Category: Design of Steel Structures Objective Questions

Design of Steel Structures Multiple Choice Questions on “Back-to-Back Connection”.

1. The slenderness ratio of each member when placed back-to-back or separated by small distance shall be
a) greater than 40
b) not greater than 40
c) 0.8 times the slenderness ratio of column as a whole
d) greater than 50
Answer: b
Clarification: Two rolled sections placed back-to-back or separated by small distance should be connected together by rivets/bolts/welds so that slenderness ratio of each member when placed back-to-back or separated by small distance is not greater than 40 or 0.6 times the slenderness ratio of column as a whole.

2. Minimum number of bolts for connecting end of strut is
a) 0
b) 3
c) 1
d) 2
Answer: d
Clarification: Ends of strut should be connected with minimum of 2 bolts/rivets or equivalent length of weld length (length must not be less than maximum width of member).

3. Which of the following is true?
a) when there is small spacing between the two sections placed back-to-back, washers and packing should be provided
b) when there is small spacing between the two sections placed back-to-back, washers and packing should not be provided
c) there should be additional connection in between along the length of member
d) when leg of angles greater than 125mm wide or web of channel is mm wide, minimum bolt is sufficient for connection
Answer: a
Clarification: There should be minimum of two additional connection in between, spaced equidistant along the length of member. When there is small spacing between the two sections placed back-to-back, washers(in case of bolts) and packing(in case of welding) should be provided to make connection. When leg of angles greater than 125mm wide or web of channel is mm wide, minimum two bolts/rivets should be for connection.

4. Minimum diameter of bolt when member is less than 16mm thick is
a) 8
b) 10
c) 22
d) 20
Answer: c
Clarification: Rivets/bolts should not be less than 16mm in diameter for member less than 10mm thick, 20mm in diameter for member less than 16mm thick and 22mm in diameter for member more than 16mm thick.

5. Which of the following is not true?
a) spacing of tack bolt should be less than 600mm
b) spacing of tack bolt should be greater than 600mm
c) if bolts are used, they should be spaced longitudinally at less than 4 times the bolt diameter
d) connection should extend at least 1.5 times the width of the member
Answer: b
Clarification: Spacing of tack bolt should be less than 600mm. If bolts are used, they should be spaced longitudinally at less than 4 times the bolt diameter. Connection should extend at least 1.5 times the width of the member.

6. Members connected back-to-back connected by bolts should be
a) not be used
b) subjected to transverse loading in plane perpendicular to bolted surface
c) subjected to twice the transverse loading in plane perpendicular to bolted surface
d) not subjected to transverse loading in plane perpendicular to bolted surface
Answer: d
Clarification: Members connected back-to-back connected by bolts should not be subjected to transverse loading in plane perpendicular to riveted/bolted/welded surface.

7. For members placed back-to-back, the spacing of bolt should not exceed
a) 12t
b) 16t
c) 18t
d) 20t
Answer: a
Clarification: For members placed back-to-back, the spacing of bolt should not exceed 12t or 200mm, where t is thickness of member.

8. Longitudinal spacing between intermittent welds used for connection should be
a) greater than 18t
b) greater than 16t
c) not greater than 16t
d) equal to 18t
Answer: c
Clarification: Longitudinal spacing between intermittent welds used for connection should not be greater than 16t, where t is thickness of thinner connection.

Design of Steel Structures Multiple Choice Questions on “Deflection & Holes in Beams”.

1. Which of the following may not occur due to excessive deflection?
a) ponding problem in roofs
b) misalignment of supporting machinery
c) cracking of plaster ceilings
d) twisting of beam
Answer: d
Clarification: Excessive deflection may cause cracking of plaster ceilings, misalignment of supporting machinery and cause excessive vibration, ponding problem in roofs, etc. Hence deflection in beam needs to be limited.

2. What is ponding?
a) excessive deflection of flat roof resulting in accumulation of rainwater
b) excessive deflection of flat roof not resulting in accumulation of rainwater
c) small deflection of flat roof resulting in accumulation of rainwater
d) small deflection of flat roof not resulting in accumulation of rainwater
Answer: a
Clarification: Excessive deflection of flat roof resulting in accumulation of water during rainstorms is called ponding and it causes damage to the roof material.

