为什么有限元法求位移偏小？

SD参谋长

为什么有限元法求位移偏小？

fengxiang84

一般认为结构为连续体，有无穷质点组成，其自由度有无穷，而通过有限元的网格离散后，其结构的自由度数与节点数有关，变成有限个，自由度的减小相当于增加了结构的刚度，从而计算所得位移偏小。

SD参谋长

呵呵   回答的不错  和课本很像

northwindgod

这个问题以前已经讨论过了，详细内容可参考如下连接：
http://forum.simwe.com/viewthrea ... p;page=1#pid2092465

tonnyw

这个问题以前已经讨论过了，详细内容可参考如下连接：
http://forum.simwe.com/viewthread.php?tid=940624&page=1#pid2092465
northwindgod 发表于 2010-12-28 18:44

This is not rigorous. On one side, we have the finite element solution which is a continuous function. On the other side, we have the exact solution which is a continuous function. How could we compare the two functions and tell which one is small or not. The two functions have to be compared by some kind of norm.

For an elliptic problem with homogeneous Dirichlet boundary condition, the approximate strain energy from the finite element solution is always smaller than the exact one. As the mesh is nest-refined and polynomial order is increased, the approximated strain energy converges monotically to the exact one. Note the element has to be compatible one.

zhounengjuan

使用位移法求解，为了保证位移连续，限定了结构的自由度，使得刚度增加，荷载不变，位移自然是减小的；如果使用力法求解，刚好相反

jing-qian

I thought above explanation is not good enough. For example, the linear displacement element can pass patch test. That means it exactly recover the linear displacement filed. In another word, the numerical solution is exactly same as real displacement. For cantilever beam bending, it happens as what you said. But, many factors can effect the stiffness of a element such as interpolation or integration method. Usually, the full integration displacement element is stiffer than reduced integration element. If you consider meshfree method or particle method too, you may find that your observation might not be always true.