Question

A copper and a tungsten plate having a thickness $\delta$ = 2mm each are riveted together so that at $0^{\circ }C$ they form a flat bimetallic plate. Find the average radius of the curvature of this plate at $t$ = ${200}^{\circ }C$. The coefficient of linear expansion for copper and tungsten are ${\alpha }_{Cu}$ = $1.7\times {10}^{-5}/K$ and ${\alpha }_W$ = $0.4\times {10}^{-5}/K$.

• A. 0.769 m
• B. 1 m
• C. 0.582 m
• D. 1.243 m

Answer 1

The correct option is A

0.769 m

The average length of the copper plate at a temperature $T$ = ${200}^{\circ }C$ is $l_C$ = $l_0(1+\alpha CuT)$, where $l_0$ is the length of copper plate at $0^{\circ }C$. The length of the tungsten plate is $l_t$ = $l_0(1+{\alpha }_{W}T)$.

We shall assume that the edges of plates are not displaced during deformation and that an increase in the plate thickness due to heating can be neglected.

From figure we have:

$l_C$ = $\varphi (R+\frac{\delta }{2})$

$\Rightarrow l_t$ = $\varphi (R-\frac{\delta }{2})$.

Consequently, $\varphi (R+\frac{\delta }{2})$ = $l_0(1+{\alpha }_{Cu}T)$, - - - - - - (1)

$\varphi (R-\frac{\delta }{2})$ = $l_0(1+{\alpha }_{W}T)$. - - - - - - (2)

To eliminate the unknown quantities, $\varphi$ and $l_0$, we divide the equation (1) by (2), term wise:

$\Rightarrow \frac{(R+\frac{\delta }{2})}{(R-\frac{\delta }{2})}$ = $\frac{(1+{\alpha }_{cu}T)}{(1+{\alpha }_{W}T)}$

$\Rightarrow R$ = $\delta \frac{[2+({\alpha }_{cu}+{\alpha }_W\left)T\right]}{\left[2\right({\alpha }_{cu}-{\alpha }_W\left)T\right]}$

$\Rightarrow R$ = $\frac{\delta }{({\alpha }_{cu}-{\alpha }_W)T}$

Neglecting $({\alpha }_{Cu}+{\alpha }_W)$ in the numerator, and substituting the values in the above relation,

we get: $R$ = 0.769 m.