA-A23.3-04, the design shear force or resistance must not
be less than the smaller of: (1) the shear force corresponding to the development of
the nominal moment capacity of the wall at its plastic hinge location and (2) shear
force at R
d
R
o
= 1.0. S-CONCRETE can make this estimation.

Design Sectional Forces at base for Walls #2 & #6 (generated by S-FRAME)


LC #1: N
f
= +477 kN, V
fy
= 411 kN, M
fz
= +1134 kNm, ∆
fy
= 2.2 mm
V
fz
= 53 kN, M
fy
= +84 kNm, ∆
fz
= 0.1 mm
LC #2: N
f
= +482 kN, V
fy
= 388 kN, M
fz
= +1062 kNm, ∆
fy
= 2.0 mm
V
fz
= 30 kN, M
fy
= -65 kNm, ∆
fz
= 0.03 mm
LC #5: N
f
= -1458 kN, V
fy
= 373 kN, M
fz
= -1209 kNm, ∆
fy
= 2.6 mm
V
fz
= 64 kN, M
fy
= +122 kNm, ∆
fz
= 0.1 mm
LC #6: N
f
= -1395 kN, V
fy
= 350 kN, M
fz
= -1137 kNm, ∆
fy
= 2.5 mm
V
fz
= 41 kN, M
fy
= +85 kNm, ∆
fz
= 0.1 mm

Note: This wall may experience small tension forces according to S-FRAME results. This
is reasonable because Wall #3 will be carrying a significant amount of shear force due
to the torsional moment which, in term, will tend to “lift” Walls #6 and #2.
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Design Sectional Forces at base for Wall #1 (generated by S-FRAME)


Load Combination #1: N
f
= -1641 kN, V
f
= 676 kN, M
f
= 9380 kNm, ∆
f
= 0.19 mm
Load Combination #2: N
f
= -1580 kN, V
f
= 195 kN, M
f
= 6448 kNm, ∆
f
= 0.08 mm
Load Combination #3: N
f
= -1882 kN, V
f
= 712 kN, M
f
= 3963 kNm, ∆
f
= 0.16 mm
Load Combination #4: N
f
= -1851 kN, V
f
= 953 kN, M
f
= 5429 kNm, ∆
f
= 0.22 mm
Load Combination #5: N
f
= -2196 kN, V
f
= 803 kN, M
f
= 8855 kNm, ∆
f
= 0.23 mm
Load Combination #6: N
f
= -2257 kN, V
f
= 322 kN, M
f
= 5923 kNm, ∆
f
= 0.11 mm
Load Combination #7: N
f
= -1955 kN, V
f
= 585 kN, M
f
= 4488 kNm, ∆
f
= 0.13 mm
Load Combination #8: N
f
= -1986 kN, V
f
= 826 kN, M
f
= 5953 kNm, ∆
f
= 0.19 mm

Note: Here, the largest moment is generated from a load combination with a significant
torsional moment (#1) which is interesting. The largest shear force is generated from
a load combination that applies the lateral loads in the “strong direction” for this wall
(#4) which is as expected.


S-FRAME results (i.e. axial force, shear force, and moment diagrams) can be directly exported
to S-CONCRETE to complete the design. This is illustrated below for Wall #7b, Wall #2 & #6,
and Wall #1. Hand calculations will also be performed to verify the results of S-CONCRETE.
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Wall #7b – Design and Detailing





S-FRAME will export sectional forces and moments
evaluated at various stations along the member. In this
case, it has evaluated sectional forces at three stations
per member segment per load combination. For this
member in the 3D model, it has been subdivided into
two segments on the first floor.

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22
Let’s assume that minimum distributed reinforcing and zone reinforcing will be sufficient to
meet all the requirements of CSA-A23.3-04. We will design the base of the wall (i.e. plastic
hinge region).

Wall Dimensions: L
w
= 4800mm, b
w
= 200mm, h
w
= 9500mm

ly)(technical WallquatS0.298.1
4800
9500
L
h
w
w
→≤==

However, for practical design purposes, we will treat it as a normal wall.
To force S-CONCRETE to not apply the squat wall provisions, we
assigned a value of 9601mm to h
w
.

