GATE CE · Rcc Structures

Working Stress And Limit State Method – GATE CE Rcc StructuresPractice Questions & PYQs

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58 questions · 20 PYQs · 0 AI practice · GATE CE 2027

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Q1GATE 2025NAT

Consider a reinforced concrete beam section of 300 mm width and 700 mm depth. The beam is reinforced with the tension steel of $2000 \mathrm{~mm}^{2}$ area at an effective cover of 50 mm . Concrete in the tension zone is assumed to be cracked. Assume the modular ratio of 12 and Young's modulus of 200 GPa for steel. When the extreme fibre in the compression zone undergoes the strain of 0.0004 due to the applied bending moment, the stress in the steel (in MPa) is _________ (rounded off to the nearest integer).

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📖 Explanation

Given: Width of beam, $b = 300 , \text{mm}$ Depth of beam, $D = 700 , \text{mm}$ Effective cover, $d' = 50 , \text{mm}$ ? Effective depth to steel, $d = 700 - 50 = 650 , \text{mm}$ Area of steel, $A_s = 2000 , \text{mm}^2$ Modular ratio, $m = 12$ Young?s modulus of steel, $E_s = 200 , \text{GPa}$ Strain at extreme compression fibre, $\varepsilon_c = 0.0004$ Step 1: Let the neutral axis be at a depth $x$ from the top. By similar triangles from the strain diagram: $\frac{0.0004}{x} = \frac{\varepsilon_s}{650 - x}$ ? $\varepsilon_s = 0.0004 \times \frac{650 - x}{x}$ Step 2: Take moment of area about NA using transformed section Concrete in tension is cracked ? Use equivalent area method: Steel area transformed = $m A_s = 12 \times 2000 = 24000 , \text{mm}^2$ Equating compressive and tensile forces for equilibrium: Compression force = $C = 0.5 \times b \times x \times \varepsilon_c E_c$ Tension force = $T = A_s E_s \varepsilon_s$ But easier to use moments using equivalent area: Step 3: Take moment of area about neutral axis to locate $x$ Moment of compression block about NA: $M_c = b x \times \left(\frac{x}{2}\right) = 300 x \times \frac{x}{2} = 150 x^2$ Moment of steel area about NA: $M_s = m A_s (d - x) = 24000 (650 - x)$ Set total moment of compression and transformed steel equal: $150 x^2 = 24000 (650 - x)$ $150 x^2 + 24000 x - 15600000 = 0$ Divide entire equation by 150: $x^2 + 160 x - 104000 = 0$ Solving using quadratic formula: $x = \frac{-160 \pm \sqrt{160^2 + 4 \times 104000}}{2} = \frac{-160 \pm \sqrt{25600 + 416000}}{2}$ $= \frac{-160 \pm \sqrt{441600}}{2} = \frac{-160 \pm 664.6}{2}$ Taking positive root: $x = \frac{504.6}{2} = 252.3 , \text{mm}$ Step 4: Strain in steel $\varepsilon_s = 0.0004 \times \frac{650 - x}{x} = 0.0004 \times \frac{650 - 252.3}{252.3}$ $= 0.0004 \times \frac{397.7}{252.3} \approx 0.0004 \times 1.576 = 0.0006305$ Step 5: Stress in steel $\sigma_s = E_s \varepsilon_s = 200 \times 10^3 \times 0.0006305 = 126.1 , \text{MPa}$ Final Answer: $\boxed{126 , \text{MPa}}$

Q2GATE 2024MCQ

The structural design method that DOES NOT take into account the safety factors on the design load is

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Q3GATE 2023MCQ

Creep of concrete under compression is defined as the

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Q4GATE 2023NAT

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, As is the total area of tension steel bars, $b$ is the width of the section, $d$ is the effective depth of the bars, $f_{c k}$ is the characteristic compressive cube strength of concrete, $f_{y}$ is the yield stress of steel, and $\mathrm{xu}$ is the depth of neutral axis.]

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📖 Explanation

Given data, Grade of concrete $=$ M25 Grade of Steel $=\mathrm{Fe} 415$ Effective depth, $d=600-45=555 \mathrm{~mm}$ $A_{s}=3 \times \frac{\pi}{4} \times (28)^{2}=1847.26 \mathrm{~mm}^{2}$ From Compression $=$ Tension Depth of neutral axis, $x_{u}=\frac{0.87 f_{y} A_{s}}{0.36 f_{c k} b}=\frac{0.87 \times 415 \times 1847.26}{0.36 \times 25 \times 300} =247.02 \mathrm{~mm}$ Limiting depth of neutral axis, $x_{u, l i m}=0.48 \mathrm{~d}$ $=0.48 \times 555=266.40 \mathrm{~mm}$ $\Rightarrow$ Section is under reinforced. Ultimate moment of resistance $=0.36 f_{c k} b x_{u}\left(d-0.42 x_{u}\right)$ $=0.36 \times 25 \times 300 \times 247.02 \times$ $(555-0.42 \times 247.02) \times 10^{-6}$ $=301.0$ (after rounding off to one decimal place).

