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

Home Browse Search

Chord1 T1=0 Kips Chord2 T2=0 Kips Wind load deflection Design shear force Vδw=FWserv×W=0.5kips Deflection limit Δ w_allow=H×12/360= 0.075 in Induced unit shear vδW=VδW/B/Cub=36.23lb/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 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))= 0.33 kips Vertical elongation at anchor ∆T = Tδ×1000/Ka= 0 in Vertical compression at chord ∆C = 0.04×Cδ*1000/(Ae×Fc-perp)= 0.002 in Total vertical deflection ∆a=(∆T+∆C)×(B/Beff)= 0.002 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) = 0.005 in 0/0.08=0.07 PASS - Shear wall deflection is less than deflection limit Seismic deflection Design shear force V δs=Eq/1000= 1.06 kips Deflection limit Δ s_allow=0.025×H×12= 0.675 in Induced unit shear v δs=Vδs/B/Cub=76.81 lb/ft Anchor tension force Tδ = max(0,(v δs/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))= 0.02 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))= 0.34 kips Vertical elongation at anchor ∆T = Tδ×1000/Ka= 0.002 in Vertical compression at chord ∆C = 0.04×Cδ*1000/(Ae×Fc -perp)= 0.002 in Total vertical deflection ∆a=(∆T+∆C)×(B/Beff)= 0.004 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) = 0.011 in Amp. seis. deflection δsws = Cdδ*δswse /Ie = 0.044 in 0.04/0.68=0.07 PASS - Shear wall deflection is less than deflection limit Page 80 of 213