Change in length due to temperature

[haruspex]

what are you trying to say ?

Earlier you appeared to be confused about compression and tension in the context of this problem. That part of the thread was my attempt to prove to you that in the end both rods will be in tension. From there, it is a matter of finding exactly how those two tensions balance out. I have been leaving that in the capable hands of Chester M.

 

[chetzread]

Earlier you appeared to be confused about compression and tension in the context of this problem. That part of the thread was my attempt to prove to you that in the end both rods will be in tension. From there, it is a matter of finding exactly how those two tensions balance out. I have been leaving that in the capable hands of Chester M.

so , due to Del(st) is a consequence of that constraint , so we could only see the del(A) as the change in length rather than del(Tst) ?

 

[Chestermiller]

it 's not i said . It's in figure , refer to green circle

In my judgment, the figure is incorrect.

 

[chetzread]

In my judgment, the figure is incorrect.

what should be the correct one ?

 

[Chestermiller]

what should be the correct one ?

I don't have time right now to draw the correct figure because we are having some relatives visiting over the next couple of days. Sometime in the next few days, I will provide the correct figure. I hope you can afford to wait.

 

[haruspex]

so , due to Del(st) is a consequence of that constraint , so we could only see the del(A) as the change in length rather than del(Tst) ?

Yes.

 

[haruspex]

it 's not i said . It's in figure , refer to green circle

The diagram in the green circle matches what I wrote in post #4:

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

it's the contraction that would occur due to the temperature change alone, yes. δA is the observed contraction in the steel. δst is the discrepancy between the two.

But as I also noted, the main diagram is very hard to read at bottom right. It shows some other δ_{some subscript} for the extension of the aluminium rod. What is the subscript there? If that is also just "A" then it is wrong. Alternatively, if δ_A is defined as the extension in the aluminium rod then the diagram in the green circle is wrong.

 

[TSny]

Is the labeling as follows?

$\delta_{st}$ is the change in length of the steel rod "due to stress alone".

$\delta_{T(st)}$ is the change in length of the steel rod "due to temperature change alone".

$\delta_A$ is the net displacement of point A (also the net change in length of the steel rod).

$\delta_{Al}$ is the change in length of the aluminum rod due to stress (also the displacement of C).

 

[chetzread]

Is the labeling as follows?
View attachment 104083

$\delta_{st}$ is the change in length of the steel rod "due to stress alone".

$\delta_{T(st)}$ is the change in length of the steel rod "due to temperature change alone".

$\delta_A$ is the net displacement of point A (also the net change in length of the steel rod).
Yes, but $\delta_st$ is due to stress? Why there's stress? What caused it?
$\delta_{Al}$ is the change in length of the aluminum rod due to stress (also the displacement of C).

Is the labeling as follows?
View attachment 104083

$\delta_{st}$ is the change in length of the steel rod "due to stress alone".

$\delta_{T(st)}$ is the change in length of the steel rod "due to temperature change alone".

$\delta_A$ is the net displacement of point A (also the net change in length of the steel rod).

$\delta_{Al}$ is the change in length of the aluminum rod due to stress (also the displacement of C).

 

[haruspex]

Merely quoting (twice) what someone else posted is not very helpful. You were asked a question. Did you mean to include an answer?

 

[chetzread]

δstδst\delta_{st} is the change in length of the steel rod "due to stress alone".

What forces will cause the stress?

 

[chetzread]

Is the labeling as follows?
View attachment 104083

$\delta_{st}$ is the change in length of the steel rod "due to stress alone".

$\delta_{T(st)}$ is the change in length of the steel rod "due to temperature change alone".

$\delta_A$ is the net displacement of point A (also the net change in length of the steel rod).

$\delta_{Al}$ is the change in length of the aluminum rod due to stress (also the displacement of C).

that's exactly same in the notes

 

[chetzread]

why wouldn't we see the(change in length) as del(Tst) ,but we will see del(A) only?

 

[chetzread]

deleted

 

[chetzread]

Is the labeling as follows?
View attachment 104083

$\delta_{st}$ is the change in length of the steel rod "due to stress alone".

$\delta_{T(st)}$ is the change in length of the steel rod "due to temperature change alone".

$\delta_A$ is the net displacement of point A (also the net change in length of the steel rod).

