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Homework Help: Mechanics of Materials: Torsion

  1. May 25, 2010 #1
    1. The problem statement, all variables and given/known data
    A shaft fixed on both ends is shown below. It is made of a steel tube, which is bonded to a brass core. A torque [tex] T_D[/tex] is applied at location shown. The tube is fixed at both ends.

    [PLAIN]http://img202.imageshack.us/img202/2940/problem2eb.jpg [Broken]

    Show that the fraction of the torque, [tex] T_D [/tex], resisted by the steel at C labeled as [tex] T_c[/tex] is given by the following expression: [tex] T_c = T_D \left[\frac{G_s J_s}{G_b J_b + G_s J_s}\right]\left[\frac{L_{ad}}{L}\right] [/tex]

    2. Relevant equations

    [tex] \phi = \frac{TL}{JG} [/tex]

    3. The attempt at a solution

    So I am thinking I should start with a free body diagram. The shaft is fixed which makes me think that the angle of twist will be zero allowing me to use this equation to solve for torques. If I am not mistaken the torque as C will be [tex] T_D [/tex] in the opposite direction.

    At point C I have [tex] \frac{T_c L}{G_sJ_s}= \frac{T_dL_{ad}}{G_bJ_b} + \frac{T_d L_{ad}}{G_sJ_s} [/tex]

    Obviously this isn't correct; it seems like they may have simply combined the modulus of rigidiy and polar moments for the part of the shaft AD. Any input, am I on the correct track? Or did I possibly miss something about combining these components when a shaft is bonded to a core of different material?
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 25, 2010 #2


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    Yes, which free body diagram?
    Where will it be 0?
    . No. If the torque at C is equal and opposite to T_D, that implies that the torque at A is 0, which seems illogical.
    Hint: The angle of twist at D is the same immediately to the left and right of D. What's the torque immediately to the left of D? What's the torque immediately to the right of D? You must draw free body diagrams of sections cut through these points and about one end to find out. And don't forget the equilibrium equation for sum of torques.
  4. May 26, 2010 #3
    Thank you very much for your reply! Earlier some students and I worked the problem together and it is much clearer now. Thanks!
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