This set of Strength of Materials Multiple Choice Questions & Answers (MCQs) focuses on “Analyse Fixed Beam”.
1. A beam which is inbuilt in at its support is called _________
a) Cantilever beam
b) Simply supported beam
c) Fixed beam
d) Continuous beam
Explanation: A beam which is built in at its support is known as a fixed beam. In a fixed beam, fixed end moments are developed at the ends. The slope at the end support is zero or (unaltered).
2. Fixed beam is also known as _______
a) Encaster beam
b) Constressed beam
c) In built beam
d) Constricted beam
Explanation: Fixed beam is also called Encaster beam or Constraint beam or Built in beam. In a fixed beam the fixed end moments develop at the end supports. In these beams, the supports should be kept at the same level.
3. In fixed beams, the slope at the supports be ___________
Explanation: The fixed beam is stronger, stiffer and more stable. The slope at the supports is zero.
Maximum bending moment at the centre is reduced because of fixing moments developed at supports.
4. _______ changes induce large stresses in a fixed beam.
Explanation: In fixed beam, sinking of any one support sets large stresses. The temperature changes induce the largest stress. The moving loads make the degree of fixity at support uncertain.
5. A beam 6 metres long is fixed at it ends. It carries a udl of 5 kN/m. Find the maximum bending moment in the beam.
a) 15 kNm
b) 20 kNm
c) 35 kNm
d) 40 kNm
Explanation: A beam carrying udl along its entire span, the maximum bending moment developed = wl2 / 12.
= 5×62 / 12.
6. Calculate the maximum deflection of a fixed beam carrying udl of 5 kN/m. The span of beam is 6 m. Take E = 200kN/m2 and I = 5×107 mm4.
a) 1.865 m
b) 2.235 m
c) 1.6875 m
d) 2.5 m
Explanation: The maximum deflection in fixed beam is wl4/384EI
= 5×64 × 109/ 384×200×5×107
= 1.6875 mm.
7. Calculate the load intensity of fixed beam if the maximum deflection shall not exceed 1/ 400 of the span. Take EI as 1010 kN mm2.
a) 40 kN
b) 35 kN
c) 45 kN
d) 60 kN
Explanation: When the maximum deflection equals to 1 / 400 of the span.
Wl4/ 384 EI = 1 /400.
W= 384 EI / 400 l3
W = 45 kN.
8. ____ is known as a serpentine curve.
a) Circular curve
b) Transition curve
c) Reverse curve
d) Leminiscate curve
Explanation: Reverse curves are provided in difficult terrain. In these curves, the simple curves have a common tangent. They consist of two simple curves of same or different radii. These curves are also known as serpentine curves.
9. The maximum super elevation to be provided is ___
a) 2 in 15
b) 1 in 15
c) 1 in 10
d) 2 in 10
Explanation: According to IRC, the maximum super elevation of 1 in 15 is to be provided. Minimum super elevation is required for proper drainage. If the super elevation calculated is less than the camber no superelevation is to be provided.
10. ______ curves are used to solve the problems of land acquisition.
a) Vertical curves
b) Horizontal curves
c) Circular curves
d) Transition curves
Explanation: A horizontal curve is the curve in plane to provide change in direction to the centre line of the alignment. It is used to preserve the certain existing amenities and to solve the problems of land acquisition.
11. The limiting gradient for mountainous terrain is ________
a) 6.00 %
b) 7.00 %
c) 8.00 %
d) 5.00 %
Explanation: The limiting gradient for mountainous terrain is 6.00%.
|Type of terrain||Ruling Gradient||Limiting Gradient||Exceptional Gradient|
12. Which of the following do not have units?
a) Specific weight
b) Specific gravity
c) Specific volume
d) Mass density
Explanation: Specific gravity is defined as the ratio of the specific weight of solids to the specific weight of an equal volume of water at the temperature. It is denoted by S. As it is a ratio, it doesn’t possess units.
13. In engineering properties of soils, the “e” denotes?
b) Water content
d) Voids ratio
Explanation: Void ratio is defined as the ratio of the total volume of voids to volume of soil solids. It is expressed as a decimal.
14. _____ is a glacier deposit of sand, gravel or clay.
Explanation: The deposits made by glaciers are called drifts. The deposits made by the melting of glaciers are called till. Till is a stratified soil.
15. The bearing capacity of laminated rocks used in foundation is ___________
a) 1450 kN/m2
b) 1620 kN/m2
c) 1785 kN/m2
d) 2125 kN/m2
Explanation: The bearing capacity of laminated rocks used in foundation is 1620 kN/m2.
|Type Of Rock||Bearing capacity in kN/m2|
Sanfoundry Global Education & Learning Series – Strength of Materials.
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