Test: RCC - Civil Engineering (CE) MCQ


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25 Questions MCQ Test GATE Civil Engineering (CE) 2025 Mock Test Series - Test: RCC

Test: RCC for Civil Engineering (CE) 2024 is part of GATE Civil Engineering (CE) 2025 Mock Test Series preparation. The Test: RCC questions and answers have been prepared according to the Civil Engineering (CE) exam syllabus.The Test: RCC MCQs are made for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: RCC below.
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Test: RCC - Question 1

Assuming the concrete is uncracked, compute Bending stress in extreme fibres of Beam for Bending moment 5kN-m & axial load 200 kN use M25, Fe45

Detailed Solution for Test: RCC - Question 1

Test: RCC - Question 2

Assuming the concrete is uncracked, compute Bending stress in extreme fibres of Beam for Bending moment 5kN-m & axial load 200 kN use M25, Fe45

In above question, determine cracking moment in kNm.

Detailed Solution for Test: RCC - Question 2

For Cracking of section - 

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Test: RCC - Question 3

Slab supported on wall on two opposite edges as given in figure. Thickness of wall is 250mm & thickness of slab is 175mm. Consider M20, Fe415 & effective cover 25mm. Find moment of Resistance of slab per meter width using LSM

Detailed Solution for Test: RCC - Question 3

Test: RCC - Question 4

The maximum area of tension reinforcement in beams shall not exceed 

Detailed Solution for Test: RCC - Question 4

If tensile reinforcement of beam should exceed 4% of total gross area then some crack will be developed in concrete.

Test: RCC - Question 5

A simply supported beam having effective length 8m subjected to a live load 6 kN/m the size of Beam is given below. Width of support is 300mm. Use M20/Fe415 use LSM. Find spacing for 2-legged 8mm φ - vertical stirrups. Use τc = 0.20 N/mm

Detailed Solution for Test: RCC - Question 5

*Answer can only contain numeric values
Test: RCC - Question 6

In the previous question calculate bend stress. (in N/mm2)

(Important - Enter only the numerical value in the answer)


Detailed Solution for Test: RCC - Question 6

Test: RCC - Question 7

For a pretensioned beam, Young’s modulus of steel and concrete are 200 GPa and 35.35 GPa, respectively If the ultimate shrinkage strain and ultimate creep coefficient are 200 microns and 1.6 respectively, what is the level of sustained stress in concrete at the level of steel if the loss due to creep is three times the loss due to
shrinkage? 

Detailed Solution for Test: RCC - Question 7

Test: RCC - Question 8

In a post tensioned pre-stressed concrete Beam. The end block zone is in between the end of the beam
and the section where

Test: RCC - Question 9

Find ultimate axial load carrying capacity of circular column having diameter 500mm. Consider clear cover 50mm. Column having 12Nos-20mm φ longitudinal Reinforcement & 8mm φ spiral Reinforcement.
Use M20/Fe415.

Detailed Solution for Test: RCC - Question 9

Test: RCC - Question 10

In above question find pitch of spiral Reinforcement

Detailed Solution for Test: RCC - Question 10

Test: RCC - Question 11

A simply supported post-tensioned prestressed concrete beam of span L is prestressed by a straight tendon
at a uniform eccentricity ‘e’ below the centroidal axis. If the magnitude of prestressing force is P and
flexural rigidity of beam is EI, maximum deflection of the beam is

Detailed Solution for Test: RCC - Question 11

Test: RCC - Question 12

For maximum hogging bending moment at suppot in a continuous RC beam, live load should be placed on

Detailed Solution for Test: RCC - Question 12

Test: RCC - Question 13

A box girder is subjected to loads as shown in the figure below. The critical section for shear in the bottom slab will occur at 

Detailed Solution for Test: RCC - Question 13

Since, the bottom slab will experience tension hence the critical section occurs at the face of wall.

Test: RCC - Question 14

An isolated T-beam is used as a walkway. The beam is simply supported with an effective span of 6 m. The effective width of flange, for the cross-section shown in the figure is

 

Detailed Solution for Test: RCC - Question 14

Test: RCC - Question 15

Shrinkage deflection in case of rectangular beams and slabs can be eliminated by putting

Detailed Solution for Test: RCC - Question 15

Shrinkage deflection in rectangular beams and slabs occurs due to the shrinkage strain in concrete, which causes differential movement between the tensile and compressive zones. This deflection can be counteracted by providing compression reinforcement.

  • When the compression steel is equal to the tensile steel, the shrinkage strain in the concrete is balanced because the additional compression steel counteracts the forces caused by shrinkage.
  • This creates an equilibrium, reducing or eliminating shrinkage deflection.