Q5GATE 2023MCQ

Two plates are connected by fillet welds of size 10 $\mathrm{mm}$ and subjected to tension, as shown in the figure. The thickness of each plate is $12 \mathrm{~mm}$ . The yield stress and the ultimate stress of steel under tension are $250 \mathrm{MPa}$ and $410 \mathrm{MPa}$ , respectively. The welding is done in the workshop (partial safety factor, $\gamma_{\mathrm{mw}}=1.25$ ). As per the Limit State Method of IS 800: 2007, what is the minimum length (in $\mathrm{mm}$ , rounded off to the nearest higher multiple of $5 \mathrm{~mm}$ ) required of each weld to transmit a factored force $P$ equal to $275 \mathrm{kN}$ ?

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📖 Explanation

Size of weld $=10 \mathrm{~mm}$ Thickness of each plate $=12 \mathrm{~mm}$ $f_{y}=250 \mathrm{MPa}, f_{u}=410 \mathrm{MPa}$

\begin{aligned} \gamma_{\mathrm{mw}} & =1.25 \\ \mathrm{P} & =275 \mathrm{kN} \text { (Factored) } \end{aligned}

Let us assume minimum length required of each weld $=l$ $\Rightarrow \frac{\mathrm{f}_{\mathrm{u}}}{\sqrt{3} \gamma_{\mathrm{mw}}} \times 2 \mathrm{l} \times 0.7 \times$ size of weld $=275 \times 10^{3}$ $\Rightarrow \quad \quad l=\frac{275 \times 10^{3} \times \sqrt{3} \times 1.25}{410 \times 2 \times 0.7 \times 10} =103.73 \mathrm{~mm}$ $$ The answer has to be round off to be nearest higher multiple of 5 and hence it shall be taken as $105 \mathrm{~mm}$ . Hence, correct option is (B).

Q6GATE 2022MCQ

In the context of elastic theory of reinforced concrete, the modular ratio is defined as the ratio of

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📖 Explanation

This is a question of working stress method i.e. elastic theory. Modular ratio $=\frac{E_s}{E_c}=\frac{\text{Young's modulus of steel}}{\text{Young's modulus of concrete}}$

Q7GATE 2020NAT

The maximum applied load on a cylindrical concrete specimen of diameter 150 mm and length 300 mm tested as per the split tensile strength test guidelines of IS 5816 : 1999 is 157 kN. The split tensile strength (in MPa, round off to one decimal place) of the specimen is _______.

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📖 Explanation

P = 157 kN D = 150 mm L = 300 mm In split tensile strength test, split tensile strength of concrete

\begin{aligned} f_{et}&=\frac{2P}{\pi DL}=\frac{2 \times 157000}{\pi \times 150 \times 300}\\ &=2.22 N/mm^2 \end{aligned}

Explanation Figure 1

Q8GATE 2020MCQ

During the process of hydration of cement, due to increase in Dicalcium Silicate ( $C_2S$ ) content in cement clinker, the heat of hydration

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📖 Explanation

Due to increase in $C_2S$ heat of hydration decreases.

Q9GATE 2020MCQ

As per IS 456:2000, the pH value of water for concrete mix shall NOT be less than

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Q10GATE 2019MCQ

In the reinforced beam section shown in the figure, the nominal cover provided at the bottom of the beam as per IS 456-2000, is

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📖 Explanation

Nominal cover = Effective cover $-\frac{\phi _m}{2}-\phi _{st}$ $=50-\frac{16}{2}-12=30mm$ Nominal cover is the distance from extreme concrete fbre to the surface of stirrup.

Q11GATE 2019NAT

When aspecimen of M25 concrete is loaded to a stress level of 12.5 MPa, a strain of $500 \times 10^{-6}$ is recorded. If this load is allowed to stand for a long time, the strain increasesto $1000 \times 10^{-6}$ . In accordance with the provisions of IS:456-2000, considering the long-term effects, the effective modulus of elasticity of the concrete (in MPa) is________

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📖 Explanation

Effective modulus of elastic

\begin{aligned} E_{ce}&=\frac{E_c}{1+\theta } \\ E_c&= 5000\sqrt{f_{ck}}\\ &= 5000\sqrt{25}\\ &=25000MPa \end{aligned}

Creep coefcient

\begin{aligned} \theta &=\frac{\text{creep strain}}{\text{elastic strain}} \\ &= \frac{\text{longterm strain - elastic strain}}{\text{elastic strain}}\\ &=\frac{(1000 \times 10^{-6})-(500 \times 10^{-6})}{(500 \times 10^{-6})}=1\\ \therefore \;\;E_{ce}&=\frac{25000}{1+1}=12500MPa \end{aligned}

Q12GATE 2019MCQ

In the context of provisions relating to durability of concrete, consider the following assertions: Assertion (1): As per IS 456-2000, air entrainment to the extent of 3% to 6% is required for concrete exposed to marine environment. Assertion (2): The equivalent alkali content (in terms of $Na_2O$ equivalent) for a cement containing 1% and 0.6% of $Na_2O$ and $K_2O$ , respectively, is approximately 1.4% (rounded to 1 decimal place). Which one of the following statements is CORRECT?

