Determining new diameter of shaft after deformation

[Rickk Gomez]

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

 

[SteamKing]

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.

 

[Rickk Gomez]

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?

 

[SteamKing]

No. Hint: look up Poisson's ratio.

 

[rezeew]

4. The new diameter can be determined using the formula for strain:

ε = (ΔL/L) = (d-d0)/d0

Where:

ε = strain
ΔL = change in length
L = original length
d = new diameter
d0 = original diameter5. Rearranging the formula, we get:

d = d0 / (1-ε)

6. Substituting the values, we get:

d = 0.5 / (1-3.2573 x 10^3)

7. Solving for d, we get the new diameter of the specimen to be approximately 0.4999 inches.

8. It is important to note that this calculation assumes the material is behaving in a linear elastic manner. If the deformation is large enough to cause plastic deformation, the material may not return to its original shape and the calculations may not be accurate. It is also important to consider any potential safety factors and to consult with a professional engineer for accurate and safe design calculations.