WGFN2FN1f2f1Q假设物重为W(变化值),AB杆直径D,弹性模量为E1,许用应力为[σ]1,钢索1的直径为a,弹性模量为E2,许用应力为[σ]2各角度如图WF2F1F3滑轮Ayx有定滑轮性质可得,F2=W,得出F3=Wcos30°/sin15°F1=FNcos15°-W(1+cos60°)为计算方便,选取如图的坐标系,列出平衡方程:F1+F2+Wcos60°-F3cos15°=0`F3sin15°-Wcos30°=0WF2F1F3滑轮AyxAB撑杆所受应力及线应变:σ1=F3/A1=F3/(πD2/4)=4Wcos30°/πD2sin15°=σ1/E1=4Wcos30°/E1πD2sin15°在AB杆许用应力范围内,最大物重为:σ1≤[σ]1化简得:W≤[σ]1πD2sin15°/4Wcos30°钢索1受到的应力及线应变:σ2=F1/A2=FNcos15°-W(1+cos60°)aπ/4=σ2/E2=FNcos15-W(1+cos60°)E2aπ/4在钢索的许用应力范围内,物重最大值为:σ2≤[σ]2化简得:W≤FNcos15°-[σ]2aπ/41+cos60°
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