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Trouble understanding stress and strain problems

  1. Oct 28, 2012 #1
    Hi everyone,

    so I'm trying to understand how to use the (Cauchy I think?) stress tensor. The definition I have is that the element [itex]\sigma_{\alpha\beta}[/itex] of the stress tensor σ is the force per unit area in the α direction on a surface that's normal to the β direction.

    I also have a definition of the strain tensor ε, where [itex]\epsilon_{\alpha\beta} = (M_{\alpha\beta} + M_{\beta\alpha})/2[/itex], and [itex]M_{\alpha\beta} = \frac{\partial u_\alpha}{\partial r_\beta}[/itex], where u is the displacement of the solid at position r.

    We also have the relation between stress and strain: [itex]\sigma_{\alpha\beta} = \sum\limits_{\gamma \delta} C_{\alpha \beta \gamma\delta} \epsilon_{\gamma\delta}[/itex]

    This all makes a little sense to me, but it's kind of convoluted and I'm having trouble using it. For example, wikipedia seems to say that the stress tensor can have different values at different points in the solid, generally. But let's say we squeeze two opposite sides of a cube. Then, the stress tensor is the same at both of those sides, right? If the cube is otherwise uniform, is the stress tensor the same everywhere in the cube?

    I found an example online but some of the stuff in it is unexplained to me. It's a cube with side A of a cubic crystal, and the faces of the cube are the faces of the crystal planes. A force per unit area σ is applied to one side of the cube away from the cube (call this direction +z) and another σ is applied to the opposite side, in the opposite direction (so the cube is basically being stretched).

    They use the Voigt notation, and say that only [itex]\sigma_3[/itex] is nonzero, which makes sense. Then they say that only [itex]\epsilon_1, \epsilon_2, \epsilon_3[/itex] will be nonzero, and thus we get the equations (from the stress/strain relation):

    [itex]\sigma_1 = C_{11} \epsilon_1 + C_{12} \epsilon_2 + C_{13} \epsilon_3 = 0[/itex]
    [itex]\sigma_2 = C_{21} \epsilon_1 + C_{22} \epsilon_2 + C_{23} \epsilon_3 = 0[/itex]
    [itex]\sigma_3 = C_{31} \epsilon_1 + C_{32} \epsilon_2 + C_{33} \epsilon_3[/itex]

    Then, because it's a cubic crystal, we have the relations [itex]C_{11} = C_{22} = C_{33}[/itex] and [itex]C_{12} = C_{23} = C_{31}[/itex], so that lets us solve for [itex]\epsilon_1[/itex], [itex]\epsilon_2[/itex], & [itex]\epsilon_3[/itex] in terms of the C's and σ.

    Then they say that the change in the length of the crystal in the z direction is [itex]A\epsilon_3[/itex]. I know that [itex]\epsilon_{3} = M_{33} = \frac{\partial u_3}{\partial r_3}[/itex], but what exactly are we doing to find this change in length? Are we really just doing:

    [itex]\epsilon_{3} = \frac{\partial u_3}{\partial r_3} \rightarrow \partial u_3 = \epsilon_{3} \partial r_3 \rightarrow u(A) = \int \partial u_3 = \int_0 ^A \epsilon_{3} \partial r_3 = A\epsilon_3[/itex]

    ? It seems like I'm doing something wrong... I saw another problem similar to this, but one side is held stationary (while the opposite side is the same as before, still stretching). The answer comes out to be the same, that the change is length is the unstretched length times the strain element in that direction. Why? It seems like the one that's being pulled from both directions should stretch twice as much. (Unless the side being held stationary is held stationary by the same force as before, so it's actually identical...?)

    Does anyone have any good resources for reading about this? I'm having trouble learning this stuff from the few sources I have.

    Thank you!
     
  2. jcsd
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