Consider a "beam". In order to perform modal analysis using the finite element method, the beam is meshed with 2nd order hexa elements. The total count of nodes is 1280. The same beam can be either clamped on one side or both sides. Clamped ends are modelled by zero displacements (dx=dy=dz=0) for all nodes on the clamped surface. There are 80 nodes per clamped surface.

The mass and stiffness matrices are generated using code_aster.

The clamped-clamped beam matrices are 4800x4800, meaning that the system has 4800 degrees of freedom (dof).

The simply-clamped beam matrices are 4320x4320, meaning that the system has 4320 degrees of freedom (dof).

I don't understand where the number of dof comes from. I would expect 6 dof per node which gives: 1280 • 6 = 7680

Worst than that, if constraints are already included, I would expect the clamped-clamped beam to have less dof than the simply-clamped.

Can anyone help me?

  • $\begingroup$ Could you verify the exact number of nodes per clamped surface? By constraining the displacement of those surfaces you remove 3 dofs, so it shouldn't be irrelevant. $\endgroup$
    – NMech
    Commented Jun 29, 2021 at 20:17
  • $\begingroup$ Why do you think hexa (3-D) elements have rotational DOFs at the nodes? Apart from some very esoteric element formulations, they don't. Also the exact value of the number of clamped nodes is very relevant, depending on when the software eliminated clamped DOFs from the system matrices. $\endgroup$
    – alephzero
    Commented Jun 29, 2021 at 20:51
  • $\begingroup$ Please tell us exactly what FE software you used. Different codes work in different ways. $\endgroup$
    – alephzero
    Commented Jun 29, 2021 at 20:55
  • $\begingroup$ Thank you for the reply, I am using code aster to produce my matrices.I'll provide the number of nodes soon! I thought they had rotational dof due to the book I am reading: Fundamentals of Finite Element Analysis by David V Hutton, where generic beam elements have rotational dofs. How can you account for moments if you only have translations? $\endgroup$
    – Worldsheep
    Commented Jun 30, 2021 at 22:05
  • $\begingroup$ I updated my question, if you need futher information, don't hesitate to ask! Thank you! $\endgroup$
    – Worldsheep
    Commented Jul 5, 2021 at 7:42

3 Answers 3


Please simulate the same problem in ANSYS. From your description, I can say you have 20-node hexahedron element.

That means in ANSYS you have SOLID186 Element. It has only 3 DOF per node. So in total, you'd obtain 1280*3= 3840 DOFs.

Now it depends on the software how they treat your beam and the elements that you use. Please, have a look at the ANSYS element reference.


the book I am reading: Fundamentals of Finite Element Analysis by David V Hutton, where generic beam elements have rotational dofs

Well, this is correct. Generic beam elements do have 6 DOF per node, 3 translation and 3 rotational.

the beam is meshed with 2nd order hexa elements.

Hex elements are not beam elements. They are very different types of elements. Hex elements typically have only 3 DOF per node.

Some other distinctions:

a 1st order beam element has only 2 nodes per element (i.e. a beam element is a line element, it has 0 volume and 0 area), and a 2nd order beam element has 3 nodes.

a 1st order hex element has 8 nodes per element (i.e. it is a solid volume element) and a 2nd order hex element has 20 nodes per element.

in between a beam element and a hex element you will find that there are an intermediate kind, like a shell element. A 1st order quadrilateral shells has 4 nodes per element (i.e. is has a finite area, but 0 volume), and a 2nd order has 8 nodes per element. Shell elements typically do have both translational and rotational elements.

How can you account for moments if you only have translations?

You apply a pair of equal and opposite forces that are separated by a finite distance.


I found the solution. First of all, I made a mistake, in the example I presented I was using 1st order hexa elements.

Code_aster uses lagrange multipliers to apply constraints (in this case clamped faces).

In order to use the $LDL^T-SP$ algorithm, it doubles the lagrange multipliers.

The additional equations are added into the matrices (mass, stiffness,...).


  • Each unconstrained node has 3 dof.

  • Each node in the clamped faces has 3 dof, plus 2 lagrange multipliers for each dof (because dx=dy=dz=0). Total = 3 + 2x3 = 9.

Finally, the total count:

  • Clamped clamped beam: 1280x3 + 80x3x2x2 = 4800

  • Clamped beam: 1280x3 + 80x3x2 = 4320

For more details, please refer to the documentation.


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