Determining new diameter of shaft after deformation

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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?