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Homework Help: ME statics Draw the moment diagram

  1. Nov 9, 2014 #1
    1. The problem statement, all variables and given/known data
    The first segment starts at -650 and ends at -1100.
    The second segment starts at 1200 and ends at 0.

    2. Relevant equations
    dM/dx = V

    3. The attempt at a solution
    I attempted to break the problem into two parts.

    For the first segment:
    Skipping a few steps:
    w = 100x

    dV/dx = -100x

    V= -50x^2

    This is where my first confusion arises.
    If dV/dx = -w

    then how can the solutions guide say

    Sum of the forces in Y = 0 = -650 - 50x^2 - V
    V= -650 - 50x^2

    Now I have two things that V is supposed to equal...?!

    Moving on:
    dV/dx = M

    M = -650x - (50/3)x^3

    So the first segment starts at 0 and ends at -2100 which I got from plugging in x=3 into my moment equation. It is a cubic function that has an increasing negative slope as it approaches x=3.

    For the second segment:

    V = -300x

    Sum of the forces in the Y = 0 = -650 +2300 -300x - V
    V = 1650 - 300x
    M= 1650x - 150x^2

    but now I'm not accounting for the first part of the beam. I'm completely lost at this point.
  2. jcsd
  3. Nov 9, 2014 #2
    We traveled an almost identical road (at great length) on your last homework problem. It would be a really good idea to try to work this one out by yourself.
  4. Nov 9, 2014 #3
    I know, I tried going back to it but it was different since it only had a uniform distributive load. This has a distributive load that changes in the middle and that messes me up.

    I've been working at this problem for almost 2hrs straight and I noticed that I wasn't cubing my X so I think I can get the correct graph now but I'm still confused about the parts that I pointed in my opening post.
  5. Nov 9, 2014 #4
    So, maybe you need to start understanding, thinking the whole process through, rather than being so focused only on getting an answer. What will you do if you meet a problem where the load varies as sin(x)?
  6. Nov 9, 2014 #5
    That's exactly what I'm trying to do my friend. I just solved this problem but that's not really what I care about. I don't understand the part where they say that dV=dx = -w and then they use the sum in the Y-direction to find V... Well, which one is it!?!?

    And what happens to the first distributive load when I solve for moments? I'm so confused. The book just adds to my confusion and my professor has a Chinese accent which makes him completely incomprehensible.
  7. Nov 9, 2014 #6


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    It's best to start off with the basics:

    The beam must be in static equilibrium in order to be analyzed. What reactions have you calculated at points A and B to put the beam in equilibrium?
  8. Nov 10, 2014 #7
    There are no forces in X.

    Ay = -650 and By= 2300.
  9. Nov 10, 2014 #8


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    Start at point A. If you accumulate the distributed load w over the interval 0 < x < 1 meter, how many newtons is this?
  10. Nov 10, 2014 #9
    300/3 = w/x

    w= 100x

    from 0 to 1, w= 100N.
  11. Nov 10, 2014 #10


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    The distributed load at x = 3 m is w = 300 N/m. By ratio, the distributed load at x = 1 m is w = 100 N/m. Note the units here. w is not the total load over the interval 0 < x < 1, it is just the ordinate of the distributed load curve. In order to calculate the total applied load on this interval, you must find the area under the distributed load curve, which is 100 N/m * 1 m * (1/2) = 50 N.

    At x = 1 m, the shear V = RA - 50 N = -650 - 50 = -700 N, which should plot on the shear diagram shown in the OP.
  12. Nov 10, 2014 #11
    So in other words:

    w= 100x gives me the slope.

    Then the integral of w should be the accumulated load over a length X.

    Accumulated Load = 50x^2

    Then, over 3m I get 9x50 = 450N from the distributed load.

    Sum of the forces in Y if we cut the beam right before point B:

    = -650 - 50x^2 - V

    Why is the book calling the accumulated load "V". Accumulated load doesn't equal shear. Shear is the accumulated load + whatever concentrated loads, correct?
  13. Nov 10, 2014 #12


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    I don't know what your book is saying exactly because I'm not able to read it.

    The shear force includes any reaction forces combined with whatever loads are applied to the beam.
  14. Nov 10, 2014 #13
    But am I correct so far?
  15. Nov 10, 2014 #14


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    Yes. All you have to do is compare your calculations with the shear diagram in the OP.
  16. Nov 11, 2014 #15
    I drew that diagram :). The shear part was pretty straight forward although I had some confusions that you cleared out.

    Now to the hard part, the moment diagram.

    How do I go about finding the moment diagram?

    It is the integral of the shear diagram so for the first part it should just be

    M = -650x - (50/3)x^3 ?

    Okay fine, but for the second part.

    M = integral of what!?

    I found that the second equation should be V = 2100 - 300x

    from y=mx + b where m= -300

    Ay V=0, x=7 therefore b=2100

    but that gives M= 2100x - 150x^2 which doesn't yield M=0 at x=7.
  17. Nov 11, 2014 #16


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    This expression for the bending moment is valid on the interval 0 < x < 3 m.
    Once you have found the bending moment at x = 3 m, you continue calculating the area under the shear curve for x > 3 m. This area is then added to the bending moment at x = 3.
    The equation M = 2100x - 150x2 is valid only for calculating the area under the shear curve in the interval 3 < x < 7 m.

    Let's call the bending moment equations as follows:

    M1(x) = -650x - (50/3)x3, for 0 < x < 3; and

    M2(x) = 2100x - 150x2, for 3 < x < 7

    Both of which are derived by integrating the shear force curve.

    In order to find the bending moment in the interval 0 < x < 3, you would simply calculate M1(x). Once x > 3, then the bending moment is calculated as follows:

    M(x) = M1(3) + M2(x) - M2(3), for 3 < x < 7.

    As an example, for x = 7 m, the bending moment is

    M(7) = M1(3) + M2(7) - M2(3), where

    M1(3) = -2400 N-m

    M2(3) = 4950 N-m

    M2(7) = 7350 N-m

    So that M(7) = -2400 +7350 - 4950 = 0 N-m, as required.
    Last edited: Nov 11, 2014
  18. Nov 11, 2014 #17
    Thank you! I completely get it now.

    If you're ever in Los Angeles I'll buy you a beer haha.
  19. Nov 11, 2014 #18
    What the heck....?! It doesn't work for this one: Untitled.png

    I was able to solve it graphically but if I use the same steps

    w= 2x

    Vx= -x^2

    Sum of the forces in Y
    V = 9 - x^2

    clearly that is not the graph, it is (x-3)^2=V
  20. Nov 11, 2014 #19


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    You're still trying to eyeball these problems and not work them out in a systematic fashion.

    Look at the distributed load for the interval 0 < x < 3 m. Is w = 2x? I don't think so. If x = 0, then w must equal 6kN/m. If x = 3, then w = 0.
    What sort of equation must you write to make these two facts true?
  21. Nov 11, 2014 #20
    I'm absolutely not. I tried going back and using the same exact method that we used on the previous one.

    I think I see my mistake now though.

    Vx = 6-2x

    So the ratio only gave me the slope and I have to find the Y-intercept.

    Sum of the forces in Y =
    V = X^2 -6x + 9

    I see. Thanks again and thanks for your patience.
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