Change in length due to temperature

[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?

 

[chetzread]

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

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

tension , am i right ?

 

[Chestermiller]

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 your arithmetic in post 47 is correct, but your final equation is incorrect, and inconsistent with your previous arithmetic.

 

[chetzread]

Then your arithmetic in post 47 is correct, but your final equation is incorrect, and inconsistent with your previous arithmetic.

why not ? i still couldn't get it

 

[Chestermiller]

why not ? i still couldn't get it

You can't have it both ways: $$\sigma_s=E_s(\epsilon_s-\alpha_s \Delta T) $$and$$\sigma_s=E_s(-\epsilon_s+\alpha_s \Delta T) $$ because the two equations produce opposite results for $\sigma_s$. Try substituting $\epsilon_s=-3$ and $\alpha_s \Delta T=-6$ into the equation that you wrote, and tell me what you get.

 

[haruspex]

tension , am i right ?

Yes.

 

[chetzread]

Yes.

so ?

 

[chetzread]

You can't have it both ways: $$\sigma_s=E_s(\epsilon_s-\alpha_s \Delta T) $$and$$\sigma_s=E_s(-\epsilon_s+\alpha_s \Delta T) $$ because the two equations produce opposite results for $\sigma_s$. Try substituting $\epsilon_s=-3$ and $\alpha_s \Delta T=-6$ into the equation that you wrote, and tell me what you get.

But , in the figure in post # 1 , we could see that $$\alpha_s\Delta T$$ > ϵs , this doesn't match with the equation that given by you

 

[Chestermiller]

But , in the figure in post # 1 , we could see that $$\alpha_s\Delta T$$ > ϵs , this doesn't match with the equation that given by you

How can you see that from the figure? It doesn't show these two quantities separately. Besides, which of the two equations gives tension for $\sigma_s$, yours or mine? You yourself said that it has to be in tension.

 

[chetzread]

How can you see that from the figure? It doesn't show these two quantities separately. Besides, which of the two equations gives tension for $\sigma_s$, yours or mine? You yourself said that it has to be in tension.

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

 

[chetzread]

Yes.

what are you trying to say ?