Question

Two metal strips, one of brass and other of steel are of equal length $1.5\, m$ and equal thickness $0.25\, cm$. These are put together riveted and clamped at one end at $0^{\circ} C$ ae shown in fig. If the temperature rises to $50^{\circ} C$, calculate the radius of curvature (in meter) of the combined strip. $\alpha$ for brass is $19 \times 10^{-6}/{ }^{\circ} C , \alpha$ for steel $=11 \times 10^{-6} /{}^{\circ} C$.

Answer 1

When temperature rises from $50^{\circ} C$ to $100^{\circ} C$ then the two strips expand, the expansions being different due to different coefficients of expansions ( $\alpha$ Brass $>\alpha$ Steel). This causes bimetallic strip as a whole to bend into arc.
Let $R$ be the mean radius of curvature
The lengths of two strips at temperature of $50^{\circ} C$ are
$L _{ A }= L _{0}\left(1+\alpha_{ A } \Delta T \right)$ ...(1)
$\left\{\Delta T =50^{\circ}\right\}$
$L _{ B }= L _{0}\left(1+\alpha_{ B } \Delta T \right)$ ...(2)
If $R_{A}$ and $R_{B}$ are radii of curvature of two strips then $L _{ A }= R _{ A } \theta, L _{ B }= R _{ B } \theta$ ... (3)
Where $\theta$ is common angle in radians intercepted by two strips.
$\therefore L _{ A }- L _{ B }=\theta\left( R _{ A }- R _{ B }\right)$
or $\theta=\frac{L_{A}-L_{B}}{R_{A}-R_{B}}=\frac{L_{0}\left(\alpha_{A}-\alpha_{B}\right) \Delta T}{R_{A}-R_{B}}$
But $R_{A}-R_{B}=d(d=$ thickness of each strip $)$
$\theta=\frac{L_{0}\left(\alpha_{A}-\alpha_{B}\right) \Delta T}{d}$ ...(4)
Adding (1) and (2),
$L _{ A }+ L _{ B }=2 L _{0}+ L _{0}\left(\alpha_{ A }+\alpha_{ B }\right) \Delta T$ ...(5)
From (3) and (5),
$\theta\left( R _{ A }+ R _{ B }\right)=2 L _{0}+ L _{0}\left(\alpha_{ A }+\alpha_{ B }\right) \Delta T$
$R _{ A }+ R _{ B }=\frac{2 L _{0}+ L _{0}\left(\alpha_{ A }+\alpha_{ B }\right) \Delta T }{\theta}$
Mean radius $=\frac{ R _{ A }+ R _{ B }}{2}$
$=\frac{2 L _{0}+ L _{0}\left(\alpha_{ A }+\alpha_{ B }\right) \Delta T }{2 \theta}$
Putting value of $\theta$ from (4) we get
$R =\frac{2 L _{0}+ L _{0}\left(\alpha_{ A }+\alpha_{ B }\right) \Delta T }{\frac{2 L _{0}\left(\alpha_{ A }-\alpha_{ B }\right) \Delta T }{ d }}$
$=\frac{\left[2+\left(\alpha_{A}+\alpha_{B}\right) \Delta T\right] d}{2\left(\alpha_{A}-\alpha_{B}\right) \Delta T}$
$=\frac{\left[2+\left(19 \times 10^{-6}+11 \times 10^{-6}\right) \times 50\right]\left(0.25 \times 10^{-2}\right)}{2\left(19 \times 10^{-6}-11 \times 10^{-6}\right) \times(50-0)}$
$=\frac{(2.0015) \times 0.25 \times 10^{-2}}{16 \times 10^{-6} \times 50}$
$=6.25\, m$