V7.8超高支模計算軟件八、立柱驗算數據中——?3、穩(wěn)定性驗算1) 不組合風荷載時立桿從左到右受力為(Nut＝γG∑NGk+1.4∑NQk)：R1=0.321+[1.2×(0.3+(24+1.1)×0.12)×0.8+1.4×(1+2)×0.8]×(0.9/2+(0.5-0.35/2)/2)
+1.35×0.15×9.8=6.311kNR2=9.427+1.35×0.15×(9.8-1.2)=11.168kNR3=9.427+1.35×0.15×(9.8-1.2)=11.168kNR4=0.321+[1.2×(0.3+(24+1.1)×0.12)×0.8+1.4×(1+2)×0.8]×(0.9/2+(1-0.5-0.35/2)/2)
+1.35×0.15×9.8=6.311kNNut=max[R1,R2,R3,R4]=max[6.311, 11.168, 11.168, 6.311]=11.168kN1.05Nut/(φAKH)=1.05×11.168×103/(0.281×424×0.972)=101.278N/mm2≤[f]=205N/mm2符合要求！2) 組合風荷載時立桿從左到右受力為(Nut＝γG∑NGk+0.85×1.4∑NQk)：R1=5.829kN, R2=10.977kN, R3=10.977kN, R4=5.829kN問題1：組合風荷載時的立桿軸力真的會比不組合風...閱讀全文