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

250+ TOP MCQs on Section Modulus and Answers

This set of Strength of Materials Multiple Choice Questions on “Section Modulus”.

1. What is the section modulus (Z) for a rectangular section?
a) bd²/6
b) a³/6
c) BD³-bd³
d) D⁴-d⁴
Answer: a
Clarification: The modulus of section may be defined as the ratio of moment of inertia to the distance to the extreme fibre. It is denoted by Z.
Z= I/y ; For rectangular section, I = bd³/12 & y = d/2.
Z= bd²/6.

2. Find the modulus of section of square beam of size 300×300 mm.

a) 4.8 × 10⁶ mm³
b) 4.5 × 10⁶ mm³
c) 5.6 × 10⁶ mm³
d) 4.2 × 10⁶ mm³
Answer: b
Clarification: Here, a = side of square section = 300 mm.
I = a⁴/12. y= a/2.
Z = I/y = a³/6
= 300³/6
= 4.5 × 10⁶ mm³.

3. _________ of a beam is a measure of its resistance against deflection.
a) Strength
b) Stiffness
c) Deflection
d) Slope
Answer: b
Clarification: A beam is said to be a strength when the maximum induced bending and shear stresses are within the safe permissible stresses stiffness of a beam is a measure of its resistance against deflection.

4. To what radius an Aluminium strip 300 mm wide and 40mm thick can be bent, if the maximum stress in a strip is not to exceed 40 N/mm². Take young’s modulus for Aluminium is 7×105 N/mm².
a) 45m
b) 52m
c) 35m
d) 65m
Answer: c
Clarification: Here, b = 300mm
d= 40mm. y= 20mm.
From the relation; E/R = f/y
R= E×y/f
=70×10³ × 20 / 40
= 35m.

5. The bending stress in a beam is ______ to bending moment.
a) Less than
b) Directly proportionate
c) More than
d) Equal
Answer: b
Clarification: As we know, the bending stress is equal to bending moment per area. Hence, as the bending (flexure) moment increases/decreases the same is noticed in the bending stress too.

6. The Poisson’s ratio for concrete is __________
a) 0.4
b) 0.35
c) 0.12
d) 0.2
Answer: d
Clarification: The ratio of lateral strain to the corresponding longitudinal strain is called Poisson’s ratio. The value of poisons ratio for elastic materials usually lies between 0.25 and 0.33 and in no case exceeds 0.5. The Poisson’s ratio for concrete is 0.20.

7. The term “Tenacity” means __________
a) Working stress
b) Ultimate stress
c) Bulk modulus
d) Shear modulus
Answer: b
Clarification: The ultimate stress of a material is the greatest load required to fracture the material divided by the area of the original cross section in the point of fracture The ultimate stress is also known as tenacity.

8. A steel rod of 25 mm diameter and 600 mm long is subjected to an axial pull of 40000. The intensity of stress is?
a) 34.64 N/mm²
b) 46.22 N/mm²
c) 76.54 N/mm²
d) 81.49 N/mm²
Answer: d
Clarification: Cross sectional area of steel rod [Circular]be 490.87 mm².
The intensity of stress = P/A = 40000/490.87
= 81.49 N/mm².

9. The bending strain is zero at _______
a) Point of contraflexure
b) Neutral axis
c) Curvature
d) Line of action of loading
Answer: b
Clarification: The neutral axis is a line of intersection of neutral plane or neutral layer on a cross section. The neutral axis of a beam passes through the centroid of the section. At the neutral axis bending stress and bending strain is zero.

10. Strength of the beam depends only on the cross section.
a) True
b) False
Answer: b
Clarification: The strength of two beams of the same material can be compared by the section modulus values. The strength of beam depends on the material, size and shape of cross section. The beam is stronger when section modulus is more, strength of the beam depends on Z.

250+ TOP MCQs on Trapezoidal Dam as Inclined Side Phase and Answers

This set of Strength of Materials Questions and Answers for Campus interviews on “Trapezoidal Dam as Inclined Side Phase”.

1. Calculate the eccentricity of a trapezoidal dam with a distance between the centre of gravity and point where the resultant cuts the base is 5m. The bottom width of the dam is 3m.
a) 2.5m
b) 3.5m
c) 4.5m
d) 5m
Answer: b
Clarification: The eccentricity (e) = Z – b/2.
Z= distance between the centre of gravity and point where the resultant cuts the base is 5m.
= 5-3/2 = 1.5m.

