三角形悬臂梁应力分析三角形悬臂梁应力分析摘要:在有限元分析软件ANSYS12.0平台上建立三角形悬臂梁的力学模型,添加约束和载荷,计算出应力分布,并与理论计算值相比较。
⒈ 引言目前,ANSYS 软件具有其强大的功能已经被广泛的应用于机械,化工,土木,交通等各个领域。
应用ANSYS 分析,可以大大减少人力物力的投入,而且可靠性高,对于三角形悬臂梁分析其应力和变形情况,分析方法和结论可作为这类设计的参考。
⒉ 计算模型Ⅰ问题描述【三角形悬臂梁忽略重力作用,∠BAC=α,AB 边上作用均布载荷q ,求应力的解析表达,计算出BC 边上的应力值并与ANSYS 计算值比较,绘出应力曲线图】选取应力函数:Ansys 计算参数值:AB=1000mm ,α=30°,厚度t=20mm2222[()sin cos cos tan ]C r r r ϕθθθθα=∂-+-Ⅱ解析解根据弹塑性平面问题的极坐标解答,利用以下公式推导:222222211111()r r r r r rr r r r r θθϕϕσθϕσϕϕϕτθθθ∂∂=+∂∂∂=∂∂∂∂∂=-=-∂∂∂∂∂ 以及2222cos sin 2sin cos sin cos 2sin cos x r r y r r θθθθσσθσθτθθσσθσθτθθ=+-=++已知2222[()sin cos cos tan ]C r r r ϕθθθθα=∂-+-, 故有以下式子成立:22222222222[2()2sin cos 2cos tan ][2()2sin cos 2cos tan ][cos 2sin 2tan ][2sin 22cos 2tan ]C r r r rC r C r r r C r r ϕαθθθθαϕαθθθθαϕθθαθϕθθαθ∂=-+-∂∂=-+-∂∂=-++∂∂=-+∂ 所以,22222222211[2()sin 22cos tan 2cos 2tan ][2()sin 22cos tan ]111()[1cos 2sin 2tan ]r r C r r r C rC r r r r r θθϕϕσαθθθαθαθϕσαθθθαϕϕϕτθθαθθθ∂∂=+=---+∂∂∂==-+-∂∂∂∂∂=-=-=--∂∂∂∂∂ 因此,222222222224cos sin 2sin cos [2()2sin 2cos 2cos tan 2cos cos 2tan sin 2cos 2sin 2tan ]sin cos 2sin cos [2()2cos sin 2cos 2sin 2tan 2tan sin cos 3tan cos ]x r r y r r C C θθθθσσθσθτθθαθθθθαθθαθθθασσθσθτθθαθθθθθααθθαθ=+-=---+++=++=-+-++- 由边界0()/y y q t σ==-,即当0θ=时,/y q t σ=-;带入y σ的表达式中可得:2(tan )q C t αα=- 将C 带入x σ表达式中得:22222222[2()2sin 2cos 2cos tan 2cos cos 2tan sin 2cos 2sin 2tan ][2()2sin 2cos 2cos tan 2(tan )2cos cos 2tan sin 2cos 2sin 2tan ]x C q t σαθθθθαθθαθθθααθθθθαααθθαθθθα=---+++=----+++Ⅲ Ansys 计算问题详细说明:取AB=1000mm ,30α=,厚度t=20mm ,50/q N mm =-。
解题过程:① 前处理,设置材料属性,E=200e3,泊松比0.3μ=,单元类型为solid 20 node 95;建立几何模型,有点生成面,再由面生成体;单元网格划分为三角形。
② 加载求解,加约束,面约束,加载荷,均布载荷50/q N mm =-,并求解。
③ 后处理⒊ 结果分析变形图计算云图等效应力曲线与理论值相比较,利用ANSYS计算的结果很接近于理论值,采用有限元法对悬臂梁进行应力分析具有很高的效率和可靠性。
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