How does thermal stress affect the length of a constrained rod?

[Physicsnb]

Homework Statement

In a lecture of thermal stress my professor defined thermal strain as ' L α ΔT/ L = α ΔT ' [since strain= ΔL/L and ΔL after application of heat is L α ΔT].
He said that on a constrained body that's not allowed to expand , a force is applied on the body .
If the constraints were removed then the new length would be equal to ' L+ L α ΔT '
However while deriving the formula for thermal strain he used the length 'L' as original length .
My question is : "Shouldn't L(new) be used to derive the equation instead of length L , since the rod is compressed from this length by the walls/constraints not length L. "

Relevant Equations

ΔL= L α ΔT
ΔL = Change in length of the body
L= Original Length of the body
α= Thermal Coefficient og linear expansion
ΔT=Change in temperature
New length after change in temp, L(new) = L+L α ΔT

My attempt at deriving the equation:
Let a temperature a rod of length L constrained by walls on both ends be changed by ΔT.
So the change in lenghth of rod under no constraints = L α ΔT
Length of rod after change in temperature without walls =L+L α ΔT= L(1+L α ΔT)
However with constraints the rod is forced to length L
Then,
Thermal strain = ΔL/L(1+L α ΔT)
Thermal strain= L α ΔT/L(1+L α ΔT) = α ΔT/(1+L α ΔT)

 

[erobz]

Homework Statement: In a lecture of thermal stress my professor defined thermal strain as ' L α ΔT/ L = α ΔT ' [since strain= ΔL/L and ΔL after application of heat is L α ΔT].
He said that on a constrained body that's not allowed to expand , a force is applied on the body .
If the constraints were removed then the new length would be equal to ' L+ L α ΔT '
However while deriving the formula for thermal strain he used the length 'L' as original length .
My question is : "Shouldn't L(new) be used to derive the equation instead of length L , since the rod is compressed from this length by the walls/constraints not length L. "
Relevant Equations: ΔL= L α ΔT
ΔL = Change in length of the body
L= Original Length of the body
α= Thermal Coefficient og linear expansion
ΔT=Change in temperature
New length after change in temp, L(new) = L+L α ΔT

My attempt at deriving the equation:
Let a temperature a rod of length L constrained by walls on both ends be changed by ΔT.
So the change in lenghth of rod under no constraints = L α ΔT
Length of rod after change in temperature without walls =L+L α ΔT= L(1+L α ΔT)
However with constraints the rod is forced to length L
Then,
Thermal strain = ΔL/L(1+L α ΔT)
Thermal strain= L α ΔT/L(1+L α ΔT) = α ΔT/(1+L α ΔT)

It's a first order approximation to the differential equation (1) about $L = L_o$:

$$ \frac{dL}{dT} = \alpha L \tag{1}$$

So we say that since $L \gg \Delta L$

$$ L \approx L_o + \left. \frac{dL}{dT}\right|_{L_o} \Delta T $$

$$ L \approx L_o + \alpha L_o \Delta T = L_o \left( 1 + \alpha \Delta T \right) $$

 

[TSny]

So the change in lenghth of rod under no constraints = L α ΔT
Length of rod after change in temperature without walls =L+L α ΔT= L(1+L α ΔT)

At the far right of the second line above, the $L$ inside the parentheses should not be there.

However with constraints the rod is forced to length L
Then,
Thermal strain = ΔL/L(1+L α ΔT)
Thermal strain= L α ΔT/L(1+L α ΔT) = α ΔT/(1+L α ΔT)

OK, except for the $L$ inside the parentheses. So, your derivation gives for the thermal strain $$\text{Thermal strain} = \alpha \Delta T/(1+\alpha \Delta T)$$
Generally, $\alpha \Delta T$ is small compared to 1. In that case, the $\alpha \Delta T$ can be neglected in the denominator. So, to a good approximation $$\text{Thermal strain} = \alpha \Delta T$$ which is the same as what your professor gets. But, I think it's good to think about it the way you did.

