PDF of MCQ
PDF of MCQ
PDF of MCQ
a) high toughness
b) reduced ductility
c) high strength
d) reduced strength
View Answer
Answer: b
Explanation: High carbon steel contains high carbon content. Hence it has reduced ductility,
toughness and weldability.
Answer: a
Explanation: High carbon steel is used in transmission lines and microwave towers where relatively
light members are joint by bolting.
3. What is the permissible percentage of micro-alloys in medium and high strength micro-alloyed
steel?
a) 0.1%
b) 0.5%
c) 0.25%
d) 1.0%
View Answer
Answer: c
Explanation: Medium and High strength micro-alloyed steel have low carbon content, but alloys such
as niobium, vanadium, titanium or boron are added to achieve high strength.
Answer: d
Explanation: Fire resistant steels are also called as thermomechanically treated steel. They perform
better than ordinary steel under fire.
5. What is the minimum percentage of chromium and nickel added to stainless steel?
a) 0.5%, 10.5%
b) 2%, 20%
c) 10.5%, 0.5%
d) 30%, 50%
View Answer
Answer: c
Explanation: Stainless steel are low carbon steels to which a minimum of 10.5% chromium
(maximum 20%) and 0.5% nickel is added.
6. Match the pair of Type of steel with its ultimate tensile strength :
Answer: c
Explanation: Ultimate tensile strength is the capacity of material to withstand loads tending to
elongate. It is the maximum stress that a material can withstand while being stretched or pulled. The
Answer: a
Explanation: Weathering steel are low-alloy atmospheric corrosion-resistant steel. They are often
left unpainted. They have an ultimate tensile strength of 480 MPa.
Answer: b
Explanation: For class 4.6, ultimate strength = 4×100 = 400 N/mm2
yield strength / ultimate strength = 0.6
yield strength = 0.6×400 = 240 N/mm2.
Answer: d
Explanation: Size of hole = nominal diameter of fastener + clearances
Clearance may be standard size, oversize, short slotted or long slotted.
Answer: d
Explanation: High strength bolt may be used for slip resistant and bearing type connection. At
serviceability, HSFG bolts do not slip and the joints are called slip resistant connections. At ultimate
4. Which of the following is advantage of HSFG bolts over bearing type bolts?
a) joints are not rigid
b) bolts are subjected to shearing and bearing stresses
c) high strength fatigue
d) low static strength
View Answer
Answer: c
Explanation: The advantages of HSFG bolts over bearing type bolts are : (i) joints are rigid, (ii) bolts
are not subjected to shearing and bearing stresses as load transfer is mainly due to friction, (iii) high
5. Which of the following is correct for pitch of the bolts and gauge?
a) pitch is measured along direction of load, gauge is measured perpendicular to direction of load
b) pitch is measured perpendicular direction of load, gauge is measured along to direction of load
c) pitch is measured along direction of load, gauge is measured along to direction of load
d) pitch is measured perpendicular direction of load, gauge is measured perpendicular to direction of
load
View Answer
Answer: a
Explanation: Pitch is centre to centre spacing of bolts in a row, measured along direction of load.
Gauge is the distance between two consecutive bolts of adjacent row measured at right angles to
Answer: d
Explanation: Pitch is centre to centre spacing of bolts in a row, measured along direction of load.
Distance between centre to centre of fasteners shall not be more than 2.5 times nominal diameter of
Answer: b
Explanation: Distance between centre of any two adjacent fasteners shall not exceed 32t or 300mm,
whichever is less where t is thickness of thinner plate.
