Seismic Desin of Reinforced Concrete Structures CASE STUDY #1 Three Storey Office Buildin Vancouver, B.C. Canada SOFTEK Services Ltd. #75 1500 Maycrest Way Richmond, BC, Canada V6V N8 Tel: (604) 7-777 Web: Copyriht Notice This document is copyriht 007 by Softek Services Ltd. All rihts reserved. No part of this publication may be reproduced, transmitted, transcribed, stored in a retrieval system, or translated into any human or computer lanuae, in any form or by any means, electronic, mechanical, optical, chemical, manual, or otherise, ithout the prior ritten 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 sinificance and limitations of its content and recommendations, and ho ill 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 includin computer softare referenced herein.
CASE STUDY #1 In this project, e ill desin various shear alls in a three storey office buildin located in a hih risk seismic zone usin SOFTEK s products: S-FRAME and S-CONCRETE TM. Key results enerated by S-FRAME and S-CONCRETE TM ill also be verified usin hand calculations. Buildin Description Desin Data (NBCC 005) Location: Loads: Commercial Buildin in Vancouver, BC, Canada (Granville & 41 st Ave), Site Soil Class C Retail on Ground Floor, Live Load 4.8 kpa Offices on Upper Levels, Live Load.4 kpa Partition Alloance 1.0 kpa Front Curtain Wall Weiht 1.0 kpa Ground Sno, S s 1.9 kpa and S r 0. kpa Seismic: S a (0.) 0.95, S a (0.5) 0.65, S a (1.0) 0.4, S a (.0) 0.17, PGA 0.47 Importance Factor I E 1.0 Force Modification Factors R d.0, R 0 1.4 (moderately ductile SFRS)
Materials: ' ' f 5 MPa, f 400 MPa, E 4500 f,500 MPa c y c c c W 400 k/m.5 kn/m Ec,500 Gc 91 MPa (1 +ν ) (1 + 0.) Sno Load for Roof Level, NBCC 005 Clause 4.1.6. S Is Ss( CbCCsCa) + Sr 1 1.9 ( 0.8 1.0 1.0 1.0) + 0. 1.8 kpa Buildin Seismic Weiht Estimation Buildin Seismic Weiht Estimation at Roof Level Item Description Weiht (kn) 1 Sno Load 0.5 x 1.8 kn/m x 1m x 4m (no hole) 11 Slab.5 kn/m x 0.m x 1m x 4m (no hole) 154 Front Curtain Wall 1.0 kn/m x 1.5m x 1m 18 4 Columns.5 kn/m x ( x 0.m x 0.m + 0.m x 0.4m) x 1.5m 1 5 Walls.5 kn/m x [1.5mx(0.mx64m + 0.5mx.4mx) x0.mx1.mx0.75m] Buildin Seismic Weiht Estimation at rd Floor 506 Sub-Total W 4,01 Item Description Weiht (kn) Area of Holes 4.8m x.4m +.4m x.4m 17.8 m Net Area 1m x 4m 17.8 m 70.7 m 1 Partition Load 0.5 kn/m x 70.7 m (max 0.5 kpa) 15 Slab.5 kn/m x 0.m x 70.7 m 17 Front Curtain Wall 1.0 kn/m x m x 1m 6 4 Columns.5 kn/m x ( x 0.m x 0.m + 0.m x 0.4m) x m 5 5 Walls.5 kn/m x [mx(0.mx64m + 0.5mx.4mx) - x0.x1.mx1.5m] 101 Buildin Seismic Weiht Estimation at nd Floor Sub-Total W,480 Item Description Weiht (kn) Area of Holes 4.8m x.4m +.4m x.4m 17.8 m Net Area 1m x 4m 17.8 m 70.7 m 1 Partition Load 0.5 kn/m x 70.7 m (max 0.5 kpa) 15 Slab.5 kn/m x 0.m x 70.7 m 17 Front Curtain Wall 1.0 kn/m x.5m x 1m 9 4 Columns.5 kn/m x ( x 0.m x 0.m + 0.m x 0.4m) x.5m 7 5 Walls.5 kn/m x [.5mx(0.mx64m + 0.5mx.4mx) x0.x1.mx1.5m] 1098 Sub-Total W,571 Total Weiht W W x W y W + W + W 4 01 + 480 + 571 707 kn
Buildin Period, NBCC 005 Clause 4.1.8.11() a ( ) ( ) T 0.05 h 0.05 9.5 0.7 sec same for each direction n 4 0.75 Accordin to Article 4.1.8.7(), Equivalent Static Force Procedure as described in Article 4.1.8.11 may be used for structures that meet the folloin criteria: reular structures that are less than 60m in heiht and have a fundamental lateral period, T a, less than sec in each of the orthoonal directions. For this buildin, it meets this criteria. Seismic Base Shear Calculation Sa(0.) For 5.6 < 8 and S (.0) a < 1.0, 1.0 Sa(0.) For 5.6 < 8 and Ta < 0.5, J 1.0 Sa (.0) No reduction in overturnin moment T a M v Table 4.1.8.11 Table 4.1.8.11 For Site C Class For Site C Class and and S (0.) 0.95, a S (1.0) 0.4, a F F a v 1.0 1.0 Table 4.1.8.4.B Table 4.1.8.4.C Minimum Lateral Earthquake Force S(Ta )M view V Article 4.1.8.11 R R S(.0)M view V R R d d 0 0 For R d 1.5, V S(0.)I R R S(T 0.s) F a S a (0.) 1.0x0.95 0.95 S(T 0.5s) F v S a (0.5) or F a S a (0.), hichever is smaller 0.65 S(T.0s) F v S a (.0) 0.17 Usin linear interpolation: S(T 0.7s) 0.88 d 0 E W V V 0.88x1.0x1.0x707 kn kn.0x1.4 x0.95x1.0x707 kn 1600 kn.0x1.4 0.17x1.0x1.0x707.0x1.4 49 kn Seismic Base Shear V 1600 kn 4
