# Homework Help: Elastic energy stored in a shaft under torsion?

1. Nov 7, 2009

### mibjkk

1. The problem statement, all variables and given/known data
Consider a hollow drive shaft of length L, inner radius b, outer radius a. A constant torque is applied (we can assume one end of the shaft is fixed because we're applying a constant torque T regardless of the angular velocity of the shaft). We know material properties of the shaft, G the shear modulus and E young's modulus, plus any others we might need.

2. Relevant equations
(i) T/K = G$$\Theta$$/L = $$\tau$$/r
(ii) K for a hollow cylinder is K = ($$\pi$$/2)($$a^{4}$$ - $$b^{4}$$)
(iii) Elastic energy stored per unit volume P = (1/2)($$\sigma^{2}$$/E) where $$\sigma$$ is the maximum stress applied to that unit area

3. The attempt at a solution
I'm not sure if these are all the relevant equations, but here is what I've tried.
- At any point distance r from the center of the cylinder, there is a shear stress $$\tau$$ = rT/K = (rT$$\pi$$/2)($$a^{4}$$ - $$b^{4}$$). If I plug this into my original equation for elastic energy per unit volume, then integrate over the volume of the cylinder, I find

E(T) = 1/(2E)$$\int\int\int$$[($$\pi$$rT/2)($$a^{4}$$ - $$b^{4}$$)]^2 dr d$$\theta$$ dL.

I get as a result

E(T) = $$((\pi^{3}LT^{2})/(12E)) (a^{4} - b^{4})^{2} (a^{3} - b^{3}))$$

This answer seems fairly reasonable, because it depends on L, but im a little worried that it would have a factor of a^8, which seems excessively high for an engineering problem. Also, I'm skeptical of being able to substitute shear stress from (i) for tensile stress in (iii). Would I use the shear modulus G instead of young's modulus E? Is this derivation mostly correct, or is there a complete different approach?

2. Nov 7, 2009

### rock.freak667

The work done is given by Uk=Tθ/2.

From the equation of three above you can find θ in terms of T and find Uk.

3. Nov 7, 2009

### mibjkk

I understand the idea behind Uk=Tθ/2, but where does the factor of 1/2 come from?

From what I understand, the work done by a torque over an angular distance is given by $$\int Td\theta$$, integrated from 0 to $$\theta$$, which results in an equation for the energy put into the shaft being:
$$U_{k} = T \theta$$

Using your method, I come to the following solution:
$$\theta=LT/KG$$
$$K = (\pi/2)(a^{4} - b^{4})$$, L, T and G are constants

then $$U_{k} = (LT^{2})/((G\pi/2)(a^{4} - b^{4}))$$, or 1/2 that using your equation of $$U_{k} = T\theta/2$$

This answer is definitely more accurate. I realize that since rotation is linearly related to torque and elastic energy is stored as the square of the angular rotation, T should be squared as the new equation.

If possible, help me understand the factor of 2 please.

4. Nov 7, 2009

### rock.freak667

Torque relates linearly with angle θ.

So on a T-θ diagram, from 0 to θ, the shape (graph) formed is a triangle of height T and base θ.

The area under that graph gives the work done Uk. So Uk=(T)*(θ)/2=Tθ/2