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Shear Stress Differential Equation

  1. Oct 26, 2009 #1
    1. The problem statement, all variables and given/known data
    Equation solved from previous problem:

    (g/[tex]\nu[/tex]) = (1/r) d/dr [ r dvz/dr]

    [tex]T[/tex] = [tex]\mu[/tex] * dvz/dr

    Boundary condtions:

    vz (r=a) = vz (r=b) = 0

    3. The attempt at a solution


    (d/dr)[d/dr r*vz = g*r / [tex]\nu[/tex]

    (d/dr) r*vz = gr2/2[tex]\nu[/tex] + A

    vz = gr2/6[tex]\nu[/tex] + A + B/r

    I am not sure if this is correct or what I am supposed to do with the B.C.s. I know that I can have 2 equations with 2 unknowns ( v = 0 @ r = a and v = 0 @ r = b, unks = A, B)

    0 = vz (r=a) = ga2/6[tex]\nu[/tex] + A

    A = -ga2 / 6[tex]\nu[/tex]

    0 = vz (r=b) = gb2/6[tex]\nu[/tex] - ga2/6[tex]\nu[/tex] + B/r

    B = -[rg(b2-a2)] / 6[tex]\nu[/tex]

    vz = gr/3[tex]\nu[/tex] - ga/3[tex]\nu[/tex] -[rg(b2-a2)] / 6[tex]\nu[/tex]

    I need to take the derivative of vz to get the torque...


    However the solution is supposed to be:

    [tex]T[/tex]=([tex]\mu[/tex]/r) [ (g(b2-a2)/4[tex]\nu[/tex]) / 2 ln(b/a) - gr2/2[tex]\nu[/tex] ]


    Obviously I messed up somewhere in there. Could someone please point out my error so I can get this thing correct?

    Thank you very much in advance!!


    I am sorry for it being a little sloppy...the [tex]\nu[/tex] kept shooting up to the top!
     
  2. jcsd
  3. Oct 27, 2009 #2
    Wow...I did butcher it. Went back and found my mistake.

    Thanks!
     
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