Category: Strength of Materials Objective Questions

250+ TOP MCQs on Introduction to Shear Stress and Answers

This set of Strength of Materials Multiple Choice Questions on “Introduction to Shear Stress”.

1. At ________ the shearing stress in a beam are maximum.
a) Extreme fibres
b) Modulus of section
c) Neutral axis
d) Along the cross-sectional area
Answer: c
Clarification: Shearing stress in a beam is maximum at the neutral axis. Shearing stress is defined as the resistance offered by the internal stress to the shear force.

2. Determine the shear stress at the level of neutral axis, if a beam has a triangle cross section having base “b” and altitude “h”. Let the shear force be subjected is F.
a) 3F/8bh
b) 4F/3bh
c) 8F/3bh
d) 3F/6bh
Answer: c
Clarification: For a triangular section subjected to shear force the shear stress in neutral axis is
Shear stress at NA = 4/3 [Average shear stress].
= 4/3 [F/0.5 bh] = 8F / 3bh.

3. The maximum shear stress is ______ times the average shear stress [For rectangular beams].
a) 2.5
b) 3
c) 1.2
d) 1.5
Answer: d
Clarification: The maximum shear stress occurs at neutral axis. Then y = 0.
Max shear stress = 3F/2bd = 3/2 [F/bd].
= 1.5 Average shear stress.

4. Shear stress in a beam is zero at ______
a) Neutral axis
b) Extreme fibres
c) Cross section
d) Junctions
Answer: b
Clarification: The resistance offered by the internal stress to shear is known as shearing stress. Shearing stress is zero at extreme fibres of the beam. The bending stresses are maximum at extreme fibres of the beam cross section.

5. Shear stress distribution over rectangular section will be _________
a) parabolic
b) elliptical
c) triangular
d) trapezoidal
Answer: a
Clarification:
strength-materials-questions-answers-introduction-shear-stress-q5
Maximum shear stress is 1.5 times that of average shear stress.
The shear stress distribution is parabolic.

6. A round Steel rod of 100 mm diameter is bent into an arc of radius 100m. What is the maximum stress in the rod? Take E = 2×105 N/mm2.
a) 150 N/mm2
b) 200 N/mm2
c) 100 N/mm2
d) 300 N/mm2
Answer: c
Clarification: D = 100m
y = 50 mm
R = 10 × 103mm
By equation of flexure; E/R = f/y
f=E/R ×y
= 2×105/ 100 × 103× 50
= 100 N/mm2.

7. For circular section, the maximum shear stress is equal to ____________ times of average shear stress.
a) 2/3
b) 3/2
c) 4/3
d) 3/4
Answer: c
Clarification: Maximum shear stress occurs at neutral axis & y = 0.
Max. Shear stress = 4/3 [ F/A ].
F/A is average shear stress.
The maximum shear stress distribution is 33% more than average shear stress.

8. A steel beam is 200 mm wide and 300 mm deep. The beam is simply supported and carries a concentrated load w. If the maximum stress are 2 N/mm2. What will be the corresponding load?
a) 50 kN
b) 80 kN
c) 40 kN
d) 85 kN
Answer: b
Clarification: For a rectangular cross section
Max. Shear stress = 3/2 [F/A ]
2 = 3/2 [w/200 × 300].
w = 80 kN.

9. Maximum shear stress in thin cylindrical shell be ___________
a) pr/2t
b) pr/3t
c) pr/4t
d) pr/ 5t
Answer: c
Clarification: Hoop stress P(h) = pr/t
Longitudinal stress P(l) = pr/2t
Thus, hoop stress is twice the longitudinal stress
Max. Shear stress = P(h) – P(l) / 2
= pr/4t.

10. Circumferential stress is same as of _________
a) Hoop stress
b) Longitudinal stress
c) Transverse stress
d) Phreatic stress
Answer: a
Clarification: In a thin cylindrical shell of internal radius r thickness t when subjected to internal fluid pressure P, the stress developed in the internal walls can be termed as circumferential stress or hoop stress.
P(h) = pr/t.

250+ TOP MCQs on Deflection of Cantilever and Answers

This set of Strength of Materials Multiple Choice Questions on “Deflection of Cantilever”.

