21. Number of shear planes
passing through shank section (ns)
for bolt are..........
a) 1 for single shear, 1 for double shear
b) 0 for single shear, 1 for double shear
c) 1 for single shear, 0 for double shear
d) 0 for single shear, 0 for double shear
Answer: b) 0 for single shear, 1 for double
shear
22. For bolt having nominal diameter 30 mm, diameter of
bolt hole to be provided should be ..........
a) 30 mm
b) 31 mm
c) 32 mm
d) 33 mm
Answer: d) 33 mm
23. Bolt value is the safest value of force
that ..........
a) Member can resist
b) Joint can resist
c) Single bolt can resist
d) Group of bolt can resist
Answer: c) Single bolt can resist
24. Almost in all cases of fillet welds which are right angle
value of k is ..........
a) 0.7
b) 0.65
c) 0.6
d) 0.55
Answer: a) 0.7
25. When two plates overlap each other at a joint, then the
overlap length should not be less than ..........
a) 4 tp
b) 5 tp
c) 6 tp
d) 7 tp
(tp = Thickness
of thinner plate)
Answer: b) 5 tp
26. If S is the size of weld then for the fillet weld value
of end return should not be less than ..........
a) S
b) 2S
c) 3S
d) 4S
Answer: b) 2S
27. Select incorrect statement from the following
a) Dead loads are constant in magnitude and fixed
in position
b) Live loads vary in magnitude and/ or in
positions
c) It is common practice of combining the wind
loads with the snow loads
d) In the design of structure, the combined effect
of external and internal pressure of wind is taken into account
Answer: c) It
is common practice of combining the wind loads with the snow loads
28. If two plates are connected in lap joint using 20
mm diameter bolt, joint is to carry a design load of 200 kN and bolt value is
50 kN, then number of bolts required are ..........
a) 4 nos.
b) 8 nos.
c) 16 nos.
d) 20 nos.
Answer: a) 4
nos.
29. If thickness of thinner plate is upto10 mm then
minimum size of weld is ..........
a) 3 mm
b) 5 mm
c) 6 mm
d) 8 mm
Answer: a) 3
mm
30. Effective length of fillet weld is equal
to ..........
a) Overall length plus twice the weld size
b) Overall length minus twice the weld size
c) Overall length plus thrice the weld size
d) Overall length minus thrice the weld size
Answer: b) Overall length minus twice the
weld size
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