Centre of Mass Calculation Since most of the eiht is distributed uniformly around the buildin, e ill place the axes at the centre of the slab (see fiure belo). The only eihts that ill influence the location of the centre of mass ill be Walls #, #4, #5, #6 and the holes in the slab for the nd and rd floor. We ill assume there are no holes at the roof level. For the roof level: WiUi.5kN / m x1.5mx0.mx.7mx.6m + WiUi 99 U 0.m Wi 01.5kN / m x1.5mx0.5mx.4mx4.8x 94 + 05 99kNm V WiVi.5kN / m x1.5mx0.mx.7mx(.05)m +.5kN / m x1.5mx0.5mx.4mx( 1. 6 8.4m) 80 0 410kNm WiVi 410 0.m Wi 01 For practical purposes, the centre of mass equals to the centre of the slab ( U V 0 ). The influence of Wall #, #4, #5, #6 and the holes in the slab is minimal because the eiht of the slab and the other alls (#1, #, and #7) dominate the centre of mass for this buildin. 5
Centre of Riidity Calculation This hand calculation is based on the assumption that alls are stron in bendin in one direction and very eak or neliible in bendin in the other direction. Lateral stiffness (k) of each all element must also be estimated assumin some form of flexural behaviour. Special attention is iven to Wall # and Wall #6. The to alls ill likely be reinforced in such a manner that they ill deflect and behave as a sinle unit an L-Shaped all. This ill be reflected in the computations belo. For the purpose of this calculation, the folloin reference axes ill be used. Walls #1 and # (use h m for calculations) A 4m x 0.m 4.8 m I 0.m x (4m) / 1 0.4 m 4 If e apply a 100kN force at the top, the lateral deflection ill be approximately: Vh 1.Vh 100,000x000 1.x100,000x000 + + 0.00017 + 0.0081 0.008 mm E I G A 1 6 c c x,500x0.4x10 91x4.8x10 (assumin a pin-fixed end condition) 100 kn k y 1,087 kn/mm and k x 0 (no stiffness in eak direction) 0.008 mm Walls #4 and #5 (use h m for calculations) A.4m x 0.5m 0.6 m I 0.5m x (.4m) / 1 0.88 m 4 If e apply a 100kN force at the top, the lateral deflection ill be approximately: Vh 1.Vh 100,000x000 + E ci Gc A x,500x0.88x10 (assumin a pin-fixed end condition) 100 kn k 490 kn/mm and k y 0.04 mm x 1 1.x100,000x000 + 0.189 + 0.0651 0.04 mm 6 91x0.6x10 0 (no stiffness in eak direction) 6
Wall # and Wall #6 as an L-Shape (use h m for calculations) S-CONCRETE L-Shape Results If e apply a 100kN force at the top in the X-direction, the lateral deflection ill be approximately: 9 4 I 678.6x10 mm, A 190x10 mm y x x Vh E I c y 0.0589 + 0.00 0.089 mm 1.Vh 100,000x000 + G A x,500x678.6x10 100 kn k x 110 kn / mm 0.089 mm c 9 1.x100,000x000 + 6 91x1.9x10 If e apply a 100kN force at the top in the Y-direction, the lateral deflection ill be approximately: 9 4 I 1785.9x10 mm, A 190x10 mm x y y Vh E I c y 0.04 + 0.00 0.057 mm 1.Vh 100,000x000 + G A x,500x1785.9x10 100 kn k y 1898 kn / mm 0.057 mm c 9 1.x100,000x000 + 6 91x1.9x10 Wall #7 (use h m for calculations) This as modelled in S-FRAME usin quadrilateral elements and a riid diaphram on the roof (see fiure belo). A force of 100kN as applied at the roof and a lateral deflection of 0.01mm as obtained. F 100 kn k x 4700 kn/mm and k y 0.01 mm 0 (no stiffness in eak direction) 7
Element Xi (m) Yi (m) Kxi (kn/m) (x 10 ) Kyi (kn/m) (x 10 ) Xi Kyi (kn x 10 ) Yi Kxi (kn x 10 ) 1 0 --- --- 1,087 0 --- & 6 10. 9.8 1,10 1,898 19,60 10,976 1 --- --- 1,087 145,044 --- 4 N/A.6 490 --- --- 1,764 5 N/A 6.0 490 --- ---,940 7 N/A 4 4,700 --- --- 11,800 Totals 6,800 6,07 164,404 18,480 Xi Kyi 164,404 Distance to Centre of Riidity Xcr 6.0 m Kyi 6,07 Small difference beteen centre of mass and centre of riidity (e x 0.m) Yi Kxi 18,480 Distance to Centre of Riidity Ycr 18.9 m Kxi 6,800 Sinificant difference beteen centre of mass and centre of riidity (e y 6.89m) 8
Distribution of Base Shear, NBCC Clause 4.1.8.11(6) F t 0 because T a < 0.7sec Wxhx Fx (V Ft ) here V 1600 kn n W h 1 i i Level Heiht h x (m) Storey Weiht W x (kn) W x h x (knm) Lateral Force F x (kn) Storey Shear V x (kn) Roof 9.5,01 19,199.5 69 69 6.5,480 16,10 58 175.5,571 8,998.5 5 1600 7,07 44,18 1600 Desin Eccentricities, NBCC Clause 4.1.8.11(10) T x F x (e x ± 0.10 D nx ) F x (0. ± 0.1 x 1) F x (0. ± 1.) knm T x F x (e y ± 0.10 D ny ) F x (6.9 ± 0.1 x 4) F x (6.9 ±.4) knm Hand Calculations Versus D Modellin in S-FRAME For this project and for one load case, hand calculations ill be used to compute the distribution of lateral forces and torsional moments to all the lateral force resistin elements in this buildin for the upper most level only. A D model usin S-FRAME ill also be created and the results