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Need help understanding cantilever beams

  1. Jun 14, 2012 #1
    Hey Guys,

    I need help understanding how to find the deflection of the following cantilever beam (in the image: capture).

    There a beam with a fixed end (point A), and a roller support halfway on the beam (point B), I need to find the deflection at the end of the beam (point C).

    Can someone just double check that I've done this right:
    To find the reaction at point B is just doing the ƩMat A=0=-P2L+RBL → RB=2P

    Now to find the deflection at C (see capture2), can I just apply: vmax=(5×RB×(2L)3)/(48EI)

    Thanks
     

    Attached Files:

  2. jcsd
  3. Jun 14, 2012 #2
    This looks like a homework type question and sorry but you are certainly on the wrong track with your current thinking.

    One of the reactions at the root of a cantilever (point A) is a fixing moment.

    How does this affect your equation

    ƩMat A=0=-P2L+RBL → RB=2P

    Consider point B. There are two properties/quantities that are zero at B. Have you heard of the slope-deflection equations?
     
  4. Jun 14, 2012 #3
    ƩMat A=0=-P2L+RBL → RB=2P
    I just used the summation of moment at point A to find the reaction a point B which was then related to finding the deflection of the beam.

    Isn't the slope-deflection equation just θ=.......? , but I don't see how that helps me find the deflection at the end of the beam.
     
  5. Jun 14, 2012 #4
    Just another thought, is it possible to use superposition principle:

    Case 1:Find the deflection between point A and C without the support of point B
    Case 2:Then find the deflection between A and B (i.e. delfection going upwards)

    The total deflection is equal to case 1 + case 2

    But this still leaves the problem as, what is the reaction at B as this equation will find the total delfection at C....

    Is this a possible way of finding the deflection at point C?
     
  6. Jun 14, 2012 #5
    But why do you think the moment at A is zero?
     
  7. Jun 14, 2012 #6
    ahh... bugger, I must of forgot moment at A, sorry.

    so it should be ƩMat A=0=M-P2L+RBL → RB=(M+2PL)/L

    Does it mean that if I was to take the moment around point B, it would be ƩMat B=0=-PL+RAL → RA=P ??
     
    Last edited: Jun 14, 2012
  8. Jun 14, 2012 #7
    Yes you can use superposition or double integration (with Macaulay brackets) or area-moment or slope-deflection.

    However you must apply all the loads correctly.

    So the beam tries to act like a see-saw about B.
    This implies that it A tries to rise so there is a downward vertical reaction at A.
    Additionally there is a moment at A.

    Since there is also a vertical reation at B the beam is redundant.
    So you have to use more than just the equations of equilibrium.

    So what quantities at B can you immediately put values to?
     
  9. Jun 14, 2012 #8
    So are you talking about the beam from the fixed end (point A) to point B?

    Do you mean like the deflection and slope at B would be zero?
     
  10. Jun 14, 2012 #9
    Exactly you are getting there.

    If you wanted the actual force values of the reactions you could also note that the deflection at A is also zero.


    To use superposition you have two cantilevers AC, one loaded with the tip load P, P. The other is loaded with the roller reaction at B. Since B is upwards everything is considered negative.

    The deflections at B are zero so you can equate this to the sum of the deflections from the individual cantilevers.

    At C one deflection is negative one is positive, and the difference is the resultant deflection.
     
  11. Jun 14, 2012 #10
    I'm still a little lost on how this would help me find the reaction at B...

    So, which slope equation should I be using?, because I am just getting confused. And do I need the deflection equation with the slope equation to find the reaction at B?



    yep, so just like what I said earlier:

    Case 1:Find the deflection between point A and C without the support of point B
    Case 2:Then find the deflection between A and B (i.e. delfection going upwards)

    The total deflection is equal to case 1 + case 2
     
  12. Jun 14, 2012 #11
    Solving the question by superposition requires you to calculate at least a figure for the reaction at B since you can then use two standard cases


    [tex]\delta = \frac{{P{L^3}}}{{3EI}}[/tex]

    for the cantilever loaded at the tip

    and


    [tex]\delta = \frac{{{R_B}{a^2}(3L - a)}}{{6EI}}[/tex]


    For a cantilever loaded at B with RB and a is the distance AB.
    Note RB is negative so deltab is negative.

    To calculate RB you need an equation with it in. This comes from one of the conditions of zero slope and deflection at A or zero slope at B. Can you see how to get there?
     
    Last edited: Jun 14, 2012
  13. Jun 14, 2012 #12
    ok, so then deflection at C is: ~ in which I get

    [tex]\delta = \frac{{{-R_B}{a^2}(3L - a)}}{{6EI}} + \frac{{P{L^3}}}{{3EI}}[/tex]

    The only part at the moment is still can't quite see how you would still find RB.

    So are you saying that there should be another part which goes into the equation above to help me find RB or do you mean there should be another separate equation?

    So say I take point B:
    [tex]\theta = 0 = \frac{{{-R_B}{a^2}}}{{2EI}}[/tex]
    For the slope at B and
    [tex]\delta = 0 = \frac{{{-R_B}{a^3}}}{{3EI}}[/tex]
    For the deflection at B

    Am I on the right track?
     
  14. Jun 15, 2012 #13
    I am continuing with superposition since you mentioned it. Further it is a good idea to understand how to derive the formulae since actual situations or problems may not match your formulae exactly.

    OK so I have done the opposite of superposition by decomposing the cantilever loadings in the sketch.

    I have drawn the BM diagrams for P alone and RB alone.

    You should check you can do these.

    Now we can use Mohr's second theorem of the area moment method to derive a formula for RB thus.

    For any two points Z and W on the beam, the deflection of Z relative to W is given by 1/EI times the first moment of area of the BM diagram between Z and W about Z.

    Now since the deflection at A = the deflection at B = 0 we can say the deflection of B relative to A = 0.

    So if we take moments of the areas of the BM diagram between A and B, about B and equate to zero we get,
    an equation between L, a, P and RB that we can solve for RB

    I have indicated sufficient dimensions and the distances of the centroids of the areas on the sketch.

    Use these to calculate a formula for RB and post it.
     

    Attached Files:

  15. Jun 16, 2012 #14
    [tex]R_B = P(a+ \frac{3}{2} (L-a))[/tex]

    So then this goes into the equation to find δ at C
     
  16. Jun 16, 2012 #15
    Yes I agree with that expression for RB.
     
  17. Jun 16, 2012 #16
    Thanks heaps Studiot for all the help :D
     
  18. Jun 16, 2012 #17
    So can you now see why you can't just cosider the part projecting over the roller as a short cantilever?

    It is also very tempting to invert the cantilever so that P becomes the prop at the end and RB becomes the load.
    If you do this you will find the formulae are different.
    Can you see why?

    This is a good problem to improve one's understanding.
    A tip here is to practise with it using real numbers from say a problem in a book.
    Using numbers reduces the algebraic manipulation and allows you to test any formulae you develop.

    As you revising for an exam?
     
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