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Homework Help: Integration of partial derivative

  1. Feb 21, 2010 #1
    1. The problem statement, all variables and given/known data
    Given a body in a state of plane stress with no body forces where
    [tex]\sigma[/tex]x=x2y
    [tex]\sigma[/tex]y=(y3-3y)/3

    Find [tex]\tau[/tex]xy


    2. Relevant equations
    For plane stress
    [tex]\partial[/tex][tex]\sigma[/tex]x/[tex]\partial[/tex]x + [tex]\partial[/tex][tex]\tau[/tex]xy/[tex]\partial[/tex]y + X = 0

    [tex]\partial[/tex][tex]\sigma[/tex]y/[tex]\partial[/tex]y + [tex]\partial[/tex][tex]\tau[/tex]xy/[tex]\partial[/tex]x + Y = 0

    No body forces: X=0 Y=0

    Simplified:
    [tex]\partial[/tex][tex]\sigma[/tex]x/[tex]\partial[/tex]x = -[tex]\partial[/tex][tex]\tau[/tex]xy/[tex]\partial[/tex]y

    [tex]\partial[/tex][tex]\sigma[/tex]y/[tex]\partial[/tex]y = -[tex]\partial[/tex][tex]\tau[/tex]xy/[tex]\partial[/tex]x

    3. The attempt at a solution

    [tex]\partial[/tex][tex]\sigma[/tex]x/[tex]\partial[/tex]x = 2xy

    [tex]\partial[/tex][tex]\sigma[/tex]y/[tex]\partial[/tex]y = y2-1

    Putting this into above equations gives:
    [tex]\partial[/tex][tex]\tau[/tex]xy/[tex]\partial[/tex]y = -2xy
    [tex]\partial[/tex][tex]\tau[/tex]xy/[tex]\partial[/tex]x = -y2+1

    [tex]\int[/tex][tex]\partial[/tex][tex]\tau[/tex]xy = [tex]\int[/tex]-2xy[tex]\partial[/tex]y
    [tex]\tau[/tex]xy = -y2x

    [tex]\int[/tex][tex]\partial[/tex][tex]\tau[/tex]xy = [tex]\int[/tex]-y2+1[tex]\partial[/tex]x
    [tex]\tau[/tex]xy = -y2x + x

    These should both be equal, but I'm not sure what I'm doing wrong. I think its my integration.
    Please help.
     
  2. jcsd
  3. Feb 21, 2010 #2

    AEM

    User Avatar

    You forgot that when you do partial integration, say with respect to y, you have to add an arbitrary function of x. Similarly when you integrate partially with respect to x you have to add an arbitrary function of y. This will give you the necessary freedom to determine your unknown function.
     
  4. Feb 22, 2010 #3
    Ok, but doing that gives me

    [tex]\tau[/tex]xy=-y2x + f(x) for the first

    and

    [tex]\tau[/tex]xy=-y2x + x + f(y)

    Is that correct? If so how do I figure out what the actual [tex]\tau[/tex]xy is?

    Thanks
     
  5. Feb 22, 2010 #4

    Mark44

    Staff: Mentor

    Don't use f as both a function of x alone and another function of y alone.

    So you have T(x, y) = -y^2x + f(x), and
    T(x, y) = -y^2x + x + g(y).
    Edit: added x in the line above.

    Does that help?

    My LaTeX doesn't seem to be rendering correctly, so I switched tau to T.
     
    Last edited: Feb 22, 2010
  6. Feb 22, 2010 #5
    How do I find f(x) and g(y)?

    I know the tau's should be equal.
     
  7. Feb 22, 2010 #6

    Mark44

    Staff: Mentor

    I omitted x from the 2nd representation by accident. Both forms of T(x, y) have -y2x. The first representation doesn't have any functions of y alone, so g(y) = ? The 2nd representation has x, so f(x) = ?
     
  8. Feb 22, 2010 #7
    So does that mean f(x)=x and g(y)=0? So Txy=-y^2x+x?
     
  9. Feb 22, 2010 #8

    Mark44

    Staff: Mentor

    Almost. g(y) = C, a constant. So T(x, y) = -xy^2 + x + C.
     
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