ill be compared to hand computed values. The D model in S-FRAME can also ive us an estimate of the torsional sensitivity, B, for this buildin, NBCC Clause 4.1.8.11(9). Numerous other load cases and load combinations ill also be enerated usin S-FRAME. D Beam Model in S-FRAME In this D model, the alls are modelled as beam elements ith a computed moment of inertia for stron axis bendin and zero for eak axis bendin. This approach should produce similar results as hand calculations because the same assumption is applied eak or non-existent in one direction and stron in the other direction. Shear areas are also provided to ive more accurate deflections for evaluation purposes. Riid members are provided to model the behavior at the ends of each all. Riid means a relatively hih moment of inertia. Accordin to Clause 4.1.8. of NBCC 005, structural modellin shall be representative of the manitude and spatial distribution of the mass of the buildin and of the stiffness of all elements of the SFRS. The model shall account for the effect of cracked sections in reinforced concrete and say effects arisin from the interaction of ravity loads ith the displaced confiuration of the structure (P-Delta). S-FRAME can perform eometric non-linear analysis (P-Delta). Furthermore, accordin to Clause 1..5..1 of CSA-A.-04, for the purpose of determinin forces in and deflections of the structure, reduced section properties shall be used. The effective property to be used as a fraction of the ross section property shall be as specified belo: 9
Couplin Beams A ve 0.15 A ; I e 0.4 I ithout diaonal reinforcement Ps Column I e α c I ; α c 0.5 + 0.6 1. 0 ' f A Ps Wall A xe α A ; I e α I ; α 0.6 + 1. 0 ' f A c c For alls, P s shall be determined at the base of the all. Preliminary calculations indicate an α value in the rane of 0.6 and α c value in the rane of 0.65 hich ill be confirmed later. Element Gross Properties Effective Section Properties Wall #1 & # Walls # & #6 Walls #4 & #5 Column 00x00 00x4,000 14 I.04x10 mm 1 A 00x4,000 4,800,000 mm 5 A v x00x4,000 4,000,000 mm 6 I 678.6x10 y x 9 I 1785.9x10 A 190x10 9 mm 4 mm mm 5 A v x190x10 1075x10 6 50x400 11 I.88x10 mm 1 A 50x400 600,000 mm 4 mm 5 A v x50x400 500,000 mm 6 4 00 8 I x Iy 6.75x10 mm 1 A 00x00 90,000 mm 4 4 4 I e 0.6 x.04 x 10 14 1.48 x 10 14 mm 4 A e 0.6 x 4,800,000,976,000 mm A ev 0.6 x 4,000,000,480,000 mm I ey 0.6 x 678.6 x 10 9 4.1 x 10 11 mm 4 I ex 0.6 x 1785.9 x 10 9 11.07 x 10 11 mm 4 A e 0.6 x 190 x 10 799,800 mm A ev 0.6 x 1075 x 10 666,500 mm I e 0.6 x.88 x 10 11 1.786 x 10 11 mm 4 A e 0.6 x 600,000 7,000 mm A ev 0.6 x 500,000 10,000 mm I e 0.65 x 6.75 x 10 8 4.87 x 10 8 mm 4 A e A 90,000 mm Column 00x400 00x400 I x 1.066x10 1 9 mm 400x00 8 I y.67x10 mm 1 A 00x400 80,000 mm 4 4 I ex 0.65 x 1.066 x 10 9 6.99 x 10 8 mm 4 I ey 0.65 x.67 x 10 8 1.75 x 10 8 mm 4 A e A 80,000 mm 10
Wall #7 is modelled as a frame ith riid members that connect the end faces of each pier to the ends of couplin beams. The couplin beams represent the sements at the openins. This is illustrated belo in elevation. Element Gross Properties Effective Section Properties Wall #7a Wall #7b Couplin Beam B1 Couplin Beam B Couplin Beam B 00x400 11 I.04x10 mm 1 A 00x400 480,000 mm 5 A v x00x400 400,000 mm 6 00x4800 I 1.84x10 1 A 00x4800 960,000 mm 1 4 mm 5 A v x00x4800 800,000 mm 6 00x700 9 I 5.717x10 mm 1 A 00x700 140,000 mm 00x1500 10 I 5.65x10 mm 1 A 00x1500 00,000 mm 00x000 I 1.x10 1 A 00x000 400,000 mm 10 4 4 mm 4 4 I e 0.6 x.04 x 10 11 1.48 x 10 11 mm 4 A e 0.6 x 480,000 97,600 mm A ev 0.6 x 400,000 48,000 mm I e 0.6 x 1.84 x 10 1 1.148x 10 1 mm 4 A e 0.6 x 960,000 595,00 mm A ev 0.6 x 800,000 496,000 mm I e 0.4 x 5.717 x 10 9.87x 10 9 mm 4 A e 0.15 x 140,000 1,000 mm I e 0.4 x 5.65 x 10 10.5x 10 10 mm 4 A e 0.15 x 00,000 45,000 mm I e 0.4 x 1. x 10 10 5.x 10 10 mm 4 A e 0.15 x 400,000 60,000 mm 11
S-FRAME Model usin Beam Type Members Only and Riid Diaphrams The S-FRAME D model of the office buildin shon here consists only of beam type members ith riid diaphrams specified for each floor level. Only the nd floor diaphram is displayed above. Special attention is iven to Walls # and #6. Walls # and #6 is modelled as one column hich ill be subjected to biaxial bendin. The properties of this column is iven the section properties of the L-Shape (i.e. I x and I y ). Note that to minimize the amount of torsion that ill be attracted to each all, the torsional constants, J, for each all ere assined neliible values. 1
