Subject :- Design of Steel Structures / DOS-I(Steel)LSM / SE-I(Steel)
Unit-II
:- Joints (Connections) in Steel structures
1. What is the yield strength of bolt of class 4.6?
a) 400 N/mm2
b) 250 N/mm2
c) 240 N/mm2
d) 500 N/mm2
Answer: c
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.
2. What is the yield strength of bolt of class 8.8?
a) 400 N/mm2
b) 480 N/mm2
c) 240 N/mm2
d) 500 N/mm2
Answer: b
Explanation: For class 8.8, ultimate strength = 8×100 = 800 N/mm2
yield strength / ultimate strength = 0.6
yield strength = 0.6×800 = 480 N/mm2.
3. Which of the following is correct?
a) size of hole = nominal diameter
of fastener – clearances
b) size of hole = nominal diameter of fastener x clearances
c) size of hole = nominal diameter of fastener / clearances
d) size of hole = nominal diameter of fastener + clearances
Answer: d
Explanation: Size of hole = nominal diameter of fastener + clearances
Clearance may be standard size, oversize, short slotted or long slotted.
4. High strength bolt is used for ____________
a) shear connection
b) both slip resistant and bearing type connection
c) slip resistant connection only
d) bearing type connection only
Answer: b
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 load, HSFG bolts do
not slip and the joints behave like bearing type connections.
5. What is the advantage of HSFG bolts over
bearing type bolts?
a) joints are not rigid
b) bolts are subjected to shearing and bearing stresses
c) low static strength
d) high strength fatigue
Answer: d
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 static strength
due to high frictional resistance, (iv) high strength fatigue since nuts are
prevented from loosening, (v) smaller number of bolts required results into
smaller length of gusset plates.
6. What is the correct statemrent for pitch of the
bolts and gauge?
a) pitch is measured perpendicular
direction of load, gauge is measured along to direction of load
b) pitch is measured along direction of load, gauge is measured perpendicular
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
Answer: b
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 the direction of load.
7. the minimum pitch distance should not be less
than…………?
a) 2.0 x nominal diameter of
fastener
b) 3.0 x nominal diameter of fastener
c) 1.5 x nominal diameter of fastener
d) 2.5 x nominal diameter of fastener
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 fasteners.
8. Maximum pitch distance shall not exceed less of
_____________0r 300mm.
a) 16 x thickness of thinner plate
b) 32 x thickness of thinner plate
c) 40 x thickness of thinner plate
d) 20 x thickness of thinner plate
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.
9. Pitch shall not be more than ___ in tension
member and _______ in compression member.
a) 16t, 12t, where t = thickness of
thinner plate
b) 12t, 16t, where t = thickness of thinner plate
c) 20t, 16t, where t = thickness of thinner plate
d) 16t, 20t, where t = thickness of thinner plate
Answer: a
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 compression member, where t is
thickness of thinner plate.
10. In case of staggered pitch, pitch may be
increased by ______
a) 20%
b) 100%
c) 50%
d) 30%
Answer: c
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 condition.
11. What is the difference between end and edge
distance?
a) Edge distance is measured
parallel to direction of stress, while end distance is measured perpendicular
to direction of stress
b) Edge distance is measured perpendicular to direction of stress, while end
distance is measured parallel to direction of stress
c) Edge distance is measured parallel to direction of stress, while end
distance is measured parallel to direction of stress
d) Edge distance is measured perpendicular to direction of stress, while end
distance is measured perpendicular to direction of stress
Answer: b
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 element.
12. Maximum gauge length is should not exceed_________
a) 100-4t, where t is thickness of
thinner plate
b) 100+4t, where t is thickness of thinner plate
c) 4t, where t is thickness of thinner plate
d) 100mm
Answer: b
Explanation: Distance between centres of any two consecutive fasteners
in line adjacent and parallel to edge of outside plate shall not exceed
(100+4t) or 200mm, whichever is less in compression and tension members.
