300+ REAL TIME Strength of Materials Objective Questions & Answers 2023

250+ TOP MCQs on Hooke’s Law and Answers

Strength of Materials Multiple Choice Questions on “Hooke’s Law”.

1. The law which states that within elastic limits strain produced is proportional to the stress producing it is known as _____________
a) Bernoulli’s law
b) Hooke’s law
c) Stress law
d) Poisson’s law
Answer: b
Clarification: Hooke’s law states that strain is directly proportional to strain produced by the stress when a material is loaded within the elastic limit.

2. For an isotropic, homogeneous and elastic material obeying Hooke’s law, the number of independent elastic constants is ____________
a) 2
b) 3
c) 9
d) 1
Answer: b
Clarification: There are 3 constants Young’s modulus, Shear modulus and Bulk modulus.

3. What is the factor of safety?
a) The ratio of stress to strain
b) The raio of permissible stress to the ultimate stress
c) The ratio of ultimate stress to the permissible stress
d) The ratio of longitudinal strain to stress
Answer: c
Clarification: Factor of safety is the ratio of ultimate stress to the permissible stress.

4. What is Hooke’s law for the 1-D system?
a) The relation between normal stress and the corresponding strain
b) The relation between shear stress and the corresponding strain
c) The relation between lateral strain and the corresponding stress
d) None of the mentioned
Answer: a
Clarification: For the 1-D system, the stress will be only in one direction. Lateral stress is for an area while normal stress is of a length.

5. Limit of proportionality depends upon ____________
a) Area of cross-section
b) Type of loading
c) Type of material
d) All of the mentioned
Answer: a
Clarification: The proportionality limit is proportional to the area of cross-section. The material type and loading type will have no influence on the proportionality limit.

6. The stress at which extension of a material takes place more quickly as compared to the increase in load is called ____________
a) Elastic point
b) Plastic point
c) Breaking point
d) Yielding point
Answer: d
Clarification: On the stress strain curve, on the elastic point the stress of a material takes place more quickly.

7. Which of these is a non-hoookean material?
a) Steel
b) Rubber
c) Aluminium
d) Copper
Answer: b
Clarification: Rubber is generally regarded as a “non-hookean” material because its elasticity is stress dependent and sensitive to temperature and loading rate.

8. Where in the stress-strain curve, the hooke’s law is valid?
a) Strain hardening region
b) Necking region
c) Elastic range
d) Valid everywhere
Answer: c
Clarification: The hooke’s law itself states that it is valid only up to the elastic range of the material I.e. only to that limit where the material is behaving elastic.

9. Highest value of stress for which Hooke’s law is applicable for a given material is called ____________
a) Stress limit
b) Strain limit
c) Proportional limit
d) Significant limit
Answer: c
Clarification: The hooke’s law is valid only when the stress is proportional to the strain, that is only in the proportionality limit.

250+ TOP MCQs on Bars of Composite Sections – 2 and Answers

This set of Strength of Materials Interview Questions and Answers for freshers on “Bars of Composite Sections – 2”.

1. A member ABCD is subjected to points load P₁=45kN, P₂, P₃=450kN and P₄=130kN. what will be the value of P necessary for equilibrium?

a) 350kN
b) 365kN
c) 375kN
d) 400kN
Answer: b
Clarification: On resolving forces P₁ + P₃ = P₂ + P₄
So P₂ = 45 + 450 – 130 I.e. P₂ = 365kN.

2. A member ABCD is subjected to points load P₁=45kN, P₂, P₃=450kN and P₄=130kN. What will be the total elongation of the member, assuming the modulus of elasticity to be 2.1x105N/mm². The cross sectional area is 625mm, 2500mm, 1250mm respectively.
a) 0.4914mm
b) 0.4235mm
c) 0.4621mm
d) 0.4354mm
Answer: a
Clarification: First of all the fores will be calculated
on resolving forces P₁ + P₃ = P₂ + P₄
So P₂ = 45 + 450 – 130 I.e. P₂ = 365kN
So forces on three sections will be 45kN, 320kN and 130kN respectively.
After that increase in length = PL/AE for all three sections will be calculated.

3. A tensile rod of 40kN is acting on a rod of diameter 40mm and of length 4m. a bore of diameter 20mm is made centrally on the rod. To what length the rod should be bored so that the total extension will increase 30% under the same tensile load if E = 2×105 N/mm²?

a) 2m
b) 2.7m
c) 3.2m
d) 3.6m
Answer: d
Clarification: The extension = PL / AE = 2/π mm
Extension after the bore is made = 1.3x 2/π mm = 2.6/π mm
The extension after the bore is made, is also obtained by finding the extension of the un bored length and bored length.
Stress = load / area
So total extension after bore is made can have two equations which can be put equal and the length the rod should be bored up is calculated.

