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Homework Help: Torsion Bar Spring in door

  1. Oct 28, 2015 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations
    G=modulus of rigidity
    J=polar moment of inertia
    L= length
    Φ=twist angle (rad)
    Direct shear = Force/Area
    Torsional Shear = Tr/J, where T is the torque and r is the radius from the center.

    3. The attempt at a solution
    G=11e6 KSI
    J=(π/16)d^4, where d= diameter of the rod
    Φ=110 degrees = 1.9199 radians


    L = (GJΦ)/T

    Just to try to see if my logic is correct I put in the given value for d (=.49 in) but I'm not getting the correct answer. I feel like my FBD is incorrect.
  2. jcsd
  3. Oct 28, 2015 #2


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    The way you have the closing torque drawn would tend to raise the door off the stop, rather than keep the door closed on the stop.
  4. Oct 28, 2015 #3
    That's the way I see it, but I see what you mean. I feel like this question is poorly worded. It's not entirely clear to me what exactly is going on.
    Either way, it wouldn't really matter, I took the absolute value of the torque. Depending on which way the torque goes would only change the side on which the maximum shear occurs so the calculations would be the same (using absolute values). I still can't figure out how to set up the problem.
    It looks to me like the center of gravity is NOT in the center of the door, and the length to the side with the stop is not given, so I can't do a sum of the moments analysis to get the total torque. Instead I assumed that the given 6 lb ft was the torque acting on the pin. I'm getting nowhere with this problem.
  5. Oct 28, 2015 #4


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    All is not lost.

    Imagine there's no torsion spring on the door hinge, and the door is closed.

    The weight of the door, 60 lbs, and the c.g. of the door, 24 inches from the hinge, would necessarily produce a torque at the hinge of 60 x (24/12) = 120 ft-lbs, which keeps the door closed. We don't know what force this produces at the door stop, since we don't know how the length of the door from the hinge to the stop.

    However, by adding the torsion bar spring to the door hinge, the effort required to open the door should be reduced, if the spring is designed carefully. According to the problem statement, instead of a torque of 120 ft-lbs keeping the door closed against the stop, the spring should reduce this torque to only 6 ft-lbs, which consequently would make it easier to open the door if a force at the stop was applied. It's the torsional stress caused by this reduction in the closing torque which should be used to design the torsion spring.
  6. Oct 28, 2015 #5
    Ah okay okay, it's making more sense now, thanks!
    Just to clarify, are you saying that the torque in the torsion bar is trying to open the door when it's shut? I guess what I'm asking is at what position the torsion bar is neutral? It seems to me that if the torsion bar is to help reduce the torque needed to lift the door once it is shut, then the torque should be acting in the direction that I drew it (I drew my FBD imagining that the door was shut).

    So if I understand correctly, the internal torque in the torsion bar, when the door is shut is equal to 114 lb-ft. But it seems to me like we're just ignoring the torque that is produces by the 12 lb force on the door stop. How come we can just ignore it?
  7. Oct 28, 2015 #6


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    I would assume that the torsion bar is in a 'neutral' position when the door is open, for what it's worth. But that's not the condition for which the torsion bar must be designed.

    Remember, from your diagram, a clockwise torque keeps the door closed. That's the direction in which the 6 ft-lb net torque must act to keep the door pressed up against the stop.
    You don't know that there is a 12 lb. force on the door stop, because you don't know how far the door stop is from the hinge, as you yourself said in a previous post.

    Going back to your diagram, the torque caused by the weight of the door, 120 ft-lbs., would be shown acting clockwise. The torsion spring provides a counteracting, counterclockwise torque of 114 ft-lbs acting at the hinge, for a net torque on the door of 6 ft-lbs, which also acts clockwise, thus holding the door shut.
  8. Oct 28, 2015 #7
  9. Oct 28, 2015 #8


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    Sounds like the same answer to me.

    Your formula for J is slightly incorrect. For a solid circular cross-section, J = πd4 / 32 in4, where d is the shaft diameter in inches.


    What about the length of the torsion bar?
  10. Oct 28, 2015 #9
    Normally I'd agree that it's close enough, but since I didn't round off anything - I feel like my answer should match the book's answer exactly.

    J = π* d^4 / 32


    Tshear= T c / J

    Simplifies to 16T/πd^3, no?

    Anyway, it's probably close enough. For the length I got 116 inches from

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