Seismic Design of Reinforced Concrete Structures CASE STUDY #1

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

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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-CONCRETETM. Key results generated by S-FRAME® and S-CONCRETETM will also be verified using hand calculations. Building Description

Design Data (NBCC 2005) Location: Commercial Building in Vancouver, BC, Canada (Granville & 41st 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, Ss = 1.9 kPa and Sr = 0.3 kPa Seismic: Sa(0.2) = 0.95, Sa(0.5) = 0.65, Sa(1.0) = 0.34, Sa(2.0) = 0.17, PGA = 0.47 Importance Factor IE = 1.0 Force Modification Factors Rd = 2.0, R0 = 1.4 (moderately ductile SFRS)

www.s-frame.com © 2007 Softek Services Ltd. 2

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

3 3cW 2400 kg/m 23.5 kN/m= =

cc

E 22,500G 92(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 rS 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/m2 x 12m x 24m (no hole) 131 2 Slab = 23.5 kN/m3 x 0.2m x 12m x 24m (no hole) 1354 3 Front Curtain Wall = 1.0 kN/m2 x 1.5m x 12m 18 4 Columns = 23.5 kN/m3 x (3 x 0.3m x 0.3m + 0.2m x 0.4m) x 1.5m 12 5 Walls = 23.5 kN/m3x [1.5mx(0.2mx64m + 0.25mx2.4mx3) –

2x0.2mx1.2mx0.75m] 506

Sub-Total W4 = 2,021 Building Seismic Weight Estimation at 3rd Floor


Area of Holes = 4.8m x 2.4m + 2.4m x 2.4m = 17.28 m2 Net Area = 12m x 24m – 17.28 m2 = 270.7 m2

1 Partition Load = 0.5 kN/m2 x 270.7 m2 (max 0.5 kPa) 135 2 Slab = 23.5 kN/m3 x 0.2m x 270.7 m2 1272 3 Front Curtain Wall = 1.0 kN/m2 x 3m x 12m 36 4 Columns = 23.5 kN/m3 x (3 x 0.3m x 0.3m + 0.2m x 0.4m) x 3m 25 5 Walls = 23.5 kN/m3x [3mx(0.2mx64m + 0.25mx2.4mx3) - 2x0.2x1.2mx1.5m] 1012

Sub-Total W3 = 2,480 Building Seismic Weight Estimation at 2nd Floor


Area of Holes = 4.8m x 2.4m + 2.4m x 2.4m = 17.28 m2 Net Area = 12m x 24m – 17.28 m2 = 270.7 m2

1 Partition Load = 0.5 kN/m2 x 270.7 m2 (max 0.5 kPa) 135 2 Slab = 23.5 kN/m3 x 0.2m x 270.7 m2 1272 3 Front Curtain Wall = 1.0 kN/m2 x 3.25m x 12m 39 4 Columns = 23.5 kN/m3 x (3 x 0.3m x 0.3m + 0.2m x 0.4m) x 3.25m 27 5 Walls = 23.5 kN/m3x [3.25mx(0.2mx64m + 0.25mx2.4mx3) – 2x0.2x1.2mx1.5m] 1098

Sub-Total W2 = 2,571 Total Weight = W = Wx = Wy = W2 + W3 + W4 = 2021 + 2480 + 2571 = 7072 kN


Building Period, NBCC 2005 Clause 4.1.8.11(3) ( ) ( )

