Reinforced Concrete Structures MCQ Quiz - Objective Question with Answer for Reinforced Concrete Structures - Download Free PDF

Last updated on Jun 20, 2025

Latest Reinforced Concrete Structures MCQ Objective Questions

Reinforced Concrete Structures Question 1:

If a simply supported concrete beam, prestressed with a force of 2500 kN , is designed by load balancing concept for an effective span of 10 m and to carry a total load of 40kN/m, the central dip of the cable profile should be

  1. 100 mm
  2. 200 mm
  3. 300 mm
  4. 400 mm

Answer (Detailed Solution Below)

Option 2 : 200 mm

Reinforced Concrete Structures Question 1 Detailed Solution

Concept:

Prestress concrete is the concrete in which internal stresses are produced due to compression or tension applied before applying external load and these stresses are counter balanced by the applied load to the desired degree.

F2 Madhuri Engineering 25.04.2022 D2

 

 

 

 

 

Central dip is given by: e=WL28P

where, P = Prestressing force, L = effective span, W = Acting UDL

Calculation:

Given:

P = 2500 kN, W = 40 kN/m, L = 10 m

e=40×1028×2500=0.2m=200mm

Reinforced Concrete Structures Question 2:

According to I.S.: 456, slabs which span in two directions with corners held down, are assumed to be divided in each direction into middle strips and edge strips such that the width of the middle strip, is

  1. Half of the width of the slab
  2. Two-third of the width of the slab
  3. Three-fourth of the width of the slab
  4. Four-fifth of the width of the slab

Answer (Detailed Solution Below)

Option 3 : Three-fourth of the width of the slab

Reinforced Concrete Structures Question 2 Detailed Solution

Explanation:

According to IS 456:2000, Annex D (Clause D-1.1), when a two-way slab is supported on all four sides with corners held down:

  1. The slab is assumed to be divided in each direction into:

    • Middle strip

    • Edge strips

  2. The middle strip in each direction has a width equal to three-fourths (3/4) of the total width of the slab in that direction.

  3. The remaining one-fourth is equally divided into two edge strips (each edge strip = one-eighth of the width).

 F1 Akhil 4.5.21 Pallavi D4

Additional Information

  • A one-way slab is a reinforced concrete slab that primarily bends in one direction only, due to the geometry and support conditions.

  • It spans between two opposite supports (usually longer in one direction), and the load is carried along the shorter span.

  • As per IS 456:2000, a slab is considered one-way if the ratio of longer span (Ly) to shorter span (Lx) ≥ 2.

  • This condition ensures that bending along the longer direction is negligible.

Reinforced Concrete Structures Question 3:

A RC beam has cross section 300mm×600mm and is subjected to the following design forces
Bending moment =115KNm
Shear force =95KNm
Torsional moment =45KNm
Determine the equivalent bending moment for which section is to be designed.

  1. 194.41KNm
  2. 205.6KNm
  3. 150.63KNm
  4. 146.56KNm

Answer (Detailed Solution Below)

Option 1 : 194.41KNm

Reinforced Concrete Structures Question 3 Detailed Solution

Concept:

To determine the equivalent bending moment for the RC beam, we can use the following formula: Meq=M+T(1+Db)1.7

where: M = bending moment, V = shear force, d = distance from the neutral axis to the point of application of the shear force, and T = torsional moment. For a rectangular section, we can assume d to be half the height of the beam.

Calculation:

Given: M = 115 kN-m; V = 95 kN-m; T = 45 kN-m; D = 0.6 m; b = 0.3 m

Meq=115+45×(1+0.60.3)1.7

Meq=115+79.41=194.41kNm

Reinforced Concrete Structures Question 4:

If the depth of neutral axis for singly reinforced rectangular section is represented by " Kd " in working stress design, then the value of " K " for balance section where ' D " is the effective depth, σst permissible stress in steel in tension and σcbc is permissible stress in steel in bending compression.

