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just as all other engineering students, I was taught in my mechanics lectures that stress tensors are always symmetrical. Back in the day I just accepted that fact, but now after some time, I wonder about the actual reason behind that. Consider the exemplary 2D stress states below: Both are capable of fulfilling equlibrium of forces and momentums and are plausbile on a macroscopic scale. Yet on an infinitesimal scale, as we all have learned, only the one on the left hand side can occur. Could anyone explain to me why that is? (sadly I'm no mathematician, so I'd be glad for an "understandable" explanantion ;)

Exemplary stress states

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  • $\begingroup$ The stress tensor is symmetric if the body is in equilibrium. In your second figure that is not the case, so it's not needed to be symmetric. $\endgroup$
    – Grego_gc
    Commented Feb 6, 2020 at 15:05
  • $\begingroup$ @Grego_gc I think the OP is saying $2τ_{yx}=σ_y$ gives equilibrium in $y$ direction. $\endgroup$
    – Bob D
    Commented Feb 6, 2020 at 15:15
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/62963/2451 , physics.stackexchange.com/q/65169/2451 and links therein. $\endgroup$
    – Qmechanic ♦
    Commented Feb 6, 2020 at 16:15
  • $\begingroup$ Per Newton's third law, all forces occur in pairs. The components of shear stresses always occur in pairs, just like normal stresses. Without any normal stresses, the rectangle should be in pure shear stress and in equilibrium. Without shear stresses the rectangle should be in pure axial stress and in equilibrium. The diagram at the left meets these criterial. The diagram on the right does not. $\endgroup$
    – Bob D
    Commented Feb 6, 2020 at 16:23
  • $\begingroup$ Surface forces and the stress tensor are connected by the relation $\sigma d\vec{s}$, I guess you are ok with that. Ok, note that on the right sketch the left and right faces of the (infinitesimal, I guess) cube have an opposite orientation so the forces on the two faces HAVE TO be opposite in sign. There is no stress tensor (not even asymmetric) that can give rise to that configuration. Those $\tau_{yx}$ just cannot point in the same direction on both sides of the cube.... not if the tensor $\sigma$ is $\approx$ constant on the infinitesimal cube. $\endgroup$
    – Ste
    Commented Jun 1, 2021 at 22:25

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The problem with the right hand diagram is that it doesn't scale properly.

Suppose this stress field was uniform over a finite sized region. Now take a rectangular element of size $\Delta x$ by $\Delta y$. You now have a direct force $\sigma_y \Delta x$ balanced by the shear forces $2 \tau_{yx} \Delta y$ and you can't eliminate $\Delta x$ and $\Delta y$ because they are both arbitrary.

In the left hand diagram, the forces balance for any shape of rectangle, not just a square.

Of course this simple argument only shows that your "wrong answer" is wrong, not the that "right answer" is the only right answer. A more general idea is to consider rotating the small element through an arbitrary angle, and considering that the forces on the edges must be. If you remember what you learned about principal stresses, you can choose an orientation for the element where all the shear stresses are zero, for example.

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  • $\begingroup$ Thank you for your answers. Newton’s third law doesn't seem sufficient to me at first glance. After all, the load state of the right body can exist on a macroscopic scale although it is not symmetric. There must be something happening during the transition from macroscopic to infinitesimal that forces symmetry of the stress tensor, something I still don’t understand. The argument about $\Delta x$ and $\Delta y$ being arbitrary seems to be part of the solution, but isn’t it a prerequisite that all dimensions are infinitesimal (and thus equal?) for the stress tensor concept to be valid? $\endgroup$
    – chicken_game
    Commented Feb 11, 2020 at 7:38

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