Question

A copper and a tungsten plate having thickness $\delta$ each are riveted together so that at $0°C$, they form a flat bimetallic plate. Find the average radius of the curvature of this plate at temperature $T$. The coefficient of linear expansion for copper and tungsten are ${\alpha }_{Cu}$ and ${\alpha }_W$.

• A. $\delta \frac{[1+({\alpha }_{Cu}+{\alpha }_W\left)T\right]}{\left[2\right({\alpha }_{Cu}-{\alpha }_W\left)T\right]}$
• B. $\delta \frac{[1+({\alpha }_{Cu}+{\alpha }_W\left)T\right]}{\left[\right({\alpha }_{Cu}-{\alpha }_W\left)T\right]}$
• C. $\delta \frac{[2+({\alpha }_{Cu}+{\alpha }_W\left)T\right]}{\left[\right({\alpha }_{Cu}-{\alpha }_W\left)T\right]}$
• D. $\delta \frac{[2+({\alpha }_{Cu}+{\alpha }_W\left)T\right]}{\left[2\right({\alpha }_{Cu}-{\alpha }_W\left)T\right]}$

Answer 1

The correct option is D $\delta \frac{[2+({\alpha }_{Cu}+{\alpha }_W\left)T\right]}{\left[2\right({\alpha }_{Cu}-{\alpha }_W\left)T\right]}$

The average length of copper plate at a Temperature $T$ is $l_{Cu}=l_o(1+{\alpha }_{Cu}T)$, where $l_o$ is the length of copper plate at $0^{\circ }C$. The length of the tungsten plate is $l_W=l_o(1+{\alpha }_{W}T)$



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 he neglected.
From figure, we have length of copper plate: $l_{Cu}=\varphi (R+\delta /2)$ and length of tungsten plate $l_W=\varphi (R-\delta /2)$
Consequently, $\begin{array}{}\varphi (R+\delta /2)=l_o(1+{\alpha }_{Cu}T)& \text{(i)}\end{array}$
$\begin{array}{}\varphi (R-\delta /2)=l_o(1+{\alpha }_{W}T)& \text{(ii)}\end{array}$
To eliminate the unknown quantities $\varphi$ and $l_o$, we divide the equation $\text{(i)}$ by $\text{(ii)}$ term wise:

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

$\Rightarrow R=\delta \frac{[2+({\alpha }_{Cu}+{\alpha }_W\left)T\right]}{\left[2\right({\alpha }_{Cu}-{\alpha }_W\left)T\right]}$ ...(use componendo - dividendo)