CSA Seismic Loads
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PROJECT SECTION 1 TITLE Canadian Seismic Design of Steel Structures - Case Stu DATE 08/15/22 FILE TIME 8:09 PM DESCRIPTION Determine the Seismic Force on Model 1 (Ductile Moment Resistance Frame) Using Quasi-Static Analysis INPUT Story height hx = 4.0 [m] Number of storey n = 4.0 Frame length L = 12.0 [m] Weight of each floor Wx = 1000.0 [kN] 25% Snow load Ws = 1000.0 [kN] Zone velocity ratio v = 0.2 NBCC Appendix C Acceleration seismic zo Za = 4.0 NBCC Appendix C Velocity seismic zone Zv = 4.0 NBCC Appendix C Seismic importance fact I = 1.0 NBCC 4.1.9.1.10 (Other Building) Foundation Factor F = 1.3 NBCC Table 4.1.9.1.C (Firm Soil) Force reduction factor R = 4.0 NBCC Table 4.1.9.1.B (Ductile MRP) Calibration factor U = 0.6 NBCC 4.1.9.1.4 CALCULATIONS Total building height h = (hx*n) = 16.0 [m] Total building weight W = (Wx*n)+Ws = 5000.0 [kN] Seismic response factor S = (1.5)/(T^0.5) = 1.8 NBCC Table 4.1.9.1.A (T>0.5) Period T = 0.085*(h^(3/4)) = 0.7 [s] NBCC 4.1.9.1.7b (Steel Frame) Base Shear Ve = (v*S*I*F*W) = 2364.7 [kN] NBCC 4.1.9.1.5 Lateral Seismic Force V = (Ve/R)*U = 354.7 [kN] NBCC 4.1.9.1.4 Extra load at roof Ft = (0.07*T*V) = 16.9 [kN] NBCC 4.1.9.1.13 (0.7s<T<3.6s) F and S requirement FS = F*S = 2.4 if(FS>3.0, "Not OK", "OK") = OK NBCC 4.1.9.1.11 (Za=Zv) Weight x Height Level 4 wh4 = (Wx+Ws)*(hx*4) = 32000.0 [kNm] Weight x Height Level 3 wh3 = Wx*(hx*3) = 12000.0 [kNm] Weight x Height Level 2 wh2 = Wx*(hx*2) = 8000.0 [kNm] Weight x Height Level 1 wh1 = Wx*(hx*1) = 4000.0 [kNm] RESULTS Lateral Force on level F4 = (V-Ft)*wh4/(wh1+wh2+wh3+wh4)+Ft= 209.9 [kN] NBCC 4.1.9.1.13 Lateral Force on level F3 = (V-Ft)*wh3/(wh1+wh2+wh3+wh4) = 72.4 [kN] Lateral Force on level F2 = (V-Ft)*wh2/(wh1+wh2+wh3+wh4) = 48.3 [kN] Lateral Force on level F1 = (V-Ft)*wh1/(wh1+wh2+wh3+wh4) = 24.1 [kN] Uniform load on roof U2 = (Wx+Ws)/L = 166.7 [kN/m] Uniform load on other l U1 = (Wx)/L = 83.3 [kN/m] SUMMARY Beam Section = W460x128 [SI] (From Cheops2000) Column Section = W760x185 [SI] (From Cheops2000)
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Transcript of CSA Seismic Loads
PROJECT SECTION 1TITLE Canadian Seismic Design of Steel Structures - Case Studies DATE 04/22/23
FILE TIME 5:58 AM
DESCRIPTION
Determine the Seismic Force on Model 1 (Ductile Moment Resistance Frame) Using Quasi-Static Analysis
INPUT
Story height hx = 4.0 [m]Number of storey n = 4.0Frame length L = 12.0 [m]Weight of each floor Wx = 1000.0 [kN]25% Snow load Ws = 1000.0 [kN]Zone velocity ratio v = 0.2 NBCC Appendix C Acceleration seismic zone Za = 4.0 NBCC Appendix C Velocity seismic zone Zv = 4.0 NBCC Appendix C Seismic importance factor I = 1.0 NBCC 4.1.9.1.10 (Other Building)Foundation Factor F = 1.3 NBCC Table 4.1.9.1.C (Firm Soil)Force reduction factor R = 4.0 NBCC Table 4.1.9.1.B (Ductile MRP)Calibration factor U = 0.6 NBCC 4.1.9.1.4
