Thermal expansion of bimetallic strip

[PeppaPig]

Homework Statement

[/B]
A bimetallic strip is made of metal which has coefficient of thermal expansion is equal to α₁ and the other's is equal to α₂ at the temperature of T₀. The temperature is increased to T₀ + ΔT (ΔT > 0). The strip curves as shown in the figure. If both strip have the same width of d. Find r₁ measured from the center of the curve to the center of the strip as shown in the figure

Homework Equations

1) radian is (Length of the curve)/(Radius)
2) Using the following estimation
(1 + x)^-1 ≈ 1-x and (1 + ax)(1+bx) ≈ 1+(a+b)x for x << 1

The Attempt at a Solution

α₁ strip expansion is L₀(1 + α₁ΔT) for the length
and d(1 + α₁ΔT) for the width
α₂ strip expansion is L₀(1 + α₂ΔT) for the length
and d(1 + α₂ΔT) for the width

Multiplying radian with radius, the result should be the curve length

r₁θ = L₀(1 + α₁ΔT)
r₂θ = L₀(1 + α₂ΔT)

r₂ = r₁ + d(2 + α₁ΔT + α₂ΔT)/2

Is that correct?

 

[Chestermiller]

What is the ratio of r2 to r1? What is the difference between r2 and r1?

 

[PeppaPig]

What is the ratio of r2 to r1? What is the difference between r2 and r1?

The difference is d(2 + α₁ΔT + α₁ΔT)/2 because of the width expansion.

Is that correct?

 

[Chestermiller]

The difference is d(2 + α₁ΔT + α₁ΔT)/2 because of the width expansion.

Is that correct?

Neglecting the change in thickness, it is just d.

 

[PeppaPig]

Neglecting the change in thickness, it is just d.

Then the difference should be d. And the ratio should be (r₁ + d)/r₁ or (1 + α₂ΔT)/(1+α₁ΔT)

Is that correct?

 

[Chestermiller]

Yes. To terms of 1st order in $\alpha \Delta T$, this gives $$\frac{d}{r_1}=(\alpha_2-\alpha_1)\Delta T$$

 

[PeppaPig]

Yes. To terms of 1st order in $\alpha \Delta T$, this gives $$\frac{d}{r_1}=(\alpha_2-\alpha_1)\Delta T$$

Thank you.
Which mean r₁ is equal to d/((α₂ - α₁)ΔT)

What about the estimation? Or do I have to estimate them before the calculation?

 

[Chestermiller]

Thank you.
Which mean r₁ is equal to d/((α₂ - α₁)ΔT)

What about the estimation? Or do I have to estimate them before the calculation?

I don't know what you mean by "the estimation." Are you talking about retaining only terms of first order in $\alpha \Delta T$?

 

[PeppaPig]

2) Using the following estimation
(1 + x)^-1 ≈ 1-x and (1 + ax)(1+bx) ≈ 1+(a+b)x for x << 1

 

[Chestermiller]

Ok. That’s what we did with the equation in post #5.

 

[PeppaPig]

Ok. That’s what we did with the equation in post #5.

Then it should be 1/(1 - α₂ΔT)(1 + α₁ΔT) and then 1/(1+(α₁ - α₂)ΔT)

Is that correct?

 

[Chestermiller]

Then it should be 1/(1 - α₂ΔT)(1 + α₁ΔT) and then 1/(1+(α₁ - α₂)ΔT)

Is that correct?

Yes. So, from that, what is d/r1?

 

[PeppaPig]

Yes. So, from that, what is d/r1?

$\frac{d}{r_1} = \frac{-(\alpha_1 - \alpha_2) \Delta T}{1 + (\alpha_1 - \alpha_2) \Delta T}$

Then $r_1 = (\frac{1}{(\alpha_2 - \alpha_1) \Delta T} - 1) d$

Is that correct?

 

[Chestermiller]

$\frac{d}{r_1} = \frac{-(\alpha_1 - \alpha_2) \Delta T}{1 + (\alpha_1 - \alpha_2) \Delta T}$

Then $r_1 = (\frac{1}{(\alpha_2 - \alpha_1) \Delta T} - 1) d$

Is that correct?

No. Check your math.

 

[PeppaPig]

No. Check your math.

From these equations
$r_1 \theta = L_0 (1 + \alpha_1 \Delta T)$
$r_2 \theta = L_0 (1 + \alpha_2 \Delta T)$

Then substitute $r_2$ with $r_1 + d$

$\frac{r_1 + d}{r_1} = \frac{1}{1+(\alpha_1 - \alpha_2) \Delta T}$

Where am I going wrong?

 

[Chestermiller]

From these equations
$r_1 \theta = L_0 (1 + \alpha_1 \Delta T)$
$r_2 \theta = L_0 (1 + \alpha_2 \Delta T)$

Then substitute $r_2$ with $r_1 + d$

$\frac{r_1 + d}{r_1} = \frac{1}{1+(\alpha_1 - \alpha_2) \Delta T}$

Where am I going wrong?

$$\frac{r_1 + d}{r_1} = 1+(\alpha_2 - \alpha_1) \Delta T$$
$$1+\frac{d}{r_1}= 1+(\alpha_2 - \alpha_1) \Delta T$$

 

[PeppaPig]

$$\frac{r_1 + d}{r_1} = 1+(\alpha_2 - \alpha_1) \Delta T$$
$$1+\frac{d}{r_1}= 1+(\alpha_2 - \alpha_1) \Delta T$$

Oh! I should estimate the equation into

$1 + (\alpha_2 - \alpha_1) \Delta T$

$r_1 = \frac{d}{(\alpha_2 - \alpha_1) \Delta T}$

Is that correct?

 

[Chestermiller]

Oh! I should estimate the equation into

$1 + (\alpha_2 - \alpha_1) \Delta T$

$r_1 = \frac{d}{(\alpha_2 - \alpha_1) \Delta T}$

Is that correct?

Yes. It all involves knowing how to make the correct mathematical approximations.

 

[PeppaPig]

Yes. It all involves knowing how to make the correct mathematical approximations.

Thank you very much and Merry Christmas.

 

[Chestermiller]

Thank you very much and Merry Christmas.

Back at ya.