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Project <br />1011 <br />Job Ref. <br />Section <br />LINE-A <br />Sheet no./rev. <br /> 3 <br />Calc. by <br />N.K <br />Date <br />15/11/2024 <br />Chk'd by Date App'd by Date <br />Wet service factor for tension – Table 4A;CMt = 1.00 <br />W et service factor for compression – Table 4A;CMc = 1.00 <br />Wet service factor for modulus of elasticity – Table 4A <br />CME = 1.00 <br />Temperature factor for tension – Table 2.3.3;Ctt = 1.00 <br />Temperature factor for compression – Table 2.3.3 <br />Ctc = 1.00 <br />Temperature factor for modulus of elasticity – Table 2.3.3 <br />CtE = 1.00 <br />Incising factor – cl.4.3.8;Ci = 1.00 <br />Buckling stiffness factor – cl.4.4.2;CT = 1.00 <br />Bearing area factor - cl. 3.10.4 ;Cb = 1.0 <br />Adjusted modulus of elasticity;Emin' = Emin  CME  CtE  Ci  CT = 580000 psi <br />Critical buckling design value;FcE = 0.822  Emin' / (h / d)2 = 584 psi <br />Reference compression design value;Fc = Fc  CD  CMc  Ctc  CFc  Ci = 2484 psi <br />For sawn lumber;c = 0.8 <br />Column stability factor – eqn.3.7-1;CP = (1 + (FcE / Fc)) / (2  c) – ([(1 + (FcE / Fc)) / (2  c)]2 - (FcE / Fc) / c) = <br />0.22 <br />From SDPWS Table 4.3.3 Maximum Shear Wall Aspect Ratios <br />Maximum shear wall aspect ratio;3.5 <br />Shear wall length;b = 18.16 ft <br />Shear wall aspect ratio;h / b = 0.459 <br />Segmented shear wall capacity <br />Maximum shear force under wind loading;Vw_max = 0.6  W = 1.2 kips <br />Shear capacity for wind loading;Vw = vw  b / 2.0 = 7.936 kips <br />Vw_max / Vw = 0.151 <br />PASS - Shear capacity for wind load exceeds maximum shear force <br />Maximum shear force under seismic loading;Vs_max = 0.7  Eq = 1.535 kips <br />Shear capacity for seismic loading;Vs = vs  b / 2.8 = 5.669 kips <br />Vs_max / Vs = 0.271 <br />PASS - Shear capacity for seismic load exceeds maximum shear force <br />Chord capacity for chord 1 <br />Shear wall aspect ratio;h / b = 0.459 <br />Effective length for chord forces ;beff = b - 3 / 2  bEndPost - eanchor = 17.72 ft <br />Load combination 6 <br />Shear force for maximum tension;V = 0.7  Eq = 1.535 kips <br />Axial force for maximum tension;P = (0.6  (D + Swt  h) - 0.7  0.2  SDS  (D + Swt  h))  b1 / 2 + 0.7  <br />Eq_ch1 = 0.264 kips <br />Maximum tensile force in chord;T = V  h / beff - P = 0.458 kips <br />Maximum applied tensile stress;ft = T / Aen = 46 lb/in2 <br />Design tensile stress;Ft' = Ft  CD  CMt  Ctt  CFt  Ci = 1380 lb/in2 <br />ft / Ft' = 0.033 <br />PASS - Design tensile stress exceeds maximum applied tensile stress <br />Page 48 of 108 <br />1011 W Second St <br />5/1/2025