S-FRAME Model Center of Riidity Evaluation To assess the accuracy of C of R calculation, e ill apply 1000 kn force at each level in the X-direction at the computed C of R. In theory, loadin the buildin at the C of R ill enerate deflections ithout rotation pure translation. The results are displayed belo (Xdeflections in mm). As you can see above, the buildin is rotatin in a clockise direction. This most likely means that e have underestimated the stiffness of the L-Shape (Walls # & #6). Usin a trial-anderror approach in S-FRAME, e discovered the true center of riidity near e y 5.5m for this buildin (as indicated belo). 1
Torsional Sensitivity Evaluation The next step is to use this D model in S-FRAME to ive us an estimate of the torsional sensitivity, B, for this buildin. Here, the primary concern ould be loadin in the X-direction creatin a tist in the buildin. Accordin to NBCC Clause 4.1.8.11(9), the equivalent static forces, F x, shall be applied at distances of ±0.10D ny ±.4m from the center of mass at each floor level. The critical load case for this evaluation ould be applyin the forces at a distance of +.4m aay from the C of M. This is implemented in S-FRAME by applyin the equivalent static forces, F x, at the center of mass at each level plus a torsional moment of F x x.4m in the appropriate direction. Torsional Sensitivity: B x δ δ max av Level Corner Deflections Total # of δ max (mm) Corners (mm) (mm) B x Roof x.85 + x.1 9.96 4.85.49 1.14 rd x1.8 + x1.7 6.0 4 1.8 1.55 1.18 nd x0.794+x0.475.54 4 0.794 0.64 1.5 Base on the results above, B 1.5 for this buildin. Accordin to NBCC Clause 4.1.8.11(10), for a buildin ith B 1.7, torsional effects can be accounted for by applyin equivalent static forces, F x, to the buildin located at ±0.10D nx and ±0.10D ny from the C of M for each principle direction. Technically, e should also evaluate the torsional sensitivity for loadin in the y-direction (N-S direction). Since lare alls (Wall #1 and #) dominate the riidity in the y-direction, it is unlikely that the torsional sensitivity parameter, B, for loadin in this direction ill be reater than that computed above. δ av 14
Distribution of Lateral Force to Walls For hand calculations in tabular form, e ill consider only one load case (E-W direction) and applied to the roof only. The results of these hand calculations ill then be compared to the results enerated by S-FRAME for the alls in the top floor. To complete the desin of this buildin, other load cases ill be enerated in S-FRAME includin loadin in the N-S direction, dead loads, and factored load combinations. V x F x 69 kn, V y 0 kn T V x (e y + 0.10D ny ) 69 x (6.9 + 0.10 x 4) 6445 knm Note: T V x (e y - 0.10D ny ) 69 x (6.9-0.10 x 4) 119 knm done in S-FRAME only 15
Distribution of E-W Lateral Force and Torsional Moment to Walls V x F x 69 kn, V y 0 kn T -V x (e y + 0.10D ny ) -69 x (6.9 + 0.10 x 4) -6445 knm Force and Torsional Moment Applied at Roof Level Wall X i (m) Y i (m) K xi (kn/m) (x 10 ) K yi (kn/m) (x 10 ) K xi K xi (kn) V x y k i J r xi (kn) T V xi (kn) #1-6. --- 0 1,087 0 0 0 0-74 -74 #/6.9-9.1 1,10 1,898 114 50 164 0 6 6 K yi K yi (kn) V y x k # 5.7 --- 0 1,087 0 0 0 0 8 8 #4 --- -15. 490 0 50 7 87 0 0 0 #5 --- -1.9 490 0 50 1 81 0 0 0 #7 --- 5.1 4700 0 479-118 61 0 0 0 6800 6,07 69 0 69 0 0 0 i J r yi (kn) T V yi (kn) J J J r r r X K + Y yi [ 6. 1,087 +.9 1,898 + 9.1 110 + 5.7 1,087 + (15. + 1.9 ) 490 + 5.1 4700] 1.1x10 i 9 i knm K xi x10 knm 16
Comparison of Hand Calculations Versus S-FRAME D Results Wall Hand Calculations S-FRAME Comments #1-74 kn -47 kn less stiff in S-FRAME # & #6 6 kn / 164 kn* 8 kn / kn* more stiff in S-FRAME # 8 kn 19 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 61 kn 68 + 187 + 71 6 kn less stiff in S-FRAME * Shear in the eak direction (V y kn) for the L-Shape (Walls # & #6) is not displayed in the above plot but can be obtained easily in a plot for y Shear. Overall, hand calculated results ive similar values to S-FRAME. Reasonable numbers ere obtained usin simple assumptions on flexural behaviour hich otherise ould be considered rather complex in the D orld. The key to structural desin is to develop a complete load path, determine the sectional forces from this load path, and reinforce the members appropriately. This has been accomplished usin both hand calculations and in S-FRAME. To complete the desin of this buildin, other load cases and load combinations ill be enerated usin S-FRAME includin earthquake loadin E-W (-0.10D ny ), earthquake loadin N- S (±0.10D nx ), and dead load. 17
