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

I am really confused about this.

It seems hourglass mode is a kind of spurious modes and produced by selective or reduced integration in computing the element stiffness matrix, but I am not clear about the detail.  And I also want to know if there are any spurious modes other than hourglass mode?  Should all spurious modes be zero energy modes? Should rigid body mode be spurious modes?

 Hope someone can explain these questions in detail. Thanks!

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 

$\displaystyle \ensuremath{\boldsymbol{\varepsilon}}= \ensuremath{\boldsymbol{B}...<br />
			...h{\boldsymbol{\sigma}}= \ensuremath{\boldsymbol{D}}\cdot\ensuremath{\mathbf{u}}$

to get 

$\displaystyle \ensuremath{\mathcal{W}}_\ensuremath{\text{int}}= \delta\ensurema...<br />

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

$\displaystyle \left[\ensuremath{\int_{\Omega}}\ensuremath{\boldsymbol{B}}^T\cdo...<br />
			...suremath{\boldsymbol{K}}\cdot\ensuremath{\mathbf{u}}= \ensuremath{\mathbf{f}}~.$

Suppose we have a four-noded two-dimensional element. Then the $ \ensuremath{\boldsymbol{B}}$matrix has size $ 3\times 8$$ \ensuremath{\boldsymbol{D}}$ matrix has size $ 3\times 3$ to give a symmetric $ \ensuremath{\boldsymbol{K}}$matrix of size $ 8 \times 8$.  If you compute $ \ensuremath{\boldsymbol{K}}$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 $ \ensuremath{\boldsymbol{K}}$ 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 $ \ensuremath{\boldsymbol{K}}$using numerical integration.  If you use $ 2\times 2$ Gaussian quadrature then that is equivalent to using an analytical solution. Which means that you still have a matrix$ \ensuremath{\boldsymbol{K}}$which 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 $ \ensuremath{\boldsymbol{K}}$is evaluated at the centerof the element. If you work out the algebra you will find that now the matrix$ \ensuremath{\boldsymbol{K}}$has 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).



T. J. R. Hughes.

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

Dover, New York, 2000. 

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

Nonlinear Finite Elements for Continua and Structures.

John Wiley and Sons, Ltd., New York, 2000.
mohamedlamine's picture

Dear Biswajit,

1/ The formula that you have cited above for Wint is a Virtual Work with δε variation.

2/ Why σ = D.u ? 

    The exact form is σ = D.ε  where D is the Material Properties matrix.

3/For the 4 noded element is it :

   * a Quadrilateral element with Membrane effet (axial stiffness) : Ux , Uy degrees of freedom (d.o.f)

      [Q4 element with 8 d.o.f]. I think this is your studied case and it is clear physically that we

      obtain 2 Rigid Body Modes where the Element has a Rigid Body Motion.

   * a Plate Bending element with Flexural effect : Uz , θx , θy  d.o.f [12 d.o.f]

   * a Shell element with the two above effects : five and then six d.o.f in general case ?

4/ How do you find your Eigenvalues to decide in which case you are ?


Mohammed lamine

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.



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.

mohamedlamine's picture

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


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