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A solid steel shaft is subjected to a torque of 45 KN/m.

  1. Jun 28, 2012 #1
    A solid steel shaft is subjected to a torque of 45 KN/m. If the angle of twist is 0.5 degree per meter length of the shaft and the shear stress is not exceed 90 MN/m^2
    (I) find the suitable diameter of the shaft . Take C = 80 GN/m^2
    (II) Maximum shear strain


    we can solve the question by torsion equation
    T/Ip = e/R = CQ/l

    T maximum twisting torque in Nm
    Ip polar moment of inertia = pid^4/32 in m^4
    e shear stress in N/m^2
    R radius of the shaft
    Q the angle of the shaft
    l length if the shaft in m

    my question know why in solving fined two diameter ?
     
  2. jcsd
  3. Jun 28, 2012 #2

    rock.freak667

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    Well from your formula, which two should you equate with what you are given in order to get the diameter? (You know one quantity as one of the terms must be related to the diameter).
     
  4. Jun 29, 2012 #3
    yes why ?

    and here I have the solution

    the fist answer is
    Diameter on the basic of twist
    = 0.16 m

    and then diameter on the basis of shear stress :
    D = 0.1365 m

    Now my question why in answer give us two diameter ?

    and which one is correct ?
     
  5. Jun 29, 2012 #4

    rock.freak667

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    Using shear stress, you got d=0.1365 m, so if you put d = 0.16 m will your shaft be able to withstand the shear stress or will it fail?

    (Hint: You don't want your shaft to fail)
     
  6. Jun 29, 2012 #5
    How I can know if the shaft It will withstand the shear stress or not ... ?

    for finding the second diameter is depend on first diameter ?
     
  7. Jun 29, 2012 #6

    rock.freak667

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    At the maximum shear stress value the diameter is 0.1356 m. (Meaning that this is the minimum diameter the shaft should be)

    For the angle of twist, the diameter is 0.16 m. Since you want your shaft to be subjected to the specified angle of twist, then d = 0.16 m will be sufficient.
     
  8. Jun 29, 2012 #7
    thanks so much ..
     
  9. Jun 29, 2012 #8

    rock.freak667

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    In essence, what you did was to get the diameter for the angle of twist.

    Then found the minimum diameter of the shaft to prove the first diameter would not exceed the shear stress.


    Alternatively, you could have used the first diameter (found using the angle of twist) to get the shear stress using

    [tex]\frac{T}{I_p} = \frac{e}{R}[/tex]
     
  10. Jun 29, 2012 #9
    thanks .. I forget ... how I can figure out the equation from above equation to solve for maximum strain ?? there is no simple for strain ?? ...
     
  11. Jun 29, 2012 #10

    rock.freak667

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    How is shear stress and shear strain related to the modulus of rigidity (your value for C)?
     
  12. Jun 29, 2012 #11
    module of rigidity = steers / strain .
     
  13. Jun 29, 2012 #12

    rock.freak667

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    so from e/R = CQ/l, what do you get when you replace e with C*strain ?
     
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