1. The ratio of maximum deflection of a beam to its ___________ is called stiffness of the beam.
a) Load
b) Slope
c) Span
d) Reaction at the support
Answer: c
Clarification: The stiffness of a beam is a measure of it’s resistance against deflection. The ratio of the maximum deflection of a beam to its span can be termed as stiffness of the beam.

2. Stiffness of the beam is inversely proportional to the _____ of the beam.
a) Slope
b) Support reaction
c) Deflection
d) Load
Answer: c
Clarification: Stiffness of a beam is inversely proportional to the deflection. Smaller the deflection in a beam due to given external load, greater is its stiffness.

3. The maximum ____ should not exceed the permissible limit to the span of the beam.
a) Slope
b) Deflection
c) Load
dl Bending moment
Answer: b
Clarification: The maximum deflection of a loaded beam should not exceed the permissible limit in relation to the span of a beam. While designing the beam the designer should be keep in mind that both strength and stiffness criteria.

4. In cantilever beam the deflection occurs at ______
a) Free end
b) Point of loading
c) Through out
d) Fixed end
Answer: a
Clarification: Deflection can be defined as the perpendicular displacement of a point on straight access to the curved axis. In cantilever beams, the maximum deflection occurs at free end.

5. The maximum deflection in cantilever beam of span “l”m and loading at free end is “W” kN.
strength-materials-questions-answers-deflection-cantilever-q5
a) Wl3/2EI
b) Wl3/3EI
c) Wl3/4EI
d) Wl2/2EI
Answer: b
Clarification: Maximum deflection occurs at free end distance between centre of gravity of bending moment diagram and free end is x = 2l/3.
As deflection is equal to the slope × “x”. The slope = Wl2/2EI radians
Maximum deflection (y) = Ax/EI = Wl3/3EI.

6. In an ideal fluid, the ____________ stresses are pretend to be absent.
a) Bending
b) Shearing
c) Tensile
d) Compressive
Answer: b
Clarification: An ideal fluid is a fluid where there is no resistance to the deformation. Ideal Fluids are those Fluids which have no viscosity surface tension. The shear stress is also absent. This fluid is also called as perfect fluid.

7. Air and water are the examples of ___________
a) Non Newtonian fluids
b) Vortex fluids
c) Real fluids
d) Ideal fluids
Answer: d
Clarification: The ideal Fluids are imaginary fluids in nature, they are incompressible. These fluids possess low viscosity. Air and water are considered as ideal fluids.

8. _______ fluids are practical fluids
a) Ideal
b) Real
c) Vortex
d) Newtonian
Answer: b
Clarification: These fluids possess properties such as viscosity, surface tension. They are compressible in nature. The certain amount of resistance is always offered by the fluids, they also possess shear stress. They are also known as practical fluids.

9. Specific weight of water at 4°C is ____________ N/m3.
a) 9810
b) 9760
c) 9950
d) 9865
Answer: a
Clarification: The specific weight (weight density) of a fluid is weight per unit volume. It is represented by symbol w & it is expressed in Newton per metre cube (N/m3). The specific weight of water at 4 degree centigrade is 9810 N/m3or 9.81 kN/m3.

10. The inverse of specific weight of a fluid is __________
a) Specific gravity
b) Specific Volume
c) Compressibility
d) Viscosity
Answer: b
Clarification: Specific volume is the volume of the fluid by Unit Weight it is the reciprocal of specific weight is denoted by “v”. SI units are m3/N.
v= 1/specific weight.

11. Calculate the specific gravity of mercury.
a) 12.5
b) 14.7
c) 13.6
d) 11.8
Answer: c
Clarification: The specific gravity of any fluid is the ratio of the specific weight of fluid by specific weight of water. For mercury, the specific weight is 133416 N/m3. For water, w = 9810 N/m3.
S = 133416/9810
S= 13.6.

12. Specific gravity of water is __________
a) 0.8
b) 1
c) 1.2
d) 1.5
Answer: b
Clarification: The specific gravity is also called as relative density. It is dimensionless quantity and it has no units. The specific gravity of water is the ratio of specific weight of fluid to specific weight of water, as both the numerator and denominator are same. The value is 1.

