1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Maximum shear stress

  1. Jun 18, 2013 #1
    1. The problem statement, all variables and given/known data

    A locked door handle is composed of a solid circular shaft AB with a diameter of b = 108mm and a flat plate BC with a force P = 69N applied at point C as shown. Let c = 532mm , d = 126mm , and e = 153mm .

    2. Relevant equations

    [itex]\tau[/itex]max = [itex]\frac{Tc}{J}[/itex]
    T = internal torque acting at cross section
    c = outer radius
    J = polar moment of inertia → J = [itex]\frac{\pi}{2}[/itex]c4

    3. The attempt at a solution

    I've converted all mm to m.
    I just wanted to see if someone could verify that I am on the right track.
    radius, c = 0.054 m

    1. The torque caused by the P force:
    69 N × 0.532 m = 36.71 N*m → acting along the length of the solid shaft.

    2. Internal torque, T:
    T = 36.71 N*m *0. 126 m = 4.62546 N*m → acting at section m-m

    3. J = [itex]\frac{\pi}{2}[/itex]0.054 m4 = 13.36×10-6m4

    [itex]\tau[/itex]max = [itex]\frac{(4.62546 N*m)*(0.054 m)}{13.36×10^{-6} m^4}[/itex]
     

    Attached Files:

  2. jcsd
  3. Jun 18, 2013 #2

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Is the problem asking for max shear stress at the section m-m? You left out the actual question.
    Yes, but I wouldn't say it acts 'along' the length: it is an internal torque that acts about the axis AB
    Why are you multiplying a torque by a distance? Your result is in N*m^2 and incorrect.
    Once you calculate the torsional shear stress, there is the vertical shear stress to consider from the vertical load P.

    Please state the question.
     
  4. Jun 19, 2013 #3

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Jay:

    The shear stress due to an applied torque on a circular shaft is T * r / J. You overlooked the units of the polar moment of inertia J, which is m^4.

    So,

    tau = T*r/J which has units of N-m * m / m^4 = N/m^2 = Pa, a unit of stress, I believe.

    tau = Tr/J is somewhat analogous to finding the bending stress in a beam, where

    sigma = M * y / I
     
    Last edited: Jun 19, 2013
  5. Jun 19, 2013 #4

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    The OP incorrectly identified the torque in item 2 of Post 1. The vertical shear in the rod was also neglected, if that is what the problem is asking.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted