Category: Strength of Materials Objective Questions

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.

250+ TOP MCQs on Pure Bending Stress and Answers

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

1. In simply supported beams, the _____ stress distribution is not uniform.
a) Bending
b) Shearing
c) Tensile
d) Compressive

Answer: a
Clarification: In a simply supported beam, there is compressive stress above the neutral axis and tensile stress below it. It bends with concavity upwards. Hence the bending stress distribution is not uniform over the section.

2. The maximum _________ stresses occur at top most fibre of a simply supported beam.
a) Tensile
b) Compressive
c) Shear
d) Bending

Answer: b
Clarification: As bending stress distribution is not uniform over the section in simply supported beams, the maximum compressive stress lies above the neutral axis. Obviously, top most fibre of beam. The maximum tensile stress occurs at bottom most fibre.

3. The stress is directly proportional to _______
a) E
b) u
c) y
d) R

Answer: c
Clarification: By two equations; we have e = y/R & e = f/E
Equating both equations; we get e = f/E = y/R
Hence stress (f) is directly proportional to the distance from neutral axis(y).

4. At the extreme fibre, bending stress is______
a) Minimum
b) Zero
c) Constant
d) Maximum

Answer: d
Clarification: Bending stress is defined as the resistance offered by internal stress to bending. In beams, stresses occurs above or below the neutral axis i.e at the extreme fibres. Hence bending stress is maximum at the extreme fibres.

5. The curvature of a beam is equal to _____
a) EI/M
b) M/E
c) M/EI
d) E/MI

Answer: c
Clarification: From the bending equation, E/R = M/I = f/y.
Where R is called “radius of curvature “
1/R is called “curvature of the beam “.
So, 1/R = M/EI.
So curvature of the beam is directly proportional to bending moment.

6. Skin stress is also called as ______
a) Shear stress
b) Bending stress
c) Lateral stress
d) Temperature stress

Answer: b
Clarification: The bending moment leads to deform or deflect the beam and internal stress resists bending. The resistance offered by internal stress to bending is called bending stress or “fibre stress” or “skin stress” or “longitudinal stress”.

7. _________ is the total Strain energy stored in a body.
a) modulus of resilience
b) impact energy
c) resilience
d) proof resilience

Answer: c
Clarification: When a load acts on a body, there is deformation of the body which causes movement of the applied load. Thus work is done is stored in the body as energy and the load is removed this stored energy which is by virtue of strain is called resilience.

8. In cantilever beams, there is _______ stress above neutral axis.
a) Compressive
b) Tensile
c) Temperature
d) Shear

Answer: b
Clarification: In a cantilever beam maximum compressive stress occurs at bottom most fibre and maximum tensile stress occurs at the top most fibre and zero at neutral axis hence the tensile stresses lies above the neutral axis.

9. The product of modulus of elasticity (E) and polar moment of inertia (J) is called torsional rigidity.
a) True
b) False

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

10. The strength of beams depend merely on________
a) Modulus section
b) Moment of inertia
c) Flexural rigidity
d) Moment of resistance

Answer: a
Clarification: The ratio of moment of inertia to the distance to the extreme fibre is called modulus of section. The Beam is stronger when section modulus is more. The strength of beam depends on section modulus. The beams of same strength mean section modulus is same for the beams.

12. The steel plate is bent into a circular path of radius 10 metres. If the plate section be 120 mm wide and 20 mm thick, then calculate the maximum bending stress. [Consider Young’s modulus = 200000 N/mm2].
a) 350 N/mm2
b) 400 N/mm2
c) 200 N/mm2
d) 500 N/mm2

Answer: c
Clarification: R = 10000 mm; y = 20/2 = 10 mm; E = 200000 N/mm2
By bending equation we have E/R = f/y
f = 200000×10 / 10000
= 200 N/mm2.

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

This set of Tricky Strength of Materials Questions and Answers on “Trapezoidal Dam as Vertical Side Phase”.

1. Calculate the horizontal water pressure acting on a dam. The total depth of water be 13m. Take specific weight of water be 10kN/m3.
a) 765 kN
b) 845 kN
c) 965 kN
d) 1175 kN
Answer: b
Clarification: Horizontal water pressure(P) = w×h2/ 2.
Where w = 10 kN/m3. h= 13m.
P = 10×132/ 2 = 845kN.