13. Compute the maximum deflection at free end of a cantilever beam subjected to udl for entire span of l metres.
a) wl4/8EI
b) wl4/4EI
c) wl3/8EI
d) wl2/6EI
Answer: a
Clarification: The slope at free end = A/EI = wl3/6EI
Maximum deflection at free end is Ax/EI; [x= ¾ l]
y= wl3/6EI × ¾ l = wl4/8EI.

14. Calculate the maximum deflection of a cantilever beam with udl on entire span of 3m the intensity of you udl be 25 kN/m. Take EI as 4000 kN/m2.
a) 0.052m
b) 0.063m
c) 0.076m
d) 0.09m
Answer: b
Clarification: For cantilever beams with udl on entire span, the maximum deflection = wl4/8EI
y = wl4/8EI = 25 × 34/ 8 × 4000 = 0.063m.

15. Which of the following is not an example of Malleability?
a) Wrought Iron
b) Ornamental silver
c) Torsteel
d) Ornamental gold
Answer: c
Clarification: Torsteel is an example of mechanical property ductility. The ductility is a property of a material by which material can be fractured into thin wires after undergoing a considerable deformation without any rupture.

250+ TOP MCQs on Composite Shaft and Answers

This set of Strength of Materials Multiple Choice Questions on “Composite Shaft”.

1. Calculate that torque, if the diameter of the shaft is 50 mm and revolutions @ 130 rpm. The maximum shear stress is 62.5 N/mm2.
a) 1564 Nm
b) 1478 Nm
c) 1534 Nm
d) 1494 Nm
Answer: c
Clarification: Diameter of shaft = 50 mm
Revolutions of shaft = 130 rpm.
Maximum shear stress = f= 62.5 N/mm2.
T = f π D3/ 16
T = 62.5 ×503× π / 16.
T = 1534 Nm.

2. What is the example for a centrifugal pump?
a) Reciprocating pump
b) Suction pump
c) Rotodynamic pump
d) Delivery pump
Answer: c
Clarification: Rotodynamic pumps have a rotating element through which as the liquid passes its angular momentum changes, due to which the pressure energy of the liquid is increased. The centrifugal pump is a rotodynamic pump.

3. Reciprocating pump is an example of ___________
a) Positive displacement pump
b) Delivery pump
c) Suction pump
d) Rotodynamic pump
Answer: a
Clarification: Positive displacement pumps are those pumps in which the liquid is sucked and then it is pushed to the thrust exerted on it by a piston. The most common example of the positive displacement pump is the reciprocating pump.

4. ____ is the difference between theoretical discharge and the actual discharge of the pump.
a) Crank
b) Hook
c) Slip
d) Centile
Answer: c
Clarification: c
Clarification: Slip of a pump is defined as the difference between the theoretical discharge and actual discharge after pump.

5. _____ is a phenomenon by which the study and continuous flow of liquid are obstructed.
a) Slip
b) Separation
c) Air vessels
d) Knockage
Answer: b
Clarification: Separation of reciprocating pump is that phenomenon by which the steady and continuous flow of liquid is affected by the presence of air and dissolved gases.

6. Negative slip occurs when the______ is more than theoretical discharge.
a) Virtual discharge
b) Actual discharge
c) Mean discharge
d) Mode discharge
Answer: b
Clarification: When the delivery valve opens before the suction stroke is completed, the actual discharge is more than the theoretical discharge. In such cases, the slip of the pump is known as a negative slip.

7. _____ slip occurs, when the delivery pipe is short and the suction pipe is long.
a) Positive
b) Critical
c) Negative
d) Zero
Answer: b
Clarification: The slip occurs when the delivery pipe is short and the suction pipe is long. The pump is running at high speeds as the delivery valve open before a suction stroke is completed, the slip of the pump is known as negative slip.

8. ________ reduces the possibility of separation.
a) Air vessels
b) Casing
c) Impeller
d) Vortex
Answer: a
Clarification: An air vessel may be a closed chamber having the compressed air in a top portion and the water at the bottom. It reduces the possibility of separation and it ensures the pump to run at high speed.

