AnsweredAssumed Answered

Simulation Question: As I refine the mesh, max displacement decreases but stress increases

Question asked by Ronan Timmins on Aug 21, 2014
Latest reply on Aug 23, 2014 by Pablo Jorge

Like • Show 0 Likes0
Comment • 6

I have been running a study on a trailer coupling ball.

For the first run i used the default mesh density (not a draft quality mesh) ;

Results: Deformation is as expected.

Max stress (node) = 729 N/mm^2

Max stress (element) = 582 N/mm^2

Max displacement = 1.605 mm

So I set a mesh control to refine the mesh in the area of interest. I increased the mesh density in that area until the difference between Node and Element stress values were within 10%.

Results:

1. Displacement

2. Node stress

3. Element stress

Can anybody explain how stress has increased if displacement decreased? This trend was seen gradually in the results as I refined the mesh.

Also, the max stress calculated is roughly 40% higher than the fatigue limit, But the component has passed a 2 million cycle physical test at this loading.

Any help would be greatly appreciated.

No one else has this question

Outcomes

6 Replies

Shaun Densberger Aug 21, 2014 2:48 PM
Mark Correct
Correct Answer
Can you post the model itself so that I can take a look at it? The only thing I can think of off of the top of my head is that there might be a highly distorted element in that area (due to CAD geometry or something else) that is causing wildly inaccurate results.

That being said, it's very odd that your displacement deceased while your stress increased. It's also weird that your displacement decreased as you increased the number of DOF associated with the model; it's typically the other way around.

Regarding the discrepancy between your FEA model and your physical test results, are you sure that your model properties (loads, constraints, material properties, analysis type, etc.) between the two are very similar?
Like • Show 1 Like1
Actions
Wayne Matus Aug 21, 2014 4:19 PM
Mark Correct
Correct Answer
The finer mesh is better capturing the geometry just under the ball that has the highest impact on deflections. Think about a 1" diameter round bar cantilever beam fixed at one end and shear force on the other. The finer mesh will better capture the true cross-sectional area than the course mesh. The finer mesh also better captures any stress concentrations. Hence less displacement and higher stress.

It looks like you have a uniform load on the ball. Most likely that is the reason for inaccuracy in results. In real world, it is not loaded like that.
Like • Show 1 Like1
Actions
- Shaun Densberger @ Wayne Matus on Aug 21, 2014 5:09 PM
  Mark Correct
  Correct Answer
  "Think about a 1" diameter round bar cantilever beam fixed at one end and shear force on the other. The finer mesh will better capture the true cross-sectional area than the course mesh."
  
  This would only result in a meaningful difference if the order of the element was low enough such that it couldn't represent the section of the curve that it divides. Since Ronan stated that he was using a non-draft quality mesh, the addition of the mid-side node should be more than enough to capture the curves of the geometry to remove any meaningful error due to geometry misrepresentation.
  
  "The finer mesh also better captures any stress concentrations. Hence less displacement and higher stress."
  
  The first part is true, but not the second part. I think I get your reasoning: finer mesh better captures geometry (cross-sectional area) => more area = stiffer geometry = lower displacements, and finer mesh = higher stresses. However, this line of thinking isn't correct because the decrease in stiffness do to inaccurate geometry representation does not compare to the increase in stiffness from too few elements. This can easily be confirmed by setting up a model like you described (a cantilever rod) and looking at how the maximum stress and displacement change as your progressively refine a draft quality mesh (I use draft quality because it will give the worse geometric representation of the circular cross-section of the rod).
  
  We can see with the lowest mesh refinement level that the rod appears almost hexagonal:
  
  And with the finest mesh that it appears very close to circular:
  
  But yet we see both the maximum displacement and stress increase as we refine the mesh (and get a better approximation of the cross-sectional area):
  
  In fact, if we think about the assumptions in a linear static analysis, then it becomes apparent that the stress should increase if the displacement increases. A linear static analysis assumes Hooke's Law:
  
  Stress = Young's Modulus * Strain
  
  and
  
  Strain = d(Displacement)
  
  where d() is the derivative.
  
  In other words, larger displacement = larger strains = larger stresses. It should be noted that taking the derivative of the displacement fields to get the strain field introduces error, hence displacement results being more accurate than strain/stress results for a given mesh.
  
  "It looks like you have a uniform load on the ball. Most likely that is the reason for inaccuracy in results. In real world, it is not loaded like that."
  
  This is possible if the statically equivalent point of the load is offset further out on the uniform load when compared to the non-uniform load.
  Like • Show 1 Like1
  Actions
  - Wayne Matus @ Shaun Densberger on Aug 21, 2014 5:17 PM
    Mark Correct
    Correct Answer
    I just ran a test and saw the same results. I was thinking that there might have been a highly localized stress concentration that was not affecting the overall displacement that much but finer mesh was better capturing the stress concentration..
    Like • Show 1 Like1
    Actions
Jared Conway Aug 21, 2014 8:47 PM
Mark Correct
Correct Answer
if the stress is in and around that region you're showing, i am not surprised it is bouncing around. my experience is that elements that are being "torn" show non convergent stress. take some values from different locations in the model, especially away from that point and let us know if they converge. also how well is the displacement converging as you improve the mesh both locally and globally?
Like • Show 0 Likes0
Actions
Pablo Jorge Jan 30, 2015 2:02 PM
Mark Correct
Correct Answer
Refined mesh means small nodes. Small nodes are able to see high stress.

Big node is like an average of several small nodes.

An average is always smaller than a peak value.

Example:
4 small nodes, with stress values 2, 2, 7 and 2. Max stress: 7
4 times biger node. Stress= ( 2+2+7+2)/4=3,25
Probably the FEM mathematics are not so simple, but it is the idea.

Have you ever work with stain gages? It is the same reasoning.

Regarding the displacement, ok, just suppose the refined mesh is the good value.
Like • Show 1 Like1
Actions

Simulation Question: As I refine the mesh, max displacement decreases but stress increases

Outcomes

Related Content

Recommended Content