This set of Strength of Materials test focuses on “Bending Stress due to Eccentric Loading in Both Directions”.
1. 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
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.
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. A column in which reinforcement is wound spiral is ___________
a) Tied column
b) Spiral column
c) Composite column
d) Short column
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. The inclined members carrying compressive loads are ________
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. Polygonal links are also known as _____________
a) Bent up bars
b) Crancked bars
c) Lateral ties
d) Anchorage 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. The pitch of the lateral ties shall not be more than the least of the ______________
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. The minimum depth of foundation in all types of soils is _________
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. In T beams, the most of the compressive force is shared by __________
d) Neutral axis
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. In T beams, maximum ______ is less.
a) Shear force
b) Bending moment
c) Bending stress
d) Shear stress
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. In continuous beams ______ moment develops at supports.
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. In continuous beams ______________ moments is always less than support moments.
b) Mid span
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. Lighter materials of construction can be used for a continuous beam.
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. ________ is a good example for malleability.
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. 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
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. Volumetric strain = 3× _____________
a) Linear strain
b) Lateral strain
c) Linear stress
d) Lateral stress
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).
Sanfoundry Global Education & Learning Series – Strength of Materials.
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