8. Pitch shall not be more than ___ in tension member and _______ in compression member.
a) 12t, 16t, where t = thickness of thinner plate
b) 20t, 16t, where t = thickness of thinner plate
c) 16t, 12t, where t = thickness of thinner plate
d) 16t, 20t, where t = thickness of thinner plate
View Answer
Answer: c
Explanation: Pitch shall not be more than 16t or 200mm, whichever is less in tension member where
t is thickness of thinner plate. Pitch shall not be more than 12t or 200mm, whichever is less in
Answer: a
Explanation: Spacing between centres of fasteners may be increased by 50% when fasteners are
staggered at equal interval and gauge does not exceed at 75mm, subjected to maximum spacing
Answer: d
Explanation: Edge distance is distance at right angles to the direction of stress from centre of hole to
adjacent edge. End distance is distance in the direction of stress from centre of hole to end of
Answer: b
Explanation: Steel tension members are those structural elements that are subjected to direct axial
tensile loads, which tend to elongate the members. A member in pure tension can be stressed up to
Answer: d
Explanation: The strength of tensile members is influenced by factors such as length of connection,
size and spacing of fasteners, size and spacing of fasteners, net area of cross section, type of
Answer: a
Explanation: Single angle section with bolted connection produce eccentricity about both planes,
whereas single angle section with welded connection may produce eccentricity about one plane
Answer: c
Explanation: Single angle members are used in towers and as web members in trusses. Double
angle sections are used as chord members in light roof trusses or in situations where some rigidity
Answer: a
Explanation: Strand consists of individual wires wound helically around a central core, wire rope is
made of several strand laid helically around a core. Wire ropes are exclusively used for hoisting
Answer: d
Explanation: Cables used as floor suspenders in suspension bridges are made from individual
strands wound together in rope like fashion. Cables in form of wires ropes and strands are used in
Answer: b
Explanation: Bars and rods are used as tension members in bracing systems, sag rods to support
purlin between trusses, to support girts in industrial buildings, where light structure is desirable.
Answer: c
Explanation: Sagging of members by built up bars and rods may be minimised by limiting length
diameter or thickness ratio or by fabricating the rod/bar short of its required theoretical length by
9. Which of the following type of tension member is not mainly used in modern practice?
a) open section such as angles
b) flat bars
c) double angles
d) circular section
View Answer
Answer: b
Explanation: Tension members were generally made of flat bars earlier. But modern practice is to
use mainly the following sections for tension members: (i)open sections such as angles, channels
10. Which among the following comparison between angle and flat bars is not true?
a) for light loads, angles are preferred over flat bars
b) flat bar tension members tend to vibrate during passage of load in light bridges
c) flat bars are used instead of angles in case of stress reversal
d) angles are used instead of flat bars in case of stress reversal
View Answer
Answer: c
Explanation: For light loads, angles are preferred over flat bars. In many light bridges, flat bar
tension members tend to vibrate during passage of load. In case of stress reversal angles are more
Answer: a
Explanation: Two angle sections can either be placed back-to-back on the same side of gusset
plate, or back-to-back on the opposite side of gusset plate. When angles are connected on the
Answer: c
Explanation: Built-up members, made up of two or more plates or shapes and connected to act as
single member, are formed primarily to meet required area which cannot be provided by single rolled