2. Calculate the self-weight of a rectangular dam of 22m high and 8m wide. It contains water upto a height of 20m. Consider the specific weight of masonry be 25 kN/m3.
a) 3560 kN
b) 5432 kN
c) 4400 kN
d) 5680 kN
Answer: c
Clarification: Consider 1m length of the dam. The total depth of dam is 22 metres.
Self weight of masonry (W)= (22×8)×1×25
= 4400kN.

3. The pressure intensity of water at free surface is ________
a) Zero
b) Maximum
c) Minimum
d) Uniform
Answer: a
Clarification: The pressure intensity of water at a free surface is always zero and increase linearly to a maximum at the base and is equal to “ wh”.

4. Self weight of dam acts in ___________ direction.
a) Vertical
b) Horizontal
c) Inclined
d) Parallel
Answer: a
Clarification: Self weight of the Dam at vertically downwards passing through centre of gravity of the damn section and Total horizontal water pressure acts horizontally at heel of the dam.

5. The maximum compressive stresses developed at the base of the dam should not exceed permissible ___________ stresses for masonry.
a) Tensile
b) Crippling
c) Compressive
d) Shear
Answer: c
Clarification: To avoid the failure of crushing, the maximum compressive stress developed at the base of the dam should not exceed the permissible compressive stress for masonry with which the dam is constructed.

6. For no _______ to develop in the dam section the resultant should always lie within the middle third.
a) Compression
b) Tension
c) Shear
d) Buckling
Answer: b
Clarification: For no tension to develop in the dam section at any condition, the eccentricity developed with the resultant should be always less than b/6. The resultant must always lie within the middle third.

7. Calculate the self weight of trapezoidal dam with top width 5m and bottom width 8m. The height of dam is 15 m. Consider specific weight of masonry be 25kN/m3.
a) 3456.5 kN
b) 2768.5 kN
c) 2437.5 kN
d) 3450 kN
Answer: c
Clarification: Consider 1m of length, area of trapezoidal dam be (a+b)/2 ×H
The self weight of trapezoidal dam (W) = (a+b)/2 × H × 1 × 25
W = 2437.5 kN.

8. The material(earth) retained by the retaining wall is called as __________
a) Surcharge
b) Turf
c) Foliate
d) Back fill
Answer: d
Clarification: The material retained by the retaining wall is called back fill. The top surface of the back fills maybe him the horizontal or inclined.

9. The inclination of surcharge to the horizontal is called ____________
a) Surcharge elevation
b) Surcharge angle
c) Surcharge factor
d) Surcharge depression
Answer: b
Clarification: The back fill lying above a horizontal plane at an elevation of the top of wall is known as a surcharge and its inclination to the horizontal is called surcharge angle.

10. Which of the following is practical pressure?
a) Active earth pressure
b) Passive earth pressure
c) Soil moisture tension
d) Horizontal water pressure
Answer: a
Clarification: The pressure exerted by back fill on retaining wall is called an active earth pressure. This is the minimum earth pressure exerted by the soil. This is also known as practical pressure.

11. The angle of internal friction for water is __________
a) 180°
b) 100°
c) 0°
d) 270°
Answer: c
Clarification: Angle of internal friction is defined as the maximum slope at which the particles of soil will come in rest due to their internal friction. It is also called an angle of repose for water it is 0°.

12. Which of the following is theoretical pressure?
a) Active earth pressure
b) Passive earth pressure
c) Soil Tension
d) Horizontal water pressure
Answer: b
Clarification: The pressure exerted by the retaining wall on the retained earth is called passive earth pressure. This is a maximum earth pressure due to maximum shear stress on the retaining wall. This is also known as theoretical pressure.

13. Which of the following is an example for plasticizer?
a) Ca
b) Mg
c) Zn
d) Hg
Answer: a
Clarification: The combination of both inorganic and organic materials which will help to reduce the water content for getting higher workability are known as plasticizers. Examples are calcium, sodium, salts of hydrocarbons etc.

14. _______ curing is adopted for columns and walls.
a) Moist curing
b) Membrane curing
c) Ponding
d) Descending stage
Answer: a
Clarification: In this curing, the exposed surface of the concrete is kept in a damp and moist condition for a long time, the vertical members like columns and walls can be adopted for this type of curing.

15. Prestressed concrete is an example of _____________
a) Malleability
b) Ductility
c) Fatigue
d) Plasticity
Answer: c
Clarification: Fatigue is the property of a material by which the material with stands to repeating, reversing or varying upcoming loads. The best example of fatigue is concrete and prestressed concrete.

