boundary conditions in 1D FEA of beam deflection 1

[RolandW]

Hello everyone. I am pogramming some software for a 1D FEA of a thin beam with multiple pinned supports and loads. It's ends are always free.

I have managed to assemble the global stiffness matrix, but I am not sure how to set the boundary conditions which is needed to make it non-singular. The matrix form is:

F1 v1
M1 u1
F2 = K * v2
M2 u2
. .
. .
. .

where K is the stiffness matrix, F is force, M momentum, v displacement and u rotation angle. Obviously I want to solve it for v and u.

A pinned support should have the deflection v = 0, so I guess that I could at least remove these columns? But what rows should I remove so that the system is not over-constrained? Does it matter?

I am a total newbie when it comes to finite elements, so forgive me if the question is badly formulated. Thanks in advance.

 

[johnhors]

Roland

Your beam has two degrees of freedom, it can deflect (v) and rotate (u) using your nomenclature. Therefore to make the stiffness matrix non-singular you require at least two restraints, one of which must be v = 0. More than two restraints and the beam is over constrained. And yes it does matter which nodes are restrained. Restraints effectively apply a set of forces/moments to balance out your applied loading. Changing position of applied loads and restraints will of course generate different deflections.

 

[RolandW]

Thanks for your swift reply John. But if I set v=0 on two supports, thereby removing two columns, I will have more columns than rows, and the matrix is then over-constrained, is it not? So I must also remove the rows that corresponds to F at the two supports? Thanks again!

 

[Denial]

The "quick and dirty" way to do this is to add a thumping big number to the corresponding terms on the leading diagonal of the overall stiffness matrix. This corresponds mathematically to the physical action of adding a very stiff linear spring at your support locations. How big is "thumping"? As a guide, scan the matrix's leading diagonal for its largest value, and use (say) 10⁸ times that value.

The matrix is now non-singular and can be solved. The actual reaction at your supported nodes is equal to the (very small) deflection at the node multiplied by the (very large) thumping stiffness value.

If you wish one of your load cases to include a defined displacement at a supported node, then you convert that to a force whose magnitude would cause the required displacement in the thumping spring.

This approach does run the risk of introducing numerical errors, so: (1) store your numbers with the highest available numerical precision; (2) do not use your program on problems with a large number of degrees of freedom; and (3) always check your results with an even larger setting on your scepticism-o-meter than you would usually use. In particular, check that your calculated reactions are in good equilibrium with your applied loads.

The more rigorous approach is to rearrange the entire equation system so that it is partioned between those degrees of freedom that are free and those that are constrained. You will find this described in numerous (advanced) books on strucural analysis.

Good luck.

 

[bkal]

If I understand correctly what you are trying to do, you need to remove both relevant rows and columns of your global stiffness matrix - it will then be square. However, if you do it in that way you will not get your reactions straight out of the solution. To do that, you should use an approach suggested by Denial.

 

[RolandW]

bkal, that's just what I'm after. The reaction forces does not matter really, it's just the deflections. I will try to set the deflection to zero (but not the rotation) at the supports and see how good the result becomes. Thanks a lot guys!

 

[rb1957]

at least you can check your results with a hand calc (as the problem is very simple).

 

[rb1957]

thinking (sorry) about johnhors response above, isn't this problem 1D ? ... aren't the rotation and displacement related (derivative/integral) to one another ?

when i first read the post i thought 1D ??? ... but then i visualised drawing a line to represent the beam (a la hand calc) and thought 1D was appropiate.