Dead Load Estimation (at base of the all) Wall #1 Tributary Area 0.75 x 6m x 4m 54.0 m Slab 0.m x 54.0 m x.5 kn/m 5.8 kn at each level Partitions 54.0 m x 1 kn/m 54.0 kn at each level (except roof) Self Weiht 9.5m x 0.m x 4m x.5 kn/m 1,071.6 kn Total at base x 5.8 + x 54.0 + 1071.6 1,941 kn Walls # & #6 Tributary Area 5m x 5m 5 m Slab 0.m x 5 m x.5 kn/m 117.5 kn at each level Partitions 5 m x 1 kn/m 5 kn at each level (except roof) Self Weiht 9.5m x 0.5m x.4m x.5 kn/m 14 kn Total at base x 117.5 + x 5 + 14 57 kn Wall #7b Tributary Area 0.8 x m x 4.8m 11.5 m Slab 0.m x 11.5 m x.5 kn/m 54.0 kn at each level Partitions 11.5 m x 1 kn/m 11.5 kn at each level (except roof) Self Weiht 9.5m x 0.m x 4.8m x.5 kn/m 14 kn Total at base x 54.0 + x 11.5 + 14 99 kn Hand calculations for dead load at the base of each all are similar to the results enerated by S- FRAME. Since S-FRAME is relatively more accurate than the hand computed values, e ill use S-FRAME results to evaluate the effective section properties as outlined in Clause 1..5..1 of CSA-A.-04. 18
Wall #1 Effective Section Properties P s 1919 kn, f c ' 5 MPa, A 00 x 4,000 4,800,000 mm Ps 1,919,000 α 0.6 + 0.6 + 0.6 ' f A 5x4,800,000 c A xe 0.6 x A,976,000 mm (effective cross-sectional area) A ve 0.6 x 5/6 x A,480,000 mm (effective shear area) I e 0.6 x I 0.6 x.04x10 14 1.48x10 14 mm 4 Walls # & #6 Effective Section Properties P s 490 kn, f c ' 5 MPa, A 1,90,000 mm Ps 490,000 α 0.6 + 0.6 + 0.6 ' f A 5x1,90,000 c A xe 0.6 x A 799,800 mm (effective cross-sectional area) A ve 0.6 x 5/6 x A 666,500 mm (effective shear area) I ey 0.6 x I y 0.6 x 678.6 x10 9 4.1x10 11 mm 4 I ex 0.6 x I x 0.6 x 1785.9 x10 9 11.1x10 11 mm 4 Wall #7b Effective Section Properties P s 440 kn, f c ' 5 MPa, A 00 x 4,800 960,000 mm Ps 440,000 α 0.6 + 0.6 + 0.6 ' f A 5x960,000 c A xe 0.6 x A 595,00 mm (effective cross-sectional area) A ve 0.6 x 5/6 x A 496,000 mm (effective shear area) I e 0.6 x I 0.6 x 1.84x10 1 1.14x10 1 mm 4 For practical purposes, all the alls in this buildin appear to have an effective moment of inertia of 0.6 x I and effective cross-sectional area of 0.6 x A. This as 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 desin of these alls. 19
Desin Load Combinations S-FRAME ill be used to enerate the load cases and load combinations for desin purposes. The folloin load combinations ill be created for the desin of Wall #1, # & #6, and #7b hich 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 #: 1.0 x E-W (-0.10Dny) + 1.0 x D Load Combination #: 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 desin 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 ith Live and Sno Loads may easily be added to the above load combinations but, in this case, they ill not likely overn the desin of this buildin. The primary purpose here is to illustrate the use and application of S-FRAME and S-CONCRETE in the analysis and desin of this office buildin. Desin Sectional Forces at base for Wall #7b (enerated by S-FRAME) Load Combination #1: Load Combination #: Load Combination #5: Load Combination #6: N f -44 kn, V f 41 kn, M f 118 knm, f 1.5 mm N f -47 kn, V f 47 kn, M f 41 knm, f 1.6 mm N f -44 kn, V f 46 kn, M f 478 knm, f 1.8 mm N f -45 kn, V f 5 kn, M f 700 knm, f 1.9 mm Note: The shear forces displayed here must be manified for desin purposes. Accordin to Clause 1.7..4.1 of CSA-A.-04, the desin shear force or resistance must not be less than the smaller of: (1) the shear force correspondin to the development of the nominal moment capacity of the all at its plastic hine location and () shear force at R d R o 1.0. S-CONCRETE can make this estimation. Desin Sectional Forces at base for Walls # & #6 (enerated by S-FRAME) LC #1: N f +477 kn, V fy 411 kn, M fz +114 knm, fy. mm V fz 5 kn, M fy +84 knm, fz 0.1 mm LC #: N f +48 kn, V fy 88 kn, M fz +106 knm, fy.0 mm V fz 0 kn, M fy -65 knm, fz 0.0 mm LC #5: N f -1458 kn, V fy 7 kn, M fz -109 knm, fy.6 mm V fz 64 kn, M fy +1 knm, fz 0.1 mm LC #6: N f -195 kn, V fy 50 kn, M fz -117 knm, fy.5 mm V fz 41 kn, M fy +85 knm, fz 0.1 mm Note: This all may experience small tension forces accordin to S-FRAME results. This is reasonable because Wall # ill be carryin a sinificant amount of shear force due to the torsional moment hich, in term, ill tend to lift Walls #6 and #. 0
