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

250+ TOP MCQs on Definition of Strain Energy and Answers

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

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

2. Strain energy stored in a body to uniform stress s of volume V and modulus of elasticity E is __________
a) s²V/2E
b) sV/E
c) sV²/E
d) sV/2E
Answer: a
Clarification: Strain energy = s²V/2E.

3. In a material of pure shear stress τthe strain energy stored per unit volume in the elastic, homogeneous isotropic material having elastic constants E and v will be:
a) τ²/E x (1+ v)
b) τ²/E x (1+ v)
c) τ²/2E x (1+ v)
d) τ²/E x (2+ v)
Answer: a
Clarification: σ₁=τ, σ₂= -τσ₃=0
U = (τ²+- τ²-2μτ(-τ))V = τ²/E x (1+ v)V.

4. PL³/3EI is the deflection under the load P of a cantilever beam. What will be the strain energy?
a) P²L³/3EI
b) P²L³/6EI
c) P²L³/4EI
d) P²L³/24EI
Answer: b
Clarification: We may do it taking average
Strain energy = Average force x displacement = (P/2) x PL³/3EI = P²L³/6EI.

5. A rectangular block of size 400mm x 50mm x 50mm is subjected to a shear stress of 500kg/cm². If the modulus of rigidity of the material is 1×10⁶ kg/cm², the strain energy will be __________
a) 125 kg-cm
b) 1000 kg-cm
c) 500 kg-cm
d) 100 kg-cm
Answer: a
Clarification: Strain energy stored = τ²V/2G = 500²/2×10⁶ x 40x5x5 = 125 kg-cm.

6. A material of youngs modulus and Poissons ratio of unity is subjected to two principal stresses σ₁ and σ₂ at a point in two dimensional stress system. The strain energy per unit volume of the material is __________
a) (σ₁² + σ₂² – 2σ₁σ₂ ) / 2E
b) (σ₁² + σ₂² + 2σ₁σ₂ ) / 2E
c) (σ₁² – σ₂² – 2σ₁σ₂ ) / 2E
d) (σ₁² – σ₂² – 2σ₁σ₂ ) / 2E
Answer: a
Clarification: Strain energy = (σ₁ε₁+ σ₁ε₁ ) / 2E
= (σ₁² + σ₂² – 2σ₁σ₂ ) / 2E.

7. If forces P, P and P of a system are such that the force polygon does not close, then the system will __________
a) Be in equilibrium
b) Reduce to a resultant force
c) Reduce to a couple
d) Not be in equilibrium
Answer: d
Clarification: The forces are not concurrent so the resultant force and couple both may be present. Thus the best choice is that forces are not in equilibrium.

8. The strain energy in a member is proportional to __________
a) Product of stress and the strain
b) Total strain multiplied by the volume of the member
c) The maximum strain multiplied by the length of the member
d) Product of strain and Young’s modulus of the material
Answer: d
Clarification: Strain energy per unit volume for solid = q² / 4G.

9. A bar of cross-section A and length L is subjected to an axial load W. the strain energy stored in the bar would be __________
a) WL / AE
b) W²L / 4AE
c) W²L / 2AE
d) WL / 4AE
Answer: c
Clarification: Deformation in the bar = WL / AE
Strain energy = W/2 x WL / AE = W²L / 2AE.

10. A tensile load of 60kN is gradually applied to a circular bar of 4cm diameter and 5m long. What is the stretch in the rod if E = 2×10⁵ N/mm²?
a) 1.1mm
b) 1.24mm
c) 2mm
d) 1.19mm
Answer: d
Clarification: Stress = Load/ area = 60,000 / (π/4 D²) = 470746 N/mm²
So stretch = stress x length / E = 1.19mm.

11. A tensile load of 50kN is gradually applied to a circular bar of 5cm diameter and 5m long. What is the strain energy absorbed by the rod (E = 200GPa)?
a) 14 N-m
b) 15.9 N-mm
c) 15.9 N-m
d) 14 N-mm
Answer: c
Clarification: Stress = 50,000 / 625π = 25.46
Strain energy = σ²V/2E = 25.46×25.46×9817477 / (2×200000) = 15909.5 N-mm = 15.9 N-m.

