Proof of Cauchy's Lemma

Question

This question is related to this one I asked some time ago: Reason for Symmetry of stress tensors. The reason behind the symmetry of the Cauchy stress tensor seems to be Cauchy's Lemma. Once you accept that the traction vectors $t^{(n)}=-t^{(-n)}$, one can derive step by step that the tensor must indeed be symmetric. Now my question is: What is the proof of Cauchy's Lemma? It is obviously related to Newton's third law, but a force on one side of a body (e. g. a cube) can also be balanced e. g. by friction created by pressing on adjacent sides rather than exerting an opposing force on the opposite side (c. f. picture in my post linked above). So how can we state that $t^{(n)}=-t^{(-n)}$?

Another thing that bugs me a bit: I sense the traction vectors (and corresponding stress components) must be the mean values over the infinitesimal areas $dA$, whose lines of effect are made to pass through the center of mass of the (infinitesimal) body, which would necessitate to also take into account the moments created by the force distribution over $dA$. However, the literature always states that the lever arms are infinitesimal and thus moments need not be regarded. Since all dimensions are infinitesimal, this claim to me seems hard to accept without proof...

Answer 1

For a small volume, with one of the vertex in a point $P = (x,y,z)$, the average forces acting on its faces can be splitted in 3 orthogonal components.

Let's take the axis $x$ and the faces normal to the axis at $P$.

The components at $x$ are:
$F_x(x) = \sigma_{xx}(x)\Delta y\Delta z$
$F_y(x) = \sigma_{xy}(x)\Delta y\Delta z$
$F_z(x) = \sigma_{xz}(x)\Delta y\Delta z$

The components at $(x+\Delta x, y, z)$ are:
$F_x(x+\Delta x) = \sigma_{xx}(x+\Delta x)\Delta y\Delta z$
$F_y(x+\Delta x) = \sigma_{xy}(x+\Delta x)\Delta y\Delta z$
$F_z(x+\Delta x) = \sigma_{xz}(x+\Delta x)\Delta y\Delta z$

For the faces normal to $y$-axis at $P$:
$F_x(y) = \sigma_{yx}(y)\Delta x\Delta z$
$F_y(y) = \sigma_{yy}(y)\Delta x\Delta z$
$F_z(y) = \sigma_{yz}(y)\Delta x\Delta z$

The components at $(x,y+\Delta y,z)$ are:
$F_x(y+\Delta y) = \sigma_{yx}(y+\Delta y)\Delta x\Delta z$
$F_y(y+\Delta y) = \sigma_{yy}(y+\Delta y)\Delta x\Delta z$
$F_z(y+\Delta y) = \sigma_{yz}(y+\Delta y)\Delta x\Delta z$

For the faces normal to $z$-axis at $P$
$F_x(z) = \sigma_{zx}(z)\Delta x\Delta y$
$F_y(z) = \sigma_{zy}(z)\Delta x\Delta y$
$F_z(z) = \sigma_{zz}(z)\Delta x\Delta y$

The components at $(x,y,z+\Delta z)$ are:
$F_x(z+\Delta z) = \sigma_{zx}(z+\Delta z)\Delta x\Delta y$
$F_y(z+\Delta z) = \sigma_{zy}(z+\Delta z)\Delta x\Delta y$
$F_z(z+\Delta z) = \sigma_{zz}(z+\Delta z)\Delta x\Delta y$

The small volume can be at rest but in a gravitational field, or can be accelerated. In any case, (or both) that effect can be represented by a body force per unit of volume $\mathbf b(x,y,z) = \mu \mathbf g - \mu \mathbf a$ acting at the center of gravity of the volume. After including the body forces, the sum of the forces in each direction must be zero. For example in $x$:

$\sigma_{xx}(x)\Delta y\Delta z + \sigma_{xx}(x+\Delta x)\Delta y\Delta z + \sigma_{yx}(y)\Delta x\Delta z + \sigma_{yx}(y+\Delta y)\Delta x\Delta z + \sigma_{zx}(z)\Delta x\Delta y + \sigma_{zx}(z+\Delta z)\Delta x\Delta y + b_x\Delta x\Delta y\Delta z = 0$

When $\Delta x -> 0$, while keeping $\Delta y\Delta z$ unchanged, only 2 terms remain:

$\sigma_{xx}(x)\Delta y\Delta z + \sigma_{xx}(x+\Delta x)\Delta y\Delta z = 0$

So it is valid that: $\sigma_{xx}(x) = -\sigma_{xx}(x+\Delta x)$ when $\Delta x -> 0$, what is the same as to say that the traction forces are equal and opposing at a surface.

Note that if we let $\Delta y -> 0$ or $\Delta z -> 0$ instead of $\Delta x$, we get the same result for the shear stresses.

The same procedure can be made for the components in $y$ and $z$ direction.