Question

A steel wire of diameter 1 mm can support a tension of 0.2 kN. A steel cable to support a tension of 20 kN should have a diameter of what order of magnitude?

Answer 1

The young modulus formula is given by

$\gamma=\frac{\textsf{Longitudinal stress}}{\textsf{Longitudinal strain}}\hspace{1cm}(1)$

Where $\gamma$ is the young modulus

$\begin{aligned}\textsf{Longitudinal stress}=\frac{F}{A}\hspace{1cm}&(2)\\\textsf{Longitudinal strain}=\frac{∆L}{L}\hspace{0.9cm}&(3)\end{aligned}$ Where $F,A,∆L\textsf{ and }L$ represent the Force, Area, change in the length and the initial length respectively.

Substituting (2) and (3) in (1),

$\gamma=\frac{\frac{F}{A}}{\frac{∆L}{L}}\\ \gamma=\frac{FL}{∆LA}\\ A=\frac{FL}{\gamma∆L}$

Let $A_1$ and $A_2$ represent the cross sectional areas of the steel wire and steel cable respectively.

Let $F_1$ and $F_2$ represent the forces on the steel wire and steel cable respectively.

$A_1=\frac{F_1L}{\gamma\Delta L}\hspace{1cm}(4)\\ A_2=\frac{F_2L}{\gamma\Delta L}\hspace{1cm}(5)$

Dividing (5) by (4)

$\frac{A_2}{A_1}=\frac{\frac{F_2L}{\gamma\Delta L}}{\frac{F_1L}{\gamma\Delta L}}\\ \frac{A_2}{A_1}=\frac{F_2}{F_1}\hspace{1.3cm}(6)\\$

$A_1=\frac{\Pi {D_1}²}{4}\hspace{1cm}(7)\\ A_2=\frac{\Pi{D_2}²}{4}\hspace{1cm}(8)$

Where $D_1$ and $D_2$ represent the diameters of the steel cable and steel wire respectively.

Substituting (7) and (8) in (6),

$\frac{\frac{\Pi{D_2}²}{4}}{\frac{\Pi {D_1}²}{4}}=\frac{F_2}{F_1}$

$\frac{{D_2}²}{{D_1}²}=\frac{F_2}{F_1}\\ {D_2}²={D_1}²×\frac{F_2}{F_1}\\ {D_2}²=(1×10^{-3}m)²×(\frac{20kN}{0.2kN})\\ {D_2}²=(10^{-6}m²)×(100)\\ {D_2}²=10^{-4}m²\\ D_2=\sqrt{10^{-4}m²}\\ D_2=10^{-2}m\\ D_2=10mm$