Panel Reinforcing: Vertical Bars - 10M @ 400 Each Face (2 curtains)
Horz Bars – 10M @ 400 Each Face (2 curtains)

21.7.3.3.1 Clause0025.00025.0
400x200
100x2
Sb
A2
w
b
hv
≥===ρ=ρ


Zone Reinforcing: 4 – 15M bars at each end of the wall (minimum requirement)
10M Ties @ 95mm (Clause 21.7.3.3.2 and 21.6.6.9)

mm100200x5.0b5.0
mm2713.11x24d24
Governsmm9616x6d6S
w
tie
b
==≤
==≤
→==≤

Axial Load and Moment Capacity:
OK0.174.0
3643
2700
M
M
nUtilizatio
r
f
→≤===


S-CONCRETE results and interaction diagram shown below:



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Overstrength Factor: M
n
= 4198 kNm (nominal moment capacity from S-CONCRETE)
M
f
= 2700 kNm (from S-FRAME)

555.1
2700
4198
M
M
f
n
w
===γ
(same as S-CONCRETE’s estimate)

Dimensional Limitations: Clause 21.7.3.1, L
u
= 3500 – 200 = 3300 mm

OKmm165
20
3300
20
L
Good Notmm235
14
3300
14
L
mm200b
u
u
w
→==≥
→==≥=


However, according to Clause 21.6.3.4, the L
u
/14 requirement may be waived if the neutral axis
depth does not exceed 4b
w
or 0.3L
w
(i.e. C ≤ 800 mm) which is the case here. S-CONCRETE
will compute the neutral axis depths for load combination where flexure is dominant and
determine if the wall meets these requirements for dimensions and ductility. This is displayed
below in the “Results Report” window of S-CONCRETE.


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Ductility Evaluation: Clause 21.7.3.2, R
d
= 2.0, R
0
= 1.4

Check #1
C = 456 mm < 0.15L
w
= 0.15x4800 = 720 mm → OK

Alternative Check #2

OKmm27
350
9500
350
h
mm 9.1andmm1584L33.0C
w
fw
→==<=∆=<


Alternative Check #3 (Clause 21.6.7)
003.0003.00003.0
2
4800
9500
55.1x9.14.1x0.2x9.1
2
L
h
RR
id
w
w
wf0df
id
=θ∴≥=


=

γ∆−∆

025.00164.0002.0
456x2
4800x0035.0
002.0
C2
L
wcu
ic
≤=−=









ε


θ
id
< θ
ic
→ OK

All ductility checks indicate that special concrete confinement
requirements will not be required. S-CONCRETE has the capability to
evaluate special concrete confinement requirements as outlined in Clause
21.6.7.4 for zone reinforcing.

Design Shear Force: According to Clause 21.7.3.4.1 of CSA-A23.3-04, the design shear force
or resistance must not be less than the smaller of: (1) the shear force
corresponding to the development of the nominal moment capacity of the
wall at its plastic hinge location and (2) shear force at R
d
R
o
= 1.0.


kN1462522x4.1x0.2VRR)design(V
kN812522x555.1VV
M
M
)design(V
)FRAMES(f0df
)FRAMES(fw)FRAMES(f
f
n
f
==≤
==γ=










−−


S-CONCRETE has the option to perform “Shear Force Magnification” in
the manner described above to determine the design shear forces.

Shear Resistance: Shear Design is based on Clauses 21.6.9.2 to 21.6.9.7 (simplified method)
Panel Reinforcing – 10M @ 400 H.E.F.

