Shear Stress Differential Equation

Wildcat04
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Homework Statement


Equation solved from previous problem:

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

T = \mu * dvz/dr

Boundary condtions:

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

The Attempt at a Solution




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

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

vz = gr2/6\nu + 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\nu + A

A = -ga2 / 6\nu

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

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

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

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


However the solution is supposed to be:

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


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 \nu kept shooting up to the top!
 
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Wow...I did butcher it. Went back and found my mistake.

Thanks!
 
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