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.

 

[DeeNos]

I would first check my equations and make sure they are correct. In this case, it seems like you may have mixed up the units for the Modulus of Elasticity. It should be in GPa (gigapascals) instead of Kn/mm^2. This changes the value for P to be 27.4 Kn, which matches the answer given in the book.

Another possibility is that the book is using a different definition for Poisson's ratio. Some sources define it as the negative ratio of lateral strain to axial strain, while others define it as the positive ratio. So it's important to make sure you are using the same definition as the book.

In general, when solving problems in science, it's important to double check your units and equations to make sure they are consistent and correct. It's also helpful to compare your answer to a known value or to the answer provided in the book to check for any mistakes. And if there are discrepancies, it's always good to ask for clarification or seek help from a teacher or tutor.