34 0.75

a nT 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, Ta, 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

aa

a

vaa

a

=<<=

=<<=

∴ No reduction in overturning moment

4.1.8.4.C Table0.1F,34.0)0.1(S and Class C Site For4.1.8.4.B Table0.1F,95.0)2.0(S and Class C Site For

va

aa

==

==

Minimum Lateral Earthquake Force

0d

E32

d0d

Ev

0d

Eva

RRWI)2.0(S

V,5.1R ForRR

WIM)0.2(SV

4.1.8.11 ArticleRR

WIM)T(SV

≤≥≥

=

S(T = 0.2s) = FaSa(0.2) = 1.0x0.95 = 0.95 S(T = 0.5s) = FvSa(0.5) or FaSa(0.2), whichever is smaller = 0.65 S(T = 2.0s) = FvSa(2.0) = 0.17 Using linear interpolation: S(T = 0.27s) = 0.88

kN 16004.1x0.2

kN 7072x0.1x95.0xV

kN 4294.1x0.2

7072x0.1x0.1x17.0kN 22234.1x0.2

kN 7072x0.1x0.1x88.0V

32

=≤

=≥==

∴Seismic Base Shear = V = 1600 kN


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 2nd and 3rd 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 =+=+=∑

m2.02021399

WiWiUi

U ===∑∑

kNm41033080)m4.862.1(mx4.2mx25.0mx5.1xm/kN5.23

m)05.3(mx7.3mx2.0mx5.1xm/kN5.23WiVi3

3

−=−−=−−−+

−=∑

m2.02021

410Wi

WiViV −=

−==

∑∑

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.


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) Ag = 24m x 0.2m = 4.8 m2 Ig = 0.2m x (24m)3 / 12 = 230.4 m4 If we apply a 100kN force at the top, the lateral deflection will be approximately:

mm 0083.00081.000017.010x8.4x92213000x000,100x2.1

10x4.230x500,22x33000x000,100

AGVh2.1

IE3Vh

612

3

gcgc

3=+=+=+=∆

(assuming a pin-fixed end condition)

direction) weakin stiffness (no 0kandkN/mm 087,12mm 0.0083

kN 100k xy ===

Walls #4 and #5 (use h = 3m for calculations) Ag = 2.4m x 0.25m = 0.6 m2 Ig = 0.25m x (2.4m)3 / 12 = 0.288 m4 If we apply a 100kN force at the top, the lateral deflection will be approximately:

mm 204.00651.01389.010x6.0x92213000x000,100x2.1

10x288.0x500,22x33000x000,100

AGVh2.1

IE3Vh

612

3

gcgc

3=+=+=+=∆

(assuming a pin-fixed end condition)

direction) weakin stiffness (no 0kandkN/mm 490mm 0.204kN 100k yx ===


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:

23g

49gy mm10x1290A,mm10x6.678I ==

mm089.00303.00589.010x29.1x92213000x000,100x2.1

10x6.678x500,22x33000x000,100

AGVh2.1

IE3Vh

x

69

3

gcgyc

3

x

=+=∆

+=+=∆

mm/kN1120mm0.089kN 100k x ==

If we apply a 100kN force at the top in the Y-direction, the lateral deflection will be approximately:

23g

49gx mm10x1290A,mm10x9.1785I ==

mm0527.00303.00224.010x29.1x92213000x000,100x2.1

10x9.1785x500,22x33000x000,100

AGVh2.1

IE3Vh

y

69

3

gcgyc

3

y

=+=∆

+=+=∆

mm/kN1898mm0.0527

kN 100k 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 4700mm 0.0213

kN 100Fk yx ===∆

=


Element Xi

(m) Yi (m)

Kxi (kN/m) (x 103)

Kyi (kN/m) (x 103)

Xi Kyi (kN x 103)

Yi Kxi (kN x 103)

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 30.6072,26404,164

KyiKyiXiX Rigidity of Centre to Distance cr ====

∑∑

∴Small difference between centre of mass and centre of rigidity (ex = 0.3m)

m 9.18800,6480,128

KxiKxiYi YRigidity of Centre to Distance cr ====

∑∑

∴Significant difference between centre of mass and centre of rigidity (ey = 6.89m)


