0 /i=(a 1 2 +b 1 2) 0.5 /i=(2304 2 +7640 2) 0.5 /93.405=85.433 ≤[λ]=100 ,查规范表得:φ 1 =0.652 满足要求! 附墙杆 2长细比: λ 2 =L 0 /i=(a 2 2 +b 2 2) 0.5 /i=(2304 2 +5540 2) 0.5 /93.405=64.236 ≤[λ]=100 ,查规范表得:φ 2 =0.784 满足要求! 附墙杆 3长细比: λ 3 =L 0 /i=(a 3 2 +b 3 2) 0.5 /i=(2372 2 +5840 2) 0.5 /93.405=67.484 ≤[λ]=100 ,查规范表得:φ 3 =0.766 满足要求! 附墙杆 4长细比: λ 4 =L 0 /i=(a 4 2 +b 4 2) 0.5 /i=(2372 2 +7640 2) 0.5 /93.405=85.646 ≤[λ]=100 ,查规范表得:φ 4 =0.65 满足要求! 附墙杆 1轴心受压稳定系数: σ 1=N 1 /(φ 1 A)=781248/(0.652 × 8293.805)=144.473N/mm 2≤[f]=215N/mm 2 满足要求! 附墙杆 2轴心受压稳定系数: σ 2=N 2 /(φ 2 A)=769769/(0.784 × 8293.805)=118.383N/mm 2≤[f]=215N/mm 2 满足要求! 附墙杆 3轴心受压稳定系数: σ 3=N 3 /(φ 3 A)=740736/(0.766 × 8293.805)=116.595N/mm 2≤[f]=215N/mm 2 满足要求! 附墙杆 4轴心受压稳定系数: σ 4=N 4 /(φ 4 A)=759105/(0.65 × 8293.805)=140.81N/mm 2≤[f]=215N/mm 2 满足要求!
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