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姓名:朱伟华学号:14102020278第一章工程结构非线性分析概述作业1:已知薄板L=30m,B=1.32m,t=0.02m,E=2.06×105MPa,ρ=7.8×103kg/m3。求自重作用下的变形。要求用ANSYS求解,并写出命令流。BL解答1:按薄板进行计算finish/clear/prep7et,1,shell63r,1,0.02mp,ex,1,206e9mp,dens,1,7.85e3k,1,k,2,30k,3,0,1.32k,4,30,1.32a,1,2,4,3esize,0.2mshape,0,2dmshkey,1amesh,allnsel,,loc,x,0d,all,all,0nsel,,loc,x,30d,all,all,0allsel,allfinish/soluacel,,,9.8nlgeom,onnsubst,100outres,all,allsolvefinish打开大位移控制开关的计算结果如下图所示,不打开大位移控制开关的计算结果如图2所示图1考虑非线性的位移计算结果图2不考虑非线性的计算结果从图中可以看出,考虑非线性的影响,跨中最大位移0.238441m、不考虑非线性影响其跨中最大位移为23.359m。如果按梁单元时行,其计算结果也具有较好的精度FNISH/CLEAR/PREP7ET,1,BEAM3R,1,0.0264,8.8E-7MP,EX,1,2.06E11MP,PRXY,1,0.3MP,DENS,1,7800K,1,0.0K,2,30L,1,2LATT,1,1,1LESIZE,1,,,30LMESH,1DK,1,ALL图3采用BEAM3的计算结果为0.241025mDK,2,ALLFINISH/SOLUACEL,0,9.8,0NLGEON,ONNSUBST,25SOLVE作业2:已知一悬臂梁,主要参数为:L=365.76mm,A=25.4×25.4mm,E=2.0684E+11Pa,G=3E/8,ρ=7.757E+3kg/m3,剪切修正系数k=2/3。试计算其振动频率,要求:1.求出其解析式;2.用计算软件计算其振动频率,并对比分析按Euler梁理论与Timoshenko梁理论的差别。根据Euler梁的振动理论,其振动方程为:02244tymxyEI(2-1)采用分离变量法,令tqxtxy,,则有:0444kx式中:42EImk上式的解为:kxDkxCkxBkxAxcoshsinhcossin(2-2)相应的转角、弯矩和剪力为:转角:kxDkkxCkkxBkkxAkxxsinhcoshsincos'弯矩:kxDkxCkxBkxAkxEIxMcoshsinhcossin2剪力:kxDkxCkxBkxAkxEIxQsinhcoshsincos3'边界为:当0x时,有0;当Lx时,有M=Q=0,将边界条件代入上式,系数A、B、C、D有解的条件是系数行列式为0.化简后有超越方程为:01coshcoskLkL解得固有频率为:mEIL21875.1、mEIL22694.4、mEIL23855.7将数据代入有圆频率为:9995.9948628.37875.121L、0054.62362、7521.174623相应的工程频率为:Hz3590.158211f、Hz4910.99222f、Hz2833.277933f相应的ANSYS命令流finish/clear/prep7et,1,beam3r,1,6.4516e-4,3.468595e-8mp,ex,1,2.0684e11mp,dens,1,7.757e3k,1,k,2,0.36576l,1,2lesize,1,,,5lmesh,alld,1,allfinish/soluantype,modalmodopt,subsp,10mxpand,5solvefinish/post1set,list计算结果第一、二、三阶振型如下:考虑剪切变形影响,相应的命令流为(ANSYS10.0)finish/clear/prep7et,1,beam3r,1,6.4516e-4,3.468595e-8,0.0254,1.5mp,ex,1,2.0684e11mp,dens,1,7.757e3k,1,k,2,0.36576l,1,2lesize,1,,,5lmesh,alld,1,allfinish/soluantype,modalmodopt,subsp,10mxpand,5solvefinish/post1set,list
本文标题:第一章工程结构非线性分析概述
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