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In the process of deriving the stress tensor of a cube which is shown in the picture below, it is concluded from the torque equilibrium equation that $\sigma_{ij}$ is equal to $\sigma_{ji}$. I am trying to prove the same thing, but for a cylinder instead of a cube. Also, the cylinder is only under radial and longitudinal stress. This is why I am using cylindrical coordinates, but without the angle coordinate.

Cube under stress

In the picture below is the cylinder under stress. I assumed that on each surface there are stresses, unlike in the cube example where stresses are only defined at three of the six cube surfaces.

Cylinder stress

For the axis m which goes through the cylinders centroid in the radial direction, I wrote the following torque equilibrium equation:

$$ \sum{}M_Q=2\cdot\Big({dz\over2}\cdot2\pi (dr^2) \cdot \sigma_{zr}\Big) - dr \cdot2\pi drdz\cdot\sigma_{rz}=0 =>\sigma_{zr}=\sigma_{rz} \tag{1}$$

From equation $(1)$ I conclueded that $\sigma_{rz}$ is equal to $\sigma_{zr}$. However, I am not sure if my way of coming to this conclusion is correct because I can not find any simular example online or in literature.

So my questions are:

  1. Is my process of coming to the conclusion $\sigma_{rz} = \sigma_{zr}$ correct?
  2. If it is, why are cube stresses defined only at three of the six cube surfaces?

Thank you for your time.

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  • $\begingroup$ Same as $\sigma_{xy} = \sigma_{yx}$. Shear stresses need to matched in connected surfaces otherwise they will cause rotations. $\endgroup$ Aug 23, 2022 at 22:05

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The problem is that the cylinder is not an element of cylindrical coordinates. You cannot subdivide a solid with cylinders.

You need a cylindrical wedge, crudely drawn below

fig1

On this element, you place the stress components on its faces and do the force balance.

fig2

See this article for the full definition of strain in cylindrical coordinates.

The stress tensor has a similar treatment to the strain tensor from above.

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  • $\begingroup$ Hello John. Your answer helped guide me in the right direction. But I still do not understand one thing. For what point of the wedge should the torque equilibrium equation be written? Is it the one through which the axes $T_{RR}$ and $T_{ZZ}$ go through? If yes, why exactly that point? Is it because that point is the center of mass or something? Thank you for your time. $\endgroup$ Aug 26, 2022 at 18:57
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    $\begingroup$ @NikolaRistic - Equilibrium should happen at ALL internal elemental volumes. This is how you develop how the stresses form for different locations. $\endgroup$ Aug 26, 2022 at 19:16
  • $\begingroup$ Thank you very much John. This helped me :D $\endgroup$ Aug 27, 2022 at 10:21

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