12. A tensile load of 60kN is gradually applied to a circular bar of 4cm diameter and 5m long. What is the strain energy in the rod if the load is applied suddenly (E = 2×10⁵ N/mm²)?
a) d143.23 N-m
b) 140.51 N-m
c) 135.145 N-m
d) 197.214 N-m
Answer: a
Clarification: Maximum instantaneous stress = 2P / A = 95.493
Strain energy = σ²V/2E = 143288N-mm = 143.238 N-m.

250+ TOP MCQs on Bending Equation and Answers

This set of Strength of Materials Multiple Choice Questions on “Bending Equation”.

1. In simple bending, ______ is constant.
a) Shear force
b) Loading
c) Deformation
d) Bending moment
Answer: d
Clarification: If a beam is undergone with simple bending, the beam deforms under the action of bending moment. If this bending moment is constant and does not affect by any shear force, then the beam is in state of simple bending.

2. If a beam is subjected to pure bending, then the deformation of the beam is_____
a) Arc of circle
b) Triangular
c) Trapezoidal
d) Rectangular
Answer: a
Clarification: The beam being subjected to pure bending, there will be only bending moment and no shear force it results in the formation of an arc of circle with some radius known as radius of curvature.

3. When a beam is subjected to simple bending, ____________ is the same in both tension and compression for the material.
a) Modulus of rigidity
b) Modulus of elasticity
c) Poisson’s ratio
d) Modulus of section
Answer: b
Clarification: It is one of the most important assumptions made in the theory of simple bending that is the modulus of elasticity that is Young’s modulus [E] is same in both tension and compression for the material and the stress in a beam do not exceed the elastic limit.

4. E/R = M/I = f/y is a bending equation.
a) True
b) False
Answer: a
Clarification: The above-mentioned equation is absolutely correct.
E/R = M/I = f/y is a bending equation. It is also known as flexure equation (or) equation for theory of simple bending.
Where,
E stands for Young’s modulus or modulus of elasticity.
R stands for radius of curvature.
M stands for bending moment
I stand for moment of inertia
f stands for bending stress
y stands for neutral axis.

5. Maximum Shearing stress in a beam is at _____
a) Neutral axis
b) Extreme fibres
c) Mid span
d) Action of loading
Answer: a
Clarification: Shearing stress is defined as the resistance offered by the internal stress to the shear force. Shearing stress in a beam is maximum at a neutral axis.

6. At the neutral axis, bending stress is _____
a) Minimum
b) Maximum
c) Zero
d) Constant
Answer: c
Clarification: Neutral axis is defined as a line of intersection of neutral plane or neutral layer on a cross section at the neutral axis of that section. At the NA, bending stress or bending strain is zero. The first moment of area of a beam section about neutral axis is also zero. The layer of neutral axis neither contracts nor extends.

7. Curvature of the beam is __________ to bending moment.
a) Equal
b) Directly proportion
c) Inversely proportion
d) Coincides
Answer: b
Clarification: From the flexural equation, we have 1/R is called as the “curvature of the beam”.
1 / R = M / EI
Hence the curvature of the beam is directly proportional to bending moment and inversely proportional to flexural rigidity (EI).

8. What are the units of flexural rigidity?
a) Nm²
b) Nm
c) N/m
d) m/N³
Answer: a
Clarification: The product of young’s modulus (E) of the material and moment of inertia (I) of the beam section about its neutral axis is called flexural rigidity.
Units for E are N/m²
Units for I are m⁴
Their product is Nm².

9. What are the units for section modulus?
a) m²
b) m⁴
c) m³
d) m
Answer: c
Clarification: The ratio of moment of inertia to the distance to the extreme fibre is called modulus of section or section modulus. It is generally denoted by the letter Z. Section modulus is expressed in m³
Z = I/y
= m⁴/ m
= m³.

10. What are the units of axial stiffness?
a) m³
b) m²
c) N/ m
d) -m
Answer: c
Clarification: Axial rigidity is a product of young’s modulus (E) and the cross-sectional area (A) of that section. Axial rigidity per unit length is known as axial stiffness the si units of axial stiffness are Newton per metre (N/m).

11. Calculate the modulus of section of rectangle beam of size 240 mm × 400 mm.
a) 5.4 × 10⁶ mm³
b) 6.2 × 10⁶ mm³
c) 5.5 × 10⁶ mm³
d) 6.4 × 10⁶ mm³
Answer: d
Clarification: b = 240 mm & d = 400 mm
Moment of inertia (I) = bd³/12; y = d/2
Section modulus (Z) = I/y = bd²/ 6
= 1/6 × 240 × 400 ×400
= 6.4 × 10⁶ mm³.

