# Help finding deflection of beam

1. Jan 10, 2010

### Dell

find the deflection of the beam at point B

i drew the free body diagram of the beam and found that
Ay=0
Ax=qL
Cy=qL
Cx=qL

M(x)= (qL)*x - (qx)*x/2 +(qL)*<x-L/2>

M(x)=(qL)x -(q/2)x2 +(qL)<x-L/2>

EIΦ(x)=(qL/2)x2 -(q/6)x3 +(qL/2)<x-L/2>2 + C1

EIY(x)=(qL/6)x3 -(q/24)x4 +(qL/6)<x-L/2>3 + C1*x + C2

EIY(0)=0= 0+ C2
===>c2=0

but i need another condition to find C1, i think i need to use the deflection of the bar DC somehow, but im not quite sure how, i tried using triangles and saying that the deflection of point C is the same as the deflection of the bar DC*sin(45) but that didnt work

2. Jan 10, 2010

### pongo38

There are 3 contributions to the deflection at B. Can you identify them? One of them is the extension of CD, as you have realized. If you know the force in CD, you should be able to work out its axial extension. By the way I don't agree your expressions for M(x). In each case one of the terms is not right. The definition of a moment is force times PERPENDICULAR distance. Look at it again. The moment at a section is the algebraic sum of moments on one side of that section. So it can always be checked by looking at the other side of the section; you should then get the same result.

3. Jan 11, 2010

### Dell

1) the deflectin of DC,