T f“20su 式中2为截面中心线所围的面积。 M=T=208mn[t]=2hbolt] =2×600x100x10-6m2x6x10-3mx60x106Pa=108x102N.m 2、确定开缝后的许用扭力知 式中号∑4=×230×3x10m+10x31x10m)=80x3x10m 800 M-T-3 5mm 33x10-12m4 -=3 3×10-3m 60x10Pa=144N.m 讨论：上述计算结果表明，开口薄壁杆比闭口薄壁杆扭转时承载能力差很多。8 [ ] 2 min max τ δ τ = ≤ Ω T 式中 Ω 为截面中心线所围的面积。 2 ( )( ) 300 100 10 m 3 10 m 60 10 Pa 108 10 N m 2 [ ] 2 [ ] -6 2 -3 6 2 min = × × × × × × × = × ⋅ M = T = Ωδ τ = hbδ τ 2、确定开缝后的许用扭力矩 [ ] 3 1 3 max max τ δ δ τ = ≤ ∑ i i h T 式中 3 3 12 4 3 12 4 3 12 4 3 10 m 3 800 2(300 3 10 m 100 3 10 m ) 3 1 3 1 − − − ∑hiδ i = × × × + × × = × × 60 10 Pa 144 N m 3 10 m 3 10 m 3 800 [ ] 3 1 6 3 3 12 4 max 3 × × = ⋅ × × × = = = − − ∑ τ δ iδ i h M T 讨论：上述计算结果表明，开口薄壁杆比闭口薄壁杆扭转时承载能力差很多