Design of Steel Structures MCQ Quiz - Objective Question with Answer for Design of Steel Structures - Download Free PDF

Last updated on May 12, 2025

Latest Design of Steel Structures MCQ Objective Questions

Design of Steel Structures Question 1:

Which of the following types of beams is used for lightly loaded beams such as roof purlins and sheeting rails?

  1. Angles sections 
  2. Castellated beams
  3. Box girders
  4. Plate girders 

Answer (Detailed Solution Below)

Option 1 : Angles sections 

Design of Steel Structures Question 1 Detailed Solution

Explanation:

Angle Sections:

  1. Lightweight: Angle sections are often used for lightly loaded beams like roof purlins, making them cost-effective for smaller, less demanding structures.

  2. Versatile: They can be used in various configurations, including horizontal, vertical, or sloped, providing flexibility in design.

  3. Economical: Due to their shape and material efficiency, angle sections are cheaper compared to more complex beam designs, making them ideal for non-heavy duty applications.

Additional Information Castellated Beams:

  1. Lightweight with High Strength: Castellated beams are designed with a web that is cut and rejoined, making them lighter while maintaining strength, ideal for longer spans.

  2. Aesthetic Appeal: The design is often chosen for its aesthetic value, as it allows for architectural uniqueness in larger buildings or structures.

  3. Not Ideal for Light Loads: Castellated beams are typically used for medium to heavy loads, making them unsuitable for lightly loaded applications like roof purlins.

Box Girders:

  1. High Load-Bearing Capacity: Box girders are typically used for bridges or heavy-duty structures due to their ability to bear significant loads.

  2. Complex Design: They are more complex in design and fabrication, making them costlier and less suited for light applications like roof purlins.

  3. Durable: Box girders offer good resistance to torsional forces, making them suitable for situations where rotational forces are present, but not for lightly loaded beams.

Plate Girders:

  1. Strong Load-Bearing Capacity: Plate girders are used for supporting large loads and are common in bridges, large buildings, and industrial structures.

  2. Costly and Heavy: Due to their large size and the heavy material used in their construction, plate girders are more expensive and unsuitable for light-load applications.

  3. Complex Fabrication: Plate girders require specialized fabrication, which makes them less efficient and practical for simpler structures like roof purlins and sheeting rails.

Design of Steel Structures Question 2:

What are beams fabricated using rolled plates called?

  1. Plate girders
  2. Castellated beams
  3. Hybrid beams
  4. Composite beams

Answer (Detailed Solution Below)

Option 1 : Plate girders

Design of Steel Structures Question 2 Detailed Solution

Explanation:

  1. Plate girders are beams fabricated using rolled plates that are welded or bolted together to form a beam.
  2. These girders are often used in large structures like bridges or buildings where the span is long and the loads are significant.
  3. The plates used in the fabrication of plate girders are usually made of steel and are shaped to form the web and flanges of the girder.

 Additional Information

  1. Castellated beams: These are beams that are created by cutting a regular beam (usually an I-beam) in a pattern and then reassembling it to form a beam with an enlarged web, which helps reduce weight and increase the depth of the beam for larger spans.

  2. Hybrid beams: Hybrid beams combine two or more materials, such as steel and concrete, to take advantage of the benefits of both materials.

  3. Composite beams: These beams combine two materials, typically steel and concrete, where the concrete works in compression and the steel works in tension, offering better load distribution and efficiency.

Design of Steel Structures Question 3:

As per IS 800: 2007, which type of beam does NOT need a lateral torsional buckling check?

  1. Plate girders
  2. Beams bending about their major axis with λLT more than 0.4
  3. Hollow rectangular or tubular sections
  4. I-section beams with a long span

Answer (Detailed Solution Below)

Option 3 : Hollow rectangular or tubular sections

Design of Steel Structures Question 3 Detailed Solution

Explanation:

As per IS 800: 2007 (General Construction in Steel — Code of Practice):

  • Lateral Torsional Buckling (LTB) is a stability concern primarily for open sections (like I-beams) when bending about their major axis and the compression flange is unrestrained.

  • Hollow rectangular or tubular sections are closed shapes and geometrically stable in torsion.

  • Their symmetry and enclosed shape resist both lateral displacement and twisting, making LTB checks generally unnecessary.

Additional Information

  • Plate girders: Require LTB checks, especially due to their large depth and slenderness.
  • Beams bending about their major axis with λLT more than 0.4: Higher slenderness ratio indicates potential for buckling; LTB check is required.
  • I-section beams with a long span: More susceptible to LTB due to unrestrained compression flange over long lengths.

Design of Steel Structures Question 4:

The collar beam in a collar beam roof performs which of the following functions?

