Category: Strength of Materials Objective Questions

250+ TOP MCQs on Elastic Constants Relationship – 1 and Answers

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

1. How many elastic constants of a linear, elastic, isotropic material will be?
a) 2
b) 3
c) 1
d) 4
Answer: a
Clarification: Isotropic materials have the same properties in all directions. The number of independent elastic constants for such materials is 2. out of E, G, K, and μ, if any two constants are known for any linear elastic and isotropic material than rest two can be derived. Examples are steel, aluminium, copper, gold.
Orthotropic materials refer to layered structure such as wood or plywood. The number of independent elastic constants for such materials is 9.
Non isotropic or anisotropic materials have different properties in different directions. They show non- homogeneous behaviour. The number of elastic constants is 21.

2. How many elastic constants of a non homogeneous, non isotropic material will be?
a) 9
b) 15
c) 20
d) 21
Answer: d
Clarification: Non isotropic or anisotropic materials have different properties in different directions. They show non- homogeneous behaviour. The number of elastic constants is 21.

3. How can be the Poissons ratio be expressed in terms of bulk modulus(K) and modulus of rigidity(G)?
a) (3K – 4G) / (6K + 4G)
b) (3K + 4G) /( 6K – 4G)
c) (3K – 2G) / (6K + 2G)
d) (3K + 2G) / (6K – 2G)
Answer: c
Clarification: There are four elastic modulus relationships. the relation between Poissons ration, bulk modulus and modulus of rigidity is given as
μ = (3K – 2G) / (6K + 2G).

4. Calculate the modulus of resilience for a 2m long bar which extends 2mm under limiting axial stress of 200 N/mm2?
a) 0.01
b) 0.20
c) 0.10
d) 0.02
Answer: c
Clarification: Modulus of resilience = f2/2E
= 200×2/2×2000
= 0.10.

5. In an experiment, the bulk modulus of elasticity of a material is twice its modulus of rigidity. The Poissons ratio of the material is ___________
a) 1/7
b) 2/7
c) 3/7
d) 4/7
Answer: b
Clarification: As we know, μ= (3K – 2G) / (6K + 2G)
Given K = 2G
Then, μ = (6G – 2G) / (12G + 2G) = 4/14 = 2/7.

6. What will be the value of the Poisson’s ratio if the Youngs modulus E is equal to the bulk modulus K?
a) 1/2
b) 1/4
c) 1/3
d) 3/4
Answer: c
Clarification: K = E / 3(1 – 2μ)
Since K = E
So (1-2μ) = 1/3
Therefore, μ = 1/3.

7. What is the expression for modulus of rigidity in terms of modulus of elasticity and the Poissons ratio?
a) G = 3E / 2(1 + μ)
b) G = 5E / (1 + μ)
c) G = E / 2(1 + μ)
d) G = E/ (1 + 2μ)
Answer: c
Clarification: The relation between the modulus of rigidity, modulus of elasticity and the Poissons ratio is given as
G = E / 2(1 + μ).

8. What is the relationship between Youngs modulus E, modulus of rigidity C, and bulk modulus K?
a) E = 9KC / (3K + C)
b) E = 9KC / (9K + C)
c) E = 3KC / (3K + C)
d) E = 3KC / (9K + C)
Answer: a
Clarification: The relationship between E, K, C is given by
E = 9KC / (3K + C).

9. What is the limiting values of Poisson’s ratio?
a) -1 and 0.5
b) -1 and -0.5
c) -1 and -0.5
d) 0 and 0.5
Answer: d
Clarification: The value of Poissons ratio varies from 0 to 0.5. For rubber, its value ranges from.45 to 0.50.

10. What is the relationship between modulus of elasticity and modulus of rigidity?
a) C = E / 2(1 + μ)
b) C = E / (1 + μ)
c) C = 2E / (1 + μ)
d) C = 2E / 2(1 + μ)
Answer: c
Clarification: The relation is given by calculating the tensile strain of square block is given by taking tensile strain in a diagonal. On equating that stains we get the relation,
C = E / 2(1 + μ).

250+ TOP MCQs on Impact Loading and Answers

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

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

2. What is the value of stress induced in the rod due to impact load?
a) P/A (1 + (1 + 2AEh/PL)1/2)
b) P/A (2 + 2AEh/PL)
c) P/A (1 + (1 + AEh/PL)1/2)
d) P/A ((1 + 2AEh/PL)1/2)
Answer: a
Clarification: The value of stress is calculated by equating the strain energy equation and the work done equation.

