Question

Consider the singly reinforced section of a cantilever concrete beam under bending, as shown in the figure (M25 grade concrete, Fe415 grade steel). The stress block parameters for the section at ultimate limit state, as per IS 456: 2000 notations, are given. The ultimate moment of resistance for the section by the Limit State Method is \underline{\hspace{2cm} kN.m (round off to one decimal place).} \includegraphics[width=0.75\linewidth]{image88.png}

Hint:

For the ultimate moment of resistance, use the formula \( M_u = 0.87 f_y A_s \left(d - \frac{x_u}{3}\right) \), where \( x_u = 0.45 d \) for a singly reinforced section.

Answer 1

Given: - \( b = 300 \, \text{mm} \) (width of the beam), - \( d = 600 \, \text{mm} \) (effective depth), - \( A_s = 3 \times \left( \pi \times \left(\frac{28}{2}\right)^2 \right) \, \text{mm}^2 \), \text{(area of reinforcement bars)} - \( f_{ck} = 25 \, \text{N/mm}^2 \) (characteristic compressive strength), - \( f_y = 415 \, \text{N/mm}^2 \) (yield strength of steel), - \( x_u = 0.45 \, d \) (depth of neutral axis).

Step 1: Calculate \( A_s \): \[ A_s = 3 \times \left(\pi \times \left(\frac{28}{2}\right)^2\right) = 3 \times \left(\pi \times 14^2 \right) \approx 3 \times 615.75 = 1847.25 \, \text{mm}^2. \]

Step 2: Calculate \( x_u \): \[ x_u = 0.45 \times 600 = 270 \, \text{mm}. \]

Step 3: Moment of resistance formula: The formula for the ultimate moment of resistance \( M_u \) is given by: \[ M_u = 0.87 f_y A_s \left(d - \frac{x_u}{3}\right) \]

Step 4: Substituting values: \[ M_u = 0.87 \times 415 \times 1847.25 \left(600 - \frac{270}{3}\right) \, \text{Nmm}. \] \[ M_u = 0.87 \times 415 \times 1847.25 \times 510 \, \text{Nmm}. \] \[ M_u \approx 295.5 \times 10^3 \, \text{Nmm}. \] \[ M_u \approx 295.5 \, \text{kN.m}. \] \[ \boxed{295.5 \, \text{kN.m}} \]