Producing a shearing force diagram and bending moment curve

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1. Dec 19, 2015

diredragon

1. The problem statement, all variables and given/known data
A box shaped vessel, of $4$ compartments and $80m$ lenght, light displacement of $800$ tonnes loads $200$ tonnes in the first compartment and $200$ tonnes in the last compartment. Produce the shearing force diagram and bending moment curve.
2. Relevant equations
3. The attempt at a solution

I dont fully understand the problem but here is what i tried to do. I calculated the total mass $M=m + 200 + 200$
So total mass divided by 4 gives me the buoyant force which acts upward opposing the gravity force. $M/4 = 300$
I uploaded the image below and got that the shearing force goes to $-100$ in the first and last compartment and $+100$ in the middle two. I dont know if this is true however and of the diagram is correct. Moreso i dont know how to draw a bending moment curve. Anyone have a slightest clue of what i should do here?
http://postimg.org/image/4kunvxm1h/

2. Dec 19, 2015

SteamKing

Staff Emeritus
You are being asked to analyze a simple beam which happens to be in the shape of a boat. Because of the uneven loading of the vessel, with 200 tonnes of cargo in each end compartment, there will be shearing forces created in the structure of the vessel, since the buoyancy must remain evenly distributed along the length of the vessel.

Because the shape of the hull is a simple box, you should be able to calculate how much buoyant force per unit length of the hull is required to keep the vessel in static equilibrium. Remember, even though the center two compartments are not loaded with cargo, there are buoyant forces applied there to keep the vessel floating. Therefore, the shearing forces you have initially calculated are incorrect, since you neglected to include the buoyant forces acting on the hull. {Edit: Strike this last sentence. Shear Forces OK; you should make a plot of shear force v. length along the barge.}

Unlike most beam problems, a floating vessel is free at the ends. What does this mean in terms of the values of shear and bending moment at these locations?

You say you don't know how to draw a bending moment curve. Have you worked any other problems involving either concentrated or distributed loads applied to a beam?

At a first step, take your initial sketch of the box and show how the loads are distributed along the lengths of the compartments.

Last edited: Dec 19, 2015
3. Dec 19, 2015

haruspex

You should know two integrals, one for the shear force, and one for bending moment. Since you got the shear force curve right I assume you know it's $F_{shear}(x)=\int_0^xf(s).ds$, where f(s) is the load per unit length at s.
Can you quote the corresponding integral for bending moment?

4. Dec 21, 2015

diredragon

I looked a little bit on the net as i have never drawn a diagram before and i tried to usd that method on the problem i posted. http://postimg.org/image/hzxk809nv/ yet i dont know how to draw a bending moment diagram whose integral should be ($F*x$) inside the integral, right?
I have a diagram of the problem that is similar and solved in the book i have so would it help to post that?

5. Dec 21, 2015

SteamKing

Staff Emeritus
You've drawn the loading diagram for the vessel, it looks like, from the image. It's hard to tell, because you haven't provided any calculations to go along with the diagram.

As far as drawing a diagram is concerned, I find it hard to believe you have reached this stage of your academic career and have never sketched a picture, plotted points to draw the curve of a function, etc.

A diagram is just a pictorial representation of the results of your calculations. No one is asking you to make a diagram without doing calculations first.

6. Dec 21, 2015

haruspex

yes, that's the right integral. F is just a step function, so it's a very easy integral. You just have to break up the integral at the step boundaries.

7. Dec 22, 2015

diredragon

So is the function then of the shear force when integrated $F_{shear}(x)=f(x)*x$ and the bending moment $B(x)=f(x)*\frac{x^2}{2}$, i have also found some calculations in the book i dont get. Theyr from here http://postimg.org/image/g9ojv8eyb/

8. Dec 22, 2015

SteamKing

Staff Emeritus
Have you analyzed a beam with a distributed load applied to it before?

The buoyancy load on the hull of the vessel is distributed by nature. Because this is a box-shaped hull, calculating the amount of buoyant force per meter should be fairly easy.

Although it is not clearly spelled out so, the load of 200 tonnes in each end compartment will be distributed over the length of the end compartments, with no cargo in the center compartments.

The difference in the distributed weight of the barge structure and cargo and the buoyancy of the hull will give you what is known as the load curve of the barge.

The integral of the load curve w.r.t. the length of the barge will result in the shear force curve.

The integral of the shear force curve w.r.t. the length of the barge will result in the bending moment curve.

You should be able to make these calculations and draw the requisite diagrams, given the foregoing. Forget looking at all these books you don't understand.

9. Dec 22, 2015

haruspex

that happens to work where f is constant. But more generally you need to plug in the actual f(x) function before you integrate.
In the present case, f is a step function, i.e. it is constant over intervals. This means you need to break up the integral into those intervals.

10. Dec 22, 2015

diredragon

I tried drawing three diagrams, ( http://postimg.org/image/41r8bxe3f/ ), the load diagram, shear force diagram and bending moment diagram. I broke the integrals into 4 parts and i got something like a parabolic curve in the bending moment diagram. I do not know what numbers are associated with the calculations though. I also set force to be uniformly distributed, so that the overall force that acts is either $-5t/m$, or $+5t/m$. Could you check this please?

11. Dec 22, 2015

SteamKing

Staff Emeritus
Your bending moment diagram is wrong. If the triangle from the shear force diagram between x = 0 m and x = 40 m is completely on one side of the x-axis or the other, then the area under that triangle will either be positive or negative, not split as you have shown.

Why not? Can't you calculate the areas under a rectangle or a triangle?

Take a look at your load curve. The net load on the barge between x = 0 m and x = 20 m is -5 tonne / meter.

The area under this portion of the load curve is simply -5 t/m × 20 m = -100 tonnes, which you can convert to Newtons if you so desire. On the shear curve, at x = 20 m, the value of shear force is -100 tonnes. To calculate intermediate values of the shear force, these will be proportional to the distance from x = 0.

You follow a similar procedure to integrate the shear force curve to calculate values of bending moment, but instead of figuring the area under a rectangle, you must calculate the area under a triangle. Remember, as x increases from zero, the total area under the shear force curve is cumulative, and it's this cumulative area which provides the values of the bending moment.

Yes, this is correct. IDK why the rest of the problem is proving so difficult.

12. Dec 22, 2015

haruspex

A pity you did not post your working for the bending moment curves. But my guess is that you made two mistakes on the middle sections:
1. You got a sign wrong
2. You need to add in the integral for the first quarter, i.e. for 20<x<40, $\int_0^{20}xf_1(x).dx+\int_{20}^xxf_2(x).dx$.

13. Dec 22, 2015

diredragon

14. Dec 22, 2015

haruspex

Yes, those curves are about right. You could make the slope at the endpoints a bit more accurate.