3. What will be the stress induced in the rod if the height through which load is dropped is zero?
a) P/A
b) 2P/A
c) P/E
d) 2P/E
Answer: b
Clarification: As stress = P/A (1 + (1 + 2AEh/PL)1/2)
Putting h=0, we get stress = 2P/A.

4. A weight of 10kN falls by 30mm on a collar rigidly attached to a vertical bar 4m long and 1000mm2 in section. What will be the instantaneous stress (E=210GPa)?
a) 149.4 N/mm2
b) 179.24 N/mm2
c) 187.7 N/mm2
d) 156.1 N/mm2
Answer: c
Clarification: As stress = P/A (1 + (1 + 2AEh/PL)1/2)
Putting P = 10,000, h = 30, L = 4000, A = 1000, E = 210,000 we will get stress = 187.7 N/mm2.

5. A load of 100N falls through a height of 2cm onto a collar rigidly attached to the lower end of a vertical bar 1.5m long and of 105cm2 cross- sectional area. The upper end of the vertical bar is fixed. What is the maximum instantaneous stress induced in the vertical bar if E = 200GPa?
a) 50.87 N/mm2
b) 60.23 N/mm2
c) 45.24 N/mm2
d) 63.14 N/mm2
Answer: b
Clarification: As stress = P/A ( 1 + ( 1 + 2AEh/PL)1/2 )
Putting P = 100, h = 20, L = 1500, A = 150, E = 200,000 we will get stress = 60.23 N/mm2.

6. A weight of 10kN falls by 30mm on a collar rigidly attached to a vertical bar 4m long and 1000mm2 in section. What will be the strain (E=210GPa)?
a) 0.00089
b) 0.0005
c) 0.00064
d) 0.00098
Answer: a
Clarification: As stress = P/A (1 + (1 + 2AEh/PL)1/2)
Putting P = 10,000, h = 30, L = 4000, A = 1000, E = 210,000 we will get stress = 187.7 N/mm2
As strain = stress / E, thus, strain = 187.7 / 210,000 = 0.00089.

7. A load of 100N falls through a height of 2cm onto a collar rigidly attached to the lower end of a vertical bar 1.5m long and of 105cm2 cross- sectional area. The upper end of the vertical bar is fixed. What is the maximum instantaneous elongation in the vertical bar if E = 200GPa?
a) 0.245mm
b) 0.324mm
c) 0.452mm
d) 0.623mm
Answer: c
Clarification: As stress = P/A ( 1 + ( 1 + 2AEh/PL)1/2 )
Putting P = 100, h = 20, L = 1500, A = 150, E = 200,000 we will get stress = 60.23 N/mm2
Elongation = stress x length / E = 60.23 x 1500 / 200,000 = 0.452mm.

8. A load of 100N falls through a height of 2cm onto a collar rigidly attached to the lower end of a vertical bar 1.5m long and of 105cm2 cross- sectional area. The upper end of the vertical bar is fixed. What is the strain energy stored in the vertical bar if E = 200GPa?
a) 2.045 N-m
b) 3.14 N-m
c) 9.4 N-mm
d) 2.14 N-m
Answer: a
Clarification: As stress = P/A ( 1 + ( 1 + 2AEh/PL)1/2 )
Putting P = 100, h = 20, L = 1500, A = 150, E = 200,000 we will get stress = 60.23 N/mm2.
Strain energy stored = stress2 x volume / 2E = 60.232 x 2525000 / (2×200,000) = 2.045 N-m.

9. The maximum instantaneous extension, produced by an unknown falling weight in a vertical bar of length 3m. what will be the instantaneous stress induced in the vertical bar and the value of unknown weight if E = 200GPa?
a) 100 N/mm2
b) 110 N/mm2
c) 120 N/mm2
d) 140 N/mm2
Answer: d
Clarification: Instantaneous stress = E x instantaneous strain = E x δL/L = 200,000x 2.1 / 3000 = 140N/mm2.

10. The maximum instantaneous extension, produced by an unknown falling weight through a height of 4cm in a vertical bar of length 3m and of cross section area 5cm2. what will be the instantaneous stress induced in the vertical bar and the value of unknown weight if E = 200GPa?
a) 1700 N
b) 1459.4 N
c) 1745.8 N
d) 1947.5 N
Answer: c
Clarification: Instantaneous stress = E x instantaneous strain = E x δL/L = 200,000x 2.1 / 3000 = 140N/mm2.
As, P( h + δL) = σ2/2E x V
So P = 1745.8 N.