3. Deflection can be reduced by
a) proving less restraints
b) increasing span
c) increasing depth of beam
d) decreasing depth of beam
Answer: c
Clarification: Deflection can be reduced by increasing depth of beam, reducing the span, providing greater end restraints or by other means such as providing camber.

4. Beam deflection is not a function of
a) loading
b) span
c) length of column
d) geometry of cross section
Answer: b
Clarification: Beam deflection is a function of loading, span, modulus of elasticity and geometry of cross section. Small deflections of beams do not cause structural problems in general except for discomfort to the users. But excessive deflections may lead to crack in plaster or ceilings and may damage material attached to or supported by beams.

5. What is the maximum vertical deflection in industrial building for purlins and girts subjected to live load/wind load for elastic cladding?
a) span/150
b) span/180
c) span/250
d) span/100
Answer: a
Clarification: The maximum deflection in industrial building for purlins and girts subjected to live load/wind load for elastic cladding is span/150 and for brittle cladding is span/180.

6. What is the maximum vertical deflection in other buildings (other than industrial buildings)for floor subjected to live load and elements not susceptible to cracking?
a) span/150
b) span/180
c) span/300
d) span/100
Answer: c
Clarification: The maximum deflection in other buildings (other than industrial buildings) for floor subjected to live load and elements not susceptible to cracking is span/300.

7. What is the maximum vertical deflection in other buildings (other than industrial buildings) for floor subjected to live load and elements susceptible to cracking?
a) span/150
b) span/360
c) span/300
d) span/100
Answer: b
Clarification: The maximum deflection in other buildings (other than industrial buildings) for floor subjected to live load and elements susceptible to cracking is span/360.

8. What is the maximum lateral deflection in other buildings (other than industrial buildings) subjected to wind load and for brittle cladding?
a) height/300
b) height/250
c) height/100
d) height/500
Answer: d
Clarification: The maximum lateral deflection in other buildings (other than industrial buildings) subjected to wind load and for brittle cladding is height /500 and for elastic cladding is height/300.

9. What is the maximum vertical deflection for a cantilever member in other buildings (other than industrial buildings) subjected to live load and elements not susceptible to cracking?
a) span/150
b) span/180
c) span/300
d) span/100
Answer: a
Clarification: The maximum vertical deflection for a cantilever member in other buildings (other than industrial buildings) subjected to live load and elements not susceptible to cracking is span/150 and for elements susceptible to cracking is span/180.

10. What is the maximum lateral deflection of column/frame in industrial buildings subjected to crane load plus wind load and for brittle cladding?
a) height/300
b) height/250
c) height/400
d) height/500
Answer: c
Clarification: The maximum lateral deflection of column/frame in industrial buildings subjected to crane load plus wind load and for brittle cladding (pendant operated) is height/400 and for elastic cladding (cab operated) is height/200.

11. The strength of steel beam depends on
a) strength of tension flange
b) strength of compression flange
c) strength of web
d) does not depend on strength of section
Answer: b
Clarification: The strength of steel beam depends on the strength of compression flange. An open hole in the compression flange affects the strength of steel beam more than a hole in tension flange.

12. A hole in flange of beam causes
a) increase in stress
b) decrease in stress
c) makes the stress to half
d) does not affect the stress
Answer: a
Clarification: A hole in flange of beam causes an increase in stress. If the hole in compression flange contains rivet or bolt, the strength reduction is lessened as fastener can transmit compression.

13. Holes in beam webs should be placed at ____ and in flanges it should be placed at ________
a) high shear, high bending moment
b) high bending moment, high shear
c) low bending moment, low shear
d) low shear, low bending moment
Answer: d
Clarification: Holes in beam webs have less effect on flexural strength than holes in the flanges. Holes in beam webs should be placed only at sections of low shear. In the flanges, the holes should be cut at points of low bending moment. If this is not possible, the effect of the holes should be accounted for design.

14. The strength of the beams with openings may be governed by plastic deformations due to
a) moment only
b) shear only
c) both moment and shear
d) does not depend on moment or shear
Answer: c
Clarification: The strength of the beams with openings may be governed by plastic deformations due to both moment and shear at the openings. The strength realised will depend upon the interaction between moment and shear. The reduction in moment capacity at the openings is small while the reduction in shear capacity may be significant.

