# Calculating Beam Deflection using Double Integration

1. Dec 26, 2012

### mm391

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

If the distributed force is removed from the beam in the picture attached and not considered, calculate the deflection at point D.

2. Relevant equations

Double integration for deflection [V]

3. The attempt at a solution

Moment = Ra*x-P(x-3*L) =

Slope = 1/EI*((7p/4)*(x^2/2)-(Px^2/2)-(3PL)+c1)
Deflection = 1/EI*((7p/4)*(x^3/6)-(Px^3/6)-(3PL)+(c1*x)+c2)

Boundary Conditions to calculate c1 and c2
x=0=4L V=0

I am not sure of another boundary condition. I thought maybe

x=2L Slope=0 but I do not think this is right.

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2. Dec 26, 2012

### SteamKing

Staff Emeritus
Your boundary conditions at A and B are very curious.

Since the beam is simply supported at both A and B, what BC must apply?
(Hint: they will both be the same)

3. Dec 26, 2012

### mm391

When x=0 V=0
or
When x=4L V=0

This is because they are simply supported so therefore there can be no deflection at either end. But don't I need a BC for the slope. As there is an unknown slope at A and B how can I find one to use?

4. Dec 26, 2012

### SteamKing

Staff Emeritus
I'm confused. V usually represents the shear force. Are you using it to denote deflection?

5. Dec 26, 2012

### SteamKing

Staff Emeritus

You can still use the BCs for deflection at A and B since you have only two unknown constants of integration.

6. Dec 27, 2012

### mm391

Sorry out lecturer started using V for deflection but I see in most text books it is U.

How can i calculate C1 if I don't have a boundary condition for a slope?

7. Dec 27, 2012

### SteamKing

Staff Emeritus
You can't specify a slope for a simply supported beam.

Your two BCs are the deflections at A and B, both of which are zero.

8. Jan 2, 2013

### mm391

From my boundary conditions I have:
c1=-PL^2/8
c2= 3PL

Using these I still have the wrong answer for the defelction at D. Can anyone see where I maybe going wrong?

9. Jan 2, 2013

### SteamKing

Staff Emeritus
What reactions did you calculate at A and B?

10. Jan 4, 2013

### mm391

I made a mistake. I worked out the reactions which included the Distributed load. The question asks us to ignore the distributed load.

In which case my reactions are:

Reactions@A= P/4
Reactions @B=3P/4

The of the beam from the left hand side of P is (P/4)*x = M1
The moment from 3L<x<4L is (p/4)*(x)-P(x-3L) = M2

Boundary conditions I can see are (now using U as the deflction):

x=0 U=0
x=3L Slope=0 (although I am not sure this is correct as we are given no info about the slope)
x=4L U=0

Is this correct?

Sorry I am struggling to see where I am going wrong.

Last edited: Jan 4, 2013
11. Jan 4, 2013

### SteamKing

Staff Emeritus
Two boundary conditions are sufficient, since there are only two constants of integration. The BCs at the ends are easily determined by inspection. Although there is a point of zero slope somewhere between A and B, it may not necessarily be at x = 3l.