Calculating Steel Reinforcement for a Rectangular Column

This problem involves determining the required steel reinforcement area for a rectangular column subjected to an axial load, based on IS 456 guidelines for concrete structures.

We are given the following information:

• Column size: 400 mm × 550 mm
• Concrete grade: M25 (which means $f_{ck} = 25 \, N/mm^2$)
• Steel grade: Fe415 (which means $f_y = 415 \, N/mm^2$)
• Axial service load: 1700 kN
• Effective length of column: 3.1 m

First, we need to calculate the factored axial load, which is the service load multiplied by a load factor of 1.5 for limit state design:

Factored Load, $P_u = \text{Service Load} \times 1.5$

$P_u = 1700 \, kN \times 1.5 = 2550 \, kN$

To use the formula for load capacity in N, we convert kN to N:

$P_u = 2550 \times 10^3 \, N$

Next, we calculate the gross area of the column cross-section, $A_g$:

$A_g = \text{Width} \times \text{Depth}$

$A_g = 400 \, mm \times 550 \, mm = 220000 \, mm^2$

According to IS 456:2000 (Clause 39.3) for a short axially loaded column, the ultimate load capacity ($P_u$) is given by the formula:

$P_u = 0.4 f_{ck} A_c + 0.67 f_y A_s$

where:

• $f_{ck}$ is the characteristic compressive strength of concrete.
• $A_c$ is the area of concrete.
• $f_y$ is the characteristic strength of steel.
• $A_s$ is the area of longitudinal steel reinforcement.

The total area of the column $A_g$ is the sum of the area of concrete $A_c$ and the area of steel $A_s$. So, $A_c = A_g - A_s$.

Substitute $A_c$ into the formula:

$P_u = 0.4 f_{ck} (A_g - A_s) + 0.67 f_y A_s$

$P_u = 0.4 f_{ck} A_g - 0.4 f_{ck} A_s + 0.67 f_y A_s$

$P_u = 0.4 f_{ck} A_g + (0.67 f_y - 0.4 f_{ck}) A_s$

We need to find the required steel area $A_s$. Rearrange the formula to solve for $A_s$:

$(0.67 f_y - 0.4 f_{ck}) A_s = P_u - 0.4 f_{ck} A_g$

$A_s = \frac{P_u - 0.4 f_{ck} A_g}{0.67 f_y - 0.4 f_{ck}}$

Now, substitute the given values into this equation:

$P_u = 2550 \times 10^3 \, N$

$f_{ck} = 25 \, N/mm^2$

$A_g = 220000 \, mm^2$

$f_y = 415 \, N/mm^2$

Calculation of $A_s$:

$A_s = \frac{2550 \times 10^3 - 0.4 \times 25 \times 220000}{0.67 \times 415 - 0.4 \times 25}$

$A_s = \frac{2550000 - 10 \times 220000}{278.05 - 10}$

$A_s = \frac{2550000 - 2200000}{268.05}$

$A_s = \frac{350000}{268.05}$

$A_s \approx 1305.76 \, mm^2$

The required area of steel reinforcement is approximately $1305.76 \, mm^2$. We compare this value with the given options.

Let's look at the options:

• 1500 mm²
• 1400 mm²
• 1605 mm²
• 1305 mm²

The calculated value $1305.76 \, mm^2$ is closest to $1305 \, mm^2$.

Therefore, the steel reinforcement required is approximately $1305 \, mm^2$. The effective length (3.1 m) is given, but for the calculation of axial load capacity using the formula $P_u = 0.4 f_{ck} A_c + 0.67 f_y A_s$, it's implicitly assumed to be a short column. A slenderness check ($L_{eff}/D$) would confirm this assumption, but the problem can be solved using the provided formula and load.