1. Which of the following equation is correct for bolt subjected to combined shear and tension?
a) (Vsb/Vdb)2 + (Tsb/Tdb)2 ≤ 1
b) (Vsb/Vdb)2 + (Tsb/Tdb)2 ≥ 1
c) (Vsb/Vdb) + (Tsb/Tdb) ≤ 1
d) (Vsb/Vdb) + (Tsb/Tdb) ≥ 1
View Answer
Answer: a
Explanation: Bolt required to satisfy both shear and tension at the same time should satisfy
(Vsb/Vdb)2 + (Tsb/Tdb)2 ≤ 1 , where Vsb= factored shear force, Vdb = design shear capacity, Tsb =
2. Shear Capacity of HSFG bolts is
a) μfnekhFo
b) μfnekhFoγmf
c) μfnekhoγmf
d) μfnekhFo/γmf
View Answer
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
View Answer
Answer: b
Explanation: 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
View Answer
Answer: c
Explanation: 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
View Answer
Answer: a
Explanation: 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, fub /fb = 400/410=0.975
kb = minimum of (e/3d0 , p/3d0 -0.25, fub /fb, 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
View Answer
Answer: c
Explanation: 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, fub /fb = 400/410=0.975
kb = minimum of (e/3d0, p/3d0 -0.25, fub /fb,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
View Answer
Answer: b
Explanation: Fo=0.7fubAnb=0.7x800x0.78x(π/4)x202x10-3=137.22 kN
Assume μf=0.5, ne=1, kh=1
Slip resistance of bolt = μf ne kh Fo/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%
View Answer
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
View Answer
Answer: a
Explanation: 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
Answer: d
Explanation: Angles if axially loaded through centroid can be designed as plates. Angles connected
to gusset plates by welding or bolting only through one of the two legs results in eccentric loading,
Answer: a
Explanation: (i) The effect of gusset plate thickness on ultimate tensile strength is not significant, (ii)
the thickness of angle has no significant influence on member strength, (iii) the net section efficiency
3. The additional factor to be added for angles for design strength of tension member corresponding
to net section rupture is given by :
a) βAg0fyγm0
b) βAg0fy/γm0
c) βAg0γm0
d) βAg0/fyγm0
View Answer
Answer: b
Explanation: The design strength of angle section governed by tearing at net section is given by
Tdn = (0.9Ancfu/γm1) + βAg0fy/γm0 , where Anc = net area of connected leg, Ag0 = gross area of
4. The constant β in βAg0fy/γm0 for tensile strength of angle section does not depend on :
a) area of unconnected leg
b) size of outstanding leg
c) ultimate stress of material
d) thickness of outstanding leg
View Answer
Answer: a
Explanation: β = 1.4 – 0.076[(bs/Lc)(w/t)(fy/fu)] , where fu and fy are ultimate and yield stress of
material, w and t are size and thickness of outstanding leg respectively, bs is the shear distance
Answer: c
Explanation: Tdn = (0.9Ancfu/γm1) + βAg0fy/γm0, where β = 1.4 – 0.076[(bs/Lc)(w/t)(fy/fu)] and β ≤
fuγm0/fyγm1, β ≥ 0.7.
6. What is the maximum value of β in βAg0fy/γm0 for tensile strength of angle section?
a) 1.2
b) 0.9
c) 1.4
d) 0.7
View Answer
Answer: d
Explanation: Tdn = (0.9Ancfu/γm1) + βAg0fy/γm0, where β = 1.4 – 0.076[(bs/Lc)(w/t)(fy/fu)] and β ≥ 0.7.
7. What is the value of partial factor of safety for material α for preliminary design for angle section
as per IS code for three bolts in connection?
a) 0.6
b) 0.7
c) 0.8
d) 1.0
View Answer
Answer: b
Explanation: As per IS code, the equation for preliminary design of angle tension member with
partial factor of safety for material is given by T dn = αAnfy/γm1, where α = 0.6 for one or two bolts, 0.7
Answer: a
Explanation: Strength of members with punched holes may be 10-15% less than the members with
drilled holes. This is due to strain hardening effect of material around punched holes and
Answer: c
Explanation: The bolt holes reduce the area of cross section available to carry tension and hence
reduce the strength of tension member.
10. Staggering of holes __________ the load carrying capacity of tension member
a) reduces
b) improves
c) does not affect
d) halves
View Answer
Answer: b
Explanation: Staggering of holes improves the load carrying capacity of tension member for given
number of bolts. The failure paths may occur along sections normal to axis of member, or they may
Answer: d
Explanation: The actual failure mode in bearing depends on end distance, bolt diameter and
thickness of the connected material.
12. The shear lag effect _____ with increase in connection length
a) increases
b) reduces
c) does not change
d) doubles
View Answer
Answer: b
Explanation: The shear lag effect increases with increase in connection length. The shear lag
reduces the effectiveness of component plates of tension member that are not connected directly to
Answer: d
Explanation: Reduction in ductility tends to reduce strength of member. An increase in ductility tends
to increase net section strength by allowing better plastic redistribution of stress concentration over
Answer: a
Explanation: Residual stress result in local early strain hardening and reduce plastic range of
member. Residual stresses have no consequences with respect to static strength of member, they
Answer: b
Explanation: Structural member which is subjected to compressive forces along its axis is called
compression member. Compression members are subjected to loads that tend to decrease their
Answer: d
Explanation: if the net end moments are zero, the compression member is required to resist load
acting concentric to original longitudinal axis of member and is called axially loaded column or
Answer: c
Explanation: If the net end moments are not zero, the member will be subjected to axial force and
bending moments along its length. Such members are called beam-columns.