4. A bar is subjected to a tensile load of 150kN. If the stress in the middle portion is limited to 160 N/mm², what will be the diameter of the middle portion of the total elongation of the bar is 0.25cm (E= 2 x 105)?

a) 3cm
b) 3.45cm
c) 3.85cm
d) 4cm
Answer: b
Clarification: Total extension = P/E x (L₁/A₁ + L₂/A₂ + L₃/A₃ )
Only variable in the equation is A₂. after getting this the diameter of the section can be calculated.

5. A rod, which tapers uniformly from 5cm diameter to 3cm diameter in a length of 50cm, is subjected to an axial load of 6000N. if E = 2,00,000 N/mm², what will be the extension of the rod?
a) 0.00114cm
b) 0.00124cm
c) 0.00127cm
d) 0.00154cm
Answer: c
Clarification: The extension in the rod = PL / Et(a-b) x loge (a/b)
Where a = 50mm, b = 30mm.

6. A bar is in two sections having equal lengths. The area of cross section of 1^st is double that of 2^nd. if the bar carries an axial load of P, then what will be the ratio of elongation in section 2nd to section 1st ?
a) 1/2
b) 2
c) 4
d) 1/4
Answer: b
Clarification: Ratio of elongation in 2nd / ratio of elongation in 1st = L₂/L₁ x A₂/A₁
Since L₁ = L₂ and A₁ = 2A₂
Therefore, ratio = 1 x 2/1 = 2.

7. A round bar made of same material consists of 4 parts each of 100mm length having diameters of 40mm, 50mm, 60mm and 70mm, respectively. If the bar is subjected to an axial load of 10kN, what will be the total elongation of the bar in mm?
a) 0.4/πE ( 1/16 + 1/25 + 1/36 + 1/49)
b) 4/πE ( 1/16 + 1/25 + 1/36 + 1/49)
c) 2/πE ( 1/16 + 1/25 + 1/36 + 1/49)
d) 40/πE ( 1/16 + 1/25 + 1/36 + 1/49)
Answer: d
Clarification: Total elongation = 4PL/πE ( 1/d₁² + 1/d₂² + 1/d₃² + 1/d₄²)
= 4x10x100/πEx100 ( 1/16 + 1/25 + 1/36 + 1/49) mm
= 40/πE ( 1/16 + 1/25 + 1/36 + 1/49).

8. A bar shown in the diagram below is subjected to load 160kN. If the stress in the middle portion Is limited to 150N/mm², what will be the length of the middle portion, if the total elongation of the bar is to be 0.2mm? Take E = 2.1 x 105 N/mm².

a) 18.45cm
b) 17.24cm
c) 16.45cm
d) 20.71cm
Answer: d
Clarification: Let L₂ and D₂ be the dimensions of the middle portion and L₁ and D₂ be the end portion dimensions.
For middle portion area = load / stress
This gives area by which diameter can be calculated.
As Total extension = P/E x (L₁/A₁ + L₂/A₂)
This gives the value of L₂.

9. A composite bar consists of a bar enclosed inside a tune of another material when compressed under a load as whole through rigid collars at the end of the bar. What will be the equation of compatibility?
a) W₁ + W₂ = W
b) W₁ + W₂ = constant
c) W₁/A₁E₁ = W₂/A₂E₂
d) W₁/A₁E₂ = W₂/A₂E₁
Answer: a
Clarification: Compatibility equation insists that the change in length of the bar must be compatible with the boundary conditions. Here W₁ + W₂ = W it is also correct but it is equilibrium equation.

Strength of Materials for Interviews,

250+ TOP MCQs on Shear Force and Bending Moment diagram and Answers

This set of Strength of Materials Multiple Choice Questions on “Shear Force and Bending Moment diagram”.

1. What is the bending moment at end supports of a simply supported beam?
a) Maximum
b) Minimum
c) Zero
d) Uniform
Answer: c
Clarification: At the end supports, the moment (couple) developed is zero, because there is no distance to take the perpendicular acting load. As the distance is zero, the moment is obviously zero.

