If a Column has been design with a cross sectional area of 120,000 mm² is subject to an axial load of 2,500 KN and a diameter of the steel to be 25mm determine the steel stress in both the steel and concrete if the elastic modulus of the steel and concrete are 200 KN/mm² and 21 KN/mm² respectively

The area of a circular bar is given by $A = \frac{\pi d^2}{4}$.

$$A_s = \frac{\pi (25 \text{ mm})^2}{4} = \frac{\pi \times 625}{4} \text{ mm}^2 \approx 490.87 \text{ mm}^2$$

The total area of the column is the sum of the steel area and the concrete area.

$$A_c = A_{total} - A_s = 120,000 \text{ mm}^2 - 490.87 \text{ mm}^2 = 119,509.13 \text{ mm}^2$$

The modular ratio is the ratio of the elastic moduli of steel to concrete.

$$n = \frac{E_s}{E_c} = \frac{200 \text{ KN/mm}^2}{21 \text{ KN/mm}^2} \approx 9.5238$$

For a composite section under axial load, the total load $P$ is distributed between the concrete and steel. The strain in both materials is equal ($\epsilon_s = \epsilon_c$).
We know that $\sigma = E \epsilon$, so $\sigma_s = E_s \epsilon$ and $\sigma_c = E_c \epsilon$.
This implies $\sigma_s = \frac{E_s}{E_c} \sigma_c = n \sigma_c$.
The total load $P = P_c + P_s = \sigma_c A_c + \sigma_s A_s$.
Substitute $\sigma_s = n \sigma_c$:

$$P = \sigma_c A_c + (n \sigma_c) A_s = \sigma_c (A_c + n A_s)$$

Now, solve for $\sigma_c$:

$$\sigma_c = \frac{P}{A_c + n A_s}$$

Substitute the values:

$$\sigma_c = \frac{2,500 \text{ KN}}{119,509.13 \text{ mm}^2 + (9.5238 \times 490.87 \text{ mm}^2)}$$ $$\sigma_c = \frac{2,500 \text{ KN}}{119,509.13 \text{ mm}^2 + 4675.00 \text{ mm}^2}$$ $$\sigma_c = \frac{2,500 \text{ KN}}{124,184.13 \text{ mm}^2}$$ $$\sigma_c \approx 0.02013 \text{ KN/mm}^2$$

Using the relationship $\sigma_s = n \sigma_c$:

$$\sigma_s = 9.5238 \times 0.02013 \text{ KN/mm}^2$$ $$\sigma_s \approx 0.1917 \text{ KN/mm}^2$$

The stresses are:
Stress in concrete: $\boxed{\text{0.02013 KN/mm}^2}$
Stress in steel stud: $\boxed{\text{0.1917 KN/mm}^2}$