Desin Sectional Forces at base for Wall #1 (enerated by S-FRAME) Load Combination #1: Load Combination #: Load Combination #: Load Combination #4: Load Combination #5: Load Combination #6: Load Combination #7: Load Combination #8: N f -1641 kn, V f 676 kn, M f 980 knm, f 0.19 mm N f -1580 kn, V f 195 kn, M f 6448 knm, f 0.08 mm N f -188 kn, V f 71 kn, M f 96 knm, f 0.16 mm N f -1851 kn, V f 95 kn, M f 549 knm, f 0. mm N f -196 kn, V f 80 kn, M f 8855 knm, f 0. mm N f -57 kn, V f kn, M f 59 knm, f 0.11 mm N f -1955 kn, V f 585 kn, M f 4488 knm, f 0.1 mm N f -1986 kn, V f 86 kn, M f 595 knm, f 0.19 mm Note: Here, the larest moment is enerated from a load combination ith a sinificant torsional moment (#1) hich is interestin. The larest shear force is enerated from a load combination that applies the lateral loads in the stron direction for this all (#4) hich is as expected. S-FRAME results (i.e. axial force, shear force, and moment diarams) can be directly exported to S-CONCRETE to complete the desin. This is illustrated belo for Wall #7b, Wall # & #6, and Wall #1. Hand calculations ill also be performed to verify the results of S-CONCRETE. 1
Wall #7b Desin and Detailin S-FRAME ill export sectional forces and moments evaluated at various stations alon the member. In this case, it has evaluated sectional forces at three stations per member sement per load combination. For this member in the D model, it has been subdivided into to sements on the first floor.
Let s assume that minimum distributed reinforcin and zone reinforcin ill be sufficient to meet all the requirements of CSA-A.-04. We ill desin the base of the all (i.e. plastic hine reion). Wall Dimensions: Panel Reinforcin: Zone Reinforcin: L 4800mm, b 00mm, h 9500mm h 9500 1.98.0 Squat Wall (technically) L 4800 Hoever, for practical desin purposes, e ill treat it as a normal all. To force S-CONCRETE to not apply the squat all provisions, e assined a value of 9601mm to h. Vertical Bars - 10M @ 400 Each Face ( curtains) Horz Bars 10M @ 400 Each Face ( curtains) A b x100 ρ v ρh 0.005 0.005 Clause 1.7...1 b S 00x400 4 15M bars at each end of the all (minimum requirement) 10M Ties @ 95mm (Clause 1.7... and 1.6.6.9) S 6d 6x16 96 mm Governs b 4d 0.5b tie 4x11. 71mm 0.5x00 100 mm Axial Load and Moment Capacity: Mf 700 Utilization 0.74 1.0 M 64 r OK S-CONCRETE results and interaction diaram shon belo:
Overstrenth Factor: M n 4198 knm (nominal moment capacity from S-CONCRETE) M f 700 knm (from S-FRAME) Mn 4198 γ 1.555 (same as S-CONCRETE s estimate) M 700 f Dimensional Limitations: Clause 1.7..1, L u 500 00 00 mm b 00 mm Lu 00 5 mm 14 14 Lu 00 165 mm 0 0 Not Good OK Hoever, accordin to Clause 1.6..4, the L u /14 requirement may be aived if the neutral axis depth does not exceed 4b or 0.L (i.e. C 800 mm) hich is the case here. S-CONCRETE ill compute the neutral axis depths for load combination here flexure is dominant and determine if the all meets these requirements for dimensions and ductility. This is displayed belo in the Results Report indo of S-CONCRETE. 4
Ductility Evaluation: Clause 1.7.., R d.0, R 0 1.4 Check #1 C 456 mm < 0.15L 0.15x4800 70 mm OK Alternative Check # C < 0.L 1584 mm and f h 9500 1.9 mm < 7 mm 50 50 OK Alternative Check # (Clause 1.6.7) frdr0 f γ 1.9x.0x1.4 1.9x1.55 θid 0.000 0.00 θid 0.00 L 4800 h 9500 ε cu L 0.005x4800 θ ic 0.00 0.00 0.0164 0.05 C x456 θ id < θ ic OK All ductility checks indicate that special concrete confinement requirements ill not be required. S-CONCRETE has the capability to evaluate special concrete confinement requirements as outlined in Clause 1.6.7.4 for zone reinforcin. Desin Shear Force: Accordin to Clause 1.7..4.1 of CSA-A.-04, the desin shear force or resistance must not be less than the smaller of: (1) the shear force correspondin to the development of the nominal moment capacity of the all at its plastic hine location and () shear force at R d R o 1.0. V V f f Mn (desin) Vf M f (desin) R R V d 0 (S FRAME) f (S FRAME) γ V f (S FRAME).0 x 1.4 x 5 146 kn 1.555 x 5 81 kn S-CONCRETE has the option to perform Shear Force Manification in the manner described above to determine the desin shear forces. Shear Resistance: Shear Desin is based on Clauses 1.6.9. to 1.6.9.7 (simplified method) Panel Reinforcin 10M @ 400 H.E.F. V 81 kn V V f r c V c λβφ o + V c s f V ' c b r max d v 0.15λφ and V ;' c c s f b d v for θ φsa v fyd S tan θ v id 0.005 θ 45 Clause 1.7..4.(c), dv 0.8L 840 mm ' b S If A v 00mm > 0.06 fc 60mm, then β 0.18 Clause 11..6.(a) f β 0.18 for θid 0.005 Clause 1.6.9.6(b) yv ' Vc λβφ c fc b dv 1 x 0.18 x 0.65 x 5 x 00 x840 N 449. kn 5
