• Support PF! Buy your school textbooks, materials and every day products Here!

Thermal Stress and a Duel rod of Steel and Al.

  • Thread starter TFM
  • Start date
  • #1
TFM
1,026
0
[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 [tex]\Delta L[/tex] when its temperature changes by [tex]\Delta T[/tex] the stress is equal to [tex] \frac{F}{A} = Y\left(\frac{\Delta L}{L_0}-\alpha \Delta T\right) [/tex].)


Homework Equations



[tex] \frac{F}{A} = -Y \alpha \Delta T [/tex]

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
 

Answers and Replies

  • #2
mgb_phys
Science Advisor
Homework Helper
7,774
12
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
TFM
1,026
0
Would you calculate the change in length by equating the two Stresses like so:

[tex]Y_{steel}(\frac{\Delta L}{L_{0 Steel}} - \alpha_{steel} \Delta T = Y_{aluminium}(\frac{\Delta L}{L_{0 aluminium}} - \alpha_{aluminium} \Delta T [/tex]

?

TFM
 
  • #4
mgb_phys
Science Advisor
Homework Helper
7,774
12
Sounds good - remember you are really balancing the forces, but since both rods have the same area that's equivalent to equal stresses.
 
  • #5
TFM
1,026
0
I get the change inlemngth to be -0.00054m. Does this look right?

TFM
 
  • #6
TFM
1,026
0
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:

[tex] Stress (P) = Y (\frac{\Delta L}{L_0} - \alpha \Delta T) [/tex]

What could I be doing wrong???

TFM
 
  • #7
272
0
Hey TFM,

I get [tex]\Delta L[/tex] to be [tex]1.4897*10^{-4} m[/tex],

How did you get 0.00054?
 
  • #8
TFM
1,026
0
I used:

[tex]Y_{steel}(\frac{\Delta L}{L_{0 Steel}} - \alpha_{steel} \Delta T = Y_{aluminium}(\frac{\Delta L}{L_{0 aluminium}} - \alpha_{aluminium} \Delta T [/tex]

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:

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

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

and rearranged to find Delta L

TFM
 
Last edited:
  • #9
TFM
1,026
0
Further Rearrangement:

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

Expand the Brackets:

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

Cancel down the fraction:

[tex]((5.71*10^{11}\Delta L - (1.44*10^{8})) = 2.8*10^{11}\Delta L - (3.024*10^{8})) [/tex]

Rearrange:

[tex] 5.71*10^{11} \Delta L - 2.8^{11} \Delta L = 3.024*10^{8}+1.44*10^{8} [/tex]

[tex] 2.91*10^{11} \Delta L = -1.584*10^{8} [/tex]

Gives Delta L to be -0.0005435


Have I made a mistake somewhere?

TFM
 
  • #10
alphysicist
Homework Helper
2,238
1
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
TFM
1,026
0
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. :redface:

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

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

TFM
 
  • #12
alphysicist
Homework Helper
2,238
1
Hi Vuldoraq,

Hey TFM,

I get [tex]\Delta L[/tex] to be [tex]1.4897*10^{-4} m[/tex],

How did you get 0.00054?
Just wondering: did you take into account the fact that [itex]\Delta L[/itex] is positive for one rod and negative for the other?
 
  • #13
TFM
1,026
0
How do you take into account that one [tex] \Delta L [/tex] is positive in one direction and negative in the other. Would you, instead of:

[tex]Y_{steel}(\frac{\Delta L}{L_{0 Steel}} - \alpha_{steel} \Delta T = Y_{aluminium}(\frac{\Delta L}{L_{0 aluminium}} - \alpha_{aluminium} \Delta T [/tex]

Instead do:

[tex]Y_{steel}(\frac{\Delta L}{L_{0 Steel}} - \alpha_{steel} \Delta T = Y_{aluminium}(\frac{- \Delta L}{L_{0 aluminium}} - \alpha_{aluminium} \Delta T [/tex]

TFM?
 
  • #14
alphysicist
Homework Helper
2,238
1
Well that's the idea, but I think it better to think of it like this. Start with:

[tex]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) [/tex]

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

But didn't you already do that when you got your answer?
 