To practice tricky questions and answers on all areas of Strength of Materials,

250+ TOP MCQs on Shear Stress and Twisting Moment and Answers

This set of Strength of Materials Multiple Choice Questions on “Shear Stress and Twisting Moment”.

1. The intensity of shear stress at a section is ______ to the distance of the section from the axis of the shaft.
a) Inversely proportional
b) Directly proportional
c) Equal
d) Parallel
Answer: b
Clarification: The intensity of shear stress at a section is directly proportional to the distance of the section from axis of the shaft. The shear stress at a distance from the centre of the shaft is given by fs/R × r.

2. The shear stress is ____________ at the axis of the shaft.
a) Minimum
b) Maximum
c) Zero
d) Uniform
Answer: c
Clarification: The shear stress is zero at the axis of the shaft and the shear stress is linearly increasing to the maximum value at the surface of the shaft.

3. The shear stress at the outer surface of hollow circular section is _________
a) Zero
b) Maximum
c) Minimum
d) Can’t determined
Answer: b
Clarification: The shear stress in a hollow circular section varies from maximum at the outer surface to a minimum (but not zero) in the inner face. The minimum value should be greater than zero.

4. The hollow shaft will transmit greater _______ then the solid shaft of the same weight.
a) Bending moment
b) Shear stress
c) Torque
d) Sectional Modulus
Answer: c
Clarification: For the same maximum shear stress, the average shear stress in a hollow shaft is greater than that in a solid shaft of the same area. Hence the hollow shaft will transmit greater torque than the solid shaft of the same weight.

5. The process of measurement of discharge and water level of a river is called _________
a) Meandering
b) River coursing
c) River gauging
d) Scouring
Answer: c
Clarification: The process of measurement of discharge and water level of a river is known as river gauge. It helps in determining the characteristics of flow different times during the year.

6. The quantity of losses in the river can be measured with an aid of ________
a) Runoff coefficient
b) Hydrograph
c) River Coursing
d) River gauging
Answer: d
Clarification: By measuring river discharge for number of years, it is possible to know the available and dependable supply. The river gauging helps in measuring discharge in the river and the quantity of losses can also be known.

7. The site for the river gauging station should not be liable to ____________
a) Silting
b) Coursing
c) Meandering
d) Runoff
Answer: a
Clarification: River gauging station site should be selected in such a way that the site should be stable and there should not be any choice of scouring and silting. At the gauge site, the river section should be at right angles to the flow of the river.

8. Stage discharge relationship method is also known as ________ method.
a) Velocity Volume
b) Velocity Area
c) Distance Area
d) Displacement Momentum
Answer: b
Clarification: Stage discharge relationship method is a direct method of computing a discharge in a stream by measuring velocity and area of flow. The place where such measurements are taken is known as velocity area station and the method is known as the velocity area method.

9. Velocity in a river flow can be calculated by using _________
a) By current meter
b) By emperical formulae
c) By infiltration method
d) By hydrograph
Answer: a
Clarification: The velocity flow at any point in an open channel or in a river can be most accurately and conveniently determined by a mechanical device called current metre in this device the velocity of flow can be read from rating table.

10. Which of the following method is not used in measuring the velocity of a stream?
a) By floats
b) By rod float
c) By hydrograph
d) By colour
Answer: c
Clarification: Hydrograph is a method of estimation of runoff. While the rest of the methods used in measuring the velocity of a stream/ river or canal. Hydrograph is a graph which shows the variations of discharge with respect to time.

11. The maximum flood discharge is also known as ___________
a) Peak flow
b) Maximum flow
c) Peak discharge
d) Peak flood
Answer: a
Clarification: The maximum rate of discharge during a period of runoff, which is caused by a storm, is called a peak flow maximum flood discharge. Estimation of maximum flood discharge is a first step in planning for flood regulation.

12. Which of the following method is used to estimate maximum flood discharge?
a) By travelling screen
b) By current meter
c) By physical indication of past floods
d) By salt velocity
Answer: c
Clarification: The results obtained by the physical indication of past floods methods are somewhat reliable. By oral enquiry in the villages situated on the banks of the river, the maximum water level attained in the past 35 years can be obtained. But this method is out-dated.

13. ________formula is used only in southern India for calculating maximum flood discharge.
a) Dickens
b) Ryve’s
c) Lacey’s
d) Francis
Answer: b
Clarification: Ryve’s (1884) formula is used only in Southern India.
Q = C(A)2/3.
The coefficient “C” depends on the maximum intensity of rainfall and other factors such as shape slopes exedra of the catchment.

