Question

Two metal strips, one of brass and other of steel are of equal length $1.5m$ and equal thickness $0.25cm$. 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

Answer: 6.25

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),
$L_A+L_B=2L_0+L_0\left({\alpha }_A+{\alpha }_B\right)\Delta T$ ...(5)
From (3) and (5),
$\theta \left(R_A+R_B\right)=2L_0+L_0\left({\alpha }_A+{\alpha }_B\right)\Delta T$
$R_A+R_B=\frac{2L_0+L_0\left({\alpha }_A+{\alpha }_B\right)\Delta T}{\theta }$
Mean radius $=\frac{R_A+R_B}{2}$
$=\frac{2L_0+L_0\left({\alpha }_A+{\alpha }_B\right)\Delta T}{2\theta }$
Putting value of $\theta$ from (4) we get
$R=\frac{2L_0+L_0\left({\alpha }_A+{\alpha }_B\right)\Delta T}{\frac{2L_0\left({\alpha }_A-{\alpha }_B\right)\Delta T}{d}}$
$=\frac{\left[2+\left({\alpha }_A+{\alpha }_B\right)\Delta T\right]d}{2\left({\alpha }_A-{\alpha }_B\right)\Delta T}$
$=\frac{\left[2+\left(19\times 10^{-6}+11\times 10^{-6}\right)\times 50\right]\left(0.25\times 10^{-2}\right)}{2\left(19\times 10^{-6}-11\times 10^{-6}\right)\times (50-0)}$
$=\frac{(2.0015)\times 0.25\times 10^{-2}}{16\times 10^{-6}\times 50}$
$=6.25m$