Answer: a
Explanation: The vertical compression members in a building supporting floors or girders are
normally called as columns. They are sometimes called as stanchions. They are subjected to heavy
Answer: c
Explanation: The compression members used in roof trusses and bracings are called as struts. They
may be vertical or inclined and normally have small lengths. the top chord members of a roof truss
Answer: b
Explanation: Short compression members at junction of columns and roof trusses or beams are
called knee braces. They are provided to avoid moment.
Answer: b
Explanation: Axial loading on columns in buildings is due to loads from roofs, floors, and walls
transmitted to the column through beams and also due to its own self weight.
Answer: d
Explanation: Wind loads in multi-storey buildings are usually applied at respective floor levels and
are assumed to be resisted by bracings. Hence in braced buildings wind loads do not cause large
Answer: c
Explanation: In industrial buildings, loads from crane and wind cause moments in columns. In such
cases, wind load is applied to the column through sheeting rails and may be taken as uniformly
Answer: a
Explanation: The strength of column depends on material of column, cross sectional configuration,
length of column, support conditions at the ends, residual stresses, imperfections.
Answer: c
Explanation: Imperfections in column include material not being isotropic and homogenous,
geometric variations of columns and eccentricity of load.
1. For the calculation of net area of flat with staggered bolts, the area to be deducted from gross
area is :
a) nd
b) n’p2t/8g
c) ndt – n’p2t/4g
d) nd + n’p2t/4g
View Answer
Answer: c
Explanation: The net area of flat with staggered hole is given by : A = (b – ndh + n’p2/4g)t, where b =
width of plate, n = number of holes in zig-zag line, n’ = number of staggered pitches, p = pitch
2. What is the net section area of steel plate 40cm wide and 10mm thick with one bolt if diameter of
bolt hole is 18mm?
a) 38.2 cm2
b) 20 cm2
c) 240 mm2
d) 480 mm2
View Answer
Answer: a
Explanation: b = 40cm = 400mm, t = 10mm, dh = 18mm
Net section area = 400×10 – 16×10 = 3820mm2 = 38.2 cm2.
3. Which section to be considered in the design for the net area of flat?
a) 1-5-6-3
b) 2-7-4
c) 1-5-7-4
d) 1-5-7-6-3
View Answer
Answer: d
Explanation: The section giving minimum area of plate is considered for design. So, section 1-5-7-6-
3 is used for net area of flat.
4. Calculate the minimum effective net area for the given section (300mm width, 10mm thick)
connected to a 10 mm thick gusset plate by 18mm diameter bolts.
a) 2796mm2
b) 2681mm2
c) 2861mm2
d) 3055mm2
View Answer
Answer: b
Explanation: B = 300mm, t = 10mm, dh = 18+2 =20mm, n = 3, n’ = 1, p = 75mm, g = 50mm
Effective net area = (B-ndh+n’p2/4g)t = (300 – 3×20 + 1×752/4×50)x10 = 2681.25 mm2.
5. What is the net area for the plate 100 x 8 mm bolted with a single bolt of 20mm diameter in case
of drilled hole ?
a) 624 mm2
b) 756 mm2
c) 800 mm2
d) 640 mm2
View Answer
Answer: d
Explanation: In case of drilled hole, dh = 20mm
Net Area An = Ag – dht = 100 x 8 – 20 x 8 = 640mm2.
