分享
验证例题13.pdf
下载文档

ID:3634549

大小:196.80KB

页数:4页

格式:PDF

时间:2024-06-26

收藏 分享赚钱
温馨提示:
1. 部分包含数学公式或PPT动画的文件,查看预览时可能会显示错乱或异常,文件下载后无此问题,请放心下载。
2. 本文档由用户上传,版权归属用户,汇文网负责整理代发布。如果您对本文档版权有争议请及时联系客服。
3. 下载前请仔细阅读文档内容,确认文档内容符合您的需求后进行下载,若出现内容与标题不符可向本站投诉处理。
4. 下载文档时可能由于网络波动等原因无法下载或下载错误,付费完成后未能成功下载的用户请联系客服处理。
网站客服:3074922707
验证 例题 13
1Example 13 Title Cantilever plate subjected to a uniform pressure load Description Determine the deflections and the bending moments of the structure.Structural geometry and analysis model 2Model Analysis Type 3-D static analysis Unit System mm,N Dimension Length 300 mm Width 100 mm Thickness 25 mm Element Plate element (Thick type)Material Modulus of elasticity E =2.1 105 N/mm2 Element Property Size a b =50 mm 50 mm Thickness t =25 mm Boundary Condition Nodes 1,3 and 10 ;Constrain all DOFs.Load Case A uniform pressure load,5 N/mm2is distributed over the entire cantilever plate in the-Z direction.Example 60 3Results Z-displacements shape of the structure(Node 2)Element Forces 4Comparison of Results Theoretical calculation of the maximum deflection and bending moment Maximum deflection(max)=ElL82+GAsL22(at the free end)=)12/25100)(10210(8)300)(1005(334+)25100(6/51052)300)(1005(5/62=8.51 mm Maximum bending moment(Mmax)=L2/2 (at the fixed end)=(5 100)(300)2/2=22.5 106 Nmm FEM analysis of the maximum deflection and bending moment Maximum deflection(max)=18.62 mm(nodes 2,16 and 4 at the free end)Maximum bending moment(Mmax)=Number of nodes pertaining to the fixed end elements the bending moment at one node (nodes 1&10 of the element 1and nodes 3&10 of the element 7)=4 5625000 Nmm =22.5 106 Nmm Unit:mm,KNm Result Theoretical SAP2000 MIDAS/GenMaximum deflection(max)18.617 18.788 18.686 Maximum bending moment(Mmax)22.50106 22.50106 22.5106 Reference Gere&Timoshenko,”Mechanics of Material”,1984

此文档下载收益归作者所有

下载文档
你可能关注的文档
收起
展开