Design of Steel Structures MCQ Quiz - Objective Question with Answer for Design of Steel Structures - Download Free PDF
Last updated on Jun 13, 2025
Latest Design of Steel Structures MCQ Objective Questions
Design of Steel Structures Question 1:
For economical spacing of roof truss, if t, p, r are the cost of truss, purlin and roof coverings respectively, then
Answer (Detailed Solution Below)
Design of Steel Structures Question 1 Detailed Solution
Explanation:
- The cost of the truss should be equal to twice the cost of purlins plus the cost of roof covering. t = 2p + r
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The total cost (t) depends on the costs of trusses, purlins, and roof covering.
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Roof covering cost usually dominates as it covers the entire roof area.
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Economical spacing balances these costs to minimize total cost.
Additional Information
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A roof truss is a structural framework composed of straight members arranged in triangular units to support roof loads efficiently.
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Trusses transfer loads from the roof covering and purlins to the main supporting structure, such as walls or beams.
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The spacing of trusses affects the quantity and cost of purlins and roof covering required.
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Closer truss spacing means more trusses (higher truss cost) but fewer and shorter purlins and less roof covering overlap, while wider spacing reduces truss cost but increases purlin and roof covering costs.
Design of Steel Structures Question 2:
The shear lag effect in beam flanges are disregarded when the outstand of the beam flange is less than or equal to
Answer (Detailed Solution Below)
Design of Steel Structures Question 2 Detailed Solution
Explanation:
IS 800(Page number 53)
The shear lag effects in flanges may be disregarded provided:
a) For outstand elements (supported along one edge), bO ≤ LO/20;
and b) For internal elements (supported along two edges), bi ≤ LO/ 10.
where LO = length between points of zero moment (inflection) in the span, bO = width of the flange with outstand, and bi = width of the flange as an internal element..
Additional InformationShear Lag
Shear lag is a structural effect where not all parts of a cross-section effectively resist stress, especially in tension or bending. This leads to non-uniform stress distribution, typically in wide flanges or tension plates.
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It occurs because stresses don’t distribute instantly across the entire section.
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Areas far from the point of load application (e.g., edges of wide flanges) respond with a delay.
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Shear lag is accounted for by reducing effective width or using effective area in calculations.
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Helps prevent overestimation of capacity, especially in critical structures.
Design of Steel Structures Question 3:
The effective length of a battened column is
Answer (Detailed Solution Below)
Design of Steel Structures Question 3 Detailed Solution
Explanation:
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In battened columns, due to the tendency of the members to shear and bend between battens, additional flexibility is introduced.
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As per IS 800:2007, for battened built-up columns, the effective length is taken as 1.05 times the actual effective length of the equivalent single section.
So, the effective length is increased by 5% to account for the additional slenderness due to battens.
Additional Information
- A battened column is a built-up compression member composed of two or more individual sections (typically channels or angles) connected together with batten plates to act as a single unit.
- The battens restrain the members from buckling independently and ensure load-sharing.
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The battens provide lateral support, but the column is still more flexible than a solid single section because of:
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Shear deformation between the individual components.
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Lateral deflections of the components between battens.
Design of Steel Structures Question 4:
For a propped cantilever subjected to a point load at mid span, the value of collapse load is
Answer (Detailed Solution Below)
Design of Steel Structures Question 4 Detailed Solution
Explanation:
By the principle of virtual work done,
External work done = Internal work done
\(W\times{L\over2}\times\theta=M_p(\theta)+M_p(\theta+\theta)\)
\(W_u= {{6M_u}\over L}\)
Design of Steel Structures Question 5:
The slenderness ratio of lacing flats is limited to
Answer (Detailed Solution Below)
Design of Steel Structures Question 5 Detailed Solution
Explanation:
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According to IS 800, the slenderness ratio
" id="MathJax-Element-24-Frame" role="presentation" style="position: relative;" tabindex="0"> (length/thickness) of lacing flats used in steel structures is limited to 145.l t \frac{l}{t} -
This limit is to ensure that the lacing flats do not buckle locally under compressive forces.
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Keeping slenderness within limits helps maintain stability and load transfer efficiency in built-up columns.
Additional Information
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Lacing flats are used to connect two main components of a built-up steel column.
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Proper slenderness ratio prevents local buckling.
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IS 800 specifies limits on slenderness ratios for different structural elements to ensure safety.
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:
Answer (Detailed Solution Below)
Design of Steel Structures Question 6 Detailed Solution
Download Solution PDFConcept:
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
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
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
Answer (Detailed Solution Below)
Design of Steel Structures Question 7 Detailed Solution
Download Solution PDFConcept:
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:
Answer (Detailed Solution Below)
Design of Steel Structures Question 8 Detailed Solution
Download Solution PDFExplanation:
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.
Answer (Detailed Solution Below)
Design of Steel Structures Question 9 Detailed Solution
Download Solution PDFConcept:
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?
Answer (Detailed Solution Below)
Design of Steel Structures Question 10 Detailed Solution
Download Solution PDFAs 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 mmIf '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:
Answer (Detailed Solution Below)
Design of Steel Structures Question 11 Detailed Solution
Download Solution PDFExplanation:
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:
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:
Answer (Detailed Solution Below)
Design of Steel Structures Question 12 Detailed Solution
Download Solution PDFConcept:
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,
\(\begin{array}{*{20}{c}} {{\eta _t} = \frac{\rm{Tearing\;strength}}{{\rm Strength\;of\;solid\;plate}}}\\ {{\eta _t} = \frac{{\left( {p - d} \right) × t × {\sigma _t}}}{{{p_t} × {\sigma _t}}} = \frac{{p - d}}{p}} \end{array}\)
\(\therefore \eta = \frac{{p - d}}{p}\ \)
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)
Answer (Detailed Solution Below)
Design of Steel Structures Question 13 Detailed Solution
Download Solution PDFExplanation:
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?
Answer (Detailed Solution Below)
Design of Steel Structures Question 14 Detailed Solution
Download Solution PDFConcept:
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:
Answer (Detailed Solution Below)
Design of Steel Structures Question 15 Detailed Solution
Download Solution PDFExplanation:
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