Engineering Questions with Answers - Multiple Choice Questions
MCQs on Bending Stress due to Eccentric Loading in Both Directions
1 - Question
As per IS 456-2000, the minimum eccentricity for columns shall be given by ________
a) l/500 + D/30
b) l/450 + D/45
c) l/400 + D/40
d) l/250 + D/25
View Answer
Answer: a
Explanation: As per IS 456-2000, clause 25.4 recommends that all columns show the design for the minimum of its eccentricity. No column will be perfectly loaded axially. There might be kind of moment acting due to improper construction.
Explanation: As per IS 456-2000, clause 25.4 recommends that all columns show the design for the minimum of its eccentricity. No column will be perfectly loaded axially. There might be kind of moment acting due to improper construction.
2. If the columns are effectively held in position and restrained against rotation at both ends. Recommend the value of effective length.
a) 0.6×l
b) 0.65×l
c) 0.77×l
d) 0.9×l
2 - Question
If the columns are effectively held in position and restrained against rotation at both ends. Recommend the value of effective length.
a) 0.6×l
b) 0.65×l
c) 0.77×l
d) 0.9×l
View Answer
Answer: b
Explanation: The effective length of column for various and conditions may be taken from IS 456 2000, for effectively held in position and restrained against rotation in both ends recommended value of effective length is 0.65 l
[ l = unsupported length of compression member ].
Explanation: The effective length of column for various and conditions may be taken from IS 456 2000, for effectively held in position and restrained against rotation in both ends recommended value of effective length is 0.65 l
[ l = unsupported length of compression member ].
3 - Question
A column in which reinforcement is wound spiral is ___________
a) Tied column
b) Spiral column
c) Composite column
d) Short column
View Answer
Answer: b
Explanation: When the main longitudinal bars of the column are enclosed within closely spaced and continuously wound spiral reinforcement, then the column is said to be a spiral column.
4 - Question
The inclined members carrying compressive loads are ________
a) Pedestal
b) Strut
c) Post
d) Winch
View Answer
Answer: b
Explanation: The inclined member carries compressive loads in case of frames and trusses are known as Struts. The Pedestal is a vertical compression member whose effective length is less than 3 times its least lateral dimension.
Explanation: The inclined member carries compressive loads in case of frames and trusses are known as Struts. The Pedestal is a vertical compression member whose effective length is less than 3 times its least lateral dimension.
5 - Question
Polygonal links are also known as _____________
a) Bent up bars
b) Crancked bars
c) Lateral ties
d) Anchorage bars
View Answer
Answer: c
Explanation: A reinforced concrete member of compression shall have transverse or helical reinforcement. It is either in the form of spiral rings capable of taking up tension or polygonal links (lateral ties) placed closely and confined with main bars.
Explanation: A reinforced concrete member of compression shall have transverse or helical reinforcement. It is either in the form of spiral rings capable of taking up tension or polygonal links (lateral ties) placed closely and confined with main bars.
6 - Question
The pitch of the lateral ties shall not be more than the least of the ______________
a) 300mm
b) 450mm
c) 500mm
d) 550mm
View Answer
Answer: a
Explanation: As per IS 456 2000; the which of the ties shall not be more than the least of the
▪ least lateral dimension of the column
▪ sixteen times the diameter of the smallest longitudinal bar
▪ 300 mm.
Explanation: As per IS 456 2000; the which of the ties shall not be more than the least of the
▪ least lateral dimension of the column
▪ sixteen times the diameter of the smallest longitudinal bar
▪ 300 mm.
7 - Question
The minimum depth of foundation in all types of soils is _________
a) 350mm
b) 680mm
c) 500mm
d) 280mm
View Answer
Answer: c
Explanation: According to IS 1080 – 1962, the minimum depth of foundation should be not less than 500 mm. However, if good rock is made it smaller depth, only removal of top soil may be sufficient.
