Calculating a) and Baffled by b): A Math Challenge

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Discussion Overview

The discussion revolves around a math challenge involving calculations related to elongation and Poisson's ratio in a material under tensile load. Participants explore the necessary parameters and formulas needed to solve the problem, particularly focusing on parts a) and b) of the challenge.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant calculated the elongation for part a) as 0.7 mm but expressed confusion regarding part b).
  • Several participants suggested that cross-sectional area is necessary for the calculations, with some assuming it was provided since the elongation was calculated.
  • There was mention of Poisson's ratio being defined as the ratio of strains, with a discussion on how the tensile load might not be relevant for calculating the maximum strain before plastic deformation.
  • Another participant indicated that the question might be poorly worded, leading to confusion about the parameters needed for the calculations.
  • A participant provided a detailed solution involving specific calculations for area and diameter, but noted that the relationship between yield strength and applied force was not clear from the question.

Areas of Agreement / Disagreement

Participants generally agree on the need for cross-sectional area in the calculations, but there is no consensus on how to interpret the question or the relevance of the given parameters. Some participants express confusion about the wording of the problem, while others provide differing interpretations of the necessary calculations.

Contextual Notes

Limitations include the lack of provided values for cross-sectional area and diameter, which are critical for solving part b). The discussion also reflects uncertainty regarding the interpretation of the problem statement and the relationship between yield strength and applied force.

menotu3169
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Homework Statement
A cylindrical steel specimen 350mm long is to be subjected to a tensile load of 10KN. The modulus of elasticity is 207GPA, Yield Strength is 415 MPa, and Poisson's ratio is 0.30
Determine
a)The maximum elastic elongation
b)the reduction in diameter
Relevant Equations
stress = Force / Area
strain = Stress / Modulus of Elasticity
Poisson's ratio = transverse stress / longitudinal stress
I was able to calculate a), and got 0.7mm
But I have no idea where to even start with b)
 
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It's been awhile since done these types of calculations, but I believe you need cross sectional area. I assume that you have this information since you were able to calculate the elongation.
 
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scottdave said:
It's been awhile since done these types of calculations, but I believe you need cross sectional area. I assume that you have this information since you were able to calculate the elongation.
I was not given the area or the diameter, which is why I am having trouble with the question
 
Could it be percentage?
 
menotu3169 said:
I was not given the area or the diameter, which is why I am having trouble with the question
What formulas did you use to get the 0.7mm ?
 
Poisson's ratio is usually defined as the ratio of strains, not stresses (although this shouldn't matter if you're in the elastic regime). I think the tensile load of 10 kN that OP listed is a red herring. The yield strength and the elastic modulus allow you to calculate the maximum strain (elongation) of the material before plastic deformation sets in. This strain will be a percentage elongation which can be used to find a length (which the OP seems to have done correctly). The Poisson's ratio is then just the ratio of transverse strain to axial strain, with a negative sign out front because most materials shrink transversely when you pull them axially.
 
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Spoke with the professor and got the solution. Seems like it is just a poorly worded question. The question was worded exactly as I have it in the original post.
Solution:
A=10000/415=24.09 mm2
d=5.54 mm
change in length = 415/207000*350=0.7 mm
change in diameter = 415/207000*(-0.3)*5.54=-0.003 mm

This seems to indicate that the Yield Strength corresponds to the applied force of 10kN, which in my opinion is not obvious from the question.

Thanks everyone for all the help!
 
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