# HELP Stress Strain Question

1. Mar 21, 2009

HELP!! Stress Strain Question

1. The problem statement, all variables and given/known data

In part (a) of the problem, we found that due to a certain stress, the amount of

elastic strain that a material undergoes is
$\epsilon_E=.0087$

and the amount of plastic strain is
$\epsilon_{pl}=.0113$.

The total strain is therefore
$\epsilon_T=.02$

We are then told that a sample of this material with original length
$l_o=610 \ mm$ undergoes that same stress involved in part (a).

What is the new length $l_f$after the stress is removed ?

So I believe the idea behind this is that we gain the elastic portion of the strain back, but the plastic elongation should be added onto the original length.

I wrote this quantitatively as:

$$l_f=l_0+\epsilon_{pl}\Delta l$$ (1)

To find the change in length we have:

$$\epsilon_T=\frac{\Delta l}{l_0} \Rightarrow \Delta l=\epsilon_Tl_0$$ (2)

Therefore (1) becomes:

$$l_f=l_0+\epsilon_{pl}(\epsilon_Tl_0)$$

$$\Rightarrow l_f=l_0(1+\epsilon_T\epsilon_{pl})$$

Plugging in numbers we have lf=.6101 mm

but the correct answer is .6167 which is waayyy off.

What am I missing here?

Last edited: Mar 22, 2009
2. Mar 22, 2009

### Mapes

Re: HELP!! Stress Strain Question

First find the length when the total load is applied (i.e., the length incorporating elastic and plastic strain). Then find the length change associated with elastic recovery of this sample, and subtract this.

3. Mar 22, 2009

Re: HELP!! Stress Strain Question

I will try that. But how is that different? What concept am I missing here? I do not see the difference between adding the plastic elongation to the initial length and subtracting the elastic elongation from the final length.

4. Mar 22, 2009

### Mapes

Re: HELP!! Stress Strain Question

The mistake you made originally was taking $\epsilon_{pl}$ as the proportion of plastic strain, rather than the amount of plastic strain. The product $\epsilon_T\epsilon_{pl}$ doesn't have any meaning in this problem.

Personally, I think an answer of $l_f=l_0(1+\epsilon_{pl})=0.6169[/tex] is good enough and simpler. But the approach I described gets you to .6167 (actually .616786), so perhaps it's the approach intended by the person who wrote the problem. 5. Mar 22, 2009 ### Saladsamurai Re: HELP!! Stress Strain Question I don't understand this statement? It was given that this was the plastic strain Edit: hold on.... let me think... 6. Mar 22, 2009 ### Saladsamurai Re: HELP!! Stress Strain Question Nope... I still don't understand. If by definition strain is [itex]\epsilon_T=\frac{\Delta l}{l_0}$ then isn't it by definition a proportion?

i.e., percent change in length.

7. Mar 22, 2009

### Mapes

Re: HELP!! Stress Strain Question

In your equation (1) you've multiplied strain by change in length. This doesn't mean anything; strain is already change in length divided by original length. The term $\epsilon\Delta L$ just isn't useful or meaningful here.

However, you could calculate that

$$\epsilon_{pl}/ \epsilon_T=\epsilon_{pl}/(\epsilon_{pl}+\epsilon_{E})=56.5%$$

of the total strain is plastic, and multiply this percentage by the total change in length to get the plastic change in length. That's what I meant by the proportion of plastic strain.

8. Mar 22, 2009

Re: HELP!! Stress Strain Question

Okay. I kind of follow you now. But what is the total change in length is it not $\epsilon_{tot}*l_o$ ?

So my original (1) should have been:

$$l_f=l_0(1+\frac{\epsilon_{pl}}{\epsilon_{tot}})$$

arrgggg goddamn Latex

9. Mar 22, 2009

### Mapes

Re: HELP!! Stress Strain Question

Yes, that's the total change in length. Did I say something different?

10. Mar 22, 2009

Re: HELP!! Stress Strain Question

I don't know :rofl: Perhaps I did! I hate this class! Thanks for the help!

11. Mar 22, 2009

### Mapes

Re: HELP!! Stress Strain Question

Just keep at it and soon it'll seem easy. Good luck.