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Minimum Shear Stress in Hollow Circular Tube

  1. Sep 30, 2014 #1
    1. The problem statement, all variables and given/known data
    A hollow pipe has an inner diameter of 80 mm and an outer diameter of 100 mm. If its end is tightened using a torque wrench using 80 N forces, determine the maximum and minimum shear stress in the material. Where are they located?

    Note: In the diagram of the picture, the left hand applies an 80 N force upward on the pipe, 200 mm from the shaft, and 80N downward with the right hand, 300 mm from the axis of the pipe. Had trouble getting the picture.

    2. Relevant equations
    \tau_{max} = \frac{T*radius}{I_p}\\
    Torque = r X F\\
    I_p = \frac{\pi}{32}((d_2)^4-(d_1)^4)

    3. The attempt at a solution

    So I know how to calculate the maximum shear stress in the pipe:
    Torque = r X F = (.200 m)*(80 N) + (.300 m)*(80 N) = 40 N*m\\
    I_p = \frac{\pi}{32}((d_2)^4-(d_1)^4) = \frac{\pi}{32}((.100 m)^4-(.080 m)^4) = 5.796e-6 m^4\\
    \tau_{max} = \frac{T*radius}{I_p} = \frac{(40 N*m)*(.050 m)}{5.796e-6 m^4} = 345051.4 N/m^2

    Therefore, tau_max takes place at the outer surface of the shaft.

    For tau_min, would I evaluate my expression for tau max at the inner diameter of the pipe? It makes sense to me, but I was hoping someone could verify this
  2. jcsd
  3. Sep 30, 2014 #2


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    Yes. Where else could you take it?
  4. Sep 30, 2014 #3
    I'm having trouble understanding the tone of your question. Are you saying there is another location?
  5. Sep 30, 2014 #4
    Sorry, I'm having trouble understanding the tone of your response. Are you implying there are other locations?
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