Compression Member MCQ Quiz - Objective Question with Answer for Compression Member - Download Free PDF

Last updated on May 5, 2025

Latest Compression Member MCQ Objective Questions

Compression Member Question 1:

For a built-up column with lacing, what percentage of the axial load is considered as the transverse shear force (Vt ) that the lacing system must resist? 

  1. 1.5%
  2. 2.5%
  3. 2.0%
  4. 3.0%

Answer (Detailed Solution Below)

Option 2 : 2.5%

Compression Member Question 1 Detailed Solution

Explanation:

  • According to IS 800: 2007 (General Construction in Steel – Code of Practice), for a built-up column with lacing, the transverse shear force (Vt) that the lacing system must resist is taken as 2.5% of the axial load.
  • This is specified in the code to account for the shear transfer between the lacing system and the main column.

Additional Information

Lacing System:

  • The lacing system consists of slanted members that connect the flanges of the built-up column, typically arranged in a criss-cross pattern.
  • It helps stabilize the column by resisting lateral forces.
  • The 2.5% figure is meant to represent the part of the axial load that is resisted by the transverse shear, which the lacing members contribute to.

  • This allows for efficient load transfer and ensures the column's stability under lateral loads such as wind or earthquake forces.

  • In actual design, this value might be adjusted depending on specific conditions, such as the column's dimensions, the material used, or the particular load scenario. However, 2.5% is the commonly used value under normal conditions as per IS 800.

Compression Member Question 2:

The flexural torsional buckling strength of single angleloaded in compression through one of its legs may beevaluated using the equivalent slenderness ratio. This slenderness ratio does not depend upon:

  1. Yield stress ratio
  2. width of the two legs of the angle
  3. Radius of gyration about the minor axis
  4. Edge to edge length of the supportingmember

Answer (Detailed Solution Below)

Option 4 : Edge to edge length of the supportingmember

Compression Member Question 2 Detailed Solution

Explanation:

Flexural Torsional Buckling Strength of Single Angle in Compression

The flexural torsional buckling strength of a single angle loaded in compression through one of its legs is evaluated using the equivalent slenderness ratio. This slenderness ratio is influenced by several factors; however, it does not depend upon the edge to edge length of the supporting member.

Analyzing the Given Options

  1. "Yield stress ratio." (Relevant Factor)

    • The yield stress ratio affects the buckling strength as it determines the material's ability to withstand stress before yielding.

  2. "Width of the two legs of the angle." (Relevant Factor)

    • The width of the legs affects the cross-sectional properties and thus influences the buckling strength.

  3. "Radius of gyration about the minor axis." (Relevant Factor)

    • The radius of gyration about the minor axis is a key geometric property that directly affects the slenderness ratio and buckling strength.

  4. "Edge to edge length of the supporting member." (Not a Relevant Factor)

    • The edge to edge length of the supporting member does not influence the slenderness ratio of the angle itself.

    • Therefore, it is not a factor that affects the flexural torsional buckling strength of the angle.

Compression Member Question 3:

Which of the following statements (S1 to S4), pertaining to the design of steel compression member, is/are correct?

S1: For the same cross-section, longer compression members have lesser tendency to buckle, and it will support greater load.

S2: An asymmetrical angle section (about centroidal axes) will bend about the principal axis for which the radius of gyration is minimum.

S3: To minimise the steel requirement in the column design (to use the material at the greatest possible stress), the slenderness ratio should be kept as large as possible.

S4: A symmetrical I-section (about both the centroidal axes) will bend about one of the centroidal axis giving lesser radius of gyration.

  1. S1, S3, and S4
  2. S1 and S3
  3. S2 and S4
  4. Only S1

Answer (Detailed Solution Below)

Option 3 : S2 and S4

Compression Member Question 3 Detailed Solution

Concept:

Analysis of Given Statements:

S1: "For the same cross-section, longer compression members have lesser tendency to buckle, and it will support greater load."  (False)

  • According to Euler’s buckling formula: \( P_{cr} = \frac{\pi^2 EI}{(KL)^2} \), the longer the column, the lower the critical buckling load.
  • Hence, longer members have a higher tendency to buckle and support a lesser load.

S2: "An asymmetrical angle section (about centroidal axes) will bend about the principal axis for which the radius of gyration is minimum."  (True)

  • Asymmetrical sections (like angle sections) have different radii of gyration along different principal axes.
  • Buckling occurs along the axis with the smallest radius of gyration, confirming this statement.

