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

Last updated on Mar 13, 2025

पाईये Column Base उत्तरे आणि तपशीलवार उपायांसह एकाधिक निवड प्रश्न (MCQ क्विझ). हे मोफत डाउनलोड करा Column Base एमसीक्यू क्विझ पीडीएफ आणि बँकिंग, एसएससी, रेल्वे, यूपीएससी, स्टेट पीएससी यासारख्या तुमच्या आगामी परीक्षांची तयारी करा.

Latest Column Base MCQ Objective Questions

Top Column Base MCQ Objective Questions

Column Base Question 1:

A steel column is restrained against both translation and rotation at one end and is restrained only against translation but free to rotate at the other end. Design (IS: 800 – 2007) values, of effective length factor of the column, are 

  1. 2.0
  2. 1.2
  3. 1.0
  4. 0.8

Answer (Detailed Solution Below)

Option 4 : 0.8

Column Base Question 1 Detailed Solution

As per IS: 800 – 2007

For column, restrained against both rotation and translation at one end and restrained only against translation but free to rotate at other ends

Design value of effective length factor = 0.8

The design value of the effective length factor for various combination is given below:

GATE CE 2019 Shift 2  images madhushri Q16

 

Column Base Question 2:

The thickness of the base plate is determined from the 

  1. Flexural strength of the plate
  2. Shear strength of the plate
  3. Bearing strength of the concrete
  4. Punching criteria

Answer (Detailed Solution Below)

Option 1 : Flexural strength of the plate

Column Base Question 2 Detailed Solution

It is assumed that the maximum bending moment occurs at the edge of the column. As a slab tends to bend simultaneously about the two principal axes, the stress caused by bending about one axis is influenced by stress due to bending about other axes. The Poisson ratio of 0.3 is used to account for this effect.

\({M_{x,net}} = \frac{{w{a^2}}}{2} - 0.3\frac{{w{b^2}}}{2} = \frac{w}{2}\left( {{a^2} - 0.3{b^2}} \right)\)      ----- (1)

Moment capacity of plate = 1.2 fyZe

\(= 1.2{f_y} \times 1 \times \frac{{t_s^2}}{6}\)  ---- (2)

Equating (1) and (2)

\(\frac{{1.2{f_y}t_s^2}}{6} = \frac{w}{2}\left( {{a^2} - 0.3{b^2}} \right)\)

Applying partial factor of safety = γm0

\(\begin{array}{l} \frac{{1.2{f_y}t_s^2}}{{6{\gamma _{mo}}}} = \frac{w}{2}\left( {{a^2} - 0.3{b^2}} \right)\\ {t_s} = \sqrt {\frac{{2.5w\left( {{a^2} - 0.3{b^2}} \right){\gamma _{mo}}}}{{{f_y}}}} \end{array}\)

Column Base Question 3:

To keep the intensity of bearing pressure between the column base and concrete compressive and to vary from zero to 2P/BL, the ratio of the moment M to the axial load P should be

  1. L/2
  2. L/3
  3. L/4
  4. L/6

Answer (Detailed Solution Below)

Option 4 : L/6

Column Base Question 3 Detailed Solution

Explanation:

For the purpose of determining the pressure under the footing, the moment M may be removed by shifting the vertical load P by the distance,\(e = \frac{M}{P}\).

To keep the intensity of bearing pressure between the column base and concrete compressive, the eccentric load could be within the middle third.

