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When the force vector is parallel to the surface, the stress is called

\[ \sigma = {F_\text{normal} \over A} \qquad \text{and} \qquad \tau = {F_\text{parallel} \over A} \]

Of course, things can get complicated in nonlinear problems with large deformations (and rotations) because the final deformed area may be different from the initial area, among other things. We will ignore all these complexities for now and assume that before-and-after differences are negligible. So there will be no need to specify whether the force and area are for undeformed or deformed conditions.

\[ \sigma_{xx} = {F_x \over A_x} \qquad \text{and} \qquad \tau_{xy} = {F_y \over A_x} \]

Note how the two subscripts on the stress variables match those on the force and area components with one subscript coming from each.

Alternately, one could (virtually) cut the object horizontally to produce a surface with an outward normal in the y-direction. This leads to

\[ \sigma_{yy} = {F_y \over A_y} \qquad \text{and} \qquad \tau_{yx} = {F_x \over A_y} \]

If a numerical example were worked out, one would notice an amazing result. It is that \(\tau_{xy} = \tau_{yx}\). This will always be true in order to maintain rotational equilibrium. This is discussed in more detail next.

First, let's look at the normal stresses, \(\sigma_{xx}\) and \(\sigma_{yy}\). Note how the

The y-normal stresses, \(\sigma_{yy}\), are also present on two surfaces, top and bottom, in order to maintain vertical equilibrium. Like \(\sigma_{xx}\), \(\sigma_{yy}\) is also drawn to represent tension, which is positive.

The difference between the left and right pictures is that \(\tau_{yx}\) in the left figure is replaced by \(\tau_{xy}\) in the right figure. The left figure contains two shear stress values, \(\tau_{xy}\), which rotates the square counter-clockwise, and \(\tau_{yx}\), which rotates the square clockwise. But if the two shear values are not equal, then the square will not be in rotational equilibrium. The only way to maintain rotational equilibrium is for \(\tau_{xy}\) to be equal \(\tau_{yx}\). So there is no need to have two separate variables. The right figure contains only one, \(\tau_{xy}\).

\[ \boldsymbol{\sigma} = \left[ \matrix{ \sigma_{11} & \sigma_{12} \\ \sigma_{21} & \sigma_{22} } \right] = \left[ \matrix{ \sigma_{xx} & \sigma_{xy} \\ \sigma_{yx} & \sigma_{yy} } \right] = \left[ \matrix{ \sigma_{xx} & \tau_{xy} \\ \tau_{yx} & \sigma_{yy} } \right] \]

But since \(\tau_{xy} = \tau_{yx}\), all the tensors can also be written as

\[ \boldsymbol{\sigma} = \left[ \matrix{ \sigma_{11} & \sigma_{12} \\ \sigma_{12} & \sigma_{22} } \right] = \left[ \matrix{ \sigma_{xx} & \sigma_{xy} \\ \sigma_{xy} & \sigma_{yy} } \right] = \left[ \matrix{ \sigma_{xx} & \tau_{xy} \\ \tau_{xy} & \sigma_{yy} } \right] \]

Setting \(\tau_{xy} = \tau_{yx}\) has the effect of making (requiring in fact) the stress tensors symmetric.

\[ \boldsymbol{\sigma} = \left[ \matrix{ \sigma_{11} & \sigma_{12} & \sigma_{13} \\ \sigma_{21} & \sigma_{22} & \sigma_{23} \\ \sigma_{31} & \sigma_{32} & \sigma_{33} } \right] = \left[ \matrix{ \sigma_{xx} & \sigma_{xy} & \sigma_{xz} \\ \sigma_{yx} & \sigma_{yy} & \sigma_{yz} \\ \sigma_{zx} & \sigma_{zy} & \sigma_{zz} } \right] = \left[ \matrix{ \sigma_{xx} & \tau_{xy} & \tau_{xz} \\ \tau_{yx} & \sigma_{yy} & \tau_{yz} \\ \tau_{zx} & \tau_{zy} & \sigma_{zz} } \right] \]

But rotational equilibrium requires that \(\tau_{xy} = \tau_{yx}\), \(\tau_{xz} = \tau_{zx}\), and \(\tau_{yz} = \tau_{zy}\). This also produces symmetric tensors.

\[ \boldsymbol{\sigma} = \left[ \matrix{ \sigma_{11} & \sigma_{12} & \sigma_{13} \\ \sigma_{12} & \sigma_{22} & \sigma_{23} \\ \sigma_{13} & \sigma_{23} & \sigma_{33} } \right] = \left[ \matrix{ \sigma_{xx} & \sigma_{xy} & \sigma_{xz} \\ \sigma_{xy} & \sigma_{yy} & \sigma_{yz} \\ \sigma_{xz} & \sigma_{yz} & \sigma_{zz} } \right] = \left[ \matrix{ \sigma_{xx} & \tau_{xy} & \tau_{xz} \\ \tau_{xy} & \sigma_{yy} & \tau_{yz} \\ \tau_{xz} & \tau_{yz} & \sigma_{zz} } \right] \]