I Definition of force over an area

AI Thread Summary
The discussion revolves around the definition of force over an area, particularly in the context of the Cauchy stress tensor. It clarifies that the contact force, represented as ΔF, is an average force over an area ΔS, defined in terms of mean surface traction, T^n. The traction vector is described as the force per unit area projected onto the surface normal, incorporating both normal and shear components. Questions arise regarding the interpretation of the stress tensor's action on vectors and the choice of normal vectors when deriving the stress tensor. The conversation emphasizes the mathematical formulation of stress and its implications in continuum mechanics.
JonnyG
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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.
 
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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.
 
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##.
 
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.
 
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@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?
 
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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.
 
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Sorry, I got busy this weekend and couldn't respond. I will second what @Chestermiller just said.
 
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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}##
 
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