2. Which of the following is not a failure of retaining wall?
a) Structural slide
b) Shear sliding
c) Crushing
d) Slope pitching
Answer: c
Clarification: Crushing Failure is related to dams. The retaining walls are the structures constructed to store earth on one side especially in case of hill or ghat roads.

3. ________ pressure which occurs commonly in dams.
a) Passive earth pressure
b) Active earth pressure
c) Soil moisture tension
d) Wind pressure
Answer: b
Clarification: Active earth pressure is exerted by backfill on retaining walls. It is also called as practical pressure. It occurs commonly in dams.

4. _________ failures contribute 40% to earthen dams.
a) Seepage
b) Structural
c) Hydraulic
d) Natural
Answer: c
Clarification: On the basis of various investigation reports and case studies, hydraulic failures contribute about 40% of failures to earthen dams. The rest of the failures is shared by seepage failures and structural failures.

5. Which of the following filters are also known as chimney drains?
a) Horizontal filter
b) Inclined filter
c) Rock toe
d) Toe drain
Answer: b
Clarification: The filters which are laid across the outer slope of the impervious core are called as inclined filters. They are also known as Chimney drains. They are provided mainly to collect the seepage emerging out of the core.

6. Zoned earthen dams are also known as ______
a) Heterogeneous dams
b) Core wall dams
c) Homogeneous dams
d) Hydraulic dam
Answer: a
Clarification: The dams are constructed on shallow pervious foundations in this dam section about outer zones are made fairly pervious material and the inner most zoning called “hearting” is done was fairly impervious material. It is also known as heterogeneous dam.

7. ___________ dams are built with key trenches.
a) Heterogeneous earth dam
b) Homogeneous earth dam
c) Earth Dam with Core wall
d) Rolled fill dam
Answer: c
Clarification: The outer zones of this dam are made of pervious material as in zoned dam. In this case, it is essentially build cut-off wall (cut-off trench) built quite deep preferably upto impervious rock layer in the foundation.

8. Line of seepage is also known as __________
a) Hydraulic gradient
b) Phreatic line
c) Seepage gradient
d) Hydraulic seepage line
Answer: b
Clarification: The line within the dam section below which there are positive hydrostatic pressures in a dam and above the line the hydrostatic pressures are negative. It gives a divide line between dry and saturated soils.

9. ____________ represents the top stream line.
a) Phreatic line
b) Hydraulic gradient line
c) Seepage gradient
d) Hydraulic seepage line
Answer: a
Clarification: Phreatic line is also called a line of seepage or saturation line. The phreatic line represents the top streamline and hence helps us in drawing the flow net.

10. The hydrostatic pressures on phreatic line are equal to____
a) Zero
b) Maximum
c) Minimum
d) Constant
Answer: a
Clarification: The hydrostatic pressure on phreatic line is equal to atmospheric pressure and hence equal to zero, the flow through the body of the dam below the phreatic line reduces the effective weight of the soil and thus reduces the shear strength of the soil.

11. Expand MWL?
a) Minimum water level
b) Maximum water level
c) Meagre water level
d) Most wind level
Answer: b
Clarification: The water level that is attained during floods is called the maximum water level. The dams and spillway sections are designed to withstand water pressure at this level.

12. ______ is the difference of Level between full reservoir level and top of the dam.
a) Net free board
b) Gross free board
c) Design free board
d) Over free board
Answer: b
Clarification: In dams, in order to prevent the overtopping during peak floods, a sufficient margin is provided between the full reservoir level and top of the dam. This is known as gross free board.

13. By keeping the phreatic line within the downstream toe, the ___________ can be avoided.
a) Piping
b) Gullying
c) Sloughing
d) Over topping
Answer: c
Clarification: If the filter at downstream side toe is choked then also the downstream too becomes saturated. In such circumstances, some erosion occurs in downstream to this causes sloughing. To avoid this phreatic line must be within the downstream toe.

14. Springs(closely coiled) are examples of _____________
a) stiffness
b) hardness
c) toughness
d) creep
Answer: a
Clarification: The property of a material or substance which offers resistance to bending action and measures the load required to be applied is called stiffness. It is denoted by s or k. Springs are the best examples of stiffness.

15. Perennial canals are also known as ________
a) Inundation canal
b) Productive canal
c) Feeder canal
d) Permanent canal
Answer: d
Clarification: The canal which is fed by a permanent source of supply is said to be a permanent Canal. It has also regulatory works. This canal is also sometimes known as a perennial canal.