005.0fordbf15.0VVVV
kN812V
idvw
';
ccmaxrscr
f
≤θλφ=≤+=
=

θ
φ
=λβφ=
tanS
dfA
VanddbfV
vyvs
svw
'
ccc


mm3840L8.0d (c),21.7.3.4.2 Clause 45
wv
===θ
o

)11.3.6.3(a Clause 0.18 then ,mm60
f
Sb
f06.0mm200 AIf
2
yv
w'
c
2
v
=β=>=


)21.6.9.6(b Clause 005.0 for 18.0
id

θ≤β


kN3.449N3840x200x25x65.0x18.0x1dbfV
vw
'
ccc
==λβφ=

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25

kN8.652N
45tanx400
3840x400x200x85.0
tanS
dfA
V
vyvs
s
==
θ
φ
=
o


kN11028.6523.449VVV
scr
=
+
=+=


OK0.1737.0
1102
812
V
V
nUtilizatio
r
f
→<===


S-FRAME Results




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26
Walls #2 & #6 – Design and Detailing




Let’s assume that minimum distributed reinforcing and zone reinforcing will be sufficient to
meet all the requirements of CSA-A23.3-04. We will design the base of the wall (i.e. plastic
hinge region).

Wall Dimensions: L
w
= 3700mm, b
w
= 250mm, h
w
= 9500mm

WallquatS a Not0.296.3
3700
9500
L
h
w
w
→>==


Panel 1 Reinforcing: Vertical Bars - 10M @ 300 Each Face (2 curtains)
Horz Bars – 10M @ 300 Each Face (2 curtains)

21.7.3.3.1 Clause0025.000267.0
300x250
100x2
Sb
A2
w
b
hv
≥===ρ=ρ


Panel 2 Reinforcing: Vertical Bars - 10M @ 400 Each Face (2 curtains)
Horz Bars – 10M @ 400 Each Face (2 curtains)

21.7.3.3.1 Clause0025.00025.0
400x200
100x2
Sb
A2
w
b
hv
≥===ρ=ρ


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Zone A Reinforcing: 4 – 10M bars at each end of the wall (minimum requirement)
10M Ties @ 65mm (Clause 21.7.3.3.2 and 21.6.6.9)

mm100200x5.0b5.0
mm2713.11x24d24
Governsmm683.11x6d6S
w
tie
b
==≤
==≤
→==≤

Zone B Reinforcing: 4 – 15M bars at each end of the wall (minimum requirement)
10M Ties @ 95mm (Clause 21.7.3.3.2 and 21.6.6.9)

mm100200x5.0b5.0
mm2713.11x24d24
Governsmm9616x6d6S
w
tie
b
==≤
==≤
→==≤

Zone C Reinforcing: 4 – 10M bars at each end of the wall (minimum requirement)
10M Ties @ 65mm (Clause 21.7.3.3.2 and 21.6.6.9)

mm100200x5.0b5.0
mm2713.11x24d24
Governsmm683.11x6d6S
w
tie
b
==≤
==≤
→==≤

Note: Emphasis was placed on minimizing the amount of vertical bars in the section including
both zone steel and distributed reinforcing. This will reduce the axial load and moment
capacity which increases the N vs M utilization. This, in turn, will reduce the design or
magnified shear forces because it will generate a smaller overstrength factor.

Axial Load and Moment Interaction Diagram (Biaxial Bending, Theta = 94°):


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28


Borderline0.1025.1
4.1109
1.1137
M
M
nUtilizatio
r
f
→≈===


Overstrength Factor for bending about z-z axis (Theta = 90°):

S-CONCRETE has determined the governing load combination for pure unixaxial bending
about the z-z axis is LC #1 which is 1.0xE-W (+0.10D
ny
) + 1.0xD.

N
f
= +477 kN, M
f
= 1134 kNm
M
n
= 1365 kNm (nominal moment capacity from S-CONCRETE)

20.1
1134
1365
M
M
f
n
w
===γ
(same as S-CONCRETE’s estimate)

Panel 1 Dimensions: Clause 21.7.3.1, L
u
= 3500 – 200 = 3300 mm

OKmm165
20
3300
20
L
mm250b
u
w
→==≥=


OKmm25752009500x25.0th25.0mm2400L
2w

=
+
=
+

=


Panel 2 Dimensions: Clause 21.7.3.1, L
u
= 3500 – 200 = 3300 mm

OKmm165
20
3300
20
L
mm200b
u
w
→==≥=


NGmm26252509500x25.0th25.0mm3700L
2w

=
+
=
+
>=




According to Clause 21.7.3.1, the flange width of Panel 2 is too long. This means that part of
Panel 2 is ineffective in the overall axial load and moment capacity of the section for bending
about the z-z axis. Technically, we should shorten the length of the panel which is unlikely.
Evaluating the nominal moment capacity in this direction using the full length will give a
conservative estimate on the required design shear force (i.e. higher overstrength factor). The
“Warning” can be ignored in this case.