The pressure under footing,

 \(q = \frac{P}{A} \pm \frac{{M\frac{L}{2}}}{I}\) 

where,  \(I = \frac{{B{L^3}}}{{12}}\)

\(q = \frac{P}{{BL}}\left( {1 \pm \frac{{6e}}{L}} \right)\)

For compressive stress, maximum pressure = minimum pressure

\(\frac{P}{{BL}}\left( {1 + \frac{{6e}}{L}} \right)\) = \(\frac{P}{{BL}}\left( {1 - \frac{{6e}}{L}} \right)\)
 
On solving, e = L / 6

Column Base Question 4:

Which the following graphs represent the allowable compressive stress (σac­) versus slenderness ratio (KL/r)

  1. RRB JE CE R 4 15Q Structure Test Part1(Hindi) - Final images Q6
  2. RRB JE CE R 4 15Q Structure Test Part1(Hindi) - Final images Q6a
  3. RRB JE CE R 4 15Q Structure Test Part1(Hindi) - Final images Q6b
  4. RRB JE CE R 4 15Q Structure Test Part1(Hindi) - Final images Q6c

Answer (Detailed Solution Below)

Option 4 : RRB JE CE R 4 15Q Structure Test Part1(Hindi) - Final images Q6c

Column Base Question 4 Detailed Solution

Allowable axial compressive stress

\({\sigma _{cc}} \propto {f_{cc}} \propto \frac{1}{{{{\left( {\frac{{KL}}{r}} \right)}^2}}}\;,\;\because\;{f_{cc}} = \frac{{{\pi ^2}EI}}{{{{\left( {\frac{{KL}}{r}} \right)}^2}}}\)

The allowable compressive stress is inversely proportional to the square of slenderness ratio.

RRB JE CE R 4 15Q Structure Test Part1(Hindi) - Final images Q6d

This shows a linear relationship between compressive strength and slenderness ratio which is wrong.

RRB JE CE R 4 15Q Structure Test Part1(Hindi) - Final images Q6e

This also shows a linear relationship between compressive strength and slenderness ratio which is wrong.

RRB JE CE R 4 15Q Structure Test Part1(Hindi) - Final images Q6f

This shows a finite compressive strength for zero slenderness ratio value, practically zero value of slenderness ratio is not possible and even if it’s possible, compressive strength should be infinite.

RRB JE CE R 4 15Q Structure Test Part1(Hindi) - Final images Q6g

∴ This option is correct as it satisfies relationship as well as boundary conditions.

Column Base Question 5:

The lacings are subjected to resist a transverse shear of

  1. 2.8% of axial force in the member
  2. 2.75% of axial force in the member
  3. 2.5% of axial force in the member
  4. 2.25% of axial force in the member

Answer (Detailed Solution Below)

Option 3 : 2.5% of axial force in the member

Column Base Question 5 Detailed Solution

The lacing shall be proportioned to resist a total transverse shear, (Vt), at any point in the member, equals to at least 2.5 per cent of the axial force in the member and shall be divided equally among all transverse lacing systems in parallel planes.

For a single lacing system on two parallel faces, the force (compressive or tensile) in each bar,

\(F = \frac{V}{{2\sin \theta }}\)

For double lacing system on two parallel planes, the force (compressive or tensile) in each bar,

\(F = \frac{V}{{4\sin \theta }}\)

Column Base Question 6:

The most economical section for a column is

  1. rectangular
  2. solid round
  3. flat strip
  4. tubular section

Answer (Detailed Solution Below)

Option 4 : tubular section

Column Base Question 6 Detailed Solution

Tubular section is best suited for the design of smaller compression member subjected to smaller load.

The various reasons are as follows:

1. Round tubes have the same radius of gyration in all directions and have a high local buckling strength. These are usually very economical unless moments are too large for the sizes available.

2. Tube has excellent torsional resistance.

3. In case of members subjected to wind, round tubes are subjected to less force than flat sections.

4. They have less surface area to paint or fireproof.

5. Tubes do not have the problem of dirt collection or cleaning.

6. The weight of tube sections usually are less than one half the weight required for open profile sections. Although tube sections cost about 25 per cent more than open sections, but about 20 per cent cost savings can still be achieved.