11. An unknown weight falls through a height of 10mm on a collar rigidly attached to a lower end of a vertical bar 500cm long. If E =200GPa what will be the value of stress?
a) 50 N/mm2
b) 60 N/mm2
c) 70 N/mm2
d) 80 N/mm2
Answer: d
Clarification: Stress = E x strain = E x δL/L = 200,000 x 2 /5000 = 80 N/mm2.

250+ TOP MCQs on Bending Stress in Unsymmetrical Sections and Answers

This set of Strength of Materials Questions and Answers for Aptitude test on “Bending Stress in Unsymmetrical Sections”.

1. Unsymmetrical bending occurs due to ______
a) The Beam cross section is unsymmetrical
b) The shear Centre does not coincide with the neutral axis
c) The Beam is subjected to trust in addition to bending moment
d) The bending moment diagram is unsymmetrical
Answer: d
Clarification: If the bending moment diagram of a beam seems to unsymmetrical, then with respect to that diagram, the bending is said to be unsymmetrical bending.

2. A body having similar properties throughout its volume is said to be _____________
a) Isotropic
b) Homogeneous
c) Continuous
d) Uniform
Answer: b
Clarification: A body having similar properties throughout its volume is said to be “homogeneous” and the material which exhibits the same elastic properties in all directions is called “isotropic”.

3. Principal plane has ____________
a) Maximum shear stress
b) Maximum tensile stress
c) Zero shear stress
d) Minimum bending stress
Answer: c
Clarification: Principal stress is a magnitude of direct stress, across a principal plane which is a particular plane having no shear stress at all.

4. Calculate the Strain energy that can be stored in a body to be pulled with 100 N/mm2 stress (f) and E = 2×105 N/mm2.
a) 0.9 kNm
b) 0.05kNm
c) 0.87kNm
d) 0.54kNm
Answer: b
Clarification: Strain energy stored in the body be “U” = f2/ 2 E × Volume.
= 1002/ 2×2×105
= 0.05 kNm.

5. Materials exhibiting time bound behaviour are known as _______
a) Isentropic
b) Reactive
c) Fissile
d) Visco elastic
Answer: d
Clarification: Materials exhibiting time bound behaviour popularly known as visco elastic and if a body having similar properties throughout its volume it is known as homogeneous and according to one assumption, the concrete is considered to be homogeneous material.

6. What are the units of true strain?
a) Kg/m2
b) Kg/ m3
c) No dimensions
d) N/mm
Answer: c
Clarification: As we know strain is the ratio of change in dimension to the original dimension. It is denoted by “e”. Metres/metres hence no dimensions.

7. Revert size is generally expressed in terms of _______
a) Shank width
b) Girder length
c) Lap length
d) Shank diameter
Answer: d
Clarification: Rivets are ductile metal pins of often used for joining structure members as in case of trusses, stanchions plate girders, cylindrical shells etc. The distance between two heads is known as shank and rivet size is generally expressed in terms of shank diameter.

8. ________ joints are necessary to keep a structure safe against shrinkage.
a) Construction
b) Functional
c) Transverse
d) Longitudinal
Answer: b
Clarification: Functional joints are necessary to keep the structures safe against shrinkage, expansion sliding and warping of concrete. These types of joints are made by forming continuous breaks in large continuous areas of structures at suitable distance apart. The joints or breaks may be 6 to 38 mm wide.

9. The specific gravity of sand is __________
a) 2.8
b) 2.25
c) 3.2
d) 2.65
Answer: d
Clarification: The specific gravity of sand is 2.65.

Materials Specific gravity
Trap 2.9
Gravel 2.66
Granite 2.8
Sand 2.65

10. To what radius a silver strip 200 mm wide and 40 mm thick can be bent if the maximum stress in the ship is 80 N/mm2. Young’s modulus for Silver is 80×103 N/mm2.
a) 20m
b) 30m
c) 15m
d) 35m
Answer: a
Clarification: Here, b = 200 mm; d = 40mm
y = 40/2 = 20 mm
f = 80N/mm2
From the relation; E/R = f/y
R = E×y / f
= 80000×20 / 80
= 20000mm = 20m.

Strength of Materials for Aptitude test,

250+ TOP MCQs on Slope and Answers

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

1. Slope in the beam at any point is measured in ____________
a) Degrees
b) Minutes
c) Radians
d) Metric tonnes
Answer: c
Clarification: The slope is defined as at any point on the bent beam is the angle measured in terms of radians to which the tangent at that point makes with the x axis.