Distribution of Base Shear, NBCC Clause 4.1.8.11(6) Ft = 0 because Ta < 0.7sec

kN1600VwherehW

hW)FV(F

n

1ii

xxtx =−=

∑

Level Height

hx (m) Storey Weight

Wx (kN) Wx hx (kNm) Lateral Force

Fx (kN) Storey Shear

Vx (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) Tx = Fx (ex ± 0.10 Dnx) = Fx (0.3 ± 0.1 x 12) = Fx (0.3 ± 1.2) kNm Tx = Fx (ey ± 0.10 Dny) = Fx (6.9 ± 0.1 x 24) = Fx (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:


Coupling Beams Ave = 0.15 Ag; Ie = 0.4 Ig without diagonal reinforcement

Column Ie = αc Ig ; 0.1Af

P6.05.0

g'c

sc ≤+=α

Wall Axe = αw Ag; Ie = αw Ig ; 0.1Af

P6.0g

'c

sw ≤+=α

For walls, Ps shall be determined at the base of the wall. Preliminary calculations indicate an αw value in the range of 0.62 and αc value in the range of 0.65 which will be confirmed later. Element Gross Properties Effective Section Properties

Wall #1 & #3

4143

g mm 10x304.212

000,24x200I == 2

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

v mm 000,000,4000,24x200x65A ==

Ie = 0.62 x 2.304 x 1014 = 1.428 x 1014 mm4 Ae = 0.62 x 4,800,000 = 2,976,000 mm2 Aev = 0.62 x 4,000,000 = 2,480,000 mm2

Walls #2 & #6

49gy mm 10x6.678I =

49gx mm 10x9.1785I =

23g mm 10x1290A =

233v mm 10x107510x1290x

65A ==

Iey = 0.62 x 678.6 x 109 = 4.21 x 1011 mm4 Iex = 0.62 x 1785.9 x 109 = 11.07 x 1011 mm4 Ae = 0.62 x 1290 x 103 = 799,800 mm2 Aev = 0.62 x 1075 x 103 = 666,500 mm2

Walls #4 & #5

4113

g mm 10x88.2122400x250I ==

2g mm 000,6002400x250A ==

2v mm 000,5002400x250x

65A ==

Ie = 0.62 x 2.88 x 1011 = 1.786 x 1011 mm4 Ae = 0.62 x 600,000 = 372,000 mm2 Aev = 0.62 x 500,000 = 310,000 mm2

Column 300x300

484

gygx mm 10x75.612

300II === 2

g mm 000,90300x300A ==

Ie = 0.65 x 6.75 x 108 = 4.387 x 108 mm4 Ae = Ag = 90,000 mm2

Column 200x400

493

gx mm 10x066.112

400x200I ==

483

gy mm 10x67.212

200x400I == 2

g mm 000,80400x200A ==

Iex = 0.65 x 1.066 x 109 = 6.929 x 108 mm4 Iey = 0.65 x 2.67 x 108 = 1.735 x 108 mm4

Ae = Ag = 80,000 mm2


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

4113

g mm 10x304.2122400x200I ==

2g mm 000,4802400x200A ==

2v mm 000,4002400x200x

65A ==

Ie = 0.62 x 2.304 x 1011 = 1.428 x 1011 mm4 Ae = 0.62 x 480,000 = 297,600 mm2 Aev = 0.62 x 400,000 = 248,000 mm2

Wall #7b

4123

g mm 10x8432.1124800x200I ==

2g mm 000,9604800x200A ==

2v mm 000,8004800x200x

65A ==

Ie = 0.62 x 1.8432 x 1012 = 1.1428x 1012 mm4 Ae = 0.62 x 960,000 = 595,200 mm2 Aev = 0.62 x 800,000 = 496,000 mm2

Coupling Beam B1 49

3

g mm 10x717.512

700x200I == 2

g mm 000,140700x200A ==

Ie = 0.4 x 5.717 x 109 = 2.287x 109 mm4 Ae = 0.15 x 140,000 = 21,000 mm2

Coupling Beam B2 410

3

g mm 10x625.5121500x200I ==

2g mm 000,3001500x200A ==

Ie = 0.4 x 5.625 x 1010 = 2.25x 1010 mm4 Ae = 0.15 x 300,000 = 45,000 mm2

Coupling Beam B3 410

3

g mm 10x333.13122000x200I ==

2g mm 000,4002000x200A ==

Ie = 0.4 x 13.333 x 1010 = 5.333x 1010 mm4 Ae = 0.15 x 400,000 = 60,000 mm2


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 2nd 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. Ix and Iy). 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.


