§91轴向拉压杆的变形及刚度计算 解 (1)受力分析:取结点A为研究对象了N ∑F,=0,→FN1=F/sin6 =2F1=80kN拉力) ∑F4=0,→F2=F31C0s=697kN(压力) (2)变形计算 EA ErEd? 4L Fn22_ FN2 cos 0 NI 4F. NI 21 048×103m 0.24×103m =048mm(伸长) =024mm(缩短)c9-1 ᅀ຿ࣸགئԅέ໹ذރէޙസ
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˄1˅ফߚ࡯ᵤ˖প㒧⚍AЎⷨおᇍ䈵 Fy = 0,
2 80 kN (ᢝ࡯˅ /sin P N1 P = = = F F F θ Fx = 0,
FN2 = FN1 cosθ = 69.7 kN (य़࡯˅
˄2˅বᔶ䅵ㅫ 0.48 mm (Ԍ䭓˅ 0.48 10 m 4 3 2 1 N1 1 1 1 N1 1 1 = = × = = − E d F l E A F l l π ∆ 0.24 mm˄㓽ⷁ˅ 0.24 10 m cos 3 2 2 N2 1 2 2 N2 2 2 = = × = = − E a F l E A F l l θ ∆