  1. It supports the ridge beam.
  2. It forms an extension to the eaves.
  3. It ties the wall plates together.
  4. It prevents rafters from bending.

Answer (Detailed Solution Below)

Option 4 : It prevents rafters from bending.

Design of Steel Structures Question 4 Detailed Solution

Explanation:

  • A collar beam in a roof structure is placed horizontally between opposing rafters, typically at a higher point of the roof.
  • Its primary function is to prevent the rafters from bending or buckling under load by tying the rafters together and providing additional support.
  • This helps to maintain the roof’s structural integrity.

 Additional InformationCoupled Roof:

  • Structure: A simple roof type where rafters are paired to form a basic triangular shape.

  • Function: The rafters support the roof covering and are typically connected at the ridge.

  • Design: Commonly used in small buildings or low-cost structures.

  • Collar Beam: Not typically used in coupled roofs; rafters act directly to support the load.

  • Application: Suitable for smaller span buildings, such as houses with simple roof designs.

Scissors Roof:

  • Structure: A roof design that uses intersecting rafters to create an angled structure resembling a scissors mechanism.

  • Function: Provides more headroom and creates an open, vaulted ceiling space within the building.

  • Design: More complex than a coupled roof, offering aesthetic value along with functionality.

  • Collar Beam: May be used in scissor roofs to support the rafters and prevent them from bending under load.

  • Application: Common in buildings requiring high ceilings or an open interior space, such as churches or large homes.

Design of Steel Structures Question 5:

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%

Design of Steel Structures Question 5 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.

Top Design of Steel Structures MCQ Objective Questions

In a double riveted double-covered butt joint, the strength of the joint per pitch length in shearing the rivets Pd and strength of one rivet in single shear Ps are related as:

  1. Pd = 4 Ps
  2. Pd = 2.5 Ps
  3. Pd = 2 Ps
  4. Pd = 3 Ps

Answer (Detailed Solution Below)

Option 1 : Pd = 4 Ps

Design of Steel Structures Question 6 Detailed Solution

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Concept:

Riveted Joint:

A riveted joint is a permanent joint which uses rivets to fasten two materials.

Types of riveted joints:

1. Lap Joint

A lap joint is that in which one plate overlaps the other and the two plates are then riveted together.

Number of shearing planes = 1

Steel Design 1 D2

2. Butt joint

A butt joint is that in which the main plates are kept in alignment butting each other and a cover plate is placed either on one side or on both sides of the main plates. The cover plate is then riveted together with the main plates.

Number of shearing plane in single cover butt joint = 1

Number of shearing plane in double cover butt joint = 2

Steel Design 1 D3

Steel Design 1 D4

 

Concept of shearing strength:

Strength of joint per pitch length in shearing, Pd = Ps × N × n

Where, Ps = Strength of one rivet in single shear, n = Number of shearing planes, N = Number of rivets

Calculation:

Given, Ps = strength of rivet in single shear, Number of rivets (N) = 2 in double riveted joints and number of shearing planes (n) = 2 for double cover butt joint

∴ \({P_d} = {P_s} \times 2 \times 2 = 4{P_s}\)

When the effect of wind or earthquake load is taken into account, the permissible stress as specified in rivets may be increased by

  1. 33.33%
  2. 50%
  3. 10%
  4. 25%

Answer (Detailed Solution Below)

Option 4 : 25%

Design of Steel Structures Question 7 Detailed Solution

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Concept:

According to clause no. 11.1.4 of IS 800: 2007, when the effect of wind or earthquake load is taken into account, the permissible stress as specified in rivets (or in anchor bolts) may be increased by 25%.

Confusion/Mistake Point:

According to clause no. 11.1.4 of IS 800: 2007, when the effect of wind or earthquake load is taken into account, the permissible stress as specified in structural steel member may be increased by 33.33%.

The maximum allowable vertical deflection under live load for a cantilever member supporting brittle cladding in an industrial building is:

  1. Span/240
  2. Span/180
  3. Span/120
  4. Span/150

Answer (Detailed Solution Below)

Option 4 : Span/150

Design of Steel Structures Question 8 Detailed Solution

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Explanation:

As per IS 800:2007, Clause No. 5.6.1:

The maximum vertical deflection for cantilever beam:
  • Supported by elastic cladding, deflection is limited to Span/120.
  • Supported by brittle cladding, deflection is limited to Span/150.

The maximum vertical deflection for simply supported beam:

  • Supported by elastic cladding, deflection is limited to Span/240.
  • Supported by brittle cladding, deflection is limited to Span/300.
  • Max deflection < L/325 of the span in general

As per IS 456:2000, Clause 23.2:

The deflection shall generally be limited to the following:

  • The final deflection due to all loads including the effects of temperature, creep, and shrinkage and measured from the as-cast level of the supports of floors, roofs, and all other horizontal members, should not normally exceed span/250.
  • The deflection including the effects of temperature, creep and shrinkage occurring after the erection of partitions and the application of finishes should not normally exceed span/350 or 20 mm whichever is less.

