A steel bar of rectangular cross-section, 3cm by 2cm, carries an axial load of 30kN. Estimate the average tensile stress over a normal cross-section of the bar.

The dimensions are $3 \text{ cm}$ by $2 \text{ cm}$. Convert these to meters.

$$A = (3 \text{ cm} \times \frac{1 \text{ m}}{100 \text{ cm}}) \times (2 \text{ cm} \times \frac{1 \text{ m}}{100 \text{ cm}})$$ $$A = 0.03 \text{ m} \times 0.02 \text{ m}$$ $$A = 0.0006 \text{ m}^2$$

The axial load is $30 \text{ kN}$.

$$P = 30 \text{ kN} \times \frac{1000 \text{ N}}{1 \text{ kN}}$$ $$P = 30000 \text{ N}$$

The formula for tensile stress ($\sigma$) is load ($P$) divided by area ($A$).

$$\sigma = \frac{P}{A}$$ $$\sigma = \frac{30000 \text{ N}}{0.0006 \text{ m}^2}$$ $$\sigma = 50000000 \text{ N/m}^2$$ $$\sigma = 50 \times 10^6 \text{ N/m}^2$$ $$\sigma = 50 \text{ MPa}$$

The average tensile stress over a normal cross-section of the bar is $\boxed{\text{50 MPa}}$.