Category: Strength of Materials Objective Questions

250+ TOP MCQs on Bending Stress due to Eccentric Loading and Answers

Tough Strength of Materials Questions and Answers on “Bending Stress due to Eccentric Loading”.

1. Eccentrically loaded structures have to be designed for _______
a) Uniaxial force
b) Biaxial force
c) Combined axial force
d) Combined biaxial force
Answer: c
Clarification: When the line of action of the resultant compressive force doesn’t coincide with the centre of gravity of the cross section of the structure, it is called eccentrically loaded structure. They have to be designed for combined axial force.

2. ______ transfer the loads from beams or slabs to footings or foundations.
a) Pedestal
b) Post
c) Rib
d) Column
Answer: d
Clarification: A vertical member whose effective length is greater than 3 times its least lateral dimension carrying compressive loads is called a column. The main function of column is to transfer the loads from the beams or slabs to the footings or foundation.

3. In long columns, the lateral deflection causes at the ______
a) Supports
b) Throughout
c) Midspan
d) Along outer periphery
Answer: c
Clarification: A long column under the action of axial loads deflects laterally causing maximum deflection at the centre. A long column fails due to buckling.

4. Short columns causes deflection in the structure.
a) True
b) False
Answer: b
Clarification: If the ratio of the effective length of the column to the least lateral dimension is less than 12. The column is called a short column. It fails by crushing (pure compression failure) and there is no chance of causing deflections.

5. The approximate percentage of reinforcement provided in a beam varies from _______
a) 1-2%
b) 1-4%
c) 2-3%
d) 3-4%
Answer: a
Clarification: The approximate percentage of reinforcement provided in a beam varies from 1-2%.

Type of Structure Approx. % of Steel
Beam 1-2%
Slabs 0.7-1%
Columns 1-4%

6. To avoid the failure of a column by buckling ___________ limits are to be recommended.
a) Slenderness
b) Effective length
c) Kernel
d) Radius of gyration
Answer: a
Clarification: The column dimensions shall be such that it fails by material failure only (crushing due to compression) and not by buckling. To avoid the failure of column buckling clause 25.3 of IS 456 recommends the slenderness limits for the column.

7. According to IS 456- 2000, the minimum eccentricity subjected to a column is __________
a) 30mm
b) 20mm
c) 45mm
d) 50mm
Answer: b
Clarification: No column can have a perfectly axial load. There may be some moments acting due to the imperfection of construction or due to actual conditions of loading when IS 456-2000, recommends that all columns Shall be designed for minimum eccentricity of 20 mm.

8. Radius of gyration is denoted by _________
a) k
b) n
c) e
d) y
Answer: a
Clarification: The radius of gyration about a given axis is defined as the effective distance from the given axis at which the whole area may be considered or located. It is denoted by “k” or “r”. The units for the radius of gyration are mm.

9. Find the moment of inertia of a rectangular section of 40 mm width and 80 mm depth about the base.
a) 632×104mm4
b) 682×104mm4
c) 734×104mm4
d) 568×104mm4
Answer: b
Clarification:
tough-strength-materials-questions-answers-q9
Moment of inertia of the rectangular section passing through the base is bd3/ 3.
I = bd/3
= 40×(80)3/ 3
= 682.66×104mm4.

10. Mild steel is an example of ______________ mechanical property of the material.
a) Malleability
b) Creep
c) Ductility
d) Elasticity
Answer: c
Clarification: Ductility is the property of a material by which material can be drawn into thin wires after undergoing a considerable deformation without rupture. The mild steel, silver, tor steel, aluminium etc. are considered as examples for ductility.

11. Which of the following are the relative properties of the material?
a) Creep
b) Fatigue
c) Hardness
d) Stiffness
Answer: c
Clarification: The hardness is the ability of a material to resist indentation (impression), scratching or surface abrasion. It is the relative property of the material. Every material has its own hardness number.

12. Rotating key of a lock is an example of ____________
a) Varignon’s Theory
b) Walton’s Theory
c) Formation of couple
d) Parallel axis theorem
Answer: c
Clarification: A set of two equal and opposite forces whose line of action is different form a couple. The effect of couple is always to produce moment on which it acts either in clockwise or anticlockwise directions. The example is rotating key of a lock.

13. The relative change in position is called ______________
a) Matter
b) Body
c) Inertia
d) Motion
Answer: d
Clarification: A body said to be in motion when it changes its position with respect to other bodies. The relative change in position is called motion. The motion involves both space and time.

14. Which of the following is not base unit?
a) Area
b) Length
c) Time
d) Temperature
Answer: a
Clarification: If the units are expressed in other units which are derived from fundamental units, such units are known as derived units. The examples are area, velocity, acceleration & pressure etc.

15. According to IS 456-2000, the minimum number of longitudinal bars to be provided in rectangular columns is ________
a) 5
b) 4
c) 6
d) 8
Answer: b
Clarification: According to IS 456-2000, the cross sectional area of longitudinal reinforcement should not be less than 0.8% and not more than 6% of gross cross-sectional area. The minimum diameter of longitudinal bars is 12 mm and minimum number of longitudinal bars to be provided is 4 for a rectangular column.

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250+ TOP MCQs on Analyse Fixed Beam and Answers

This set of Strength of Materials Multiple Choice Questions 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

Answer: c
Clarification: 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

Answer: a
Clarification: 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 ___________
a) Minimum
b) Zero
c) Maximum
d) Throughout

Answer: b
Clarification: 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.
a) Lateral
b) Deflection
c) Temperature
d) Slope

Answer: c
Clarification: 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

Answer: a
Clarification: A beam carrying udl along its entire span, the maximum bending moment developed = wl2 / 12.
= 5×62 / 12.
15 kNm.

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

Answer: c
Clarification: 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

Answer: c
Clarification: 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

Answer: c
Clarification: 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

Answer: b
Clarification: 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

Answer: b
Clarification: 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 %

Answer: a
Clarification: The limiting gradient for mountainous terrain is 6.00%.

Type of terrain Ruling Gradient Limiting Gradient Exceptional Gradient
Plain 3.30% 5.00% 6.70%
Mountainous 5.00% 6.00% 7.00%

12. Which of the following do not have units?
a) Specific weight
b) Specific gravity
c) Specific volume
d) Mass density

Answer: b
Clarification: 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?
a) Compressibility
b) Water content
c) Porosity
d) Voids ratio

Answer: d
Clarification: 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.
a) Till
b) Tull
c) Loess
d) Mart

Answer: a
Clarification: 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

Answer: b
Clarification: The bearing capacity of laminated rocks used in foundation is 1620 kN/m2.

Type Of Rock Bearing capacity in kN/m2
Granite 3240
Laminated 1620
Residual 880
Soft 440