Reinforced Concrete Structures MCQ Quiz - Objective Question with Answer for Reinforced Concrete Structures - Download Free PDF
Last updated on May 27, 2025
Latest Reinforced Concrete Structures MCQ Objective Questions
Reinforced Concrete Structures Question 1:
For design of concrete structures, failure strain of concrete under direct compression and flexure respectively is ______ and ______.
Answer (Detailed Solution Below)
Reinforced Concrete Structures Question 1 Detailed Solution
Explanation:
Following assumptions have been made in IS 456: 200 regarding maximum strain in compression members:
1. As per, Clause 38.1, the maximum compressive strain in concrete in bending compression is taken as 0.0035.
2. As per, clause 39.1, the maximum compressive strain in concrete in axial compression is taken as 0.002.
Important Points
The maximum compressive strain at the highly compressed extreme fibre in concrete subjected to axial compression and bending and when there is no tension on the section shall be 0.0035 minus 0.75 times the strain at the least compressed extreme fibre.
Reinforced Concrete Structures Question 2:
Which of the following square slab will behave as one-way slab ?
Answer (Detailed Solution Below)
Reinforced Concrete Structures Question 2 Detailed Solution
Concept:
One-way slab:
(i) If a slab is supported only on two opposite supports, it is called a one-way slab.
(ii) If the slab is opposite on all four sides and span ratio 'ly/lx > 2', it is called a one-way slab.
(iii) Main reinforcement is provided along a shorter span.
(iv) Distribution reinforcement is provided along a longer span.
(v) maximum moment in the slab is along a shorter span direction.
(vi) When simply supported along two opposite edges square slab will behave as the one-way slab.
The maximum moment for the simply supported slab is given by,
\({M_x} = \frac{{wl_x^2}}{8}\)
Where lx = shorter span length
Reinforced Concrete Structures Question 3:
For reinforced brick work the modular ratio is generally taken as -
Answer (Detailed Solution Below)
Reinforced Concrete Structures Question 3 Detailed Solution
Concept:
Reinforced brickwork
- It is a composite structural material consisting of load-bearing brickwork masonry into which lengths of a suitable metal (normally steel) are introduced and so bonded as to render the resultant composite capable of resisting not only the compressive stresses but also the tensile and shear stresses which obtain in a structure.
- The reinforcing bars are embedded in the gap between the bricks, which is filled with cement mortar.
- For reinforced brickwork, the modular ratio is generally taken as 40.
Reinforced Concrete Structures Question 4:
For mild exposure, for main reinforcement steel bars up to 12 mm diameter, the nominal cover may be reduced by:
Answer (Detailed Solution Below)
Reinforced Concrete Structures Question 4 Detailed Solution
Explanation:
As per IS 456 (2000), Table 16, For main reinforcement, up to 12 mm diameter bar for mild exposure the nominal cover may be reduced by 5 mm.
Additional Information
Nominal Cover to Meet Durability Requirements:
Exposure |
Nominal concrete cover (in mm) not less than |
Mild |
20 |
Moderate |
30 |
Severe |
45 |
Very Severe |
50 |
Extreme |
75 |
Nominal Cover to Meet Fire Resistance:
The nominal cover is the design depth of concrete cover to all steel reinforcements.
It shall be not less than the diameter of the bar.
Minimum values of the nominal cover of normal-weight aggregate concrete to be provided to all reinforcement to meet the specified period of fire resistance shall be given in the table below
Member |
Minimum Cover |
Slab |
20 mm |
Beam |
20 mm |
Column |
40 mm for d > 12 mm 25 mm for d < 12 mm |
Footing |
50 mm |
Reinforced Concrete Structures Question 5:
The IS code that deals with modular coordination in prefabricated buildings is:
Answer (Detailed Solution Below)
Reinforced Concrete Structures Question 5 Detailed Solution
Explanation:
IS 15916 deals with modular coordination in prefabricated buildings. It provides guidelines for the dimensional coordination of building components such as walls, windows, doors, and other prefabricated elements. The purpose of modular coordination is to standardize the dimensions of building components to improve the ease and efficiency of construction, especially when using prefabricated materials.
Key aspects covered by IS 15916:
-
Standardized dimensions for building components.
-
Coordination between different components to ensure ease of assembly and minimize errors in prefabricated construction.
-
Design guidelines for maintaining uniformity across various elements in a modular building system.
Additional Information
-
IS 456: This is the Indian Standard for Plain and Reinforced Concrete, focusing on the design and construction of concrete structures, not specifically for modular coordination.
-
IS 875: This code deals with Design Loads (Other than Earthquake) for buildings and structures, covering wind loads, dead loads, live loads, etc.
-
IS 4926: This code covers the Production and Supply of Ready-Mixed Concrete, specifying the requirements for ready-mixed concrete production and delivery, not modular coordination.
