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determine the required area of reinforcing steel

To determine the required area of reinforcing steel (\(A_s\)) for the reinforced concrete column, we'll use the following steps based on the given information:

### Given Data:
- Diameter of the column = 300 mm
- Axial compressive load \(P\) = 500 kN
- Allowable stress for concrete \(S_c\) = 6 MPa
- Allowable stress for steel \(S_s\) = 120 MPa
- Modulus of elasticity for concrete \(E_c\) = 14 GPa
- Modulus of elasticity for steel \(E_s\) = 200 GPa

### Step 1: Compute the cross-sectional area of the column

The cross-sectional area \(A_c\) of the concrete column is given by:

\[
A_c = \frac{\pi d^2}{4}
\]

Where \(d\) is the diameter of the column.

\[
A_c = \frac{\pi (0.3)^2}{4} = 0.0706858 \, \text{m}^2 = 70685.8 \, \text{mm}^2
\]

### Step 2: Apply the Load Sharing Between Concrete and Steel

The total compressive load \(P = P_c + P_s\), where:
- \(P_c = S_c \cdot A_c'\) (Load carried by the concrete)
- \(P_s = S_s \cdot A_s\) (Load carried by the steel)

Here, we will first determine the modified area for concrete (\(A_c'\)) after considering the modular ratio, which accounts for the difference in stiffness between steel and concrete.

The modular ratio \(n\) is defined as:

\[
n = \frac{E_s}{E_c} = \frac{200 \, \text{GPa}}{14 \, \text{GPa}} = 14.29
\]

Thus, the effective area for the concrete portion is reduced by the steel contribution:

\[
A_c' = A_c - A_s
\]

### Step 3: Set up the Equation for Load Balance

Using the equation for total load:

\[
500 \, \text{kN} = S_c \cdot (A_c - A_s) + S_s \cdot A_s
\]

Convert the load into Newtons:

\[
500 \, \text{kN} = 500 \times 10^3 \, \text{N} = 500,000 \, \text{N}
\]

Substitute the allowable stresses:

\[
500,000 = 6 \cdot (A_c - A_s) + 120 \cdot A_s
\]

Substitute \(A_c = 70685.8 \, \text{mm}^2\):

\[
500,000 = 6 \cdot (70685.8 - A_s) + 120 \cdot A_s
\]

### Step 4: Solve for \(A_s\)

Now, solve the equation for \(A_s\):

\[
500,000 = 424114.8 - 6 A_s + 120 A_s
\]

Simplify:

\[
500,000 = 424114.8 + 114 A_s
\]

\[
500,000 - 424114.8 = 114 A_s
\]

\[
75885.2 = 114 A_s
\]

\[
A_s = \frac{75885.2}{114} = 665.66 \, \text{mm}^2
\]

### Final Answer:
The required area of reinforcing steel \(A_s\) is approximately **665.66 mm²**.

---

Would you like more details or have any questions? Here are five related questions to explore:

1. How does the modular ratio affect the load distribution between concrete and steel?
2. What happens if we use higher-strength concrete with a higher allowable stress?
3. How is the area of the column affected if we change its diameter?
4. What is the significance of using the allowable stress method in this calculation?
5. How does the modulus of elasticity of steel and concrete impact structural design?

**Tip:** Always check if the stresses for both concrete and steel are within the allowable limits after calculating the required areas.

Determine the required area of reinforcing steel for a reinforced concrete column with a diameter of 300 mm designed to carry an axial compressive load of 500 kN. Given allowable stresses: concrete = 6 MPa, steel = 120 MPa, modulus of elasticity for concrete = 14 GPa, steel = 200 GPa.

Determine the required area of reinforcing steel for a 300mm diameter reinforced concrete column under 500kN axial load. Using allowable stresses (concrete = 6 MPa, steel = 120 MPa), modulus of elasticity for concrete = 14 GPa, steel = 200 GPa.

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