Finding the new diameter of a metal rod after tensile loading

AI Thread Summary
A metal rod with a 0.5 inch diameter is subjected to a 2000 lb tensile load, and the goal is to find the new diameter after loading, given the modulus of elasticity, Poisson's ratio, and yield strength. Initial calculations for stress and strain are discussed, with corrections suggested for subscripts and significant figures. After refining the calculations, the new diameter is determined to be approximately 0.49983 inches. The importance of maintaining significant digits in calculations is emphasized, particularly for this problem due to the small changes involved. Overall, the discussion focuses on the correct application of formulas and attention to detail in engineering calculations.
Jim4592
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Homework Statement


A metal rod with a 0.5 inch diameter is subjected to a 2000 lb tensile load. Calculate the resulting diameter of the rod after loading. Assume that the modulus of elasticity is 10,000,000 psi, Poisson's ratio is 0.33, and the yield strength is 21,000 psi.


Homework Equations



I'm not really sure how to go about this problem, below is how I started going about it but I'm not sure if I'm on the right track or not.

The Attempt at a Solution



σ = P/A = 2000 psi / ( (Pi(0.5^2)/4)) = 1.02 (10^4) psi

y = (σy - v(σz + σx) ) / E
= (1.02(10^4) - 0.33 * 0) 10,000,000 psi

y = 0.00102 in/in

Now I'm not really sure what next, or even if what I did to this poing is correct.

Any help will be appreciated!
 
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Jim4592: Put a subscript on sigma in your first formula. The subscripts in your second formula are incorrect. Correct those, and try again. Also, always maintain four or five significant digits throughout all your intermediate calculations. Round only the final answer to three (or four) significant digits.
 
nvn: thanks for responding, I do use the full value during intermediate calculations but just for the purpose of posting here i shortened it some to save time.

After reading your response I went back and thought about it some more, here's my new attempt at a solution:

σx = P/A = 2000 psi / ((Pi * 0.52)/4)) = 1.018591636 (104) psi

x = ∆L/L = σx/E = 1.01859(104) psi/10,000,000 psi = 0.0010185916

∆d = d*v(∆L/L) = 0.5 in * 0.33 * 0.0010185916 = 1.6806762 (10-4) in

d = do - ∆d = 0.5 in - 1.68067 (10-4) = 0.4998319324 in

What do you think about my solution now?
 
Nice work, Jim4592. Regarding significant digits, generally don't list more than four or five (maximum six) significant digits on paper, in your intermediate calculations, even though you use the exact value in your calculator. Then round your final answer to three or four significant digits.

However, this particular problem might be somewhat of an exception, because of the minute quantities involved. I can see why you might want to round your final answer to up to five or six significant digits here. Nice work.
 
alright thanks for all your help!
 
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