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Torsion in a beam

  1. Jan 3, 2009 #1
    1. The problem statement, all variables and given/known data

    The question is shown in the attached picture. I know how to do the first part fine,and i know how to do the second part but i keep getting the wrong answer (answers are on the back of the sheet)

    2. Relevant equations

    tau/r=T/J (tau = shear stress, r = radial distance from centre, T = Torque, J = Polar second moment of area)

    J = (pi/2)*r^4 for a solid circular cross section

    3. The attempt at a solution

    T = 12k*0.75 = 9k Nm

    J = (pi/2)*(37.5x10^-3)^4 = 3.106x10^-6 m^4

    tau = (9x10^3*37.5x10^-3)/3.106x10^-6
    tau = 108.7 MPa (answer given is tau = 45.8 MPa)

    sigma = My/I

    I = (pi/4)*r^4 = 1.553x10^-6

    M = 18kNm
    y = 37.5mm
    sigma = 18x10^3*37.5x10^-3/1.553x10^-6
    sigma = 434.6 MPa (answer given is sigma = 188 MPa)

    both answers are wrong but if you use r and y = 15.8mm (but not when calculating second moments) the right answer comes out. Where does this 15.8mm come from or am I going wrong somewhere else?

    Thanks for any help

    Attached Files:

    Last edited: Jan 3, 2009
  2. jcsd
  3. Jan 3, 2009 #2


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    Well, first of all, isn't J given with [tex]J = \frac{d^4 \pi}{32} = \frac{r^4 \pi}{2}[/tex]?
  4. Jan 3, 2009 #3
    oh yeah thats a typo
  5. Jan 3, 2009 #4


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    Very interesting, I get the same answer for the shear stress due to the torsion moment. I'll have to consult my mechanics of materials book, I'll be back later.
  6. Jan 3, 2009 #5


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    jaderberg: Nice work. You did not do anything wrong. You got all answers correct. Both of the given answers from the back of the sheet are wrong.
  7. Jan 3, 2009 #6
    cheers man, was beginning to wonder whether that was the case :p
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