CALCULATIONS
Total building height h = (hx*n) = 16.0 [m]Total building weight W = (Wx*n)+Ws = 5000.0 [kN]Seismic response factor S = (1.5)/(T^0.5) = 1.8 NBCC Table 4.1.9.1.A (T>0.5)Period T = 0.085*(h^(3/4)) = 0.7 [s] NBCC 4.1.9.1.7b (Steel Frame)Base Shear Ve = (v*S*I*F*W) = 2364.7 [kN] NBCC 4.1.9.1.5Lateral Seismic Force V = (Ve/R)*U = 354.7 [kN] NBCC 4.1.9.1.4Extra load at roof Ft = (0.07*T*V) = 16.9 [kN] NBCC 4.1.9.1.13 (0.7s<T<3.6s)F and S requirement FS = F*S = 2.4
if(FS>3.0, "Not OK", "OK") = OK NBCC 4.1.9.1.11 (Za=Zv)
Weight x Height Level 4 wh4 = (Wx+Ws)*(hx*4) = 32000.0 [kNm]Weight x Height Level 3 wh3 = Wx*(hx*3) = 12000.0 [kNm]Weight x Height Level 2 wh2 = Wx*(hx*2) = 8000.0 [kNm]Weight x Height Level 1 wh1 = Wx*(hx*1) = 4000.0 [kNm]
RESULTS
Lateral Force on level 4 F4 = (V-Ft)*wh4/(wh1+wh2+wh3+wh4)+Ft = 209.9 [kN] NBCC 4.1.9.1.13Lateral Force on level 3 F3 = (V-Ft)*wh3/(wh1+wh2+wh3+wh4) = 72.4 [kN]Lateral Force on level 2 F2 = (V-Ft)*wh2/(wh1+wh2+wh3+wh4) = 48.3 [kN]Lateral Force on level 1 F1 = (V-Ft)*wh1/(wh1+wh2+wh3+wh4) = 24.1 [kN]
Uniform load on roof U2 = (Wx+Ws)/L = 166.7 [kN/m]Uniform load on other level U1 = (Wx)/L = 83.3 [kN/m]
SUMMARYBeam Section = W460x128 [SI] (From Cheops2000)Column Section = W760x185 [SI] (From Cheops2000)
PROJECT SECTION 2TITLE Canadian Seismic Design of Steel Structures - Case Studies DATE 04/22/23
FILE TIME 5:58 AM
DESCRIPTION
Determine the Seismic Force on Model 2 (Ductile Concentrically Braced Frame) Using Quasi-Static Analysis
INPUT
Story height hx = 4.0 [m]Number of storey n = 4.0Frame length L = 12.0 [m]Weight of each floor Wx = 1000.0 [kN]25% Snow load Ws = 1000.0 [kN]Zone velocity ratio v = 0.2 NBCC Appendix C Acceleration seismic zone Za = 4.0 NBCC Appendix C Velocity seismic zone Zv = 4.0 NBCC Appendix C Seismic importance factor I = 1.0 NBCC 4.1.9.1.10 (Other Building)Foundation Factor F = 1.3 NBCC Table 4.1.9.1.C (Firm Soil)Force reduction factor R = 3.0 NBCC Table 4.1.9.1.B (Ductile CBF)Calibration factor U = 0.6 NBCC 4.1.9.1.4
CALCULATIONS
Total building height h = (hx*n) = 16.0 [m]Total building weight W = (Wx*n)+Ws = 5000.0 [kN]Seismic response factor S = (1.5)/(T^0.5) = 1.8 NBCC Table 4.1.9.1.A (T>0.5)Period T = 0.085*(h^(3/4)) = 0.7 [s] NBCC 4.1.9.1.7b (Steel Frame)Base Shear Ve = (v*S*I*F*W) = 2364.7 [kN] NBCC 4.1.9.1.5Lateral Seismic Force V = (Ve/R)*U = 472.9 [kN] NBCC 4.1.9.1.4Extra load at roof Ft = (0.07*T*V) = 22.5 [kN] NBCC 4.1.9.1.13 (0.7s<T<3.6s)F and S requirement FS = F*S = 2.4
if(FS>3.0, "Not OK", "OK") = OK NBCC 4.1.9.1.11 (Za=Zv)