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Ductility Evaluation: Clause 21.7.3.2, R
d
= 2.0, R
0
= 1.4, C = 310 mm from S-CONCRETE

Check #1
C = 310 mm < 0.15L
w
= 0.15x2400 = 380 mm → Not OK

Alternative Check #2

OKmm27
350
9500
350
h
mm 6.2andmm792L33.0C
w
fw
→==<=∆=<


Alternative Check #3 (Clause 21.6.7)
003.0003.00005.0
2
2400
9500
20.1x6.24.1x0.2x6.2
2
L
h
RR
id
w
w
wf0df
id
=θ∴≥=


=

γ∆−∆

025.00115.0002.0
310x2
2400x0035.0
002.0
C2
L
wcu
ic
≤=−=









ε


θ
id
< θ
ic
→ OK

Design Shear Force:
kN494411x203.1VV
M
M
)design(V
)FRAMES(fw)FRAMES(f
f
n
f
==γ=









−−


kN1151411x4.1x0.2VRR)design(V
)FRAMES(f0df
=
=




S-CONCRETE has the option to perform “Shear Force Magnification” in
the manner described above to determine the design shear forces.

Shear Resistance: S-CONCRETE results



For this wall, the section may be subjected to tension forces. Here, the
General Method of Shear Design must be used to evaluate the shear
resistance.


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Wall #1 – Design and Detailing (Sq uat Wall)




Let’s assume that minimum distributed reinforcing and zone reinforcing will be sufficient to
meet all the requirements of CSA-A23.3-04. We will design the base of the wall.

Wall Dimensions: L
w
= 24,000mm, b
w
= 200mm, h
w
= 9500mm

21.7.4) (Clause WallquatS0.2396.0
000,24
9500
L
h
w
w
→<==


Panel Reinforcing: Vertical Bars - 10M @ 300 Each Face (2 curtains)
Horz Bars – 10M @ 300 Each Face (2 curtains)

)21.7.4.5(a Clause003.000333.0
300x200
100x2
Sb
A2
w
b
hv
≥===ρ=ρ

Maximum Bar Spacing = 300mm Clause 21.7.4.5(a)

Zone Reinforcing: According to Clause 21.7.4.6, tied vertical reinforcement shall be provided
at each end of the wall. The minimum reinforcement ratio of 0.005 shall
be provided over a minimum wall length of 300mm. A minimum of four
bars shall be provided and tied as a column in accordance with Clause 7.6.
The ties shall be detailed as hoops.

6 – 15M bars at each end of the wall (spaced at 150 mm apart)

)21.7.4.5(a Clause005.00133.0
150x200
200x2
Sb
A2
w
b
≥===ρ


10M Ties @ 200 mm (Clause 7.6.5.2)

Governsmm200b
mm5423.11x48d48
mm25616x16d16S
w
tie
b
→=≤
==≤
==≤

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Axial Load and Moment Capacity: (S-CONCRETE Results)

OK0.111.0
83860
9379
M
M
nUtilizatio
r
f
→<===






According to Clause 21.7.4.7, the vertical tension force required to resist overturning at the base
of the wall shall be provided by zone reinforcing and panel reinforcing in addition to the amount
required by Clause 21.7.4.8 to resist the shear corresponding to the applied bending moment.