15. Which of the following are correct regarding design of beams with openings?
a) web opening should be away from support by twice the beam depth
b) hole should be eccentrically placed in web
c) hole should not be placed within middle one third of the span
d) clear spacing between openings should be less than beam depth
Answer: a
Clarification: General guidelines for design of beams with openings are as follows : (i)The hole should be centrally placed in web and eccentricity should be avoided, (ii) The best location for the opening is within the middle one third of the span, (iii) Web opening should be away from support and it should be away by twice the depth of beam, (iv) Clear spacing between openings should be more than depth of beam.

16. Which of the following are not correct regarding design of beams with openings?
a) diameter of circular opening should be restricted to 0.5D
b) for rectangular stiffened openings depth should be less than 0.7D and length less 2D
c) for rectangular unstiffened openings, depth should be less than 0.5D and length less than 1.5D
d) point load should be applied within a distance d from adjacent opening
Answer: d
Clarification: General guidelines for design of beams with openings are as follows : (i) The diameter of circular opening should be restricted to 0.5D, where D is depth of beam, (ii) For rectangular unstiffened openings, depth should be less than 0.5D and length less than 1.5D, where D is depth of beam (iii) For rectangular stiffened openings, depth should be less than 0.7D and length less 2D, where D is depth of beam (iv) Point loads should not be applied within a distance d from the adjacent opening.

Design of Steel Structures Interview Questions and Answers for freshers on “Bolted Connections – III”.

1. Which of the following equation is correct for bolt subjected to combined shear and tension?
a) (V_sb/V_db)2 + (T_sb/T_db)2 ≤ 1
b) (V_sb/V_db)2 + (T_sb/T_db)2 ≥ 1
c) (V_sb/V_db) + (T_sb/T_db) ≤ 1
d) (V_sb/V_db) + (T_sb/T_db) ≥ 1
Answer: a
Clarification: Bolt required to satisfy both shear and tension at the same time should satisfy (V_sb/V_db)2 + (T_sb/T_db)2 ≤ 1 , where V_sb= factored shear force, V_db = design shear capacity, T_sb = factored tensile force, T_db= design tensile capacity.

2. Shear Capacity of HSFG bolts is
a) μ_fn_ek_hF_o
b) μ_fn_ek_hF_oγ_mf
c) μ_fn_ek_h? /F_oγ_mf
d) μ_fn_ek_hF_o/γ_mf
Answer: d
Clarification: Shear Capacity of HSFG bolts is μ_fn_ek_hF_o/γ_mf, where μ_f = coefficient of friction(0.55), n_e = number of frictional interfaces offering frictional resistance to slip, k_h = 1 for fasteners in clearance holes, 0.85 for fasteners in over sized and short slotted holes, γ_mf = 1.1 (slip resistance designed at service load), 1.25 (slip resistance designed at ultimate load), F_o = minimum bolt tension = A_nbf₀ , where A_nb = net area of bolt, f₀ = 0.7fub , f_ub = ultimate tensile stress of bolt.

3. The maximum number of bolts of diameter 25mm that can be accomodated in one row in 200mm wide flat are:
a) 2
b) 3
c) 4
d) 5
Answer: b
Clarification: Minimum end distance = 2.5×25 = 62.5mm
Number of bolts that can be accommodated = (200-2×62.5)/25 = 3 bolts.

4. Calculate strength in shear of 16mm diameter of bolt of grade 4.6 for lap joint
a) 50 kN
b) 40 kN
c) 29 kN
d) 59 kN
Answer: c
Clarification: Bolts will be in single shear. Diameter of bolt = 16mm. Net area = 0.78x(π/4)x16²=156.83mm².
Strength of bolt in shear = Anbfub/(√3 x 1.25) = 156.83x400x10^-3/1.25x√3 = 28.97kN.

