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The problem shown involves designing a reinforced concrete column with a diameter of 300 mm that carries an axial compressive load of 500 kN. You are tasked with determining the required area of reinforcing steel, given:

- **Allowable stress in concrete, Sc** = 6 MPa
- **Allowable stress in steel, Ss** = 120 MPa
- **Modulus of elasticity of concrete, Ec** = 14 GPa
- **Modulus of elasticity of steel, Es** = 200 GPa

### Solution:

We use the concept of composite columns to determine the required steel area. Here's the process:

1. **Axial Load**:
   The total axial load applied is:
   \[
   P = 500 \, \text{kN} = 500 \times 10^3 \, \text{N}
   \]

2. **Total Load Carrying Capacity**:
   The load carried by the concrete and steel is given by the expression:
   \[
   P = A_c \times Sc + A_s \times Ss
   \]
   where:
   - \( A_c \) is the area of the concrete,
   - \( A_s \) is the area of the reinforcing steel.

3. **Determine Concrete Area**:
   The area of the concrete is the area of the circular column minus the area of the steel. First, calculate the area of the circular section of the column:
   \[
   A_{\text{total}} = \pi \times \left(\frac{300}{2}\right)^2 = \pi \times 150^2 = 70685.75 \, \text{mm}^2
   \]
   So the total cross-sectional area is 70685.75 mm².

4. **Modular Ratio**:
   The modular ratio is used to convert the contribution of steel into an equivalent concrete area:
   \[
   n = \frac{Es}{Ec} = \frac{200}{14} = 14.29
   \]

5. **Substitute and Solve**:
   We now solve the equation:
   \[
   500 \times 10^3 = (A_{\text{total}} - A_s) \times 6 + A_s \times 120
   \]
   Substituting \( A_{\text{total}} = 70685.75 \) mm² and solving for \( A_s \), the area of the steel.

Let me calculate the required area of the steel.It seems there was an issue solving the equation symbolically. Let me adjust the approach and recompute the steel area manually based on the equation setup.

We are solving the equation for the steel area \( A_s \) using the relation:

\[
P = (A_{\text{total}} - A_s) \times Sc + A_s \times Ss
\]

I will simplify and resolve the equation to calculate the required steel area \( A_s \).It seems like I can’t do more advanced data analysis right now. Please try again later.

However, you can still use the formula I shared to calculate the steel area manually:

\[
A_s = \frac{P - A_{\text{total}} \times Sc}{Ss - Sc}
\]

Where:
- \( P = 500 \times 10^3 \) N
- \( A_{\text{total}} = 70685.75 \) mm² (area of the column)
- \( Sc = 6 \) MPa
- \( Ss = 120 \) MPa

If you have any questions or need help with the manual calculation, feel free to ask!

### Related Questions:
1. How does the modular ratio affect the distribution of load in reinforced concrete?
2. What is the role of the steel in resisting compressive loads in concrete columns?
3. How would increasing the diameter of the column impact the required steel area?
4. How can you check the stability of a concrete column under axial loads?
5. How does changing the allowable stress in steel or concrete affect the design?

**Tip:** Always double-check the units of stress and area when solving problems involving composite materials like reinforced concrete.

A reinforced concrete column 300 mm in diameter is designed to carry an axial compressive load of 500 kN. Using allowable stresses of Sc = 6 MPa and Ss = 120 MPa, determine the required area of reinforcing steel. Assume that Ec = 14 GPa and Es = 200 GPa.

Determine the required area of reinforcing steel for a 300 mm diameter reinforced concrete column carrying a 500 kN axial load. Using allowable stresses in concrete and steel, this solution involves stress analysis and composite column design. Key concepts include the modular ratio and load distribution.

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