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