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Homework Help: Increase of radius of circular rod due to torque

  1. Feb 15, 2010 #1
    1. The problem statement, all variables and given/known data

    I am trying to understand how will the radius of a circular beam grow when a torque is applied.
    There is a beam with length l, radius r, and a moment M1 applied on it. Its radius will grow with dr.
    What is the Young modulus of the beam?
    M= 15 N m
    r= 0.008 m
    l= 0.2 m
    dr= 0.00064 m


    2. Relevant equations


    Torsional shearing stress
    [tex]\tau=\frac{M r}{J}[/tex]
    Polar moment of inertia
    [tex]J=\frac{1}{2} \pi r^{4}[/tex]
    Young modulus definition
    [tex]G=\frac{l \tau}{dx}[/tex]
    Contraction due to shear
    [tex]dl=l - \sqrt{l^{2} - dx^{2}}[/tex]

    3. The attempt at a solution


    expressing [tex]dx[/tex] with [tex]G[/tex]
    [tex]dx=\frac{l \tau}{G}[/tex]
    substituing [tex]dx[/tex] to [tex]dl[/tex]
    [tex]dl=l - \sqrt{l^{2} - \frac{l^{2} \tau^{2}}{G^{2}}}[/tex]
    substituing [tex]\tau[/tex] to [tex]dl[/tex]
    [tex]dl=l - \sqrt{l^{2} - \frac{M^{2} l^{2} r^{2}}{G^{2} J^{2}}}[/tex]
    substituing [tex]J[/tex] to [tex]dl[/tex]
    [tex]dl=l - \sqrt{- 4 \frac{M^{2} l^{2}}{\pi^{2} G^{2} r^{6}} + l^{2}}[/tex]
    dr is square root of dl (multiplying with [tex]\sqrt{m}[/tex] to keep units)
    [tex]dr=- \sqrt{dl} \sqrt{m}[/tex]
    substituing [tex]dl[/tex] to [tex]dr[/tex]
    [tex]dr=- \sqrt{m} \sqrt{l - \sqrt{- 4 \frac{M^{2} l^{2}}{\pi^{2} G^{2} r^{6}} + l^{2}}}[/tex]
    solving for G
    [tex]G=- 2 \mathbf{\imath} m \sqrt{M^{2}} \sqrt{l^{2}} \sqrt{\frac{1}{\pi^{2} dr^{4} r^{6} - 2 l m \pi^{2} dr^{2} r^{6}}}[/tex]
    [tex]G = 9215558692.57753 \frac{N}{m^{2}}[/tex]
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
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