$\delta_{Al}$ is the change in length of the aluminum rod due to stress (also the displacement of C).

the change in length that we will notice is ∂ A , am i right? why we wouldn't be able to see ∂ A + ∂ (st ) = ∂ (Tst) ?

 

[haruspex]

that's exactly same in the notes

I told you twice that I could not read the subscript on the δ at C. If you had told me earlier it said δ_Al that would have been a great help.

What forces will cause the stress

You agreed earlier that both rods are stretched under tension because otherwise they would not be long enough for both to reach the rigid bar.

 

[chetzread]

I told you twice that I could not read the subscript on the δ at C. If you had told me earlier it said δ_Al that would have been a great help.

You agreed earlier that both rods are stretched under tension because otherwise they would not be long enough for both to reach the rigid bar.

ok, that's the tension that caused the stress.

 

[Chestermiller]

what should be the correct one ?

Here is the figure I was referring to:

It is a before and after picture of the steel bar. After cooling, the new unextended length of the steel bar is now equal to $L_0(1+\alpha \Delta T)$. This would be the new length of the bar if it were not under stress, and is shorter than the original length because $\Delta T$ is negative. The actual new length of the bar is equal to $L_0-\delta=L_0(1+\epsilon_s)$. The amount of stretching that at bar has experienced relative to its new unextended length is
$$(L_0-\delta)-L_0(1+\alpha \Delta T)=L_0(\epsilon_s-\alpha \Delta T)$$Thus, the effective strain in the bar which gives rise to tensile stress is $\epsilon_s-\alpha \Delta T$, and the tensile stress in the bar is given by: $\sigma=E(\epsilon_s-\alpha \Delta T)$

 

[chetzread]

Here is the figure I was referring to:
View attachment 104102
It is a before and after picture of the steel bar. After cooling, the new unextended length of the steel bar is now equal to
$L_0(1+\alpha \Delta T)$. This would be the
new length of the bar if it were not under stress, and is shorter than the original length because $\Delta T$ is negative. The
actual new length of the bar is equal to
$L_0-\delta=L_0(1+\epsilon_s)$. The amount of stretching that at bar has experienced relative to its new unextended length is
$$(L_0-\delta)-L_0(1+\alpha \Delta T)=L_0(\epsilon_s-\alpha \Delta T)$$Thus,
the effective strain in the bar which gives rise to
tensile stress is $\epsilon_s-\alpha \Delta T$, and the tensile stress in the bar is given by: $\sigma=\epsilon_s-\alpha \Delta T$

So, what you are saying is without the constrain(aluminum rod), the steel rod will contract to the unstretched length... But,
there's aluminum rod to prevent the steel rod to
contact too much.
So what we notice is delta only?

 

[Chestermiller]

So, what you are saying is without the constrain(aluminum rod), the steel rod will contract to the unstretched length... But,
there's aluminum rod to prevent the steel rod to
contact too much.
So what we notice is delta only?

Yes and yes.

 

[TSny]

the change in length that we will notice is ∂ A , am i right?

Yes

why we wouldn't be able to see ∂ A + ∂ (st ) = ∂ (Tst) ?

$\delta_{T(st)}$ is not observable because $\delta_{T(st)}$ represents how much the length of the rod would change due to cooling if the rod were not constrained by the rest of the system. But the constraint keeps the rod from actually changing by $\delta_{T(st)}$. So you never observe $\delta_{T(st)}$.

 

[chetzread]

if the steel rod expands with no constraint (i.e., with σs=0\sigma_s = 0), the strain in the rod is ϵs=αsΔT\epsilon_s = \alpha_s \Delta T, but, if the stress in the rod is greater than 0, the strain will be greater than αsΔT\alpha_s \Delta T. If δs\delta_s is the downward displacement of the end of the steel rod, then ϵs=δs/Ls\epsilon_s=\delta_s/L_s, and

you mean contract ( in the case in post 1 , the steel contract) ?

when u said stress > 0 , you mean the steel contract with contract with constraint ?

 

[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?

so , by referring to post #1 ,
in equation of σs=Es(ϵs-αsΔT) ,
-αsΔT has positive value and ϵs has positive value , so σs = positive ?

