Calculate force required to reduce diameter of bar?

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Discussion Overview

The discussion revolves around calculating the force required to reduce the diameter of a mild steel bar from 40 mm to 39.99 mm under tensile stress. Participants explore the application of Young's modulus, Poisson's ratio, and relevant equations in the context of this problem.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant expresses confusion about the use of the x-y coordinate system in the problem.
  • Another participant provides hints regarding the equations for Poisson's ratio and axial elongation, emphasizing the need for the original poster to derive the relevant equations themselves.
  • A participant calculates transverse strain and axial strain, leading to an axial stress and subsequently a force, but questions whether they are on the right track.
  • One participant critiques the arithmetic of the previous calculation, suggesting an incorrect answer was reached.
  • Another participant shares their own calculations for axial strain, axial stress, and force, arriving at a different numerical result.

Areas of Agreement / Disagreement

There is no consensus on the correct answer, as participants have provided different calculations and results. Disagreement exists regarding the accuracy of the arithmetic and the final force value.

Contextual Notes

Participants have not fully resolved the mathematical steps involved in the calculations, and there are indications of missing assumptions or potential errors in arithmetic that have not been clarified.

bensm0
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A mild stell bar 40 mm diameter and 100 mm long is subjected to a tensile force along its axis.
Young's modulus of elasticity = 200 GN m -2. Poisson's ratio is 0.3.
Calculate the force required to reduce the diameter to 39.99 mm

It says use the x-y coordinate system?? This i don't understand?

Thank you
 
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bensm0: Hint 1: What is the equation for the definition of Poisson's ratio, in terms of transverse strain εy, and axial strain εx? Hint 2: What is the equation for the bar axial elongation, in terms of the applied tensile force P?

We are not allowed to give you the relevant equations for your homework. You must list relevant equations yourself, and show your work. And then someone might check your math.

Hint 3: The y-axis is the bar transverse direction. The y-axis is parallel to the bar diameter.

By the way, GN*m^-2 is called GPa. Always use the correct, special name for a unit. E.g., 200 GPa, not 200 GN*m^-2.
 
ok so i think i may have the answer:

poisson's ratio = - transverse strain/axial strain
Transverse strain = (39.99 - 40)/40 = -0.25x10^-3
axial strain = -(-0.25x10^-3/0.3) = 833.333x10^-6
Axial stress = 833.333x10^-6 x 200x10^9 = 166.6666x10^6
Force = stress x area
Force = 166.6666x10^6 x (0.25∏x0.04^2) = 5.236x10^6 N

Am i on the right track?
 
bensm0: Nice work, except you did not check your arithmetic, and got the wrong answer.
 
Hi there, i am also stuck on this question along with the rest of the assignment to be honest. Bensm0 contact me on here if you want to try and tackle this togeter?
 
Hi there,

Ive been working on this for most of today and i have come up with the following :

Axial Strain = 8.333 x 10^-4

Axial Stress = 166.66 x 10^6

Force = Stress x Area
= (166.66 x 10^6) x (1.257 x 10^-3)
= 209491.62 N
 

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