Change in length due to temperature

[chetzread]

Homework Statement

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) ?

Homework Equations

The Attempt at a Solution

I think the change length of steel rod that we can 'see' is ∂ st ?

 

[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 $\delta_{T(st)}$ ?

 

[chetzread]

It clearly says 'the movement of A'
Has nothing to do with the resistance ??

the author didnt say that. He stated that the steel rod cannot contract freely because resistance of aluminium rod.

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

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)

 

[haruspex]

I think the δT(st) is contraction of steel rod due to drop in temperature...

It's the contraction that would have occurred due to temperature if the aluminium rod were not there.

the change in length, the length that we gt is δA or) δT(st)?

δ_A is the observed contraction in the steel. δ_st 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.

 

[chetzread]

It's the contraction that would have occurred due to temperature if the aluminium rod were not there.

δ_A is the observed contraction in the steel. δ_st 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.

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

 

[Chestermiller]

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, E_s 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.

 

[haruspex]

What is δT(st), we wouldn't see it with naked eyes?

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

 

[chetzread]

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, E_s 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.

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

 

[chetzread]

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

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

 

[haruspex]

what does δ(st) means ?

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

 

[chetzread]

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

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

 

[haruspex]

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

See your own response to Chet

strain , which is the change in length / original length ,

 

[chetzread]

See your own response to Chet

can you explain further ?

 

[haruspex]

can you explain further ?

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

 

[chetzread]

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

When steel contract, aluminum extend...

 

[haruspex]

When steel contract, aluminum extend...

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

 

[chetzread]

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

The steel contract, the forces act upwards?

 

[haruspex]

The steel contract, the forces act upwards?

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?

 

[Chestermiller]

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

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

 

[chetzread]

The stress in the steel rod is given by:

σs=Es(ϵs−αsΔT)​

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) ?

 

[chetzread]

steel under compression

steel under compression

 

[Chestermiller]

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) ?

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.

 

[haruspex]

steel under compression

Why?

 

[chetzread]

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.

but, in post#6 ,

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, E_s 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 $$\sigma_s=E_s\left(\frac{\delta_s}{L_s}-\alpha_s \Delta T\right)$$

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?

 

[chetzread]

Why?

steel contract

 

[haruspex]

steel contract

The steel contracts because it is cold. That is not compression. Compression would be contraction because of an externally applied force.
What force is being externally applied?

 

[chetzread]

The steel contracts because it is cold. That is not compression. Compression would be contraction because of an externally applied force.
What force is being externally applied?

do you mean steel contract is tension? why? i still didnt get that...

What force is being externally applied?

external compression force

 

[Chestermiller]

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?

No. $\alpha \Delta T$ is negative, so $-\alpha \Delta T$ is positive and, even if $\epsilon_s$ is negative, the stress is still positive.

Here is a thought experiment for you. You take a rod of steel and cool it (without constraining it) so that it contracts to a new length. The stress in the rod after cooling is still zero, but it has experienced a negative strain. Now you stretch the rod at constant temperature back to nearly (but not quite) its original length. Is it now under tension or compression?

 

[chetzread]

No. $\alpha \Delta T$ is negative, so $-\alpha \Delta T$ is positive and, even if $\epsilon_s$ is negative, the stress is still positive.

Here is a thought experiment for you. You take a rod of steel and cool it (without constraining it) so that it contracts to a new length. The stress in the rod after cooling is still zero, but it has experienced a negative strain. Now you stretch the rod at constant temperature back to nearly (but not quite) its original length. Is it now under tension or compression?

do you mean in $$\sigma_s=E_s(\epsilon_s-\alpha_s \Delta T)\tag{key}$$
, εs is negative(due to contraction) , and -σsΔT has positive value (-σs(-T) =positive) , so σs has positive value?

 

[chetzread]

Here is a thought experiment for you. You take a rod of steel and cool it (without constraining it) so that it contracts to a new length. The stress in the rod after cooling is still zero, but it has experienced a negative strain. Now you stretch the rod at constant temperature back to nearly (but not quite) its original length. Is it now under tension or compression?

tension

 

[Chestermiller]

tension

Right. So, even though it is a little shorter, it is in tension. In your problem, even though the steel bar is shorter, it is in tension.

 

[chetzread]

Right. So, even though it is a little shorter, it is in tension. In your problem, even though the steel bar is shorter, it is in tension.

You mean pull the contracted steel , the force is tension?
So, how does it relate to the question I asked?

 

[Chestermiller]

You mean pull the contracted steel , the force is tension?

Yes.

do you mean in $$\sigma_s=E_s(\epsilon_s-\alpha_s \Delta T)\tag{key}$$

Yes.

, εs is negative(due to contraction) , and -σsΔT has positive value (-σs(-T) =positive) , so σs has positive value?

Yes.

So, how does it relate to the question I asked?

The answer to your question is Yes.

 

[chetzread]

Yes.

Yes.

Yes.

The answer to your question is Yes.

so, σs=Es(-ϵs+αsΔT)​

where
αsΔT >ϵs , so σs = positive?

 

[Chestermiller]

so, σs=Es(-ϵs+αsΔT)​

where
αsΔT >ϵs , so σs = positive?

No. The equation I gave is correct.

 

[chetzread]

No. The equation I gave is correct.

I refer to the question that i posted in post#1, the steel contracted,
original equation is
$$\sigma_s=E_s(\epsilon_s-\alpha_s \Delta T)$$

so, εs is negative(due to contraction) , and -σsΔT has positive value (-σs(-T) =positive) , so σs has positive value?
thus, leading the equation to become
σs=Es(-ϵs+αsΔT) ?

