# Elastic Recovery-need help

1. Nov 17, 2006

### maha

Elastic Recovery-need help plz

Hiya
Im doing material science as part of my dental technology course. I've got a question, and i'm fidning it hard to tackle.If anyone can help or set me in the right direction, i would be most grateful

Wire, of diameter 0.46 mm, length 100mm.it is subjected to tensile force of 2356 N, taking it beyond its yield point.
Calculate, in mm, the elastic recovery that would occur upon removal of the tensile load

Info given: Modulas of elasticity 67 GPa
Yield Strength 698 MPa
Tensile Strength 1379 MPa

2. Nov 17, 2006

### Pyrrhus

How do you think you should start?

3. Nov 18, 2006

### maha

Ive drawn a diagram so i have a visual representation of whats going on and i've worked out tensile force, by re-arranguing the tensile strenght=max load/ Area.
My answer for that is 720 N

but then i just get stuck as where to go from there :(
Has it got anything to do with resilence?

4. Nov 18, 2006

### Pyrrhus

are you sure with the problem statement?, because the wire fails on that load.

The Max load that can be applied before reaching failure is about 229 N.

5. Nov 18, 2006

### maha

im a bit confused with what you mean? the question i wrote in the beginning is right, that is what is on my question sheet. how did you work out that the max load is 229N ?

6. Nov 18, 2006

### maha

sorry, just understood what you ment, i've worked out tensile stress by re-arranging the stress=force/area formula.

I am really confused as what to do, its baffling me and i dont like it!

7. Nov 18, 2006

### Pyrrhus

Well the solution for elastic recovery will be work out from the Yield Strength, remember assuming the material is linear elastic (Hooke's Law applies) the proportional limit will coincide with its elastic limit, which will be at the yield point (actually close to it). Therefore by using Hooke's Law you can calculate the recovery, but of course there won't be any permanent deformation (residual due to plastic deformation) on the material because under that load the material fails, it breaks.

I got the number by using

$$\sigma_{TensileStrength} = \frac{P}{A}$$

Last edited: Nov 18, 2006
8. Nov 18, 2006

### maha

Right, i kinda understand you. but we havent used hooke's law in out work, its been mentioned in notes but no actually formula. but ive found one on the net and tried to use it and got an answer of 2.12 x 10 -5 mm. Any ideas if im right??

9. Nov 18, 2006

### maha

oh is hookes law, modulas of elasticity = stress/strain?
I really appreciate your help :)

10. Nov 18, 2006

### Pyrrhus

Yes, and good luck!

11. Nov 18, 2006

### maha

i've worked it out!!! yay, thank you so much :)

12. Nov 18, 2006

### PhanthomJay

How did you arrive at that answer? You had the right equation for the stress-strain relationship, but you must look within the elastic range only when determining elastic recovery. Since
$$\epsilon_{yield} = \sigma_{yield}/E$$ solve
$$\epsilon_{yield} = 698MPa/67GPa = .010$$
and since elongation = $$\epsilon(L) =\delta(L) = .010(100)$$ then the elastic recovery is 1mm. The strain beyond the yield point is unrecoverable. Only the elastic part is recoverable.

Last edited: Nov 18, 2006
13. Nov 18, 2006

### Pyrrhus

Hey maha, after i left i didn't check your work, it should be like phantom's with a minor fix.

14. Nov 21, 2006

### maha

Okies, im still a little confused now, so i dont need to calculate tensile strain? So i need to work out the elastic yield, by the equation that phatom jay wrote. But then why has he wrote 'and since elongation = LaTeX graphic is being generated. Reload this page in a moment. then the elastic recovery is 1mm. The strain beyond the yield point is unrecoverable. Only the elastic part is recoverable.'
I'm sorry if i sound a bit silly :\$ i'm usually really good at these

15. Nov 21, 2006

### maha

okies, i think i got it. so, to calculate elastic recovery, i can only use the elastic range, so in order to work out elastic strain, i use yield strength/modulas of elasticiy. that gives me an answer of 1.041791045 N
Then, strain = extension/original, so to work out extension, i caluclate, strain x original length, which gives me 1.041791045 x 10-3 m
So answer is 1.04 mm of elastic recovery?

So then , i do not need to calulate the cross-sectional area or the stress value?

Last edited: Nov 21, 2006
16. Nov 21, 2006

### maha

okies, i think i got it. so, to calculate elastic recovery, i can only use the elastic range, so in order to work out elastic strain, i use yield strength/modulas of elasticiy. that gives me an answer of 1.041791045 N
Then, strain = extension/original, so to work out extension, i caluclate, strain x original length, which gives me 1.041791045 x 10-3 m
So answer is 1.04 mm of elastic recovery?

17. Nov 21, 2006

### maha

okies, i think i got it. so, to calculate elastic recovery, i can only use the elastic range, so in order to work out elastic strain, i use yield strength/modulas of elasticiy. that gives me an answer of 1.041791045 N
Then, strain = extension/original, so to work out extension, i caluclate, strain x original length, which gives me 1.041791045 x 10-3 m
So answer is 1.04 mm of elastic recovery?

So then , i do not need to calulate the cross-sectional area or the stress value?

18. Nov 21, 2006

### maha

okies, i think i got it. so, to calculate elastic recovery, i can only use the elastic range, so in order to work out elastic strain, i use yield strength/modulas of elasticiy. that gives me an answer of 1.041791045 N
Then, strain = extension/original, so to work out extension, i caluclate, strain x original length, which gives me 1.041791045 x 10-3 m
So answer is 1.04 mm of elastic recovery?

So then , i do not need to calulate the cross-sectional area or the stress value?

19. Nov 21, 2006

### Pyrrhus

No if you have the stress already.

20. Nov 21, 2006

### Pyrrhus

Yes, and remember you had all you needed, unless there was more to the problem than what was posted.