HISTORIC-606 S Van Ness Ave Unit #2 - Plan

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Maximum shear force under wind loading V w_max=0.6×W/1000 =0.38 Kips Shear capacity for wind loading V w=(Vn×B/2000) =1.18 Kips ratio V w-max / Vw =0.32 PASS - Shear capacity for wind load exceeds maximum shear force Maximum shear force under seismic loading V s_max=0.7×Eq/1000 =0.42 Kips Shear capacity for seismic loading V s=(Vn×B/(2.8×1000)) =0.84 Kips ratio V s_max / Vs =0.5 PASS - Shear capacity for seismic load exceeds maximum shear force Chord capacity for chord1 Load Combination : 0.6DL - 0.7EQ Maximum tensile force in chord T=387.2 lb Maximum applied tensile stress f t=T/Aen=31.61 lb/in2 Design tensile stress f t*=Ft×CD×CMt×Ctt×CFt×Ci=1380 lb/in2 31.61/1380=0.02 PASS - Design tensile stress exceeds maximum applied tensile stress Load Combination : DL + 0.7EQ Maximum compressive force in chord P=467.46 lb Maximum applied compressive stress f c=P/Ae=38.16 lb/in2 Design compressive stress f c*=Fc×CD×CMc×Ctc×CFc×Ci×Cp=2310.12 lb/in2 38.16/2310.12=0.02 PASS - Design compressive stress exceeds maximum applied tensile stress Chord capacity for chord2 Load Combination : 0.6DL - 0.7EQ Maximum tensile force in chord T=387.2 lb Maximum applied tensile stress f t=T/Aen=31.61 lb/in2 Design tensile stress ft*=Ft×CD×CMt×Ctt×CFt×Ci=1380 lb/in2 31.61/1380=0.02 PASS - Design tensile stress exceeds maximum applied tensile stress Load Combination : DL + 0.7EQ Maximum compressive force in chord P=467.46 lb Maximum applied compressive stress f c=P/Ae=38.16 lb/in2 Design compressive stress f c*=Fc×CD×CMc×Ctc×CFc×Ci×Cp=2310.12 lb/in2 38.16/2310.12=0.02 PASS - Design compressive stress exceeds maximum applied tensile stress bearing compr. stress, bottom plate For (maximum compressive force) fc-perp*=Fc-perp×CMc×Ctc×Cb×Ci=625 lb/in2 38.16/625=0.06 PASS - bearing compr. stress, bottom plate Hold down force Page 54 of 213