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

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. 

Explanation:

• A slab is considered one-way slab if the longer span is more than twice the shorter span (i.e., ratio of long span to short span > 2).

• In one-way slabs, loads are primarily carried by the shorter span, and reinforcement is provided mainly in that direction.

• If the ratio is less than or equal to 2, the slab behaves more like a two-way slab, distributing loads in both directions.

• This classification helps in deciding the design approach and reinforcement detailing for slabs.

Additional InformationOne-Way Slab

• Load is primarily carried in one direction (shorter span).

• Occurs when the ratio of long span to short span is greater than 2.

• Reinforcement is provided mainly in the short span direction.

• Supported on two opposite sides only (usually beams on two sides).

• Simpler design and less reinforcement compared to two-way slabs.

Two-Way Slab

• Load is carried in both directions (long and short spans).

• Happens when the ratio of long span to short span is less than or equal to 2.

• Reinforcement is provided in both directions.

• Supported on all four sides (beams or walls).

• More complex design, with bending moments distributed in two directions.

Explanation:

• As per IS 456:2000, the minimum concrete cover to the ties or spirals in reinforced concrete columns should not be less than 25 mm.

• This cover protects the reinforcement against corrosion and fire and ensures proper bond between concrete and steel.

 Additional InformationMinimum Cover to Ties or Spirals — IS 456:2000

• Concrete cover protects the reinforcement against corrosion, fire, and mechanical damage.

• It ensures bond strength between concrete and steel for proper composite action.

• Proper cover improves durability and longevity of the structure.

• Clause 26.4.2 of IS 456:2000 specifies minimum covers for reinforcement in different exposure conditions and types of elements.

• For tied columns and spirally reinforced columns, the minimum cover to ties or spirals should not be less than 25 mm.

• This applies for normal exposure conditions (not severe or aggressive environments).

Explanation:

• According to Whitney’s stress block theory and IS 456, the depth of the neutral axis $x_{u, \text{max}}$ for a balanced section (where steel yields just as concrete reaches its ultimate strain) is limited to about 0.53 to 0.537 times the effective depth $d$.

• This ensures the maximum permissible depth of the stress block for safe design before concrete crushing dominates.

Additional Information

• Whitney simplified the actual nonlinear stress distribution in concrete to an equivalent rectangular block.

• The factor 0.53d is key in limiting the depth of compression zone in design calculations.

• It helps distinguish between under-reinforced, balanced, and over-reinforced sections in concrete beam design.

Explanation:

• In pre-stressed concrete, the concrete is pre-compressed by applying a compressive force using tendons or cables.

• However, concrete is weak in tension, so reinforcement (steel tendons or wires) is provided to resist tensile stresses that occur when the structure is loaded.

• The reinforcement carries tensile forces and prevents cracking in the concrete.

• The initial compressive stress is imparted by the pre-stressing process itself (not the reinforcement alone).

• Bond stress develops between concrete and steel to transfer forces, but the primary purpose of reinforcement is to resist tension.

Additional Information

• Pre-stressed concrete combines concrete's strength in compression with steel's strength in tension.

• Reinforcement ensures durability and serviceability by controlling cracks.

• It allows longer spans and thinner sections compared to conventional reinforced concrete.

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