GATE CE 2023 Set 2 — Question 55

The problem asks for the ultimate moment of resistance ($M_u$) of the given reinforced concrete section. The diagram has some misleading information. The dimension line for 'd' (effective depth) points to the overall depth of 600 mm. However, an effective cover of 45 mm is also specified. The correct interpretation that leads to the answer is to consider 600 mm as the overall depth (D) and 45 mm as the effective cover.Step 1: List the given parameters

• Concrete grade: M25 $\implies f_{ck} = 25$ N/mm²
• Steel grade: Fe415 $\implies f_y = 415$ N/mm²
• Width of the beam, $b = 300$ mm
• Overall depth, $D = 600$ mm
• Effective cover = 45 mm
• Reinforcement: 3 bars of 28 mm diameter

Step 2: Calculate the effective depth (d)
Effective depth is the distance from the compression face to the centroid of the tension reinforcement.
$d = \text{Overall depth} - \text{Effective cover}$
$d = 600 - 45 = 555$ mmStep 3: Calculate the area of tension steel ($A_{st}$)
$A_{st} = 3 \times (\frac{\pi}{4} \times d_{bar}^2) = 3 \times (\frac{\pi}{4} \times 28^2)$
$A_{st} = 3 \times 615.75 = 1847.26$ mm²Step 4: Determine the depth of the neutral axis ($x_u$)
By equating the total compressive force ($C_u$) and total tensile force ($T_u$):
$C_u = T_u$
$0.36 f_{ck} b x_u = 0.87 f_y A_{st}$
$0.36 \times 25 \times 300 \times x_u = 0.87 \times 415 \times 1847.26$
$2700 x_u = 666807.4$
$x_u = \frac{666807.4}{2700} = 246.97$ mmStep 5: Check if the section is under-reinforced
Calculate the limiting depth of the neutral axis ($x_{u,max}$) for Fe415 steel.
As per IS 456:2000, for $f_y = 415$ N/mm², $x_{u,max} = 0.48 d$.
$x_{u,max} = 0.48 \times 555 = 266.4$ mmSince $x_u (246.97 \text{ mm}) < x_{u,max} (266.4 \text{ mm})$, the section is under-reinforced, which is desirable. The steel will yield before the concrete crushes.Step 6: Calculate the ultimate moment of resistance ($M_u$)
The ultimate moment of resistance is calculated by taking the moment of either the compressive or tensile force about the other.
$M_u = 0.87 f_y A_{st} \times (\text{lever arm})$
Lever arm, $z = d - 0.42 x_u$
$z = 555 - 0.42 \times 246.97 = 555 - 103.73 = 451.27$ mm$M_u = (0.87 \times 415 \times 1847.26) \times 451.27$
$M_u = 666807.4 \times 451.27 = 300,902,159$ N.mmTo convert to kN.m, divide by $10^6$.
$M_u = \frac{300,902,159}{10^6} = 300.9$ kN.mRounding to one decimal place, the ultimate moment of resistance is 300.9 kN.m.