 

[Chestermiller]

I refer to the question that i posted in post#1, the steel contracted,
original equation is
$$\sigma_s=E_s(\epsilon_s-\alpha_s \Delta T)\$$

so, εs is negative(due to contraction) , and -σsΔT has positive value (-σs(-T) =positive) , so σs has positive value?
thus, leading the equation to become
σs=Es(-ϵs+αsΔT) ?

You need to brush up on algebra.

 

[chetzread]

You need to brush up on algebra.

Which part is wrong?

 

[chetzread]

It's the contraction that would have occurred due to temperature if the aluminium rod were not there.

δ_A is the observed contraction in the steel. δ_st 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.

So, the change in length that we can notice is del (A) ? del (st) is the constraint?

 

[Chestermiller]

Which part is wrong?

The equation as I wrote it is correct and the equation as you wrote it is incorrect. If $\Delta T=0$, your equation predicts that $\sigma=-E\epsilon$. Does that make sense to you?

 

[haruspex]

del (st) is the constraint?

I would not have called it a constraint. That is not quite the right word. You could say the constraint is that the horizontal bar remains straight and attached to the two rods.
Del(st) is a consequence of that constraint. Another consequence is the stretching of the aluminium rod.

It might help to think of the process in three stages:

1. The steel rod is cooled with no aluminium rod present. It contracts by δ_T(st). It is not under any tension or compression. Correspondingly, the horizontal bar rotates clockwise a bit.
2. The aluminium rod is now to be attached to the bar, but because the bar has rotated it is not quite long enough. The bar is forcibly rotated back to the horizontal so that the aluminium rod can be attached. What has happened to the steel rod?
3. The bar is now released. What will happen to the bar and each rod?

 

[chetzread]

What has happened to the steel rod?

The steel rod extend so that the bar can be rotated back to it's original condition...

The bar is now released. What will happen to the bar and each rod?

Steel rod shorten, aluminum rod extend?

The bar is forcibly rotated back to
the horizontal so that the aluminium rod can be attached

Do you mean, the aluminium will tried to pull the bar back to it's horizontal conditions, but the bar may not be perfectly horizontal as before rotation?

 

[chetzread]

The equation as I wrote it is correct and the equation as you wrote it is incorrect. If $\Delta T=0$, your equation predicts that $\sigma=-E\epsilon$. Does that make sense to you?

no . But , how does the equation you wrote relate to the question i asked in post #1?
$$\sigma_s=E_s(\epsilon_s-\alpha_s \Delta T)$$
to get σs = positive , ϵs must be > αsΔT ,
But , in the picture in the notes , i noticed that the change due to temperature (αsΔT or ∂ T(st) ) is more than ϵs(∂ A) , how can the equation given by you true ?

 

[Chestermiller]

no . But , how does the equation you wrote relate to the question i asked in post #1?
$$\sigma_s=E_s(\epsilon_s-\alpha_s \Delta T)$$
to get σs = positive , ϵs must be > αsΔT ,

What if $\alpha_s\Delta T$ is negative, and $\epsilon_s$ is also negative, but smaller in absolute magnitude than $\alpha_s\Delta T$? Then, under these circumstances, $\epsilon_s>\alpha_s\Delta T$, and $\sigma>0$

But , in the picture in the notes , i noticed that the change due to temperature (αsΔT or ∂ T(st) ) is more than ϵs(∂ A) , how can the equation given by you true ?

Look at the picture again...carefully.

 

[chetzread]

What if αsΔT\alpha_s\Delta T is negative, and ϵs\epsilon_s is also negative, but smaller in absolute magnitude than αsΔT\alpha_s\Delta T? Then, under these circumstances, ϵs>αsΔT\epsilon_s>\alpha_s\Delta T, and σ>0

Let ϵs = -3 , αsΔT = -6 , so
$$\sigma_s=E_s(\epsilon_s-\alpha_s \Delta T) = Es(-3-(-6) ) = +3 $$
so , we can rewrite the equation as $$\sigma_s=E_s(\-epsilon_s+\alpha_s \Delta T)$$ , am i right ?

 

[chetzread]

Look at the picture again.

can you explain further ?

 

[Chestermiller]

Let ϵs = -3 , αsΔT = -6 , so
$$\sigma_s=E_s(\epsilon_s-\alpha_s \Delta T) = Es(-3-(-6) ) = +3 $$
so , we can rewrite the equation as $$\sigma_s=E_s(\-epsilon_s+\alpha_s \Delta T)$$ , am i right ?

No. This is algebraically incorrect.

 

[Chestermiller]

can you explain further ?

Whichever of the dotted lines you look at, the decrease in length of the steel rod is half the increase in length of the aluminum rod.

 

[chetzread]

No. This is algebraically incorrect.

What if $\alpha_s\Delta T$ is negative, and $\epsilon_s$ is also negative, but smaller in absolute magnitude than $\alpha_s\Delta T$?

 

[haruspex]

The steel rod extend so that the bar can be rotated back to it's original condition

Yes. Is it now under compression, under tension, or neutral?

Steel rod shorten, aluminum rod extend?

Yes.

Do you mean, the aluminium will tried to pull the bar back to it's horizontal conditions, but the bar may not be perfectly horizontal as before rotation?

yes. So, for each rod, is it now under compression, under tension, or neutral?