Change in length due to temperature

[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 ?

 

[Chestermiller]

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 ?

Heating should be even easier to analyze than cooling. If you heat it, and you allow it to expand unconstrained (no stress), the strain must be positive and equal to $\alpha \Delta T$. If you constrain it such that the strain is $0<\epsilon<\alpha \Delta t$, the strain will still be positive, but the rod will be under compression, and the stress will be negative. If apply a strain over and above the amount that it would stretch unconstrained, $\epsilon>\alpha \Delta T$, the rod will be under tension, and the stress will be positive. All this is contained in the equation we have been working with if you are sufficiently disciplined mathematically to work with the signs correctly.

 

[chetzread]

but the rod will be under compression, and the stress will be negative.

here's the main point,thanks,initially , i was assuming the stress must be positive all the times...

 

[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 stress in the steel rod is given by: $$\sigma_s=E_s(\epsilon_s-\alpha_s \Delta T)\tag{key}$$

can you show how to derive this formula ?
can i derive it as $$\sigma_s=-E_s(\epsilon_s+\alpha_s \Delta T)\$$

we just want to show that εs = αT , so that stress = 0 , right ?

 

[Chestermiller]

The stress in the steel rod is given by: $$\sigma_s=E_s(\epsilon_s-\alpha_s \Delta T)\tag{key}$$

can you show how to derive this formula ?
can i derive it as $$\sigma_s=-E_s(\epsilon_s+\alpha_s \Delta T)\$$

we just want to show that εs = αT , so that stress = 0 , right ?

If we have a rod to which we apply a temperature change while at the same time partially constraining it, we can model this as a two step process.

Step 1. Apply the temperature change to the rod without constraining it
Step 2. Apply a stress $\sigma$ to the rod to arrive at the final constrained length

Let $\epsilon_1$ represent the strain of the rod in Step 1 and let $\epsilon_2$ represent the strain in Step 2. Then:
$$\epsilon_1=\alpha \Delta T$$
$$\epsilon_2=\frac{\sigma}{E}$$
The total strain for the combined process is the sum of the strains for each of the two steps:
$$\epsilon=\epsilon_1+\epsilon_2=\alpha \Delta T+\frac{\sigma}{E}$$
What do you get if you solve this equation for $\sigma$ as a function of $\alpha \Delta T$ and $\epsilon$?

 

[chetzread]

If we have a rod to which we apply a temperature change while at the same time partially constraining it, we can model this as a two step process.

Step 1. Apply the temperature change to the rod without constraining it
Step 2. Apply a stress $\sigma$ to the rod to arrive at the final constrained length

Let $\epsilon_1$ represent the strain of the rod in Step 1 and let $\epsilon_2$ represent the strain in Step 2. Then:
$$\epsilon_1=\alpha \Delta T$$
$$\epsilon_2=\frac{\sigma}{E}$$
The total strain for the combined process is the sum of the strains for each of the two steps:
$$\epsilon=\epsilon_1+\epsilon_2=\alpha \Delta T+\frac{\sigma}{E}$$
What do you get if you solve this equation for $\sigma$ as a function of $\alpha \Delta T$ and $\epsilon$?

total strain = 0 , which means
i could write it as σs = -E α(T) , so ?

 

[Chestermiller]

total strain = 0 , which means
i could write it as σs = -E α(T) , so ?

So, the equation I derived applies to all values of the total strain, and not just to total strain = 0.

 

[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.

here's the full notes , δst is caused by force Pst( in upward) ?
if it' so , we will be able to 'see' the δst , am i right ?

 

[Chestermiller]

here's the full notes , δst is caused by force Pst( in upward) ?

It is caused by the force and the temperature change. The force on the rod is in the upward direction at the top of the rod and in the downward direction at the bottom of the rod. The force on the lever is in the upward direction at the connection with the steel rod.

if it' so , we will be able to 'see' the δst , am i right ?

δst is defined as the displacement that you see.

 

[chetzread]

downward direction at the bottom of the rod. The force on the lever is in the upward direction at the connection with the steel rod.

The force at point A is pointed upwards or downwrads?

 

[Chestermiller]

The force at point A is pointed upwards or downwrads?

Well, you tell me. What is the direction of the force that the rod applies to the lever at point A? What is the direction of the force that the lever applies to the rod at point A?

 

[chetzread]

Well, you tell me. What is the direction of the force that the rod applies to the lever at point A? What is the direction of the force that the lever applies to the rod at point A?

Upwards?
If it's upwards the change in length that we can see with naked eyes is del (st) + del (a), rather than del (a) only, right?

 

[Chestermiller]

Upwards?
If it's upwards the change in length that we can see with naked eyes is del (st) + del (a), rather than del (a) only, right?

I asked you two separate questions. What are your answers to each of these questions?

 

[chetzread]

Well, you tell me. What is the direction of the force that the rod applies to the lever at point A? What is the direction of the force that the lever applies to the rod at point A?

The force that the rod applies to lever is upwards, the force that lever applies to rod at A is downwards?

 

[Chestermiller]

The force that the rod applies to lever is upwards, the force that lever applies to rod at A is downwards?

It sounds like you're not sure. Please let me know when you are sure.

 

[chetzread]

It sounds like you're not sure. Please let me know when you are sure.

Huh? If the forces opposite to each other, they will cancel each other, right? If so, the force P(st) wouldn't exist, right?