# A Stress-strain; strain-displacement in 2-D; uniqueness

1. Mar 24, 2017

### pyroknife

I am working on a 2-D planar problem in the x-y direction, dealing with stresses, strain, displacements. Under the linear elastic relation and after substitution I can write the following:

$\begin{bmatrix} \sigma_{xx} & \sigma_{xy} \\ \sigma_{xy} & \sigma_{yy} \end{bmatrix} = \mu \begin{bmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{bmatrix} + \mu \begin{bmatrix} \frac{\partial u}{\partial x} & \frac{\partial v}{\partial x}\\ \frac{\partial u}{\partial y} & \frac{\partial v}{\partial y} \end{bmatrix} + \lambda\begin{bmatrix} \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} & 0\\ 0 & \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} \end{bmatrix}$

I need to obtain the unknowns, which are the 4 displacement gradients
$\frac{du}{dx}; \frac{du}{dy}; \frac{dv}{dx}; \frac{dv}{dy}$

However, since the stress matrix is symmetrical, I only have 3 knowns. In structural problems, where does this 4th condition come from?

2. Mar 24, 2017

### Staff: Mentor

In structural problems, you don't solve for the displacement gradients. You solve for the displacements.

3. Mar 24, 2017

### pyroknife

This is for a finite difference numerical solution. I need the displacement gradient in order to compute the displacement. The stress is being specified at a boundary face (and each side of the face is a cell). SO I need to compute the gradient at that face and then knowing the gradient and one of the cell values, I can compute the other cell's value.

But either way, doesn't the solving require one more condition to make the system unique?

4. Mar 25, 2017

### Staff: Mentor

Maybe you could tell us the specific problem you're solving?

5. Mar 25, 2017

### pyroknife

I'm trying to simulate linear elastic deformation on a square flat plate using the finite difference method. The plate is of size 1x1x0.1. So it has 2 square faces and 4 long rectangular faces. The 2 square sides are oriented in the x-y direction. Assume the 0.1 dimension is in the z direction. 2 of the square and 2 of the rectangular faces are fixed with 0 displacement boundary condition. Then let's say there's a compressive load applied to the plate on 2 of the rectangular faces; 1 in the -x direction and one in the -y direction.

This compressive load acts as a boundary condition on those 2 rectangular sides. The square plate is discretized into 10 cells of size 0.1x0.1x0.1. I am solving for displacements.

Since this compressive load is applied to the face as a stress BC; I need solve the system in the OP to obtain the displacement gradients at the face, which I will use to compute the displacements in the ghost cells, and the ghost cell displacements are in turn used to compute the cell centroid gradient of each cell of the real domain.

6. Mar 25, 2017

### pyroknife

So for example, let's consider the load applied in the -x direction's face.
$\begin{bmatrix} -1 & 0 \\ 0 & \ 0 \end{bmatrix} = \mu \begin{bmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{bmatrix} + \mu \begin{bmatrix} \frac{\partial u}{\partial x} & \frac{\partial v}{\partial x}\\ \frac{\partial u}{\partial y} & \frac{\partial v}{\partial y} \end{bmatrix} + \lambda\begin{bmatrix} \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} & 0\\ 0 & \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} \end{bmatrix}$

Due to the symmetry of the stress tensor, this is NOT a linear independent system of equations. We have 3 knowns (3 known stresses) and 4 unknowns (4 unknown displacements), which would yield a non-unique solution for the displacement gradients. So I was thinking that I need to impose an addition condition such that a unique solution can be obtained.

7. Mar 25, 2017

### Staff: Mentor

So the problem is this:
:
with the arrows distributed uniformly over each of their rectangular faces, correct? What are the boundary conditions on the other two rectangular faces?

8. Mar 25, 2017

### pyroknife

Yes! I actually just drew one out on paper. The boundary conditions on the other 2 rectangular faces AND the square faces are fixed with 0 displacement in all directions.

It's not a very realistic problem.

9. Mar 25, 2017

### pyroknife

10. Mar 25, 2017

### Staff: Mentor

OK. Let's assign coordinates. The square faces are at z = +h/2 and z = - h/2, where h = 0.1 m. The rectangular faces are at x = 0, x = L, y = 0, and y = L where L = 1 meter. So your are saying that,

at x = 0, u = v = 0,
at x = L, v = 0,
at y = 0, u = v = 0,
at y = L, u = 0,
at z = + h/2, u = v = 0
at z = - h/2, u = v = 0

Do you realize that the last two conditions require that u and v must be functions of z?

11. Mar 25, 2017

### Staff: Mentor

12. Mar 25, 2017

### pyroknife

Regarding the last question: hmmm, so I'm really only trying to simulate a 2-D problem for this case. If u and v must be functions of z, then everything in the OP turns into a 3-D and all the matrices become 3x3. Maybe I can make the assumption that the thickness in the z-direction is infinite, and not 0.1m, so that I can constrain this problem to 2-D physics only.

As for the BCs, 4/6 of those are what I'm imposing. Which are
at x = 0, u = v = 0,
at y = 0, u = v = 0,
at z = + h/2, u = v = 0
at z = - h/2, u = v = 0

However, at x=L and y = L, I plan to impose only the stress boundary conditions, and solve for all the displacements. So u&v are unknowns for both of these faces.

13. Mar 25, 2017

### Staff: Mentor

OK. I think what you want is the following:

at z = +h/2, $\sigma_{xz} = \sigma_{yz} = 0$, w = 0
at z = - h/2, $\sigma_{xz} = \sigma_{yz} = 0$, w = 0

These will be equivalent to the thickness in the z direction being infinite

Also,

at x = 0, u = v = 0,
at y = 0, u = v = 0,
at x = L, $\sigma_{xx}=-\sigma_1,\ \ \sigma_{xy}=0$
at y = L, $\sigma_{yy}=-\sigma_2, \ \ \sigma_{xy}=0$

Is this what you had in mind?

14. Mar 25, 2017

### pyroknife

Yes! Exactly.

15. Mar 25, 2017

### Staff: Mentor

Gotta go now. Be back later.

By the way, at x = 0 and y = 0, are you sure you wouldn't rather have

at x = 0, u = 0, $\sigma_{xy}=0$
at y = 0, v = 0, $\sigma_{xy}=0$

16. Mar 25, 2017

### pyroknife

hmmm. Trying to figure out what this means. So instead of "fixing" both displacements to be 0, we fix one displacement and have the shear stress be 0.
Is this equivalent to constraining both u&v AND more? Because, e.g., v displacement on the x=0 face can only be caused by shear, and without shear, I expect v=0.

I'm going to be back in a bit too. Thanks for all the help.

17. Mar 25, 2017

### Staff: Mentor

No. Zero shear stress is less constraining than zero tangential displacement. The points on the boundary are free to slide along the boundary in order to relieve any shear stress. It's like the boundary being a shower curtain rod, and the rings of the shower curtain being allowed to slide freely along the rod. At DuPont, we used to call this a "shower curtain boundary condition." So, if the shear stress is zero along the boundary, v is not equal to zero.

18. Mar 25, 2017

### pyroknife

Ah I see that is interesting. So how would you obtain v along the boundary if you specify the shear stress? I guess this is the same issue as the other 2 compression boundary conditions.

19. Mar 25, 2017

### Staff: Mentor

We'll get to all that. But first, which of the two boundary conditions do you prefer using.

20. Mar 25, 2017

### pyroknife

Hmm Let me stick with my original one with u=v=0 for now. Setting shear stress=0 as opposed to both displacements=0 makes the computation harder. After figuring out the compressive BC computations, I would think that I would be able to solve the shear=0 BC.