# Compute the elastic modulus of hypothetical alloy

1. Dec 9, 2008

### physicsnnewbie

1. The problem statement, all variables and given/known data
Consider a cylindrical specimen of some hypothetical metal alloy that has a diameter of 10mm. A tensile force of 1500 N produces an elastic reduction in diameter of 6.7e-4mm. Compute the elastic modulus of this alloy, given that poisson's ratio is 0.35.

2. Relevant equations
stress = eleastic modulus*strain
poisson's ratio = lateral strain/axial strain

3. The attempt at a solution
Calculate pressure in Pa:
1500/(5e-3^2*3.14159) = 1.90986e+7

Convert to GPa:
1.90986e+7/1e6 = 1.90986e+1

Calculate lateral strain:
6.7e-4/10 = 6.7e-5

Calculate axial strain:
6.7e-5/.35 = 1.91429e-4

Calculate Elastic Modulus:
1.90986e+1/1.91429e-4 = 9.97686e+4

=100GPa

The correct answer is 100MPa. I'm not sure where i've made the mistake though.

2. Dec 9, 2008

### timmay

What does the prefix 'giga' mean? And what does the prefix 'mega' mean? And what unit are you normalising by in this step? Add in the fact that you're a factor of 103 out and you know there is going to be an error in your consideration of units somewhere in your solution...

3. Dec 9, 2008

### physicsnnewbie

Well I originally converted it to MPa but decided to convert it to GPa instead (I'm not sure why). Anyway this shouldn't be the cause of the magnitude difference because it just means i will get the answer in GPa instead of MPa right?

So where have I gone wrong with the magnitudes?

4. Dec 10, 2008

### timmay

Sorry, was on autopilot there and spotted a difference between what you said you were expressing it in terms of and what you actually have. I agree with your calculations and answer. I know it's not a reliable guideline in this case (especially as the original problem states 'hypothetical') but most metals and their alloys do exhibit moduli upwards of 50 GPa, so the given answer does seem erroneous.

5. Dec 10, 2008

### physicsnnewbie

Ok thanks.

6. Dec 11, 2008

### FredGarvin

It looks like you are right and whoever did the calculations for the answer improperly added a factor of 1e-3 to the diameter measurement when computing lateral strain. Since the deflection number was already stated in mm, you are correct in your approach.