# Thermal Stress and a Duel rod of Steel and Al.

1. May 13, 2008

### TFM

[SOLVED] Thermal Stress and a Duel rod of Steel and Al.

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

A steel rod 0.350 m long and an aluminum rod 0.250 m long, both with the same diameter, are placed end to end between rigid supports with no initial stress in the rods. The temperature of the rods is now raised by 60.0 degrees Celsius.

What is the stress in each rod? (Hint: The length of the combined rod remains the same, but the lengths of the individual rods do not. If the length is permitted to change by an amount $$\Delta L$$ when its temperature changes by $$\Delta T$$ the stress is equal to $$\frac{F}{A} = Y\left(\frac{\Delta L}{L_0}-\alpha \Delta T\right)$$.)

2. Relevant equations

$$\frac{F}{A} = -Y \alpha \Delta T$$

3. The attempt at a solution

Not quite sure. Firstly, are they assuming that one bar is conmtracting and the other expanding, if the overall length of the combined bar doesn't change, but the length of each individual rod does?

The question also fails to give the area oif the rod?

TFM

2. May 13, 2008

### mgb_phys

Yes the rod with the greater exapnsivity will expand compressing the other rod.
The equilibrium is when the pressure in the two rods is the same - so they exert an equal force on each other.

You don't need the area because it cancels - think about making both rods larger diameter, would it change the result?

3. May 13, 2008

### TFM

Would you calculate the change in length by equating the two Stresses like so:

$$Y_{steel}(\frac{\Delta L}{L_{0 Steel}} - \alpha_{steel} \Delta T = Y_{aluminium}(\frac{\Delta L}{L_{0 aluminium}} - \alpha_{aluminium} \Delta T$$

?

TFM

4. May 13, 2008

### mgb_phys

Sounds good - remember you are really balancing the forces, but since both rods have the same area that's equivalent to equal stresses.

5. May 13, 2008

### TFM

I get the change inlemngth to be -0.00054m. Does this look right?

TFM

6. May 13, 2008

### TFM

I tried putting in my value of length change into the equations, but I appear not to get rught answers? Assuming I am using the right equation:

$$Stress (P) = Y (\frac{\Delta L}{L_0} - \alpha \Delta T)$$

What could I be doing wrong???

TFM

7. May 13, 2008

### Vuldoraq

Hey TFM,

I get $$\Delta L$$ to be $$1.4897*10^{-4} m$$,

How did you get 0.00054?

8. May 14, 2008

### TFM

I used:

$$Y_{steel}(\frac{\Delta L}{L_{0 Steel}} - \alpha_{steel} \Delta T = Y_{aluminium}(\frac{\Delta L}{L_{0 aluminium}} - \alpha_{aluminium} \Delta T$$

Where for Steel,

Y: 2 x 10^11
L_0: 0.35
Alpha: 0.000012
temp. change: 60

For Aluminium:

Y: 7 x 10^10
L_0: 0.25
Alpha: 0.000072
temp. change: 60

and inserted them into the equation:

$$(2*10^{11})(\frac{\Delta L}{0.035} - (0.000012) (60) = (7*10^10)(\frac{\Delta L}{0.035} - (0.000072)(60)$$

{Edit: For the second equation, that should be 7*10^10}

and rearranged to find Delta L

TFM

Last edited: May 14, 2008
9. May 14, 2008

### TFM

Further Rearrangement:

$$(2*10^{11})(\frac{\Delta L}{0.035} - (0.00072)) = (7*10^{10})(\frac{\Delta L}{0.025} - (0.00432))$$

Expand the Brackets:

$$((2*10^{11}\frac{\Delta L}{0.035} - (1.44*10^{8})) = \frac{7*10^{10}\Delta L}{0.35} - (3.024*10^{8}))$$

Cancel down the fraction:

$$((5.71*10^{11}\Delta L - (1.44*10^{8})) = 2.8*10^{11}\Delta L - (3.024*10^{8}))$$

Rearrange:

$$5.71*10^{11} \Delta L - 2.8^{11} \Delta L = 3.024*10^{8}+1.44*10^{8}$$

$$2.91*10^{11} \Delta L = -1.584*10^{8}$$

Gives Delta L to be -0.0005435

Have I made a mistake somewhere?

