1
$\begingroup$

I was looking at the equation for the differential force $dF$ caused from a stress tensor $\sigma$ acting on a differential surface $dS$. In mathematical terms,

$dF=\sigma \cdot dS$

Taking the integral of both sides yields.

$$F=\int \sigma \cdot dS$$

What is the nature of this integral? How does on go about calculating it? I would like to see an example given a non-trivial $\sigma$.

Can this be generalized to $F=\int \sigma \cdot dV$?

My idea was to integrate with respect to the variables $x$ and $y$ and represent $dS$ as a normal vector multiplied by the $dx$ and $dy$ differentials. Taking the product of the right hand side, you then integrate with respect to the differentials.

$\endgroup$
7
  • $\begingroup$ In this case, the dot product helps because it gets rid of one rank of the tensor. But in general, I think the opposite of $\frac{\partial}{\partial x^\mu}$ is the obvious $\int\,dx^\mu$, which would be approached the normal tensor way. $\endgroup$ – Jim Aug 19 '15 at 16:41
  • $\begingroup$ @Jim I meant matrix product by $\sigma \cdot dS$. Still, it does yield a vector, I just don't know what to do with it :/ $\endgroup$ – Zach466920 Aug 19 '15 at 16:43
  • $\begingroup$ I'd double check with math.se, but I think you just do the integral for each component of the vector and the result will still be a vector. Honestly, I'm more used to tensor form than matrix form $\endgroup$ – Jim Aug 19 '15 at 16:45
  • 1
    $\begingroup$ You probably do know what to do with it and just don't recognize the fact. Here's a problem that you know involving integrals and vectors: given a particle with initial position $\vec{x}_0$, initial velocity $\vec{v}_0$ and constant acceleration $\vec{a}$, find the position as a function of time. The extension to higher rank tensors is similar. $\endgroup$ – dmckee --- ex-moderator kitten Aug 19 '15 at 16:47
  • $\begingroup$ @dmckee yes, but $dS$ has to be a vector. I don't actually know how to pick $dS$ so integration can be done under something like $dx \ dy$. With your example the equation already comes with $dt$ in a nice form. $\endgroup$ – Zach466920 Aug 19 '15 at 16:53
5
$\begingroup$

You can imagine three equations one for each component of the force and then the tensor can be thought of as three vectors.

For instance $F_x=\int \sigma_{xx}n_x+\sigma_{xy}n_y+\sigma_{xz}n_zdA$ where $d\vec S=\hat n dA.$ So now it is a regular flux surface integral. You just have three of them:

$F_x=\int \sigma_{xx}n_x+\sigma_{xy}n_y+\sigma_{xz}n_zdA$, $F_y=\int \sigma_{yx}n_x+\sigma_{yy}n_y+\sigma_{yz}n_zdA,$ and $F_z=\int \sigma_{zx}n_x+\sigma_{zy}n_y+\sigma_{zz}n_zdA.$

See how it almost looks like a vector on the left and a matrix times a vector on the right? It's not more complicated than that.

$$F=\int \sigma dS=\int \sigma n dA.$$

$\endgroup$
0
$\begingroup$

Seems as though my idea is the correct one. Here's an example.

Since, $dS$ is normal to the surface and has a magnitude equal to the area of a differential. $dS$ can be represented by a cross product of the differentials of the relevant surface.

Given,

$$\sigma= \begin{bmatrix} 1 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 1 \\ \end{bmatrix} $$

We want to find the force over the surface of a square parallel to the x-y plane. The cross product of the differentials yields $[0,0,dx \cdot dy]$. Taking the transpose of this we get $dS$. Putting this into the formula yields,

$$\sigma \cdot dS=[dx \cdot dy,0, dx \ dy]^t$$

Taking the integral over the square gives,

$$F=[s^2,0,s^2]^t$$

Where $s$ is the side length of the square. This method can be generalized for other surfaces of course.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.