V7.8计算软件计算
八、立柱验算数据中：
3、稳定性验算
1) 不组合风荷载时立杆从左到右受力为(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=6.311kN
R2=9.427+1.35×0.15×(9.8-1.2)=11.168kN
R3=9.427+1.35×0.15×(9.8-1.2)=11.168kN
R4=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=6.311kN
Nut=max[R1,R2,R3,R4]=max[6.311, 11.168, 11.168, 6.311]=11.168kN
1.05Nut/(φAKH)=1.05×11.168×103/(0.281×424×0.972)=101.278N/mm2≤[f]=205N/mm2
符合要求！
2) 组合风荷载时立杆从左到右受力为(Nut＝γG∑NGk+0.85×1.4∑NQk)：
R1=5.829kN, R2=10.977kN, R3=10.977kN, R4=5.829kN
组合风荷载时的立杆轴力为什么比不
组合风荷载时立杆轴力小？