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Project <br />1011 <br />Job Ref. <br />Section <br />LINE-B <br />Sheet no./rev. <br /> 6 <br />Calc. by <br />N.K <br />Date <br />15/11/2024 <br />Chk'd by Date App'd by Date <br />Load combination 2 <br />Shear force for maximum compression;V = 0.7  Eq = 1.35 kips <br />Axial force for maximum compression;P = ((D + Swt  h) + 0.7  0.2  SDS  (D + Swt  h))  s / 2 + -1  0.7  Eq_ch3 <br />+ DC_ch3 + 0.7  0.2  SDS  DC_ch3 = 1.253 kips <br />Maximum compressive force in chord;C = V  (k2 / sum(k1,k2))  h / b2_eff + P = 3.601 kips <br />Maximum applied compressive stress;fc = C / Ae = 187 lb/in2 <br />Design compressive stress;Fc' = Fc  CD  CMc  Ctc  CFc  Ci  CP = 552 lb/in2 <br />fc / Fc' = 0.339 <br />PASS - Design compressive stress exceeds maximum applied compressive stress <br />Design bearing compr. stress, bottom plate;Fc_perp' = Fc_perp  CMc  Ctc  Ci  Cb = 625 lb/in2 <br />fc / Fc_perp' = 0.299 <br />PASS - Design bearing compressive stress exceeds maximum applied bearing compressive stress <br />Chord capacity for chord 4 <br />Load combination 6 <br />Shear force for maximum tension;V = 0.7  Eq = 1.35 kips <br />Axial force for maximum tension;P = (0.6  (D + Swt  h) - 0.7  0.2  SDS  (D + Swt  h))  b2 / 2 + -1  0.7  <br />Eq_ch4 = -0.217 kips <br />Maximum tensile force in chord;T = V  (k2 / sum(k1,k2))  h / b2_eff - P = 2.564 kips <br />Maximum applied tensile stress;ft = T / Aen = 162 lb/in2 <br />Design tensile stress;Ft' = Ft  CD  CMt  Ctt  CFt  Ci = 1380 lb/in2 <br />ft / Ft' = 0.117 <br />PASS - Design tensile stress exceeds maximum applied tensile stress <br />Load combination 2 <br />Shear force for maximum compression;V = 0.7  Eq = 1.35 kips <br />Axial force for maximum compression;P = ((D + Swt  h) + 0.7  0.2  SDS  (D + Swt  h))  s / 2 + 0.7  Eq_ch4 = <br />0.459 kips <br />Maximum compressive force in chord;C = V  (k2 / sum(k1,k2))  h / b2_eff + P = 2.806 kips <br />Maximum applied compressive stress;fc = C / Ae = 146 lb/in2 <br />Design compressive stress;Fc' = Fc  CD  CMc  Ctc  CFc  Ci  CP = 552 lb/in2 <br />fc / Fc' = 0.264 <br />PASS - Design compressive stress exceeds maximum applied compressive stress <br />Design bearing compr. stress, bottom plate;Fc_perp' = Fc_perp  CMc  Ctc  Ci  Cb = 625 lb/in2 <br />fc / Fc_perp' = 0.233 <br />PASS - Design bearing compressive stress exceeds maximum applied bearing compressive stress <br />Page 56 of 108 <br />1011 W Second St <br />5/1/2025