Strength of Materials for Campus Interviews,

250+ TOP MCQs on Polar Moment of Inertia and Answers

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

1. The moment of inertia of a plane area with respect to an axis ____________ to the plane is called a polar moment of inertia.
a) Parallel
b) Perpendicular
c) Equal
d) Opposite
Answer: b
Clarification: The moment of inertia of a plane area with respect to an axis perpendicular to the plane of the figure is called a polar moment of inertia with respect to a point, where the axis intersects a plane.

2. What are the units of Polar modulus?
a) mm³
b) mm²
c) mm
d) mm⁴
Answer: a
Clarification: The ratio of polar moment of inertia (J) to the radius of section(R) is known as polar modulus or torsional section modulus. Its units are mm³.

3. What is the polar modulus for solid shaft?
a) π/16 D²
b) π/12 D³
c) π/ 16 D³
d) π/16 D
Answer: c
Clarification: For solid shaft Z = J/R = π/32 × D⁴/ D/2.
Z = π/16 D³.

4. Calculate the polar moment of inertia for a solid circular shaft of 30 mm diameter.
a) 76m⁴
b) 79.5m⁴
c) 81m⁴
d) 84m⁴
Answer: b
Clarification: Diameter of the shaft = 30 mm
Polar moment of inertia = J = π/32 × (30)⁴ mm⁴
J = 79.52 m⁴.

5. A hollow shaft outside diameter 120 mm and thickness 20 mm. Find polar moment of inertia.
a) 16.36 × 10⁶ mm⁴
b) 18.45 × 10⁶ mm⁴
c) 21.3 × 10⁶ mm⁴
d) 22.5 × 10⁶ mm⁴
Answer: a
Clarification: For hollow circular shaft, outside diameter = 120 mm; d = 120-2×20 = 80 mm
the polar moment of inertia = π/32 × (120⁴– 80⁴).
J = 16.36 × 10⁶ mm⁴.

6. Determine the maximum flood discharge from a catchment area of 40.25 km2 and it is situated in the Western Ghats.
a) 350 cumecs
b) 375 cumecs
c) 400 cumecs
d) 425 cumecs
Answer: c
Clarification: Since the catchment area is situated in the Western Ghats, the formula best suited is Dicken’s formula and the coefficient of Dicken’s may be taken as 25.
Q = CA^3/4
Q = 25×(40.25)^3/4
Q = 400 cumecs.

7. Which of the following is known as “under sluices”?
a) Scouring Sluices
b) Divide wall
c) Fish ladder
d) Head Regulator
Answer: a
Clarification: The openings provided in a body wall of the weir almost at the bed level of the river are known as scouring sluices. These are also known as under sluices.

8. _______ provides straight approach to the scouring sluices.
a) Head regulator
b) Silt Excluder
c) Divide wall
d) Guide banks
Answer: c
Clarification: A divide wall is a long solid wall constructed perpendicular to the axis of weir. It provides a straight approach to the scouring sluices. By preventing the formation of cross currents, it protects the body wall of weir.

9. __________ is provided for the easy movement of fish from upstream to downstream.
a) Fish ladder
b) Silt excluder
c) Marginal bunds
d) Marginal embankments
Answer: a
Clarification: A passage provided just by the side of a divide wall for the movement of fish from upstream to downstream or vice versa is known as a fish ladder.

10. __________ is used as measuring device.
a) Head regulator
b) Divide wall
c) Cross regulator
d) Scouring sluices
Answer: a
Clarification: A structure constructed at the head of the canal to regulate the supply of water into the canal is called “Head Regulator”. The functions:
i. It is used as a measuring device.
ii. It controls the entry of silt into the canal.

11. __________ is provided to prevent the river from outflanking the work.
a) Guide banks
b) Marginal bunds
c) Silt excluder
d) Divide wall
Answer: a
Clarification: Guide banks are provided on either side of the diversion head works in alluvial soils for a smooth non -tortuous approach to the diversion head works and prevent the river from outflanking the work.

12. ____________ are provided to protect the land and property with is likely to be submerged.
a) Weir
b) Divide wall
c) Marginal bunds
d) Fish ladder
Answer: c
Clarification: Marginal Bunds or marginal embankments are provided on either bank of the river upstream side of diversion head works in alluvial soils in order to protect the land and property which is likely to be submerged during ponding of water during floods.

