# I Definition of force over an area

1. Jun 4, 2016

### JonnyG

I am reading the wikipedia article on the Cauchy stress-tensor. The article says that given some object, let $P$ be a point in the object and let $S$ be a plane passing through that point. Then "an element of area $\Delta S$ containing $P$, with normal vector $n$, the force distribution is equipollent to a contact force $\Delta F$ and surface moment $\Delta M$. In particular, the contact force is given by $\Delta F = T^n \Delta S$".

Now, I have always thought of force as a vector field, meaning that in this case, each individual point in $S$ would be assigned a vector which represents the force at that point. It seems that $\Delta F$ is some kind of average force though. It is defined in terms of $T^n$, which is the "mean surface traction". What is the definition of mean surface traction? I have googled it and cannot find a definition.

2. Jun 4, 2016

Are you sure that those $\Delta$s aren't supposed to be $d$s? In other words, given the plane $S$, there is some continuous force field, $F(s)$ acting on that plane, and given a differential element of that plane, $dS$, the differential force, $dF$, and moment, $dM$, are acting on it.

3. Jun 4, 2016

### JonnyG

In the article, $dF = \lim\limits_{\Delta S \rightarrow 0} \frac{\Delta F}{\Delta S}$. So the definition of $dF$ depends on the definition of $\Delta F$.

4. Jun 4, 2016

Ah, well then it would appear to me that $\Delta F$ is the average force over $\Delta S$. The average would just be an average over the area,
$$\Delta F = \dfrac{1}{\Delta S}\int\int_{\Delta S} F(s) dS.$$
A traction is essentially the force field per unit area projected onto the surface normal.

EDIT: It's been a while since I studied continuum mechanics, but I believe I misspoke here slightly and I feel I should amend my previous statement. It is a force per unit area projected onto a surface normal, but traction generally specifically refers to the projection of the Cauchy stress tensor onto that normal, so it is still a three-component vector with one normal component and two shear components.

Last edited: Jun 4, 2016
5. Jun 4, 2016

### JonnyG

@boneh3ad Thank you for the clarification. I have three more questions, if you don't mind answering them:

1) Letting $\sigma$ denote the stress-tensor, what is the best way to think of the action of $\sigma$ on its two arguments? For example, suppose $p$ is a point in the object and $n$ is a unit normal to some plane passing through $p$, then $\sigma(n, \cdot) = T^n$ where $T^n$ is the traction vector. But this seems to imply that $T^n$ is a dual-vector. Am I correct in thinking this?

2) Given two arbitrary vectors $v,w$, how can I physically interpret $\sigma(v,w)$?

3) Given a plane, there are always two normals to that plane. So given a point $P$ and a plane passing through $P$, then we could choose two distinct normals when deriving the stress-tensor. There doesn't seem to be a natural choice of normal vector. Which one do we take? Or is the tensor actually independent of the choice of normal?

6. Jun 6, 2016

bump...

7. Jun 6, 2016

### Staff: Mentor

The way I learned it is that, if $\vec{\sigma}$ is the stress tensor, and $\vec{n}$ is a unit vector normal to an element of surface area dS within the material, the force per unit area exerted by the material on the side of dS toward which $\vec{n}$ is pointing and on the material on the side of dS from which $\vec{n}$ is pointing is given by: $$\vec{T}=\vec{\sigma}\cdot \vec{n}$$where the force per unit area $\vec{T}$ is called the traction vector on the element of surface area.

8. Jun 6, 2016

Sorry, I got busy this weekend and couldn't respond. I will second what @Chestermiller just said.

9. Jun 7, 2016

### wrobel

I will be the third who says the same (but by little bit another words)
Assume that a domain $D\subset\mathbb{R}^3$ is filled with a continuous media. Let $x=(x^i)$ be any right curvilinear coordinates in $D$ with coordinate vectors $\boldsymbol e_i=\boldsymbol e_i(x)$ and $g_{ij}=(\boldsymbol e_i,\boldsymbol e_j),\quad g=det(g_{ij})$.
Now take any volume $W\subset D$ and any surface $\Sigma$ that is a part of $\partial W,\quad \Sigma\subset\partial W$ . Postulate that the media outside $W$ acts on $\Sigma$ with the force
$\boldsymbol F=\int_\Sigma \sqrt g \boldsymbol e_i\otimes(p^{i1}dx^2\wedge dx^3+p^{i2}dx^3\wedge dx^1+p^{i3}dx^1\wedge dx^2)$
or symbolically
$d\boldsymbol F=\sqrt g \boldsymbol e_i\otimes(p^{i1}dx^2\wedge dx^3+p^{i2}dx^3\wedge dx^1+p^{i3}dx^1\wedge dx^2).$
This defines the Cauchy stress tensor $p^{ij}$. Note also that there are situations (not classical problems) such that $p^{ij}\ne p^{ji}$

Last edited: Jun 7, 2016