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Reaction Force help

  1. Jun 5, 2015 #1
    1. The problem statement, all variables and given/known data
    Hi guys, I'm currently working on a homework question but I'm a little bit stuck although I do know what I'm meant to be doing. My question asks me to find the torque being caused by a beam pulling down on another. I'll explain a little more the context of it - there is one beam (beam 1) spanning across from point A to point B, and one end is attached at the top of point A and the other is attached at the top of point B . Right beside that beam (beam 1) which spans across from point A to point B is another beam (beam 2). This beam however only spans half the way out and one end is attached at the top of point A. Beam 2 is securely attached all the way along beam 1 and is therefore causing it to twist due to the downwards forces.

    2. Relevant equations

    So far I know that I need to use the formula: Torque=Force x Radius x sinQ
    I think that should be all that is involved although I might have to use rotational equilibrium

    3. The attempt at a solution

    I have worked out the radius like this: (looking at a cross section of the end of the beams - beam 2 on left and beam 1 on right) I have assumed the pivot is in the centre of beam 1 and the radius is from there to the furthest edge of beam 2. Is this correct or do I need to work it out using rotational equilibrium or something?
    sinQ: Would be sin90
    Force: I think that I would need to find the reaction force being applied at each end of beam 2 (secured side will be acting up and the side secured to the edge of beam 1 will be acting down). By finding the reaction force that the beam is creating on point A I can then find the reaction force at the other end of beam 2 (I'm not sure how to do that??? Or is it equal except negative?) This is what I'm not sure how to do though. I have the weight and length of all of the beams too.

    Any help regarding finding the reaction forces and radius would be greatly appreciated.

    Thanks, Alex
     
  2. jcsd
  3. Jun 5, 2015 #2

    haruspex

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    The only loads are the weights of the beams, right?
    How are you defining pivot point? Are you saying that you think the midline of beam 1 will not be deflected by the weight of beam 2?
    Or are you just choosing it as the point to take moments about?
    The 'radius' is for the torque beam 2 exerts about the midline of beam 1, yes? Why are you taking it to the far edge of beam 2? Through what vertical line does the weight of beam 2 act?
     
  4. Jun 5, 2015 #3

    SteamKing

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    My head started hurting reading the description of the beams in the OP.

    Can you provide any kind of sketch showing how these beams are arranged?
     
  5. Jun 5, 2015 #4

    haruspex

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    As I read it, the two beams (same x-section I assume) are lashed together for the entire length of beam 2. They both sit at one end on support A, but only beam 1 reaches support B. Beam 2 only reaches half way. I'm guessing that the load is the (uniform) weight of the beams, and what we are concerned with is the torsion on beam 1 from the weight of beam 2.
    Seems a bit strange, so I could well be wrong.
     
  6. Jun 5, 2015 #5
    Thanks for the replies. Haruspex - Yes that's almost what I mean. I'm not sure if this is what you already thought, but beam 2 is connected and right beside beam 1 except beam 2 only goes halfway out. Beam 2 will therefore be twisting and pulling beam 1 down causing it to fail. Also as you said the only load being applied is the weight of the beams. Oh and yes, the pivot point was where I was planning to take moments about. And now I think about it I should only be going to the middle of beam 2 to the middle of beam 1 to find the radius.

    Thanks, Alex
     
    Last edited: Jun 5, 2015
  7. Jun 5, 2015 #6

    haruspex

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    Are both beams resting on support A, or is beam 2 only touching beam 1? If the second, beam 1 cannot be merely resting on support A but must be fixed to it so that it cannot rotate.
    In practice, beam 1 would be twisted to different extents along its length. That could mean the torque per unit length is not the same all the way along. Not sure how hard the question is supposed to be. Is it just the total torque we're after?
    Yes, go from beam centre to beam centre.
     
  8. Jun 5, 2015 #7
    Yes both beams are on support A. It was meant to be a challenging question I think. Yes, it's just torque that we are trying to find but obviously there's a few things to find before that. Could I just say that because the beam is uniform, that by finding the centre point of beam 2 and using mass x g that will be the force I can use in my calculations? Then sub that in to find torque? Or would I need to find the reaction force at one end of beam 2 and then the reaction force at the other end?