Column Base Question 7:

The design compressive stress of an axially loaded compression member in IS: 800 – 2007 is given by

  1. Rankine formula
  2. Secant formula
  3. Merchant – Rankine formula
  4. Perry Robertson formula

Answer (Detailed Solution Below)

Option 4 : Perry Robertson formula

Column Base Question 7 Detailed Solution

The design compressive stress of axially loaded compression member in IS 800: 2007 is given by Perry-Robertson formula.

IS 800:2007 proposes multiple columns curves in non-dimensional form based on Perry-Robertson approach.

Also pay attention that:

IS 800: 1984 recommended the use of Merchant-Rankine formula.

Column Base Question 8:

The fasteners in gusseted base plate are designed for 

  1. 25% column load
  2. 50% column load
  3. 75% column load
  4. 100% column load

Answer (Detailed Solution Below)

Option 2 : 50% column load

Column Base Question 8 Detailed Solution

Explanation

If the column ends and gusset materials are not faced/machined for complete bearing, the fasteners are designed for complete bearing, the fasteners are designed for the total forces to be transferred. If they are faced/machined for complete bearing, 50 % of the forces are transferred directly by the column and 50 % through the fasteners.

Column bases are structural elements used in the design of steel structures to transfer the column load to the footings.

Types of Column bases

Slab base

F4 Chandramouli 20-01-21 Savita D2

F4 Chandramouli 20-01-21 Savita D3

Gusseted base

F4 Chandramouli 20-01-21 Savita D4

F4 Chandramouli 20-01-21 Savita D5

Note:

Slab bases are used where the columns have independent concrete pedestals. A thick steel base plate and two-cleat angles connecting the flanges of the column to the base plate. In addition to these, web cleats are provided to connect the web of the column to the base plate. 

These web cleats guard against the possible dislocation of the column during erection. The ends

of the column and also the base plate should be mechanized so that the column load is wholly transferred to the base plate.

Gusseted bases are provided for columns carrying heavier loads requiring large base plates. A gusseted base consists of a base of reduced thickness and two gusseted plates are attached one to each flange of the column.

Column Base Question 9:

Maximum permissible slenderness ratio of compressive members which carry dead and superimposed load is

  1. 350
  2. 250
  3. 180
  4. 80

Answer (Detailed Solution Below)

Option 3 : 180

Column Base Question 9 Detailed Solution

Maximum slenderness ratios for tension members:

Type of Member

λ

A tension member in which a reversal of direct stress occurs due to loads other than wind or seismic forces.

180

A member normally acting as a tie in roof truss or a bracing system but subjected to a possible reversal of stresses resulting from the action of the wind or earthquake forces.

350

Member always under tension

400

Maximum slenderness ratios for compression members:

Type of Member

λ

A tension member in which a reversal of direct stress occurs due to loads other than wind or seismic forces.

180

A member carrying compressive loads resulting from dead loads and imposed loads.

180

A member subjected to compressive forces resulting only from combination with wind/earthquake actions, provided the deformation of such members does not adversely affect the stress in any part of the structure.

250

Compression flange of a beam restrained against lateral-torsional buckling.

300

A member normally acting as a tie in a roof truss or a bracing system not considered effective when subjected to a possible reversal of stresses resulting from the action of wind or earthquake forces.

350

Column Base Question 10:

If the unsupported length of a stanchion is 4 meters and least radius of gyration of its cross-section is 5 cm, the slenderness ratio of the stanchion is

  1. 60
  2. 70
  3. 80
  4. 90

Answer (Detailed Solution Below)

Option 3 : 80

Column Base Question 10 Detailed Solution

Concept:

A stanchion is a structural steel member, usually larger than a strut or a column whose main function is to withstand axial compressive stresses. e.g. an upright metal post on a ship.

The slenderness ratio is the ratio of the length of the member to the radius of gyration of the cross-section.

Calculation:

Slenderness ratio (SR) = l/r,

Where

l = effective length of the column and r = minimum radius of gyration

Slenderness ratio (SR) = 4 m/5 cm = 400 cm/5 cm = 80
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