For bending simulation, both standard linear-triangle element (in 3D is tetrahedral) and standard linear-quad element (in 3D is hexahedral) show a stiff behavior known as shear locking. The reason is the interpolation functions of these elements contains inconsistent terms. However the quad element usually shows better behavior than the triangle one.

In fact in the bending cases shear strains should be zero or negligible. But the inconsistent terms in the interpolation functions of linear elements make shear strains much different from zero. The non-zero artificial shear strains absorb much energy making the element stiffer. To alleviate shear locking you can: 1) use finer mesh of standard linear elements; 2) use standard high-order elements; 3) use incompatible (non standard) elements designed for anti shear locking.

locking is one of the problems using triangular elements - which can partially be overcome using higher grade triangular elements.

more literally explained a signle triangle forms a stable structure whereas a quad is allways less rigid as it forms a mechanism if no bending stiffness is introduced at the corner points.

in most numerical meshing tools the introduction of triangles is straigth forward whereas the introduction of quads only can be challenging - mixed elements can be a practical approach. going to 3d tetrahedrons are easily applicable to allmost every structure.

Thank you very very much!! Philipe !!!

It helped me a lot!!!  I was trying to figure out why this happens

I'm very grateful :)

But can I ask one more question?

In 3-dimensional case (3D solid elements)

does the  4-node tetrahedral show more stiffness than the 8-node cubic ones?

Thanks in advance!!

Dear Giseok,

You may also want to refer to the book

"Concepts and applications of finite element analysis, 3rd edition" 

by Robert D. Cook, David S. Malkus, and Micheal E. Plesha.

On page 156-157, investigation on the linear triangle element under pure bending is conducted.

And for your question about 3D elements, my understanding is: if you are using the same mesh, then the tetrahedral will usually be stiffer than the high ordered element. You could conduct a convergence test to the two case by refining the mesh step by step, and they may give you the same results finally. Only difference is that the 8-node cubic meshed body will converge much faster than the 4-node tetrahedral one.

Best Regards,

Fang

The convergence properties to the "exact" solution is very poor. In view of large strain problems, see, for example, http://onlinelibrary.wiley.com/doi/10.1002/zamm.201200040/Abstract

See references, cited therein, as well.

Sincerely Yours

Stefan Hartmann

Bending requires variable stresses in elements, while simple trianguler element CST (constant stress) only offers constant stress element wise. 

For bending problems, second order elements are generally recommended. In addition, the use of a reduced integration scheme improve the bending behavior. With that kind of elements (for example 8-node elements with 4 Gauss points in 2D or 20-node elements with 8 Gauss points in 3D) accurate results can be obtained with quite coarse meshes.

For bending simulation, both standard linear-triangle element (in 3D is tetrahedral) and standard linear-quad element (in 3D is hexahedral) show a stiff behavior known as shear locking. The reason is the interpolation functions of these elements contains inconsistent terms. However the quad element usually shows better behavior than the triangle one.

In fact in the bending cases shear strains should be zero or negligible. But the inconsistent terms in the interpolation functions of linear elements make shear strains much different from zero. The non-zero artificial shear strains absorb much energy making the element stiffer. To alleviate shear locking you can: 1) use finer mesh of standard linear elements; 2) use standard high-order elements; 3) use incompatible (non standard) elements designed for anti shear locking.