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Question 1
ప్రశ్న 1
For a three-dimensional isotropic linear-elastic material, how many independent elastic constants are required to completely describe its elastic constitutive behaviour?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
An isotropic linear-elastic material requires only two independent elastic constants to define its constitutive behaviour. Once any suitable two constants are known, the remaining conventional elastic constants can be calculated from established relationships. Common choices include E and ν, E and G, or K and G. This reduction to two independent constants follows from material isotropy and linear elasticity. Therefore Option B is correct.
వివరణ:
Question 2
ప్రశ్న 2
Which relationship between Young's modulus E, shear modulus G and Poisson's ratio ν is valid for an isotropic linear-elastic material?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For isotropic linear elasticity, Young's modulus and shear modulus are related by E = 2G(1 + ν). This relationship follows from the constitutive equations for an isotropic elastic solid. It should not be applied indiscriminately to an anisotropic material, for which elastic properties can depend on direction. The factor of 2 is essential and is a common source of numerical errors in objective examinations. Therefore Option B is the correct relationship.
వివరణ:
Question 3
ప్రశ్న 3
An isotropic linear-elastic material has Young's modulus E = 200 GPa and Poisson's ratio ν = 0.25. What is its shear modulus G?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For an isotropic linear-elastic material, G = E/[2(1 + ν)]. Substituting E = 200 GPa and ν = 0.25 gives G = 200/[2 × 1.25]. The denominator is 2.50, so G = 80 GPa. The units remain those of stress because both E and G are elastic moduli. Therefore Option C is correct.
వివరణ:
Question 4
ప్రశ్న 4
An isotropic linear-elastic material has E = 210 GPa and ν = 0.30. What is its bulk modulus K?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For an isotropic linear-elastic material, K = E/[3(1 - 2ν)]. With ν = 0.30, the factor 1 - 2ν equals 0.40. Therefore K = 210/[3 × 0.40] = 210/1.20. This gives K = 175 GPa. Hence Option C is the correct bulk modulus.
వివరణ:
Question 5
ప్రశ్న 5
For an isotropic linear-elastic solid, E = 200 GPa and G = 80 GPa. What is Poisson's ratio?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The applicable relation is E = 2G(1 + ν). Rearranging gives ν = E/(2G) - 1. Substitution gives ν = 200/(2 × 80) - 1 = 1.25 - 1. Therefore ν = 0.25 and Option B is correct.
వివరణ:
Question 6
ప్రశ్న 6
An isotropic linear-elastic material has Young's modulus E = 180 GPa and bulk modulus K = 150 GPa. What is its Poisson's ratio?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For isotropic linear elasticity, E = 3K(1 - 2ν). Rearranging gives ν = (3K - E)/(6K). Substituting E = 180 GPa and K = 150 GPa gives ν = (450 - 180)/900. This evaluates to 270/900 = 0.30. Therefore Option C is correct.
వివరణ:
Question 7
ప్రశ్న 7
For an isotropic linear-elastic material, K = 100 GPa and G = 60 GPa. Which pair correctly gives Young's modulus E and Poisson's ratio ν?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Young's modulus can be obtained from E = 9KG/(3K + G). Substitution gives E = 9 × 100 × 60/(300 + 60) = 54000/360 = 150 GPa. Poisson's ratio can then be obtained from ν = E/(2G) - 1. Thus ν = 150/120 - 1 = 0.25. Therefore Option B gives the correct pair.
వివరణ:
Question 8
ప్రశ్న 8
For an isotropic linear-elastic material having Poisson's ratio ν = 0.25, what is the ratio K/G?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For isotropic elasticity, K = E/[3(1 - 2ν)] and G = E/[2(1 + ν)]. Dividing these equations gives K/G = 2(1 + ν)/[3(1 - 2ν)]. For ν = 0.25, this becomes 2 × 1.25/[3 × 0.50] = 2.50/1.50. The ratio is therefore 5/3. Hence Option D is correct.
వివరణ:
Question 9
ప్రశ్న 9
Which statement concerning the dimensions of the conventional elastic constants is correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Young's modulus is a ratio of normal stress to normal strain, so it has the dimensions of stress. Shear modulus is a ratio of shear stress to shear strain and therefore also has stress dimensions. Bulk modulus is hydrostatic stress divided by volumetric strain, so it likewise has stress dimensions. Poisson's ratio is a ratio of two normal strains and is therefore dimensionless. Consequently Option A is correct.
వివరణ:
Question 10
ప్రశ్న 10
Which definition most accurately describes the bulk modulus K for an isotropic linear-elastic material subjected to uniform hydrostatic loading?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Bulk modulus measures resistance to uniform volumetric deformation under hydrostatic loading. In magnitude form, K is the hydrostatic pressure divided by the corresponding volumetric strain magnitude. A large bulk modulus therefore indicates that a relatively large pressure is needed to produce a given fractional change in volume. Young's modulus instead relates uniaxial normal stress to longitudinal strain, while shear modulus relates shear stress to shear strain. Therefore Option C correctly defines the bulk modulus.
వివరణ:
Question 11
ప్రశ్న 11
A material with bulk modulus K = 160 GPa is subjected to a uniform hydrostatic compressive pressure of 80 MPa. What is the magnitude of the resulting volumetric strain within the linear-elastic range?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The magnitude of volumetric strain under hydrostatic pressure is εv = p/K. The bulk modulus is 160 GPa = 160000 MPa. Therefore εv = 80/160000 = 0.00050. Under a tension-positive strain convention the actual volumetric strain would be negative because the volume decreases, but the question asks for its magnitude. Hence Option B is correct.