14. A catchment area of 30.5 km2 is situated in Central India calculate the maximum discharge coming from the catchment area.
a) 253.08 cumecs
b) 341.06 cumecs
c) 457.88 cumecs
d) 485.66 cumecs
Answer: a
Clarification: As the catchment area is situated in central India. Dicken’s formula is suitable and a maximum value of Dickens Coefficient is taken as 19.5
Q = CA3/4
Q = 19.5 × (30.5)3/4
Q = 253.08 cumecs.

15. If the catchment area is situated in north India, then what is the flood coefficient?
a) 10.45
b) 11.37
c) 12.6
d) 19.4
Answer: b
Clarification: Dicken’s formula (1865)

Region Value of C
North India 11.37
Central India 11.77 – 19.28
Western India 22.04

250+ TOP MCQs on Strain Constants – 1 and Answers

Strength of Materials Multiple Choice Questions on “Strain Constants – 1”.

1. What will be the elastic modulus of a material if the Poisson’s ratio for that material is 0.5?
a) Equal to its shear modulus
b) Three times its shear modulus
c) Four times its shear modulus
d) Not determinable
Answer: b
Clarification:
Clarification: Elastic modulus = E
Shear modulus = G
E = 2G ( 1 + μ )
Given, μ= 0.5, E = 2×1.5xG
E = 3G.

2. A rigid beam ABCD is hinged at D and supported by two springs at A and B as shown in the given figure. The beam carries a vertical load P and C. the stiffness of spring at A is 2K and that of B is K.
strength-materials-questions-answers-strain-constants-1-q2
What will be the ratio of forces of spring at A and that of spring at B?
a) 4
b) 3
c) 2
d) 1
Answer: b
Clarification: The rigid beam will rotate about point D, due to the load at C.
strength-materials-questions-answers-strain-constants-1-q2-1
From similar triangle,
δa/2a = δb/3b
Force in spring A/Force in spring B = Pa/Pb
= 2k/k x 3/2 = 3.

3. A solid metal bat of uniform diameter D and length L is hung vertically from a ceiling. If the density of the material of the bar is 1 and the modulus of elasticity is E, then the total elongation of the bar due to its own weight will be ____________
a) L/2E
b) L2/2E
c) E/2L
d) E/2L2
Answer: b
Clarification: The elongation of bar due to its own weight is δ= WL/2AE
Now W = ρAL
There fore δ= L2 / 2E.

4. A bar of diameter 30mm is subjected to a tensile load such that the measured extension on a gauge length of 200mm is 0.09mm and the change in diameter is 0.0045mm. Calculate the Poissons ratio?
a) 1/3
b) 1/4
c) 1/5
d) 1/6
Answer: a
Clarification: Longitudinal strain = 0.09/200
Lateral strain = – 0.0045/30
Poissons ratio = – lateral strain/ longitudinal strain
= 0.0045/30 x 200/0.09
= 1/3.

5. What will be the ratio of Youngs modulus to the modulus of rigidity of a material having Poissons ratio 0.25?
a) 3.75
b) 3.00
c) 1.5
d) 2.5
Answer: d
Clarification: Modulus of rigidity, G = E/2(1 + μ)
Therefore, E/G = 2x(1+0.25) = 2.5.

6. An experiment was done and it was found that the bulk modulus of a material is equal to its shear modulus. Then what will be its Poissons ratio?
a) 0.125
b) 0.150
c) 0.200
d) 0.375
Answer: a
Clarification: We know that, μ = (3K – 2G) / (6K + 2G)
Here K = G
Therefore, μ = 3-2 / 6+2 = 0.125.

7. A bar of 40mm dia and 40cm length is subjected to an axial load of 100 kN. It elongates by 0.005mm. Calculate the Poissons ratio of the material of bar?
a) 0.25
b) 0.28
c) 0.30
d) 0.33
Answer: d
Clarification: Longitudinal strain = 0.150/400 = 0.000375
Lateral strain = – 0.005/40 = -0.000125
Poissons ratio = – lateral strain/longitudinal strain
= 0.33.

