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

  1. Nov 7, 2009 #1
    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[tex]\Theta[/tex]/L = [tex]\tau[/tex]/r
    (ii) K for a hollow cylinder is K = ([tex]\pi[/tex]/2)([tex]a^{4}[/tex] - [tex]b^{4}[/tex])
    (iii) Elastic energy stored per unit volume P = (1/2)([tex]\sigma^{2}[/tex]/E) where [tex]\sigma[/tex] 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 [tex]\tau[/tex] = rT/K = (rT[tex]\pi[/tex]/2)([tex]a^{4}[/tex] - [tex]b^{4}[/tex]). 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)[tex]\int\int\int[/tex][([tex]\pi[/tex]rT/2)([tex]a^{4}[/tex] - [tex]b^{4}[/tex])]^2 dr d[tex]\theta[/tex] dL.

    I get as a result

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

    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. jcsd
  3. Nov 7, 2009 #2


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    Homework Helper

    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.
  4. Nov 7, 2009 #3
    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 [tex]\int Td\theta[/tex], integrated from 0 to [tex]\theta[/tex], which results in an equation for the energy put into the shaft being:
    [tex]U_{k} = T \theta[/tex]

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

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

    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.
  5. Nov 7, 2009 #4


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    Homework Helper

    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
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