5. What is the value of kb in nominal bearing strength for a bolt of 20mm diameter of grade 4.6?
a) 0.5
b) 1
c) 0.97
d) 2
Answer: a
Clarification: diameter of bolt = 20mm, diameter of hole = 20+2 =22mm
e=1.5×22=33mm, p=2.5×20=50mm
e/3d₀ = 33/(3×22) = 0.5, p/3d₀ -0.25 = 50/(3×22) -0.25=0.5, f_ub /f_b = 400/410=0.975
k_b = minimum of (e/3d₀ , p/3d₀ -0.25, f_ub /f_b, 1) = 0.5.

6. Calculate bearing strength of 20mm diameter bolt of grade 4.6 for joining main plates of 10mm thick using cover plate of 8mm thick using single cover butt joint.
a) 70.26 kN
b) 109.82 kN
c) 50.18 kN
d) 29.56 kN
Answer: c
Clarification: diameter of bolt = 16mm, diameter of hole =16+2 =18mm
e=1.5×18=27mm, p=2.5×16=40mm
e/3d₀ = 27/(3×18) = 0.5, p/3d₀ -0.25 = 40/(3×18) -0.25=0.49, f_ub /f_b = 400/410=0.975
k_b = minimum of (e/3d₀, p/3d₀ -0.25, f_ub /f_b,1) = 0.49
bearing strength = 2.5kbdtfu/1.25 = 2.5×0.49x16x8x400x10-3/1.25 = 50.18 kN.

7. Find the number of HSFG bolts of diameter 20mm, grade 88 for connection of member carrying factored tensile load of 200kN when no slip is permitted.
a) 5
b) 4
c) 3
d) 2
Answer: b
Clarification: F_o=0.7f_ubA_nb=0.7x800x0.78x(π/4)x20²x10^-3=137.22 kN
Assume μ_f=0.5, n_e=1, k_h=1
Slip resistance of bolt = μ_f n_e k_h F_o/1.25 = 0.5x1x1x137.22/1.25 =54.88 kN
Number of bolts required = 200/54.88 = 3.64 = 4(approximately).

8. What is the efficiency of joint when strength of bolt per pitch length is 60kN and strength of plate per pitch length is 150kN?
a) 25%
b) 30%
c) 35%
d) 40%
Answer: d
Clarification: Efficiency = (strength of bolt per pitch length/ strength of plate per pitch length)x100 = 60×100/150 = 40%.

9. Strength of bolt is
a) minimum of shear strength and bearing capacity of bolt
b) maximum of shear strength and bearing capacity of bolt
c) shear strength of bolt
d) bearing capacity of bolt
Answer: a
Clarification: Strength of bolt is minimum of shear strength and bearing capacity of bolt. Design shear strength = nominal shear capacity/1.25, Design bearing strength = nominal bearing capacity/1.25.

10. Prying forces are
a) friction forces
b) shear forced
c) tensile forces
d) bending forces
Answer: c
Clarification: In connections subjected to tensile stresses, the flexibility of connected parts can lead to deformations that increases tension applied to bolts. This additional tension is called prying force.

Design of Steel Structures Multiple Choice Questions on “Plastic Theory”.

1. Structures designed using elastic analysis may be ______ than those designed using plastic analysis
a) lighter
b) heavier
c) of same weight
d) almost half times the weight
Answer: b
Clarification: In elastic design structures are designed for allowable stress level well below the elastic limit, whereas in plastic design structures are designed using ultimate load rather than yield stress, Hence, structures designed using elastic analysis may be heavier than those designed using plastic analysis.

2. Both elastic and plastic methods neglect ________
a) live load acting on structure
b) dead load acting on structure
c) deformations due to load
d) influence of stability
Answer: d
Clarification: Both elastic and plastic methods neglect the influence of stability, which may significantly affect the load carrying capacity of structures or elements which are slender and subjected to compressive stresses.

3. What is buckling?
a) Structural behaviour in which a deformation develops in direction of plane perpendicular to that of load which produced it
b) Structural behaviour in which a deformation does not develop in direction of plane perpendicular to that of load which produced it
c) Structural behaviour in which a deformation develop in direction of plane parallel to that of load which produced it
d) Structural behaviour in which a deformation develops in direction of plane along that of load which produced it
Answer: a
Clarification: Buckling may be defined as structural behaviour in which a deformation develops in direction of plane perpendicular to that of load which produced it. This deformation changes rapidly with variations in the applied load.

