Beam Deflection Equations and Boundary Conditions for Solving Homework Problems

[___]

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 $$R_B$$

 

[___]

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

 

[___]

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.

 

[DeeNos]

and the moment at x=3L?As a fellow scientist, I would like to provide some guidance and tips for solving beam deflection problems with boundary conditions.

Firstly, it is important to understand that the deflection of a beam is a function of its length, shape, and the applied loads. Therefore, to solve a beam deflection problem, we need to consider the following:

1. Determine the boundary conditions: As you mentioned, boundary conditions are essential in solving beam deflection problems. They define the end conditions of the beam, such as the support and loading conditions. In your example, the boundary conditions are that the deflection at points x=0 and x=3L must be zero. These conditions will help us determine the constants of integration when we integrate the moment equation twice.

2. Write the moment equation: The moment equation for a beam is M=EI(d^2y/dx^2), as you have mentioned. This equation relates the bending moment M to the curvature of the beam, which is represented by the second derivative of the deflection y with respect to x. In your example, the moment equation would be M=R_B(3L-x) - W(4L-x).

3. Integrate the moment equation: Once we have the moment equation, we can integrate it twice to get the deflection equation. The first integration gives us the slope equation, and the second integration gives us the deflection equation. In your example, the deflection equation would be y = -R_B(3L-x)^3/6EI + W(4L-x)^3/6EI + Cx + D.

4. Apply the boundary conditions: Now we can apply the boundary conditions to solve for the constants of integration. In your example, we have two boundary conditions, which means we will have two equations to solve for the two unknown constants C and D. By setting x=0 and x=3L in the deflection equation, we can solve for C and D.

5. Solve for the reaction and moment: Once we have the deflection equation, we can use it to solve for the reaction at point B (R_B) and the moment at x=3L. By substituting x=3L in the deflection equation, we can solve for R_B, and by differentiating the deflection equation and substituting x=3L, we can solve for the moment at x=3L