Solid Mechanics(Stress Question)

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Homework Statement


In the picture below

Homework Equations


Stress=P/A

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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  • Fig. P1.3: diagram of two solid cylindrical rods AB and BC welded at B, with force P pulling left at end A, 30 kips applied at B in opposite directions, and a fixed support at C. Dimensions: AB 30 in. long, 2 in. diameter; BC 40 in. long, 3 in. diameter.
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Does anybody know whether my answer is right or wrong?

Thank you
 
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!
 
But for my homework, I got this question wrong and in the book the answer is 28.2kips
 
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..
 
zack7: Nice work. I currently agree with your answer in post 1, and with the answer by PhanthomJay[/color]. The answer in the book currently appears wrong.
 
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