4. Which of the following relation about plastic moment is correct?
a) M_p = Z_p /f_y
b) M_p = Z_p + f_y
c) M_p = Z_pf_y
d) M_p = Z_p – f_y
Answer: c
Clarification: When every fibre of the section has a strain equal to or greater than ε_y=f_y/E_s , nominal moment strength is referred as plastic moment and is given by M_p = Z_pf_y , where Z_p = ∫ydA is plastic section modulus and f_y = yield stress.

5. What is plastic moment of resistance?
a) maximum moment in stress strain curve, the point where the curvature can increase indefinitely
b) maximum moment in stress strain curve, the point where the curvature can decrease indefinitely
c) minimum moment in stress strain curve, the point where the curvature can increase indefinitely
d) minimum moment in stress strain curve, the point where the curvature can decrease indefinitely
Answer: a
Clarification: In stress strain curve, the maximum moment is reached at a point where the curvature can increase indefinitely, neglecting the strain hardening benefits. This maximum moment is called plastic moment of resistance and the portion where this moment occurs is called as plastic hinge.

6. In elastic stage, equilibrium condition is achieved when neutral axis ___________ and in fully plastic stage, it is achieved when neutral axis ___________
a) is above centroid of the section, divides the section into two parts of one-third area and two-third area
b) is below centroid of the section, divides the section into two parts of one-third area and two-third area
c) is above centroid of the section, divides the section into two equal areas
d) passes through centroid of the section, divides the section into two equal areas
Answer: d
Clarification: In elastic stage, when bending stress varies from zero at neutral axis to maximum at extreme fibres, equilibrium condition is achieved when neutral axis passes through centroid of the section. In fully plastic stage, because the stress is uniformly equal to yield stress, equilibrium condition is achieved when neutral axis divides the section into two equal areas.

7. Which of the following relation is correct for plastic section modulus, Zo ?
a) Z_p = 2A(y₁+y₂)
b) Z_p = A(y₁+y₂)/2
c) Z_p = A(y₁+y₂)/4
d) Z_p = 4A(y₁+y₂)
Answer: b
Clarification: Z_p = A(y₁+y₂)/2, where A= area of cross section, y₁ and y₂ are centroids of portion above and below neutral axis respectively. Plastic modulus is defined as combined statical moment of cross sectional area above and below the equal-area axis.

8. Which of the following relation is correct about shape factor, v?
a) v = Z_p+Z_e
b) v = Z_pZ_e
c) v = Z_p/Z_e
d) v = Z_e/Z_p
Answer: c
Clarification: Shape factor, v = Z_p/Z_e = M_p/M_y , where Z_p and Z_e are plastic and elastic section modulus respectively, M_p and M_y are plastic and elastic moments respectively.

9. The shape factor does not depend on ___
a) material properties
b) cross sectional shape
c) moment of resistance
d) section modulus
Answer: a
Clarification: Shape factor, v = Z_p/Z_e = M_p/M_y, where Z_p and Z_e are plastic and elastic section modulus respectively, Mp and My are plastic and elastic moments respectively. This ratio is the property of cross-sectional shape and is independent of material properties.

10. Match the pairs with correct shape factor

		Cross section				Shape factor (average or maximum)
	A) Circular					(i) 1.8
	B) Rectangular					(ii) 1.14
	C) wide flange I-section (about major axis)	(iii) 1.7
	D) Channels (about minor axis) 			(iv) 1.5

a) A-i, B-ii, C-iii, D-iv
b) A-iv, B-iii, C-ii, D-i
c) A-iii, B-iv, C-ii, D-i
d) A-iii, B-ii, C-iv, D-i
Answer: c
Clarification: Shape factor, v = Z_p/Z_e = M_p/M_y. The shape factor for various cross section are (i) for circular = 1.7, (ii) for rectangular = 1.5, (iii) wide flange I-section (about major axis) = 1.09-1.18, average is 1.14, (iv) wide flange I-section (about minor axis) = 1.67, (v) channels (about major axis) = 1.16-1.22, average is 1.18, (vi) channels (about minor axis) = 1.8 .