Weight x Height Level 4 wh4 = (Wx+Ws)*(hx*4) = 32000.0 [kNm]Weight x Height Level 3 wh3 = Wx*(hx*3) = 12000.0 [kNm]Weight x Height Level 2 wh2 = Wx*(hx*2) = 8000.0 [kNm]Weight x Height Level 1 wh1 = Wx*(hx*1) = 4000.0 [kNm]
RESULTS
Lateral Force on level 4 F4 = (V-Ft)*wh4/(wh1+wh2+wh3+wh4)+Ft = 279.9 [kN] NBCC 4.1.9.1.13Lateral Force on level 3 F3 = (V-Ft)*wh3/(wh1+wh2+wh3+wh4) = 96.5 [kN]Lateral Force on level 2 F2 = (V-Ft)*wh2/(wh1+wh2+wh3+wh4) = 64.3 [kN]Lateral Force on level 1 F1 = (V-Ft)*wh1/(wh1+wh2+wh3+wh4) = 32.2 [kN]
Uniform load on roof U2 = (Wx+Ws)/L = 166.7 [kN/m]Uniform load on other level U1 = (Wx)/L = 83.3 [kN/m]
SUMMARYBeam Section = W460x128 [SI] (From Cheops2000)Column Section = W460x128 [SI] (From Cheops2000)Brace Section = HS203x203x11 (From Cheops2000)
PROJECT SECTION 2TITLE Canadian Seismic Design of Steel Structures - Case Studies DATE 04/22/23
FILE TIME 5:58 AM
DESCRIPTION
Determine the Seismic Force on Model 2 (Ductile Concentrically Braced Frame) Using Quasi-Static Analysis
INPUT
Story height hx = 4.0 [m]Number of storey n = 4.0Frame length L = 12.0 [m]Weight of each floor Wx = 1000.0 [kN]25% Snow load Ws = 1000.0 [kN]Zone velocity ratio v = 0.2 NBCC Appendix C Acceleration seismic zone Za = 4.0 NBCC Appendix C Velocity seismic zone Zv = 4.0 NBCC Appendix C Seismic importance factor I = 1.0 NBCC 4.1.9.1.10 (Other Building)Foundation Factor F = 1.3 NBCC Table 4.1.9.1.C (Firm Soil)Force reduction factor R = 4.0 NBCC Table 4.1.9.1.B (Ductile CBF)Calibration factor U = 0.6 NBCC 4.1.9.1.4
CALCULATIONS
Total building height h = (hx*n) = 16.0 [m]Total building weight W = (Wx*n)+Ws = 5000.0 [kN]Seismic response factor S = (1.5)/(T^0.5) = 1.8 NBCC Table 4.1.9.1.A (T>0.5)Period T = 0.085*(h^(3/4)) = 0.7 [s] NBCC 4.1.9.1.7b (Steel Frame)Base Shear Ve = (v*S*I*F*W) = 2364.7 [kN] NBCC 4.1.9.1.5Lateral Seismic Force V = (Ve/R)*U = 354.7 [kN] NBCC 4.1.9.1.4Extra load at roof Ft = (0.07*T*V) = 16.9 [kN] NBCC 4.1.9.1.13 (0.7s<T<3.6s)F and S requirement FS = F*S = 2.4
if(FS>3.0, "Not OK", "OK") = OK NBCC 4.1.9.1.11 (Za=Zv)
Weight x Height Level 4 wh4 = (Wx+Ws)*(hx*4) = 32000.0 [kNm]Weight x Height Level 3 wh3 = Wx*(hx*3) = 12000.0 [kNm]Weight x Height Level 2 wh2 = Wx*(hx*2) = 8000.0 [kNm]Weight x Height Level 1 wh1 = Wx*(hx*1) = 4000.0 [kNm]
RESULTS
Lateral Force on level 4 F4 = (V-Ft)*wh4/(wh1+wh2+wh3+wh4)+Ft = 209.9 [kN] NBCC 4.1.9.1.13Lateral Force on level 3 F3 = (V-Ft)*wh3/(wh1+wh2+wh3+wh4) = 72.4 [kN]Lateral Force on level 2 F2 = (V-Ft)*wh2/(wh1+wh2+wh3+wh4) = 48.3 [kN]Lateral Force on level 1 F1 = (V-Ft)*wh1/(wh1+wh2+wh3+wh4) = 24.1 [kN]
Uniform load on roof U2 = (Wx+Ws)/L = 166.7 [kN/m]Uniform load on other level U1 = (Wx)/L = 83.3 [kN/m]
SUMMARYBeam Section = W460x128 [SI] (From Cheops2000)Column Section = W460x128 [SI] (From Cheops2000)Brace Section = HS203x203x11 (From Cheops2000)
