Consider a reinforced concrete beam section of 350 mm width and 600 mm depth. The beam is reinforced with the tension steel of 800 mm\(^2\) area at an effective cover of 40 mm. Consider M20 concrete and Fe415 steel. Let the stress block considered for concrete in IS 456:2000 be replaced by an equivalent rectangular stress block, with no change in (a) the area of the stress block, (b) the design strength of concrete (at the strain of 0.0035), and (c) the location of neutral axis at flexural collapse. The ultimate moment of resistance of the beam (in kN.m) is ___________ (round off to the nearest integer).
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For under-reinforced sections, the moment of resistance can be calculated by using the stress block approach and considering the location of the neutral axis.
Given,
\[
B = 350 \text{ mm}, \quad d = 600 - 40 = 560 \text{ mm}, \quad f_{ck} = 20 \text{ N/mm}^2, \quad f_{y} = 415 \text{ N/mm}^2, \quad A_{st} = 800 \text{ mm}^2
\]
Step 1: Limiting depth of neutral axis (\( x_{lim} \))
The limiting depth of the neutral axis is given by:
\[
x_{lim} = 0.48 \times d = 0.48 \times 560 = 268.8 \text{ mm}
\]
Step 2: Actual depth of neutral axis (\( x_u \))
Using the formula:
\[
x_u = \frac{0.87 \times f_y \times A_{st}}{0.36 \times f_{ck} \times B}
\]
Substituting the given values:
\[
x_u = \frac{0.87 \times 415 \times 800}{0.36 \times 20 \times 350} = 114.619 \text{ mm}
\]
Since \( x_u < x_{lim} \), the section is under-reinforced.
Step 3: Calculation of ultimate moment of resistance (\( M_u \))
The ultimate moment of resistance is given by:
\[
M_u = C \times L \times A
\]
Where \( C = 0.36 f_{ck} B \) and \( L = \left(d - \frac{x_u}{2}\right) \). Therefore:
\[
M_u = 0.36 \times 20 \times 350 \times \left(560 - \frac{114.619}{2}\right)
\]
Calculating this:
\[
M_u = 0.36 \times 20 \times 350 \times 114.619 = 145.197 \times 10^6 \text{ N-mm}
\]
Converting to kN.m:
\[
M_u = 145.2 \text{ kN.m}
\]
Thus, the ultimate moment of resistance of the beam is:
\[
\boxed{148} \text{ kN.m}
\]
