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发表于 2011-5-11 10:05:22
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来自 江苏南京
谢谢啊,用你的方法我能得到总体刚度矩阵了,但是得到的刚度矩阵怎么看呢,帮助中说The assembled sparse matrix operator data are written to the text file as a series of comma-separated lists. Each row in the file represents a single matrix entry; a row is written as a comma-separated list with the following elements: - Row node label
- Degree of freedom for row node
- Column node label
- Degree of freedom for column node
- Matrix entry
还是看不明白啊,我把得到的部分结果放上来,怎么处理结果,我想得到整体的一个刚度矩阵
1,1,1,1,1.000000000000000e+036
1,2,1,1,3.389541568774197e+013
1,4,1,1,1.084202172485504e-019
1,5,1,1,-9.947078621878498e-004
1,6,1,1,2.398547867806398e+012
5,1,1,1,-3.151208340922268e+013
5,2,1,1,-1.631111851503475e+012
5,4,1,1,-1.953125000000000e-003
5,5,1,1,-1.035420731107725e-003
5,6,1,1,-9.817130099410553e+010
34,1,1,1,2.966445285045047e+013
34,2,1,1,1.581788142830688e+012
34,4,1,1,1.953125000000000e-003
34,5,1,1,1.035420731107725e-003
34,6,1,1,2.496719168800504e+012
22# yf881031 |
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