Beam Deflection Equations and Boundary Conditions for Solving Homework Problems

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Homework Statement

Homework Equations

For small deflections:
$$M=EI \frac{d^2y}{dx^2}$$

The Attempt at a Solution

To solve a problem like this, I think I was told I need to study the deflection and displacement of the beam.
If I said that the deflection at points x=0 and x=3L have to be 0 then I have two boundary conditions. Would that be enough to solve it? When I integrate twice, I will have Cx + D. Which I can find by setting x=0 to get D and x=3L to get C. But I am not being able to form an equation for the moment.
So far I have:

$$M(x) = R_B(3L-x) - W(4L-x)$$
but it doesn't look right. If x = 3L then the moment is -W(4L-x) which is fine. But if x = 4L I get moment = -$$R_B$$L which doesn't look right. What should I do?

How would I then go on to get the reaction at B $$R_B$$

 

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So, I have found some equations for deflection of beams that I'm assuming come from solving the above differential equation for basic scenarios:
http://www.advancepipeliner.com/Resources/Others/Beams/Beam_Deflection_Formulae.pdf

So, my question now is, can the principle of superposition be applied to get the deflections for more complicated situations? If so, why? I mean, surely the point of making a list like that is so that someone could put them together like this right?

If I said that the above question is a combination of case 1 (force $$R_B - W$$)and 5 moment of (WL at the end) in opposite directions then the total displacement because of these should add up to 0 at B right?

So could I say $$- \frac{(R_B - W)(3L)^2}{3EI} + \frac{(WL)(3L)^2}{2EI} =0$$?

 

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Ok, never mind.
I tried solving the $$M=EI \frac{d^2y}{dx^2}$$
equation for the scenario 1 on the link, using boundary conditions y(0)=0 and y'(0)=0 and I got the given result so I guess it is how they got all the results.

I used the result I got on my second post and it gave the right answer.

The only question I have now is:
Why can we use the principle of superposition in these situations? Why is it a linear system? Is it because differentiation and integration are linear operations?

 

[pongo38]

You are missing the point that the left hand support has a moment reaction. One way to solve it is to release the right hand support so that the structure is a cantilever, and work out the deflection at B due to the applied load. Then ask yourself what value of RB upwards would reduce that deflection to zero at B, with no other loads on the cantilever. In that case you have enough information to solve for RB and the problem is now statically determinate.