Design of Steel Structures Multiple Choice Questions on “Design of Compression Members – I”.

1. A column that can support same load in compression as it can in tension is called
a) intermediate column
b) long column
c) short column
d) cannot be determined
Answer: c
Clarification: A column that can support same load in compression as it can in tension is called short column. Short column usually fail by crushing.

2. The strength of compression members subjected to axial compression is defined by curves corresponding to _______ classes
a) a, b, c and d
b) a, d
c) b, e, f
d) e, f, g
Answer: a
Clarification: The strength of compression members subjected to axial compression is defined by curves corresponding to a, b, c and d classes. The value of imperfection factor depends on type of buckling curve.

3. Which of the following is not a compression member?
a) strut
b) boom
c) tie
d) rafter
Answer: c
Clarification: Strut, boom and rafter are compression members, whereas tie is a tension member.

4. The best compression member section generally used is
a) single angle section
b) I-section
c) double angle section
d) channel section
Answer: b
Clarification: Generally, ISHB sections are used as compression members.

5. The best double-angle compression member section is
a) unequal angles with short leg connected
b) unequal angles with long leg connected
c) unequal angles on opposite side of gusset plate
d) unequal angles on same side of gusset plate
Answer: a
Clarification: Unequal angles with short leg connected are preferred as compression member section.

6. The flange is classified as semi-compact if outstand element of compression flange of rolled section is less than
a) 8.4ε
b) 10.5ε
c) 15.7ε
d) 9.4ε
Answer: c
Clarification: The flange is classified as semi-compact if outstand element of compression flange of rolled section is less than 15.7ε and for a welded section, less than 13.6ε.

7. The flange is classified as plastic if outstand element of compression flange of rolled section is less than
a) 8.4ε
b) 9.4ε
c) 10.5ε
d) 15.7ε
Answer: b
Clarification: The flange is classified as plastic if outstand element of compression flange of rolled section is less than 9.4ε and for a welded section, less than 8.4ε.

8. The outstand element of compression flange of a rolled section is 10.2 (ε=1). The flange will be classified as
a) compact
b) plastic
c) semi-compact
d) slender
Answer: a
Clarification: The flange is classified as compact if outstand element of compression flange of rolled section is less than 10.5ε and for a welded section, less than 9.4ε.

9. The design compressive stress of compression member in IS 800 is given by
a) Rankine Formula
b) Euler Formula
c) Perry-Robertson formula
d) Secant-Rankine formula
Answer: c
Clarification: The design compressive stress of axially loaded compression member in IS 800 is given by Perry-Robertson formula. IS 800:2007 proposes multiple columns curves in nin-dimensional form based on Perry-Robertson approach.

Design of Steel Structures Multiple Choice Questions on “Castellated Beams & Lintels”.

1. What is castellated beam?
a) beam with no openings in web
b) beam with number of regular openings in web and flange
c) beam with number of regular openings in web
d) beam with number of regular openings in flange
Answer: c
Clarification: A beam with number of regular openings in its web is called castellated beam. A castellated beam is formed by flame cutting a single rolled wide flange beam in a definite predetermined pattern and then rejoining the segments by welding to form a regular pattern of holes in the web.

2. The new rolled section of castellated beam will have depth
a) 50% more than original section
b) 50% less than original section
c) 25% less than original section
d) depth does not change
Answer: a
Clarification: The new rolled section of castellated beam will have depth at least 50% more and its section modulus increases by 2.25 times the original section. This allows the beam to span further than parent rolled section.

3. Castellated beams have ______ shear capacity than original beams
a) shear capacity does not change
b) twice
c) increased
d) reduced
Answer: d
Clarification: Castellated beams have reduced shear capacity. It has reduced shear capacity due to stress concentrations near the openings.

4. Which of the following measures can be taken to improve shear capacity of castellated beams?
a) openings can be made away from neutral axis
b) openings can be made close to neutral axis
c) making cuts in straight manner
d) by not using stiffenings
Answer: b
Clarification: Shear capacity of castellated beams can be improved by making openings close to neutral axis and making cuts in a wavy manner. Stiffening can be provided at load concentrations and reaction points to improve its shear carrying capacity.