12. What is the product of force and radius?
a) Twisting shear
b) Turning shear
c) Turning moment
d) Tilting moment
Answer: c
Clarification: Twisting moment will be equal to the product of the perpendicular force and existing radius. Denoted by letter T and SI units are Nm.

13. Determine section modulus for beam of 100mm diameter.

a) 785 × 10³ mm³
b) 456 × 10³ mm³
c) 87 × 10³ mm³
d) 98 × 10³ mm³
Answer: d
Clarification: d = 300mm
For circular sections; I = π / 64 × d⁴
y= d/2
Z = π/32 × d³ (d = 100 mm)
Z = 98.17 × 10³mm³.

250+ TOP MCQs on Rectangular Dam Analysis and Answers

This set of Strength of Materials Multiple Choice Questions on “Rectangular Dam Analysis”.

1. Calculate the self-weight of the masonry of the rectangle dam of 10 m height and 4 m wide. Consider specific weight of masonry as 20kN/m³.
a) 600 kN
b) 500 kN
c) 800 kN
d) 1000 kN
Answer: c
Clarification: Self weight of masonry = W = Area of cross-section × 1× Specific weight of masonry.
= (10×4)×1×20 = 800 kN.

2. Water-cement ratio varies normally from ______________ to __________
a) 0.42 – 0.45
b) 0.45 – 0.48
c) 0.42 – 0.48
d) 0.45 – 0.5
Answer: c
Clarification: The ratio to which the required amount of water is added to weight of cement to obtain desired consistency and workability of concrete mix is known as water cement ratio. It varies from 0.42 to 0.48.

3. Calculate the resultant force of dam with given self weight 800kN and water pressure be 500kN.
a) 943.4 kN
b) 956.7 kN
c) 948.6 kN
d) 939.1 KN
Answer: a
Clarification: Resultant force (R) = (P² + W²)^1/2 = 500² + 800².
= (500² + 800²)^1/2
= 943.39 ~ 943.4 kN.

4. When the reservoir is empty tension occurs at ___________
a) Toe
b) Heel
c) Top width
d) Bottom width
Answer: a
Clarification: For no tension to develop in the damn section in any condition, the eccentricity should be less than b/6. When the reservoir is empty, tension occurs at toe and compression occurs at heel.

5. What is the mix proportion for M15 grade concrete?
a) 1:1:2
b) 1:2:4
c) 1:3:6
d) 1:4:8
Answer: b
Clarification: Mix proportion for M15 grade concrete is 1:2:4.

Grade of concrete	Mix proportionate
M10	1:3:6
M15	1:2:4
M20	1:1.5:3

6. Laterite is an example of ___________ rock.
a) Siliceous
b) Argillaceous
c) Calcareous
d) Metamorphic
Answer: b
Clarification: The Rocks having (Gneiss) aluminium or clay as the main component, such rocks are known as argillaceous rocks. Example: Slate, Laterite.

7. ________ is crystalline and compact in structure.
a) Marbles
b) Granite
c) Kadapa slabs
d) Shahabad stones
Answer: a
Clarification: Marble is a metamorphic rock and is made from limestone, this is a very costly stone. It is less durable. It is crystalline and compact in structure. So it can take a fine polish. It is not very hard.

8. As per IS, the standard dimensions for a brick is _______ (in cm).
a) 19×8×8
b) 19×9×8
c) 19×9×9
d) 19×8×9
Answer: c
Clarification: The shape of a brick should be uniform with rectangular surface and its size should be standard with 19×9×9 cm. They should have a uniform red colour and it should be well burnt.

9. Formation of white patches on the surface on the bricks is ____________
a) Tempering
b) Porosity
c) Shrinkage
d) Efflorescence
Answer: d
Clarification: A good brick should not contain excess alkaline soils when the bricks are exposed to the atmosphere. It should not absorb moisture. If it attracts moisture then dampness occurs and results in the formation of white patches. Hence the brick begins decaying.

10. Kiln burning involves 90% of first class bricks.
a) True
b) False
Answer: a
Clarification: Kiln burning is a permanent structure. There is complete control on fire. It produces a large scale of manufactured bricks, it takes only 24 hours in burning the bricks and 12 days for cooling. It produces 90% of burnt bricks.