6. Determine the effective net area for angle section ISA 100 x 75 x 12 mm, when 100mm leg is
connected to a gusset plate using weld of length 140mm.
a) 1795 mm2
b) 1812 mm2
c) 1956 mm2
d) 2100 mm2
View Answer
Answer: c
Explanation: Net area of connected leg, Anc = (100 – 12/2) x 12 = 1128 mm2
Net area of outstanding leg, Ago = (75 – 12/2) x 12 = 828 mm2
Total net area = 1128 + 828 = 1956 mm2.
7. Calculate the value of β for the given angle section ISA 150x115x8mm of Fe410 grade of steel
connected with gusset plate : Length of weld = 150mm
a) 0.89
b) 0.75
c) 0.5
d) 1
View Answer
Answer: a
Explanation: w=115mm, t=8mm, b=115mm, Lc=150mm, fy=250MPa, fu=410MPa
β = 1.4 – [0.076 (w/t)(fy/fu)(bs/Lc)] = 1.4 – [ 0.076 x (115/8) x (250/410) x (115/150)] = 0.89 (>0.7) .
8. Calculate the tensile strength due to gross section yielding of an angle section 125 x 75 x 10mm
of Fe410 grade of steel connected with a gusset plate.
a) 780 kN
b) 586.95 kN
c) 432.27 kN
d) 225.36 kN
View Answer
Answer: c
Explanation: fu = 410MPa, fy = 250MPa, γm0 = 1.1, γm1 = 1.25,
For ISA 125 x 75 x 10mm : gross area Ag = 1902 mm2
Tensile strength due to gross section yielding, Tdg = Agfy/γm0 = 1902 x 250 x 10-3 / 1.1 = 432.27 kN.
9. A single unequal angle 100 x 75 x 10 of Fe410 grade of steel is connected to a 10mm thick
gusset plate at the ends with six 16mm diameter bolts with pitch of 40mm to transfer tension. Find
a) 526.83 kN
b) 385.74 kN
c) 450.98 kN
d) 416.62 kN
View Answer
Answer: d
Explanation: dh=18 mm, fu = 410MPa, fy = 250MPa, γm0 = 1.1, γm1 = 1.25
Anc = (100 – 10/2 – 18) x 10 = 770 mm2, Ag0 = (75 – 10/2) x 10 = 700 mm2
β = 1.4 – 0.076(w/t)(fy/fu)(bs/Lc)
= 1.4 – 0.076 [(75-5)/10] [250/410] [{(75-5)+(100-40)}/{40×5}] = 1.19 > 0.7 and <
1.44[(410/250)(1.1/1.25)] (=2.07)
Tdn = 0.9fuAnc /γm1 + βAg0fy /γm0
= [0.9x410x770/1.25 + 1.19x700x250/1.1] x 10-3 = 416.62 kN.
10. Determine block shear strength of tension member shown in figure if grade of steel is Fe410.
a) 309.06 kN
b) 216.49 kN
c) 258.49 kN
d) 326.54 kN
View Answer
Answer: b
Explanation: fu = 410 MPa, fy = 250 MPa, γm0 = 1.1, γm1 = 1.25
Avg = ( 1×100 + 50 ) x 8 = 1200 mm2
Avn = (1×100 + 50 – (2 – 1/2)x20 ) x 8 = 1440 mm2
Atg = 35 x 8 = 280 mm2
Atn = (35 – 1/2 x 20) x 8 = 200 mm2
Tdb1 = (Avgfy/√3 γm0)+(0.9Atnfu/γm1) = [(1200×250/ 1.1x√3) + (0.9x200x410 / 1.25) ] x 10-3 = 216.49
Tdb2 = (Atgfy/ γm0)+(0.9Avnfu/√3 γm1) = [ (0.9x1440x410 / 1.25x√3) + (280×250/1.1) ] x 10-3 = 309.06
kN
kN
Block shear strength of tension member is 216.49 kN.
Answer: c
Explanation: Effective length of compression member is distance between points of contraflexure. It
should be derived from actual length and end conditions.
Answer: b
Explanation: Magnitude of effective length depends upon rotational restraint supplied at end of
compression member and upon resistance to lateral movement provided.
Answer: d
Explanation: Smaller the effective length of particular compression member, smaller is the danger of
lateral buckling and greater is the load carrying capacity.
