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Statically indeterminate beam

  1. Sep 27, 2013 #1
    1.
    How do I find an expression for the reaction force (Xc) in C? The beam is fixed in A and there is a roller connection in C. The roller connection is placed at the center of the beam. Length of beam is L. The beam is loaded with a triangular load from A to B (Q kN per meter).

    Illustration of problem: http://goo.gl/xJ6u7x [Broken]

    2.
    http://www.engineering.uiowa.edu/~design1/StructuralDesignII/Chapter5-ForceMethod.pdf

    3.
    The problem is that there is four unknown reaction forces, so I can't solve by applying the three equations of equilibrium. I tried using the force method for statically indeterminate beams to find the reaction force in C. I don't know if it's correct, but I got Xc to be QL/40.
    Attempt at a solution: http://goo.gl/oD7rPn [Broken]
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Sep 28, 2013 #2

    nvn

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    raymanmusic: Your answer for fcc on p. 2 is correct. Your answer for delta_qc on p. 1 is currently wrong. Therefore, let us focus on only p. 1. First, we need to mention something that is making your work unclear, and hard to check. It causes people to be baffled, and therefore, they might not respond.

    You are failing to label the points on your free-body diagrams (FBDs). E.g., your last two FBDs on p. 1, and your last two FBDs on p. 2, have no points labeled on them. Therefore, we do not know exactly where the FBD is located. (For your other FBDs, we can guess the points, although we should not be forced to guess; it is easy to label the points, regardless.)

    Note that if a point is at a varying section cut, only then can you omit that varying section cut point label.

    Secondly, you are failing to show your coordinate system on each of your FBDs. E.g., on your last two FBDs on p. 1, you give a dimension labeled with a domain of x, such as 0 < x < L. This, by itself, is not too useful. You need to show the origin of your x axis, and the direction of x. It might look something like as shown in my attached file.

    Arbitrary examples are given in my attached diagram. Notice, in both figure 1 and figure 2, any known points are labeled (points A and C). And in both figures, my coordinate system origin and direction are defined and shown. In figure 2, I also show relevant dimensions, in case they were not previously shown in an earlier FBD.

    Therefore, if you could first label the points in your FBDs, and show your chosen coordinate system on each FBD, and repost your attached file, then we might be able to understand what you are talking about. And then we might be able to find where you went wrong in your solution.
     

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    Last edited: Sep 28, 2013
  4. Sep 29, 2013 #3

    nvn

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    raymanmusic: Here is another tip. We usually write summation of moment about point A as summation(MA), as shown in my attached diagram, below. A second example in my diagram shows summation of moment about, e.g., the section cut located at coordinate x. Or, you could change x to x1, if you have multiple coordinate systems.

    Some people sometimes prefer to change the second equal sign, in my diagram, to a colon.

    Your current notation states, "Summation of moment about the dot located in the adjacent FBD." The problem with this notation is, every summation of moment equation has the same name. Let's say you have a mistake in a summation of moment equation. Then one would write, "There is a mistake in the second term of your summation of moment about dot equation." Which summation of moment about dot equation? First, second, third, ...? It is just unclear nomenclature.

    If your summation of moment equation nomenclature were instead something like as shown in my diagram, then one would write, "There is a mistake in the second term of your summation(MA) equation."
     

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    Last edited: Sep 29, 2013
  5. Sep 29, 2013 #4
    thanks for answering nvn. I labeled my FBD and added x-coordinate system. I'm struggling with finding an expression for the deflection at c with the triangular load. The triangular load is from 0 to L, but my point load is from 0 to L/2. I had to look in my book to find the elastic curve formula for triangular load on a cantilever beam and put x=L/2. I then got an expression for the deflection at C: (Q(L^4)*49)/(3840EI). My final expression for reaction force at C is then: (QL*49)/(160). Do you know how you calculate the deflection for a triangular load on a cantilever beam at midspan without using the the elastic curve formula?

    Attempt at a solution: http://goo.gl/ueOAX6 [Broken]
    I used beam type nr. 4 and used the elastic curve formula under "deflection at any section in terms of x": http://goo.gl/esY5D
     
    Last edited by a moderator: May 6, 2017
  6. Sep 29, 2013 #5

    nvn

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    raymanmusic: Your answer is correct.

    Now going back to your original approach, see my attached diagram, below, to compute delta_qc. Hint: Sum moments about section cut s1 in figure 3, then solve for M1, to obtain an expression for M1. Next, sum moments about section cut s2 in figure 4, then solve for m1, to obtain an expression for m1. Now see if you can finish it, and obtain delta_qc.
     

    Attached Files:

  7. Sep 29, 2013 #6
    Thank you so much. I followed your hints, I think I got it right now. I got my expression for deflection at C to be:[itex]\frac{QL^4}{80EI}[/itex]

    I don't know why, but my new answer for the reaction force at C [itex]\frac{3}{10}QL[/itex] is a bit less than my previous answer [itex]\frac{49}{160}QL[/itex]. It' only 0.00625 difference so I don't think it matters that much. Thanks for the help!

    Attempt at a solution: http://goo.gl/ac2dQg [Broken]
     
    Last edited by a moderator: May 6, 2017
  8. Sep 29, 2013 #7

    nvn

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    raymanmusic: Your expressions for M1 and m1 are correct. Nice work. But your answer for delta_qc is currently wrong, because your limits of integration are wrong. Your upper limit (L) is correct, but your lower limit is currently wrong. Try again. I.e., try computing delta_qc again.
     
  9. Sep 29, 2013 #8
    Of course, it's from [itex]\frac{L}{2}[/itex] to [itex]L[/itex]. Deflection at C is then: [itex]\frac{49QL^4}{3840EI}[/itex] and reaction force at C is: [itex]\frac{49QL}{160}[/itex]
     
  10. Sep 29, 2013 #9

    nvn

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    Nice work, raymanmusic. Your answer is correct.
     
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