V V φsa v fydv 0.85x00x400x840 N S tan θ o 400x tan 45 V + V 449. + 65.8 110 kn s r c s Vf Utilization V r 81 110 S-FRAME Results 0.77 < 1.0 65.8 kn OK 6
Walls # & #6 Desin and Detailin Let s assume that minimum distributed reinforcin and zone reinforcin ill be sufficient to meet all the requirements of CSA-A.-04. We ill desin the base of the all (i.e. plastic hine reion). Wall Dimensions: L 700mm, b 50mm, h 9500mm h 9500.96 >.0 Not a Squat Wall L 700 Panel 1 Reinforcin: Vertical Bars - 10M @ 00 Each Face ( curtains) Horz Bars 10M @ 00 Each Face ( curtains) A b x100 ρ v ρh 0.0067 0.005 Clause 1.7...1 b S 50x00 Panel Reinforcin: Vertical Bars - 10M @ 400 Each Face ( curtains) Horz Bars 10M @ 400 Each Face ( curtains) A b x100 ρ v ρh 0.005 0.005 Clause 1.7...1 b S 00x400 7
Zone A Reinforcin: 4 10M bars at each end of the all (minimum requirement) 10M Ties @ 65mm (Clause 1.7... and 1.6.6.9) S 6d b 4d 0.5b 6x11. 68 mm tie 4x11. 71mm 0.5x00 100 mm Governs Zone B Reinforcin: 4 15M bars at each end of the all (minimum requirement) 10M Ties @ 95mm (Clause 1.7... and 1.6.6.9) S 6d b 4d 0.5b 6x16 96 mm tie 4x11. 71mm 0.5x00 100 mm Governs Zone C Reinforcin: 4 10M bars at each end of the all (minimum requirement) 10M Ties @ 65mm (Clause 1.7... and 1.6.6.9) S 6d b 4d 0.5b 6x11. 68 mm tie 4x11. 71mm 0.5x00 100 mm Governs Note: Emphasis as placed on minimizin the amount of vertical bars in the section includin both zone steel and distributed reinforcin. This ill reduce the axial load and moment capacity hich increases the N vs M utilization. This, in turn, ill reduce the desin or manified shear forces because it ill enerate a smaller overstrenth factor. Axial Load and Moment Interaction Diaram (Biaxial Bendin, Theta 94 ): 8
M Utilization M f r 117.1 1.05 1.0 1109.4 Borderline Overstrenth Factor for bendin about z-z axis (Theta 90 ): S-CONCRETE has determined the overnin load combination for pure unixaxial bendin about the z-z axis is LC #1 hich is 1.0xE-W (+0.10D ny ) + 1.0xD. N f +477 kn, M f 114 knm M n 165 knm (nominal moment capacity from S-CONCRETE) Mn 165 γ 1.0 (same as S-CONCRETE s estimate) M 114 f Panel 1 Dimensions: Clause 1.7..1, L u 500 00 00 mm Lu 00 b 50 mm 165 mm OK 0 0 L 400mm 0.5h + t 0.5x9500 + 00 575mm OK Panel Dimensions: Clause 1.7..1, L u 500 00 00 mm Lu 00 b 00 mm 165 mm OK 0 0 L 700mm > 0.5h + t 0.5x9500 + 50 65mm NG Accordin to Clause 1.7..1, the flane idth of Panel is too lon. This means that part of Panel is ineffective in the overall axial load and moment capacity of the section for bendin about the z-z axis. Technically, e should shorten the lenth of the panel hich is unlikely. Evaluatin the nominal moment capacity in this direction usin the full lenth ill ive a conservative estimate on the required desin shear force (i.e. hiher overstrenth factor). The Warnin can be inored in this case. 9
Ductility Evaluation: Clause 1.7.., R d.0, R 0 1.4, C 10 mm from S-CONCRETE Check #1 C 10 mm < 0.15L 0.15x400 80 mm Not OK Alternative Check # C < 0.L 79 mm and f h 9500.6 mm < 7 mm 50 50 OK Alternative Check # (Clause 1.6.7) frdr0 f γ.6x.0x1.4.6x1.0 θid 0.0005 0.00 θid 0.00 L 400 h 9500 ε cu L 0.005x400 θ ic 0.00 0.00 0.0115 0.05 C x10 θ id < θ ic OK Desin Shear Force: Mn f (desin) Vf (S FRAME) γ Vf (S FRAME 1.0 x 411 M f (desin) R R V.0 x 1.4 x 411 1151kN V ) V f d 0 f (S FRAME) 494 kn S-CONCRETE has the option to perform Shear Force Manification in the manner described above to determine the desin shear forces. Shear Resistance: S-CONCRETE results For this all, the section may be subjected to tension forces. Here, the General Method of Shear Desin must be used to evaluate the shear resistance. 0