4. What is the effective length when both ends of compression member are fixed?
a) 0.65L
b) 0.8L
c) L
d) 2L
View Answer
Answer: a
Explanation: The effective length of compression member when both ends of compression member
are fixed is 0.65L (i.e. L/√2), where L is the length of the member.
5. What is the effective length when both ends of compression member are hinged?
a) 0.65L
b) 0.8L
c) L
d) 2L
View Answer
Answer: c
Explanation: The effective length of compression member when both ends of compression member
are hinged is L, where L is the length of the member.
6. What is the effective length when one end of compression member is fixed and other end is free?
a) 0.65L
b) 0.8L
c) L
d) 2L
View Answer
Answer: d
Explanation: The effective length of compression member when one end is fixed and other end is
free is 2L, where L is the length of the member.
7. What is the effective length when one end of compression member is fixed and other end is
hinged?
a) 0.65L
b) 0.8L
c) L
d) 2L
View Answer
Answer: b
Explanation: The effective length of compression member when one end is fixed and other end is
hinged is 0.8L, where L is the length of the member.
Answer: a
Explanation: The tendency of member to buckle is usually measured by its slenderness ratio.
Slenderness ratio of member is ratio of effective length to appropriate radius of gyration (λ = kL/r).
Answer: c
Explanation: Maximum radius of gyration is obtained when material of section is farthest from
centroid i.e. away from its axis
Answer: a
Explanation: Battens are plates or any other rolled section used to connect the main components of
compression members. Battens should be placed opposite to each other on the two parallel faces of
Answer: c
Explanation: Battens should be designed to resist transverse shear force which is 2.5% of total axial
force on whole compression member. This transverse shear force is divided equally in all parallel
Answer: b
Explanation: Battens should be designed to resist longitudinal shear equal to V = Vt L0 / ns , where
Vt is transverse shear force, L0 is distance between centre-to-centre of battens longitudinally, s is
Answer: d
Explanation: Battens should be designed to resist moment equal to Vt L0/2n , where Vt = transverse
shear force, L0 = distance between centre-to-centre of battens longitudinally, n = number of parallel
Answer: c
Explanation: When plates are used for battens, effective depth between end bolts/rivets or welds
should not be less than twice the width of one member in plane of batten, not less than
Answer: a
Explanation: Thickness of batten plates shall be not less than 1/50th of distance between innermost
connecting transverse bolts/rivets or welds perpendicular to main member i.e. t > (1/50)(s + 2g),
Answer: d
Explanation: Length of weld connecting each end of batten shall be more than half the depth of
plate. Length of weld connecting each end of batten should be such that at least one third of its
8. Effective slenderness ratio of battened column shall be ____ of actual slenderness ratio of
column
a) 0.5 times
b) 1.1 times
c) 2 times
d) 2.5 times
View Answer
Answer: b
Explanation: Effective slenderness ratio of battened column shall be 1.1 times the maximum actual
slenderness ratio of column to account for shear deformation effects.
9. Maximum spacing of batten should be such that slenderness ratio of component member should
be
a) not greater than 50
b) greater than 50
c) greater than 0.7 times slenderness ratio of member as a whole
d) greater than slenderness ratio of member as a whole
View Answer
Answer: a
Explanation: Maximum spacing of batten should be such that slenderness ratio of component
member should be not greater than 50 or 0.7 times slenderness ratio of member as a whole, about
10. Which of the following is true about effective depth of end batten?
a) it should be less than distance between centre of gravity of component
b) it should be half the distance between centre of gravity of component
c) it should be less than twice the width of component member
d) it should be greater than twice the width of component member
View Answer
Answer: d
Explanation: Effective depth of end batten should not be less than distance between centre of
gravity of component and should be greater than twice the width of component member.
Answer: b
Explanation: Depth of intermediate batten is taken as three fourth of the effective depth of end
batten and should be more than twice the width of component member.
Answer: b
Explanation: Thickness of batten should not be less than 1/50th of distance between innermost
connecting lines of rivets/bolts/welds perpendicular to main member.
