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

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

## Homework Statement

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 $$\frac{F}{A} = Y\left(\frac{\Delta L}{L_0}-\alpha \Delta T\right)$$.)

## Homework Equations

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

## 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

mgb_phys
Homework Helper
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?

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

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

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

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

Hey TFM,

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

How did you get 0.00054?

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:
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

alphysicist
Homework Helper
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.

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

alphysicist
Homework Helper
Hi Vuldoraq,

Hey TFM,

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

How did you get 0.00054?

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

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?

alphysicist
Homework Helper
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).

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

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

alphysicist
Homework Helper
I din't actually do that, I just kept with the two being equal Anyway, now I get:

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

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.

rearrange:

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

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

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.

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

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

alphysicist
Homework Helper
That looks like the right procedure to me. It would probably be a good idea to check the stress for both steel and aluminum like you suggest, but of course they should turn out to be the same since the cross sectional area is the same.

Does this look right?

Steel:

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

$$Stress (P) = (2*10^{11}) (\frac{5.07*10^{-5}}{0.35} - (0.000012) (60))$$

$$Stress (P) = (2*10^{11}) ((0.000145) - (0.00072))$$

$$Stress (P) = (2*10^{11}) (-0.000575)$$

Giving Stress for Steel: -1.150*10^8

Aluminium:

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

$$Stress (P) = (7*10^{10}) (\frac{5.07*10^{-5}}{0.25} - (0.000024) (60))$$

$$Stress (P) = (7*10^{10}) ((0.000203) - (0.00144))$$

$$Stress (P) = (7*10^{10}) (-0.00123)$$

Giving Stress for Aluminium: -8.659*10^7

Do these look right?

TFM

alphysicist
Homework Helper
When you solved for $\Delta L$, that was the length change for the steel. When you calculate the stress for the aluminum, you need to use the $\Delta L$ for the aluminum.

The stresses you calculate should turn out to be the same. When you solved for $\Delta L_{\rm steel}$, you were finding the change in length of steel that would make the stresses be the same (if the sum of the changes in lengths of both bars equal zero) since they have equal cross sectional areas. So when you plug back in the change of lengths for both, the stresses should turn out to be the same, since you forced them to be the same in the first place.

Does this mean that the answer to both should be my value for the Steel's Stress?

TFM

alphysicist
Homework Helper
Yes; back in your post #13 (and my post #14) you can see that in finding $\Delta L$ we were finding the length changes that would cause the stresses to be equal.

For example, if the steel were exerting a greater force than the aluminum on the junction between the rods, then the rods would just continue expanding/contracting until the forces were equal.

Should One be positive and the other negative, since one is expanding, theother contracting?

TFM

alphysicist
Homework Helper
They are both in a state of compression, and so they'll both have the same sign for the stress. Their "natural length" is longer than what they started out to be because they were heated up.

It's true that the steel rod is longer that what it started out, but if it wasn't for the wall and the aluminum rod it would be even longer still.

But as for the actual signs, it's possible that they just want the magnitudes of the stress. But whatever sign one has, the other will too.

Yeah that was the right answer, Thanks Everybody who helped Out of interest, what would this represent if it ins't the stress?

Giving Stress for Aluminium: -8.659*10^7

Thanks,

TFM

alphysicist
Homework Helper
That would be the stress for the aluminum if it lengthened by $\Delta L$ from its original length instead of shortening.

That makes Sense, Thanks,

TFM