Statically Indeterminate Torsion Members

[aznkid310]

Homework Statement

A solid circular bar ABCD with fixed supports at ends A and D is acted upon by two equal and oppositely directed torques T_0. The torques are applied at points B and C, each of which is located at a distance x from one end of the bar. (The distance x may vary from zero to L/2).

a) For what distance x will the angle of twist at points B and C be a maximum

b) What is the corresponding angle of twist W_max?

Sorry I can't upload a picture, but basically there is a torque at B which is a distance x from end A of the bar, and another torque at C which is also a distance x to the other end D.

Homework Equations

Since it is fixed, then the sum of angles should be zero, correct?

So i would have to sum up the angles as follows: W_ab + W_bc + W_cd = W_ad ?

Also, the sum of torques should be zero.

Do i need to take an integral? If so, how should i approach this?

For each angle, I can make an imaginery cut and use the angle of twist formula, but this is where I am having trouble. Could someone walk me through this?

The Attempt at a Solution

Sum of torques: T_a - T_0 + T_0 - T_d = 0

T_a = T_d
W_ab = [(T_a)x]/GI
W_bc = [(T_a - T_0)]/GI
W_cd = [(T_a - T_0 + T_0)]/GI

where I = (pi/32)(d^4)

I don't believe this is correct

 

[mathmate]

> Since it is fixed, then the sum of angles should be zero, correct?
Correct, you could deduce that from symmetry (or anti-symmetry)

> Also, the sum of torques should be zero.
Correct, you could deduce that from
> ...two equal and oppositely directed torques T_0...If you look further using symmetry, what can you deduce for the angle of rotation at mid-length? Does that simplify your problem?

What can you deduce for the answer to the question:
> a) For what distance x will the angle of twist at points B and C be a maximum