2. Elastic curve is also known as __________
a) Refraction curve
b) Reflection curve
c) Deflection curve
d) Random curve
Answer: c
Clarification: An elastic curve is defined as the line to which the longitudinal axis of a beam deviates under given load. It is also called a deflection curve.

3. Which of the following method is not used for determining slope and deflection at a point?
a) Moment area method
b) Double integration method
c) Isoheytal method
d) Macaulay’s method
Answer: c
Clarification: The method “Isoheytal” can be used for calculating run-off over an area. The remaining methods are effectively adopted to calculate the slope and deflection at a point in any type of beam.

4. The slope is denoted by _______
a) k
b) y
c) i
d) c
Answer: c
Clarification: The slope at any section in a deflection beam is defined as the angle measured in radians to the tangent at the section makes with the original axis of the beam.
•It is denoted by “i”.

5. Calculate the slope at supports, if the area is 180kNm2. Take flexural rigidity as 50000.
a) 0.0054 radians
b) 0.0072 radians
c) 0.0036 radians
d) 0.108 radians
Answer: c
Clarification: Maximum slope at supports be i = A/EI
= 180/50000
i = 0.0036 radians.

6. In cantilever beams, the slope is _____________ at fixed end.
a) Maximum
b) Zero
c) Minimum
d) Uniform
Answer: b
Clarification: The slope in cantilever beam is zero at the fixed end of the cantilever and the slope is maximum at it’s free end. The slope is determined in the moment area method through Mohr’s theorems.

7. Slope is maximum at _______ in simply supported beams.
a) Mid span
b) Through out
c) Supports
d) At point of loading
Answer: a
Clarification: In case simply supported beams, the slope is maximum at the end supports of the beam and relatively zero at midspan of a symmetrically loaded beam.

8. Mohr’s theorem- 1 states ________
a) E/AI
b) I/EA
c) A/EI
d) A=EI
Answer: c
Clarification: According to Mohr’s theorem-1, the change of slope between any of the two points on and Elastic axis is equal to the net area of bending moment diagram (A) between these two points divided by flexural rigidity(EI).

9. Using Mohr’s theorem, calculate the maximum slope of a cantilever beam if the bending moment area diagram is 90kNm2. Take EI = 4000 kNm2.
a) 0.0225 radians
b) 0 0367 radians
c) 0.0455 radians
d) 0.066 radians
Answer: a
Clarification: The maximum slope at free support (in cantilever beam) = i = A/EI
= 90/4000
= 0.0225 radians.

10. Contour canals are also called as ______
a) Single bank canal
b) Ridge canal
c) Side slope canal
d) Watershed canal
Answer: a
Clarification: In this method, the canal is aligned along the falling contour. A generally higher side is left without bank. So it is also called a single bank canal. The contour canal cuts across the natural drainage courses.

11. ______________ provides employment to the cultivators at the time of famine.
a) Productive canal
b) Link canal
c) Protective canal
d) Inundation canal
Answer: c
Clarification: The construction of protective canals and their development may be started during summer in hence they provide employment to the farmers at the time of drought and famine. Protective canals are not remunerative as productive canals.

12. ______________ bricks are used in the lining of blast furnaces.
a) Magnesia
b) Dolomite
c) Bauxite
d) Fly ash
Answer: b
Clarification: Dolomite bricks are made especially from dolomite it contains nearly 30% lime and 22% of magnesium these bricks are inferior to magnesite bricks. They are generally used in the lining of blast furnaces.

13. _____________ bricks are resistant to corrosion.
a) silica bricks
b) magnesia bricks
c) bauxite bricks
d) fire bricks
Answer: c
Clarification: Bauxite bricks contain nearly 75% of aluminium and it is mixed with fire clay 15 to 30% and added some water to mould. High alumina bricks are resistant to corrosion.

14. _____________ bricks are used in the lining of electric furnace.
a) Frosterite
b) Spinel
c) Chrome
d) Basic
Answer: b
Clarification: The spinal bricks belong to neutral bricks. The spinel bricks mainly consist of alumina and magnesia. These bricks are widely used in the lining of electric furnace.

15. The finished product after burning magnesite is named as ___________
a) Perillax
b) Hellyx
c) Pyrolytaex
d) Syrilax
Answer: a
Clarification: The heating of magnesia bricks is continued in the same kiln after reaching the temperature of 1950°C, and then some amount of iron oxide is mixed. The finished product after burning magnesite is named as perillax.

