Understanding the Beam Moment of Resistance Problem

The question asks us to find the working moment of resistance for a reinforced concrete beam. We are given the beam dimensions, the amount of tension steel reinforcement, and the grades of concrete and steel. The term "working moment of resistance" in this context, especially with M25 concrete and Fe415 steel, usually refers to calculating the ultimate moment capacity of the section based on Limit State Method (LSM) as per IS 456:2000.

Given Data:

• Beam width, $b = 300 \, mm$
• Effective depth, $d = 450 \, mm$
• Concrete grade, $M25 \implies f_{ck} = 25 \, N/mm^2$
• Steel grade, $Fe415 \implies f_y = 415 \, N/mm^2$
• Tension reinforcement, $A_s$: 3 bars of 25 mm diameter

Step-by-Step Calculation of Moment of Resistance

1. Calculate the Area of Tension Steel ($A_s$)

The area of one steel bar of 25 mm diameter is given by:

\( Area = \frac{\pi}{4} \times (diameter)^2 \)

For 3 bars of 25 mm diameter:

\( A_s = 3 \times \left(\frac{\pi}{4} \times 25^2\right) \)
\( A_s = 3 \times \left(\frac{3.14159}{4} \times 625\right) \)
\( A_s = 3 \times (0.7854 \times 625) \)
\( A_s = 3 \times 490.87 \, mm^2 \)
\( A_s = 1472.61 \, mm^2 \)

So, the total area of tension steel is \(1472.61 \, mm^2\).

2. Determine the Depth of Neutral Axis ($x_u$)

According to IS 456:2000 (Limit State Method), the depth of the neutral axis ($x_u$) is found by equating the total compressive force in concrete to the total tensile force in steel.

Compressive Force ($C$) = \(0.36 f_{ck} b x_u\)

Tensile Force ($T$) = \(0.87 f_y A_s\)

Equating C and T:

\( 0.36 f_{ck} b x_u = 0.87 f_y A_s \)

Solving for \(x_u\):

\( x_u = \frac{0.87 f_y A_s}{0.36 f_{ck} b} \)

Substitute the given values:

\( x_u = \frac{0.87 \times 415 \, N/mm^2 \times 1472.61 \, mm^2}{0.36 \times 25 \, N/mm^2 \times 300 \, mm} \)
\( x_u = \frac{531387.945}{2700} \)
\( x_u \approx 196.81 \, mm \)

The actual depth of the neutral axis is approximately \(196.81 \, mm\).

3. Determine the Limiting Depth of Neutral Axis ($x_{u,lim}$)

For Fe415 steel, the limiting depth of the neutral axis for a balanced section, as per IS 456:2000, is:

\( x_{u,lim} = 0.48d \)

Substitute the effective depth \(d = 450 \, mm\):

\( x_{u,lim} = 0.48 \times 450 \, mm \)
\( x_{u,lim} = 216 \, mm \)

The limiting depth of the neutral axis is \(216 \, mm\).

4. Check the Type of Section

We compare the actual depth of the neutral axis ($x_u$) with the limiting depth ($x_{u,lim}$).

\( x_u = 196.81 \, mm \)
\( x_{u,lim} = 216 \, mm \)

Since \(x_u < x_{u,lim}\), the section is under-reinforced. In an under-reinforced section, the steel yields before the concrete reaches its ultimate strain. The moment of resistance is governed by the yielding of steel.

5. Calculate the Moment of Resistance ($M_u$)

For an under-reinforced section, the ultimate moment of resistance ($M_u$) is calculated using the formula based on the tensile force in steel:

\( M_u = 0.87 f_y A_s (d - 0.42 x_u) \)

Substitute the calculated values:

\( M_u = 0.87 \times 415 \, N/mm^2 \times 1472.61 \, mm^2 \times (450 \, mm - 0.42 \times 196.81 \, mm) \)
\( M_u = 531387.945 \, N \times (450 \, mm - 82.66 \, mm) \)
\( M_u = 531387.945 \, N \times 367.34 \, mm \)
\( M_u \approx 195104671.5 \, Nmm \)

To convert this to KNm, divide by \(10^6\):

\( M_u \approx \frac{195104671.5}{10^6} \, KNm \)
\( M_u \approx 195.10 \, KNm \)

Conclusion

The calculated ultimate moment of resistance for the given beam section is approximately \(195.10 \, KNm\).

Comparing this value to the given options:

The calculated value of approximately 195.10 KNm is closest to 195 KNm.