# A titanium shaft has a starting diameter = 19.50 mm. It is to be inserted into a hole in an...

## Question:

A titanium shaft has a starting diameter = 19.50 mm. It is to be inserted into a hole in an expansion fit assembly operation. To be readily inserted, the shaft must be reduced in diameter by cooling. Determine the temperature to which the shaft must be reduced from room temperature (20 degrees C) in order to reduce its diameter to 19.48 mm. State any assumptions.

## Coefficient of thermal expansion:

Each material will alter its configuration when it is subjected to varying temperatures. The material changes the shape of the change in the energy of the molecules. So the ratio of strain to the temperature change is known as the coefficient of thermal expansion.

Given data

• Diameter of the shaft is {eq}{d_i} = 19.50\;{\rm{mm}} {/eq}
• Final diameter of the shaft is {eq}{d_f} = 19.48\;{\rm{mm}} {/eq}
• Initial temperature of the shaft is {eq}{T_i} = 20^\circ {\rm{C}} {/eq}
• Thermal coefficient of expansion of the titanium is {eq}\alpha = 8.5 \times {10^{ - 6}}\;{\rm{/^\circ C}} {/eq}

The expression for the change in diameter of the shaft is

{eq}\Delta D = \alpha \Delta T{d_i} {/eq}

Substituting the values,

{eq}\begin{align*} \left( {19.48 - 19.50} \right) &= 8.5 \times {10^{ - 6}} \times \left( {{T_f} - 20^\circ } \right)\left( {19.50} \right)\\ {T_f} &= - 120.66 + 20\\ {T_f} &= - 100.66^\circ {\rm{C}} \end{align*} {/eq}

Therefore, temperature at which rod must be cooled is {eq}- 100.66^\circ {\rm{C}}. {/eq} 