# Homework Help: Change in length due to temperature

1. Jul 28, 2016

1. The problem statement, all variables and given/known data
in the notes , i was told that ∂A is the resistance of aluminium rod....I'm wondering the change length of steel rod that we can 'see' is ∂ st or ∂T(st) ?

2. Relevant equations

3. The attempt at a solution
I think the change length of steel rod that we can 'see' is ∂ st ?

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2. Jul 28, 2016

### BvU

This time they use $\delta$, not $\partial$ .
It clearly says 'the movement of A'
Has nothing to do with the resistance !?!?
I don't see $\delta A$, only $\delta_A$.

And under 'contraction of steel rod' I can't distinguish what it says, something like $\delta_{T(st)}$ ?

3. Jul 28, 2016

the author didnt say that. He stated that the steel rod cannot contract freely because resistance of aluminium rod.
I think the δT(st) is contraction of steel rod due to drop in temperature...
So, when we want to measure the change in length, the length that we gt is δA or) δT(st)?
P/s : the diagram beside the figure of rod is δ(st) + δT(st) = δT(st)

Last edited: Jul 28, 2016
4. Jul 28, 2016

### haruspex

It's the contraction that would have occurred due to temperature if the aluminium rod were not there.
δA is the observed contraction in the steel. δst is the discrepancy between the two.
But I do not understand what the diagram shows at C. It's a bit fuzzy, but it looks like it says the expansion of the aluminium rod equals the observed contraction of the steel rod. That is clearly not the case.

5. Jul 28, 2016

so, δA is the observed contraction in the steel?
What is δT(st), we wouldn't see it with naked eyes?

6. Jul 28, 2016

### Staff: Mentor

The stress in the steel rod is given by: $$\sigma_s=E_s(\epsilon_s-\alpha_s \Delta T)\tag{key}$$
where $\epsilon_s$ is the strain in the steel rod, Es is the Young's modulus of the steel rod, and $\alpha_s$ is the coefficient of linear expansion of the steel rod. This equation says that, if the steel rod expands with no constraint (i.e., with $\sigma_s = 0$), the strain in the rod is $\epsilon_s = \alpha_s \Delta T$, but, if the stress in the rod is greater than 0, the strain will be greater than $\alpha_s \Delta T$. If $\delta_s$ is the downward displacement of the end of the steel rod, then $\epsilon_s=\delta_s/L_s$, and $$\sigma_s=E_s\left(\frac{\delta_s}{L_s}-\alpha_s \Delta T\right)$$
Geometrically, the downward displacement of the end of the steel rod is related to the downward displacement of the end of the aluminum rod by:$$\frac{\delta_s}{0.6}=-\frac{\delta_A}{1.2}\tag{1}$$
The tensile stress in the aluminum rod is given by$$\sigma_A=E_A\frac{\delta _A}{L_A}$$
The tension in the steel rod is given by$$P_s=A_s\sigma_s=A_sE_s\left(\frac{\delta_s}{L_s}-\alpha_s \Delta T\right)\tag{2}$$
Similarly, the tension in the aluminum rod is given by $$P_A=A_AE_A\frac{\delta _A}{L_A}\tag{3}$$
If we combine Eqns. 1-3 with the moment balance on the bar ABC, we can solve for all the displacements, strains, stresses, and tensions. What is the moment balance on the bar ABC in terms of the tensions $P_s$ and $P_A$?

The key to this whole analysis is the equation labeled "key." This equation takes into account the thermal expansion strain experienced by the rod plus whatever extra strain experienced by the rod to give the overall stress.

7. Jul 28, 2016

### haruspex

No, you would not, but you can calculate it from the temperature change.

8. Jul 29, 2016

the sigma s is strain , which is the change in length / original length , right , why you said it is constraint ?

9. Jul 29, 2016

δA is the observed contraction in the steel? the change in length that we can notice is δA ? then , what does δ(st) means ?

10. Jul 29, 2016

### haruspex

As I posted, it is the difference between δA and δT(st). What is the immediate cause of that difference?

11. Jul 29, 2016

sorry , i really have no idea what will cause the difference of δA and δT(st) . Will we be able to see that difference with naked eyes?

12. Jul 29, 2016

### haruspex

See your own response to Chet

13. Jul 29, 2016

can you explain further ?

14. Jul 29, 2016

### haruspex

How does the aluminium rod affect the contraction of the steel rod?

15. Jul 29, 2016

When steel contract, aluminum extend....

16. Jul 29, 2016

### haruspex

Yes, but think about the forces. What is the consequence for the steel?

17. Jul 29, 2016

The steel contract, the forces act upwards?

18. Jul 29, 2016

### haruspex

The contraction of the steel is because of the temperature change. How is this altered by the aluminium rod? Is the steel under compression or under tension?

19. Jul 29, 2016

### Staff: Mentor

No. As I said, it is the stress (not the strain).

20. Jul 29, 2016

from the figure , we could see that the change in length due to temperature is more than the δA , so it should be
σs=Es(-ϵs+αsΔT) ?

21. Jul 29, 2016

steel under compression

22. Jul 29, 2016

### Staff: Mentor

Forget about A. We're just focusing on the steel. Even though the steel rod is shorter than its original length (because of its decrease in temperature), it is still in tension because it is not being allowed to contract in length as much as it would have liked to. So the original equation I gave is correct, and captures both the effect of the decreased temperature and the constraint of not being able to contract (as much). We are using the sign convention that tensile stress is positive.

23. Jul 29, 2016

### haruspex

Why?

24. Jul 29, 2016

but, in post#6 ,
since there's constraint, so $$\sigma_s >0$$ ?
if so, εs > αΔT ?
But, that's not the case...In the notes in the post#1, (st) > A , which means αΔT> εs , Am i right?

25. Jul 29, 2016