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 ey = 5.5m for this building (as indicated below).


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, Fx, shall be applied at distances of ±0.10Dny = ±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, Fx, at the center of mass at each level plus a torsional moment of Fx x 2.4m in the appropriate direction.

Torsional Sensitivity:

avg

maxxB

δδ

=

Level Corner Deflections Total (mm)

# of Corners

maxδ (mm)

avgδ (mm)

Bx

Roof 2x2.85 + 2x2.13 = 9.96 4 2.85 2.49 1.14 3rd 2x1.83 + 2x1.27 = 6.20 4 1.83 1.55 1.18 2nd 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, Fx, to the building located at ±0.10Dnx and ±0.10Dny 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.


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. Vx = Fx = 693 kN, Vy = 0 kN T = Vx (ey + 0.10Dny) = 693 x (6.9 + 0.10 x 24) = 6445 kNm Note: T = Vx (ey - 0.10Dny) = 693 x (6.9 - 0.10 x 24) = 3119 kNm done in S-FRAME only


Distribution of E-W Lateral Force and Torsional Moment to Walls Vx = Fx = 693 kN, Vy = 0 kN T = -Vx (ey + 0.10Dny) = -693 x (6.9 + 0.10 x 24) = -6445 kNm Force and Torsional Moment Applied at Roof Level

Wall Xi

(m) Yi

(m)

Kxi

(kN/m) (x 103)

Kyi

(kN/m) (x 103)

xxi

xi VK

K∑

(kN)

TJky

r

xii

(kN)

Vxi

(kN) y

yi

yi VK

K

∑

(kN)

TJkx

r

yii−

(kN)

Vyi

(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

9r

32222222r

xi2iyi

2ir

=

⋅+⋅++⋅+⋅+⋅+⋅=

+= ∑∑


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 (Vy = 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 (-0.10Dny), earthquake loading N-S (±0.10Dnx), and dead load.


Dead Load Estimation (at base of the wall) Wall #1 Tributary Area ≈ 0.375 x 6m x 24m = 54.0 m2

Slab = 0.2m x 54.0 m2 x 23.5 kN/m3 = 253.8 kN at each level Partitions = 54.0 m2 x 1 kN/m2 = 54.0 kN at each level (except roof) Self Weight = 9.5m x 0.2m x 24m x 23.5 kN/m3 = 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 m2 Slab = 0.2m x 25 m2 x 23.5 kN/m3 = 117.5 kN at each level Partitions = 25 m2 x 1 kN/m2 = 25 kN at each level (except roof) Self Weight = 9.5m x 0.25m x 2.4m x 23.5 kN/m3 = 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 m2 Slab = 0.2m x 11.5 m2 x 23.5 kN/m3 = 54.0 kN at each level Partitions = 11.5 m2 x 1 kN/m2 = 11.5 kN at each level (except roof) Self Weight = 9.5m x 0.2m x 4.8m x 23.5 kN/m3 = 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.