    Alex
     
  9. Jun 5, 2015 #8

    haruspex

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    Since beam 2 rests on support A it is more complex. It will exert no torque on beam 1 at that end, but progressively more as you move away from A.
    If we assume beam 2 is arbitrarily stiff it will remain straight, but will slope down slightly from A due to the twisting in beam 1. What does that suggest for how the torque on beam 1 will vary with distance from A?
     
  10. Jun 5, 2015 #9
    The further from A the greater the torque being applied I guess. So that would mean my force at the end of beam 2 would be greater. This is where I am stuck in finding that force.

    Thanks,
    Alex
     
  11. Jun 5, 2015 #10

    haruspex

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    Yes, but as an algebraic function, how will the torque per unit length vary with distance x from A?
    Think of the beams as rigid lengthwise, so straight, but allow that beam 1 can twist. How does the extent of twisting at distance x from A depend on the torque per unit length there?
    (This looks like quite a hard problem to me, so maybe I'm missing some simpler view.)
     
  12. Jun 5, 2015 #11

    haruspex

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    No, forget all that, we don't need to care how the torque is distributed.
    There are three forces from the supports, a normal force from B on 1 and a normal force from A on each beam. Use the usual equations to find these three forces, treating the two beams as a single rigid object. After that we should be able to find the torque from 1 required to support 2.
    Edit:
    That's not working for me either. The problem is that we do not know exactly where the normal forces at A and B act on the beam ends. If we suppose they are point supports at the midlines of the beams then by taking moments about 1's midline we find that the normal force at A on 2 equals 2's weight. It follows that there in no net torque from 2 on 1! Along the length of 2, the torque on 1 will start off in one direction then gradually shift around to the other, the whole cancelling out.
    If you have stated the problem exactly as given to you then I feel the problem is not well posed.
     
    Last edited: Jun 5, 2015
  13. Jun 5, 2015 #12
    Instead of me explaining is this how you find the normal forces? http://bendingmomentdiagram.com/tutorials/calculating-reactions-at-supports/
    If use this to find my normal force on each end of beam 1, how does this help me to find the force at the end of beam 2?

    Just to confirm about finding the torque, we are needing to find the torque of beam 2 pulling down on beam 1. The answer should give a result that means that the beams collapsed due to the torque.
     
  14. Jun 5, 2015 #13

    haruspex

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    Ok, maybe the confusion is that there are two torque directions here. I was thinking in terms of the torque about beam 1's midline (and you were too, right?). But judging from that link, we are interested in torques about horizontal axes perpendicular to the beams. That reduces it from a 3D problem to a 2D problem. And there is only one normal force at A.
    It's quite easy to find the two normal forces from the supports.
    Pick a point on 1 distance x from B (x being less than half the length). What is the torque about the point?
     
  15. Jun 5, 2015 #14
    Ok I think I get that but how does finding the normal forces help me find the force used in the formula T=f x r x sinQ? Or am I missing something?
     
  16. Jun 5, 2015 #15

    haruspex

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    Just as in the link. Take the midpoint of beam 1. To find the torque on it from support B you need to know the normal force at B.
     
  17. Jun 5, 2015 #16
    I think I understand now. Once I find the torque how do I know if this will cause the structure to collapse though? Sorry about all the questions! You are extremely helpful!
     
  18. Jun 5, 2015 #17

    haruspex

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    If you are supposed to deduce that it will collapse from only the information given then I am back to the first interpretation.
    Note what I wrote in post #6:
    It now sounds as though beam 1 is merely resting on support A and beam 2 doesn't touch A at all. So the torque from 2 is in danger of twisting 1 sideways off support A.
    Now the question is whether support A is merely a point at the midline of 1 (in which case collapse is obvious) or a flat surface spanning the width of beam 1.
    Since the second is the more interesting, let's go with that.
    If it's about to topple, exactly where on beam 1 will the normal force from A act? What about the normal force at B?
    Anyway, you still need to work out the normal force magnitudes at A and B. They're the same whichever interpretation is used.

    Edit: I give up. I cannot think of an interpretation that makes the question interesting but yields the given result. It is either impossible, because of inadequate information, or trivial. I'll invite some others to take a look.
     
    Last edited: Jun 5, 2015
  19. Jun 6, 2015 #18
    No problems. Thanks for all your help!

    Alex
     
  20. Jun 6, 2015 #19

    mfb

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    Same here.

    Please give the exact and full problem statement. Maybe we are missing something important.
     
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