 

[Physicsnb]

It's a first order approximation to the differential equation (1) about $L = L_o$:

$$ \frac{dL}{dT} = \alpha L \tag{1}$$

So we say that since $L \gg \Delta L$

$$ L \approx L_o + \left. \frac{dL}{dT}\right|_{L_o} \Delta T $$

$$ L \approx L_o + \alpha L_o \Delta T = L_o \left( 1 + \alpha \Delta T \right) $$

Whats L_o? Also , my question was that why is strain =α Δ T .
Also can you please elaborate your derivation a bit more (specifically the second step)

 

[Physicsnb]

At the far right of the second line above, the $L$ inside the parentheses should not be there.OK, except for the $L$ inside the parentheses. So, your derivation gives for the thermal strain $$\text{Thermal strain} = \alpha \Delta T/(1+\alpha \Delta T)$$
Generally, $\alpha \Delta T$ is small compared to 1. In that case, the $\alpha \Delta T$ can be neglected in the denominator. So, to a good approximation $$\text{Thermal strain} = \alpha \Delta T$$ which is the same as what your professor gets. But, I think it's good to think about it the way you did.

Thanks for your explanation .

 

[erobz]

Whats L_o? Also , my question was that why is strain =

α​

Δ T .

$L_o$ is the initial length of the body.

The strain is $\epsilon = \frac{ \Delta L}{ L_o}$

$$ \epsilon = \frac{L - L_o}{L_o} = \frac{L_o( 1 + \alpha \Delta T ) - L_o}{ L_o} = \alpha \Delta T $$

 

[Physicsnb]

$L_o$ is the initial length of the body.

The strain is $\epsilon = \frac{ \Delta L}{ L_o}$

$$ \epsilon = \frac{L - L_o}{L_o} = \frac{L_o( 1 + \alpha \Delta T ) - L_o}{ L_o} = \alpha \Delta T $$

So ,
What's L here? New length?

 

[erobz]

So ,
What's L here? New length?

It's the length of the freely expanded body when thermal equilibrium has been reached.

 

[Physicsnb]

$L_o$ is the initial length of the body.

The strain is ## \epsilon = \frac{ \Delta L}{ L_o}

It's a first order approximation to the differential equation (1) about $L = L_o$:

$$ \frac{dL}{dT} = \alpha L \tag{1}$$

So we say that since $L \gg \Delta L$

$$ L \approx L_o + \left. \frac{dL}{dT}\right|_{L_o} \Delta T $$

$$ L \approx L_o + \alpha L_o \Delta T = L_o \left( 1 + \alpha \Delta T \right) $$

Could you please elaborate how you got to second step?

##

It's a first order approximation to the differential equation (1) about $L = L_o$:

$$ \frac{dL}{dT} = \alpha L \tag{1}$$

So we say that since $L \gg \Delta L$

$$ L \approx L_o + \left. \frac{dL}{dT}\right|_{L_o} \Delta T $$

$$ L \approx L_o + \alpha L_o \Delta T = L_o \left( 1 + \alpha \Delta T \right) $$

Could you please explain how you got to the send equation?

 

[erobz]

Could you please explain how you got to the send equation?

It's a first order Taylor Expansion about the fixed length $L_o$. A Taylor Expansion is an approximation of a function about a given point over some interval.

That second step is the length function (which would be found by integrating (1) ) truncated to the linear terms of the expansion.

So the first term is what we expect with no change in temp. Its just $L_o$. Then we extend it by saying over some change in temp$\Delta T$, that $L$ approximately changes like the $\frac{dL}{dT}|_{L_o} \Delta T = \alpha L_o \Delta T$

 

[Chestermiller]

You are working with small (linearized) strain theory. At this level of approximation, $\alpha \Delta T$ is negligible compared to unity. So the thermal strain is $\alpha \Delta T$ and the superimposed compressive strain is $\sigma /E$, where E is Young's modulus and $\sigma$ is the compressive stress..