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

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.

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 

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.

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:

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 

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.

Concept:

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

A_st = 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, A_st 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 A_st = 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

\(S_2 \ ={\left( {\frac{{{d_2}}}{{{d_1}}}} \right)^2} \times {S_1} = {\left( {\frac{{12}}{{10}}} \right)^2} \times 10 = 14.4\;cm\)

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.

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

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 = \(\frac{L}{{500}} + \frac{B}{{30}} \)

\(e_{min}= \;\frac{{5000}}{{500}} + \frac{{400}}{{30}} = 23.33\;mm\;\)

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

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

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.

Explanation:

As 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.

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 = \(\frac{W\ × \ l}{2}\) = \(\frac{100\ × \ 5}{2}\) = 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:

V_u = 250 - 100 × 0.5 = 200 KN

Concept:

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