Wall #1 – Effective Section Properties Ps = 1919 kN, fc' = 25 MPa, Ag = 200 x 24,000 = 4,800,000 mm2

62.0000,800,4x25

000,919,16.0Af

P6.0

g'c

sw ≈+=+=α

Axe = 0.62 x Ag = 2,976,000 mm2 (effective cross-sectional area) Ave = 0.62 x 5/6 x Ag = 2,480,000 mm2 (effective shear area) Ie = 0.62 x Ig = 0.62 x 2.304x1014 = 1.428x1014 mm4 Walls #2 & #6 – Effective Section Properties Ps = 490 kN, fc' = 25 MPa, Ag = 1,290,000 mm2

62.0000,290,1x25

000,4906.0Af

P6.0

g'c

sw ≈+=+=α

Axe = 0.62 x Ag = 799,800 mm2 (effective cross-sectional area) Ave = 0.62 x 5/6 x Ag = 666,500 mm2 (effective shear area) Iey = 0.62 x Igy = 0.62 x 678.6 x109 = 4.21x1011 mm4 Iex = 0.62 x Igx = 0.62 x 1785.9 x109 = 11.1x1011 mm4 Wall #7b – Effective Section Properties Ps = 440 kN, fc' = 25 MPa, Ag = 200 x 4,800 = 960,000 mm2

62.0000,960x25

000,4406.0Af

P6.0

g'c

sw ≈+=+=α

Axe = 0.62 x Ag = 595,200 mm2 (effective cross-sectional area) Ave = 0.62 x 5/6 x Ag = 496,000 mm2 (effective shear area) Ie = 0.62 x Ig = 0.62 x 1.84x1012 = 1.14x1012 mm4 For practical purposes, all the walls in this building appear to have an effective moment of inertia of 0.62 x Ig and effective cross-sectional area of 0.62 x Ag. This was used in the S-FRAME model to compute the factored lateral deflections (∆f) and the factored sectional forces (Nf, Vf, and Mf) used for analysis and design of these walls.


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: Nf = -424 kN, Vf = 413 kN, Mf = 2118 kNm, ∆f = 1.5 mm Load Combination #2: Nf = -437 kN, Vf = 472 kN, Mf = 2341 kNm, ∆f = 1.6 mm Load Combination #5: Nf = -443 kN, Vf = 463 kN, Mf = 2478 kNm, ∆f = 1.8 mm Load Combination #6: Nf = -425 kN, Vf = 522 kN, Mf = 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 RdRo = 1.0. S-CONCRETE can make this estimation.

Design Sectional Forces at base for Walls #2 & #6 (generated by S-FRAME) LC #1: Nf = +477 kN, Vfy = 411 kN, Mfz = +1134 kNm, ∆fy = 2.2 mm Vfz = 53 kN, Mfy = +84 kNm, ∆fz = 0.1 mm LC #2: Nf = +482 kN, Vfy = 388 kN, Mfz = +1062 kNm, ∆fy = 2.0 mm Vfz = 30 kN, Mfy = -65 kNm, ∆fz = 0.03 mm LC #5: Nf = -1458 kN, Vfy = 373 kN, Mfz = -1209 kNm, ∆fy = 2.6 mm Vfz = 64 kN, Mfy = +122 kNm, ∆fz = 0.1 mm LC #6: Nf = -1395 kN, Vfy = 350 kN, Mfz = -1137 kNm, ∆fy = 2.5 mm Vfz = 41 kN, Mfy = +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.


Design Sectional Forces at base for Wall #1 (generated by S-FRAME) Load Combination #1: Nf = -1641 kN, Vf = 676 kN, Mf = 9380 kNm, ∆f = 0.19 mm Load Combination #2: Nf = -1580 kN, Vf = 195 kN, Mf = 6448 kNm, ∆f = 0.08 mm Load Combination #3: Nf = -1882 kN, Vf = 712 kN, Mf = 3963 kNm, ∆f = 0.16 mm Load Combination #4: Nf = -1851 kN, Vf = 953 kN, Mf = 5429 kNm, ∆f = 0.22 mm Load Combination #5: Nf = -2196 kN, Vf = 803 kN, Mf = 8855 kNm, ∆f = 0.23 mm Load Combination #6: Nf = -2257 kN, Vf = 322 kN, Mf = 5923 kNm, ∆f = 0.11 mm Load Combination #7: Nf = -1955 kN, Vf = 585 kN, Mf = 4488 kNm, ∆f = 0.13 mm Load Combination #8: Nf = -1986 kN, Vf = 826 kN, Mf = 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.


