Linear expansion? Or area expansion?

[Patdon10]

Homework Statement

Steel rails for a train track are laid in a region subject to extremes of temperature. The distance from one juncture to the next is 5.4 m, and the cross-sectional area of the rails is 60 cm2. If the rails touch each other without buckling at the maximum temperature, 51 °C, how much space will there be between the rails at -17 °C? (The linear expansion coefficient for steel is 1.3 10-5 °C−1.)

Homework Equations

ΔL = α*L*ΔT
ΔA = 2αAΔT = 2L(αLΔT)

The Attempt at a Solution

Giving the cross-sectional area is confusing. I don't understand why you would need it? From what I understand of this problem, it is simply two "sections" of rails connected. At 51 degress, they are touching. I need to find how much they shrink at -17 degrees. It doesn't matter if it's a wire or a rail, it will shrink or lengthen accordingly (correct?)

ΔL = αLΔT
ΔL = (1.3*10^-5)(5.4m)(-17-51)
ΔL = -0.00477 m
Because two rails are shrinking I need to double that to 0.009547 m.
That's how much space I'm calculating and it's saying it's the wrong answer. Anyone know what I'm doing wrong?

Thanks in advance.

 

[Doc Al]

Because two rails are shrinking I need to double that to 0.009547 m.

Careful. You've calculated the shrinking of the entire rail, not just one end.

 

[Rick88]

I would say you have to use the cross-section as well.
If they gave it to you, it's probably because you have to use it.

R.

 

[Patdon10]

Careful. You've calculated the shrinking of the entire rail, not just one end.

right. because I would need to see the spacing between the two rails. I would assume both rails would shrink, correct?

I would say you have to use the cross-section as well.
If they gave it to you, it's probably because you have to use it.

R.

yeah, I'm scared of that too. Do you know how i could implement it into an equation?

 

[Rick88]

It almost seems to me like you should calculate the volumetric expansion.

 

[Doc Al]

right. because I would need to see the spacing between the two rails. I would assume both rails would shrink, correct?

Correct. So why double your answer?

 

[Patdon10]

It almost seems to me like you should calculate the volumetric expansion.

That would seem to make sense. Didn't work though. What I did:

ΔV = 3L²(αLΔT)
ΔV = 3(S.A)²(αLΔT)
ΔV = -5.1555 e^-4 * 2 for both rails
ΔV = 0.00010311 m.

That's not right either. I'm only going to get one more shot at this. I think I messed up in assuming that L^2 is equal to Surface Area. Should I try it at L^3?

 

[Patdon10]

Correct. So why double your answer?

Because I was only looking at one rail.
ΔL = αLΔT = (1.3*10^-5)(5.4m)(-68).
there are 2 rails sitting end at end. the equation is for one rail that is going to shrink by that amount. However, the temperature change is going to happen everywhere, I have to assume that the other rail is going to shrink by the same amount, causing double the distance between the two rails.

 

[Doc Al]

how much space will there be between the rails at -17 °C?

Admittedly, the word "space" is ambiguous here. But I'd bet they just want the separation distance between the rails.

 

[Patdon10]

Admittedly, the word "space" is ambiguous here. But I'd bet they just want the separation distance between the rails.

I agree. But what are you saying? Should I change anything?

 

[Doc Al]

Because I was only looking at one rail.
ΔL = αLΔT = (1.3*10^-5)(5.4m)(-68).
there are 2 rails sitting end at end. the equation is for one rail that is going to shrink by that amount. However, the temperature change is going to happen everywhere, I have to assume that the other rail is going to shrink by the same amount, causing double the distance between the two rails.

Draw yourself a diagram with two rails touching at the higher temp. Then draw a diagram with the rails contracted. Then figure out how the gap between the rails relates to your calculation.

 

[Rick88]

i get something different.

What I thought was
if \DeltaL = \alpha_L*L*\DeltaT,

then \DeltaV = \alpha_V*V*\DeltaT

I think we can approximate \alpha_V to be 3\alpha_L, and V = \sigma*L where \sigma is the cross-section.

I found \DeltaV to be 8.59e-5.

(I haven't done this stuff in ages, though!)

 

[Patdon10]

i get something different.

What I thought was
if \DeltaL = \alpha_L*L*\DeltaT,

then \DeltaV = \alpha_V*V*\DeltaT

I think we can approximate \alpha_V to be 3\alpha_L, and V = \sigma*L where \sigma is the cross-section.

I found \DeltaV to be 8.59e-5.

(I haven't done this stuff in ages, though!)

nope. that's not it :(

 

[Rick88]

will you know what the correct answer is?

 

[Patdon10]

No. but I'm going to email my teacher right now, ask for more submissions and a hint as to what it's looking for. I'll report back.

 

[Doc Al]

No. but I'm going to email my teacher right now, ask for more submissions and a hint as to what it's looking for. I'll report back.

Did you draw the diagram I suggested? Do so and you'll likely find your error.

 

[Patdon10]

Did you draw the diagram I suggested? Do so and you'll likely find your error.

Are you saying I'm assuming that the shrinking occurs in one side, when it really occurs in both? So I shouldn't double it?

 

[Doc Al]

Are you saying I'm assuming that the shrinking occurs in one side, when it really occurs in both?

The shrinking occurs throughout the entire rod. You found how much the entire rod shrinks. So how does that relate to the gap between the rods. Don't try to reason it out 'in your head'. Draw a diagram and you'll see for yourself whether you should have doubled that answer or not.

Hint: Assume the center of each rod doesn't move.

 

[Patdon10]

Yeah. So I shouldn't have doubled it. So linear expansion was the right way to go? Hopefully I'll get another submission and be able to see if it's the right answer. The surface area was just added to confuse me?

 

[Doc Al]

Yeah. So I shouldn't have doubled it.

Right.

So linear expansion was the right way to go?

I'd say so.