Columns MCQ Quiz - Objective Question with Answer for Columns - Download Free PDF

Explanation:

Slenderness Ratio:

• Slenderness Ratio (λ) = Effective Length (L) / Radius of Gyration (r)

• It is used to classify columns:

  • Short Column: Low slenderness ratio (generally λ < 12)

  • Long Column: High slenderness ratio (λ > 12)

Failure Modes:

• Short columns primarily fail due to crushing (material failure).

• Long columns fail due to buckling (instability under axial load).

Concept:

The effective length of compression members is given by

Explanation:
As per IS 456:2000, clause 25.1.1
A column or strut is a compression member in which the effective length of the column or strut exceeds three times the least lateral dimension.

i.e. Leff > (3x Least Lateral Dimension)

A compression member may be considered as short when both the slenderness ratio Lex/D and Ley/b are less than 12.

where
Lex = effective length with respect to the major axis
D = depth with respect to the major axis
Ley = effective length with respect to the minor axis
b = width of the member

Explanation:

• According to IS 456:2000, the maximum compressive strain in concrete under axial loading at the limit state of collapse is taken as 0.002.
• This is a standard value used to define the ultimate strain in concrete at the failure point under axial loads.

 Additional Information

• Strain Compatibility: The value of 0.002 is used in conjunction with the stress-strain curve for concrete, particularly in the design of reinforced concrete beams and columns. This strain value helps in determining the ultimate strength of concrete elements, ensuring that structures can carry the required loads safely.

• Safety Considerations: By using this limit state of collapse strain, the design ensures that the concrete remains within the elastic limit for most of the service life, only reaching failure (at 0.002 strain) under extreme loading conditions.

Explanation:

As per IS 456:2000, clause 25.1.1

A column or strut is a compression member in which the effective length of the column or strut exceeds three times the least lateral dimension.

\(i.e. \;L_{eff}>(3\times Least \;Lateral\;Dimension)\)

A compression member may be considered as short when both the slenderness ratio \(L_{ex}\over D\) and \(L_{ey}\over b\) are less than 12.

where

Lex = effective length with respect to the major axis

D = depth with respect to the major axis

Ley = effective length with respect to the minor axis

b = width of the member

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.

Explanation:

Longitudinal Reinforcement:

(i) CI. 26.5.3 of IS 456:2000, specifies that the total area of longitudinal bars in a column section must NOT be less than 0.8% of the gross column area. This limit on minimum reinforcement is imposed because of the following reasons:

• In order to ensure that a minimum flexural resistance of the column exists due to unexpected eccentricities in the column loading.
• In compression members, creep under sustained loading is very predominant, especially at low percentages of steel. Thus, the resulting creep stress (due to creep strain) tries to yield the bars. 

(ii) Maximum Reinforcement: The maximum area of cross-section of longitudinal bars must NOT exceed 6% of gross column area. However, in practice, a maximum of 4% is recommended.

Additional Information

Diameter and Number of bar:

(i) The diameter of longitudinal bars in column NOT be less than 12 mm. These bars must NOT be spaced more than 300 mm apart on the column perimeter.

(ii) For rectangular columns. a minimum of 4 bars is provided.

(iii) For Circular columns, a minimum of 6 bars be provided.

Cover = 40 mm or bar diameter

The permissible design stresses for an RCC Column under different conditions are given below:

1. If minimum eccentricity is considered to be Zero i.e. e_min = 0

a. Permissible axial compressive stress in concrete = 0.45 f_ck

b. Permissible axial compressive stress in steel = 0.75 f_y­

2. If minimum eccentricity effect is considered: 

a. Permissible axial compressive stress in concrete = 0.4 f_ck

b. Permissible axial compressive stress in steel = 0.67 f_y­

Note:

In RCC beam subjected to bending, shear and torsion, the permissible values in steel and concrete are:

a. Design strength of concrete in flexure = 0.45 f_ck

b. Design strength of Steel in tension = 0.87f_y

Concept:

As per IS 456:2000, clause 26.5.3.2 (c), 

Pitch:

The pitch of transverse reinforcement shall be not more than the least of the following distances:

1. the least lateral dimension of the compression members
2. 16 times the smallest diameter of the longitudinal reinforcement bar to be tied (i.e. 16 φmin)
3. 300 mm

The diameter of the Lateral Ties: 

The diameter of the lateral ties shall be not less than one-fourth of the diameter of the largest longitudinal bar and in no case less than 6 mm.