TFM

10. May 14, 2008

### alphysicist

Hi TFM,

Is your value for alpha for Al correct? I think it should be more like 24 x 10^-6 per degree Celsius.

Also, I don't think you have taken into account that the change in length is positive for one rod and negative for the other.

11. May 14, 2008

### TFM

Does 2.4x10^-5 Sound Better, They don't give the values on MP, so you have to use the tables in the book, and I got the alpha and beta mixed up.

Tht gives the change in length as -4.94 x 10^-5

But this still doesnt agree with Vuldoraq's answer???

TFM

12. May 14, 2008

### alphysicist

Hi Vuldoraq,

Just wondering: did you take into account the fact that $\Delta L$ is positive for one rod and negative for the other?

13. May 14, 2008

### TFM

How do you take into account that one $$\Delta L$$ is positive in one direction and negative in the other. Would you, instead of:

$$Y_{steel}(\frac{\Delta L}{L_{0 Steel}} - \alpha_{steel} \Delta T = Y_{aluminium}(\frac{\Delta L}{L_{0 aluminium}} - \alpha_{aluminium} \Delta T$$

$$Y_{steel}(\frac{\Delta L}{L_{0 Steel}} - \alpha_{steel} \Delta T = Y_{aluminium}(\frac{- \Delta L}{L_{0 aluminium}} - \alpha_{aluminium} \Delta T$$

TFM?

14. May 14, 2008

### alphysicist

Well that's the idea, but I think it better to think of it like this. Start with:

$$Y_{steel}(\frac{\Delta L_{\rm steel} }{L_{0 Steel}} - \alpha_{steel} \Delta T )= Y_{aluminium}(\frac{\Delta L_{\rm aluminum}}{L_{0 aluminium}} - \alpha_{aluminium} \Delta T)$$

and then use that $\Delta L_{\rm steel}+\Delta L_{\rm aluminum}=0$ to eliminate $\Delta L_{\rm aluminum}$ (or $\Delta L_{\rm steel}$ if you prefer).

15. May 14, 2008

### Vuldoraq

Hey Alphysicist,

Your right I didn't take into account the fact that one $$\Delta L$$ is the negative of the other, I merely solved for $$\Delta L$$ on both sides of the equation . Sorry if I confused you TFM .

Vuldoraq

16. May 14, 2008

### TFM

I din't actually do that, I just kept with the two being equal

Anyway, now I get:

$$-2.8*10^{11}\Delta L - 3.024*10^8 = 5.714*10^{11} \Delta l - 1.44*10^8$$

rearrange:

$$5.714*10^{11} \Delta L + 2.8*10^{11} \Delta L = -1.44*10^8 + 3.024*10^8$$

Gives:

$$8.514*10^11 \Delta L = 1.58*10^8$$

This has given me a $$\Delta L$$ as 0.000186 m, which still isn't quite the same???

Any ideas if I am nealry there?

TFM

17. May 14, 2008

### alphysicist

I think the term 3.024 * 10^8 is wrong; I think you might have used the beta value (volume expansion coefficient) for aluminum again instead of the alpha value.

It looks to me like this doesn't quite follow from the previous equation; I think there are a couple of sign errors.

18. May 14, 2008

### Vuldoraq

Hi,

I would disregard my value, I think I got my calculations muddled somewhere. I did check it, but sometimes this happens. Sorry again. I would redo it but I don't have a calculator where I am. Will try and post my redone value tomorrow.

19. May 14, 2008

### TFM

I now get:

$$5.71*10^{11}\Delta L - 1.44*10^8 = -2.8*10^{11} \Delta L - 1.008*10^8$$

then:

$$5.71*10^{11} \Delta L + 2.8*10^{11} \Delta L = -1.008*10^8 + 1.44*10^8$$

$$8.51*10^{11} \Delta L = 4.32*10^7$$

thus,

$$\Delta L = 5.07*10^-5$$

???

TFM

20. May 14, 2008

### TFM

Assuming what I have is the right answer, should I now put the values into:

$$Stress (P) = Y (\frac{\Delta L}{L_0} - \alpha \Delta T)$$

For first Steel and then Aluminium? Would one of the delta L's need to be negative again, since one is stretched, the other compressed?

TFM