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Homework Help: 3 unknown forces acting on a beam

  1. Jul 28, 2010 #1
    1. The problem statement, all variables and given/known data
    I need to determine the forces developed in each bar as shown in the attached drawing.

    2. Relevant equations

    3. The attempt at a solution
    I have taken moments about the points A, C and E and come up with 3 equations and 3 unknowns.




    Whenever I try to solve for one of the unknowns, the other unknown cancels out meaning I cannot get an answer.

    Is there a simpler way of solving this which I am overlooking?

    Thanks in advance.


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  3. Jul 28, 2010 #2


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    If you eliminate E from these two, you will get an equation in A and C which you can solve simultaneously with
    to get A and C.
  4. Jul 28, 2010 #3
    Thanks for your fast reply.

    This may be a silly question but my mind has gone blank at the moment. How do I eliminate E from the two equations?
  5. Jul 28, 2010 #4


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    Try multiplying the second equation by 2 and then adding it to the first one. :wink:
  6. Jul 28, 2010 #5


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    Heloo a_hargy,

    Welcome to Physics Forums!

    I think I see what's going on. (And RockFreak, I don't think your approach is going to work by itself this time, for reasons I shall describe below).

    It's obvious that you have 3 equations and 3 unkowns. But what's not so obvious is that the three equations are not linearly independent.

    You can see that

    [tex] \left[ \begin{array}{ccc}0 & 0.4 & 0.8 \\ 0.4 & 0 & -0.4 \\ 0.8 & 0.4 & 0 \end{array} \right] \left[ \begin{array}{c} A \\ C \\ E \end{array} \right] = \left[ \begin{array}{c} 3000 \\ 3000 \\ 9000 \end{array} \right] [/tex]

    But the determinant of the matrix is zero

    [tex] \left| \begin{array}{ccc}0 & 0.4 & 0.8 \\ 0.4 & 0 & -0.4 \\ 0.8 & 0.4 & 0 \end{array} \right| = 0 [/tex]

    The matrix has no inverse. Which indicates that the equations that you chose are not linearly independent of one another. You need some other equation to solve this problem that is linearly independent of the other two.

    It turns out that you can throw away any one of the original equations, and the other two are linearly independent of each other. But you will still need a new third equation (more on that below)

    Just a example of what linearly independent means. Suppose we have the equation,

    [tex] x + y = 1 [/tex]

    and we need another equation to solve for x and y. Well we could just say,

    [tex] 2x + 2y = 2 [/tex]

    That gives us 2 equations and 2 unkowns, so we can solve for x and y right? Wrong. The equations are not linearly independent. The second is just the first, multiplied by 2. [Edit: By that I mean that the second equation gives no new information, that is not already present in previous equations.]

    Anyway, back to problem of this thread. Throw away one of the three equations you have now. You'll need to find a new third. And don't bother taking the moment around some different point, because that won't help either this time. But there is an easy equation that you may have forgotten to use, that doesn't have anything to do with moments (Hint: what is the sum of A, C and E? :wink:)
    Last edited: Jul 28, 2010
  7. Jul 28, 2010 #6
    Thanks heaps for that collinsmark. Thanks for taking the time to post that!

    I'm still a bit stuck on what other equation to use though.

    I tried using A+C+E=15000 but still get the same results :confused:
  8. Jul 28, 2010 #7


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    By golly, you're right. A+C+E = 15000, isn't linearly independent with any two of the other three either (I probably should have checked that before posting). :confused:

    [Edit: Previous comment about assuming E = 0, and the structure having negligible mass deleted. I'd rather not speculate on such things without being able to prove them.]

    It's been a very long times since I've done these sorts of problems. Let me think about this...
    Last edited: Jul 28, 2010
  9. Jul 28, 2010 #8


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    Hello a_hargy,

    I'm still not getting anywhere. Are there any more details regarding the problem that were left out of the problem statement in your original post? Anything about material properties or some-such? Some other piece of information?
  10. Jul 28, 2010 #9
    Hi collinsmark

    The full problem definition is:
    The three steel bars shown in the figure are pin connected to a horizontal rigid member. If the applied load on the rigid member is 15kN, determine the force developed in each bar. Bars EF and AB each have a cross-sectional area of 25mm2 and bar CD has a cross-sectional area of 15mm2.

    Thanks so much for all your help!
  11. Jul 28, 2010 #10


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    In retrospect, I think that's pretty important information that was left out! :tongue2:

    Anyway, I think this problem is a lot more involved than I originally thought. I might not be able to help you through to a final solution, myself. I'm running late.

