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

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Chord1 T1=0.39 Kips Chord2 T2=0.39 Kips Wind load deflection Design shear force V δw=FWserv×W=0.29kips Deflection limit Δ w_allow=H×12/360= 0.075 in Induced unit shear v δW=VδW/B/Cub=107.41lb/ft Anchor tension force Tδ = max(0 ,(v δW /1000)×H×B/Beff- 0.6×(D + Swt×H)×B/2000 - 0.6*min((DT-ch1)/1000,(DT- ch2)/1000) + 0.45×max(abs(W-ch1)/1000, abs(W-ch2)/1000))= 0.24 kips Chord compression force Cδ = max(0,(vδW/1000)×H×B/Beff+0.6×(D+Swt×H)×B/2000 + 0.6×max((Dc -ch1)/1000,(Dc- ch2)/1000) + 0.45×max(abs(W-ch1)/1000, abs(W-ch2)/1000))= 0.34 kips Vertical elongation at anchor ∆T = Tδ×1000/Ka= 0.024 in Vertical compression at chord ∆C = 0.04×Cδ*1000/(Ae×Fc -perp)= 0.002 in Total vertical deflection ∆a=(∆T+∆C)×(B/Beff)= 0.031 in Shear wall deflection δsww =2×(vδW/12000)×((H×12)^3)/(3×(E/1000)×Ae×B×12)+(vδW/12000)×H×12/(Ga)+ H×12×∆a /(B×12) = 0.041 in 0.04/0.08=0.55 PASS - Shear wall deflection is less than deflection limit Seismic deflection Design shear force V δs=Eq/1000= 0.61 kips Deflection limit Δ s_allow=0.025×H×12= 0.675 in Induced unit shear v δs=Vδs/B/Cub=225.93 lb/ft Anchor tension force Tδ = max(0,(vδs/1000)*H*B/Beff-(0.6-0.2*SDS)*(DL+ Swt*H)*B/2000 -(0.6-0.2*SDS )* min((DT-ch1)/1000, (DT-ch2)/1000)+max(abs(Eq-ch1)/1000, abs(Eq-ch2)/1000))= 0.58 kips Chord compression force Cδ = max(0,(vδs/1000)*H*B/Beff+(0.6- 0.2*SDS)*(DL+Swt*H)*B/2000 + (0.6-0.2*SDS )* max((Dc-ch1)/1000, (Dc-ch2)/1000)+max(abs(Eq-ch1)/1000, abs(Eq- ch2)/1000))= 0.64 kips Vertical elongation at anchor ∆T = Tδ×1000/Ka= 0.058 in Vertical compression at chord ∆C = 0.04×Cδ*1000/(Ae×Fc -perp)= 0.003 in Total vertical deflection ∆a=(∆T+∆C)×(B/Beff)= 0.073 in Shear wall elastic deflection δswse =2×(vδs/12000)×((H×12)^3)/(3×(E/1000)×Ae×B×12)+(vδss/12000)×H×12/(Ga)+ H×12×∆a /(B×12) = 0.093 in Amp. seis. deflection δsws = Cdδ*δswse /Ie = 0.372 in 0.37/0.68=0.55 PASS - Shear wall deflection is less than deflection limit Page 55 of 213