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

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.

Central dip is given by: \(e=\frac{WL^2}{8P}\)

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

Calculation:

Given:

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

\(e=\frac{40 \times10^2}{8\times2500}=0.2m=200mm\)

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).

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.

Concept:

To determine the equivalent bending moment for the RC beam, we can use the following formula: \(M_{eq}=M+\frac{T(1+\frac{D}{b})}{1.7}\)

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

\(M_{eq}=115+\frac{45\times (1+\frac{0.6}{0.3})}{1.7}\)

\(M_{eq}=115+79.41=194.41kN-m\)

Explanation:

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

• 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 σ_st

\(k=\frac{280}{3\sigma_{st}+280}\)

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

Explanation:

Doubly reinforced beam

• A doubly reinforced beam has steel reinforcement in both tension and compression zones, typically at the bottom and top respectively.

• It is adopted when the beam section is restricted in size and cannot accommodate all the tension with steel at the bottom alone.

• Also used when moment reversal is expected (like in earthquake-prone zones or multi-storey frames) where both faces of the beam undergo tension alternatively.

• Compression reinforcement enhances ductility, resists reversal of stresses, and controls long-term deflection and cracking.

 Additional InformationSingly Reinforced Beam 

• A singly reinforced beam has tension reinforcement only, placed in the bottom zone (in simply supported beams), where tensile stresses develop.

• These beams are used where bending moment causes tension in only one zone, typically in cases of simply supported or cantilever beams.

• Concrete resists compression, and steel resists tension; hence only the tension side is reinforced in this design.

• It is economical and sufficient when the depth of the beam is adequate to resist the moment using only tension steel.

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.

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.

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