2. What is the maximum shear force, when a cantilever beam is loaded with udl throughout?
a) w×l
b) w
c) w/l
d) w+l
Answer: a
Clarification: In cantilever beams, the maximum shear force occurs at the fixed end. In the free end, there is zero shear force. As we need to convert the udl in to load, we multiply the length of the cantilever beam with udl acting upon. For maximum shear force to obtain we ought to multiply load and distance and it surely occurs at the fixed end (w×l).

3. Sagging, the bending moment occurs at the _____ of the beam.
a) At supports
b) Mid span
c) Point of contraflexure
d) Point of emergence
Answer: b
Clarification: The positive bending moment is considered when it causes convexity downward or concavity at top. This is sagging. In simply supported beams, it occurs at mid span because the bending moment at the supports obviously will be zero hence the positive bending moment occurs in the mid span.

4. What will be the variation in BMD for the diagram? [Assume l = 2m].

a) Rectangular
b) Trapezoidal
c) Triangular
d) Square
Answer: c
Clarification: At support B, the BM is zero. The beam undergoes maximum BM at fixed end.
By joining the base line, free end and maximum BM point. We obtain a right angled triangle.

5. What is the maximum bending moment for simply supported beam carrying a point load “W” kN at its centre?
a) W kNm
b) W/m kNm
c) W×l kNm
d) W×l/4 kNm
Answer: d
Clarification: We know that in simply supported beams the maximum BM occurs at the central span.
Moment at A = Moment at B = 0
Moment at C = W/2 × l/2 = Wl/ 4 kNm (Sagging).

6. How do point loads and udl be represented in SFD?
a) Simple lines and curved lines
b) Curved lines and inclined lines
c) Simple lines and inclined lines
d) Cant represent any more
Answer: c
Clarification: According to BIS, the standard symbols used for sketching SFD are
Point load = ———–
Udl load =

7. ________ curve is formed due to bending of over hanging beams.
a) Elastic
b) Plastic
c) Flexural
d) Axial
Answer: a
Clarification: The line to which the longitudinal axis of a beam bends or deflects or deviates under given load is known as elastic curve on deflection curve. Elastic curve can also be known as elastic line or elastic axis.

8. The relation between slope and maximum bending moment is _________
a) Directly proportion
b) Inversely proportion
c) Relative proportion
d) Mutual incidence
Answer: b
Clarification: The relationship between slope and maximum bending moment is inversely proportional because, For example in simply supported beams slope is maximum at supports and zero at midspan of a symmetrically loaded beam where as bending moment is zero at supports and maximum at mid span. Hence we conclude that slope and maximum bending moment are inversely proportional to each other in a case of the simply supported beam.

9. What is the SF at support B?

a) 5 kN
b) 3 kN
c) 2 kN
d) 0 kN
Answer: d
Clarification: Total load = 2×2 = 4kN
Shear force at A = 4 kN ( same between A and C )
Shear force at C = 4 kN
Shear force at B = 0 kN
Maximum SF at A = 4 kN.

10. Where do the maximum BM occurs for the below diagram.

a) -54 kNm
b) -92 kNm
c) -105 kNm
d) – 65 kNm
Answer: c
Clarification: Moment at B = 0
Moment at C = – (10 × 3) × (3/2)
= – 45 kNm
Moment at A = – (10 × 3) × (1.5 + 2 )
Maximum BM at A = – 105 kNm
= 105 Nm (hogging).

This set of Strength of Materials Multiple Choice Questions on “Definition of Torque”.

1. Torque is __________ moment.
a) Twisting
b) Shear
c) Bending
d) Couple
Answer: a
Clarification: A cylindrical shaft is subjected to twisting moment or torque when a force is acting on the member tangentially at some radius in a plane of its cross section.

2. Twisting moment is a product of __________ and the radius.
a) Direction
b) Velocity
c) Force
d) Acceleration
Answer: c
Clarification: Twisting moment will be equal to the product of force and radius. When a shaft is subjected to a twisting moment, every cross section of the shaft will surely experience shear stress.

3. Torsion is denoted by __________
a) R
b) Q
c) T
d) N
Answer: c
Clarification: If the moment is applied in a plane perpendicular to the longitudinal axis of the beam (or) shaft it will be subjected to torsion. Torsion is represented or denoted by T.

4. The SI units for torsion is __________
a) N m
b) N
c) N/m
d) m
Answer: a
Clarification: As torsion is a product of perpendicular force and radius, the units will be N m.
Torque is also known as torsion or twisting moment or turning moment.