S3: "To minimise the steel requirement in the column design (to use the material at the greatest possible stress), the slenderness ratio should be kept as large as possible."  (False)

  • The slenderness ratio (λ=Lr" id="MathJax-Element-7-Frame" role="presentation" style="position: relative;" tabindex="0">λ=Lr ) should be minimized to reduce the chances of buckling.
  • A higher slenderness ratio reduces load-carrying capacity, making the column weaker.

S4: "A symmetrical I-section (about both the centroidal axes) will bend about one of the centroidal axes giving lesser radius of gyration."  (True)

  • For symmetrical I-sections, buckling occurs about the axis with the least radius of gyration, as per Euler’s formula.

Compression Member Question 4:

What is the maximum slenderness ratio for a member carrying compressive loads resulting from dead loads and imposed loads? 

  1. 250
  2. 180
  3. 300
  4. 350

Answer (Detailed Solution Below)

Option 2 : 180

Compression Member Question 4 Detailed Solution

Explanation:

Slenderness ratio limits for a different type of members:

Types of member

Maximum slenderness ratio

A member carrying compressive loads resulting from the dead load and superimposed loads

180

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

250

A member carrying tension loads resulting from the dead load and superimposed load

400

A member normally carrying tension but subjected to reversal of stress due to wind or earthquake forces

350

For a tension member subjected to reversal of stress due to live loads

180      

 

Compression Member Question 5:

The permissible maximum deflection of a purlin of 4 m that is subjected to live load supporting GI metal sheet and supported by elastic cladding is:

  1. 22.22 mm
  2. 26.66 mm
  3. 16.66 mm
  4. 33.33 mm

Answer (Detailed Solution Below)

Option 2 : 26.66 mm

Compression Member Question 5 Detailed Solution

Concept: 

Vertical Deflection limits for industrial buildings as per IS 800:2007 are:

a) For Purlins and Girts subjected to live load/wind load supported on elastic cladding, maximum deflection is limited to span / 150.

b) For Purlins and Girts subjected to live load/wind load supported on Brittle cladding, maximum deflection is limited to span / 180.

Calculation:

∴ Permissible maximum deflection = Span/150 = 4000/150 = 26.66 mm

Top Compression Member MCQ Objective Questions

The maximum value of effective slenderness ratio as per IS 800 for a tension member in which a reversal of direct stress occur due to loads other than wind or seismic forces is _________

  1. 350
  2. 250
  3. 400
  4. 180

Answer (Detailed Solution Below)

Option 4 : 180

Compression Member Question 6 Detailed Solution

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Explanation:

As per IS 800:2007, Table no. 3 , Maximum values of effective slenderness ratio:

Member Maximum effective slenderness ratio (kL/r)
A member carrying compressive loads resulting from dead loads and imposed loads 180
A tension member in which a reversal of direct stress occurs due to loads other than wind or seismic forces 180 
A member subjected to compression forces resulting only from combination with wind/earthquake actions, provided the deformation of such member does not adversely affect the stress in any part of the structure 250 
Compression flange of a beam against lateral torsional buckling  300
A member normally acting m a tie in a roof truss or a bracing system not considered effective when subject to possible reversal of stress into compression resulting from the action of wind or earthquake forces  350
Members always under tension’) (other than pre-tensioned members)   400

Calculate the effective length of column having c/c length of 4m and both ends are fixed as per IS 800-2007.

  1. 2.5 m
  2. 2 m
  3. 2.6 m
  4. 4 m

Answer (Detailed Solution Below)

Option 3 : 2.6 m

Compression Member Question 7 Detailed Solution

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Explanation:

As per IS: 800 - 2007

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

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Given both ends are fixed

So the Case becomes Restrained in all movements

So effective length = 0.65 × L = 0.65 × 4 = 2.6 m

A battened column subjected to an axial force of 1100 kN. Battens are placed at 1.25 m centre to centre. Depth of batten plate is 300 mm, length is 363 mm, thickness is 6 mm and on two planes. The bending moment in the batten is:

  1. 9.6 kNm
  2. 8.6 kNm
  3. 10.6 kNm
  4. 7.6 kNm

Answer (Detailed Solution Below)

Option 2 : 8.6 kNm

Compression Member Question 8 Detailed Solution

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Concept:

The design shear and moments for batten plates is given by,

\({{\rm{V}}_{\rm{b}}} = \frac{{{{\rm{V}}_{\rm{t}}}{\rm{C}}}}{{{\rm{NS}}}}\) and \({\rm{M}} = \frac{{{{\rm{V}}_{\rm{t}}}{\rm{C}}}}{{2{\rm{N}}}}\)

Where,

Vt = Transverse shear force (2.5% of axial load)

C = Distance between centre to centre of battens longitudinally

N = Number of parallel planes

S = Minimum transverse distance between the centroid of the fasteners connecting batten to the main member.

