# Two straight metallic stripes each of thickness t and length l are riveted together. Their...

## Question:

Two straight metallic stripes each of thickness {eq}t {/eq} and length {eq}l {/eq} are riveted together. Their coefficients of linear expansion are {eq}\alpha_1 {/eq} and {eq}\alpha_2 {/eq}. If they are heated through a temperature {eq}\Delta t {/eq}, the bimetallic stripe will bend to form an arc of radius:

a. {eq}\frac{t}{(\alpha_1+\alpha_2)\Delta t} {/eq}

b. {eq}\frac{t}{\Delta t(\alpha_1+\alpha_2)} {/eq}

c. {eq}\frac{t(2+(\alpha_1+\alpha_2)\Delta t)}{2(\alpha_1-\alpha_2)\Delta t} {/eq}

d. {eq}t(\alpha_1-\alpha_2)\Delta t {/eq}

## Thermal Expansion:

In the context of heating, due to a change in temperature, the expansion occurs in the material. By the extension, the length, volume, or area of a material increases. This property of change in length is known as the linear thermal expansion. The {eq}\alpha {/eq} denotes the linear thermal expansion coefficient.

Given data

• The thickness of Rod (1) and Rod (2) is {eq}t {/eq}.
• The length of Rod (1) and Rod (2) is {eq}l {/eq}
• The thermal expansion of Rod (1) is {eq}\alpha = {\alpha _1} {/eq}.
• The thermal expansion of Rod (2) is {eq}\alpha = {\alpha _2} {/eq}.
• The rods heated to temperature {eq}\Delta T {/eq}.

On heating at {eq}\Delta T {/eq} and if, {eq}{\alpha _2} > {\alpha _1} {/eq}, then both the rods converted into an arc.

For rod (1) change in length can be calculated as,

{eq}d{l_1} = l\left( {1 + {\alpha _1} \times \Delta T} \right) {/eq}

For rod (2) change in length can be calculated as,

{eq}d{l_2} = l\left( {1 + {\alpha _2} \times \Delta T} \right) {/eq}

The change in radius of rod (1) can be calculated as,

{eq}d{r_1} = r - \dfrac{t}{2} {/eq}

Here, r is radius of arc.

The change in radius of rod (2) can be calculated as,

{eq}d{r_1} = r + \dfrac{t}{2} {/eq}

By a sector formula angle of an arc of rod (1) can be calculated as,

{eq}\theta = \dfrac{{d{l_1}}}{{d{r_1}}} {/eq}

Substitute the known values.

{eq}\theta = \dfrac{{l\left( {1 + {\alpha _1} \times \Delta T} \right)}}{{r - \dfrac{t}{2}}}..(i) {/eq}

By a sector formula angle of an arc of rod (2) can be calculated as,

{eq}\theta = \dfrac{{d{l_2}}}{{d{r_2}}} {/eq}

Substitute the known values.

{eq}\theta = \dfrac{{l\left( {1 + {\alpha _2} \times \Delta T} \right)}}{{r + \dfrac{t}{2}}}..(ii) {/eq}

On comparing equation (i) and (ii),

{eq}\begin{align*} \dfrac{{l\left( {1 + {\alpha _1} \times \Delta T} \right)}}{{r - \dfrac{t}{2}}} &= \dfrac{{l\left( {1 + {\alpha _2} \times \Delta T} \right)}}{{r + \dfrac{t}{2}}}\\ r + \dfrac{t}{2}\left( {l\left( {1 + {\alpha _1} \times \Delta T} \right)} \right) &= r - \dfrac{t}{2}\left( {l\left( {1 + {\alpha _2} \times \Delta T} \right)} \right)\\ rl + r \times l \times {\alpha _1} \times \Delta T + \dfrac{t}{2}l + \dfrac{t}{2}l \times {\alpha _1} \times \Delta T &= rl + r \times l \times {\alpha _2} \times \Delta T - \dfrac{t}{2}l - \dfrac{t}{2}l \times {\alpha _2} \times \Delta T\\ rl({\alpha _1} - {\alpha _2})\Delta T &= - 2 \times \dfrac{t}{2}l - \dfrac{t}{2}l \times \left( {{\alpha _2} + {\alpha _1}} \right) \times \Delta T\\ r({\alpha _1} - {\alpha _2})\Delta T &= - t - \dfrac{t}{2} \times \left( {{\alpha _2} + {\alpha _1}} \right) \times \Delta T\\ r({\alpha _1} - {\alpha _2})\Delta T &= \dfrac{{ - 2t - t\left( {{\alpha _2} + {\alpha _1}} \right) \times \Delta T}}{2}\\ r &= \dfrac{{t\left( {2 + \left( {{\alpha _2} + {\alpha _1}} \right) \times \Delta T} \right)}}{{2({\alpha _1} - {\alpha _2})\Delta T}} \end{align*} {/eq}

Therefore, the radius of arc is {eq}\dfrac{{t\left( {2 + \left( {{\alpha _2} + {\alpha _1}} \right) \times \Delta T} \right)}}{{2({\alpha _1} - {\alpha _2})\Delta T}} {/eq}. Thus, the option (c) is correct answer.