250+ TOP MCQs on Combined Bending and Torsion and Answers

This set of Strength of Materials Multiple Choice Questions on “Combined Bending and Torsion”.

1. A solid shaft of circular in section is subjected to torque which produces maximum shear stress in a shaft. Calculate the diameter of the shaft.
a) (16T/πf)3/2
b) (16f/πT)1/2
c) (16f/π) 1/2
d) (πT/16f) 1/2
Answer: a
Clarification: From torsional equation
T/J = f/R
T = f.Z
T = f×π/16d3.
D= (16T/πf) 3/2.

2. When two dissimilar shafts are connected together, then the shaft is __________
a) Integrated shafts
b) Composite shafts
c) Differential shafts
d) Combined shafts
Answer: b
Clarification: When two dissimilar shafts are connected together to form one shaft then the shaft can be termed as composite shaft.

3. __________ torque occurs along with maximum shear stress due to combined bending and torsion.
a) Equipment
b) Coaxial
c) Biaxial
d) Lateral
Answer: a
Clarification: Equipment torque is the twisting moment which acts along producing maximum shear stress due to the combined bending as well as torsion.

4. When a shaft is subjected to pure twisting then the type of stress developed is ________
a) Bending
b) Axial
c) Shear
d) Normal
Answer: c
Clarification: Shear stress is produced when the shaft is subjected to pure twisting (torsion). The shear stress due to twisting moment is zero at the axis of the shaft.

5. The torque which produces unit twist per unit length is ________
a) Torsional rugosity
b) Torsional rigidity
c) Torsional viscosity
d) Torsional mean radius
Answer: a
Clarification: The product of shear modulus(C) and polar moment of inertia (J) is called torsional rigidity. Torsional rigidity produces a twist of 1 radian in a shaft of unit length.

6. The level of top of weir can be termed as __________
a) Talus
b) Curtain walls
c) Crest
d) Shutter
Answer: c
Clarification: The level of the top of weir is known as a crest. The shutters are provided on the crest and can be raised or laid flat during the time of floods.

7. ________ possesses less silting and scouring.
a) Weir
b) Barrage
c) Dams
d) Reservoir
Answer: b
Clarification: The barrage is a low obstructive barrier constructed across the river. There will be less silting and better control over the levels due to low set crest.

8. In __________ there will be no means for silt disposal.
a) Weir
b) Barrage
c) Reservoir
d) Dams
Answer: a
Clarification: The weir may be defined as a solid obstruction/wall built across the river to raise the water level. Raised crest causes silting at upstream and there is no means silt disposal.

9. _________ is a pure water pressure.
a) Uplift
b) Percolation
c) Scour
d) Flood bank
Answer: a
Clarification: Uplift occurs when pore water pressure under a structure or a low permeability confining layer becomes larger than the mean overburden pressure.

10. __________ causes of uplift of structure.
a) Percolation
b) Scour
c) Critical Velocity
d) Slope Failure
Answer: a
Clarification: The effect of percolation on an irrigation structure like a weir to cause uplift pressure on the structures and topple the structure at any moment.

11. ________ protects the weir from erosive forces during floods.
a) Talus
b) Curtain walls
c) Shutter
d) Upstream solid apron
Answer: d
Clarification: Upstream solid apron is a concrete bed which is provided on the upstream side of weir to protect the weir from erosive forces during floods. The length of apron depends upon maximum discharge of the river.

12. Gross storage – Dead storage is _______
a) Live storage
b) Virtual storage
c) Excessive storage
d) Free storage
Answer: a
Clarification: It is also called as available or effective storage. It is the difference between gross storage and dead storage. It is the amount of water available from FRL to the sill of the lowest sluice.

13. Which of the following is not sound proof?
a) G I sheets
b) A C sheets
c) PVC sheets
d) Fabric sheets
Answer: a
Clarification: Galvanised iron sheets are commonly used as a roofing material. These are very durable and fire proof. The main disadvantage is they are not sound proof.

14. Which of the following is not affected by temperature?
a) Fabric sheets
b) G I sheets
c) AC sheets
d) Flat roofs
Answer: c
Clarification: Asbestos cement sheets are cheaper in the initial cost. They are fire resisting. They are heavy in weight and they are not affected by temperature.

15. Which of the following possess good insulation properties?
a) Battened roofs
b) Wooden roofs
c) Jack arch roofs
d) Flat roofs
Answer: d
Clarification: Flat roofs are easier in construction and maintenance. A flat roof is more stable against high wards. It has a better architectural appearance and it has good insulation properties.

