Change in length due to temperature

[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

 

[chetzread]

Yes

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 $\epsilon$?

if it's opposite , it will become $$\epsilon=\epsilon_1+\epsilon_2=\alpha \Delta T-\frac{\sigma}{E}$$ , right ?
then it wil become
σs=Es(-ϵs+αsΔT) ?

is it correct ?

 

[Chestermiller]

if it's opposite , it will become $$\epsilon=\epsilon_1+\epsilon_2=\alpha \Delta T-\frac{\sigma}{E}$$ , right ?
then it wil become
σs=Es(-ϵs+αsΔT) ?

is it correct ?

NO. Do the algebra. The first term on the right hand side of my equation is the strain resulting from thermal expansion. The second term is the result of added tensile stress. These add up to the total strain.

 

[chetzread]

NO. Do the algebra. The first term on the right hand side of my equation is the strain resulting from thermal expansion. The second term is the result of added tensile stress. These add up to the total strain.

but , you agreed that in $$\epsilon=\epsilon_1+\epsilon_2=\alpha \Delta T+\frac{\sigma}{E}$$
the σ is applied to the opposite of αT ?
so, $$\epsilon=\epsilon_1+\epsilon_2=\alpha \Delta T-\frac{\sigma}{E}$$

so , σs=Es(-ϵs+αsΔT) ?

 

[Chestermiller]

but , you agreed that in $$\epsilon=\epsilon_1+\epsilon_2=\alpha \Delta T+\frac{\sigma}{E}$$
the σ is applied to the opposite of αT ?
so, $$\epsilon=\epsilon_1+\epsilon_2=\alpha \Delta T-\frac{\sigma}{E}$$

so , σs=Es(-ϵs+αsΔT) ?

That has nothing to do with the algebra. Everything automatically comes out to be the right sign if you let the math do the work for you. If you are not happy with what the algebra predicts, do it your own way, but you will get the wrong answer.

 

[chetzread]

That has nothing to do with the algebra. Everything automatically comes out to be the right sign if you let the math do the work for you. If you are not happy with what the algebra predicts, do it your own way, but you will get the wrong answer.

if the σ is applied in opposite direction , then the strain should be in negative value , right ? so , total strain =
ϵ=ϵ1+ϵ2=αΔT-σE
Why am i wrong ?

 

[Chestermiller]

if the σ is applied in opposite direction , then the strain should be in negative value , right ? so , total strain =
ϵ=ϵ1+ϵ2=αΔT-σE
Why am i wrong ?

What we are dealing with here is a mathematics issue, NOT A PHYSICS ISSUE.

Suppose you have a general equation $$A+B=C$$where A, B, and C are real numbers, both positive and negative. Suppose you have n combinations of numbers $((A_i,B_i,C_i),i=1,...,n)$ all of which satisfy the general equation. Suppose I told you that, for one specific combination (i = j), $B_j$ has the opposite sign of $A_j$ (for example (-7.5, +4.85, -2.65). According to your rationale, for a combination like this, the general equation should be changed to $$A-B=C$$ Would any of the combinations now satisfy this new equation?

 

[chetzread]

What we are dealing with here is a mathematics issue, NOT A PHYSICS ISSUE.

Suppose you have a general equation $$A+B=C$$where A, B, and C are real
numbers, both positive and negative. Suppose you have n combinations of numbers $((A_i,B_i,C_i),i=1,...,n)$ all of which satisfy the general equation. Suppose
I told you that, for one specific combination (i = j), $B_j$ has the opposite sign of $A_j$ (for example (-7.5, +4.85, -2.65). According to your rationale, for a
combination like this, the general equation
should be changed to $$A-B=C$$ Would any of the combinations now satisfy this new equation?

No, still A + B = C... but, in the case that we discussed so far, we have to take the sign into consideration, rite? That is
ϵ=ϵ1+ϵ2=αΔT-σE , right ?

 

[Chestermiller]

No, still A + B = C... but, in the case that we discussed so far, we have to take the sign into consideration, rite? That is
ϵ=ϵ1+ϵ2=αΔT-σE , right ?

It's exactly the same situation. The general equation (not just for this problem) is $$\epsilon=\alpha \Delta T+\frac{\sigma}{E}$$ If doesn't matter whether $\Delta T$ is positive or negative and it doesn't mater whether $\sigma$ is positive or negative.

 

[chetzread]

It's exactly the same situation. The general equation (not just for this problem) is $$\epsilon=\alpha \Delta T+\frac{\sigma}{E}$$ If doesn't matter whether $\Delta T$ is positive or negative and it doesn't mater whether $\sigma$ is
positive or negative.

OK, since we already know that the σ/ E is applied in opposite to αLT, so, the σ/ E should has negative sign, thus ϵ=ϵ1+ϵ2=αΔT-σE , right ?

 

[Chestermiller]

OK, since we already know that the σ/ E is applied in opposite to αLT, so, the σ/ E should has negative sign, thus ϵ=ϵ1+ϵ2=αΔT-σE , right ?

No. You're doing the same thing again. You need to go back and review algebra.

 

[Chestermiller]

OK, since we already know that the σ/ E is applied in opposite to αLT, so, the σ/ E should has negative sign, thus ϵ=ϵ1+ϵ2=αΔT-σE , right ?

What would this equation predict for the relationship between stress and strain if $\Delta T$ were zero?

Here is a math problem for you to using with the so-called "SUVAT" equation: $$v_x=v_{x0}+at$$
Suppose I have a body moving with an initial velocity at time t=0 of $v_{x0}=-10$ m/s (i.e., the initial velocity is in the negative x direction). At time t = 0, I apply a constant force to the body to give it an acceleration "a" in the positive x direction (i.e., opposite to the direction of the initial velocity). What acceleration value do I have to apply so that, at t = 4 seconds, the velocity of the body is $v_x=-2$ m/s?

 

[chetzread]

What would this equation predict for the relationship between stress and strain if $\Delta T$ were zero?

Here is a math problem for you to using with the so-called "SUVAT" equation: $$v_x=v_{x0}+at$$
Suppose I have a body moving with an initial velocity at time t=0 of $v_{x0}=-10$ m/s (i.e., the initial velocity is in the negative x direction). At time t = 0, I apply a constant force to the body to give it an acceleration "a" in the positive x direction (i.e., opposite to the direction of the initial velocity). What acceleration value do I have to apply so that, at t = 4 seconds, the velocity of the body is $v_x=-2$ m/s?

-2 = -10 +4(a) , a = 2(m/s^2)

 

[Chestermiller]

-2 = -10 +4(a) , a = 2(m/s^2)

But, according to the rationale you have been using on the thermal expansion problem, you first should have rewritten the SUVAT equation as:
$$v_x=v_{x0}-at$$because a is in the opposite direction of $v_{x0}$. Do you see the point I'm trying to make? You just can't arbitrarily change the sign of a term in an equation and expect to get the right answer.