Change in length due to temperature

[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?

 

[Chestermiller]

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

Are you familiar with Newton's 3rd law and the concept of action-reaction pairs?

 

[haruspex]

δst is defined as the displacement that you see.

As I read it, $\delta_A$ is the observed displacement. $\delta_{st}$ is the difference between that and the displacement that would have occurred if not attachd to the crossbeam.

 

[Chestermiller]

As I read it, $\delta_A$ is the observed displacement. $\delta_{st}$ is the difference between that and the displacement that would have occurred if not attachd to the crossbeam.

Oops. It's been along time since this thread was active. I guess I forgot. Sorry about that.

 

[chetzread]

δst

What causes ∂st ?
It's the force Pst , right ?

 

[haruspex]

What causes ∂st ?
It's the force Pst , right ?

Yes.

 

[chetzread]

Yes.

so , the Pst point downwards at the point A , causing the steel rod can't contract by del (Tst) , but only contract by del (A) ?

 

[haruspex]

so , the Pst point downwards at the point A , causing the steel rod can't contract by del (Tst) , but only contract by del (A) ?

Yes.

 

[chetzread]

Yes.

sorry , i made a typo in my previous post ...
Just to verify my concept , do you mean the Pst cause the rigid bar ABC to move downwrads at A , so the contraction that we could observe is ∂A ( ∂(Tst) - ∂st = ∂A ) ?
This is due the original contraction with no aluminium rod constraint is ∂(Tst) ( cause end of bar ABC to go up at A ) , but the Pst cause the steel rod to move downwards at A , so the contraction that we could observe is ∂A ( ∂(Tst) - ∂st = ∂A ) only ?

 

[chetzread]

Yes.

can you verify my concept ?

 

[haruspex]

sorry , i made a typo in my previous post ...
Just to verify my concept , do you mean the Pst cause the rigid bar ABC to move downwrads at A , so the contraction that we could observe is ∂A ( ∂(Tst) - ∂st = ∂A ) ?
This is due the original contraction with no aluminium rod constraint is ∂(Tst) ( cause end of bar ABC to go up at A ) , but the Pst cause the steel rod to move downwards at A , so the contraction that we could observe is ∂A ( ∂(Tst) - ∂st = ∂A ) only ?

I find this a bit hard to answer because I cannot read the subscripts on the two P forces shown in the diagram at post #1. I assume P_st is the left hand one, and represents (in the diagram) the force exerted at the top of the steel rod by a restraint. Correspondingly, the bar at the bottom will exert an equal and opposite force. So yes, that downward force on the steel rod at A is what will reduce the contraction from ∂(Tst) to ∂A. But it is not quite right to say that P_st causes this. It is the pair of equal and opposite forces putting the rod under tension that causes it.

 

[chetzread]

But it is not quite right to say that Pst causes this. It is the pair of equal and opposite forces putting the rod under tension that causes it.

if so , then the downward and upwards direction forec will cancel out each other , right ? How to reduce the contraction from ∂(Tst) to ∂A ?

 

[chetzread]

I find this a bit hard to answer because I cannot read the subscripts on the two P forces shown in the diagram at post #1. I assume P_st is the left hand one, and represents (in the diagram) the force exerted at the top of the steel rod by a restraint. Correspondingly, the bar at the bottom will exert an equal and opposite force. So yes, that downward force on the steel rod at A is what will reduce the contraction from ∂(Tst) to ∂A. But it is not quite right to say that P_st causes this. It is the pair of equal and opposite forces putting the rod under tension that causes it.

Pst is actually caused by the aluminium rod ?

 

[haruspex]

if so , then the downward and upwards direction forec will cancel out each other , right ? How to reduce the contraction from ∂(Tst) to ∂A ?

They cancel out in the sense that there is no net force on the rod (otherwise, the rod would accelerate) but the combination produces a tension in the rod, thereby stretching it.

Pst is actually caused by the aluminium rod

That is the ultimate cause, but the connection to the bar at one end and to some restraint at the other end is the proximate cause.

 

[chetzread]

Step 2. Apply a stress σσ\sigma to the rod to arrive at the final constrained length

do u mean apply it to the opposite direction of the ϵ1=αΔT ?

 

[Chestermiller]

do u mean apply it to the opposite direction of the ϵ1=αΔT ?

Yes