Wall #1 Desin and Detailin (Sq uat Wall) Let s assume that minimum distributed reinforcin and zone reinforcin ill be sufficient to meet all the requirements of CSA-A.-04. We ill desin the base of the all. Wall Dimensions: L 4,000mm, b 00mm, h 9500mm h 9500 0.96 <.0 Squat Wall L 4,000 (Clause 1.7.4) Panel Reinforcin: Zone Reinforcin: Vertical Bars - 10M @ 00 Each Face ( curtains) Horz Bars 10M @ 00 Each Face ( curtains) A b x100 ρ v ρh 0.00 0.00 Clause 1.7.4.5(a) b S 00x00 Maximum Bar Spacin 00mm Clause 1.7.4.5(a) Accordin to Clause 1.7.4.6, tied vertical reinforcement shall be provided at each end of the all. The minimum reinforcement ratio of 0.005 shall be provided over a minimum all lenth of 00mm. A minimum of four bars shall be provided and tied as a column in accordance ith Clause 7.6. The ties shall be detailed as hoops. 6 15M bars at each end of the all (spaced at 150 mm apart) A b x00 ρ 0.01 0.005 Clause 1.7.4.5(a) b S 00x150 10M Ties @ 00 mm (Clause 7.6.5.) S 16d 16x16 56 mm 48dtie 48x11. 54 mm b 00 mm Governs b 1
Axial Load and Moment Capacity: (S-CONCRETE Results) M Utilization M f r 979 8860 0.11 < 1.0 OK Accordin to Clause 1.7.4.7, the vertical tension force required to resist overturnin at the base of the all shall be provided by zone reinforcin and panel reinforcin in addition to the amount required by Clause 1.7.4.8 to resist the shear correspondin to the applied bendin moment. Mf Let ρ m Estimated Vertical Steel Ratio Required for Moment ρv 0.00 x 0.11 0.0007 M r
Mn 97757 Desin Shear Force: γ 10.4 > RdR0.0 x 1.4. 8 M 979 f γ.8 Overstrenth Factor V f (desin) γ V f (sframe).8 x 95,668 kn (Load Combination #4) Shear Desin: Clause 1.7.4.8 d v 0.8 L 0.8 x 4,000 19,00 mm V f 0.15 λ φ c f c ' b d v 0.15 x 1 x 0.65 x 5 x 00 x 19,00 N 960 kn β 0 V c 0 φsa vfydv φsfydv A v Vs ρhb here ρhb S tanθ tanθ S V f V r V c + V s V s assume θ 45 φsfydv 0.85x400x19,00 Vr Vs ρhb x0.00x00 N 45 kn tanθ o tan 45 668 Utilization 0.61 < 1 OK 45 Let ρ vs vertical steel ratio required to resist shear A 00 x 4,000 4,800,000 mm For load combination #4, V f 668 kn, P s N f 1851 kn ρ ρ h 0.00 ρ ρ tan hs P Vf tanθ,668,000x tan 45 ρh(req' d) 0.0004 OK φ f d b 0.85x400x19,00x00 s y v 0.0004 1,851,000 o tan 45 0.85x400x4,800,000 hs s vs θ φsfya o 0.0011
For another load combination, V f 668 kn, P s N f 1448 kn ρ ρ h 0.00 ρ ρ tan Total Vertical Steel Ratio Required: hs P Vf tanθ,668,000x tan 45 ρh(req' d) 0.0004 OK φ f d b 0.85x400x19,00x00 s y v 0.0004 1,448,000 o tan 45 0.85x400x4,800,000 hs s vs θ φsfya o 0.00115 ρ v 0.00 ρ v (required) ρm + ρvs 0.0007 + 0.00115 0.0015 OK Accordin to Clause 1.7.4.7, all vertical reinforcement required at the base of the all shall be extended the full heiht of the all. S-FRAME Results (Panel and Zone Reinforcin): 4
Conclusions When desinin alls that intersect ith other alls (Wall #6), e have nelected the influence of Wall # on Wall #6. A portion of Wall # should be included in the calculation for moment capacity hich, in turn, ill likely increase the desin shear force. Overall, nelectin the intersection of Wall #6 ith Wall # ill not chane the reinforcin confiuration very much if at all. Hoever, as alays, careful consideration of all the parameters should be iven nevertheless. Some enineers may have considered a different approach to the desin of Wall #7. In our model, e have assumed a coupled all system hich may be inappropriate for such a short all. In fact, a finite element model of the same buildin appears to contradict the sectional forces produced by this beam model version. For more information on the finite element model, refer to Case Study #. The finite element model suests that beam theory of plane sections remainin plane does not apply to Wall #7 and Wall #1. In Case Study #, you ill find sinificant differences in the sectional forces enerated for each all. This suests that Wall #7 should be desined as a squat all and vie the indo openins as havin little influence on the overall behaviour of the all. Hand calculations may ive you reasonable desin values for the lateral load resistin elements in a iven buildin but D modellin ill ive you a better representation of the overall performance of the buildin provided the model truly represents its behaviour in an earthquake. The key to any desin is to ensure that a load path has been defined and carried throuh to all the lateral load and ravity load resistin elements in the buildin. Minimizin the tist in the buildin and detailin the members carefully ill help ensure that the loads reach the beams, columns and alls as desined. 5