Why Other Options Are Incorrect:

  • b) Compression steel more than tensile steel: While this may provide additional shrinkage resistance, it is not the most efficient design and leads to unnecessary cost and weight.
  • c) Compression steel less than tensile steel: This would not adequately counteract the shrinkage forces, leading to residual shrinkage deflection.
  • d) Compression steel 25% greater than tensile steel: Although this may reduce shrinkage deflection, it is not required for elimination.

Thus, the ideal solution is equal compression and tensile steel to achieve balanced behavior.

Test: RCC - Question 16

The limits of percentage ‘p’ of the longitudinal reinforcement in a column is

Detailed Solution for Test: RCC - Question 16

Minimum reinforcement in Column → 0.8%
Maximum reinforcement in Column → 6%

Test: RCC - Question 17

A reinforced cantilever beam of span 7m, has a cross-sectional area of 250 mm X 500 mm (effective)
if checked for lateral stability and deflection, the beam will

Detailed Solution for Test: RCC - Question 17

Hence the beam will fail in
deflection

(b) Check for lateral stability :

For cantilever beam clear distance between the free end of cantilever and lateral restraint shall not

Hence beam will fail in lateral stability.
Beam will fail in both deflection & lateral stability

Test: RCC - Question 18

In an RCC beam of breadth ‘b’ and overall depth exceeding 750 mm, side face reinforcement required and allowable area of maximum tension reinforcement shall be respectively.

*Answer can only contain numeric values
Test: RCC - Question 19

Effective cover – 40 mm

Effective depth – 560 mm

Width of beam – 300 mm

Grade of concrete – M20

Grade of Steel – Fe 415

Take m = 13, σcbc = 7 N/m2, σst = 230 N/mm2.

Find moment of resistance of beam using working stress method __________

(Important - Enter only the numerical value in the answer)


Detailed Solution for Test: RCC - Question 19

Test: RCC - Question 20

The equivalent bending moment for the beam having loading as follows is
BM = 300 kNm
SF = 120 kN
Torsional moment = 72 kNm
Use M 25/ Fe 415,

b = 400 mm,
d = 450 mm.

Effective cover = 50 mm

Detailed Solution for Test: RCC - Question 20

Test: RCC - Question 21

Linked Question 21 & 22

In the design of beams for unit limit state of collapse in flexure (as per IS : 456 : 2000), let the maximum strain in concrete be limited to 0.0025 (in place of 0.0035). For this situation, consider a rectangular beam section with breadth as 300 mm, effective depth as 450 mm, Area of tension steel as 1500 mm2, and characteristic
strength of concrete and steel as 30 Mpa and 250 Mpa respectively. 

Q. The depth of neutral axis for the balanced failure is

Detailed Solution for Test: RCC - Question 21

Test: RCC - Question 22

At the limiting state of collapse in flexure the force acting on the compression zone of the section is

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Test: RCC - Question 23

If characteristic compressive strength of concrete fck is defined as strength below which not more than 95% of the test result are expected to fall, the expression of fck in term of mean strength fm and standard deviation σ would be

Detailed Solution for Test: RCC - Question 23

The characteristic compressive strength of concrete, denoted fckf_{ck}, is defined as the strength below which no more than 5% of the test results are expected to fall. In statistical terms, this is the strength corresponding to the 5th percentile of the strength distribution.

In most cases, the compressive strength of concrete follows a normal distribution. The relationship between the characteristic strength fckf_{ck}, the mean strength fmf_m, and the standard deviation σ\sigma can be derived from the properties of the normal distribution.

Step 1: Understanding the Concept

The characteristic strength fckf_{ck} is defined such that only 5% of the test results fall below it. This corresponds to the 5th percentile of the normal distribution.

For a normal distribution:

  • The mean strength is fmf_m.
  • The standard deviation is σ\sigma.
  • The 5th percentile is located at a value corresponding to zz-score = -1.645 (this value is obtained from standard normal distribution tables for a cumulative probability of 5%).

Step 2: Relation between fckf_{ck}, fmf_m, and σ\sigma

The general equation for a normal distribution is:

fck=fm1.645σf_{ck} = f_m - 1.645 \cdot \sigma

This equation means that the characteristic strength fckf_{ck} is the mean strength fmf_m minus 1.645 times the standard deviation σ\sigma.

Final Expression:

Thus, the expression for the characteristic compressive strength fckf_{ck} in terms of the mean strength fmf_m and the standard deviation σ\sigma is:

fck=fm1.645σf_{ck} = f_m - 1.645 \cdot \sigma

Test: RCC - Question 24

What is the minimum percentage of steel required in tension reinforcement in a beam where Fe 415 steel is used?

Test: RCC - Question 25

Find Area of tensile steel for following section. If required consider section as doubly
Reinforced

If span of beam is 5m & external load is 20 kN/m. Use LSM. Beam is simply supported

Detailed Solution for Test: RCC - Question 25

 

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