How to Calculate Work Done in Extending a Material to Breaking Point?

[EskShift]

Homework Statement

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?

Homework Equations

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

The Attempt at a Solution

Therefore: 0.5 x 60,000 x 0.4 = 12,000 J

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

 

[EskShift]

surely someone can help!

 

[LowlyPion]

Maybe this will help?
http://en.wikipedia.org/wiki/Young's_Modulus#Elastic_potential_energy

 

[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.

 

[benk99nenm312]

Ohhh... nevermind. I posted to late.

 

[EskShift]

Maybe this will help?
http://en.wikipedia.org/wiki/Young's_Modulus#Elastic_potential_energy

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?

 

[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...

 

[EskShift]

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...

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.

 

[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.

 

[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.

 

[djeitnstine]

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.

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.

 

[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.

 

[Dadface]

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.

And a strain of 0.4? That is large and will result in a large change of cross sectional area.