Don't apply concentrated force load to surfaces

Tips & Tricks
1

It’s definitely the most common mistake made by new users (especially those who don’t have experience with Abaqus) so let me explain this in detail.

TL;DR: Never apply a concentrated force load to a surface unless you are sure that you know how it works and you have a good reason to do it (it’s rarely needed though).

Why ? Let me show you. Here’s a simple cantilever beam. Assume that we want the top surface to be loaded with a force of 200 N. Let’s use the concentrated force load for that:

concentrated force
concentrated force1612×738 116 KB

Now let’s check the results:

deformation
deformation892×692 39.5 KB

This is true scale deformation. And the maximum stress is around 3.6e4 MPa. A bit too much for a 10x10x100 mm steel cantilever beam loaded with 200 N. It should be around 0.14 mm and 60 MPa according to hand calcs (recommended whenever possible).

So what happened ? Let’s check the total reaction force (also recommended). It happens to be 117200 N instead of 200 N. Why this value ? If you divide it by 200, you will get 586. How many nodes are there on the surface of the beam where load was applied ? 593… Does it ring a bell ? Let’s check the input file. There we can see:

*Cload
Internal_Selection-1_Concentrated_Force-1, 2, -200

The Internal_Selection-1_Concentrated_Force-1 entry is a name of an internal node set. And *Cload means that each node in that node set receives the specified force.

Now if we use the surface traction load instead:

trac
trac1625×750 100 KB

we will get the following results:

  • max displacement: 0.143 mm
  • max stress: 60.05 MPa
  • total reaction force: 198.9683 N

Much better, right ? And it’s in agreement with hand calcs. Case solved :slight_smile:

P.S. Surface traction also uses the *Cload keyword but it divides the load to apply it uniformly to all nodes:

*Cload
1413, 2, -5.05281853E-001
1342, 2, -5.03713604E-001
1414, 2, -5.04257172E-001
1392, 2, -5.04347942E-001
1391, 2, -5.04430560E-001
1316, 2, -5.04075864E-001
1214, 2, -5.01590558E-001
1309, 2, -5.02944832E-001
1273, 2, -5.01578748E-001
1104, 2, -5.02055236E-001
1103, 2, -2.50253643E-001
...

The first entry is the node number, followed by a degree of freedom and force magnitude. If you sum it all up, it should give 200 N.

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