  1. Depends on σst only
  2. Depends on σcbc only
  3. Depends on Both σst and σcbc
  4. Is independent of Both σst and σcbc

Answer (Detailed Solution Below)

Option 1 : Depends on σst only

Reinforced Concrete Structures Question 4 Detailed Solution

Explanation:

  • The depth of neutral axis (K D) in a balanced section depends on the limiting tensile stress in steel (σst).

  • The permissible stress in concrete (σcbc) affects overall capacity but not the position of neutral axis in balanced section.

  • The strain compatibility forces the neutral axis to be positioned based on how much the steel can be stressed.

  • Therefore, K depends only on σst

F10 Chandramouli 19-1-2021 Swati D4

k=2803σst+280

Additional InformationA balanced section is a condition where steel reaches its permissible tensile stress and concrete reaches its permissible compressive stress simultaneously when subjected to bending.

In a bending section, the neutral axis is the horizontal axis where no tensile or compressive stress occurs.

  • Above this axis → compression zone

  • Below this axis → tension zone

Reinforced Concrete Structures Question 5:

Which of the following losses of pre-stress occurs only in pre-tensioning and not in post-tensioning?

  1. Elastic shortening of concrete
  2. Shrinkage of concrete
  3. Loss due to friction
  4. Creep of concrete

Answer (Detailed Solution Below)

Option 1 : Elastic shortening of concrete

Reinforced Concrete Structures Question 5 Detailed Solution

Explanation:

  • Elastic Shortening of Concrete occurs when the pre-stressing force is transferred to the concrete, causing the concrete to shorten elastically.

  • In pre-tensioning, tendons are tensioned before casting concrete, so when concrete is cast and the force transfers, elastic shortening happens immediately.

  • In post-tensioning, tendons are tensioned after concrete has hardened, so the elastic shortening loss in concrete is negligible or does not occur at the same stage.

  • Other losses like shrinkage, creep, and friction can occur in both systems (though friction loss specifically applies only to post-tensioning).

 Additional Information

  • Prestressed concrete involves introducing internal stresses in concrete using high-strength steel tendons to counteract tensile stresses during use.
  • It enhances the load-carrying capacity and reduces cracking by keeping concrete mainly under compression.
  • There are two main methods: pre-tensioning, where tendons are stressed before concrete casting, and post-tensioning, where tendons are stressed after the concrete hardens.
  • This technique is commonly used in bridges, beams, and slabs that require longer spans and higher strength.
  • Compared to ordinary reinforced concrete, prestressed concrete results in lighter, more durable structures with better control over deflections and cracks. 

Top Reinforced Concrete Structures MCQ Objective Questions

The height of a retaining wall is 5.5 m. It is to be designed as

  1. Cantilever type
  2. Counterfort type
  3. Cantilever or counterfort type
  4. None of the above

Answer (Detailed Solution Below)

Option 1 : Cantilever type

Reinforced Concrete Structures Question 6 Detailed Solution

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Explanation:

Retaining wall:

  • A retaining wall or retaining structure is used for maintaining the ground surfaces at different elevations on either side of it.
  • Whenever embankments are involved in construction, retaining walls usually necessary.

Types of retaining wall:

  • Depending upon the mechanisms used to carry the earth's pressure, These are classified into the following types.
  1. Gravity retaining wall.
  2. Cantilever retaining wall.
  3. Butters wall.

Gravity retaining wall:

  • It is not used for heights of more than 3.0 m.
  • In it, the resistance to the earth's pressure is generated by the weight of the structure.

Cantilever retaining wall:

  • It is the most common type of retaining wall and its height ranges up to 10-25 feet (3 to 8m).
  • Counterfort retaining walls are economical for height over about 6 m.
  • A cantilever retaining wall resists the earth pressure horizontal & another, by the cantilever bending action.

The main reinforcement of an RC slab consists of 10 mm bars at 10 cm spacing. If it is desired to replace 10 mm bars by 12 mm bars, then the spacing of 12 mm bars should be

  1. 10 cm
  2. 12 cm
  3. 14.40 cm
  4. 16 cm

Answer (Detailed Solution Below)

Option 3 : 14.40 cm

Reinforced Concrete Structures Question 7 Detailed Solution

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Concept:

Spacing between the bars for 1 m(1000 mm) width:

s=aAst×1000

Where, a = Area of one bar = π × ϕ24

ϕ = Diameter of bar

s = spacing of bars

Ast = Area of total main reinforcement

Calculations:

Given, 

Case 1: when ϕ = 10 mm then spacing(s1) = 10 cm = 100 mm

Case 2: when ϕ = 12 mm then spacing(s2) = ?

