# Solid Mechanics(Stress Question)

1. Feb 23, 2012

### zack7

1. The problem statement, all variables and given/known data
In the picture below

2. Relevant equations
Stress=P/A

3. The attempt at a solution
I equated stress AB=-stress BC
p/pi=-(p-60 kips)/2.25pi

P=18.4615kips
This is not the right answer according to the book

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Last edited: Feb 23, 2012
2. Feb 23, 2012

### zack7

Does anybody know whether my answer is right or wrong?

Thank you

3. Feb 23, 2012

### PhanthomJay

Well you have the correct equations....so let's check the math;;;
P/pi = (60-P)/2.25(pi)
2.25P = 60 -P
3.25P = 60
P = 18.5 K

Looks Good!

4. Feb 23, 2012

### zack7

But for my homework, I got this question wrong and in the book the answer is 28.2kips

5. Feb 23, 2012

### PhanthomJay

I don't know where that answer comes from...If P =28.2 K, then the force in the fat shaft would be woul be 31.8 K compression; and the tensile stress in the thin shaft would be 9 ksi, while the compressive stress in the fat one would be 4.5 ksi...far from equal in magnitude.... off by a factor of 2..

6. Feb 24, 2012

### nvn

zack7: Nice work. I currently agree with your answer in post 1, and with the answer by PhanthomJay. The answer in the book currently appears wrong.

Last edited: Feb 24, 2012