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A wooden wheel of radius $R$ is made of two semicircular part (see figure). The two parts are held together by a ring made of a metal strip of cross section area $S$ and length $L$. $L$ is slighly less than $2\pi R$. To fit the ring on the wheel, it is heated so that its temperature rises by $\Delta T$ and it just steps over the wheel.As it cools down to surronding temperature, it presses the semicircular parts together. If the coefficint of linear expansion of the metal is $\alpha$, and its young's modulus is $Y$, the force that one part of wheel applies on the other part is
$2SY\alpha \Delta T$
$2\pi SY\Delta T$
$SY\alpha \Delta T$
$\pi SY\alpha \Delta T$
Solution
$Y = \frac{{F/S}}{{\Delta L/L}} \Rightarrow \Delta L = \frac{{FL}}{{SY}}$
$\therefore L\alpha \Delta T = \frac{{FL}}{{SY}}$ $\left[ {\Delta L = L\alpha \Delta T} \right]$
$\therefore F = SY\alpha \Delta T$
$\therefore $ The ring is pressing the wheel from both sides,
$\therefore {F_{net}} = 2F = 2YS\alpha \Delta T$
