Columns MCQ Quiz in मराठी - Objective Question with Answer for Columns - मोफत PDF डाउनलोड करा

Explanation:

Minimum eccentricity to be considered along a lateral dimension as per Clause 25.4 of IS 456:2000 is equal to the unsupported length of columns/500 plus lateral dimension/30, subject to a minimum of 20 mm.

Minimum eccentricities:

\({e_{\;x\;min}} = {\rm{max}}\left\{ {\begin{array}{*{20}{c}} {\frac{{{\ell _x}}}{{500}} + \frac{D}{{30}}}\\ {20\;mm} \end{array}} \right.\)

\({e_{y\;min}} = {\rm{max}}\left\{ {\begin{array}{*{20}{c}} {\frac{{{\ell _y}}}{{500}} + \frac{b}{{30}}}\\ {20\;mm} \end{array}} \right.\)

ℓx, ℓy = Unsupported length of column

Explanation:

Given data:

Width of square column (b) = 500 mm

Service load (P) = 2000 kN

Ultimate load (P_u) =?

The minimum area of longitudinal reinforcement (Ast_min) = ?

As per IS 456:2000 for the short axially loaded columns -

The minimum area of longitudinal reinforcement or steel (Ast_min)

= 0.8% of the gross area of the column

The minimum area of longitudinal reinforcement or steel (Ast_min)

= \({0.8\over 100}× 500^2 \)

The minimum area of longitudinal reinforcement or steel (Ast_min)

= 2000 mm²

\(Ultimate\, load\,(P_u)=FOS× Service\, load\,(P)\)

Where FOS = Factor of safety = 1.5

Ultimate load (P_u) = 1.5 × 2000 = 3000 kN

Ultimate load (Pu) = 3000 kN

Concepts:

The ultimate load-carrying capacity of short column is given by:

P_u = 0.40f_ckA_c+0.67f_yA_sc

Where

A_c= Gross area of cross-section (A_g) – Steel area (A_sc)

f_ck = Characteristic compressive strength of concrete

f_y = Yield Strength of Steel

Calculation:

Given: 

Size: 300 × 300 mm; 4 bars of dia 20 mm; Fe-415 and M20

For Fe 415 Grade; f_y = 415 MPa

For M20 Grade: f_ck = 20 MPa

A_sc=4 × π/4 × 20 × 20=1256.64 mm²

 A_g = 300 × 300 = 90000 mm²

A_c = A_g - A_sc

A_c = 90000 – 1256.64 = 88743.36 mm²

The axial load carrying capacity of column is given as:

P_u = 0.40 × 20 × 88743.36 + 0.67 × 415 × 1256.64

P_u = 1059.56 kN

Explanation:

As per IS 456:2000, the design of transverse reinforcement is as:

The diameter of transverse reinforcement:

i) \(\rm{\frac{1}{4}{\phi _{main}}}\)

ii) 6 mm

whichever is maximum,

The spacing of Transverse reinforcement:

i) Least lateral dimension

ii) 16 times diameter of main reinforcement bar

iii) 300 mm

Concept:

According to Clause 26.5.3 of IS 456: 2000 the cross-sectional area of longitudinal reinforcement in a column shall not be less than 0.8 % nor more than 6 % of the gross c/s area of the column.

Calculation:

Given: Gross c/s area = 300 × 300 mm² = 90000 mm2 

Minimum reinforcement = \(90000 \times \dfrac{0.8}{100} \) = 720 mm2 

Maximum reinforcement = \(90000 \times \dfrac{6}{100}\) = 5400 mm2 
Important Points

•  Maximum area of tension reinforcement = 0.04 bD [b = width of the beam, D = Depth of the beam]
• The minimum area of tension reinforcement shall not be less than = \(\dfrac{0.85 \times b D}{f_y}\) [f_y = characteristic tensile strength of reinforcement]
• Side face reinforcement shall not be less than 0.1% of the web area for deep beams

Concept:

The column is classified based on its slenderness ratio

Slenderness ratio, \( λ= \frac{{{{\text{L}}_{{\text{eff}}}}}}{{{\text{Least lateral dimension}}}}\)

λ ≤ 3  for Pedestal

3 < λ < 12 for short column

12 ≤ λ for long column

Calculation:

Unsupported length, l_o = 4m

Since RCC column is rigidly held on both sides, so it behaves like a fixed column:

Effective length, l_eff = 0.65 l_o

\({l_{eff}} = 0.65 \times 4 = 2.6\;m\)

Slenderness Ratio, λ = l_eff/least lateral dimension

\( λ = \frac{{2600}}{{400}} = 6.5\)

∴ Slenderness ratio (λ) < 12, so it behaves as a short RCC column.

Explanation:

Transverse reinforcing bars are provided in forms of circular rings, polygonal links (lateral ties) with internal angles not exceeding 135^o or helical reinforcement.

The transverse reinforcing bars are provided to ensure that every longitudinal bar nearest to the compression face has effective lateral support against buckling. Clause 26.5.3.2 stipulates the guidelines of the arrangement of transverse reinforcement. The salient points are: 

Transverse reinforcement shall only go round corner and alternate bars if the longitudinal bars are not spaced more than 75 mm on either side 

The maximum number of sides this polygonal ring can have is 8 

As per IS 456: 2000, clause 26.5.3.1,

Longitudinal reinforcement in a column:

Percentage of steel:

The cross-sectional area of longitudinal reinforcement shall be not less than 0.8 percent nor more than 6 percent of the gross cross-sectional area of the column.

Minimum diameter of longitudinal reinforcement bars:

The main longitudinal reinforcement bars used in the column shall not be less than 12 mm in diameter.

Minimum number of longitudinal bar:

The minimum number of the longitudinal bar provided in column shall be

a) Four in rectangular columns

b) Six in circular columns

Helical reinforcement:

A reinforced concrete column having helical reinforcement shall have at least six-bar of longitudinal reinforcement within the helical reinforcement.

Spacing:

The spacing of longitudinal bars measured along the periphery of the column shall not exceed 300 mm.

Concept:

The effective length of compression members is given by

Key Points

Every condition has fixed support at one end except the fifth conditions

1. Fixed - Fixed = 0.5,0.65 (Theoretical, Recommended)
2. Fixed - hinged = 0.7.0.8 (Theoretical, Recommended)
3. Fixed - Guided = 1
4. Fixed - Free = 2
5. Hinged - Guided = 2

Concept:

As per IS 456: 2000, Table 28

The recommended value of the effective length of the column effectively held in position and restrained in directions (or against rotation) at both ends is 0.65 times the length of the column.

Calculation:

Given data:

Rectangular column size = 300 mm × 300 mm

Length of column (L) = 5 m

The effective length of a given column (L_eff) = 0.65 × L

The effective length of a given column (Leff) = 0.65 × 5 = 3.25 m

The effective length of a given column (Leff) = 3.25 m

The effective length of the column is 3.25 m.