Select the incorrect statement from the following.

  1. Purlin is subjected to biaxial bending.
  2. The span of purlin is center to center of truss; purlin is located at the panel point of the truss.
  3. Purlin runs perpendicular to truss.
  4. Purlin is designed as a tension member.

Answer (Detailed Solution Below)

Option 4 : Purlin is designed as a tension member.

Design of Steel Structures Question 9 Detailed Solution

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Concept:

Purlin:

It is a member of truss which are supported on the principle rafter and which transverse loads to the truss. These are some important properties of purlin- 

  • It is a biaxial bending member.
  • The span of purlin is center to center of truss, purlin is located at the panel point of the truss.
  • Maximum spacing between purlins is less than 1.4 m. 
  • Angle, Channel, I-sections and Z-sections are used for purlins and girders to support the cladding. 

A steel rod of 20 mm diameter is used as a tie member in the roof bracing system and may be subjected to possible reversal of stress due to wind load. What is the maximum permissible length of the member?

  1. 1750 mm
  2. 2000 mm
  3. 3000 mm
  4. 2500 mm

Answer (Detailed Solution Below)

Option 1 : 1750 mm

Design of Steel Structures Question 10 Detailed Solution

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As per IS 800:2007, The maximum slenderness ratio for a member normally acting as a tie in a roof truss or a bracing system but subjected to possible reversal of stresses resulting from the action of wind or earthquake forces will be 350.

Slenderness ratio “λ” is given by:

\({\rm{\lambda }} = \frac{{{{\rm{l}}_{{\rm{eff}}}}}}{{{{\rm{r}}_{{\rm{min}}}}}}\)

Where,

rmin = minimum radius of gyration of the member, and ℓeff = effective length of the member

\({{\rm{r}}_{{\rm{min}}}} = \sqrt {\frac{{\rm{I}}}{{\rm{A}}}} = \sqrt {\frac{{\frac{{\rm{\pi }}}{{64}}{{\rm{D}}^4}}}{{\frac{{\rm{\pi }}}{4}{{\rm{D}}^2}}}} = \frac{{\rm{D}}}{4} = \frac{{20}}{4} = 5{\rm{\;mm}}\)

Where,

A = Area of the member and I = Moment of the inertia of the member about its center of gravity

λ = 350

∴ ℓeff = 350 x 5 = 1750 mm

If 'd' is depth of web and 'tw' is thickness of web of a plate girder such that \(\dfrac{d}{t_w}\le 400 \ \varepsilon_w\), where \(\varepsilon_w=\sqrt{\dfrac{250}{f_y}}\), then:

  1. end bearing stiffeners, intermediate transverse stiffeners and longitudinal stiffeners at 0.2d from compression face are needed
  2. intermediate transverse stiffeners are needed in addition to end bearing stiffeners
  3. end bearing stiffeners, intermediate transverse stiffeners, longitudinal stiffeners at 0.2d from compression face and at neutral axis are needed
  4. only end bearing stiffeners are needed

Answer (Detailed Solution Below)

Option 3 : end bearing stiffeners, intermediate transverse stiffeners, longitudinal stiffeners at 0.2d from compression face and at neutral axis are needed

Design of Steel Structures Question 11 Detailed Solution

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Explanation:

Transverse Stiffeners are provided to increase buckling resistance of the web due to inclined compressive stress due to shear, It is provided vertically along the span.

Horizontal Stiffener / Longitudinal Stiffener is designed to prevent web buckling due to bending compression.

Both traverse and longitudinal stiffeners are provided to check the web buckling.

End Bearing Stiffeners are provided at the supports & Load Bearing Stiffeners are provided at the points of concentrated loads.

Important Point:

Eg 17

If \(\frac{d}{{{t_W}}}\)< 67ϵ ⇒ unstiffened girder can be designed i.e. No girder required.

If 85 ϵ < \(\frac{d}{{{t_W}}}\)< 200 ϵ ⇒ Vertical stiffener (C1 and C2) may be provided.

If 200 ϵ < \(\frac{d}{{{t_W}}}\) < 250 ϵ ⇒ Vertical stiffener along with longitudinal stiffener at 0.2 d may be provided.

If 250 ϵ < \(\frac{d}{{{t_W}}}\) < 345 ϵ ⇒ Vertical stiffeners along with two longitudinal stiffeners at 0.2 d and 0.5 d respectively may be provided.