13. Minimum edge distance and end distance for
rolled, machine flame cut is
a) 1.7 x hole diameter
b) 1.2 x hole diameter
c) 2.0 x hole diameter
d) 1.5 x hole diameter
Answer: d
Explanation: Minimum edge distance and end distance from centre of any
hole to nearest edge of plate shall not be (i) less than 1.7 x hole diameter,
in case of sheared or hand flame cut edge, (ii) less than 1.5 x hole diameter,
in case of rolled, machine flame cut.
14. Maximum edge distance should not exceed ______
a) 10tε, where ε = √(250/fy), t = thickness of thinner outer plate
b) 12tε, where ε = √(250/fy), t = thickness
of thinner outer plate
c) 20tε, where ε = √(250/fy), t = thickness
of thinner outer plate
d) 16tε, where ε = √(250/fy), t = thickness
of thinner outer plate
Answer: b
Explanation: Maximum edge distance should not exceed 12tε, where ε =
√(250/fy), t = thickness of thinner outer plate. If members
are exposed to corrosive influence, it shall not exceed (40+4t), where t =
thickness of thinner connected plate.
15. Tacking fasteners are used when _______
a) minimum distance between centre
of two adjacent fasteners is exceeded
b) maximum distance between centre of two adjacent fasteners is not exceeded
c) for aesthetic appearance
d) maximum distance between centre of two adjacent fasteners is exceeded
Answer: d
Explanation: Tacking fasteners are used when maximum distance between
centre of two adjacent fasteners is exceeded. These are not subjected to
calculated stress.
16. Spacing of tacking fasteners when exposed to
weather should not exceed ______
a) 32t, where t= thickness of
outside plate
b) 16t, where t= thickness of outside plate
c) 25t, where t= thickness of outside plate
d) 20t, where t= thickness of outside plate
Answer: b
Explanation: Spacing of tacking fasteners in a line should not exceed
(i)32t or 300mm, whichever is less when not exposed to weather, where t=
thickness of outside plate, (ii)16t or 200mm, whichever is less when not
exposed to weather, where t= thickness of outside plate.
17. Design Shear strength of bolt is given by ____
a) fu(nnAnb+ nsAsb)/(√3 x 1.1)
b) fy(nnAnb+ nsAsb)/(√3 x 1.1)
c) fu(nnAnb+ nsAsb)/(√3 x 1.25)
d) fy(nnAnb+ nsAsb)/(√3 x 1.25)
Answer: c
Explanation: Shear strength of bolt = fu(nnAnb+ nsAsb)/(√3 x 1.25),
where fu=ultimate strength of bolt, nn=number of shear planes with thread intercepting
shear plane, ns=number of shear planes without
thread intercepting shear plane, Anb=nominal plain
shank area of bolt, Asb=net shear area of bolt at
threads.
18. Nominal bearing strength of bolt is 2.5kbdtfu where kb depends on
(i) end distance, (ii)pitch distance, (iii)ultimate tensile
stress of bolt, (iv)shank area of bolt, (v)yield stress of bolt, (vi)diameter
of hole
a) i, ii, iii, vi
b) i, ii, iv, v
c) ii, iii, iv, v
d) iii, iv, v, vi
Answer: a
Explanation: Nominal bearing strength of bolt is 2.5kbdtfu, where kb is smaller of e/3d0,
p/3d0 -0.25, fub/fu, 1 ; where e, d = end and pitch distances, d0= diameter of hole, fub and fu = ultimate tensile stress of bolts and plate,
d = nominal diameter of bolt.
20. Tensile strength of bolt is given by____________
a) 0.9fubAn/1.1
b) 0.9fybAn/1.1
c) 0.9fubAn/1.25
d) 0.9fybAn/1.25
Answer: c
Explanation: Tensile strength of bolt is given by 0.9fubAn/1.25, where fub=ultimate tensile stress of bolt, An= net tensile area.