9. If the absolute pressure falls below ___________ m, the pump prone to separation.
a) 3 m
b) 2 m
c) 1.5 m
d) 2.5 m
Answer: d
Clarification: If the absolute pressure falls below 2.5 metres of water, the dissolved gases will be appearing in a liquid and continuous flow will be chocked. This phenomenon can be termed as separation.

10. The phenomenon of separation can also be known as ___________
a) Cavitation
b) Priming
c) Positive head
d) Pulsate
Answer: a
Clarification: Separation is a phenomenon of obstructing the flow by the presence of dissolved gases when the absolute pressure falls below 2.5 metres of water. This phenomenon of separation can also be known as knocking (or) cavitation in the reciprocating pump.

11. The work done against friction is reduced due to _____________
a) Impeller
b) Priming
c) Air vessel
d) Vortex
Answer: c
Clarification: An air vessel is fitted to the suction and delivery pipes at a point close to the cylinder of a single acting reciprocating. The pump increases the length of the suction pipe and reduces the work done against friction.

12. Volute is a type of _____________
a) Delivery pipe
b) Casing
c) Impeller
d) Suction pipe
Answer: b
Clarification: Casing is an airtight chamber covering the impeller. The different types of casing
i.Volute casing
ii.Vortex casing
iii.Casing with guide blades.

13. ______ pumps, the torque is uniform.
a) Reciprocating pump
b) Suction pump
c) Delivery pump
d) Centrifugal pump
Answer: d
Clarification: Centrifugal pump is used for lifting highly viscous liquids such as oils, muddy and sewage water, paper pulp etc. In centrifugal pump, torque is uniform and no air vessels are required.

14. What is the practical maximum suction lift in a reciprocating pump?
a) 3.5 m
b) 4.5 m
c) 5 m
d) 6.5 m
Answer: d
Clarification: Reciprocating pump can handle only pure water or less viscous liquids free from impurities. It can be operated at low speeds only. The practical maximum section lift is 6.5 metres.

15. _____ pumps give a larger discharge.
a) Suction
b) Reciprocating
c) Centrifugal
d) Positive displacement
Answer: c
Clarification: Centrifugal pump are an example of rotodynamic pump the basic principle of centrifugal pump is that “when a certain mass of liquid is rotated by an external force, then the centrifugal head is impressed which enables it to rise to a higher level”. A centrifugal pump discharges a larger quantity when compared to other pumps.

250+ TOP MCQs on Bending Stress and Answers

Strength of Materials Multiple Choice Questions on “Bending Stress”.

1. A beam is said to be of uniform strength, if ____________
a) B.M. is same throughout the beam
b) Shear stress is the same through the beam
c) Deflection is the same throughout the beam
d) Bending stress is the same at every section along its longitudinal axis
Answer: d
Clarification: Beam is said to be uniform strength if at every section along its longitudinal axis, the bending stress is same.

2. Stress in a beam due to simple bending is ____________
a) Directly proportional
b) Inversely proportional
c) Curvilinearly related
d) None of the mentioned
Answer: a
Clarification: The stress is directly proportional to the load and here the load is in terms of bending. So the stress is directly proportional to bending.

3. Which stress comes when there is an eccentric load applied?
a) Shear stress
b) Bending stress
c) Tensile stress
d) Thermal stress
Answer: b
Clarification: When there is an eccentric load it means that the load is at some distance from the axis. This causes compression in one side and tension on the other. This causes bending stress.

4. What is the expression of the bending equation?
a) M/I = σ/y = E/R
b) M/R = σ/y = E/I
c) M/y = σ/R = E/I
d) M/I = σ/R = E/y
Answer: a
Clarification: The bending equation is given by M/I = σ/y = E/R
where
M is the bending moment
I is the moment of inertia
y is the distance from neutral axis
E is the modulus of elasticity
R is the radius.

5. On bending of a beam, which is the layer which is neither elongated nor shortened?
a) Axis of load
b) Neutral axis
c) Center of gravity
d) None of the mentioned
Answer: b
Clarification: When a beam is in bending the layer in the direction of bending will be In compression and the other will be in tension. One side of the neutral axis will be shortened and the other will be elongated.

