Solve Stress-Strain Question: 3in Copper Rod Reduces to 2in Diameter

[Eastonc2]

[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.

 

[Achy47]

Thank you for posting your work and question. It seems like you have a good understanding of the concepts and equations involved in this problem. However, there are a few things that could be corrected to get a more accurate answer.

Firstly, when relating the change in area to the change in length, the equation should be Ao/A = (Lo+ΔL)/(Lo). This takes into account the initial length (Lo) and the change in length (ΔL).

Secondly, when calculating the final length under stress, the equation should be Lf = Lo + ΔL, where ΔL is the change in length due to the stress. In your calculation, you used 1.25 as the value for ΔL, but this should be the change in length at the yield strength, which in this case is 40,000psi.

Finally, when relating the change in area to the change in length, the equation should be Ao/A = (Lo+ΔL)/(Lo). This takes into account the initial length (Lo) and the change in length (ΔL).

By correcting these errors, the final diameter should be 1.995 in, as listed in the book. I hope this helps and good luck with your studies!