21. 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
Answer: b
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 = factored tensile force, Tdb= design tensile capacity.
22. The maximum number of bolts of diameter 25mm
that can be accomodated in one row in 200mm wide flat are:
a) 2
b) 4
c) 3
d) 5
Answer: c
Explanation: Minimum end distance = 2.5×25 = 62.5mm
Number of bolts that can be accommodated = (200-2×62.5)/25 = 3 bolts.
23. Calculate design strength in shear of 16mm
diameter of bolt of grade 4.6 for lap joint
a) 29 kN
b) 50 kN
c) 40 kN
d) 59 kN
Answer: a
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.
24. What is the value of kb in nominal bearing
strength for a bolt of 20mm diameter of grade 4.6?
a) 1
b) 0.97
c) 0.5
d) 2
Answer: c
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.
25. Calculate bearing strength of 16mm 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) 50.18 kN
b) 70.26 kN
c) 109.82 kN
d) 29.56 kN
Answer: a
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.
26. 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) 40%
d) 35%
Answer:
c
Explanation: Efficiency = (strength of bolt per pitch length/ strength
of plate per pitch length)x100 = 60×100/150 = 40%.
27. Strength of bolt is taken as minimum
of______________
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
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 capacity/1.25.
28. Prying forces are
a) friction forces
b) shear forced
c) bending forces
d) tensile forces
Answer: d
Explanation: 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.
29. Simple connections are used to transmit ______
a) forces
b) moments
c) stresses
d) both force and moment
Answer: a
Explanation: Simple Connection is required to transmit force only and
there may not be any moment acting on the group of connectors. This connection
may be capable of transmitting some amount of moment. Simple connections are
also called flexible connections.
30. Which of the following statement is true?
a) lap joint eliminates eccentricity
of applied load, butt joint results in eccentricity at connection
b) lap joint and butt joint eliminates eccentricity at connection
c) lap joint and butt joint results in eccentricity of applied load
d) lap joint results in eccentricity of applied load, butt joint eliminates
eccentricity at connection
Answer: d
Explanation: Lap joints and butt joints are used to connect plates or
members composed of plate elements. Lap joint results in eccentricity of
applied load, butt joint eliminates eccentricity at connection.
31. In a lap joint, at least __________ bolts
should be provided in a line.
a) 0
b) 1
c) 2
d) 3
Answer: c
Explanation: In lap joint, members to be connected are simply overlapped
and connected together by means of bolts and welds. To minimize the effect of
bending due to eccentricity in a lap joint, at least two bolts in a line should
be provided.
32. Use of lap joints is not recommended because
a) stresses are distributed unevenly
b) eccentricity is eliminated
c) bolts are in double shear
d) no bending is produced
Answer: a
Explanation: In lap joint the centre of gravity of load in one member
does not coincide with centre of gravity of load in other member. It results in
eccentricity of applied loads and bending. Due to eccentricity, stresses are
also not evenly distributed, Hence lap joint is not recommended.
33. Why is double cover butt joint preferred over
single cover butt joint or lap joint?
a) bending in bolts
b) bolts are in single shear
c) eliminates eccentricity
d) shear force is not transmitted
Answer: c
Explanation: Double cover butt joint preferred over single cover butt
joint or lap joint because (i) eccentricity of load is eliminated, hence no
bending in bolts, (ii) total shear force to be transmitted is split into two
parts, hence bolts are in double shear. Shear capacity of double cover butt
joint is double the shear capacity of single cover butt joint or lap joint.
34. Clip and seating angle connection is provided
for
a) lateral support
b) bending support
c) frictional support
d) hinged support
Answer: a
Explanation: Clip and seating angle connection transfer reaction from
beam to column through angle seat. The cleat angle is provided for lateral or
torsional support to the top flange of the beam and bolted to the top flange.