  • #15
272
0
Hey Alphysicist,

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

Vuldoraq
 
  • #16
TFM
1,026
0
I din't actually do that, I just kept with the two being equal:redface:

Anyway, now I get:

[tex] -2.8*10^{11}\Delta L - 3.024*10^8 = 5.714*10^{11} \Delta l - 1.44*10^8 [/tex]

rearrange:

[tex]5.714*10^{11} \Delta L + 2.8*10^{11} \Delta L = -1.44*10^8 + 3.024*10^8 [/tex]

Gives:

[tex] 8.514*10^11 \Delta L = 1.58*10^8 [/tex]

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

Any ideas if I am nealry there?

TFM
 
  • #17
alphysicist
Homework Helper
2,238
1
I din't actually do that, I just kept with the two being equal:redface:

Anyway, now I get:

[tex] -2.8*10^{11}\Delta L - 3.024*10^8 = 5.714*10^{11} \Delta l - 1.44*10^8 [/tex]
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:

[tex]5.714*10^{11} \Delta L + 2.8*10^{11} \Delta L = -1.44*10^8 + 3.024*10^8 [/tex]
It looks to me like this doesn't quite follow from the previous equation; I think there are a couple of sign errors.
 
  • #18
272
0
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
TFM
1,026
0
I now get:

[tex] 5.71*10^{11}\Delta L - 1.44*10^8 = -2.8*10^{11} \Delta L - 1.008*10^8[/tex]

then:

[tex] 5.71*10^{11} \Delta L + 2.8*10^{11} \Delta L = -1.008*10^8 + 1.44*10^8 [/tex]

[tex] 8.51*10^{11} \Delta L = 4.32*10^7 [/tex]

thus,

[tex] \Delta L = 5.07*10^-5 [/tex]

???

TFM
 
  • #20
TFM
1,026
0
Assuming what I have is the right answer, should I now put the values into:

[tex] Stress (P) = Y (\frac{\Delta L}{L_0} - \alpha \Delta T) [/tex]

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
 
  • #21
alphysicist
Homework Helper
2,238
1
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.
 
  • #22
TFM
1,026
0
Does this look right?

Steel:

[tex] Stress (P) = Y (\frac{\Delta L}{L_0} - \alpha \Delta T) [/tex]

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

[tex] Stress (P) = (2*10^{11}) ((0.000145) - (0.00072)) [/tex]

[tex] Stress (P) = (2*10^{11}) (-0.000575) [/tex]

Giving Stress for Steel: -1.150*10^8


Aluminium:

[tex] Stress (P) = Y (\frac{\Delta L}{L_0} - \alpha \Delta T) [/tex]

[tex] Stress (P) = (7*10^{10}) (\frac{5.07*10^{-5}}{0.25} - (0.000024) (60)) [/tex]

[tex] Stress (P) = (7*10^{10}) ((0.000203) - (0.00144)) [/tex]

[tex] Stress (P) = (7*10^{10}) (-0.00123) [/tex]

Giving Stress for Aluminium: -8.659*10^7

Do these look right?

TFM
 
  • #23
alphysicist
Homework Helper
2,238
1
When you solved for [itex]\Delta L[/itex], that was the length change for the steel. When you calculate the stress for the aluminum, you need to use the [itex]\Delta L[/itex] for the aluminum.

The stresses you calculate should turn out to be the same. When you solved for [itex]\Delta L_{\rm steel}[/itex], 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.
 
  • #24
TFM
1,026
0
Does this mean that the answer to both should be my value for the Steel's Stress?

TFM
 
  • #25
alphysicist
Homework Helper
2,238
1
Yes; back in your post #13 (and my post #14) you can see that in finding [itex]\Delta L[/itex] 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.
 

Related Threads for: Thermal Stress and a Duel rod of Steel and Al.

Replies
2
Views
328
  • Last Post
Replies
10
Views
12K
Replies
4
Views
6K
Replies
3
Views
332
  • Last Post
Replies
5
Views
3K
Replies
11
Views
1K
Replies
1
Views
9K
  • Last Post
Replies
2
Views
2K
Top