11. ______ bricks can withstand up to a temperature of 1800°C.
a) Refractory
b) Fly ash
c) Clay
d) Cement
Answer: a
Clarification: The bricks made from refractory clay are called refractory bricks. The weight of these bricks is 2 kN/m³. The standard size is 230× 65× 113 mm. These bricks can withstand up to the temperature of 1800°C.

12. According to IS, the minimum expansion joint in construction should be ________
a) 18mm to 30 mm
b) 15mm to 24mm
c) 18mm to 25mm
d) 22mm to 30mm
Answer: c
Clarification: According to Indian standards 456- 2000, it is desirable to provide 18 mm to 25 mm thick expansion joints after every 30 to 45 m construction of length.

13. The edges formed by the intersection of plane surfaces of a brick are known as _________
a) Arises
b) Stretcher
c) Header
d) Frog
Answer: a
Clarification: In any bond, the edges formed by the intersection of plane surfaces of a brick are termed as arises. These are straight and sharp in case of good bricks or brick tiles.

14. The depression made in the face of brick during its manufacture is _________
a) Brick tile
b) Bat
c) Frog
d) Quoin closer
Answer: c
Clarification: The depression provided on any face of the brick during its manufacture can be termed as frog. A hand moulded bricks has one frog. A pressed brick has two frogs.

15. Calculate the eccentricity of a rectangular dam of width 4 m. Take the distance between the water face and point where resultant cuts the base as 5.25 m.
a) 2.25m
b) 3.25m
c) 4.35m
d) 5.35m
Answer: b
Clarification: Eccentricity = e = Z – b/ 2
Where Z = 5.25 m & b = 4m ; Now, e = Z- b/2
e = 5.25 – 4/2
= 3.25m.

250+ TOP MCQs on Torsion Equation and Answers

Strength of Materials Multiple Choice Questions on “Torsion Equation”.

1. Torsional sectional modulus is also known as _________
a) Polar modulus
b) Sectional modulus
c) Torsion modulus
d) Torsional rigidity

Answer: a
Clarification: The ratio of polar moment of inertia to radius of section is called Polar modulus or Torsional section modulus. Its units are mm³ or m³ (in SI).

2. ________ is a measure of the strength of shaft in rotation.
a) Torsional modulus
b) Sectional modulus
c) Polar modulus
d) Torsional rigidity

Answer: c
Clarification: The polar modulus is a measure of the strength of shaft in rotation. As the value of Polar modulus increases torsional strength increases.

3. What are the units of torsional rigidity?
a) Nmm²
b) N/mm
c) N-mm
d) N

Answer: a
Clarification: The product of modulus of rigidity (C) and polar moment of inertia (J) is called torsional rigidity. Torsional rigidity is a torque that produces a twist of one radian in a shaft of unit length.

4. The angle of twist can be written as ________
a) TL/J
b) CJ/TL
c) TL/CJ
d) T/J

Answer: c
Clarification: The angle of Twist = TL/CJ
Where T = Torque in Nm
L = Length of shaft
CJ = Torsional rigidity.

5. The power transmitted by shaft SI system is given by __________
a) 2πNT/60
b) 3πNT/60
c) 2πNT/45
d) NT/60 W

Answer: a
Clarification: In SI system, Power (P) is measured in watts (W) ; P = 2πNT/60
Where T = Average Torque in N.m
N = rpm
= 2πNT/ 45 1 watt = 1 Joule/sec = 1N.m/s.

6. Area of catchment is measured in ___________
a) mm³
b) Km²
c) Km
d) mm

Answer: b
Clarification: Catchment area can be defined as the area which contributes the surplus water present over it to the stream or river. It is an area which is responsible for maintaining flow in natural water bodies. It is expressed in square kilometres.

7. ______ catchment area is a sum of free catchment area and intercepted catchment area.
a) Total
b) Additional
c) Combined
d) Overall

Answer: c
Clarification: Combined catchment area is defined as the total catchment area which contributes the water in to stream or a tank. Combined Catchment area = Free catchment area + intercepted catchment area.

8. ___________ has steep slopes and gives more run off.
a) Intercepted Catchment Area
b) Good Catchment Area
c) Combined Catchment Area
d) Average Catchment Area

Answer: b
Clarification: Good catchment area consists of hills or rocky lands with steep slopes and little vegetation. It gives more run off.

9. How many number of rain gauge stations should be installed an area between 250 to 500 km².
a) 2
b) 4
c) 3
d) 5

Answer: c
Clarification: 3 number of rain gauge stations should be installed an area between 250 to 500 km².

