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Bending Moment and Shear Force Problem

  1. Mar 29, 2014 #1
    1. The problem statement, all variables and given/known data

    Hi all, I have a problem determining the moment and shear force at a location. The moment, I think, I have already solved. But I get stuck at calculating the shear force at the same location. It is for a element that is suspended in water, thus the weight of the element needs to be compensated for the buoyancy also. Additionally, there is buoyancy attached. See the attached email for all the details.

    2. Relevant equations

    Wsubmerged (Ws) = Wdry (W) * (ρsteel (ρs) - ρwater (ρw))/ ρsteel

    Buoyancy force (Bu)= Buoyancy (B) * g

    Tangential component of weight (Wx) = Ws * sin(α*(pi/180))

    Normal component of weight (Wy) = Ws * cos(α*(pi/180))

    Tangent component of buoyancy (Bx) = Bu * sin(α*(pi/180))

    Normal component of buoyancy (By) = Bu * cos(α*(pi/180))

    Bending moment at end (Mb) = Wx * Yc + Wy * Xc - (Bx * Yb + By * Xb)

    Shear force (V) = ?

    3. The attempt at a solution

    see attachment, could someone help me with this problem? Many thanks in advance!

    Attached Files:

  2. jcsd
  3. Mar 29, 2014 #2


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    It's not clear what is going on. Please post the original problem so that we can see how your calculations have reached the point shown in the OP.
  4. Mar 29, 2014 #3
    Hi Steamking, sorry, but this is all I have got. It is a supplemental case which was given to me in the sketch stating that I need to derive M and V. I only have the sketch which I attached to the post. What do you need specifically?
  5. Mar 29, 2014 #4


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    Are these forces for the weight and buoyancy derived from some sort of vessel which is floating at an unusual attitude in the water? If so, the distribution of the weight and the buoyancy along the length determine the shear force and bending moment values. For example, if you had a floating box which was evenly loaded from end to end, there would be no shear force and no bending moment experienced by the box.

    When you said in the OP that you had already solved the bending moment but were uncertain about the shear, that made no sense to me, since floating objects react like any other beam, and you must know the shear force before you can calculate the bending moment. For floating objects, the difference in distribution of the weight and the buoyancy affect the magnitude of the shear force and the bending moment.
  6. Mar 29, 2014 #5
    Hi Steamking, that is correct. The weight and buoyancy are derived from a vessel which is submerged in the water. It is a basic element which is a beam at an attitude with the horizontal. I do understand your reply about the equally distributed load.

    What I did is, I ignored the shear and derived the bending moment from the buoyancy force and weight of the beam. However, this is conservative I think. However, since I do not know how to calculate the shear force at the end, I can't recalculate the bending moment with the shear force component included. Could you help me with that? I am sorry for incorrectly addressing my problem.
  7. Mar 29, 2014 #6


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    It's not clear how you were able to calculate bending moment without first calculating shear.

    Looking at the figures for W and B, it doesn't appear that the vessel is floating in equilibrium. Is one end grounded or supported in some fashion? Like all other beam problems, the beam (vessel) must be in equilibrium in order to calculate the shear force and bending moment, which of course, means that W must equal B. I'm not sure if you have calculated a bending moment on the vessel or the trimming moment, which results from not having W and B aligned vertically.

    In any event, I don't see how to proceed without further description of the vessel and the details of this problem.
  8. Mar 29, 2014 #7
    Steamking, I asked some more information from my mentor and what I got was the following:

    - The force T is at a hinged connection (can generate no moment).
    - V, M and F are a pinned connection. Hence, the beam is fixed at this location.

    Does this help?
  9. Mar 29, 2014 #8
    The notation given in the pdf attachment needs clarification:

    1) I assume V, M, and F are the forces necessary to keep the object in equilibrium?

    2) is XW and YW supposed to be XC and YC?

    3) is Mb your calculation of the bending moment M shown in the diagram?