Explanation: According to IS 1080 – 1962, the minimum depth of foundation should be not less than 500 mm. However, if good rock is made it smaller depth, only removal of top soil may be sufficient.
8 - Question
In T beams, the most of the compressive force is shared by __________
a) Web
b) Flange
c) Rib
d) Neutral axis
View Answer
Answer: b
Explanation: As the slab being Monolithic with the beam is also compressed and shares the compressive force with the flange, the depth of beam required is less and hence the maximum deflections also less.
Explanation: As the slab being Monolithic with the beam is also compressed and shares the compressive force with the flange, the depth of beam required is less and hence the maximum deflections also less.
9 - Question
In T beams, maximum ______ is less.
a) Shear force
b) Bending moment
c) Bending stress
d) Shear stress
View Answer
Answer: b
Explanation: In T beams, the maximum bending moment is less because of the sagging moment is effectively resisted. The maximum deflections are also less in these beams. They are preferred for larger spans when compared to simply supported beams.
Explanation: In T beams, the maximum bending moment is less because of the sagging moment is effectively resisted. The maximum deflections are also less in these beams. They are preferred for larger spans when compared to simply supported beams.
10 - Question
In continuous beams ______ moment develops at supports.
a) Hogging
b) Sagging
c) Couple
d) Static
View Answer
Answer: a
Explanation: When the slab beam is continuous over several supports, hogging bending moment is induced over the support developing tension at the top surface. The continuous beams and slabs I design for maximum bending moment and shear forces.
Explanation: When the slab beam is continuous over several supports, hogging bending moment is induced over the support developing tension at the top surface. The continuous beams and slabs I design for maximum bending moment and shear forces.
11 - Question
In continuous beams ______________ moments is always less than support moments.
a) Upward
b) Mid span
c) Downward
d) Sagging
View Answer
Answer: b
Explanation: The mid span moment in continuous beams and slabs is always less than the support moment and hence weight of the beam doesn’t affect the stresses induced.
Explanation: The mid span moment in continuous beams and slabs is always less than the support moment and hence weight of the beam doesn’t affect the stresses induced.
12 - Question
Lighter materials of construction can be used for a continuous beam.
a) True
b) False
View Answer
Answer: a
Explanation: Lighter materials are preferred in construction of continuous beam because as the bending moment developed in a continuous beam is less, the bending moment to be resisted is also less.
Explanation: Lighter materials are preferred in construction of continuous beam because as the bending moment developed in a continuous beam is less, the bending moment to be resisted is also less.
13 - Question
________ is a good example for malleability.
a) Glass
b) Concrete
c) Copper
d) Lead
View Answer
Answer: c
Explanation: Malleability is that property of a material by which it can be beaten or rolled into thin sheets without any rupture. The best example considered for malleability is copper. Other materials include ornamental gold, wrought iron & ornamental silver.
Explanation: Malleability is that property of a material by which it can be beaten or rolled into thin sheets without any rupture. The best example considered for malleability is copper. Other materials include ornamental gold, wrought iron & ornamental silver.
14 - Question
Determine the working stress in the factor of safety is 3 and ultimate load is what 127.32N?
a) 46 N/m2
b) 55 N/m2
c) 48 N/m2
d) 42 N/m2
View Answer
Answer: d
Explanation: We know that working stress is the ratio of ultimate load to factor of safety
Given F.O.S = 3 & W = 127.32N
Working stress = 127.32/3
= 42.44 N/m2.
Explanation: We know that working stress is the ratio of ultimate load to factor of safety
Given F.O.S = 3 & W = 127.32N
Working stress = 127.32/3
= 42.44 N/m2.
15 - Question
Volumetric strain = 3× _____________
a) Linear strain
b) Lateral strain
c) Linear stress
d) Lateral stress
View Answer
Answer: a
Explanation: The volumetric strain is the algebraic sum of all the linear or axial strains that are
€v = €xx + €yy + €zz
The volumetric strain will be 3 times the linear strain in any of three axis and €v = 3e
Where (e = linear strain).