5. Which of the following is not an advantage of castellated beam?
a) light in weight
b) can be assembled fast
c) cheaper
d) high fire resistance than original rolled section
Answer: d
Clarification: Castellated beams are light in weight, cheaper, they have relatively high resistance and can be assembled fast at the construction site. They are less fire resistant than normal rolled sections. Castellated beams can very easily be cambered and cranked.

6. In which of the following cases are castellated beam desirable?
a) when more span to be covered than rolled section
b) when beam subjected to substantial concentrated loads
c) when beam to be used as continuous beam
d) when higher fire resistance than rolled section required
Answer: a
Clarification: The section of castellated beam will have more depth and section modulus than original rolled section. This allows the beam to span further than parent rolled section. Castellated beams may not be desirable when beam is subjected to substantial concentrated loads, or when castellated beam is used as a continuous beam across several supports. Castellated beams are less attractive when very high requirements for fire resistance are required because the fire resistant coating has to be around 20% thicker than for rolled sections in order to obtain the same fire resistance as rolled section.

7. Match the pairs

   (A) Vierendeel mechanism		  (i) caused by heavy loading and short span
   (B) Lateral torsional buckling of web  (ii) caused due to excessive horizontal shear
   (C) Rupture of welded joint in web	  (iii) due to excessive deformation across openings in web
   (D) Web Buckling			  (iv)caused by large shear

a) A-i, B-ii, C-iii, D-iv
b) A-iv, B-iii, C-ii, D-i
c) A-iii, B-iv, C-ii, D-i
d) A-i, B-iv, C-iii, D-ii
Answer: c
Clarification: There are number of possible modes of failure for castellated beams. Some of them are as follows: (i) Vierendeel mechanism – occurs due to excessive deformation across one of the openings in web and formation of hinges in corners of castellation, (ii) Lateral torsional buckling of web – caused by large shear at welded joint, (iii) Rupture of welded joint in web – caused due to excessive horizontal shear at welded joint in the web, (iv) Web Buckling – caused by heavy loading and short span of beam, this may be avoided at support by filling firt castellation by welding plate in the hole.

8. What are lintels?
a) beams provided in foundation
b) beams on roof of building
c) columns above openings in wall
d) beams above openings in wall
Answer: d
Clarification: Beams provided above the openings in walls to support masonry that comes in between the opening and slab above are called as lintels. It is desirable that lintel is built flush from both the sides of the walls.

9. _____ section is suitable for small openings and _____ section is suitable for large openings
a) flat, I-section
b) I-section, flat
c) angles, flat
d) angles, angles
Answer: a
Clarification: Flats and plate sections are used for small openings. For openings of moderate dimension, back-to-back angles and inverted T-sections are best options. For large openings, channels, I-sections or built-up sections are preferred. If there is any doubt about lateral support from the wall, I-section with plates can be used.

10. Design of lintel is carried out for
a) weight of slab
b) no load is considered from masonry load above the opening
c) small portion of masonry load above the opening
d) large portion of masonry load above the opening
Answer: c
Clarification: Design loads for lintels are not well defined because it is not certain as how much load from masonry will come over lintel. It is assumed that after setting of mortar, load from masonry is distributed by arch action. Design of lintel is carried out for small portion of masonry load above the opening.

11. When the slab over lintel is above apex of equilateral triangle formed on lintel, load of masonry is considered as
a) rectangular load
b) triangular load
c) trapezoidal load
d) no load is considered
Answer: b
Clarification: When the slab over lintel is above apex of equilateral triangle formed on lintel, the load of masonry in the triangle thus formed is assumed to act over it. When the design load is from triangular portion of masonry , the maximum moment will be Wl/6, where W = triangular load from masonry and l = effective span of lintel.

12. When the slab over lintel is below apex of equilateral triangle formed on lintel, load of masonry is considered as
a) rectangular load
b) triangular load
c) trapezoidal load
d) no load is considered
Answer: a
Clarification: When the slab over lintel is below apex of equilateral triangle formed on lintel, the load of masonry in the rectangle is considered. The load of masonry in the rectangle is assumed to act over by taking length equal to span of lintel and height equal to clear height of slab above the lintel.