13. _________ is provided to reduce the kinetic energy of falling water in weir.
a) Body wall
b) Curtain walls
c) Downstream apron
d) Shutter
Answer: c
Clarification: The downstream apron is a concrete bed which is provided on the downstream side of a weir in order to reduce the kinetic energy of falling water. It should have sufficient thickness to resist uplift pressure.

14. Curtain walls are provided to increase ________
a) Creep depth
b) Creep area
c) Creep length
d) Creep volume
Answer: c
Clarification: Curtain walls are provided under the upstream and downstream apron at the ends. We are provided to increase the length of creep and thereby to reduce exit gradient.

15. Which of the following are also known as upstream and downstream piles?
a) Talus on upstream and downstream
b) Curtain walls on upstream and downstream
c) Solid apron on upstream and downstream
d) Shutters on crest of weir
Answer: b
Clarification: Curtain walls are provided especially under the upstream and downstream aprons at the respective ends. They are also called as upstream and downstream piles. The length of the curtain wall depends on the nature of subsoil.

250+ TOP MCQs on Strain Constants – 2 and Answers

This set of Strength of Materials Interview Questions and Answers on “Strain Constants – 2”.

1. A circular rod of dia 30 mm and length 200mm is extended to 0.09mm length and 0.0045 diameters through a tensile force. What will be its Poissons ratio?
a) 0.30
b) 0.31
c) 0.32
d) 0.33
Answer:d
Clarification: Poissons ratio = lateral strain / longitudinal strain
= δD/D x L/δL
= 0.0045/30 x 200/0.09
= 0.33.

2. The Poissons ratio of a material is 0.3. what will be the ratio of Youngs modulus to bulk modulus?
a) 1.4
b) 1.2
c) 0.8
d) 0.6
Answer: b
Clarification: As we know E = 3k(1- 2μ)
So E/K = 3(1-2×0.3) = 1.2.

3. What is the bulk modulus of elasticity?
a) The ratio of shear stress to shear strain
b) The ratio of direct stress to direct strain
c) The ratio of volumetric stress to volumetric strain
d) The ratio of direct stress to volumetric strain
Answer: d
Clarification: When a body is subjected to the mutually perpendicular like and equal direct stresses, the ratio of direct stress to the corresponding volumetric strain strain is found to be constant for a given material when the deformation is within a certain limit. This ratio is known as the bulk modulus.

4. For a material, Youngs modulus is given as 1.2 x 10⁵ and Poissons ratio 1/4. Calculate the bulk modulus.
a) 0.7 x 10⁵
b) 0.8 x 10⁵
c) 1.2 x 10⁵
d) 1.2 x 10⁵
Answer: b
Clarification: The bulk modulus is given as K = E/3(1 – 2μ)
= 1.2 x 10⁵/3(1 – 2/4)
= 0.8 x 10⁵.

5. Determine the Poissons ratio and bulk modulus of a material, for which Youngs modulus is 1.2 and modulus of rigidity is 4.8.
a) 7
b) 8
c) 9
d) 10
Answer: b
Clarification: As we know, E = 2C(1 + μ)
μ= 0.25
K = E / 3(1 – 2μ)
= 8.

6. The Youngs modulus of elasticity of a material is 2.5 times its modulus of rigidity. Then what will be its Poissons ratio?
a) 0.25
b) 0.33
c) 0.50
d) 0.60
Answer: a
Clarification: As we know E = 2G(1 + μ) so putting the values of E = 2.5G then we get μ= 0.25.

7. How the elastic constants E and K are related?
a) E = 2K(1 – 2μ)
b) E = 3K(1 – 2μ)
c) E = 2K(1 – μ)
d) E = K(1 – 2μ)
Answer: b
Clarification: As E = 2G(1 + μ) = 3K(1 – 2μ).

8. How many elastic constants does an isotropic, homogeneous and linearly elastic material have?
a) 1
b) 2
c) 3
d) 4
Answer: b
Clarification: E, G, K represents the elastic modulus, shear modulus, bulk modulus and poisson’s ratio respectively of a linearly elastic, isotropic and homogeneous material. To express the stress-strain relations completely for this material at least any two of the four must be known, E = 2G(1 + μ) = 3K(1 – 2μ) = 9KG / (3K + G).

9. The modulus of rigidity and the modulus of elasticity of a material are 80 GPa and 200 GPa. What will be the Poissons ratio of the material?
a) 0.25
b) 0.30
c) 0.40
d) 0.50
Answer: a
Clarification: As E = 2G(1 + μ) putting E = 200 and G = 80 we get μ = 0.25.

