# Shear Stress Differential Equation

1. Oct 26, 2009

### Wildcat04

1. The problem statement, all variables and given/known data
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

3. 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!

2. Oct 27, 2009

### Wildcat04

Wow...I did butcher it. Went back and found my mistake.

Thanks!