Question

A machinist bores a hole of diameter 1.35 cm in a steel plate that is at 25.0ºC. What is the cross-sectional area of the hole (a) at 25.0ºC and (b) when the temperature of the plate is increased to 175ºC? Assume that the coefficient of linear expansion remains constant over this temperature range.

Short answer

(a) The cross sectional area of the hole at 25.0ºC is $1.431\hspace{0.33em}c{m}^2$.

(b) The cross sectional area of the hole at increased temperature of 175.0ºC is $1.436\hspace{0.33em}c{m}^2$.

Step-by-step solution

Concept of Area Expansion in solids.

Area expansion is the expansion of the dimensions of an object with respect to temperature changes and can be expressed as,

$\mathbf{\Delta A}\mathbf{=}\mathbf{2}{\mathbf{\alpha A}}_{\mathbf{0}}\mathbf{\Delta T}$ ...(i)

Here, A is the expansion in area, A₀ is the unexpanded area, α is the coefficient of volume expansion and T is the absolute temperature.

 Determination of the diameter value of the rivet following the conditions mentioned.

For steel, the coefficient of area expansion,

$\alpha =1.2\times {10}^{-5}\hspace{0.33em}(C^o{\left)}^{\right(}-1)$

The unchanged original area is,

$$\begin{gathered} A_0={{\mathrm{\pi r}}_0}^2 \\ =\mathrm{\pi }{\left(\frac{1.350\mathrm{cm}}{2}\right)}^2 \\ =1.431{\mathrm{cm}}^2 \end{gathered}$$

Calculate the area change for the steel using equation (i),

$\mathbf{\Delta A}\mathbf{=}\mathbf{2}{\mathbf{\alpha A}}_{\mathbf{0}}\mathbf{\Delta T}$

Substitute all the values in the above equation,

$$\begin{gathered} ∆{\mathrm{A}}_{\mathrm{s}}=2\left(1.2\times {10}^{-5}{\left({\mathrm{C}}^0\right)}^{-1}\right)\left(1.431{\mathrm{cm}}^2\right)\left(175{\mathrm{C}}^0-25{\mathrm{C}}^0\right) \\ =5.15\times {10}^{-3}{\mathrm{cm}}^2 \end{gathered}$$

Thus, cross sectional area at 175.0ºC

$\begin{array}{l}A=A_0+∆A\\ =1.436{\mathrm{cm}}^2\end{array}$