Find the deflection of the following points

[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 that's 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

 

[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

 

[Dell]

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

 

[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.

 

[Dell]

i got it, thankls for the help