请教平板壳单元程序技巧

cjhero

弱弱地问一下：我自己编制了平板壳单元程序(如DKT之类单元)。该程序在计算所有单元在一个平面内的情形时(无论该平面取向如何或几何形状如何)都能取得正确结果。但在对于单元不在一个平面内，特别是单元间夹角较大时，算出来的结果就完全不对了，结果显示结构过于刚硬。不知这里有啥技巧？
望有经验者指点一二，或提供问题可能在哪里，或建议看某方面或某一篇文献。
谢谢！同道者多交流交流！

cjhero

再问一下编过此类程序的大侠们、斑竹们：我用自己编的程序算过比较经典的平板弯曲、切口圆环变形问题，结果都很好。我把平板转不同角度，然后再计算，发现结果与abaqus的差不多。但在对于半球壳拉压、半圆壳拉等问题时结果就不对了。
我尝试了面内z向转角刚度取很大或很小值、采用zienkiewicz的增加虚拟扭转刚度法、甚至也用面内变形插值时引入z向旋转自由度。这些办法对于对所有单元在一个平面内的情况结果都对，但对于非单元不在同一面内的情况结果就差得太大了。
不知道问题到底出在哪儿？闷了好些时日了，还没找出原因。渴望帮助和建议！
再次谢谢大侠和斑竹们！

hillyuan

It seems like the problem of the element you adopted.
1. Increase the nodes number and see if it converges to a right answer.
2. Considering introduction of other high accurate element

cjhero

It seems that the integration points strongly influence the numerical resluts. If more integration points are chosen, the element becomes more stiff, especially for the non-coplanar problem.
It is strange. For the plane problem or 3-D problem, the houreglass phenemone frequently appears if few integration points are chosen. However, the plane shell element (membran element+bending element) is more stiff with more integration points. For many cases, one integration point seems better than more. This problem seldom comes for the degenerated solid shell element.
Did other passerbys meet the similar problem for the membran element+bending element?

hillyuan

Following article maybe helpful

Bathe et al:A evaluation of the MITC shell elements, Comput & Struct, 75(2000), 1-30

tonnyw

Check if you get the correct volume by setting the integrand equal to 1 and compare with your hand calculation.

cjhero

Thanks for superiors' comments.
I just touch the program of shell FEM and mainly concentrated on the classical shell FEM. Today, I search K.J. Bathe's publications. K.J. Bathe is really a man in FEM. Of course, T. Belythshko is another superman.
By the wa, which kind of shell FEM is the best choice for solving both the thick and thin shell problem? It is best that it can be used in the explicit algorithm. So, only corner nodes can be used in the final formulation. In addition, it is also good if it can consider the deformation and stress along the thickness direction.
Any suggestion for the experienced superiors is helpful for  the green learner to chosse a right way.
Millions thanks again.

tonnyw

Bascially two shell models are employed: Love-Kirchhof model and Reissner-Mindlin model. The first one requires C1 which is not easy to construct in higher dimension while the second one is C0 problem. The drawback of the second one is that when the shell becomes thinner locking would happen which can be overcomed by reduced integration and other methods which you may find in references. I would say the current prevalent shell model is the second one applicable to thick and thin shell.

Four-node shell element is not a good idea. Dealing with shell structure which has curved surface, you need higher order element which could be eight or nine nodes in order to approximate the geometry correctly.

cchang

1楼提到的问题是用flat plate element解决折叠板和曲面壳时不容忽视的一个问题。
首先做曲面壳的问题，最好是用曲面壳单元。因为平板壳不好解决夹角问题。

在用平板壳单元做折叠板和曲面壳(曲率较缓)时，相连单元基本共面，2楼提到的方法其实原则上是形成平板壳的刚度矩阵时必须要考虑的。增加虚拟扭转刚度是合理的，其实相当于在单元的相交线上加了弹簧单元，但它的刚度不能为零，而且取多大是关键（可参考bathe的finite element procedures一书，有段文字解释讲的言简意赅。96版的在208～209页）。不知道你所说得”我尝试了面内z向转角刚度取很大或很小值“中的”很大或很小值“是多大，bathe推荐的是刚度矩阵最小对角单元的1/1000左右。相信读了之后会有启发的。

如果做explicit dynamics分析，一般要采用足够小的时间步以满足explicit算法数值上条件稳定的要求，也就意味着同样长一段时间里面要用更多的时间步（也即对一个积分点要循环更多次），这样以来4结点壳单元比高阶单元更有竞争力。关于4结点壳单元，这篇文章你可能会有兴趣。
T.J.R. Hughes, R.L. Taylor and W. Kanoknukulchai, " A simple and efficient finite element for plate bending", International Journal for Numerical methods in Engineering, 11, 1529-1547, 1977

但是这种单元有2个零能模态，如果在explicit dynamics里面用，需要必要的hourglass control，关于这个可参考belytschko的文章或书籍。通过这样的完善措施之后，它在通用的商业有限元（abaqus）中才得以广泛的应用。
有一点值得注意的是，有的程序可能对这种4结点单元使用uniformly reduced integration，那么只有平面内一个高斯积分点，所以在变形或应力梯度大的地方需要多划分几个单元。

另一方面，由于中间结点的存在，高阶壳单元可以很好地模拟曲面壳的问题，还因为它可定义双向曲率。但在使用前要确定问题不包括刚体模态的运动，由于弯曲和膜应力的组合使得它不好满足此时应变能为零的要求。

至于4楼提到的“If more integration points are chosen, the element becomes more stiff, especially for the non-coplanar problem.“ 这里，十之有八九是由于厚度问题引起的：用厚壳单元模拟薄板或壳问题，会出现剪切锁闭的问题。在厚壳单元的刚度矩阵数值积分过程中，积分点的数量从另一方面等价于施加”kirchhhoff constraints“的数量，所以用的越多变形越刚。

snowwave02

ls说的很有道理，只是有一点，如果仅仅是第六自由度给一个小量的刚度，依然还会存在楼主的问题，而且还会带来很多其它问题，商软一般在K11/K16/K66等都取一个小量