Calculation:

Given: C = 1.25 m, and N = 2

Vt = 2.5% of axial force = 1100 × 2.5/100 = 27.5 kN

∴ Bending moment, \({\rm{M}} = \frac{{27.5 \times 1.25}}{{2 \times 2}} = 8.6{\rm{\;kNm}}\)

Which of the following assumptions is used for the design of an axially loaded compression member (steel)?

  1. The modulus of elasticity is not assumed to be constant.
  2. Secondary stresses are neglected.
  3. 25% to 40% of secondary stresses are taken into account.
  4. The ideal column is not straight and has crookedness.

Answer (Detailed Solution Below)

Option 2 : Secondary stresses are neglected.

Compression Member Question 9 Detailed Solution

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Explanation:

Compression member:

  • A compression member is a structural member which is straight and subjected to two equal and opposite compressive forces applied at its ends.

Assumptions made while designing a compression member (or column):

  • The ideal column is assumed to be absolutely straight having no crookedness, which never occurs in practice.
  • The modulus of elasticity is assumed to be constant in a built-up column.
  • Secondary stresses(which may be of the order of even 25%-40% of primary stresses) are neglected.

What is the criteria to decide the clear distance between two channel sections in a build up column?

  1. The moment of inertia about minor axis should be 5 times the moment of inertia about major axis.
  2. The moment of inertia about major axis should be 5 times the moment of inertia about minor axis.
  3. The moment of inertia about major and minor axis should be same.
  4. The length and width of the built up section should be same.

Answer (Detailed Solution Below)

Option 3 : The moment of inertia about major and minor axis should be same.

Compression Member Question 10 Detailed Solution

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Explanation:

The clear distance between two channel sections in a built up column is determined by the moment of inertia, where the moment of inertia about the major axis is equal to the moment of inertia about the minor axis.

The length and width of the built-up section, as well as the ratio of the moments of inertia, are important factors in determining the clear distance.

According to IS 800:2007, the design compressive strength of an axially loaded compression member is based on the: 

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

Answer (Detailed Solution Below)

Option 3 : Perry Robertson formula

Compression Member Question 11 Detailed Solution

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Explanation:

As per clause no 7.1, IS 800:2007, The design compressive strength Pd of an axially loaded member is based on the Perry Robertson Formula and is given by;

\(P_{d}= A_{e}f_{cd}\) , 

where Ae = Effective sectional area , fcd = Design compressive stress 

The design compressive stress of axially loaded compression member shall be calculated using following equation:

 \(f_{cd} = \frac{\frac{f_{y}}{\gamma _{mo}}}{\Phi +\left [ \Phi ^{2}-\lambda ^{2} \right ]^{0.5}}=\chi \frac{f_{y}}{\gamma _{mo}}\leq \frac{f_{y}}{\gamma _{mo}}\),

where \(\phi= 0.5\left [ 1+\alpha \left ( \lambda -.2 \right )+ \lambda^{2}\right ]\) , \(\lambda\)= non-dimensional effective slenderness ratio \(=\sqrt{\frac{f_{y}}{f_{cc}}} \) , where fcc = Euler's buckling class\(=\frac{\pi ^{2}E}{(\frac{KL}{r})^{2}} \)

Where \(\frac{KL}{r}\)= Effective slenderness ratio , \(\alpha\) = Imperfection factor , \(\chi\) = stress reduction factor , \(\gamma_{mo}\)= partial safety factor for material strength

Where the horizontal stiffeners are provided 'd' in mm shall be taken as a clear distance between the horizontal stiffeners and tension flange. These vertical stiffeners shall be designed so that I is not less than

  1. \(1.3 \frac{d^{3} t^{3}}{C^{2}}\)
  2. \(1.5 \frac{d^{3} t^{3}}{C^{2}}\)
  3. \(0.87 \frac{d^{3} t^{3}}{C^{2}}\)
  4. \(0.85 \frac{d^{3} t^{3}}{C^{2}}\)

Answer (Detailed Solution Below)

Option 2 : \(1.5 \frac{d^{3} t^{3}}{C^{2}}\)

Compression Member Question 12 Detailed Solution

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Example:

RRB JE CE R 4 15Q Structure Test Part1(Hindi) - Final images Q7

Vertical Stiffeners:

  • Vertical stiffeners are provided to prevent buckling of the web due to diagonal compression.