If \(\frac{d}{{{t_W}}}\) ≤ 400 ϵ ⇒ End bearing stiffeners, intermediate transverse stiffeners, longitudinal stiffeners at 0.2d from compression face and at neutral axis are needed.

As per IS 800, the efficiency of a riveted joint having the minimum pitch is:

  1. 60%
  2. 70%
  3. 40%
  4. 50%

Answer (Detailed Solution Below)

Option 1 : 60%

Design of Steel Structures Question 12 Detailed Solution

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Concept:

Tensile Stresses in the main plates might result in tearing of the main plate or cover plate across a row of the bolt.

The resistance offered by the plate against tearing is known as the tearing strength or tearing value of the plate.

Tearing efficiency of joints is given by,

\(\begin{array}{*{20}{c}} {{\eta _t} = \frac{\rm{Tearing\;strength}}{{\rm Strength\;of\;solid\;plate}}}\\ {{\eta _t} = \frac{{\left( {p - d} \right) × t × {\sigma _t}}}{{{p_t} × {\sigma _t}}} = \frac{{p - d}}{p}} \end{array}\)

\(\therefore \eta = \frac{{p - d}}{p}\ \)

RRB JE 30thAug 2015 10Q Shift3 Nitesh Hindi - Final images nita Q6

Calculation:

Given, Minimum pitch of rivet joint(p) = 2.5d

 \(\therefore \eta = \frac{{p - d}}{p}\ ×100\)

\(\eta = \frac{{2.5d - d}}{2.5d}\ × 100\) = \(\frac{1.5d}{2.5d}\) × 100 = 0.6 × 100 = 60%

For simply supported beams, the maximum permitted deflection is (As pe IS 800:2007)

  1. 1/325 of the span
  2. 1/300 of the span
  3. 1/350 of the span
  4. None of these

Answer (Detailed Solution Below)

Option 2 : 1/300 of the span

Design of Steel Structures Question 13 Detailed Solution

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Explanation:

Maximum permissible deflection in simply supported steel beam other than industrial building is given by = \(\frac{1}{{300}} \times {\bf{Span}}\)

Some of the reason for limiting deflection is:

  • Excessive deflection may create problems for floor or roof drainage.
  • Excessive deflection may lead to crack in the plaster of ceilings & may damage the material attached to or supported by the beam.
  • There may cause undesirable twisting and distortion of connections and connected materials.

Confusion Points

325 is as per IS 800: 1984, Question is asking as per IS 800: 2007, according to which 300 is the correct answer.

The plastic modulus of a section is 5 × 10-4 m3. Its shape factor is 1.2 and the plastic moment capacity is 120 kNm, what is the value of the yield stress of the material?

  1. 100 N/mm2
  2. 200 N/mm2
  3. 240 N/mm2
  4. 288 N/mm2

Answer (Detailed Solution Below)

Option 3 : 240 N/mm2

Design of Steel Structures Question 14 Detailed Solution

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Concept:

Shape Factor: It is defined as the ratio of moment carrying capacity of the plastic section over that of the elastic section when the yielding just starts.

\(\text{Shape Factor} = \frac{{{Z_P}{\sigma _y}}}{{{Z_e}{\sigma _y}}} = \frac{{{M_P}}}{{{M_y}}} = \frac{{{Z_P}}}{{{Z_e}}}\)

Shape factor values for different sections:

Shape Factor

Section

1.5

Rectangular

1.7

Circular

2.34

Triangular

 

Calculation:

Plastic modulus of section ZP = 5 × 10-4 m3

SF = 1.2

Plastic moment Capacity MP = 120 kNm

Plastic moment capacity is given by,

\({M_P} = {Z_P}{\sigma _y}\)

Yield Stress is given by,

\({\sigma _y} = \frac{{{M_P}}}{{{Z_P}}} = \frac{{120\ \times\ 1000}}{{5\ \times\ {{10}^{ - 4}}}} = 240 \times {10^6}\;N/{m^2} = 240\;MPa\)

For 4.6 type bolt conforming to IS 1367, respective ultimate tensile strength and yield strength will be:

  1. 400 MPa and 240 MPa
  2. 240 MPa and 410 MPa
  3. 410 MPa and 240 MPa
  4. 240 MPa and 400 MPa

Answer (Detailed Solution Below)

Option 1 : 400 MPa and 240 MPa

Design of Steel Structures Question 15 Detailed Solution

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Explanation:

M means bolt is of Metric Thread, and 20 refers to its diameter. It's cap diameter is 1.5 D (1.5D is for hexagonal bolt). Strength is determined from grade of bolt. Here grade 4.6

So ultimate strength, fu = 4 x 100 = 400 N/mm2

Yield strength, fy = 0.6 x ultimate strength = 0.6 x 400 = 240 N/mm2
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