6. The bending stress is ____________
a) Directly proportional to the distance of layer from the neutral layer
b) Inversely proportional to the distance of layer from the neutral layer
c) Directly proportional to the neutral layer
d) Does not depend on the distance of layer from the neutral layer
Answer: a
Clarification: From the bending equation M/I = σ/y = E/R
Here stress is directly proportional to the distance of layer from the neutral layer.

7. Consider a 250mmx15mmx10mm steel bar which is free to expand is heated from 15C to 40C. what will be developed?
a) Compressive stress
b) Tensile stress
c) Shear stress
d) No stress
Answer: d
Clarification: If we resist to expand then only stress will develop. Here the bar is free to expand so there will be no stress.

8. The safe stress for a hollow steel column which carries an axial load of 2100 kN is 125 MN/m2. if the external diameter of the column is 30cm, what will be the internal diameter?
a) 25 cm
b) 26.19cm
c) 30.14 cm
d) 27.9 cm
Answer: b
Clarification: Area of the cross section of column = π/4 (0.302 – d2) m2
Area = load / stress.
So, π/4 ( 0.302 – d2) m2 = 21 / 125
d = 26.19cm.

250+ TOP MCQs on Moment of Inertia and Answers

This set of Strength of Materials Multiple Choice Questions on “Moment of Inertia”.

1. The axis about which moment of area is taken is known as ____________
a) Axis of area
b) Axis of moment
c) Axis of reference
d) Axis of rotation
Answer: c
Clarification: The axis of reference is the axis about which moment of area is taken. Most of the times it is either the standard x or y axis or the centeroidal axis.

2. Point, where the total volume of the body is assumed to be concentrated is ____________
a) Center of area
b) Centroid of volume
c) Centroid of mass
d) All of the mentioned
Answer: b
Clarification: The centroid of the volume is the point where total volume is assumed to be concentrated. It is the geometric centre of a body. If the density is uniform throughout the body, then the center of mass and center of gravity correspond to the centroid of volume. The definition of the centroid of volume is written in terms of ratios of integrals over the volume of the body.

3. What is MOI?
a) ml2
b) mal
c) ar2
d) None of the mentioned
Answer: c
Clarification: The formula of the moment of inertia is, MOI = ar2 where
M = mass, a = area, l = length, r = distance.

4. What is the formula of radius of gyration?
a) k2 = I/A
b) k2 = I2/A
c) k2 = I2/A2
d) k2 = (I/A)1/2
Answer: a
Clarification: The radius of gyration of a body about an axis is a distance such that its square multiplied by the area gives moment of inertia of the area about the given axis. The formula of radius of gyration is given as k2 = I/A.

5. What is the formula of theorem of perpendicular axis?
a) Izz = Ixx – Iyy
b) Izz = Ixx + Ah2
c) Izz – Ixx = Iyy
d) None of the mentioned
Answer: c
Clarification: Theorem of perpendicular axis stares that if IXX and IYY be the moment of inertia of a plane section about two mutually perpendicular axis X-X and Y-Y in the plane of the section then the moment of inertia of the section IZZ about the axis Z-Z, perpendicular to the plane and passing through the intersection of X-X and Y-Y is given by the formula
Izz – Ixx = Iyy.

6. What is the formula of theorem of parallel axis?
a) IAD = IG + Ah
b) IAB = Ah2 + IG
c) IAB = IG – Ah2
d) IAB = IG + Ixx
Answer: b
Clarification: The theorem of parallel axis states that if the moment of inertia of a plane area about an axis in the plane of area theough the C.G. of the plane area be represented by IG, then the moment of the inertia of the given plane area about a parallel axis AB in the plane of area at a distance h from the C.G. is given by the formula
IAB = Ah2 + IG.

7. What is the unit of radius of gyration?
a) m4
b) m
c) N
d) m2
Answer: b
Clarification: The radius of gyration = (length4/length2)1/2 = length
So its unit will be m.

8. What will be the the radius of gyration of a circular plate of diameter 10cm?
a) 1.5cm
b) 2.0cm
c) 2.5cm
d) 3cm
Answer: c
Clarification: The moment of inertia of a circle, I = πD4/64 = 491.07 cm4
The area of circle = 78.57 cm,
Radius of gyration = (I/A)1/2 = 2.5 cm.

250+ TOP MCQs on Shear Stress Distribution in Various Sections and Answers

Basic Strength of Materials Questions and Answers on “Shear Stress Distribution in Various Sections”.