8. What will be the approximate value of shear modulus of a material if the modulus of elasticity is 189.8 GN/m2 and its Poissons ratio is 0.30?
a) 73 GN/m2
b) 80 GN/m2
c) 93.3 GN/m2
d) 103.9 GN/m2
Answer: a
Clarification: The relationship between E, G, and μ is given by
is given by
E = 2G (1 + μ)
G = 189.8 / 2(1 + 0.30)
G = 73 GN/m2.

9. What will be the modulus of rigidity if the value of modulus of elasticity is 200 and Poissons ratio is 0.25?
a) 70
b) 80
c) 125
d) 250
Answer: b
Clarification: The relationship between E, G and μ is E = 2G (1 + μ)
G = 200 / 2(1 + 0.25)
G = 80.

250+ TOP MCQs on Sudden Loading and Answers

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

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

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

3. A tensile load of 60kN is suddenly applied to a circular bar of 4cm diameter. What will be the maximum instantaneous stress induced?
a) 95.493 N/mm2
b) 45.25 N/mm2
c) 85.64 N/mm2
d) 102.45 N/mm2
Answer: a
Clarification: Maximum instantaneous stress induced = 2P/A = 2×60000/400π = 95.49 N/mm2.

4. A tensile load of 60kN is suddenly applied to a circular bar of 4cm and 5m length. What will be the strain energy absorbed by the rod if E=2×105 N/mm2?
a) 140.5 N-m
b) 100 N-m
c) 197.45 N-m
d) 143.2 N-m
Answer: d
Clarification: Maximum instantaneous stress induced = 2P/A = 2×60000/400π = 95.49 N/mm2
Strain energy = σ2V/2E = 95.492 x 2×106π / (2x2x105) = 143238 N-mm = 143.23 N-m.

5. A tensile load of 100kN is suddenly applied to a rectangular bar of dimension 2cmx4cm. What will be the instantaneous stress in bar?
a) 100 N/mm2
b) 120 N/mm2
c) 150 N/mm2
d) 250 N/mm2
Answer: d
Clarification: Stress = 2x load / area = 2×100,000/ (20×40) = 250 N/mm2.

6. 2 tensile load of 100kN is suddenly applied to a rectangular bar of dimension 2cmx4cm and length of 5m. What will be the strain energy absorbed in the bar if E=1×105 N/mm2?
a) 312.5 N-m
b) 314500 N-mm
c) 1250 N-m
d) 634 N-m
Answer: c
Clarification: Stress = 2xload / area = 2×100,000/ (20×40) = 250 N/mm2
Strain energy = σ2V/2E = 250x250x20x40x5000/ (2×100,000) = 1250000 N-mm = 1250 N-m.

7. A steel rod is 2m long and 50mm in diameter. A axial pull of 100kN is suddenly applied to the rod. What will be the instantaneous stress induced in the rod?
a) 101.89 N/mm2
b) 94.25 N/mm2
c) 130.45 N/mm2
d) 178.63 N/mm2
Answer: a
Clarification: Area = π/4 d2 = 625π
Load = 100kN = 100×1000 N
Stress = 2 x load / area = 2x100x1000 / (625π) = 101.86 N/mm2.

8. A steel rod is 2m long and 50mm in diameter. An axial pull of 100kN is suddenly applied to the rod. What will be the instantaneous elongation produced in the rod if E=22GN/m2?
a) 0.0097 mm
b) 1.0754 mm
c) 1.6354 mm
d) 1.0186 mm
Answer: d
Clarification: Area = π/4 d2 = 625π
Load = 100kN = 100×1000 N
E=22GN/m2 = 200 x 109 / 106 = 200,000 N/mm2
Stress = 2 x load / area = 2x100x1000 / (625 π )
Elongation = stress x length / E = 101.86×2000 / 200000 = 1.0186 mm.

9. What will be the amount of axial pull be applied on a a 4cm diameter bar to get an instantaneous stress value of 143 N/mm2?
a) 50kN
b) 60kN
c) 70kN
d) 80kN
Answer: b
Clarification: Instantaneous stress = 2 x load / area
Load = instantaneous stress x area / 2
= 143 x 400×3.14 / 2 = 60kN.

10. What will be the instantaneous stress produced in a bar 10cm2 in area ans 4m long by the sudden application of tensile load of unknown magnitude, if the extension of the bar due to suddenly applied load is 1.35mm if E = 2×105 N/mm2?
a) 67.5 N/mm2
b) 47 N/mm2
c) 55.4 N/mm2
d) 78.5 N/mm2
Answer: a
Clarification: The value of stress = load / area where area is 10cm2 and load can be calculated by stress strain equation.