# Calculate the total amount of strain at fracture?

1. Sep 1, 2011

### TyErd

1. The problem statement, all variables and given/known data
The question is: calculate the total amount of strain at fracture? How much of this is recoverable strain? Explain your answer

2. Relevant equations
strain = (change in length)/ (initial length)

3. The attempt at a solution
I'm not sure about my answer but i just simply read of the graph.
material a - 3%
material b - 15%

but that's in percentage form, how would I get the actual strain?

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2. Sep 1, 2011

### PeterO

Re: Strain

Strain is [change in length] / [length] is it not?

If that fraction is multiplied by 100 does it make it percentage strain?

3. Sep 1, 2011

### TyErd

Re: Strain

yeah it does. but what is the change in length. the final length is the fracture but is the initial length that you use to find change of length?

4. Sep 1, 2011

### PeterO

Re: Strain

When you exceed the elastic limit of a piece of steel, it deforms, but is still steel.

The amount of strain up to the elastic limit is how much you get back out of the material - even if the sample is fractured.
You do not always get it back in a useful form, but the strain energy put in, does come back out.

EDIT: Not sure why you changed from strain [in the question] to Length??

5. Sep 1, 2011

### TyErd

Re: Strain

Okay so the answer to the second part of the question is simply the yield strength. But the first part where it asks to calculate the strain at fracture confuses me. for material a, the stress at fracture is 3%. what do i do with that?

6. Sep 1, 2011

### PeterO

Re: Strain

Actually for the fist sample, the stain at fracture is 3%, the stress was 300 odd if I recall correctly

Once you have found the fracture point, reading from one axis will give you the stress at fracture - which is also called strength - while the other axis will tell you the strain at the time.

If you look at (b) and find the yield point, that will tell you the stress at yield point, which I believe is called the yield strength and strain at the time can be found from the other axis.

Strain energy was not part of this question - but may be coming into your thoughts.

The area under the stress strain curve gives you the strain energy per unit volume absorbed by the material. The larger the area the tougher the material.
If we consider the area under the graph only as far as the yield point, then the greater the area, the more resilient the material - the energy it absorbs but is still able to spring back.

With sample (b) the amount of strain energy coming back after fracture is approximately the same as that absorbed up to yield point - all the energy absorbed during deformation is gone for good.

7. Sep 1, 2011

### PeterO

Re: Strain

Any chance you were actually after strain energy, not merely strain as you stated?

8. Sep 1, 2011

### TyErd

Re: Strain

I'm pretty sure its just strain.

9. Sep 2, 2011

### PeterO

Re: Strain

If you want strain not as a percentage, I think it is just 0.03 and 0.15 in each case.

For part 2 the strain at yield is something like 5% or 6% - i can't look at the graph and type this at the same time - and that is what is recoverable.

10. Oct 23, 2012

### E_Q

Re: Strain - percentage form?

I have this exact issue too. In my physics questions (A-level) often in a stress/strain graph the strain is given in percent. My teacher has never mentioned that this changes anything, but surely if:

stress (e.g. 0.02[no units])= extension (e.g. 0.1m)/original length(e.g. 5m)

I understand that m/m gives no units, but if stress is given in percentage form (as it seems to often be), the strain would equal 0.02*100=2%?

Hence when i read the value of strain (e.g. 2%) i should first divide by 100 to get the actual value? (as peter0 suggests)

However, what if the original equation was extension 5m and original length 100m? Then the strain "strain is a measure of how much an object is being stretched" must be invariably be in percentage "A fraction with 100 understood as the denominator", equalling 5% - in this case to then divide the answer by 100 would be incorrect. I therefore suspect that the questions i have been doing are incorrect in their useage of defining strain in %.

Any feedback would be appreciated, sorry for taking over the thread :]

-Tom (AS physics student)

Last edited: Oct 23, 2012
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