Question

Figure 18-59 shows a composite bar of length L=L₁+L₂ and consisting of two materials. One material has length L₁ and coefficient of linear expansion α₁; the other has length L₂ and coefficient of linear expansion α₂. (a) What is the coefficient of linear expansion for the composite bar? For a particular composite bar, Lis 52.4cm, material 1 is steel, and material 2 is brass. If$\mathit{\alpha }\mathbf{}\mathbf{=}\mathbf{1}\mathbf{.3}\mathbf{}\mathbf{\times }\mathbf{}{\mathbf{10}}^{\mathbf{-}\mathbf{5}}{\mathbf{\text{/}}}^{\mathbf{\text{0}}}\mathbf{\text{C}}$, what are the lengths (b) L₁ and (c) L₂?

Short answer

1. Coefficient of linear expansion α for composite bar, $\frac{{\alpha }_1{L}_1+{\alpha }_2{L}_2}{L}$.
2. Length, L₁ is 39.3cm.
3. Length, L₂ is 13.1cm.

Step-by-step solution

Stating the given data

1. Length of composite bar, L = 52.4cm or 0.524m
2. Coefficient of linear expansion for composite bar, $\alpha =1.3\times {10}^{\text{- 5}}{\text{/}}^{\text{0}}\text{C}$
3. Coefficient of linear expansion of steel, ${\text{α}}_{\text{1}}=\text{11}\times {\text{10}}^{\text{-6}}{\text{/}}^{\text{0}}\text{C}$
4. Coefficient of linear expansion of brass, ${\alpha }_2=19\times {10}^{-6}{\text{/}}^{\text{0}}\text{C}$.

Understanding the concept of thermal expansion

The propensity of matter to alter its form, area, volume, and density in reaction to a change in temperature is known as thermal expansion. As we see in the given problem, there are two different metals whose coefficients are known to us, and from the given length of the bar, we can find the lengths of the two metals in the strip. Also, we can derive the formula for the coefficient of linear expansion of composite bar.

Formula:

Linear expansion of the body due to thermal expansion, $\Delta L=\alpha L\Delta T$…(i)

Where $\alpha$is coefficient of thermal expansion, L is the length and is the change in temperature.

(a) Calculation of the coefficient of linear expansion

The bar is made of two metals of length L₁ & L₂ and their expansions are $\Delta L_1\&\Delta L_2$respectively.

So, the total expansion of the length using equation (i) is given as follows. Further, solving for the coefficient of the linear expansion, we have

$\begin{array}{c}\Delta L=\Delta L_1+\Delta L_2\\ \alpha L\Delta T={\alpha }_1{L}_1\Delta T+{\alpha }_2{L}_2\Delta T\\ \alpha =\frac{{\alpha }_1{L}_1+{\alpha }_2{L}_2}{L}\mathrm{.................................}\text{(ii})\end{array}$

Hence, the value of the coefficient of linear expansion is $\frac{{\alpha }_1{L}_1+{\alpha }_2{L}_2}{L}$.

(b) Calculation of the length L1

From the values of coefficient of linear expansion of steel and brass, using equation (ii), we can get the relation of lengths as

$$\begin{gathered} 13\times {10}^{-6}{\text{/}}^{\text{0}}\text{C}=\frac{\left(11\times {10}^{-6}{\text{/}}^{\text{0}}\text{C}\right)L_1+\left(19\times {10}^{-6}{\text{/}}^{\text{0}}\text{C}\right)L_2}{L_1+L_2} \\ 13L_1+13L_2=11L_1+19L_2 \\ 2L_1=6L_2 \\ {L}_1=3L_2..........................\left(\text{iii}\right) \end{gathered}$$

Since we know L₁+L₂=52.4cm,we can write the value of the length, L₂ as

role="math" localid="1663070205639" $\begin{array}{c}{L}_2=52.4-L_1\\ L_2=52.4-3L_2\text{(fromequation(iii))}\\ 4L_2=52.4\\ L_2=13.1\text{\hspace{0.17em}cm}\end{array}$

Substituting the value of length L₂ in equation (iii), we can get the length L₁ as

$$\begin{gathered} L_1=(3\times 13.1) \text{cm} \\ =39.3 \text{cm} \end{gathered}$$

Hence, the value of the required length is 39.3cm.

(c) Calculation of the length L2

From the calculations of part (b), we get the value of the length L₂ as 13.1cm.