Top Reinforced Concrete Structures MCQ Objective Questions
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
Answer (Detailed Solution Below)
Reinforced Concrete Structures Question 6 Detailed Solution
Download Solution PDFConcept:
Spacing between the bars for 1 m(1000 mm) width:
\({s} = \frac{a}{{{A_{st}}}} \times 1000\)
Where, a = Area of one bar = \(\frac{\pi \ \times \ ϕ^2}{4}\)
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
\({s_1} = \frac{a}{{{A_{st}}}} \times 1000\)
\(\Rightarrow {A_{st}} = \frac{{\frac{{\pi \times 10^2 }}{4}}}{{100}} \times 1000 = 785.4\;m{m^2}\)
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
\({s_2} = \frac{a}{{{A_{st}}}} \times 1000\)
\(\Rightarrow \;{s_2} = \frac{{\frac{{\pi \times 12^2}}{4}}}{{785.4}} \times 1000 = 14.40\;cm\)
Shortcut Trick
\(S_2 \ ={\left( {\frac{{{d_2}}}{{{d_1}}}} \right)^2} \times {S_1} = {\left( {\frac{{12}}{{10}}} \right)^2} \times 10 = 14.4\;cm\)
In case of one-way continuous slab, maximum bending moment will be at:
Answer (Detailed Solution Below)
Reinforced Concrete Structures Question 7 Detailed Solution
Download Solution PDFOne-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.
- Slabs have much small depth than beams.
- 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:
Answer (Detailed Solution Below)
Reinforced Concrete Structures Question 8 Detailed Solution
Download Solution PDFAs 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:
Answer (Detailed Solution Below)
Reinforced Concrete Structures Question 9 Detailed Solution
Download Solution PDFConcept:
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 = \(\frac{L}{{500}} + \frac{B}{{30}} \)
\(e_{min}= \;\frac{{5000}}{{500}} + \frac{{400}}{{30}} = 23.33\;mm\;\)
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):
Answer (Detailed Solution Below)
Reinforced Concrete Structures Question 10 Detailed Solution
Download Solution PDFThe 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:
Answer (Detailed Solution Below)
Reinforced Concrete Structures Question 11 Detailed Solution
Download Solution PDFGiven:
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 (150, 105, 144 kN-m) = 150 kN-mThe minimum stripping time of soffit formwork to beams (props to be refixed immediately after removal of formwork) is:
Answer (Detailed Solution Below)
Reinforced Concrete Structures Question 12 Detailed Solution
Download Solution PDFExplanation:
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 | |
|
7 days |
|
14 days |
Props to beams | |
|
14 days |
|
21 days |
For M20 Grade of concrete, modular ratio would be:
Answer (Detailed Solution Below)
Reinforced Concrete Structures Question 13 Detailed Solution
Download Solution PDFAs per IS 456: 2000, ANNEX B
This value of the modular ratio partially takes into account the long-term effects of creep.
\({\sigma _{cbc}}\) for M20 is 7 MPa and the modular ratio comes out to be 13.
The modular ratio is given by
\(m = \frac{{280}}{{3{\sigma _{cbc}}}}\)
For M20 concrete
σcbc = 7 N/mm2
\(\therefore m = \frac{{280}}{{3 \times 7}} = 13.33\)
Note: It is expected from students to know value of \({\sigma _{cbc}}\) 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
Answer (Detailed Solution Below)
Reinforced Concrete Structures Question 14 Detailed Solution
Download Solution PDFExplanation:
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 = \(\frac{W\ × \ l}{2}\) = \(\frac{100\ × \ 5}{2}\) = 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
If design bond stress = 1.5 N/mm2 is assumed, then the development length of an Fe 500 HYSD bar of nominal diameter 12 mm - which is fully stressed in tension - will be:
Answer (Detailed Solution Below)
Reinforced Concrete Structures Question 15 Detailed Solution
Download Solution PDFConcept:
Development length:
(i) The cal culated tension or compression in any bar at any section shall be developed on each side of the section by an appropriate development length or end anchorage or a combination thereof.
(ii) Development length can be calculated as:
\({L_d} = \frac{{ϕ × 0.87{f_y}}}{{4{τ _{bd}}}}\)
Where, ϕ = Diameter of bar
τbd = Design bond stress = Permissible value of average bond stress
The value of bond stress is increased by 60% for a deformed bar in tension and a further increase of 25% is made for bars in compression.
Calculation:
Given,
ϕ = 12 mm
τbd = 1.5 N/mm2
So, τbd = 1.5 × 1.6 N/mm2
Development length, \({L_d} = \frac{{ϕ × 0.87{f_y}}}{{4{τ _{bd}}}}\)
\({L_d} = \frac{{12 × 0.87 × 500}}{{4×1.6 × 1.5}}\) = 543.75 mm ≈ 544 mm