1. A beam has a triangular cross-section, having altitude ”h” and base “b”. If the section is being subjected to a shear force “F”. Calculate the shear stress at the level of neutral axis in the cross section.
a) 4F/5bh
b) 4F/3bh
c) 8F/3bh
d) 3F/4bh

Answer: c
Clarification: For a triangular section subjected to a shear force, the shear stress at neutral axis is
= 4/3 × average shear stress
= 4/3 × F/A/2 ; A = bh
= 8F/3bh.

2. The maximum shear stress in the rectangular section is ______________ times the average shear stress.
a) 3/4
b) 3/7
c) 5/3
d) 3/2

Answer: d
Clarification: The maximum shear stress occurs at the neutral axis. So, y = 0.
Maximum shear stress = 3/2 × F/bd (• Average shear stress = F/bd ).
= 3/2 × average shear stress.

3.The modular ratio for M20 grade concrete is _____________
a) 16
b) 13
c) 11
d) 07

Answer: b
Clarification: According to Indian Standards 456-2000 The modular ratio (m) = 280/3× cbc, For M20; compressive bearing capacity in concrete = 7 N/mm2 & tensile strength = 330 N/mm2.
Modular ratio (m) = 280/21 = 13.33.

4. In doubly reinforced beam, the maximum shear stress occurs ______________
a) along the centroid
b) along the neutral axis
c) on the planes between neutral axis and tensile reinforcement
d) on the planes between neutral axis and compressive reinforcement

Answer: d
Clarification: In continuous beam the moments developed at the supports are greater than a moment’s developed at the mid span, show the maximum bending moment occurs at the supports.
For continuous beams, the maximum shear stress occurs at the planes intersecting the compressive reinforcement and the neutral axis.

5. A cylindrical section having no joint is known as _______________
a) Proof section
b) Seamless section
c) Target section
d) Mown section

Answer: b
Clarification: A cylindrical section having no joint is known as seamless section. Built up section is not that strong as a seamless section of the same thickness.

6. The efficiency of cylindrical section is the ratio of the strength of joint to the strength of _______________
a) Solid plate
b) Boilerplate
c) Circumferential plate
d) Longitudinal plate

Answer: a
Clarification: The strength of plate or strength of rivet whichever is less is called the strength of joint. The ratio of the strength of joint to the strength of steel plate is called the efficiency of the cylinder.

7. Calculate the modulus of section for a hollow circular column of external diameter 60 mm and 10 mm thickness.
a) 170 m
b) 190 m
c) 250 m
d) 300 m
Answer: a

8. Determine the modulus of a section for an I section, given the distance from neutral axis is 50 mm and moment of inertia is 2.8×106 mm4.
a) 59m
b) 51m
c) 58m
d) 63m

Answer: c
Clarification: The modulus of section is the ratio of the moment of inertia to the distance of the neutral axis.
Given y = 50 mm
I = 2.8×106 mm4.
& Z = I/y = 2.8×106 / 50
= 57.76 ×103 mm
= 57.7 ~ 58 m.

9. A circular Beam of 0.25 m diameter is subjected to you shear force of 10 kN. Calculate the value of maximum shear stress. [Take area = 176 m2].
a) 0.75 N/mm2
b) 0.58 N/mm2
c) 0.73 N/mm2
d) 0.65 N/mm2

Answer: a
Clarification: Given diameter = 0.25 m
Area (A) = 176 m2
Shear Force (F) = 10 kN ~ 10000 N.
For circular cross section the maximum shear stress is equal to 4/3 times of average shear stress
Maximum shear force = 4/3 × F/A
= 4/3 × 10000/176
= 0.75 N/mm2.

10. The maximum shear stress distribution is _____________ percentage more than average shear stress in circular section.
a) 54 %
b) 60 %
c) 33 %
d) 50 %

Answer: c
Clarification: Maximum shear stress occurs at neutral axis; y =0
Maximum shear stress = 16/3 × average shear stress
But 4F / A is the average shear stress.
So, the maximum shear stress = 4/3 times the average shear stress.
Hence the maximum shear stress is 33% more than the average shear stress in a circular section.