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I'm hoping someone can clear up this confusion I have with the stress tensor. So here is what a stress tensor looks like as described by many authors:

enter image description here

I understand that the shear stresses acting on the positive face are taken as positive, but is there any reason why the reason why the shear stresses acting on the negative faces are in the negative direction? For example, why doesn't it look something like this? (please excuse the bad diagram, I did it on MS paint):

Stress Tensor scheme B

Let's call this scheme B. The negative faces are denoted by the blue shear vectors and red for positive faces. I can come to terms with the idea that it's convention, however if I use scheme B and re-derive the Cauchy momentum equations, the shear stress derivative terms will be missing as they cancel. So the final equations are dependant on the sign convention used, with that in mind, why is the scheme in the first image used instead of something like the second one?

Edit: I don't think it's necessary to go through a whole derivation, but let's do a force balance in the X direction.

First, using the first image for our sign convention:

$\sum F_x = $ Gravitational forces + forces from shear stresses

Taking the centre of the material element as the (x,y,z), just consider the x direction normal forces:

At some point in the force balance we are going to encounter the following:

On the positive face:

$\sigma_{x_+} = \tau_{xx} + \frac{\partial \tau_{xx}}{\partial x} \frac{\delta x}{2} $

Similarly on the negative face:

$\sigma_{x_-} = \tau_{xx} - \frac{\partial \tau_{xx}}{\partial x} \frac{\delta x}{2} $

Combining the two (notice the subtraction as they are going off in different directions):

$\sigma_{x_+} - \sigma_{x_-} = \frac{\partial \tau_{xx}}{\partial x} \delta x $

And obviously you would do this for all the shear stresses point in the x direction for all faces.

NOW, if we use 'MY' sign convention, the normal stresses look like this:

$\sigma_{x_+} = \tau_{xx} + \frac{\partial \tau_{xx}}{\partial x} \frac{\delta x}{2} $

$\sigma_{x_-} = \tau_{xx} - \frac{\partial \tau_{xx}}{\partial x} \frac{\delta x}{2} $

Now, both are facing in the same direction so instead of subtraction, they will be added:

$\sigma_{x_+} - \sigma_{x_-} = 2\tau_{xx} $

The derivative terms are gone! This will be the case the complete derivation.

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  • $\begingroup$ Could you include your derivation? Perhaps something goes awry when you use the oriented form representing the normal surface. $\endgroup$
    – Emil
    Commented Dec 5, 2017 at 23:26
  • $\begingroup$ @Emil Okay, hold on $\endgroup$ Commented Dec 5, 2017 at 23:27
  • $\begingroup$ @Emil See edit, it's not the full derivation but you'll get the idea? $\endgroup$ Commented Dec 5, 2017 at 23:47
  • 2
    $\begingroup$ Stress is a second order tensor, and it is set up in this way so that it has a bi-directional character. If you exert the same force (in opposite directions) on both ends of a rod, the rod will be in equilibrium. The force on each end will be equal to the normal stress times the cross sectional area. But, in order for the forces (tensions) to be pointing in opposite directions, the normal stresses on the left must be pointing to the left, and the (identical) normal stress to the right must be pointing to the right. This is all automatically guaranteed by the Cauchy stress relationship. $\endgroup$ Commented Dec 5, 2017 at 23:55
  • $\begingroup$ @ChesterMiller Is the material element (small cube) is always in equilibrium? $\endgroup$ Commented Dec 6, 2017 at 0:06

1 Answer 1

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Stress tensor always carries a convention:

If we take a surface ($\mathbf{\Sigma_{12}}$) and a corresponding normal vector ($\hat{\mathbf{n}})$, We know there is a 2nd order tensor ($\bar{\mathbf{A}}$) such that $\vec{\mathbf{F}}=\bar{\mathbf{A}}\cdot\hat{\mathbf{n}}$ is a force.

$\mathbf{1}$ and $\mathbf{2}$ are the volumes separated by $\mathbf{\Sigma_{12}}$.

If $\hat{\mathbf{n}}$ is pointing from $\mathbf{1}$ to $\mathbf{2}$,

  1. $\vec{\mathbf{F}} $ is force on $\mathbf{1}$ by $\mathbf{2}$, it is one convention with the stress tensor usually denoted using $\bar{\sigma}$. This is the Cauchy stress tensor convention.

(e.g. - mechanical engineering books)

  1. $\vec{\mathbf{F}} $ is force on $\mathbf{2}$ by $\mathbf{1}$, it is one convention with the stress tensor usually denoted using $\bar{\mathbf{T}}$. This comes from momentum flux rate, to which stress is related.

(e.g. Kip Thorne "Modern Classical Physics" Volume)

Now sign of $\bar{\mathbf{A}}$ is determined if

  • within the convention of choice
  • with the gives axis

, $\hat{\mathbf{n}}_i>0$ leads to $\vec{\mathbf{F}}_j>0$

for the same sign of $A_{ij}$,

if $n_i>0 \implies F_i >0$

$n_i<0 \implies F_i <0 $

That's why to ensure same sign of $A_{ij}$, the direction of $F_i$ flips with $n_j$.

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