Question

(II) To make a secure fit, rivets that are larger than the rivet hole are often used and the rivet is cooled (usually in dry ice) before it is placed in the hole. A steel rivet 1.872cm in diameter is to be placed in a hole 1.870cm in diameter in a metal at 22°C. To what temperature must the rivet be cooled if it is to fit in the hole

Answer 1

Given:

Rivet diameter, $d_{r}$ = 1.872 cm

Hole diameter, $d_{h}$ = 1.870 cm

Temperature,  $T_{2}$ = 22 °C

Formula Used:

$\alpha = \frac{\Delta d}{d\times \Delta T}$

where,

$\alpha$ = coefficient of linear expansion

$\Delta d$ = change in diameter =   $d_{h} - d_{r}$

$\Delta T}$ = change in temperature =  $T_{2} - T_{1}$

Solution:

we know that coefficient of linear expansion of steel,  $\alpha$ = $12\times 10^{-6}/^{\circ}C$

Using the above formula :

$\alpha = \frac{\Delta d}{d\times \Delta T}$

$12\times 10^{-6}/^{\circ}C$ = \frac{1.870 - 1.872}{1.872\times \T_{2} - T_{1}}[/tex]

$T_{2} - 20/^{\circ}C$ =   \frac{1.870 - 1.872}{12\times 10^{-6}}}[/tex]

$T_{2} = -67.03/^{\circ}C$

Therefore, the rivet must be cooled to  $-67.03/^{\circ}C$