Case 1:

When the main reinforcement of an RC slab consists of 10 mm bars at 10 cm spacing, Ast will be

s1=aAst×1000

Ast=π×1024100×1000=785.4mm2

Case 2:

⇒ If 10 mm bars is to be replaced by 12 mm bars, then the spacing of 12 mm bars

As the Area of the main reinforcement will be the same so Ast = 785.4 mm2

s2=aAst×1000

s2=π×1224785.4×1000=14.40cm

Shortcut Trick

S2 =(d2d1)2×S1=(1210)2×10=14.4cm

In case of one-way continuous slab, maximum bending moment will be at:

  1. interior support other than next to end support
  2. mid of end span
  3. end support
  4. a support next to end support

Answer (Detailed Solution Below)

Option 4 : a support next to end support

Reinforced Concrete Structures Question 8 Detailed Solution

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One-way slab:

Design of one way RCC slab is similar to design of beam but the only difference is during the design of one way slab we take unit width of slab as a beam width.

Slabs are designed for bending and deflection and not designed for shear.

  1. Slabs have much small depth than beams.
  2. Most of slabs subjected to uniformly distributed loads.

Note:
In one way slab, the maximum bending moment at a support next to end support.

The recommended imposed load on staircase in residential buildings as per IS 875 is:

  1. 5.0 kN/m2
  2. 3.0 kN/m2
  3. 1.5 kN/m2
  4. 1.3 kN/m2

Answer (Detailed Solution Below)

Option 2 : 3.0 kN/m2

Reinforced Concrete Structures Question 9 Detailed Solution

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As per IS 875 Part 2, clause 3.1, Imposed Floor load for residential building are:

S.No

Residential Buildings (Dwelling houses)

U.D.L (kN/m2)

1.

All rooms and kitchens

2.0

2.

Toilet and Bath rooms

2.0

3.

Corridors, passages, staircases including tire escapes and store

rooms

3.0

4.

Balconies

3.0

The minimum eccentricity to be considered for an axially loaded RCC column of size 400 mm × 400 mm with unsupported length of 5 m is:

  1. 15.6 mm
  2. 20.5 mm
  3. 23.3 mm
  4. 30.6 mm

Answer (Detailed Solution Below)

Option 3 : 23.3 mm

Reinforced Concrete Structures Question 10 Detailed Solution

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Concept:

As per IS 456: 2000, clause 25.4,

Minimum Eccentricity

All columns shall be designed for minimum eccentricity, equal to the addition of the unsupported length of column divided by 500 and lateral dimensions divided by 30, subject to a minimum of 20 mm.

Where biaxial bending is considered, it is sufficient to ensure that eccentricity exceeds the minimum about one axis at a time.

Calculation:

Unsupported length = 5000 mm

Size of the column = 400 mm

Minimum eccentricity = L500+B30

emin=5000500+40030=23.33mm

But, in no case, the minimum eccentricity should be less than 20 mm.

In mix design for M25 concrete, the assumed standard deviation for estimation of target mean strength of concrete mix, as recommended by IS 456 ∶ 2000 is (in N/mm2):

  1. 4.5
  2. 4.0
  3. 5.0
  4. 3.5

Answer (Detailed Solution Below)

Option 2 : 4.0

Reinforced Concrete Structures Question 11 Detailed Solution

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The assumed value of standard deviation for initial calculations as per IS 456 – 2000 are given below in the table:

Sl. No

Grade of Concrete

Characteristic compressive strength (N/mm2)

Assumed standard deviation (N/mm2)

1.

M10

10

3.5

2.

M15

15

3.

M20

20

4.0

4.

M25

25

5.

M30

30

5.0

6.

M35

35

7.

M40

40

8.

M45

45

9.