250+ TOP MCQs on Elastic Constants Relationship – 2 and Answers

This set of Strength of Materials Questions and Answers for Freshers on “Elastic Constants Relationship – 2”.

1. What is the ratio of Youngs modulus E to shear modulus G in terms of Poissons ratio?
a) 2(1 + μ)
b) 2(1 – μ)
c) 1/2 (1 – μ)
d) 1/2 (1 + μ)
Answer: a
Clarification: As we know G = E / 2(1 +μ) so this gives the ratio of E to G = 2(1 + μ).

2. The relationship between Youngs modulus E, bulk modulus K if the value of Poissons ratio is unity will be __________
a) E = -3K
b) K = -3E
c) E = 0
d) K = 0
Answer: a
Clarification: As E = 2G(1 + μ) putting μ=1 we get E = -3K.

3. A rod of length L and diameter D is subjected to a tensile load P. which of the following is sufficient to calculate the resulting change in diameter?
a) Youngs modulus
b) Poissons ratio
c) Shear modulus
d) Both Youngs modulus and shear modulus
Answer: a
Clarification: For longitudinal strain we need Youngs modulus and for calculating transverse strain we need Poisson’s ratio. We may calculate Poissons ratio from E = 2G(1 + μ) for that we need shear modulus.

4. E, G, K and μ elastic modulus, shear modulus, bulk modulus and Poisson’s ratio respectively. To express the stress strain relations completely for this material, at least __________
a) E, G and μmust be known
b) E, K and μmust be known
c) Any two of the four must be known
d) All the four must be known
Answer: c
Clarification: As E = 2G(1 + μ) = 3K(1 – 2μ) = 9KG / (3K + G), if any two of these four are known, the other two can be calculated by the relations between them.

5. Youngs modulus of elasticity and Poissons ratio of a material are 1.25 x 102 MPa and 0.34 respectively. The modulus of rigidity of the material is __________
a) 0.9469 MPa
b) 0.8375 MPa
c) 0.4664 MPa
d) 0.4025 MPa
Answer: c
Clarification: As E = 2G(1 + μ)
1.25 x 102 = 2G(1 + 0.34)
G = 0.4664 x 102 MPa.

6. If E,G and K have their usual meanings, for an elastic material, then which one of the following be possibly true?
a) G = 2K
b) G = K
c) K = E
d) G = E = K
Answer: c
Clarification: As E = 2G(1 + μ) = 3K(1 – 2μ) = 9KG / (3K + G)
The value of μ must be between 0 to 0.5, so as E never equal to G but if μ = 1/3, then E=K.

7. If a material had a modulus of elasticity of 2.1 kgf/cm2 and a modulus of rigidity of 0.8 kgf/cm2 then what will be the approximate value of the Poissons ratio?
a) 0.26
b) 0.31
c) 0.47
d) 0.43
Answer: b
Clarification: On using E = 2G(1 + μ) we can put the values of E and G to get the Poissons value.

8. Consider the following statements:
X. Two-dimensional stresses applied to a thin plater in its own plane represent the plane stress condition.
Y. Normal and shear stresses may occur simultaneously on a plane.
Z. Under plane stress condition, the strain in the direction perpendicular to the plane is zero.
Which of the above statements are correct?
a) 2 only
b) 1 and 2
c) 2 and 3
d) 1 and 3
Answer: d
Clarification: Under plane stress condition, the strain in the direction perpendicular to the plane is not zero. It has been found experimentally that when a body is stressed within the elastic limit, the lateral strain bears a constant ratio to the linear strain.

9. What is the relationship between the linear elastic properties Youngs modulus, bulk modulus and rigidity modulus?
a) 1/E = 9/k + 3/G
b) 9/E = 3/K + 1/G
c) 3/E = 9/K + 1/G
d) 9/E = 1/K + 3/G
Answer: d
Clarification: We can use E = 2G(1 + μ) = 3K(1 – 2μ) = 9KG / (3K + G) to get the relation between E, K and G.

10. Which of the relationship between E, G and K is true, where E, G and K have their usual meanings?
a) E = 9KC / (3K + C)
b) E = 9KC / (9K + C)
c) E = 3KC / (9K + C)
d) E = 3KC / (3K + C)
Answer: a
Clarification: As we know E = 2G(1 + μ) = 3K(1 – 2μ) = 9KG / (3K + G).

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