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Simple Mechanics of Materials: Deformation and Torsion

  1. Mar 22, 2009 #1
    1. The problem statement, all variables and given/known data

    Shaft AB is fixed at point A and is subjected to a torque T
    = 2.5 kN·m applied at point B and an axial force P = 30 kN applied at
    point C.

    The actual shaft specifications (type of material, prismatic / changing cross sections are left up to us and is NOT the point of my question)




    2. Relevant equations

    [tex]\delta[/tex] = PL/AE
    P = force, L = length, A = cross-sectional area, E = elastic modulus

    [tex]\phi[/tex] = TL/GIp
    T = torque, L = length, G = shear modulus, Ip = polar moment of inertia


    3. The attempt at a solution

    I am basically treating this as 1 deformation problem and 1 torsion problem

    For the deformation / axial part:
    After cutting the sections, I find that RA = P and RB = P also

    And thus,

    [tex]\delta[/tex]total = [tex]\delta[/tex]1 + [tex]\delta[/tex]2

    Where
    [tex]\delta[/tex]1 = PLAB/AABEAB
    And
    [tex]\delta[/tex]2 = PLBC/ABCEBC

    For the torsion part:
    Here is where I am very unsure of what to do. I know there will be a reaction torque at A, but will point C have one also?
    Right now I've got this -
    To + TC + TA = 0
    => To = TA - TC

    And

    [tex]\phi[/tex]1 = [tex]\phi[/tex]2

    Such that

    [tex]\phi[/tex]1 = TABLAB/GABIP,AB

    And

    [tex]\phi[/tex]1 = (TAB-To)LBC/GBCIP,BC

    Am I on the right track? Where is my analysis flawed ?
     

    Attached Files:

  2. jcsd
  3. Mar 22, 2009 #2

    Mapes

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    1) No, there won't be a reaction torque at point C, because there's no constraint there.

    2) I don't agree with To = TA - TC or the idea that there's an additional angular displacement between B and C.
     
  4. Mar 22, 2009 #3
    Ok, I understand that my torsion analysis is all off (I've definitely had problems with this part of the class)

    A few things:
    - Is my axial deformation analysis correct?
    - How should I go about the torsion analysis? We have to find the angle of twist at point C.
    We can't assume that there's only the reaction torque at A because otherwise there would be no angular displacement?

    I'm pretty confused with the whole torsion part overall
     
  5. Mar 22, 2009 #4

    Mapes

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    Your axial calculations look fine. I'd suggest getting something compliant, like an eraser or strip of rubber, and twisting and holding it in different spots to build up your intuition on angle of twist and reaction torque. If you hold the sample at one end and twist it in the middle, what's the angle of twist at the other end? Does the angle keep increasing?
     
  6. Mar 22, 2009 #5
    Yes the angle does keep increasing, but I'm still having difficulty in figuring out how to calculate the angle of twist directly at that end based on the single reaction torque and the external torque... how would you, quantitatively, do this?
     
  7. Mar 22, 2009 #6

    Mapes

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    Did you really try it? The angle doesn't keep increasing from B to C (theoretically or experimentally), because there's no torque on this section.

    Try plotting the torque as a function of distance along the rod.
     
  8. Mar 22, 2009 #7
    Sorry, that isn't what I meant, I just meant that there will be an angle of twist from B to C due to the applied torque at point B.

    What I'm having trouble figuring out though is exactly how to analyze this situation, because (as you said) there is no reaction torque at point C.

    Does that mean that the reaction torque at point A is equal to the applied torque? [TA = To]

    As far as the torque variation along the rod, would it be To from point A to point B, and then decrease linearly until it gets to zero at point C ?
     
  9. Mar 22, 2009 #8

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    Sure, you can show this by adding up the torques and setting the sum to zero (since the rod isn't accelerating).

    It would be To on the left side, but it wouldn't decrease gradually. Try drawing a free-body diagram from just right of B to C. The lack of reaction torque at C (we know this because there's nothing marked there) means that no positive torque can exist in this section.
     
  10. Mar 22, 2009 #9
    If I do a FBD from just after B to C, I just have the reaction torque at point B (TB) and since there is no other torque for that part, TB = 0.

    Would that mean [tex]\phi[/tex]C = ToLBC / GBCIP,BC?
     
  11. Mar 22, 2009 #10

    Mapes

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    Section BC?
     
  12. Mar 22, 2009 #11
    The shaft section from point B to point C, I get that TB = 0

    Any of this sound correct?
     
  13. Mar 22, 2009 #12

    Mapes

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    I like it up to the point where you calculate the angle between A and B by using the properties from B to C!
     
  14. Mar 22, 2009 #13
    Well I'm supposed to calculate the angle of twist at point C, so I figure to use the applied torque at point B and use the properties of the shaft from B to C (shear rigidity, polar moment of inertia, and length)

    Where did I go wrong?
     
  15. Mar 22, 2009 #14

    Mapes

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    But the angle increases from A to B; the section from B to C is just along for the ride.
     
  16. Mar 22, 2009 #15
    I'm taking my eraser out again :confused:


    Ok I think I see what you mean, the angle of twist from point A to point B is the same as the total angle of twist at point C (...?)

    This is what I'm seeing with the eraser right now, and that would mean:
    [tex]\phi[/tex]AB = ToLAB/GABIP,AB = [tex]\phi[/tex]C

    But if this is true, then the total rotational properties of the shaft part from point B to point C are irrelevant.

    Would all of this be true?
     
  17. Mar 22, 2009 #16

    Mapes

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    Exactly. Did playing with a physical model help? It always helps me.
     
  18. Mar 22, 2009 #17
    Absolutely. I never would have figured that out without playing with my eraser; thank you so much for your help! I'm just assuming that the twist on point C has to be influenced by the material of the shaft part B to part C.

    Thank you so much for your help! This is for part of a bigger project for my class and the book we're using seems to be over complicating everything. I'm glad I know how this works now... It's always easier to relate to things when you can 'see' them somehow (like you were talking about)
     
  19. Mar 22, 2009 #18

    Mapes

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    Glad to help. Good luck!
     
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