10. Which of the following is true if the value of Poisson’s ratio is zero?
a) The material is rigid
b) The material is perfectly plastic
c) The longitudinal strain in the material is infinite
d) There is no longitudinal strain in the material
Answer: a
Clarification: If the Poissons ratio is zero then the material is rigid.

Strength of Materials for Interviews,

250+ TOP MCQs on Gradual Loading and Answers

This set of Strength of Materials Multiple Choice Questions on “Gradual Loading”.

1. What is the strain energy stored in a body when the load is applied gradually?
a) σE/V
b) σE²/V
c) σV²/E
d) σV²/2E
Answer: d
Clarification: Strain energy in gradual loading = σ²V/2E.

2. What is strain energy?
a) The work done by the applied load In stretching the body
b) The strain per unit volume
c) The force applied in stretching the body
d) The stress per unit are
Answer: a
Clarification: The strain energy stored in a body is equal to the work done by the applied load in stretching the body.

3. What is the relation between maximum stress induced due to gradual load to maximum stress the sudden load?
a) Maximum stress in gradual load is equal to the maximum stress in sudden load
b) Maximum stress in gradual load is half to the maximum stress in sudden load
c) Maximum stress in gradual load is twice to the maximum stress in sudden load
d) Maximum stress in gradual load is four times to the maximum stress in sudden load
Answer: b
Clarification: Maximum stress in gradual loading = P/A
Maximum stress in sudden loading = 2P/A.

4. A tensile load of 60kN is gradually applied to a circular bar of 4cm diameter and 5cm long. What will be the stress in the rod if E=1×10⁵ N/mm²?
a) 47.746 N/mm²
b) 34.15 N/mm²
c) 48.456 N/mm²
d) 71.02 N/mm²
Answer: a
Clarification: Stress = Load/ area = 60,000 / (π/4 D²) = 47.746 N/mm².

5. A tensile load of 60kN is gradually applied to a circular bar of 4cm diameter and 10m long. What will be the stress in the rod if E=1×10⁵ N/mm²?
a) 1.19mm
b) 2.14mm
c) 3.45mm
d) 4.77mm
Answer: d
Clarification: Stress = Load/ area = 60,000 / (π/4 D²) = 47.746 N/mm²
So stretch = stress x length / E = 4.77mm.

6. A tensile load of 100kN is gradually applied to a rectangular bar of dimension 2cmx4cm. What will be the stress in bar?
a) 100 N/mm²
b) 120 N/mm²
c) 125 N/mm²
d) 150 N/mm²
Answer: c
Clarification: Stress = load / area = 100,000/ (20×40) = 125 N/mm².

7. A tensile load of 100kN is gradually applied to a rectangular bar of dimension 2cmx4cm and length of 5m. What will be the strain energy in the bar if E=1×105 N/mm²?
a) 312.5 N-m
b) 314500 N-mm
c) 245.5 N-m
d) 634 N-m
Answer: a
Clarification: Stress = load / area = 100,000/ (20×40) = 125 N/mm²
Strain energy = σ²V/2E = 125x125x20x40x5000/ (2×100,000) = 312500 N-mm = 312.5N-m.

8. A tensile load of 60kN is gradually applied to a circular bar of 4cm diameter and 10m long. What will be the strain energy absorbed by the rod if E=1×105 N/mm²?
a) 100 N-m
b) 132 N-m
c) 148 N-m
d) 143.2 N-m
Answer: d
Clarification: Stress = 60,000 / 400π = 47.746
Strain energy = σ²V/2E = 47.746×47.746×12,566,370 / (2×100000) = 143,236.54 N-mm = 143.2N-m.

9. A uniform bar has a cross sectional area of 700mm and a length of 1.5m. if the stress at the elastic limit is 160 N/mm, what will be the value of gradually applied load which will produce the same extension as that produced by the suddenly applied load above?
a) 100kN
b) 110kN
c) 112kN
d) 120kN
Answer: c
Clarification: For gradually applied load, stress = load / area
Load = stress x area = 160 x 700 = 112000 N = 112kN.

10. A tension bar 6m long is made up of two parts, 4m of its length has cross sectional area of 12.5cm while the remaining 2m has 25cm. An axial load 5tonnes is gradually applied. What will be the total strain energy produced if E = 2 x 106 kgf/cm²?
a) 240kgf/cm
b) 242kgf/cm
c) 264kgf/cm
d) 270kgf/cm
Answer: b
Clarification: First stress = load /area, then the strain energy will be calculated as
Strain energy = σ²V/2E.

