# Torsion and angle of twist

1. Oct 16, 2008

### aznkid310

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

Unfortunately, I dont have a picture to upload, so I'll describe it the best that I can.

A prismatic bar AB of length L and solid circular cross section (diameter d) is loaded by a distributed torque of constant intensity t per unit distance. Determine the angle of twist W between the ends of the bar.

2. Relevant equations

3. The attempt at a solution

d(Torque) = tdx --> Torque T = integral (from 0 to L) [tdx] = tL

W = int(0 to L) [T(x)dx/GI(x)] , where G = shear modulus, I = polar moment of inertia

Is my T(x) equal to T = tL ?

How do I find I? I realize that I = int over the area [x^2 dA], where x is the distance from the center to dA

Can i just use the formula I = [(pi)r^4]/4 for a solid bar?

2. Oct 16, 2008

### Pyrrhus

if the bar is prismatic it means its cross section is constant along its length.

For this case a differential twist angle will be

$$d \phi = \frac{T(x)dx}{GI_{polar}}$$

No, T = f(x). It's not constant for this case.

Yes, but that will just make your work longer.

That formula is wrong. That's the area moment of inertia of the section. Look up the polar moment of inertia.

Last edited: Oct 16, 2008
3. Oct 17, 2008

### aznkid310

Sorry i typed it in wrong. It should be I = (pi/2)(r^4)

Could you get me started on finding T(x)?

4. Oct 17, 2008

### Pyrrhus

What do you understand by "... a distributed torque of constant intensity t per unit distance..." ?

5. Oct 17, 2008

### aznkid310

that the torque varies linearly with distance?

6. Oct 17, 2008

### Pyrrhus

so T(x) = ?

7. Oct 18, 2008

### aznkid310

T(x) = tx?

W = (t/GI)int(0 to L) [ xdx]

= (tL^2)/2GI

= [16tL^2]/[(pi)Gd^4]

Last edited: Oct 18, 2008
8. Feb 25, 2010

### diyana

hi...how do i get the angle of twist (in radian) if the question give the revolutions number?

for more understanding, here is the question :

the ship at the surface,A has just started to drill for oil on the ocean floor at a depth of 1500m. knowing that the top of the 200-mm-diameter steel drill pipe (G=77.2GPa) rotates two complete revolutions before the drill bit at the bottom,B starts to operate, determine the maximum shear stress caused in the pipe by torsion.

i know that angle of twisT,@ = TL/JG where J= (pi/2)(c^4) .....c=outermost radius.

and when i got the T (torsion), i can use the equation ; shear stress = Tc/J.

am i correct here? i just got confuse how do i get the T..