00037.0112.0x00333.0
M
M
Moment for Required Ratio Steel VerticalEstimated Let
r
f
vm
==⋅ρ≈=ρ


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Design Shear Force:
8.24.1x0.2RR4.10
9379
97757
M
M
0d
f
n
w
==>===γ



γ
w
= 2.8 Overstrength Factor
V
f (design)
= γ
w
V
f (sframe)
= 2.8 x 953 = 2,668 kN (Load Combination #4)

Shear Design: Clause 21.7.4.8
d
v
= 0.8 L
w
= 0.8 x 24,000 = 19,200 mm
V
f
≤ 0.15
λ

φ
c
f
c
' b
w
d
v
= 0.15 x 1 x 0.65 x 25 x 200 x 19,200 N = 9360 kN

β
= 0 → V
c
= 0

wh
v
wh
vysvyvs
s
b
S
A
whereb
tan
df
tanS
dfA
V ρ=ρ⋅
θ
φ
=
θ
φ
=

V
f


V
r
= V
c
+ V
s
= V
s
assume θ = 45°


kN4352N200x00333.0x
45tan
200,19x400x85.0
b
tan
df
VV
wh
vys
sr
==ρ⋅
θ
φ
==
o


OK161.0
4352
2668
nUtilizatio →<==





Let
ρ
vs
= vertical steel ratio required to resist shear
A
g
= 200 x 24,000 = 4,800,000 mm2

For load combination #4, V
f
= 2668 kN, P
s
= N
f
= 1851 kN
OK00204.0
200x200,19x400x85.0
45tanx000,668,2
bdf
tanV
)d'req(00333.0
wvys
f
hhsh
→==
φ
θ
=ρ=ρ≥=ρ
o
00113.0
000,800,4x400x85.0
000,851,1
45tan
00204.0
Af
P
tan
2
gys
s
2
hs
vs
=−=
φ

θ
ρ

o


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For another load combination, V
f
= 2668 kN, P
s
= N
f
= 1448 kN
OK00204.0
200x200,19x400x85.0
45tanx000,668,2
bdf
tanV
)d'req(00333.0
wvys
f
hhsh
→==
φ
θ
=ρ=ρ≥=ρ
o
00115.0
000,800,4x400x85.0
000,448,1
45tan
00204.0
Af
P
tan
2
gys
s
2
hs
vs
=−=
φ

θ
ρ

o


Total Vertical Steel Ratio Required:


OK00152.000115.000037.000333.0
vsm(required) vv
→=
+
=
ρ
+
ρ
=
ρ
≥=
ρ

According to Clause 21.7.4.7, all vertical reinforcement required at the base of
the wall shall be extended the full height of the wall.

S-FRAME Results (Panel and Zone Reinforcing):


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Conclusions


When designing walls that intersect with other walls (Wall #6), we have neglected the influence
of Wall #3 on Wall #6. A portion of Wall #3 should be included in the calculation for moment
capacity which, in turn, will likely increase the design shear force. Overall, neglecting the
intersection of Wall #6 with Wall #3 will not change the reinforcing configuration very much – if
at all. However, as always, careful consideration of all the parameters should be given
nevertheless.

Some engineers may have considered a different approach to the design of Wall #7. In our
model, we have assumed a “coupled wall system” which may be inappropriate for such a short
wall. In fact, a finite element model of the same building appears to contradict the sectional
forces produced by this beam model version. For more information on the finite element model,
refer to “Case Study #2”. The finite element model suggests that “beam theory of plane sections
remaining plane” does not apply to Wall #7 and Wall #1. In Case Study #2, you will find
significant differences in the sectional forces generated for each wall. This suggests that Wall #7
should be designed as a “squat wall” and view the window openings as having little influence on
the overall behaviour of the wall.

Hand calculations may give you reasonable design values for the lateral load resisting elements
in a given building but 3D modelling will give you a better representation of the overall
performance of the building provided the model “truly” represents its behaviour in an
earthquake. The key to any design is to ensure that a “load path” has been defined and carried
through to all the lateral load and gravity load resisting elements in the building. Minimizing the
twist in the building and detailing the members carefully will help ensure that the loads reach the
beams, columns and walls as designed.

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