    But here's what I'm guessing is going to be involved for you to solve this (but keep in mind it's been a long time for me, so take this with a grain of salt if you wish):
    (i) I think you can assume that the "rigid member" (beam at the bottom) does not bend.
    (ii) The bars are steel, implying that they bend a little, acting sort of like very stiff springs (sort of).
    (iii) You can probably look up elastic modulus (or perhaps Young's modulus or tensile elasticity) of steel in your textbook (I'm guessing).
    (iv) Using the length and cross sectional area of the bars, along with the elastic modulus properties, you should have enough information to model them like springs, more-or-less.
    (v) When the 15 kN load is attached, each rod will deform by a tiny amount, and the force on each rod is a function of this deformation. And also, the deformations on different bars are related because you know the "rigid member" does not bend -- so the deformations are related via the system's geometry.
    (vi) I'm speculating that that might be enough to keep resulting equations linearly independent of one another.

    Good luck!
  12. Jul 28, 2010 #11


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    I think you are on the right track. The fact that the picture shows the length of the vertical member implies that it's important. I think this is a statically indeterminate structure that needs deflection compatibility to solve.
  13. Jul 28, 2010 #12
    Thanks for all your help collinsmark.

    Assuming that the beam is rigid and does not bend, does this mean that the elongation of the steel bar AB would be greater then that of the steel bar CD assuming that E=0?

    Or would E infact not equal zero and the elongation of all three steel bars be equal?
  14. Jul 28, 2010 #13


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    My previous comment (before I deleted it in post #7) about guessing that force E = 0, was based on my assumption that all the bars were the same (and I also assumed they were all completely rigid) -- and it may not have been a worthwhile assumption anyway, which is why I deleted it.

    However, given the new information you posted, bars AB and EF have a cross sectional area of 25 mm2, but bar BC has a cross sectional area of only 15 mm2 (and none of the vertical bars are rigid). So the whole assumption about the force E = 0 is thrown out the window! Bar BC will have a greater elongation than bar AB for the same, given force!

    So the horizontal rigid member will/might tilt, just a tiny bit. But it's enough that bar EF will pull some weight too!

    [Edit: So the horizontal rigid member does not bend. But it might tilt every so slightly one way or the other, so you shouldn't assume that the elongations of all the bars are the same. However, because the horizontal rigid member does not bend, the elongations of the bars are related to each other (think triangle or trapezoid).]
    Last edited: Jul 28, 2010
  15. Jul 29, 2010 #14


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    Hello a_hargy,

    Just so you know, I came up with a solution to the problem. Again, it's been a very long time since I've worked with these sorts of problems, but I think it's correct. I can't give you my answer, but here is a general outline.

    (i) You already have a few equations you derived in your first post. Pick any two of them. You can still use them later.
    (ii) I made the assumption that any stretching of the bars is so small, that it does not have any significant impact on first two equations, so we can use them the way they are.
    (iii) You can use the new information about the cross sectional area of each bar to find the third equation.
    (iv) When the load is applied, and the bars elongate slightly, the elongation pattern forms a trapezoid. Think of a rectangle, except with a slanted bottom. The bottom left corner is A, the bottom right corner is E, and C is right in the middle. Using simple geometry, derive an equation which shows the relationship between dA, dC, and dE, the vertical displacements at each point.
    (v) Use the cross sectional area information to model proportional "spring constants" to bars AB, CD, and EF. The way I did it, you don't need to calculate the actual spring constant values. Rather you just need relative values. The differences in cross sectional areas is the key here (at least the way I did it).
    (vi) Using that you can find a new relationship between the forces A, C and E (noting that d = F/k, for displacement d, Force F, and spring constant k). That's your new third equation! (And it is linearly independent of the first two.)
  16. Jul 29, 2010 #15
    Thanks so much for all your help collinsmark.

    I think I finally worked it out.

    I used the assumption that the elongation of AB/0.8m was equal to the elongation of CD/0.4m then used my previous equations to solve for the unknowns.

    My results for the forces were:

    The results seem fairly realistic to me. Were your results similar?
  17. Jul 29, 2010 #16


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    I don't follow what you mean here. Wouldn't that make the elongation of EF 0.0 m ([Edit: or would that be ∞ m?])?

    Don't make the assumption that the elongation of E is anything in particular. All you need to do is find a relationship between the 3 different elongations. Perhaps this figure will help:
    Realize that bar CD is directly in the middle of the other two (this distance from bar CD to bar AB is exactly the same distance from bar CD to EF).
    My answers were different. They were not too terribly different, but different enough to certainly be different.

    Attached Files:

    Last edited: Jul 29, 2010
  18. Jul 31, 2010 #17
    A simple approach to this is to 'move' the 15 kN force to C, but also impose there a moment of 15 x 0.2 kNm anticlockwise. The 15 force in the middle stretches the 3 bars equal amounts. The moment leaves the force at C unchanged and is resisted by forces in A and E. This will act as a check on the way you did it otherwise.
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