M50

50

10.

M55

55

A reinforced concrete beam is subjected to the following bending moments.

Moment due to dead load = 50 kNm

Moment due to live load = 50 kNm

Moment due to seismic load = 20 kNm

The design bending moment for limit state of collapse is:

  1. 180 kNm
  2. 150 kNm
  3. 120 kNm
  4. 144 kNm

Answer (Detailed Solution Below)

Option 2 : 150 kNm

Reinforced Concrete Structures Question 12 Detailed Solution

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Given:

Moment due to Dead load (DL)= 50 kNm

Moment due to live load (LL) = 50 kNm

Moment due to Seismic load (EL) = 20 kNm

Computation:

Design bending moment is Maximum of the following

1) Mu considering moment due to dead load and live load

Mu = 1.5 (DL+LL) = 1.5 x (50+50) = 150 kN-m

2) Mu considering moment due to dead load and Seismic load

Mu = 1.5 (DL+EL) = 1.5 x (50+20) = 105 kN-m

3) Mu considering moment due to dead load, live load and seismic load together

Mu = 1.2 (DL+LL+EL) = 1.2 x (50+50+20) = 144 kN-m.

So the answer is Max of (150105, 144 kN-m) = 150 kN-m

The minimum stripping time of soffit formwork to beams (props to be refixed immediately after removal of formwork) is:

  1. 14 days
  2. 3 days
  3. 7 days
  4. 21 days

Answer (Detailed Solution Below)

Option 3 : 7 days

Reinforced Concrete Structures Question 13 Detailed Solution

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Explanation:

Type of formwork Minimum period before sticking formwork
Vertical formwork to columns, beams, and walls 16 - 24 hour
Soffit formwork to slabs (props to be refixed immediately after removal of formwork) 3 days
Soffit formwork to beams (props to be refixed immediately after removal of formwork) 7 days
Props to slab  
  • spanning up to 4.5 m
7 days
  • spanning over 4.5 m
14 days
Props to beams  
  • spanning up to 6 m
14 days
  • spanning over 6 m
21 days

For M20 Grade of concrete, modular ratio would be:

  1. 13.33
  2. 15.54
  3. 12.89
  4. 11.56

Answer (Detailed Solution Below)

Option 1 : 13.33

Reinforced Concrete Structures Question 14 Detailed Solution

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As per IS 456: 2000, ANNEX B

This value of the modular ratio partially takes into account the long-term effects of creep.

σcbc for M20 is 7 MPa and the modular ratio comes out to be 13.

The modular ratio is given by

m=2803σcbc

For M20 concrete

σcbc = 7 N/mm2

m=2803×7=13.33

Note: It is expected from students to know value of σcbc which is nearly 1/3rd of characteristics compressive strength. Please don't report questions for no data or wrong question.

Additional InformationThe permissible stresses under bending and direct compression as per IS 456 for different grades of concrete are given below in the tabulated form.

Grade of Concrete

Permissible Stress in Compression

Bending  σ cbc (N/mm2)

Direct σcc (N/mm2)

M15

5.0

4.0

M20

7.0

5.0

M25

8.5

6.0

M30

10.0

8.0

M35

11.5

9.0

M40

13.0

10.0

M45

14.5

11.0

A reinforced concrete beam, supported on columns at ends, has a clear span of 5 m and 0.5 m effective depth. It carries a total uniformly distributed load of 100 kN/m. The design shear force for the beam is

  1. 250 kN
  2. 200 kN
  3. 175 kN
  4. 150 kN

Answer (Detailed Solution Below)

Option 2 : 200 kN

Reinforced Concrete Structures Question 15 Detailed Solution

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Explanation:

Given,

Clear span = 5 m, Effective depth = 0.5 m
Uniformly distributed load(W) = 100 KN/m

The shear force of the beam in case of uniformly distributed load,

V = W × l2 = 100 × 52 = 250 KN

The location of critical section for shear design is determined based on the conditions at the supports. The location of critical shear is at a distance of effective depth d.

Design shear force for the beam:

Vu = 250 - 100 × 0.5 = 200 KN

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