# Stresses due to change in temperature change

1. Aug 13, 2012

### ShawnCohen

Say you have a member of original length L and after a change in temperature T it changes in length to L'=L(1+αT). If we apply a force so that the length is returned back to L, the change in length is LαT and the strain is ε=LαT/L=αT.
My question is why is the denominator in the strain L and not L' since the new lenth of the member is L' and it is that length that we are shrinking.

2. Aug 13, 2012

### Studiot

Hello Shawncohen

You have a good question, if a slightly garbled title and description.

Where did your information come from?

Strictly speaking if we heat something so it expands and then squash it back to its original size it is different from loading it so that does not expand in the first place.

We regard the original length as the basis for the strain in the second process.

However, you are correct that strictly speaking the basis length for strain in the first example is L'.

However we are normally talking about millistrain (parts per thousand) for usual materials so to a first order of approximation the result is the same.

3. Aug 13, 2012

### ShawnCohen

Yeah sorry about that, my original problem in a textbook asked for stress but the problem I had in the solutions was with strain.

Thanks a lot though, great help!

Can you also recommend anything for me to read on why there is a difference?
Thanks again

4. Aug 13, 2012

### Studiot

Not quite sure what you mean - difference in what?

5. Aug 13, 2012

### ShawnCohen

Why if we heat something so it expands and then squash it back to its original size is different from loading it so that it does not expand in the first place.

6. Aug 13, 2012

### Studiot

OK shawn.

It is a fundamental principle of elasticity that the behaviour of a body under load does not depend upon its past (stress) history.

So if we strain a body by loading it then let go it returns to its original size.
We can do this as many times as we like and each time the result is the same. (So long as we remain in the elastic region).

Similarly we can heat a body without restraint and it will expand to new size.
Then if we cool it it will return to its original size.
Again we can do this as many times as we like.

Now what happens if we load something or heat it and then wait a thousand years?

If after a thousand years someone else comes back and loads or heats it again what will they use as the base size of the body?

7. Aug 13, 2012

### ShawnCohen

they will use the length they see, so the extended length surely?

my question is more that when we load the member to prevent it expanding before it is heated it isnt actually subjected to any stress, it only experiences stress when we heat it and I'm finding it hard to get my head around the fact we use an 'extension length' that hasnt actually happened to work out the strain and therefore the stress if this is clearer

8. Aug 13, 2012

### Studiot

But we don't do that.

If you heat a body so it expands from L to (L+δL), then walk away and wait five minutes or five centuries you are starting again effectively with a new situation if you then compress it back to L, whilst maintaining its higher temperature.

So your strain on thermal expansion is δL/L

and on physical compression is δL/(L+δL).

However the difference is a second order (δL)2quantity so we ignore it

$$\frac{{\delta L}}{L} - \frac{{\delta L}}{{\left( {L + \delta L} \right)}} = \frac{{{{\left( {\delta L} \right)}^2}}}{{L\left( {L + \delta L} \right)}}$$

However we don't usually consider this.

What we usually consider is partial restraint where we equate the strain in one material to the strain in another withich is joined to the first. Then we equate stains to get the stresses.

9. Aug 13, 2012

### ShawnCohen

so how do you deduce that the strain on thermal expansion is δL/L if it hasnt actually expanded by δL, that's whats confusing me

10. Aug 13, 2012

### Studiot

If you heat a body without restraint it expands from L to (L+δL).

Sorry I should have made the without restraint clearer.

11. Aug 13, 2012

### ShawnCohen

so there is strain on a body due to pure heat expansion with no restraint?

12. Aug 13, 2012

### Studiot

Yes of course.

Thermal strain is the main way of obtaining strain without stress.

Of course it may be partly restrained in which case there is some stress and some strain, but neither are as great as they would be if the other were zero.

13. Aug 13, 2012

### ShawnCohen

Thanks a lot, this has helped me a great deal!!

14. Aug 13, 2012

### Studiot

Here is a typical example.

Two rods, one of steel and one of copper are fixed between rigid walls as shown

The steel rod is heated with a blowtorch so it expands.

However it is restrained by the copper rod.

The expansion of the steel rod = the contraction of the copper rod

But the strains are not equal.

Since the x section areas are not equal the stresses will not be equal either.

But using the above piece of information we can calculate the the position where the expansion force in the steel just balances the restraint force in the copper.

Can you see this?

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15. Aug 13, 2012

### ShawnCohen

Yes I did examples like that.
The one I had problems with was having just one metal bar fixed by rigid walls and therefore no expansion was allowed what so ever

16. Aug 13, 2012

### Studiot

I suppose the easiest way to regard the fully restrained case would be like this:-

Imagine a fully restrained bar between two walls.

Heat the bar so that it is subject to a compressive force from the walls as it tries to expand.

Now imagine one wall suddenly removed so that the bar extends to L(1+αΔT) ie to L+δL

If we now apply a force compressive F so that the bar is recompressed to L the stress is given by

stress = F/A = modulus x strain = EδL/(L+δL) ≈ EδL/L

By the discussion earlier.

Note that really if you look closely at what I said earlier it is not just that (δL)2 is small but that the fraction is really (δL/L)2.

For normal materials δL/L is nomally of the order of 1 in 1000 to the square is of the order of 1 in 1 million.