250+ TOP MCQs on Strength of Section due to Section Modulus and Answers

This set of Strength of Materials Questions and Answers for Experienced people on “Strength of Section due to Section Modulus”.

1. The moment which resists the external bending is called ______
a) Moment of shear
b) Tolerating moment
c) Moment of resistance
d) Maximum bending moment
Answer: c
Clarification: The tensile and compressive stresses developed in the beam section from a couple whose moment is equal to the external bending moment. The moment of this couple which resists the external bending is known as moment of resistance [MR].

2. ______ strength is caused by a moment of resistance offered by a section.
a) Shear
b) Flexural
c) Axial
d) Longitudinal
Answer: b
Clarification: The moment of couple with resists action of bending moment is a moment of resistance and the flexural strength possessed by section is the moment of resistance offered by it.

3. A Steel rod 200 mm diameter is to be bent into a circular arc section. Find radius of curvature. Take f = 120N/mm² & E = 2×10⁵ N/mm².
a) 134m
b) 166m
c) 162m
d) 174m
Answer: b
Clarification: Diameter of Steel rod = 200mm; y = d/2 = 100mm.
f= 120N/mm².
E= 2×10⁵N/mm².
By flexural equation we have f/y = E/R
R = 2×10⁵/ 120 ×100
= 166.6m.

4. The hoop stress is also known as ______
a) Parametrical stress
b) Surface stress
c) Circumferential stress
d) Lateral stress
Answer: c
Clarification: The stress which is developed in the walls of the cylinder due to internal fluid pressure and which acts tangential to the circumference is called hoop stress or circumferential stress.
Total pressure = p × A.

5. The ____ of strongest beam that can be cut out of a circular section of diameter D.
a) Load
b) Size
c) material
d) cross section
Answer: b
Clarification: The size of the strongest Beam that can be cut out of a circular section of diameter D is
Depth; d = Square root of 2/3
b = D / square root of 3.
Among the given sections for the same depth I section gives maximum strength.

6. The moment resisting capacity of the cross section of a beam is termed as ______ of the beam.
a) Stiffness
b) Strength
c) Modulus
d) Inertia
Answer: b
Clarification: The moment resisting capacity of the cross section of a beam is termed as the strength of the beam. The bending stress is maximum at the extreme fibres of the cross section. The strength of the two beams of same material can be compared by the sectional modulus values.

7. Find the moment of resistance of rectangular beam off grid to 40 mm depth 400 mm if the bending stress is 15 N/mm².
a) 78 kNm
b) 84 kNm
c) 96 kNm
d) 132 kNm
Answer: c
Clarification: Moment of resistance (MR) = Z × f
= bd² / 6 × 15
= 96 ×10⁶ Nmm.

8. A rectangular beam 100 mm wide is subjected to a maximum shear force and 50 kN. Find the depth of the beam.
a) 350 mm
b) 185 mm
c) 200 mm
d) 250 mm
Answer: d
Clarification: Let the depth of the beam be d
Maximum shear stress = 3/2 (Average Shear stress)
d= 3×5000/ 3×2×100.

9. What is the approximate value of coefficient of linear expansion for steel?
a) 13 × 10^-66 /°C
b) 11.5 × 10^-6 /°C
c) 12 × 10^-6 /°C
d) 16 × 10^-6 /°C
Answer: b
Clarification: The increase in length of body per unit rise of temperature in original name is termed as coefficient of linear expansion and it is denoted by Greek letter alpha. Coefficient of linear expansion for steel is 11.5 × 10^-6 /°C. For copper it is 17 × 10^-6 /°C.

10. A hollow shaft has outside diameter 120 mm and thickness 20 mm. Find the polar moment of inertia (J).
a) 16.36 × 10⁶ mm⁴
b) 14.65 × 10⁶ mm⁴
c) 10.32 × 10⁶ mm⁴
d) 23.18 × 10⁶ mm⁴
Answer: a
Clarification: D = 120mm
t= 20mm & d = D – 2t = 80mm.
Polar moment of inertia (J) is π/32 ×[ D⁴– d⁴].
π/32 × [ 120⁴– 80⁴ ].
16.36 × 10⁶ mm⁴.

Strength of Materials for Experienced people,