Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Definition of force over an area

  1. Jun 4, 2016 #1
    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. jcsd
  3. Jun 4, 2016 #2

    boneh3ad

    User Avatar
    Science Advisor
    Gold Member

    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.
     
  4. Jun 4, 2016 #3
    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##.
     
  5. Jun 4, 2016 #4

    boneh3ad

    User Avatar
    Science Advisor
    Gold Member

    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,
    [tex]\Delta F = \dfrac{1}{\Delta S}\int\int_{\Delta S} F(s) dS.[/tex]
    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
  6. Jun 4, 2016 #5
    @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?
     
  7. Jun 6, 2016 #6
    bump...
     
  8. Jun 6, 2016 #7
    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.
     
  9. Jun 6, 2016 #8

    boneh3ad

    User Avatar
    Science Advisor
    Gold Member

    Sorry, I got busy this weekend and couldn't respond. I will second what @Chestermiller just said.
     
  10. Jun 7, 2016 #9
    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Definition of force over an area
  1. Definition of force. (Replies: 22)

  2. Force-area graph (Replies: 1)

Loading...