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📖 Explanation

(1) As per clause 8.2.2.3 of IS 456-2000, entrained air percentage of 3 to 6% is required to resist freezing and thawing, i.e. not for marine environment. Hence, Assertion (1) is wrong Equivalent alkali content is terms of $Na_2O$ $= [Na_2 O] + 0.685 [K_2 O]$ $= 1 + 0.685 \times 0.6 = 1.41 \%$ Hence, Assertion (2) is correct

Q13GATE 2019NAT

The characteristic compressive strength of concrete required in a project is 25MPa and the standard deviation in the observed compressive strength expected at site is 4MPa. The average compressive strength of cubes tested at different water-cement (w/c) ratios using the same material as is used for the project is given in the table. The water-cement ratio (in percent, round off to the lower integer) to be used in the mix is _____

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📖 Explanation

Target mean strength

\begin{aligned} &=f_{ck}+1.65\sigma \\ &= 25+1.65 \times 4.0\\ &= 31.6 \end{aligned}

Water content required, $=50-\frac{50-45}{35-25}\times (31.6-25)=46.7\%$ say 46% (round off to the lower integer)

Q14GATE 2018NAT

As per IS 456 : 2000, the minimum percentage of tension reinforcement (up to two decimal places) required in reinforced-concrete beams of rectangular cross-section (considering effective depth in the calculation of area) using Fe500 grade steel is ______

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📖 Explanation

Minimum percentage of steel (for Fe 500 ) $=\frac{85}{f_{y}} \%=\frac{85}{500} \%=0.17 \%$

Q15GATE 2018MCQ

The deformation in concrete due to sustained loading is

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Q16GATE 2018NAT

The frequency distribution of the compressive strength of 20 concrete cube specimens is given in the table. If $\mu$ is the mean strength of the specimens and $\sigma$ is the standard deviation, the number of specimens (out of 20) with compressive strength less than $\mu -3\sigma$ is ______

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📖 Explanation

Average strength,

\begin{aligned} \mu &=\frac{(4 \times 23)+(2 \times 28)+(5 \times 22.5)+(5 \times 31)+(4 \times 29)}{20} \\ &=26.575 \mathrm{MPa} \\ \sigma &=\sqrt{\frac{\Sigma(\mu-f)^{2}}{n-1}} \\ &=\sqrt{\frac{\begin{array}{c} (26.575-23)^{2} \times 4+(26.575-28)^{2} \times 2\\ +(26.575-22.5)^{2} \times 5+(26.575-31)^{2} \\ \times 5+(26.575-29)^{2} \times 4 \end{array}}{(20-1)}} \\&=3.7 \end{aligned}

Now, $\mu-3 \sigma=26.575-3 \times 3.7=15.48$ Thus, no specimen is having compressive strength less than $\mu-3 \sigma.$

Q17GATE 2017MCQ

Let the characteristic strength be defined as that value, below which not more than 50% of the results are expected to fall. Assuming a standard deviation of 4 MPa, the target mean strength (in MPa) to be considered in the mix design of a M25 concrete would be

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📖 Explanation

If $f_{c k}$ is the value below which not more than 50% of test results are expected then, $f_{m}=f_{c k}$ [Mean strength = Characteristics strength] So target mean strength of concrete to be considered in design mix. = Mean strength $=f_{m}=f_{c k}=25 \mathrm{MPa}$

Explanation Figure 1

Q18GATE 2017MCQ

According to IS 456-2000, which one of the following statements about the depth of neutral axis $\chi _{u,bal}$ for a balanced reinforced concrete section is correct?

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📖 Explanation

$x_{u, \text { bal }}=\frac{700}{1100+0.87 f_{y}} \times d$ So it depends upon grade of steel only.

Q19GATE 2016NAT

For M25 concrete with creep coefficient of 1.5, the long-term static modulus of elasticity (expressed in MPa) as per the provisions of IS:456-2000 is __________.

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📖 Explanation

Long term static modulus of elasticity,

\begin{aligned} E_{\theta} &=\frac{E_{C}}{1+\theta}=\frac{5000 \sqrt{f_{c k}}}{1+\theta}=\frac{5000 \sqrt{25}}{1+1.5} \\ &=10,000 \end{aligned}

Q20GATE 2015MCQ

Consider the following statements for air-entrained concrete. (i) Air-entrainment reduces the water demand for a given level of workability (ii) Use of air-entrained concrete is required in environments where cyclic freezing and thawing is expected Which of the following is TRUE?

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