35. In flexible end plate design, beam is designed
for the
a) maximum bending moment
b) shear force
c) torsional moment
d) zero end moment
Answer: d
Explanation: In flexible end plate design. beam is designed for the zero
end moment and the end plates augment the web shear and bending capacity of
beams.
36. _____________ condition is true for web side
plate connection?
a) HSFG bolts should be used
b) Bolts should be designed to fail by bearing of connected plies
c) Bolts should be designed to fail by shear of bolt
d) Edge distances must be less than two times the bolt diameter
Answer: b
Explanation: The following condition must be considered for web side
plate connection (i) only ordinary bolts should be used, (ii) bolts should be
designed to fail by bearing of connected plies ad not by shear of bolt, (iii)
edge distances must be greater than two times the bolt diameter.
37. In bolted moment end plate connection, bending
moment , axial force and shear force are transferred by
a) tension and compression
b) tension only
c) compression only
d) friction
Answer: a
Explanation: In bolted moment end plate connection, bending moment,
axial force and shear force are transferred by tension and compression or shear
through flange welds and by shear through the web welds to the end plate.
38. What is eccentric shear?
a) shear effects caused by
concentric load on a bolt group
b) shear effects caused by moment on a bolt group
c) shear effects caused by eccentric load on a bolt group
d) shear effects caused by torsion load on a bolt group
Answer: c
Explanation: When bolt groups are subjected to shear and moment in shear
plane, the load that is eccentric is eccentric with respect to centroid of bolt
group can be replaced with a force acting through the centroid of bolt group
and a moment (Magnitude = Pxe). Both the moment and the force result in shear
effects in the bolts of the group and are called as eccentric shear.
39. Which of the following parameters control the
quality of weld?
a) composition of electrode
b) size of metal plate
c) composition of metal plate
d) size of electrode
Answer: d
Explanation: The parameters control the quality of weld are size of
electrode and the current that produces sufficient heat to melt the base metal
and minimizes electrode splatter.
40. ___________ welding process is preferred for
field application?
a) Submerged arc welding
b) Shielded metal arc welding
c) Gas-shielded metal arc welding
d) Flux core arc welding
Answer: b
Explanation: Submerged arc welding, gas-shielded metal arc welding, flux
core arc welding, electro slag welding can be used when welding is done in
fabrication shop. For field applications, shielded metal arc welding is
preferred.
41. Arrange the following welds in ascending order
as per their usage in structural engineering applications.
a) slot and plug weld, groove weld,
fillet weld
b) fillet weld, groove weld, slot and plug weld
c) fillet weld, slot and plug weld, groove weld
d) groove weld, fillet weld, slot and plug weld
Answer: a
Explanation: Fillet welds are used extensively (about 80%) followed by
groove welds (15%). Slot and plug welds are rarely used (less than 5%) in
structural engineering applications.
42. __________ type of weld is most suitable for
lap and T-joints?
a) Groove weld
b) Fillet weld
c) Plug weld
d) Slot weld
Answer: b
Explanation: Fillet welds are suitable for lap and T-joints and groove
welds are suitable for butt, corner, and edge joints.
43. The size of root gap and root face for groove
weld does not depend on :
a) volume of deposited material
b) type of welding process
c) type of metal plate
d) welding position
Answer: c
Explanation: For groove weld, the root opening or gap is provided for
the electrode to access the base of the joint. The size of root gap and root
face depends on the following : (i) type of welding process, (ii) welding
position, (iii) volume of deposited material, (iv)cost of preparing edges,
(v)access for arc and electrode, (vi)shrinkage and distortion.
44. Which of the following is true about slot and
plug welds?
a) The inspection of these welds is
easy
b) They are extensively used in steel construction
c) They are normally used to connect members carrying tensile loads
d) They are assumed to fail in shear
Answer: d
Explanation: Slot and plug welds are not extensively used in steel construction.
They are used to fill up holes in connections. They are assumed to fail in
shear. The inspection of these welds is difficult. They are useful in
preventing overlapping parts from buckling.
