Shear Stress Differential Equation

In summary, the conversation discusses solving an equation from a previous problem involving velocity and torque. The solution is found to be T=(\mu/r) [ (g(b2-a2)/4\nu) / 2 ln(b/a) - gr2/2\nu ] with boundary conditions vz (r=a) = vz (r=b) = 0.
  • #1
Wildcat04
34
0

Homework Statement


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

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!
 
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  • #2
Wow...I did butcher it. Went back and found my mistake.

Thanks!
 

1. What is a shear stress differential equation?

A shear stress differential equation is a mathematical formula used to describe the relationship between shear stress and shear strain in a material. It is commonly used in the study of fluid mechanics and solid mechanics to understand the behavior of materials under shear stress.

2. How is the shear stress differential equation derived?

The shear stress differential equation is derived from the fundamental principles of mechanics, specifically Newton's second law of motion and Hooke's law. It is based on the assumption that the shear stress is directly proportional to the shear strain in a material.

3. What are the applications of the shear stress differential equation?

The shear stress differential equation is used in a variety of engineering and scientific fields, including fluid mechanics, solid mechanics, and materials science. It is essential in the design and analysis of structures such as bridges, buildings, and aircraft, as well as in the study of fluid flow in pipes and channels.

4. How is the shear stress differential equation solved?

The shear stress differential equation can be solved using various mathematical methods, such as separation of variables, Laplace transform, and numerical methods. The specific method used depends on the complexity of the problem and the desired level of accuracy.

5. What factors affect the solution of the shear stress differential equation?

The solution of the shear stress differential equation is affected by various factors, such as the material properties of the object under shear stress, the shape and size of the object, and the applied load or force. Other factors, such as temperature and environmental conditions, can also influence the solution.

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