Can anyone explain the meaning of spurious mode,zero energy mode and hourglass mode in FEM?

The stiffness matrix of an element is computed from the sum of the internal energies at each material point inside that element using the integral

We convert the relations

into Voigt form 

to get 

The issue of zero energy modes arises when we try to solve a system of equations of the form 

Suppose we have a four-noded two-dimensional element. Then the matrix has size matrix has size to give a symmetric matrix of size .  If you compute analytically, you will find that the number of linearly independent equations that the system of equations represents is only 5 and not 8 (i.e., the matrix has rank 5).

If you directly try to invert the system of equations you will get an infinite number of solutions. To get a unique solution you will have toget rid of the three rigid body modes (2 translations and 1 rotation) by applying some boundary conditions. These rigid body modes do not contribute to the internal energy of the system and are zero energy modes.

Now, suppose that you try to calculate using numerical integration.  If you use Gaussian quadrature then that is equivalent to using an analytical solution. Which means that you still have a matrixwhich has rank 5.

Sometimes, for speed or to avoid locking, a lower order numerical integration method is used. For the four noded 2-D element we can use a 1-point Gaussian quadrature where the integral for is evaluated at the centerof the element. If you work out the algebra you will find that now the matrixhas rank 3, i.e., only three rows of the matrix are linearly independent.

In that case, unless you apply boundary conditions such that only three degrees of freedom need to be solved for, you will get an infinite number of solutions. Recall that for a fully integrated element there are three rigid body modes which contribute zero energy to the element. For the 1-point integrated element, there are now two more zero energy modes. These modes must be spurious because they have no physical basis.

The effect of these extra modes is that a number of different configurations of the four noded element lead to the same internal energy. The element shapes
look like hourglasses and hence they are also called hourglass modes.

Suppose that you have a number of underintegrated elements in a mesh. Even if you don't apply a sufficient number of boundary conditions you might end up with a global stiffness matrix that is not singular. However, if the stiffness matrix is close to singular, the hour glass modes will show up in your results as strangely deformed elements.

You can find detailed and very clear discussions of these issues inHughes book [1] (p. 239) and in Belytschko, Liu, and Moran [2] (p. 492).

Bibliography

1

T. J. R. Hughes.

The Finite Element Method: Linear Static and Dynamic Finite
Element Analysis.

Dover, New York, 2000. 

2

T. Belytschko, W. K. Liu, and B. Moran.

Nonlinear Finite Elements for Continua and Structures.

John Wiley and Sons, Ltd., New York, 2000.

Thank you Biswajit. I have been looking for hourglass modes in the web and you gave a very nice explanation. I still have to work out the details in your post and convince myself. Thanks for the references also.

Pradeep

Dear Biswajit,

              That was quite useful for me too. I see that such spurious instabilities occur in Meshless methods too, for example in the case of Smoothed Particle Hydrodynamics (SPH). May I have any information from you on this? 

Siva Prasad AVS.

Dear Zhang,

Zero Energy Modes or Rigid Body Modes correspond to the case where:

Lambda = {PhiTr}*[K]*{Phir} = 2U = 0  with normalized eigenvectors. Lambda: Eigenvalue(s),

{Phir}: Rigid Body Eigenvector, K: Stiffness matrix, U: Strain Energy (produced by stresses from internal loads). We have then zero Frequencie(s) in this case.

In general case : Lambda = ({PhiT}[K]{Phi}) / ({PhiT}{Phi}). If {Phi} is normalized then {PhiT}.{Phi}=1

For a Plate Bending we have three Rigid Body Modes. 

Mohammed lamine