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.


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: Lw = 4800mm, bw = 200mm, hw = 9500mm

ly)(technical WallquatS0.298.148009500

Lh

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 hw.

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

21.7.3.3.1 Clause0025.00025.0400x200

100x2Sb

A2

w

bhv ≥===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)

mm100200x5.0b5.0mm2713.11x24d24

Governsmm9616x6d6S

w

tie

b

==≤

==≤

→==≤

Axial Load and Moment Capacity:

OK0.174.036432700

MMnUtilizatio

r

f →≤===

S-CONCRETE results and interaction diagram shown below:


Overstrength Factor: Mn = 4198 kNm (nominal moment capacity from S-CONCRETE) Mf = 2700 kNm (from S-FRAME)

555.127004198

MM

f

nw ===γ (same as S-CONCRETE’s estimate)

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

OKmm165

203300

20L

Good Notmm23514

330014L

mm200b

u

uw

→==≥

→==≥=

However, according to Clause 21.6.3.4, the Lu/14 requirement may be waived if the neutral axis depth does not exceed 4bw or 0.3Lw (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.


Ductility Evaluation: Clause 21.7.3.2, Rd = 2.0, R0 = 1.4 Check #1 C = 456 mm < 0.15Lw = 0.15x4800 = 720 mm → OK Alternative Check #2 OKmm27

3509500

350h mm 9.1andmm1584L33.0C w

fw →==<=∆=<

Alternative Check #3 (Clause 21.6.7)

003.0003.00003.0

248009500

55.1x9.14.1x0.2x9.1

2Lh

RRid

ww

wf0dfid =θ∴≥=

−

−=

−

γ∆−∆=θ

025.00164.0002.0456x2

4800x0035.0002.0C2Lwcu

ic ≤=−=⎟⎟⎠

⎞⎜⎜⎝

⎛−

ε=θ

θid < θic → 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 RdRo = 1.0.

kN1462522x4.1x0.2VRR)design(V

kN812522x555.1VVMM)design(V

)FRAMES(f0df

)FRAMES(fw)FRAMES(ff

nf

==≤

==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=

tanSdfA

VanddbfV vyvssvw

'ccc

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

)11.3.6.3(a Clause 0.18 then ,mm60f

Sbf06.0mm200 AIf 2

yv

w'c

2v =β=>=

)21.6.9.6(b Clause 005.0 for 18.0 id ≤θ≤β

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


kN8.652N45tanx400

3840x400x200x85.0tanS

dfAV vyvs

s ==θ

φ=

o

kN11028.6523.449VVV scr =+=+=

OK0.1737.01102812

VVnUtilizatio

r

f →<===

S-FRAME Results


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: Lw = 3700mm, bw = 250mm, hw = 9500mm

WallquatS a Not0.296.337009500

Lh

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 Clause0025.000267.0300x250

100x2Sb

A2

w

bhv ≥===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 Clause0025.00025.0400x200

100x2Sb

A2

w

bhv ≥===ρ=ρ


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)

mm100200x5.0b5.0mm2713.11x24d24

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)

mm100200x5.0b5.0mm2713.11x24d24

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)

mm100200x5.0b5.0mm2713.11x24d24

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°):


Borderline0.1025.14.11091.1137

MMnUtilizatio

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 (+0.10Dny) + 1.0xD.

Nf = +477 kN, Mf = 1134 kNm Mn = 1365 kNm (nominal moment capacity from S-CONCRETE)

20.111341365

MM

f

nw ===γ (same as S-CONCRETE’s estimate)

Panel 1 Dimensions: Clause 21.7.3.1, Lu = 3500 – 200 = 3300 mm OKmm165

203300

20L

mm250b uw →==≥=

OKmm25752009500x25.0th25.0mm2400L 2w →=+=+≤= Panel 2 Dimensions: Clause 21.7.3.1, Lu = 3500 – 200 = 3300 mm OKmm165

203300

20L

mm200b uw →==≥=

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.


