# Find the deflection of the following points

1. Nov 18, 2009

### Dell

in this question

A1=8*10-4m2
A2=5*10-4m2
E=70*109Pa
F1=-100*103N
F2=75*103N
F3=50*103N

$$\sigma$$=F/A
$$\epsilon$$=$$\sigma$$/E = $$\frac{F}{A*E}$$
$$\delta$$=$$\epsilon$$*L = $$\frac{F*L}{A*E}$$

$$\delta$$B = $$\frac{F1*1.75}{A1*E}$$ + $$\frac{F2*3}{A1*E}$$ + $$\frac{F3*3}{A2*E}$$ = 5.1785*10-3m

but thats not right

even looking at the second answer
i thought

$$\delta$$D=$$\delta$$B + $$\frac{F3*1.5}{A2*E}$$
but if i plug in THEIR answer for $$\delta$$D i get 2.924mm and not the 5.7 they say

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Last edited: Nov 18, 2009
2. Nov 18, 2009

### djeitnstine

The forces used in the second and third terms of your equation for $$\delta_B$$ are wrong. You have to take a cut at each point, draw a free body diagram and sum the forces for equilibrium

3. Nov 19, 2009

### Dell

if so, then why not for the 1st term as well?

4. Nov 19, 2009

### Dell

for a similar problem, but where the diameter was constant and the E was different for the 2 parts, i did exactly that and it worked.

5. Nov 19, 2009

### Dell

i got it, thankls for the help