45. Which of the following are not the assumptions
made in the analysis of welded joints?
a) welds connecting various joints
are homogenous, isotropic
b) parts connected by weld are rigid
c) only stresses due to internal forces are considered
d) effects of residual stresses are neglected
Answer: c
Explanation: The following are the assumptions made in analysis of
welded joints: (i) welds connecting various joints are homogenous, isotropic
and elastic elements, (ii) parts connected by weld are rigid and their
deformations are neglected, (iii) only stresses due to external forces are
considered. Effects of residual stresses, stress concentration and shape of
welds are neglected.
46. The minimum size of fillet weld should _______
a) be less than 3mm
b) not be less than 3mm
c) greater than thickness of thinner part joined
d) be less than 2mm
Answer: b
Explanation: The minimum size of fillet weld should not be less than 3mm
and not more than thickness of thinner part joined.
47. For members having sharp machine cut edges, the
maximum size of fillet weld is obtained by _______
a) subtracting 1.5mm from thickness
of thinner member to be jointed
b) adding 1.5mm to thickness of thinner member to be jointed
c) adding 3mm to thickness of thinner member to be jointed
d) subtracting 3mm from thickness of thinner member to be jointed
Answer: a
Explanation: The maximum size of fillet weld is obtained by subtracting
1.5mm from thickness of thinner member to be jointed. The maximum size of weld
should not be more than 3/4 of the thickness of section at toe when welds are
applied to round toe of steel sections.
48. As per IS code the minimum specified length of
fillet weld is ___________
a) two times the size of weld
b) four times the size of weld
c) six times the size of weld
d) half the size of weld
Answer: b
Explanation: As per IS code, the actual length of fillet weld should not
be less than four times the size of weld. If this requirement is not met, the
size of weld should be one fourth of the effective length.
49. Effective length of fillet weld is _______
a) twice the overall length plus
twice the weld size
b) twice the overall length minus twice the weld size
c) equal to overall length minus twice the weld size
d) equal to overall length plus twice the weld size
Answer: c
Explanation: Effective length of fillet weld is taken equal to overall
length minus twice the weld size. The deduction is made to allow for craters to
be formed at the ends of welded length.
50. End returns are made ________
a) equal to half the size of weld
b) equal to the size of weld
c) equal to thrice the size of weld
d) equal to twice the size of weld
Answer: d
Explanation: End returns are made equal to twice the size of weld to
relieve the weld lengths from high stress concentrations at the ends.
51. Which of the following is not true regarding
effective throat thickness of weld?
a) It should not exceed 0.7t or 1t,
where t is thickness of thinner plate of elements being welded
b) Effective throat thickness = K x size of weld, where K is a constant
c) Effective throat thickness = K x (size of weld)2 ,
where K is a constant
d) Effective throat thickness should not be less than 3mm
Answer: c
Explanation: Effective throat thickness is the shortest distance from
the root of fillet weld to face of diagrammatic weld(line joining the toes).
The effective throat thickness should not be less than 3mm and it should not
exceed 0.7t or 1t, where t is thickness of thinner plate of elements being
welded. Effective throat thickness = K x size of weld, where K is a constant
which depends on angle between fusion faces.
52. The effective throat thickness is K times the
size of weld. What is the value of K when angle between fusion faces is 80˚?
a) 0.65
b) 0.70
c) 0.55
d) 0.50
Answer: b
Explanation: The value of K varies with angle between fusion faces.
Values of K for different angles between fusion faces are :
|
Angle between fusion faces |
60˚-90˚ |
91˚-100˚ |
101˚-106˚ |
107˚-113˚ |
114˚-120˚ |
|
K |
0.7 |
0.65 |
0.6 |
0.55 |
0.5 |
53. In lap joint the length of overlap of plates
to be fillet welded ____
a) should not be less than 4 times
the thickness of thinner part
b) should not be less than 2 times the thickness of thinner part
c) should be less than 4 times the thickness of thinner part
d) should be less than 2 times the thickness of thinner part
Answer: a
Explanation: The length of overlap of plates to be fillet welded in lap
joint should not be less than 4 times the thickness of thinner part.
