Changing active forces for enforced deformations and vice versa

I have seen many mistakes where people wrongly changed enforced deformations with active forces. In some cases, this is a simple task, which gives beginners the feeling that they can always do it easily. This, in turn, leads to various mistakes in analysis.

I think it is worth discussing this, to point out common mistakes in FEA models.

How can you load your model

In structural design the 2 most common possibilities for loading models are:

  • Forces/moments (and pressures etc.)
  • enforced deformations (displacements and rotations)

In some simple cases, it is irrelevant if you use forces of enforced deformation. In such cases, outcomes (stresses etc.) will be the same, regardless of load. It is really easy to prove with a simple beam, loaded with concentrated force and enforced deformation:

Simple beam with active forces and enforced deformations

If the enforced deformation delta on the middle is equal to deformation under force F, then the outcomes of the analysis will be the same. Since this is a simple example it is also simple to prove it’s true.

Step 1: Static schematic

I have created 2 beams with the same cross-section and length. One has an additional support in the middle with enforced deformation downward. Other have a concentrated force (also in the middle).

Simple beam with active forces and enforced deformations: static

Step 2: Static calculation

Note that deformation under force matches the value of enforced deformation. Also, reaction force in the middle support (green arrow) match the value of a concentrated force in the other beam.

Simple beam with active forces and enforced deformations: outcomes 1

Step 3: Outcomes comparison

Outcomes (in this case bending moment) are compared to check if everything is in order:

Simple beam with active forces and enforced deformations: outcomes 2

As you can see the outcomes are the same.

Can I change enforced deformation for active loads?

This is a great question and one that is so easy to misunderstand.

Yes, they are… but sometimes it is so difficult to find the correct pattern to do so, that it makes little sense to try.

Think about a beam loaded with uniform load (or something loaded with pressure).

Active forces and enforced deformations in uniformly loaded beam

You can, of course, find a set of supports that when displaced with enforced deformation will give you the “correct” shape of the beam. Then reaction forces in the supports will be the equivalent of loads as in the previous example and outcomes will be the same.

However, there is so much work in searching, that it rarely make any sense to even try to find equivalent enforced deformation to any continuous load.

It is best to use “real” loads

The best advice I can give you: use the correct load type. In some cases, you will be able to find the equivalent force or displacement, but usually, there is no reason to do so. The only situation I think where this could be useful is convergence problem with model loaded with forces. Switching load type to displacement may help with the convergence.

This makes it important to understand, when you are dealing with enforced deformations and when with loads.

My experience is, that loads are usually active forces. Sometimes this is not true, like in metal forming or laboratory tests. Let’s analyze the possibility of “real” enforced deformations.

Think about a shell loaded in a vertical direction in a testing machine. Setup is as shown below (small icons show an isometry, so it is easier to see the shell):

Active forces and enforced deformations in shells

With certain simplification (everything is more difficult than it seems!) we can say that in the case of the left model it is easy to find the equivalent load. Since the plate of laboratory machine is rigid and moves downward, thanks to uniform support at the bottom stress will be evenly distributed. This means that uniform load on the edge of the shell will do the trick.

In the second example (on the right) this is not the case. When you use machine enforcing deformations the top edge of the shell remains flat. However, if you would use a uniformly distributed force along circumference the top edge will not be flat. This is caused by local supports. Deformation difference will be minimal, but it will be there. Below you see a stress distribution in the case of uniform enforced displacement and uniform loads on the circumference:Active forces and enforced deformations - shell example

The differences in most places are small, but those can be crucial in many cases.

To avoid such problems you can, of course, load your model with displacement first, read the reaction forces (they won’t be uniform!) and apply them as a load. But this means you will have to solve the problem twice!

Bringing this all together

The difference between enforced deformations and active forces is slippery. Sometimes it is easy to find an equivalent enforced deformation (or equivalent load). We did that in the first example. However, in many cases, it is not as simple as was seen in the other cases I presented.

Load vs Enforced displacement

  • In case of concentrated forces and enforced displacement on point supports it is easy to find the equivalent load/displacement.
  • For uniform loads (and pressures), finding equivalent enforced deformation will be time consuming.
  • With enforced deformations on certain areas/along lines it is equally hard to find equivalent loads.
  • The easiest way to find the equivalent force is to make a model with enforced deformation and read reaction forces (and use those as loads in the second model!).
  • The same can be done in search for equivalent enforced deformation.
  • The fact that the load is uniform on area / along a line does not mean that the enforced deformation should be constant on this area / along this line. Most likely this is FALSE!
  • It is always the easiest to use the loading just as it appears in nature. This way there is no need to seek for the equivalent forced or displacements.
  • If the relation between load and displacement is no longer linear, equivalent displacements/forces loose sense at all! This is suitable approach only to linear analysis.
  • It is theoretically possible to find an equivalent deformations for nonlinear analysis. However the amount of work would be unreasonable! Also such approach would require you to know the correct solution at the start. Treat this more or less as a “theoretical” possibility.

Takeaway

To summarize everything is one short paragraph I think it is worth saying that:

In some simple cases it is possible to change active loads with enforced displacements. Most of the time however such actions require a lot of work, and are simply unreasonable. It is best to use the type of loading that is expected.

There is also one conclusion I already pointed out, but it is so important I will do it again:

The fact that the load is uniform does not mean that the enforced displacement should be uniform as well. The same goes for uniform enforced displacements which very rarely can be change for uniform loads!

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4 Comments

  1. a May 24, 2017 at 8:06 am - Reply

    it is possible to apply exactly the displacements equivalent to a load,but the coupling terms of the stiffness matrix must be considered also

    • Łukasz Skotny May 24, 2017 at 12:32 pm - Reply

      Hey A 🙂

      Thank you for the comment!

      Have a great day
      Ł

  2. Angus Ramsay May 25, 2017 at 12:03 am - Reply

    Interestingly Lukasz, convergence of models with displacement loading will converge from the opposite direction as to those with force loading, i.e., one converges from above the theoretical solution whereas the other from below – you could try this for the simply supported beam example…

    • Łukasz Skotny May 25, 2017 at 4:21 am - Reply

      Hey Angus!

      This is a really great point! I will look into it for sure 🙂 Thank you for pointing this out!

      Have a great day!
      Ł

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