Chapter 13, Problem 12P

To make a secure fit. rivets that are larger than the rivet hole are often usedand the rivet is cooled (usually in dry ice) before it is placed in the hole. A steel rivet 1.872 cm in diameter is to be placed in a hole 1.870 cm in diameter in a metal at 22⁰C. To what temperature must the rivet be cooled if it is to fit in the hole?

To determine

The temperature to which the rivet has to be cooled to fit in the rivet hole

Answer to Problem 12P

Solution:

The steel rivet has to be cooled to -67°C.

Explanation of Solution

Given:

Original diameter of the rivet, $l_r=1.872cm$

Diameter of the rivet hole, $l_h=1.870cm$

Initial temperature, $T_i=22°C$

The rivet is made of steel

Formula used:

$\Delta l=l\times \alpha \times \Delta T$, where α is the coefficient of linear expansion of a material.

Calculation:

In order to fit in a rivet hole, the rivet is cooled so that its diameter decreases to that of the rivet hole

The relation between change in length of a substance and change in temperature is given by,

$\Delta l=l_0\times \alpha \times \Delta T$---(i),

Where, ΔT is the change in temperature and α is the coefficient of linear expansion of a material.

For steel, $\alpha =12\times {10}^{-6}/°C$

$\Delta T=T_i-T_f$, where, T_f is the temperature to which the rivet has to be cooled

$\Delta l=l_r–l_h=1.872cm–1.870cm=0.002cm$

$l_0=l_r$

Substituting the known values in equation (i),

$0.002cm=\left(1.872cm\right)\times (12\times {10}^{-6}/°C)\times \left(22°C–T_f\right)$

or $\left(22°C–T_f\right)=$$\frac{0.002cm}{(1.872cm)x(12x{10}^{-6}/°C)}$

or $\left(22°C–T_f\right)=89°C$

or $T_f=22°C-89°C$

or $T_f=-67°C$