    4) it's hard to tell what values were given and what values you calculated

    5) is γ supposed to be the direction of T only with α=30.2 deg. still?

    6) it looks like you got your normals and tangentials mixed up

    7) you seem to have taken moments about the other end of the object since L is mysteriously omitted

    8) W appears to be the weight of the object yet it is out of synch with the dimension L

    9) the center of pressure of the buoyant force will not be the same as the center of mass
  10. Mar 29, 2014 #9


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    It helps a little, but I'm still not sure if it will be enough.

    The beam still needs to be in static equilibrium in order to calculate SF and BM.

    If there are no further details on the longitudinal distribution of the weight and the buoyancy, then calculating SF and BM won't be possible.

    If M is at a pinned connection, there must be something special about the connection, because typical pinned connections are not usually capable of developing a moment.
  11. Mar 29, 2014 #10
    Ok, I have discussed it with my mentor and he hasn't got all the details. He therefore provided the attached model to calculate M and V. However, since I am most likely looking to wrong way, could you help me please with M and V (shear force)?

    Attached Files:

  12. Apr 1, 2014 #11
    I intercede in this discussion only to point out that in practice Civil Engineers often obtain the bending moment function before they obtain the shear force function, because bending moment so often determines the size of the member. If M is a f(x), then differentiating it wr to x gives the shear force function. That is common practice.
  13. Apr 6, 2014 #12
    Hi, sorry, but it does not help me much :(.
  14. Apr 8, 2014 #13


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    Martijn79: Notice, in my attached diagram, how each point is labeled (points A, B, C, D, E). You must always label each point in your structure, so people will know which point you are referring to.

    Also notice the constraint (support) symbols at points A and D. (These are the constraints you seemed to describe in your posts.) If you do not draw accurate constraints, reaction forces, and applied loads on your FBD, then no one will know what the given problem is.

    For clarity, I want to rename your V reaction force to Vd, and your M reaction moment to Md, and your F reaction force to Rdx, and your T reaction force to Rax, and your B applied force to Fb, as shown in my attached diagram.

    Notice, at point D, which is a roller constraint, reactions Rdx and Md are zero, by inspection. The mass of your structure (mc) is highly asymmetric, but that is OK. The center of gravity of your structure is located at point C. Gravity and buoyancy force act on mass mc. There is also an applied force Fb, applied at point B.

    Your attempted solution was somewhat close, but contained a few mistakes. To show you some of your mistakes, and to show you how to compute shear reaction force Vd, here is an example problem.

    mc = 28.256 tonne
    Fb = 220.650 kN
    xb = 4040 mm
    yb = 1328 mm
    xc = 3639 mm
    yc = 856 mm
    L = 12 200 mm
    theta = 20 deg
    g = 9.80665 m/s^2
    rho_s = 7850 kg/m^3
    rho_w = 1025 kg/m^3​

    Now determine Vd, as follows.
    Ws = mc*g*(rho_s - rho_w)/rho_s = 240.915 kN
    Wsx = Ws*sin(theta) = 82.398 kN
    Wsy = Ws*cos(theta) = 226.386 kN
    Fbx = Fb*sin(theta) = 75.467 kN
    Fby = Fb*cos(theta) = 207.343 kN

    Therefore, summation of moment about point A gives,

    summation(Ma) = 0 = -xc*Wsy - yc*Wsx + xb*Fby + yb*Fbx + L*Vd.

    Solving for Vd gives,

    Vd = (xc*Wsy + yc*Wsx - xb*Fby - yb*Fbx)/L
    = -3.568 kN​

    Now see if you can solve your given problem.

    PS: If Rdx and Md are instead applied loads, instead of reaction forces, then they would need to be given in the given parameters.

    (1) By the way, always leave a space between a numeric value and its following unit symbol. E.g., 4040 mm, not 4040mm. See the international standard for writing units (ISO 31-0).

    Attached Files:

    Last edited by a moderator: Apr 12, 2014
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