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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14

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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15

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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16

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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17

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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18

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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21

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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23

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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24

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∆−∆

=θ

025.00164.0002.0

456x2

4800x0035.0

002.0

C2

L

wcu

ic

≤=−=

⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

−

ε

=θ

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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27

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∆−∆

=θ

025.00115.0002.0

310x2

2400x0035.0

002.0

C2

L

wcu

ic

≤=−=

⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

−

ε

=θ

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

=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

=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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