Method of joints, force in each member

  1. Dec 25, 2011 #1
    1. The problem statement, all variables and given/known data
    Using method of joints, determine the force in each member of the truss. State whether each member is in compression or tension.
    [​IMG]

    2. Relevant equations
    joint method, I'm not sure how to describe it.


    3. The attempt at a solution
    [​IMG]

    The answer is of course
    Fab = 261.8 lb T
    Fbc = 78.9 lb C
    Fac = 39.4 lb T
     
  2. jcsd
  3. Dec 27, 2011 #2
    did you guys actually remove his post? It's very difficult to get anyone to help you here, why would you ever do that. Its the only resource I know of.
     
  4. Dec 27, 2011 #3

    SammyS

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    Do realize how hard it is to figure out what you did by looking at that image?

    It looks like there is some work relating to other problems.

    I'll try to take a closer look soon.
     
  5. Dec 27, 2011 #4
    alright so I checked out the example again and can't quite see whats wrong with my method (force triangles). Isn't it possible to break down that angled force into horizontal and vertical component, two separate forces, then use a force triangle for each force and add up the resultants somehow?
     
  6. Dec 27, 2011 #5
    I'm not asking you to decipher what I did. I pretty much just posted it out of desperation.

    and yes there is calculus on the top of the page
     
  7. Dec 27, 2011 #6
    [​IMG]

    This is the example question I base my method off. To find Fab & Fad:

    200lb/4 = Fab/3 = Fad/5

    (although the previously erased solution is making sense to me, I'm not sure if a question asks you to solve method of joints, that you are allowed to use moments?

    http://img291.imagevenue.com/img.php?image=875779209_1_122_404lo.jpg

    http://img158.imagevenue.com/img.php?image=875790153_2_122_257lo.jpg

    http://img255.imagevenue.com/img.php?image=875803217_3_122_240lo.jpg

    http://img240.imagevenue.com/img.php?image=875815914_4_122_538lo.jpg)
     
    Last edited: Dec 27, 2011
  8. Dec 28, 2011 #7
    I'm the guy who gave the other solution.

    It had later occurred to me about using moments. I realize that it would have been easier to do a force triangle at joint B. Unfortunately, I can't quite follow your photographed page
    I think that it's still method of joints so long as you use joint method to find each member. I just used moments to find one of the reactions first.

    If you have the time to post a better picture of how you tried to solve it then I'm sure that someone will 'lawfully' tell you what you did wrong (or I will try to).
     
  9. Dec 28, 2011 #8
    again thanks for your help, the question is not typical of the kind coming out of this chapter.. (ch6 of mechanics for engineers, statics, 5th ed) as the forces in the other questions are all perpendicular. So I'm moving on to them.
     
  10. Dec 28, 2011 #9

    SammyS

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    I get that Fcv=68.297 lb (up).

    However, that is based on moments, which I believe is what you did. That's not the method of joints.

    By the method of joints, joint B is in equilibrium (as are all the joints), so
    ƩFx=0 → (FAC)cos(50°) - (FBC)cos(60°)+240=0

    ƩFy=0 → (FAC)sin(50°) + (FBC)sin(60°)+100=0

    (I chose the signs to assume each beam was under compression.)​
     
  11. Dec 28, 2011 #10
    one more question though, is there a way to know which members are in tension or compression?
     
  12. Dec 28, 2011 #11

    SammyS

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    Yes, of course.

    Use a sign convention. In my suggestion, I assumed compression was positive.

    BTW: You should be able to check your solution by comparing what you got using moments.
     
  13. Jan 1, 2012 #12
    You know I'm not quite seeing this, would it be possible for you to do a free body diagram? thanks

    (shouldn't Fac in your equation actually be Fba?)
     
    Last edited: Jan 1, 2012
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