Some criteria regarding the design of vertical stiffness:

  • The spacing between the vertical stiffness should be between 0.33 × d and 1.5 × d, d is the distance between flange angles,s, and where there is no flange angle, the distance between flanges ignoring fillets is taken.
  • The greater unsupported clear dimension of the web panel should not be greater than 270 times the thickness of the web and the lesser unsupported clear dimension of the same web panel should not be greater than 180 times the thickness of the web.
  • The length of the outstanding leg of vertical stiffness may be taken equal to 1/30 of the clear depth of girder plus 50 mm and the outstanding of stiffener from web shall not be more than \(\frac{256\times t}{\sigma _{y}^{1/2}}\), where t is the thickness of the section )
  • The length of the connected leg of vertical stiffness should be sufficient to accommodate the rivets connecting the stiffness to the web. The amount of moment `of inertia I of the stiffness selected should not be less than

I < \( \frac{1.5\times d^{3}\times t _{w}^{3}}{C^{2}}\)

Where I is the moment of inertia of pair of stiffness about the center of the web

tw = Minimum required thickness of the web

C = Maximum permitted distance between the vertical stiffeners

Which of the following is NOT a type of lacing?

  1. Triple
  2. Welded
  3. Double
  4. Single

Answer (Detailed Solution Below)

Option 1 : Triple

Compression Member Question 13 Detailed Solution

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Explanation:

Lacing:

  • When the component of a column connecting by a system of generally flat plates.
  • The table below the type of lacing and Effective length

 

Type of lacing Effective length (le)
Single lacing, riveted at ends Length between inner end rivets on lacing bar = l
Double lacing, riveted at ends and intersection 0.7 times the length between inner end rivets on lacing (0.7 × l)
Welded lacing

0.7 times the distance between inner ends o effective lengths of welds at the end (0.7 × l)

 

The table provided above shows the type of lacing, triple is not the type of lacing

According to IS 800 ∶ 2007, if a built-up column is subjected to axial force, the lacing for compression members should be proportioned to resist a total transverse shear Vt equal to ________ percentage of the axial load on the column.

  1. 1
  2. 1.5
  3. 2
  4. 2.5

Answer (Detailed Solution Below)

Option 4 : 2.5

Compression Member Question 14 Detailed Solution

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Explanation:

As per CI. 7.6.6.1 of IS 800:2007, recommends that the lacing for compression members are designed for a transverse shear force equal to 2.5% of the axial load on the column.Important Points 

(i) For laced columns, the radius of gyration of the column cross-section about an axis normal to the plane of lacing shall not be less than the radius of gyration about an axis parallel to the plane of the lacing.

(ii) Lacing on two parallel planes should be the shadow of each other and not the crossed one. This provision prevents torsion about the column axis.

(iii) The angle θ in single lacing system can be increased from 40° to 70° only. even if the component column is found to be unsafe then a double lacing system remains the only resource.

(iv) CI. 7.6.15 of IS 800:2007, states that effective slenderness ratio of laced column should be 5% more than the actual maximum slenderness ratio so that shear deformation effects are accounted for. 

Splices for a compression member are designed as?

  1. Pedestal
  2. Long column
  3. Short column
  4. Compression block

Answer (Detailed Solution Below)

Option 3 : Short column

Compression Member Question 15 Detailed Solution

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Explanation:

Column splices:

(i) A splice is a joint provided in the length of the member.

(ii) In case of column splice, if the load is truly concentric then theorectically no splice is required since compression wili be directly transfered through bearing. But truly axial load in column never occurs.

(iii) Also columns are most of the times also subjected to bending. This raise the necessity of column splices. Column sections are required to be spliced for the following cases

  • When the available length of structural steel section is less than the required length of the column.
  • In case of multi-storey buildings, the section of column required for the various storeys may be different.
  • In multi-storeyed buildings, for convenience of fabrication it is kept at about 5 m lengths. So, splicing of column is necessary to join the fabrication along the length.

Specification for the design of splices:

(i) Where the ends of the compression members are faced for complete bearing over the whole area there the splices are designed to hold the members accurately in position and to resist any tension where bending is also there.

(ii) In case the connecting members are not faced for complete bearing then splices are designed to transmit all the forces to which they are subjected to.

(iii) Splices are designed as short columns.

Important Points

(i) Ideally a splice plate should be located at a place where flexural moment in the column is zero i.e. at the location of point of contra flexure.

(ii) Due to direct load, there are two points of contra flexure varying from middle of the column to the points above or below the middle. 

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