# Homework Help: Deflection of beams problem

1. Apr 12, 2009

### Aerstz

1. The problem statement, all variables and given/known data

Deflection of simply-supported beam problem. Please see the attached image of an example problem from a textbook:

http://img14.imageshack.us/img14/619/hearnbeamproblem.png [Broken]

I have absolutely no idea why

A = - (wL^3)/24

and why

0 = (wL^4)/12 - (wL^4)/24 +AL

is used in the determination of A.

I especially do not know why (wL^4)/12 is used in the above equation.

I would have thought that A would represent the left beam support, where I also would have thought that x = 0. But, according to the example in the attached image, x at A = L.

Last edited by a moderator: May 4, 2017
2. Apr 12, 2009

### Aerstz

Another example of a deflection problem:

http://img18.imageshack.us/img18/2868/beerbeamproblem.png [Broken]

I am not sure why C1 = 1/2PL^2, but I have no idea why C2 = -1/3PL^3.

My maths is very weak; I think I just need some kind soul to gently walk me through this!

Last edited by a moderator: May 4, 2017
3. Apr 12, 2009

### Aerstz

And a third example:

http://img133.imageshack.us/img133/9210/be3amproblemthree.png [Broken]

I have no idea how 52.08 came to equal A.

Last edited by a moderator: May 4, 2017
4. Apr 13, 2009

### PhanthomJay

These terms A, B, C1 in the problems above refer to the constants of integration of the differential equation, as determined from the boundary conditions, and do not in any way refer to the support reactions. Boundary conditions are established at the ends of the beams based on the support condition. For example, if there is no deflection at a left end support, then the vertical deflection, y, equals 0 , when x, the horizontal distance from the left end, is 0. The values of the constants of integration are derived by carefully following the given steps in the examples.

5. Apr 13, 2009

### Aerstz

That's the problem; I am unable to follow the steps in the examples. The steps are too big; I need smaller steps to bridge the gaps.

To me, the examples seem to go from A straight to Z in one giant leap. I need to know B,C,D...etc., in between. Currently I am completely blind to what these intermediate steps are.

For example, and as I asked above in the first post: Why does A = - (wL^3)/24? What I mean to ask is, how was the (wL^3)/24 arrived at? I am extremely challenged with this 'simple' mathematics and I really need a kind soul to guide me through it very gently and slowly!

6. Apr 13, 2009

### PhanthomJay

I hear you. Looking at part of the first problem, step by step, inch by inch:

1. $$EI(y) = wLx^3/12 -wx^4/24 + Ax + B$$

Now since at the left end, at x = 0, we know there is no deflection at that point; thus, y = 0 when x =0, so substitute these zero values into Step 1 to obtain

2. $$0 = 0 - 0 + 0 + B$$, which yields

3. $$B = 0$$, thus Eq. 1 becomes

4. $$EI(y) = wLx^3/12 - wx^4/24 + Ax$$

Now since at the right end, at x = L, we also know that y = 0 , substitute X=L and y=0 into Eq. 4 to yield

5. $$0 = wL(L^3)/12 - wL^4/24 + AL$$ or

6. $$0 = w(L^4)/12 - wL^4/24 + AL$$ .

Now since the first term in Eq. 6 above, $$wL^4/12$$, can be rewritten as $$2wL^4/24$$, then

7. $$0 = (2wL^4/24 - wL^4/24) +AL$$, or

8. $$0 = wL^4/24 + AL$$. Now divide both sides of the equation by L, and thus

9. $$0 = wL^3/24 + A = 0$$,

and now solve for A by subtracting $$(wL^3/24)$$ from both sides of the equation to get

10. $$0 -wL^3/24 = (wL^3/24 -wL^3/24) + A$$, or

11. $$-wL^3/24 = 0 + A$$

12. $$A = -wL^3/24$$

7. Apr 13, 2009

### Aerstz

Thank you very much, Jay. I appreciate you taking the time to lay the process out as you did!

It is much clearer now, so hopefully I should be able to get past this quagmire and actually progress with some work!