Hooke's Law/Poisson's Ratio and a bar

[Autanimous]

Homework Statement

A round bar of 10 mm diameter is made of
aluminum alloy 7075-T6 (see figure). When the bar is
stretched by axial forces P, its diameter decreases by
0.016 mm.
Find the magnitude of the load P. (Obtain the material
properties from Appendix H.)

From Appendix H:
Modulus of Elasticity (E) = 72 Gpa (Kn/mm^2)
Poisson's Ratio (v) = .33

Homework Equations

ε = axial strain
ε' = lateral strain
σ = axial stress
L = Length of bar
δ = change in length
A = Area

σ = E*ε
ε = δ/L
σ = P/A
v = ε' / ε

The Attempt at a Solution

A = pi*(10 mm )^2 = 314.159 mm^2
ε = (.016mm) ----------------that's right isn't it?
P = σ*A = ε*E*A = (.016mm)*(314.159 mm^2) * (72 Kn/mm^2)
= 361.911 Kn

However, the book says the answer is 27.4 Kn

I'm lost and sad ;-; Where am I going horribly horribly wrong? I figure it's the fact I'm not using the Poisson's ratio (as this section of homework was titled Hooke's Law and Poisson's ratio after a chapter in my book), however it didn't seem to come into play. This could stem from a mis-reading of one of my equations... however I've looked at them multiple times, so either my book is just confusing me or something else is wrong. Thank you for your help *bows*

 

[Autanimous]

Is this a more 'upper level' problem that should've been in the other section? I should get a chance to ask my teacher about it later today, though he won't be able to give me much time on the subject the way the schedules work out.

 

[pongo38]

"its diameter decreases by
0.016 mm." is epsilon dash.
Now you can work out epsilon.

 

[Autanimous]

"its diameter decreases by
0.016 mm." is epsilon dash.
Now you can work out epsilon.

Yes, thanks. I'm sorry I never responded here but I did an overhaul reading of the material and I got it right. I for some reason continued to have a small power of ten problem, but otherwise the numbers matched up with what was expected.

It is for the very reason you described.