Problem

A cylindrical steel rod $0.50 \, \mathrm{m}$ long and $1 \, \mathrm{cm}$ in radius is subjected to a tensile force of $1.0 \times 10^{4} \, \mathrm{N}$. Calculate:

1. The tensile stress.
2. The change in length ($\Delta L$) of the rod.

Data

• Length of the rod: $L = 0.50 \, \mathrm{m}$
• Radius: $r = 1.0 \, \mathrm{cm} = 1.0 \times 10^{-2} \, \mathrm{m}$
• Force applied: $F = 1.0 \times 10^{4} \, \mathrm{N}$
• Young’s modulus of steel: $Y = 20 \times 10^{10} \, \mathrm{Nm}^{-2}$

Prerequisite Concepts

1. Tensile Stress:

$$\text{Tensile Stress} = \frac{F}{A}$$

where $A = \pi r^2$ is the cross-sectional area of the rod.

2. Young’s Modulus:

$$Y = \frac{\text{Tensile Stress}}{\text{Tensile Strain}}$$

Rearranging for tensile strain:

$$\text{Tensile Strain} = \frac{\text{Tensile Stress}}{Y}$$

3. Tensile Strain and Change in Length:

$$\text{Tensile Strain} = \frac{\Delta L}{L} \quad \implies \quad \Delta L = \text{Tensile Strain} \times L$$

Solution

Step 1: Calculate Tensile Stress

The cross-sectional area of the rod is:

$$A = \pi r^2 = \pi (1.0 \times 10^{-2})^2 = 3.14 \times 10^{-4} \, \mathrm{m}^2$$

The tensile stress is:

$$\text{Tensile Stress} = \frac{F}{A} = \frac{1.0 \times 10^{4}}{3.14 \times 10^{-4}} = 3.18 \times 10^{7} \, \mathrm{Nm}^{-2}$$

Step 2: Calculate Tensile Strain

Using the formula for tensile strain:

$$\text{Tensile Strain} = \frac{\text{Tensile Stress}}{Y} = \frac{3.18 \times 10^{7}}{20 \times 10^{10}} = 1.59 \times 10^{-4}$$

Step 3: Calculate Change in Length

Using the relationship between tensile strain and change in length:

$$\Delta L = \text{Tensile Strain} \times L = (1.59 \times 10^{-4}) \times 0.50 = 7.95 \times 10^{-5} \, \mathrm{m}$$

Answer

1. Tensile Stress: $3.18 \times 10^{7} \, \mathrm{Nm}^{-2}$
2. Change in Length ($\Delta L$): $7.95 \times 10^{-5} \, \mathrm{m}$