Determining new diameter of shaft after deformation

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Discussion Overview

The discussion revolves around determining the new diameter of a shaft after it undergoes deformation due to an applied axial load. Participants are exploring the relationship between tension, strain, and the resulting dimensional changes in the specimen, specifically focusing on an aluminum alloy under a 10 kip load.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant presents a calculation involving the deformation of a specimen, using the formula δL = PL/AE, but expresses uncertainty about how to find the new diameter after deformation.
  • Another participant questions whether the area (A) or Young's modulus (E) has been determined in the initial calculation, suggesting that the focus should be on calculating the change in diameter (δD) instead of length (δL).
  • A subsequent post inquires if the formula used for δL is applicable for finding δD, indicating a need for clarification on the appropriate approach.
  • Another participant hints at the relevance of Poisson's ratio in the context of calculating changes in diameter due to axial loading.

Areas of Agreement / Disagreement

Participants do not appear to reach a consensus on the correct approach to calculate the new diameter after deformation, with multiple viewpoints and methods being discussed.

Contextual Notes

There is uncertainty regarding the values of area (A) and Young's modulus (E) needed for the calculations. Additionally, the applicability of the formula for δD remains unresolved, and the role of Poisson's ratio is mentioned but not fully explored.

Who May Find This Useful

This discussion may be useful for students or professionals dealing with material deformation, particularly in mechanical engineering or materials science contexts, who are interested in the effects of axial loads on specimen dimensions.

Rickk Gomez
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Hey guys, I've been having an issue with my approach to this problem. I'm stuck after I get the deformation... :

1. The proportional portion of the tension-strain diagram for an aluminum alloy is shown in the figure (attached). The specimen used for the test gauge length of 2in and a diameter of 0.5in. If the applied load is 10 kip, determine the new diameter of the specimen. The shear modulus is G = 38 x 10 ^3 ksi.2. L = PL / AE3. (10)(2)/(1)(0.00614) = 3.2573 x 10^3
 

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Rickk Gomez said:
Hey guys, I've been having an issue with my approach to this problem. I'm stuck after I get the deformation... :

1. The proportional portion of the tension-strain diagram for an aluminum alloy is shown in the figure (attached). The specimen used for the test gauge length of 2in and a diameter of 0.5in. If the applied load is 10 kip, determine the new diameter of the specimen. The shear modulus is G = 38 x 10 ^3 ksi.2. L = PL / AE3. (10)(2)/(1)(0.00614) = 3.2573 x 10^3

It's not clear from your calculation if you have determined either A or E for this specimen. Regardless, calculating δL = PL/AE is not really what the problem is looking for, which is to calculate δD for the bar when an axial load of 10 kip is applied.
 
SteamKing said:
It's not clear from your calculation if you have determined either A or E for this specimen. Regardless, calculating δL = PL/AE is not really what the problem is looking for, which is to calculate δD for the bar when an axial load of 10 kip is applied.
Is the formula the same for finding the δD?
 
No. Hint: look up Poisson's ratio.
 

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