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Force analysis, tell me if I am correct.

  1. Jul 23, 2010 #1

    rock.freak667

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    1. The problem statement, all variables and given/known data
    I have this sort of design basically

    http://img834.imageshack.us/img834/6650/14820321.jpg [Broken]

    I'd like to a force analysis on it to determine if it is a good design as a point load is applied at A (we want to reduce AB from bending).

    2. Relevant equations

    Force analysis

    3. The attempt at a solution

    I was not given the dimensions as yet so I am trying to do it with symbols initially.

    Forces at the points are F (e.g. Force at A= FA)
    Reaction forces are R
    Weights are W
    Moments are M
    If needed angle CBD=θ, BCS=90°
    Lengths are just the two points (e.g. Length of AB = AB)

    Weight of ABC and BD are downwards (WABC, WBD) and there is a force at A, FA and a force in BD, FBD (at θ).

    This induces a reaction at C and D, RCx,RDx,RCy and RDy

    Since there is a force acting at a distance there is moment due to FA and FBD (MA and MBD).

    Now I either will need to get an expression for deflection or bending stress. I believe I'd need to check both. Now I know how to do this with just the cantilever part alone, but the member BD is confusing me as to how to formulate the bending stress, σbending and deflection δ.

    So firstly, is my free body diagram correct? (well mainly the forces I typed out, sorry I couldn't draw it in paint)
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jul 24, 2010 #2
    Can we suppose that BC also has weight? Can you define the reactions at C and D as pin-jointed? Is the member ABC continuous across the joint B. How is BD connected to ABC at B? A free body diagram should have ALL the forces put on it.
     
  4. Jul 24, 2010 #3

    rock.freak667

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    Sorry, I meant member ABC has weight WABC. BD is sort of welded onto ABC.

    EDIT: I put in all of the force's directions now, but I did not label them since it is quite tedious in paint. But I did identify the forces in the first post


    http://img716.imageshack.us/img716/8066/55122304.jpg [Broken]

    EDIT2: Assuming I have my forces and moment correct.

    ∑Fy= 0 ⇒WABC+WBD+FA=RCy+RDy+FBDsinθ.

    ∑Fx= 0 ⇒ FBDcosθ = RCx+RDx

    Now I am not sure if my reaction moments at C and D are correct else I will not have enough equations to solve for all my variables I believe.
     
    Last edited by a moderator: May 4, 2017
  5. Jul 25, 2010 #4
    If the connection of BD to ABC at B is fully welded, then the bending moment at B arising from the load at A will be shared between BC and BD in proportion to the flexural stiffnesses. Your latest diagram shows rigid (that is, moment resisting) supports at D and at C. To achieve this is practically quite difficult, requiring a good embedment of the members say 8 times their thickness. Your arrows confuse external forces (vertical) with internal forces (in BD). The other end of BD has an internal force going in the direction DB. You need to be rigorous about fb diagrams. If this is a practical problem, best to assume pin supports at C and D, with ABC continuous, and the connection of BD to ABC at B as a pin (even if it isn't). In that case, if you know the perpendicular distance from C to BD, you can the force in BD by taking moments about C, and move on from there.
     
  6. Jul 25, 2010 #5

    rock.freak667

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    Therefore, I just need to add an arrow from D to B and remove the resisting moments (I just looked up fixed end supports and saw it gave those reactions).

    For the forces in member BD, when taking moments, I just need to consider the force from either D to B, or from B to D and not both right?

    How would I also find the deflection produced now?

    The support at C is the most important I believe since BD was only added to reduce bending. So C is the most heavily stressed point. The bending moment at C would just be FA*)ABC) right? Or is affected by member BD?

    Also, if at B is a pin joint, I'll have two extra forces correct(RBx and RBy), so my system is statically determinate right?
     
  7. Jul 25, 2010 #6
    "For the forces in member BD, when taking moments, I just need to consider the force from either D to B, or from B to D and not both right?"

    There is only one force in BD. Do you suppose that is tension or compression?

    To find an internal force, you cut the member and 'expose' the unknown force, in this case FBD. As I said, taking moments about A for the cut structure (not involving joint D at all) will give an equation from which you can deduce FBD.

    If BD has pins at the ends, can you see what direction the reaction must be at D? This is a structure subjected to 3 external forces. For equilibrium, those 3 lines of forces must meet at a single point. Can you identify that point, and then conclude the direction of the reaction at A? I won't answer the rest of your questions until you have made a bit more progress with FBD and the directions of the reactions.
     
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