Poisson Ratio -- Finding a corresponding analytical solution for the strain

  • #1
3
0
Summary:
Find a corresponding analytical solution for the strain in the x-direction caused by the poisson ratio.
Hi, I ran into problems using the poisson ratio.
For a FE simulation I created a simple 2D 1mm x 1mm block, and prescribed a 0.1 mm displacement at the top edge.
Furthermore, the bottom edge is constraint in the y-dir, and the left edge in the x-dir.
The material parameters are E = 100, and v (poisson ratio) = 0.3.
Note the simulation is executed for a plane strain assumption!

To verify the results I would like to solve the analytical solution for this problem.
This is quite simple tbh, I use the 2D strain-stress relations for the plane strain problem.
However, the simulation shows that the strain_yy = -0.1 (as expected) but the strain_xx = 0.04285714.

I really cannot figure out the analytical solution for the strain_xx, I would expect this to be 0.03 (strain_xx = -v * strain_yy).

I know that the 0.0428.. value should be the correct one, because I tried the simulation in different simulation software.
Hopefully someone can explain me how to get this value analytically?

Thanks in advance!

Situation_sketch.PNG Displacement_x.PNG Strain_xx.PNG
 
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Answers and Replies

  • #2
224
102
Just by looking at the numbers: strain_xx = -v * strain_yy / (1-v) = 0.0485714....
 
  • #3
21,156
4,672
In the plain strain situation, the strain in the 3rd direction is zero. This results in a larger strain in the transverse direction. Try the calculation with plain stress and see what you get. For plain strain, you get $$\epsilon_y=-\frac{\nu}{(1-\nu)}\epsilon_x=-0.04286$$
 
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  • #4
3
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In the plain strain situation, the strain in the 3rd direction is zero. This results in a larger strain in the transverse direction. Try the calculation with plain stress and see what you get. For plain strain, you get $$\epsilon_y=-\frac{\nu}{(1-\nu)}\epsilon_x=-0.04286$$
Thank you very much for your answer, that's the relation I was looking for. (Note, there is a small typo: $$\epsilon_x$$ and $$\epsilon_y$$ should be the other way around)

To make the story complete, I expect this relation is derived by:
$$\sigma_x = \frac{E}{(1+\nu)(1-2\nu)} [(1-\nu) \epsilon_x + \nu \epsilon_y]$$

Because the block is not constraint in the right x-dir, it can expand freely -> conclusion: $$\sigma_x = 0$$
Therefore, we can write the previous eq. for the strain in the x-dir as :
$$\epsilon_x = -\frac{\nu}{1-\nu} \epsilon_y $$

Correct me if I am wrong.
 
  • #5
21,156
4,672
Thank you very much for your answer, that's the relation I was looking for. (Note, there is a small typo: $$\epsilon_x$$ and $$\epsilon_y$$ should be the other way around)

To make the story complete, I expect this relation is derived by:
$$\sigma_x = \frac{E}{(1+\nu)(1-2\nu)} [(1-\nu) \epsilon_x + \nu \epsilon_y]$$

Because the block is not constraint in the right x-dir, it can expand freely -> conclusion: $$\sigma_x = 0$$
Therefore, we can write the previous eq. for the strain in the x-dir as :
$$\epsilon_x = -\frac{\nu}{1-\nu} \epsilon_y $$

Correct me if I am wrong.
That's not exactly the way I did it, but it is correct. I hope you also noticed that ##\sigma_y## is not equal to ##E\epsilon_y##, but rather $$\frac{E\epsilon_y}{(1-\nu^2)}$$
 
  • #6
3
0
That's not exactly the way I did it, but it is correct. I hope you also noticed that ##\sigma_y## is not equal to ##E\epsilon_y##, but rather $$\frac{E\epsilon_y}{(1-\nu^2)}$$
Thank you for your help! Yes I've noticed this.
 

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