 

[Chestermiller]

so , by referring to post #1 ,
in equation of σs=Es(ϵs-αsΔT) ,
-αsΔT has positive value and ϵs has positive value , so σs = positive ?

No. $\epsilon_s$ has a negative value, but -αsΔT is positive and exceeds this negative value, so the net effect is that σs = positive.

 

[Chestermiller]

you mean contract ( in the case in post 1 , the steel contract) ?

when u said stress > 0 , you mean the steel contract with contract with constraint ?

Contraction is just negative expansion. If $\Delta T$ is negative, and, if the rod were free to contract, $\epsilon_s$ would be negative, and equal to $\alpha \Delta T$, and the stress would be zero. But, $\Delta T$ were negative, and, if the rod were partially constrained from contracting, $\epsilon_s$ could still be negative, but, relative to the new unextended length of the rod, there would be extension, and the stress would be positive.

 

[chetzread]

Here is the figure I was referring to:
View attachment 104102
It is a before and after picture of the steel bar. After cooling, the new unextended length of the steel bar is now equal to $L_0(1+\alpha \Delta T)$. This would be the new length of the bar if it were not under stress, and is shorter than the original length because $\Delta T$ is negative. The actual new length of the bar is equal to $L_0-\delta=L_0(1+\epsilon_s)$. The amount of stretching that at bar has experienced relative to its new unextended length is
$$(L_0-\delta)-L_0(1+\alpha \Delta T)=L_0(\epsilon_s-\alpha \Delta T)$$Thus, the effective strain in the bar which gives rise to tensile stress is $\epsilon_s-\alpha \Delta T$, and the tensile stress in the bar is given by: $\sigma=E(\epsilon_s-\alpha \Delta T)$

why u want to make -αTL and εsL negative ?
They just represent magnitude , right ?
if i omit the negative sign ,
so the unstretched length will be = L - αTL , instead of L + αTL

 

[Chestermiller]

why u want to make -αTL and εsL negative ?
They just represent magnitude , right ?
if i omit the negative sign ,
so the unstretched length will be = L - αTL , instead of L + αTL

They don't just represent the magnitudes. They represent the actual values. You can see visually, from the figure that $\epsilon_sL$ is negative. And you can see from the figure that $-\alpha \Delta T L$ is positive, so that the unstretched length is $L_0+\alpha \Delta T L_0$ is shorter than the initial length $L_0$

 

[chetzread]

They don't just represent the magnitudes. They represent the actual values. You can see visually, from the figure that $\epsilon_sL$ is negative. And you can see from the figure that $-\alpha \Delta T L$ is positive, so that the unstretched length is $L_0+\alpha \Delta T L_0$ is shorter than the initial length $L_0$

ok, for equation
$\sigma=E(\epsilon-\alpha \Delta T)$
it's only applicable for the steel is being cooled?
It's not applicable for steel that is heated?
I said so because in $\sigma=E(\epsilon-\alpha \Delta T)$, εs is positve, and - αTL is negative and εs < - αTL ,Am i right?
, so in $\sigma=E(\epsilon-\alpha \Delta T)$ ,
let's say εs = 3 , αTL = -5
so, σ = E(3-5)= -2E , which is negative, how could that possible?

 

[Chestermiller]

ok, for equation
$\sigma=E(\epsilon-\alpha \Delta T)$
it's only applicable for the steel is being cooled?
It's not applicable for steel that is heated?
I said so because in $\sigma=E(\epsilon-\alpha \Delta T)$, εs is positve, and - αTL is negative and εs < - αTL ,Am i right?
, so in $\sigma=E(\epsilon-\alpha \Delta T)$ ,
let's say εs = 3 , - αTL = -5
so, σ = E(3-5)= -2E , which is negative, how could that possible?

This equation is applicable to both heating and cooling. The reason you are having so much trouble with this is that you are refusing to let the mathematics do the bookkeeping for you and, instead, you are trying to "wild ass" it ( I.e., refusing to be mathematically disciplined).

 

[chetzread]

This equation is applicable to both heating and cooling. The reason you are having so much trouble with this is that you are refusing to let the mathematics do the bookkeeping for you and, instead, you are trying to "wild ass" it ( I.e., refusing to be mathematically disciplined).

I understood the cooling part, I'm confused about the heating part, which part of my idea is incorrect for the heating part in post#88 ?