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Chord capacity for chord2 Load Combination : 0.6DL - 0.7EQ Maximum tensile force in chord T=256.911 lbs Maximum applied tensile stress f t=T/Aen=25.533 lb/in2 Design tensile stress F t'=Ft×CD×CMt×Ctt×CFt×Ci=1380 lb/in2 Ratio25.53/1380=0.02 PASS - Design tensile stress exceeds maximum applied tensile stress Load Combination : DL + 0.7EQ Maximum compressive force in chord P=1002.158 lbs Maximum applied compressive stress f c=P/Ae=81.809 lb/in2 Design compressive stress F c'=Fc×CD×CMc×Ctc×CFc×Ci×Cp=596.16 lb/in2Ratio81.81/596.16=0.14 PASS - Design compressive stress exceeds maximum applied compressive stress Design bearing compr. stress, bottom plate For maximum compressive force F c-perp'=Fc-perp×CMc×Ctc×Cb×Ci=625 lb/in2 Ratio81.81/625=0.13 Pass - Design bearing compr. stress, bottom plate exceeds Maximum applied compressive stress Hold down force Chord1 T1=0.257 Kips Chord2 T2=0.257 Kips Wind load deflection Design shear force V δw=FWserv×WL/1000=0.533kips Deflection limit Δ w_allow=H×12/360= 0.267 in Induced unit shear v δW=(VδW×1000/(Co×∑bi))/Cub=38.526 lb/ft Anchor tension force Tδ = max(0 ,(vδW /1000)×H- 0.6×(D + Swt×H)×B/2000 - 0.6×min((DLT-ch1)/1000,(DLT- ch2)/1000) + 0.45×max(abs(W-ch1)/1000, abs(W-ch2)/1000))= 0 kips Chord compression force Cδ = max(0,(vδW/1000)×H+0.6×(DL+Swt×H)×B/2000 + 0.6×max((DLC-ch1)/1000,(DLC- ch2)/1000) + 0.45×max(abs(W-ch1)/1000, abs(W-ch2)/1000))= 1.314 kips Vertical elongation at anchor ∆T = Tδ×1000/Ka= 0 in Vertical compression at chord ∆C = (0.04 in)×Cδ×1000/(Ae×Fc -perp)= 0.007 in Total vertical deflection ∆a=(∆T+∆C)×(B/Beff)= 0.007 in Shear wall deflection δsww =2×(vδW /12000)×((H×12)^3)/(3×(E/1000)×Ae×∑bi×12)+(vδW /12000)×H×12/(Ga)+ H×12×∆a /(∑bi×12) = 0.023 inRatio0.023/0.27=0.09 PASS - Shear wall deflection is less than deflection limit Seismic deflection Page 58 of 290