54. Which of the following Statement is incorrect?
a) Effective length of intermittent
weld should have a minimum length 80mm
b) Effective length of groove weld should be less than 4 times the weld size
c) Effective length of groove weld should not be less than 4 times the weld
size
d) Effective length of intermittent weld should not be less than 4 times the
weld size
Answer: b
Explanation: Effective length of groove weld should not be less than 4
times the weld size. Effective length of intermittent weld should not be less
than 4 times the weld size, with a minimum of 40mm.
55. The design nominal strength of fillet weld is
given by ____________
a) fu
b) √3 fu
c) fu/√3
d) fu/(1.25 x √3)
Answer: c
Explanation: Design nominal strength of fillet weld = fu/√3, where fu is smaller
of ultimate stress of weld or parent metal.
56. When welds are subjected to compressive or
tensile or shear force alone, the stress in weld is given by :
a) Ptt/lw
b) Pttlw
c) Plw/tt
d) P/ttlw
Answer: d
Explanation: When welds are subjected to compressive or tensile or shear
force alone, the stress in weld is given by q = P/ttlw , where q=shear stress in N/mm2, P = force transmitted, tt = effective throat thickness of weld in mm, lw = effective length of weld in mm.
57. When fillet welds are subjected to combination
of normal and shear stress, the equivalent stress is given by :
a) √(fa2+2q2)
b) √(3fa2+q2)
c) √(fa2+3q2)
d) √(fa2+q2)
Answer: c
Explanation: When fillet welds are subjected to combination of normal
and shear stress, the equivalent stress is given by fe = √(fa2+3q2), where fa = normal
stresses, compression or tension, due to axial force or bending moment, q =
shear stress due to shear force or tension.
58. What is the strength of weld per mm length
used to connect two plates of 10mm thickness using a lap joint?
a) 487.93 N/mm
b) 795.36 N/mm
c) 295.5N/mm
d) 552.6 N/mm
Answer: b
Explanation: Minimum size of weld = 3mm,
Maximum size of weld = 10-1.5 = 8.5mm,
Assume weld size = 6mm
Effective throat thickness, te = 0.7 x 6 =
4.2mm
Strength of weld = te [fu/(√3 x 1.25)] = 410 x 4.2 /(√3 x 1.25) = 795.36
N/mm.
59. What is the overall length of fillet weld to
be provided for lap joint to transmit a factored load of 100kN? Assume site
welds and width and thickness of plate as 75mm and 8mm respectively, Fe410
steel.
a) 201mm
b) 500mm
c) 468mm
d) 382mm
Answer: a
Explanation: Minimum size of weld = 3mm
Maximum size of weld = 8-1.5 = 6.5mm
Assume size of weld = 5mm
Effective throat thickness = 0.7 x 5 = 3.5mm
Strength of weld = 3.5×410/(√3 x1.5) = 552.33 N/mm
Required length of weld = 100 x 103/552.33 = 181.05
mm
length to be provided on each side = 181/2 = 90.5mm
End return = 2×5 = 10mm
Overall length = 2 x (90.5 + 2 x 5) = 201mm.
60. As per IS code, the clear spacing between
effective lengths of intermittent welds should not be ______
a) less than 20t in case of
compression joint, where t is thickness of thinner plate
b) less than 16t in case of tension joint, where t is thickness of thinner
plate
c) less than 12t in case of compression joint, where t is thickness of thinner
plate
d) less than 20t in case of tension joint, where t is thickness of thinner
plate
Answer: b
Explanation: The clear spacing between effective lengths of intermittent
welds should not be less than 16 times and 12 times the thickness of thinner
plate jointed in case of tension joint and compression joint respectively, and
should never be more than 200mm.
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