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

  1. Mar 24, 2017 #1
    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. jcsd
  3. Mar 24, 2017 #2
    In structural problems, you don't solve for the displacement gradients. You solve for the displacements.
     
  4. Mar 24, 2017 #3
    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?
     
  5. Mar 25, 2017 #4
    Maybe you could tell us the specific problem you're solving?
     
  6. Mar 25, 2017 #5
    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.
     
  7. Mar 25, 2017 #6
    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.
     
  8. Mar 25, 2017 #7
    So the problem is this:
    : Stress square.PNG
    with the arrows distributed uniformly over each of their rectangular faces, correct? What are the boundary conditions on the other two rectangular faces?
     
  9. Mar 25, 2017 #8
    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.
     
  10. Mar 25, 2017 #9
  11. Mar 25, 2017 #10
    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?
     
  12. Mar 25, 2017 #11
  13. Mar 25, 2017 #12
    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.
     
  14. Mar 25, 2017 #13
    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?
     
  15. Mar 25, 2017 #14
    Yes! Exactly.
     
  16. Mar 25, 2017 #15
    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##
     
  17. Mar 25, 2017 #16
    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.
     
  18. Mar 25, 2017 #17
    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.
     
  19. Mar 25, 2017 #18
    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.
     
  20. Mar 25, 2017 #19
    We'll get to all that. But first, which of the two boundary conditions do you prefer using.
     
  21. Mar 25, 2017 #20
    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.
     
  22. Mar 25, 2017 #21
    OK. So the stresses are:
    $$\sigma_{xx}=(2\mu +\lambda)\frac{\partial u}{\partial x}+\lambda \frac{\partial v}{\partial y}$$
    $$\sigma_{yy}=(2\mu +\lambda)\frac{\partial v}{\partial y}+\lambda \frac{\partial u}{\partial x}$$
    $$\sigma_{xy}=\mu\left(\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x}\right)$$

    The stress equilibrium equations are given by:
    $$\frac{\partial \sigma_{xx}}{\partial x}+\frac{\partial \sigma_{xy}}{\partial y}=0$$
    $$\frac{\partial \sigma_{xy}}{\partial x}+\frac{\partial \sigma_{yy}}{\partial y}=0$$

    Now, what do you get when you substitute the equations for the stresses into the stress equilibrium equations?
     
  23. Mar 25, 2017 #22
    After substitution, I get the following:
    $$ (2\mu+\lambda)\frac{\partial^2 u}{\partial x^2} +(\mu+\lambda) \frac{\partial^2 v}{\partial x\partial y} + \mu \frac{\partial^2 u}{\partial y^2} = 0$$
    $$ (2\mu+\lambda)\frac{\partial^2 v}{\partial y^2} +(\mu+\lambda) \frac{\partial^2 u}{\partial x\partial y} + \mu \frac{\partial^2 v}{\partial x^2} = 0$$
     
    Last edited by a moderator: Mar 26, 2017
  24. Mar 25, 2017 #23
    Excellent. So now you have two partial differential equations in the 2 unknown displacements, u and v. These are the equations that you can solve numerically. Next, work out the boundary conditions in terms of u, v, and their derivatives. Let's see what you come up with.

    Incidentally, for the case in which the shear stress is zero on all the boundaries, you get a very simple analytic solution. See if you can derive it. It might be a good test case for your numerical method.
     
    Last edited: Mar 26, 2017
  25. Mar 25, 2017 #24
    So for now, let me just consider the x=L boundary condition. Aren't the equations that I need on post #6 on the previous page? However that post is written in terms of displacement gradients.
     
  26. Mar 25, 2017 #25
    The x=L BC gives the following:
    $$
    \begin{bmatrix}
    -1 & 0 \\
    0 & \ 0
    \end{bmatrix} = \mu
    \begin{bmatrix}
    \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\
    \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}
    \end{bmatrix}
    + \mu
    \begin{bmatrix}
    \frac{\partial u}{\partial x} & \frac{\partial v}{\partial x}\\
    \frac{\partial u}{\partial y} & \frac{\partial v}{\partial y}
    \end{bmatrix}
    + \lambda\begin{bmatrix}
    \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} & 0\\
    0 & \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y}
    \end{bmatrix}
    $$

    Or
    $$
    (2\mu+\lambda)\frac{\partial u}{\partial x} + \lambda \frac{\partial u}{\partial y} = -1 \\
    \mu (\frac{\partial u}{\partial y} + \frac{\partial v}{\partial x} = 0 \\
    (2\mu+\lambda) \frac{\partial v}{\partial y} + \lambda \frac{\partial u}{\partial x} = 0 \\
    $$

    But these are only 3 equations with 4 unknowns.

    Using a central difference, I get the following:
    $$
    \frac{\partial u}{\partial x} = \frac{u_{i+1,j} - u_{i-1,j}}{2\Delta x}
    $$

    where i&j are the indices for the x&y directions, respectively.
     
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