5. _____________ torsion is produced when twisting couple coincides with the axis of the shaft.
a) Exact
b) Pure
c) Nominal
d) Mild
Answer: b
Clarification: When a member is subjected to the equal and opposite twisting moment at its ends, then the member is said to be subjected under pure torsion. Pure Torsion is often produced when the axis of the twisting couple coincides with the axis of the shaft.

6. Which of the following is known as Re-entrant mouthpiece?
a) External Mouthpiece
b) Convergent Mouthpiece
c) Internal Mouthpiece
d) Cylindrical Mouthpiece
Answer: c
Clarification: According to the position, mouthpieces are classified as an external mouthpiece and internal mouthpiece. If the tube projects inside the tank, it is called an internal mouthpiece or re-entrant or borda’s mouthpiece.

7. Micrometre contraction gauge is used to determine ___________
a) Cv
b) Cc
c) Ca
d) Cd
Answer: b
Clarification: The coefficient of contraction may be determined experimentally by measuring the radius of jet as vena contact with the help of micro meter contraction gauge. This method is not accurate because it is very difficult to measure the correct radius of jet.

8. What is the general value for coefficient of contraction?
a) 0.64
b) 0.67
c) 0.66
d) 0.7
Answer: a
Clarification: The ratio of the area of a jet at vena contracta to the area of orifice is known as the coefficient of contraction. The value of Cc varies from 0.61 to 0.69 for different orifices. Generally, for sharp edged orifice the value of Cc may be taken as 0. 64.

9. The Cd value for internal mouthpiece running free is __________
a) 0.6
b) 0.5
c) 0.7
d) 0.8
Answer: b
Clarification: The Cd value for internal mouthpiece running free is 0.5.

Type Of Mouthpiece	Value of Cd
External cylindrical mouthpiece	0.855
Internal mouthpiece running free	0.5
Internal mouthpiece running full	0.707

10. _______ is the velocity with which water reaches the notch or before it flows over it.
a) Velocity of contact
b) Velocity of moment
c) Velocity of approach
d) Velocity of head
Answer: c
Clarification: The velocity of approach is defined as the velocity with which water reaches the notch or weir before it flows over it. This velocity of approach creates an additional head “ha” equal to Va2 / 2g and effect head over the notch is increased to H+ha.

11. Which of the following formula was proposed by Bazin?
a) m(2g)^1/2×LH^3/2
b) m(2g) ^1/2×H^3/2
c) n(2g)^1/2×LH^4/3
d) n(2g)^1/2×LH^3/2
Answer: a
Clarification: Bazin proposed the following formula for the discharge over rectangular weir:
Q = m(2g) ^1/2× L H^3/2.
Where m = 0.405 + 0.003/H.

12. For measuring low discharges _____________ notch is preferred.
a) Rectangular
b) Stepped
c) Trapezoidal
d) Triangular
Answer: d
Clarification: A triangular notch is preferred to a rectangular notch due to
i. The nappe emerging from a triangular notch has the same shape for all heads. As such the value for the triangular notch is constant for all heads.
ii. The expression for discharge for right angle triangle law not is very simple.

13. Which of the following is also known as V notch?
a) Trapezoidal
b) Stepped
c) Triangular
d) Sharp edged
Answer: c
Clarification: A triangular notch also called a v notch is of triangle shape with apex down. The expression of the discharge over triangular notch or weir is Q = 8/15 Cd (2g)^1/2 × H^5/2.

14. Calculate the discharge over rectangular Weir of 3 metres length under the head of 400mm.Use Francis formula.
a) 1.268 m³/s
b) 1.396 m³/s
c) 1.475 m³/s
d) 1.528 m³/s
Answer: b
Clarification: Francis formula for discharge Q = 1.84 LH³/2.
Given L = 3m & H = 0.4m
Q = 1.84 × 3 × (0.4)³/2.
Q = 1.396 m³/s.

15. _____ converts mechanical energy into hydraulic energy.
a) Dynamo
b) Pump
c) Turbine
d) Generator
Answer: b
Clarification: A pump is a mechanical device which converts the mechanical energy into hydraulic energy. The hydraulic energy is in the form of pressure energy. The pumps are generally used for lifting liquid from a lower level to a higher level.

Category: Strength of Materials Objective Questions

250+ TOP MCQs on Hooke’s Law and Answers

250+ TOP MCQs on Bars of Composite Sections – 2 and Answers

250+ TOP MCQs on Shear Force and Bending Moment diagram and Answers

250+ TOP MCQs on Rectangular Dam and Answers

250+ TOP MCQs on Definition of Torque and Answers

250+ TOP MCQs on Stress & Strain Curve and Answers