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Shear force and bending moment diagrams question

  1. Dec 3, 2015 #1

    diredragon

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    1. The problem statement, all variables and given/known data
    In the image uploaded below, a boxed shaped object is shown. It is floating in water so the forces that act on it are buoyancy and gravity. The first and the last compartments are filled with evenly distributed weight. The middle 2 compartments are empty.
    Specifics are:
    Lenght of total box: 20m
    Lenght of each compartment: 5m
    Weight of empty comp: 20t
    Weight of 10t evenly distributed at conpartments by 2t/m
    My question here is how were the diagrams posted below drawn? I dont understand the bottom one at all.
    http://postimg.org/image/9q6xrouer/
    2. Relevant equations
    3. The attempt at a solution
     
  2. jcsd
  3. Dec 3, 2015 #2

    haruspex

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    The shear force is shown by the zigzag line. Pick some point at distance x along from the left. Add up all the vertical forces acting between x and one end. It doesn't matter which end, the only difference being that one will give you a negative number and the other a positive.
    In the diagram, looks like they take the forces on the left of x. At x=0 the result is zero. As you move along, negative (downward) force is being added linearly. One you cross the quarter point, the trend reverses.

    Bending moment is shown by the curved line. Bending moment is similar, but in this case we also multiply by the distance from x to the applied force, and only consider from x to one end. If the vertical force at point y is F(y) then the bending moment at x is ##\int_{y=0}^xF(y)(y-x).dy##.
    (Since the system is static, it doesn't matter which end we integrate to, the moment will be the same except, again, for the sign.)
     
    Last edited: Dec 3, 2015
  4. Dec 3, 2015 #3

    diredragon

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    I get the first pictures but on the last i dont get the numbers, on the zigzag line which you say is shear force it goes from -5 to 5. How did they get that? And also how does this sort of like quadratic function reversed represent bending moment? Any calculation to prove it?
     
  5. Dec 4, 2015 #4

    haruspex

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    The total mass is 100t. All compartments have same displacement, so the buoyant force is a uniform 5tonnes-weight/m.
    Across the first compartment we have a total weight of 20+10=30t weight, and a buoyancy of 25 t weight, for a net downward force of 5 t weight, i.e. a shear force of 0-5t weight.
    Bending moment is the integral I posted, by definition. If you perform the integral for a uniformly distributed load you will get a quadratic.
     
  6. Dec 4, 2015 #5

    diredragon

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    When i integrate the force is considered constant because it doesnt change right? So i just integrate (y - x) is it so?
     
  7. Dec 4, 2015 #6

    haruspex

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    The force per unit length is constant across each compartment. This produces a different quadratic curve for each.
     
  8. Dec 4, 2015 #7

    diredragon

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    How exactly are x and y chosen? Y is the point of application of force but x is some distance from the force? The point where we calculate moment right?
     
  9. Dec 4, 2015 #8

    SteamKing

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    You're also overlooking the Load Diagram, which is shown as the second diagram just below the diagram of the loaded vessel.

    The Load Diagram shows the net load applied to the vessel's hull. It is calculated from the Buoyancy/Weight Diagram located immediately above.

    The Load, L(x), at a particular location x along the length of the hull is simply the difference between the Buoyancy, B(x), and the Weight, W(x), at that location.
    In other words, L(x) = B(x) - W(x).

    For this particular vessel, B(x) = 2 tonne / m for all points x such that 0 ≤ x ≤ 20 m.
    The Weight curve, W(x), is defined as follows:
    Code (Text):

      W(x) = 3 tonne / m for 0 ≤ x ≤ 5 m
             1 tonne / m for 5 m < x < 15 m
             3 tonne / m for 15 m ≤ x ≤ 20 m
     
    The Load Curve is thus:
    Code (Text):

      L(x) = B(x) - W(x)

      L(x) = 2 tonne / m - 3 tonne / m for 0 ≤ x ≤ 5 m
             2 tonne / m - 1 tonne / m for 5 m < x < 15 m
             2 tonne / m - 3 tonne / m for 15 m ≤ x ≤ 20 m

    Simplifying:

      L(x) = -1 tonne / m for 0 ≤ x ≤ 5 m
             +1 tonne / m for 5 m < x < 15 m
             -1 tonne / m for 15 m ≤ x ≤ 20 m
     
    The shear force, V(x), is the integral of the Load Curve along the length of the vessel:

    ##V (x) = \int L(x)\, dx##

    which, given the nature of L(x), should be integrated in a piece-wise fashion over the three intervals. That is, the shear force is the area under the load curve from the end of the vessel to location x.

    Similarly, the bending moment, M(x), is the integral of the Shear Force Curve along the length of the vessel:

    ##M(x) = \int V(x)\,dx##

    To save space, the Shear Force and the Bending Moment curves are plotted on the same diagram at the bottom using the same x-axis but different y-axes to show the values of V(x) and M(x) versus length, x. The value of the Shear Force V(x) is read from the left, while the value of the Bending Moment M(x) is read from the right.

    Two other characteristics stand out here:
    1. The Shear Force must begin at V(x) = 0 for x = 0 and end at V(x) = 0 for x = 20. IOW, there is no net difference between the total weight and the total buoyancy of the vessel.
    2. The Bending Moment curve is also 0 at the ends, since the vessel is a free-free beam and cannot support a moment at the free ends.
     
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