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Homework Help: 2 tubes of identical length, the larger placed inside the smaller

  1. Dec 30, 2009 #1
    in my question i have 2 tubes of identical length, the larger placed inside the smaller, both welded to a wall at one end and a rigid plate at the other, a moment of torsion Mt is applied to the outer tube at a distance d from the wall,

    in the question i am asked to find the maximum stress in each tube, i know how to do this but have only ever had questions where the torsion was applied at the end. since the inner tube will obviously twist differently to the outer tube i have a problem here, this is statically undefinded, i know that θ1=θ2 (tha angle of twist) due to the rigid plate. but i need some other condition.

    what i have tried to do is the following. since the inner tube resists the twistiong, i say that the torsion the outer tube applies to the inner is T1 and the torsion the inner tube applies on the outer tube is T2. as far as directions go T1 is the same as Mt and T2 is opposite.
    now what can i say? is T1+T2=Mt??? that is what i had done in a case of 2 tubes glued together with torsion applied at their end, but here they twist differently.

    i hope this question is clear and someone can help me
     
  2. jcsd
  3. Dec 30, 2009 #2

    Q_Goest

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    Re: torsion

    Hi Dell. If T1 is the torque exerted by the larger tube on the smaller tube, then shouldn't that be the same torque the smaller tube exerts on the larger one? I don't see any reason to suggest T2 (torque applied to inner) is different than the torque exerted by the outer on the inner. Note that T1 isn't the same as the moment being applied to the outer tube though (Mt).

    You can consider the moment applied as being a summation of two torques. Consider a free body diagram for example of the inner tube from the wall to the weld and then back along the outer tube to the location where the moment is applied. Consider a second free body diagram of the outer tube from the wall to the location of the moment. In other words, imagine the outer tube being broken exactly where the torque is being applied. The moment being applied can then be thought of as being the summation of two separate moments. One of these two moments is being transmitted down the larger tube to the wall. The second of these two moments is being transmitted down the larger tube to the welded end, and then back along the smaller tube to the wall. These two moments must add up to the total moment being applied.
     
  4. Dec 31, 2009 #3
    Re: torsion

    i thought that might be the case but wasnt sure since the 2 tubes are of different materials, with different properties, i thougt the torque may be divided differently than T1=T2
     
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