Finding the new diameter of a metal rod after tensile loading

Click For Summary

Discussion Overview

The discussion revolves around calculating the new diameter of a metal rod after it has been subjected to a tensile load. The problem involves applying principles of mechanics of materials, specifically focusing on stress, strain, and the effects of tensile loading on dimensions of the rod. Participants explore the calculations necessary to determine the resulting diameter, considering factors such as modulus of elasticity and Poisson's ratio.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • The initial calculation of stress (σ) was presented, but the poster expressed uncertainty about its correctness and the subsequent steps.
  • One participant suggested correcting subscripts in the formulas and emphasized the importance of maintaining significant digits throughout calculations.
  • A revised calculation was provided, detailing the stress and strain, along with the method for calculating the change in diameter (∆d) and the final diameter (d).
  • Another participant acknowledged the work done and reiterated the importance of significant digits, while also noting that the problem's context might allow for more precision in the final answer due to the small quantities involved.

Areas of Agreement / Disagreement

Participants generally agree on the approach to solving the problem, but there is no consensus on the correctness of the initial calculations or the final answer, as some uncertainty remains regarding the intermediate steps and significant digits.

Contextual Notes

There are unresolved aspects regarding the accuracy of the initial calculations, the handling of significant digits, and the implications of using different levels of precision in the context of this problem.

Jim4592
Messages
49
Reaction score
0

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!
 
Physics news on Phys.org
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!
 

Similar threads

  • · Replies 10 ·
Replies
10
Views
12K
Replies
2
Views
6K
Replies
1
Views
4K
Replies
6
Views
5K
  • · Replies 15 ·
Replies
15
Views
54K