GATE 2023 CE (CE2) – Question 55

Concrete Structures · Design and detailing of beams and slabs · 2 marks · Numerical answer

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 ______ kN.m (round off to one decimal place). [Note: Here, $A_s$ is the total area of tension steel bars, b is the width of the section, d is the effective depth of the bars, $f_{ck}$ is the characteristic compressive cube strength of concrete, $f_y$ is the yield stress of steel, and $x_u$ is the depth of neutral axis.]

section width 300 mm, total depth 600 mm, three 28 mm top tension bars at effective cover 45 mm; compression force 0.36fck b xu acts 0.42xu from the bottom.

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Correct answer: 299.49 to 302.51

Explanation

**Step 1: steel area and effective depth.** Three 28 mm bars:
$$A_{st}=3\times\frac{\pi}{4}(28)^2=1847.3\text{ mm}^2,\qquad d=600-45=555\text{ mm}.$$

**Step 2: depth of the neutral axis** from equilibrium of the stress block ($C=T$), with $f_y=415$ MPa and $f_{ck}=25$ MPa (IS 456):
$$0.36f_{ck}\,b\,x_u=0.87f_yA_{st}\;\Rightarrow\;x_u=\frac{0.87(415)(1847.3)}{0.36(25)(300)}=247.0\text{ mm}.$$

**Step 3: check the section is under-reinforced.** For Fe415, $x_{u,max}=0.48d=0.48(555)=266.4$ mm. Since $247.0<266.4$, the steel yields before the concrete crushes, so the assumption is valid.

**Step 4: ultimate moment of resistance.** The lever arm is $d-0.42x_u$:
$$M_u=0.87f_yA_{st}(d-0.42x_u)=0.87(415)(1847.3)(555-0.42\times247.0)$$
$$M_u=666{,}965\times451.3=3.0096\times10^8\text{ N mm}=\mathbf{301.0\ kN\,m}.$$