Calculations:

The pitch of transverse reinforcement shall be

1. The lest lateral dimension = 400 mm
2. 16 × 20 mm = 320 mm
3. 300 mm

whichever is smaller

Hence, the pitch of lateral ties shall be kept as 300 mm.

Explanation:

As per IS 456:2000, clause 39.4,

Compression Member with Helical Reinforcement

\(Pitch < \begin{Bmatrix} \frac{Core diameter}{6}\\75 mm\end{Bmatrix} \)

In the given question, the core diameter is not given, so we have to take 75 mm only

Important Points

The strength of compression members with helical reinforcement shall be taken as 1.05 times the strength of similar members with lateral ties provided the condition

 \({Volume\; of\; helical\; reinforcement\over Volume\; of\; the\; core}\nless 0.36\left( {\frac{{{A_g}}}{{{A_c}}} - 1} \right)\frac{{{f_{ck}}}}{{{f_y}}}\;\)should be satisfied.

where

Ag = gross area of the section

Ac = area of the core of the helically reinforced column measured to the outside diameter of the helix

fck = characteristic compressive strength of the concrete

fy = characteristic strength of the helical reinforcement but not exceeding 415 N/mm2

Hence, the load-carrying capacity of the column with helical reinforcement is increased by 5% due to increased strength as a result of better confinement.

Explanation:

As per IS 456:2000, clause 25.1.1,

A column or struts is a compression member in which the effective length of column or struts exceeds three times the least lateral dimension.

i.e. L_eff > (3 × Least Lateral Dimension) 

A compression member may be considered as short when both the slenderness ratio \(L_{ex}\over D\) and \(L_{ey}\over b\) are less than 12.

where

Lex = effective length with respect to the major axis

D = depth with respect to the major axis

Ley = effective length with respect to the minor axis

b = width of the member

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:

Given,

Unsupported length = 3500 mm

Size of the column = 350 mm

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

\(e_{min}= \;\frac{{3500}}{{500}} + \frac{{350}}{{30}} = 18.667\;mm\;\)

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

as e _min < 20 mm.

The Minimum eccentricity = 20 mm

Concept:

In the design of long columns considering the factor of buckling, lower value of working stresses in steel and concrete is adopted, by multiplying the general working stresses by the reduction coefficient.

 \(C_r= 1.25-\frac{{{l_{eff}}}}{{48D}} \ or\ 1.25-\frac{{{l_{eff}}}}{{48B}}\) 

where,

C_r = Reduction coefficient

B and D = Least lateral dimension of the column

l_eff = Effective Length of the column

Calculation:

Given,

l_eff = 4.5 m

B = 200 mm

D = 250 mm

C_r = 1.25 - 4500/(48 x 200) = 0.78  or 1.25 - 4500/(48 x 250) = 0.875

Lesser value adopted, so Cr = 0.78

• If Effective Length / Least Lateral Dimension > 12, Long column otherwise short.

Concepts:

As per IS 456: 2000, clause 25.4, all columns shall be designed for minimum eccentricity, equal to the summation of the unsupported length of column divided by 500 and lateral dimensions divided by 30, subject to a minimum of 20 mm. i.e.

 The eccentricity is the maximum of the following:

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

→  e = 20 mm
Where L is the effective length of the column and B is the lateral dimension.

Further, where biaxial bending is considered, it is sufficient to ensure that eccentricity exceeds the minimum about one axis at a time.

Calculation:

Given:

Effective Length of Column, L = 3.5 m

Lateral Dimension, B = 300 mm

Minimum eccentricity, e is calculated as:

\(e = \frac{3500}{500} + \frac{300}{30}\)

e = 17 mm < e_min = 20 mm

Therefore, the minimum eccentricity for an axially loaded column should be 20 mm.

Concept: 

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.

The minimum number of longitudinal bars:

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

• four in rectangular columns
• six in circular columns

Helical reinforcement:

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

Spacing:

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

Calculation:

Given data,

Size of RCC column = 200 mm x 400 mm

Hence, The minimum area of longitudinal reinforcement in an RCC column is 

= 0.008 × 200 × 400 = 640 mm²