Area of Basin(Km²)	Number of Rain gauge stations
	1
125 – 250	2
250 – 500	3

10. Trend of rainfall can be studied from _______
a) Rainfall graphs
b) Rainfall records
c) Rainfall curves
d) Rainfall cumulatives

Answer: b
Clarification: Rainfall records are useful for calculating run off over a basin. By using rainfall records estimate of design parameters of irrigation structures can be made. The maximum flow due to any storm can be calculated and predicted.

11. Estimation of run off “R” is 0.85P-30.48.
The above formula was coined by _____
a) Lacey
b) Darcy
c) Khosla
d) Ingli

Answer: d
Clarification: Run off can be estimated by
R= 0.85P-30.48
Where R = annual runoff in mm
P = annual rainfall in mm.

12. Monsoon duration factor is denoted by ________
a) P
b) S
c) F
d) T

Answer: c
Clarification: Monsoon duration factor is denoted by F.

Class of Monsoon	Monsoon Duration Factor (F)
Very Short	0.5
Standard length	1.0
Very long	1.5

13. Runoff coefficient is denoted by _______
a) P
b) N
c) K
d) H

Answer: c
Clarification: The runoff coefficient can be defined as the ratio of runoff to rainfall. Rainfall and runoff can be interrelated by runoff coefficient.
R = KP
K = R/P [K = is a runoff Coefficient depending on the surface of the catchment area].

14. _________ is a graph showing variations of discharge with time.
a) Rising limb graph
b) Crest graph
c) Hydraulic graph
d) Gauge graph

Answer: c
Clarification: Hydrograph is a graph showing variations of discharge with time at a particular point of the stream. The hydrograph shows the time distribution of total run off at a point of measurement. Maximum flood discharge can also be calculated by using hydrograph.

15. Calculate the torque which a shaft of 300 mm diameter can safely transmit, if the shear stress is 48 N / mm².
a) 356 kNm
b) 254 kNm
c) 332 kNm
d) 564 kNm

Answer: b
Clarification: Given, the diameter of shaft D = 300 mm
Maximum shear stress fs = 48 N/mm².
Torque = T = π/16 fs D³
= 254469004.9 Nmm
= 254 kNm.

250+ TOP MCQs on Properties of Strain and Answers

Strength of Materials Multiple Choice Questions on “Properties of Strain”.

1. The property of a material by which it can be drawn into thin wires is?
a) Malleability
b) Plasticity
c) Ductility
d) Elasticity
Answer: c
Clarification: The ductile material can be drawn into wires because it can resist large deformation. Malleability is the property by which it can be made into thin sheets.

2. If the material has identical elastic properties in all directions, it is called ____________
a) Elastic
b) Isotropic
c) Plastic
d) Homogeneous
Answer: b
Clarification: An homogeneous material is that with uniform composition. An elastic and plastic are different on the criteria.

3. Why is the strain the fundamental property but not the stress?
a) Because it is dimensionless
b) Because it is a ratio
c) Because it’s value is calculated in the laboratory
d) No stress is the fundamental property
Answer: c
Clarification: The stress is the fundamental property because it is calculated in the laboratory. It is a non dependable value.

4. The material in which large deformation is possible before absolute failure by rupture is called ____________
a) Plastic
b) Elastic
c) Brittle
d) Ductile
Answer: d
Clarification: The ductile material can be drawn into wires because it can resist large deformation before it fails.

5. What is a creep?
a) Gradual increase of plastic strain with time at constant load
b) Gradual increase of elastic strain with time at constant load
c) Gradual increase of plastic strain with time at varying load
d) Gradual increase of elastic strain with time at varying load
Answer: a
Clarification: Creep is the property by virtue of which a metal specimen undergoes additional deformation with the passage of time under sustained loading within elastic limit. It is permanent in nature and cannot be recovered after removal of load, hence is plastic in nature.

6. If the material has different elastic properties in perpendicular directions, it is called ____________
a) Elastic
b) Isotropic
c) Orthotropic
d) Plastic
Answer: c
Clarification: Isotropic material has the same elastic properties but ortho tropic material has the same.