Ductility Evaluation: Clause 21.7.3.2, Rd = 2.0, R0 = 1.4, C = 310 mm from S-CONCRETE Check #1 C = 310 mm < 0.15Lw = 0.15x2400 = 380 mm → Not OK Alternative Check #2 OKmm27

3509500

350h mm 6.2andmm792L33.0C w

fw →==<=∆=<

Alternative Check #3 (Clause 21.6.7)

003.0003.00005.0

224009500

20.1x6.24.1x0.2x6.2

2Lh

RRid

ww

wf0dfid =θ∴≥=

−

−=

−

γ∆−∆=θ

025.00115.0002.0310x2

2400x0035.0002.0C2Lwcu

ic ≤=−=⎟⎟⎠

⎞⎜⎜⎝

⎛−

ε=θ

θid < θic → OK

Design Shear Force: kN494411x203.1VVMM)design(V )FRAMES(fw)FRAMES(f

f

nf ==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.


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: Lw = 24,000mm, bw = 200mm, hw = 9500mm

21.7.4) (Clause WallquatS0.2396.0000,24

9500Lh

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 Clause003.000333.0300x200

100x2Sb

A2

w

bhv ≥===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 Clause005.00133.0150x200

200x2Sb

A2

w

b ≥===ρ

10M Ties @ 200 mm (Clause 7.6.5.2)

Governsmm200b

mm5423.11x48d48mm25616x16d16S

w

tie

b

→=≤

==≤

==≤


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

OK0.111.0838609379

MMnUtilizatio

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.

00037.0112.0x00333.0MM Moment for Required Ratio Steel VerticalEstimated Let

r

fvm ==⋅ρ≈=ρ


Design Shear Force: 8.24.1x0.2RR4.10937997757

MM

0df

nw ==>===γ

∴γw = 2.8 Overstrength Factor Vf (design) = γw Vf (sframe) = 2.8 x 953 = 2,668 kN (Load Combination #4) Shear Design: Clause 21.7.4.8 dv = 0.8 Lw = 0.8 x 24,000 = 19,200 mm Vf ≤ 0.15 λ φc fc' bw dv = 0.15 x 1 x 0.65 x 25 x 200 x 19,200 N = 9360 kN β = 0 → Vc = 0

whv

whvysvyvs

s bS

A wherebtan

dftanS

dfAV ρ=ρ⋅

θ

φ=

θ

φ=

Vf ≤ Vr = Vc + Vs = Vs assume θ = 45°

kN4352N200x00333.0x45tan

200,19x400x85.0btan

dfVV wh

vyssr ==ρ⋅

θ

φ==

o

OK161.043522668nUtilizatio →<==

Let ρvs = vertical steel ratio required to resist shear Ag = 200 x 24,000 = 4,800,000 mm2 For load combination #4, Vf = 2668 kN, Ps = Nf = 1851 kN

OK00204.0200x200,19x400x85.0

45tanx000,668,2bdf

tanV)d'req(00333.0wvys

fhhsh →==

φθ

=ρ=ρ≥=ρo

00113.0000,800,4x400x85.0

000,851,145tan

00204.0Af

Ptan 2

gys

s2

hsvs =−=

φ−

θ

ρ=ρ

o


For another load combination, Vf = 2668 kN, Ps = Nf = 1448 kN

OK00204.0200x200,19x400x85.0

45tanx000,668,2bdf

tanV)d'req(00333.0wvys

fhhsh →==

φθ

=ρ=ρ≥=ρo

00115.0000,800,4x400x85.0

000,448,145tan

00204.0Af

Ptan 2

gys

s2

hsvs =−=

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):


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.


Documents

Seismic Design of Reinforced Concrete Structures CASE STUDY #1