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Statics problem -- Frame supported by pin connections...

  1. Jul 18, 2015 #1
    Problem is here,

    http://s24.postimg.org/3y6wpyncl/frame.png

    I'm looking at this problem with the top and bottom beams being separate members and doing a FBD of both separately. However I'm unsure about the points where we have tension/compression. Like at point B, do I include a y component and the tension, or just tension? My first instinct was to not include the y component by itself, but the problem asks to find the reactions at point B, so wouldn't that need to be broken down into x and y?

    Doing the latter, what I was thinking is taking a moment about D and going from there.

    MD = (0.8)TBEsin51.3 + (1.2)TFCsin68.2

    I would then solve for one of those variables, take the moment about A in the other diagram, and and plug in. Would that be correct?

    Also, for distributed force, would it be 400(1.4), with the distance being it is acting on the top member be [400(1.4)][0.7]?

    Thanks
     
    Last edited by a moderator: Aug 2, 2015
  2. jcsd
  3. Jul 18, 2015 #2

    Merlin3189

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    I'm not an expert here, but my first thoughts are:

    As far as tension or compression goes, I try removing that link (eg. BE) and ask how the ends would move (eg. would B and E become closer or further apart?) From the geometry and load one can often see what would happen.

    Looking at this diagram, I ask myself, can they both be in compression? Or both in tension? How would that affect the bottom beam?

    Looking at this problem, I would think about x and y components because you will be able to ignore one or other in some calculations: for eg. when you take moments about D, you can ignore y at A and x at E and F. It will also remove sines and cosines from some calculations.

    Perhaps you can share more of your thoughts, then people are able to make more pertinent comments.

    BTW I think there may be a typo in the question, as there does not seem to me to be a "member ABD". I am assuming they mean ABC.
    P#3 is worded strangely for me. I would work out the horizontal and vertical components, then the magnitude and direction of the total force.
     
  4. Jul 19, 2015 #3

    PhanthomJay

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    Yeah correct the typo.

    You should note that the pinned diagonals which are subject to forces at either end with no forces in between are thus 2-force members, subject to resultant axial forces along the member in tension or compression. Therefore, the x and y components of the tension or compression forces are trig related sines or cosines of the axial force. Draw free body diagrams of each beam and solve for unknowns.
     
  5. Jul 19, 2015 #4
    I solved the problem the way I think it's suppose to be solved, but the numbers I'm getting seem way too low, so I screwed up somewhere. Is my trigonometry correct in this equation?

    MD = (0.8)TBEsin51.3 + (1.2)TFCsin68.2 = 0
     
  6. Jul 22, 2015 #5

    PhanthomJay

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    Equation looks good but remember that in writing it you have assumed that both vert components of BE and FC forces act in the same direction. That's ok. Now when you look at the top beam, you must assume that those vert force components being equal but opposite per Newton act in the opposite direction. Then when you solve for them, if a result comes out negative, you assumed the wrong direction, and must make appropriate adjustments. Please show your full calcs for better assistance. The minus sign and it's interpretation will bury you if you are not careful.
     
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