Stress-strain question

  • Thread starter Eastonc2
  • Start date
20
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[1] Problem Statement:
A 3-in. diameter copper rod is to be reduced to 2-in dameter by being pushed through an opening. to account for elastic strain, what should the diameter of the opening be? Modulus of Elasticity (E) =17e6 psi, yield strength (YS) = 40,000psi.

[2] My work:
I've tried to relate the change in area to the change in length assuming constant volume, (Ao/A = L/Lo) and got final length = 2.25*Lo. using 1 for Lo, Lf = 2.25 in. I then just used the modulus of elasticity to calculate how much strain would be present before stress was removed, using YS=E(x-1.25), giving me x = ~1.2524 in. Adding that to Lo, gives me a final length under stress of 2.2524 in. I then related this back to (Ao/A = L/Lo), and got A= ~3.13825. diving by pi, and taking the sqrt, I get r= ~.99947 in., so Diameter = ~1.9989 in.

The book lists an answer of 1.995 in.

I am very confident that I'm doing something wrong here but these are all the methods discussed in the book, and in class thus far, so if anyone can point me in the right direction i'd appreciate it.
 

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