13. A laced column is_____ than battened column for same load
a) equally strong
b) weaker
c) stronger
d) cannot be compared
View Answer
Answer: c
Explanation: A laced column is stronger than battened column for same load, unsupported length
and end conditions
1. The radius of gyration of combined column about axis perpendicular to plane of lacing should be
_____ than about axis parallel to plane of lacing.
a) cannot be compared
b) smaller
c) greater
d) equal to
View Answer
Answer: c
Explanation: The radius of gyration of combined column about axis perpendicular to plane of lacing
should be greater than about axis parallel to plane of lacing.
Answer: a
Explanation: Lacing system should be uniform throughout length of column. Single and double laced
systems should not be provided on opposite sides of same member. Lacings and battens should not
3. Lacing shall be designed to resist a total transverse shear equal to ____ of axial force in member
a) 5%
b) 1%
c) 4.3%
d) 2.5%
View Answer
Answer: d
Explanation: Lacing can be designed to resist a total transverse shear at any point in the member
equal to 2.5% of axial force in member. This shear shall be divided among lacing systems in parallel
Answer: b
Explanation: Slenderness ratio is the ratio of effective length by radius of gyration. Slenderness ratio
of lacing shall not exceed 145.
Answer: a
Explanation: Effective length shall be taken as length between inner end bolts/rivets of bars for
single lacings and 0.7 times length between inner end bolts/rivets of bars for double lacings. For
6. Minimum width of lacing bars shall _______
a) be less than 3 times diameter of connecting bolt/rivet
b) be less than 5 times diameter of connecting bolt/rivet
c) not be less than 3 times diameter of connecting bolt/rivet
d) be less than 2 times diameter of connecting bolt/rivet
View Answer
Answer: c
Explanation: Minimum width of lacing bars shall not be less than approximately 3 times the diameter
of connecting bolt/rivet.
Answer: b
Explanation: Thickness of lacing member should not be less than 1/40th of the effective length for
single lacing and not less than 1/60th of the effective length for double lacing.
Answer: b
Explanation: The spacing of lacing bars should be such that maximum slenderness ratio of
component of main members between two consecutive lacing connection is not greater than 50. It
Answer: d
Explanation: When welded lacing bars overlap main members, amount of lap should not be less
than 4 times thickness of bar. Welding is to be provided along each side of bar for full length of lap.
10. lacing bars shall be inclined at an angle of ___ to axis of built up member.
a) 20o
b) 35o
c) 50o
d) 90o
View Answer
Answer: c
Explanation: Lacing bars shall be inclined at an angle of 40o to 70o to axis of built up member.
Answer: b
Explanation: Effective slenderness ratio of laced column shall be taken as 5% more than the
maximum slenderness ratio of column i.e. 1.05 times the maximum slenderness ratio of column, to
Answer: d
Explanation: Compressive strength in lacing bars is equal to (Vt /N) cosecΘ for single lacing system
and (Vt/2N) cosecΘ for double lacing system, where N = number of shear resisting elements.
Answer: a
Explanation: Minimum radius of gyration for lacing flats is t/√12 , where t is thickness of flat.
14. The load on rivet/bolt when two lacing flats are connected at same point is
a) (Vt / N) cotΘ
b) 2(Vt / N) cotΘ
c) 2Vt N cotΘ
d) Vt NcotΘ
View Answer
Answer: b
Explanation: Strength of bolt /rivet should be greater than load coming over rivet/bolt. The load on
rivet/bolt when two lacing flats are connected at same point is 2(Vt/N)cotΘ.
15. The load on rivet/bolt when two lacing flats are connected at different point is
a) (Vt / N) cotΘ
b) 2(Vt / N) cotΘ
c) 2Vt N cotΘ
d) Vt NcotΘ
View Answer
Answer: a
Explanation: Strength of bolt /rivet should be greater than load coming over rivet/bolt. The load on
rivet/bolt when two lacing flats are connected at different point is (Vt/ N) cotΘ