# Stress - Strain Question

1. Jun 7, 2009

### EskShift

1. The problem statement, all variables and given/known data
A material has a stress of 60,000Pa and a strain of 0.4 at breaking point.
Given that it is in the shape of a 2m long cylinder of radius 20cm, what work had to be done to extend the material to the point of fracture?

2. Relevant equations
Well i thought work done was equal to the area under the graph.

3. The attempt at a solution
Therefore: 0.5 x 60,000 x 0.4 = 12,000 J

Apparently answer is 0.6 J, so im fairly off. any ideas?

2. Jun 7, 2009

### EskShift

surely someone can help!

3. Jun 8, 2009

4. Jun 8, 2009

### benk99nenm312

All I can tell you is that in order to make it equal to .6 J, you need to multiply the given stress and strain by .000025. But, you may have already known that. As for how to get that... no idea. Sorry.

5. Jun 8, 2009

### benk99nenm312

Ohhh... nevermind. I posted to late.

6. Jun 8, 2009

### EskShift

thats still fairly confusing!
the change in length is 0.8m, the original length is 2m, stress is 60,000Pa at point of fracture and Strain is 0.4 at point of fracture, radius of the cylinder is 20cm and i need to find the work done to extend the materials to the point of fracture.

Work = Fx right?

but what values do i use?

7. Jun 8, 2009

### djeitnstine

Eskshift look at the article again. You have all of the information needed. What else do you need. And work is most accurately described as $$\int F ds$$ Where F in this case is $$\frac{EA}{L_0}$$ and is done over the change in length....

8. Jun 8, 2009

### EskShift

Well that gave me about 3000+, and the answer says 0.6 J, so either the answer is a mistype or its the wrong equation.

9. Jun 8, 2009

### PhanthomJay

Looks like you or the answer slipped a few decimal points. Like maybe the answer is 6 kJ? Also, the answer, if 6 kJ, appears to be off by a factor of 2, since I get 3 kJ, which agrees with your 2nd answer. It seems like they took work as equal to Fx, whereas, since you must calculate Work = the integral of Fx, where F varies from 0 to F, this implies W=F/2(x). The work done is NOT the area under the stress strain graph. It is the area under the force-displacemnt graph.

And if Wiki confuses you, welcome to the club.

Last edited: Jun 8, 2009
10. Jun 8, 2009

### Mapes

The volumetric work done is the area under the stress-strain graph, though. EskShift made a units error in the first post: $\mathrm{Pascals}\neq\mathrm{Joules}$.

By multiplying the volumetric work by the volume, I get 3.02 kJ if the stress is assumed to have increased linearly up to the failure point.

11. Jun 8, 2009

### djeitnstine

Think about this, have you ever tried to stretch 2m long circular rod? The only way it would ever take 0.6J is if it were made of paper and in that case the sigma yield would not be 60000Pa.

Its always good to think about the question and what the answer implies.

12. Jun 8, 2009

### Mapes

A hydrogel fits $\epsilon = 0.4$ and $\sigma_\mathrm{fail}=60\,\mathrm{kPa}$ pretty well. But I agree that the 0.6 J value doesn't fit at all.

13. Jun 8, 2009