Calculating Beam Deflection using Double Integration

[mm391]

Homework Statement

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

Homework Equations

Double integration for deflection [V]

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.

 

[SteamKing]

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)

 

[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?

 

[SteamKing]

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

 

[SteamKing]

I reread your OP.

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

 

[mm391]

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

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

 

[SteamKing]

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.

 

[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?

 

[SteamKing]

What reactions did you calculate at A and B?

 

[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.

 

[SteamKing]

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.