7. Which one of the following pairs is NOT correctly matched?
a) Visco-elastic – small plastic zone
b) Orthotropic material – different properties in three perpendicular directions
c) Strain hardening material – stiffening effect felt at some stage
d) Isotropic material – same physical property in all directions at a point
Answer: a
Clarification: Visco-elastic material exhibit a mixture of creep and elastic after effects at room temperature. Thus their behaviour is time dependent. Materials with different properties in different directions are called anisotropic. Orthotropic material is a special case of an anisotropic material in three mutually perpendicular directions. However, these are symmetric about any axis.

8. The phenomenon of slow extension of materials having a constant load, I.e. increasing with the time is called
a) Creeping
b) Yielding
c) Breaking
d) None of the mentioned
Answer: a
Clarification: The creeping is the phenomenon of deformation in materials which have been under load for several time. When the load is put on the material, initially it deforms but when the load is not removed, it causes a small amount of deformation which increases with time.

250+ TOP MCQs on Resilience and Answers

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

1. The ability of a material to absorb energy when elastically deformed and to return it when unloaded is called __________
a) Elasticity
b) Resilience
c) Plasticity
d) Strain resistance
Answer: b
Clarification: Resilience is the ability of a material to absorb energy when elastically deformed and to return it. Elasticity is the property by which any body regain its original shape.

2. The strain energy stored in a specimen when stained within the elastic limit is known as __________
a) Resilience
b) Plasticity
c) Malleability
d) Stain energy
Answer: a
Clarification: Resilience is the ability of a material to absorb energy when elastically deformed and to return it. Elasticity is the property by which any body regain its original shape. Malleability is the property by which any material can be beaten into thin sheets.

3. The maximum strain energy stored at elastic limit is __________
a) Resilience
b) Proof resilience
c) Elasticity
d) Malleability
Answer: b
Clarification: Proof resilience is the maximum stored energy at the elastic limit. Resilience is the ability of material to absorb energy when elastically deformed and to return it. Elasticity is the property by which any body regain its original shape. Malleability is the property by which any material can be beaten into thin sheets.

4. The mathematical expression for resilience ‘U’ is __________
a) U = σ²/E x volume
b) U = σ²/3E x volume
c) U = σ²/2E x volume
d) U = σ/2E x volume
Answer: c
Clarification: The resilience is the strain energy stored in a specimen so it will be
U = σ²/2E x volume.

5. What is the modulus of resilience?
a) The ratio of resilience to volume
b) The ratio of proof resilience to the modulus of elasticity
c) The ratio of proof resilience to the strain energy
d) The ratio of proof resilience to volume
Answer: d
Clarification: The modulus of resilience is the proof resilience per unit volume. It is denoted by σ.

6. The property by which an amount of energy is absorbed by material without plastic deformation is called __________
a) Toughness
b) Impact strength
c) Ductility
d) Resilience
Answer: d
Clarification: Resilience is the ability of a material to absorb energy when elastically deformed and to return it when unloaded.

7. Resilience of a material plays important role in which of the following?
a) Thermal stress
b) Shock loading
c) Fatigue
d) Pure static loading
Answer: b
Clarification: The total strain energy stored in a body is commonly known as resilience. Whenever the straining force is removed from the strained body, the body is capable of doing work. Hence the resilience is also define as the capacity of a strained body for doing work on the removal of the straining force.

8. A steel has its yield strength of 200N/mm² and modulus of elasticity of 1x105MPa. Assuming the material to obey hookes law up to yielding, what will be its proof resilience?
a) 0.8 N/mm²
b) 0.4 N/mm²
c) 0.2 N/mm²
d) 0.6 N/mm²
Answer: c
Clarification: Proof resilience = σ²/2E = (200)² / (2 x 10⁵) = 0.2 N/mm².

9. A 1m long bar of uniform section extends 1mm under limiting axial stress of 200N/mm². What is the modulus of resilience for the bar?
a) 0.1 units
b) 1 units
c) 10units
d) 100units
Answer: a
Clarification: Modulus of resilience, u = f²/2E, where E = fL/δL
Therefore, u = 200×1 / 2×1000 = 0.1units.

10. A square steel bar of 10mm side and 5m length is subjected to a load whereupon it absorbs a strain energy of 100J. what is its modulus of resilience?
a) 1/5 N-mm/mm³
b) 25 N-mm/mm³
c) 1/25 N-mm/mm³
d) 5 N-mm/mm³
Answer: a
Clarification: Modulus of resilience is the strain energy stored in the material per unit volume.
u = U/v
= ( 100 x 1000 ) / ( 10 x 10x 5x 1000)
= 1/5 N-mm/mm³.