వివరణ:
Question 12
ప్రశ్న 12
A body having an initial volume of 0.50 m³ is subjected to a uniform hydrostatic pressure of 120 MPa. If its bulk modulus is 160 GPa, what is the approximate decrease in volume?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The volumetric strain magnitude is p/K = 120/160000 = 0.00075. The volume decrease is therefore 0.00075 × 0.50 = 0.000375 m³. Since 1 m³ equals 1000000 cm³, this corresponds to 375 cm³. The calculation assumes a uniform hydrostatic stress state and linear-elastic volumetric response. Therefore Option C is correct.
వివరణ:
Question 13
ప్రశ్న 13
A material is subjected to simple shear stress τ within its linear-elastic range. Which relationship correctly defines the shear modulus G in terms of engineering shear strain γ?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Shear modulus is the proportionality constant between shear stress and engineering shear strain in the linear-elastic range. The governing relation is τ = Gγ. Rearranging gives G = τ/γ. Engineering shear strain γ represents the small angular distortion between initially perpendicular material lines. Therefore Option B is correct.
వివరణ:
Question 14
ప్రశ్న 14
For an isotropic linear-elastic material having E = 210 GPa and ν = 0.30, which pair of moduli is approximately correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The shear modulus is G = E/[2(1 + ν)] = 210/[2 × 1.30]. This gives G = approximately 80.77 GPa. The bulk modulus is K = E/[3(1 - 2ν)] = 210/[3 × 0.40] = 175 GPa. Both values arise from the same isotropic linear-elastic constants E and ν. Therefore Option A is correct.
వివరణ:
Question 15
ప్రశ్న 15
Consider the following statements for a stable isotropic linear-elastic material:
1. Poisson's ratio is defined as minus the ratio of lateral strain to longitudinal strain under uniaxial loading.
2. A positive Poisson's ratio normally gives lateral contraction under longitudinal tension.
3. A negative Poisson's ratio is mathematically admissible within part of the stable isotropic elastic range.
4. Any value ν greater than 0.5 is admissible provided Young's modulus is positive.
Which statements are correct?
1. Poisson's ratio is defined as minus the ratio of lateral strain to longitudinal strain under uniaxial loading.
2. A positive Poisson's ratio normally gives lateral contraction under longitudinal tension.
3. A negative Poisson's ratio is mathematically admissible within part of the stable isotropic elastic range.
4. Any value ν greater than 0.5 is admissible provided Young's modulus is positive.
Which statements are correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Statement 1 is correct because ν = -εlateral/εlongitudinal for the uniaxial definition. Statement 2 is correct because positive ν makes the lateral strain opposite in sign to a tensile longitudinal strain. Statement 3 is correct because stable isotropic linear elasticity permits negative Poisson's ratios greater than -1. Statement 4 is incorrect because positive bulk and shear moduli require ν to remain below 0.5. Therefore Statements 1, 2 and 3 are correct and Option B is the unique answer.
వివరణ:
Question 16
ప్రశ్న 16
An isotropic linear-elastic material has Poisson's ratio ν = -0.20. If a uniaxial tensile load produces a longitudinal strain of +0.0010, what lateral strain is predicted?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Poisson's ratio is defined by ν = -εlateral/εlongitudinal. Rearranging gives εlateral = -νεlongitudinal. With ν = -0.20 and εlongitudinal = +0.0010, the lateral strain is +0.00020. Thus this material expands laterally while being stretched longitudinally, which is characteristic of a negative Poisson-ratio response. Therefore Option B is correct.
వివరణ:
Question 17
ప్రశ్న 17
For a stable isotropic linear-elastic solid, what happens as Poisson's ratio approaches the incompressible limit of 0.5 from below while shear rigidity remains finite?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For isotropic elasticity, K = E/[3(1 - 2ν)], so the denominator tends to zero as ν approaches 0.5 from below. The material therefore becomes increasingly resistant to volumetric deformation and K tends to infinity in the incompressible limit. The relation E = 2G(1 + ν) gives E approaching 3G as ν approaches 0.5. This limit should be interpreted carefully because perfect incompressibility requires special treatment of the volumetric constraint. Therefore Option B correctly describes the limiting behaviour.
వివరణ:
Question 18
ప్రశ్న 18
A proposed isotropic linear-elastic material is reported to have E = 200 GPa and G = 60 GPa. What conclusion follows from these values?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For isotropic elasticity, ν = E/(2G) - 1. Substituting the reported values gives ν = 200/(120) - 1 = approximately 0.667. A stable isotropic linear-elastic material with positive shear and bulk moduli requires -1 < ν < 0.5. The computed value therefore violates the admissible upper bound and indicates that the reported E and G cannot simultaneously represent such a material. Hence Option C is correct.
వివరణ:
Question 19
ప్రశ్న 19
Match each elastic constant with the deformation mode most directly associated with its basic definition:
1. Young's modulus E
2. Shear modulus G
3. Bulk modulus K
4. Poisson's ratio ν
P. Hydrostatic stress and volumetric strain
Q. Uniaxial normal stress and longitudinal strain
R. Lateral strain relative to longitudinal strain
S. Shear stress and angular distortion
Which matching is correct?
1. Young's modulus E
2. Shear modulus G
3. Bulk modulus K
4. Poisson's ratio ν
P. Hydrostatic stress and volumetric strain
Q. Uniaxial normal stress and longitudinal strain
R. Lateral strain relative to longitudinal strain
S. Shear stress and angular distortion
Which matching is correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Young's modulus relates uniaxial normal stress to longitudinal normal strain, so 1 matches Q. Shear modulus relates shear stress to engineering shear strain or angular distortion, so 2 matches S. Bulk modulus measures resistance to volumetric deformation under hydrostatic stress, so 3 matches P. Poisson's ratio relates lateral strain to longitudinal strain under uniaxial loading, so 4 matches R. Therefore the correct matching is 1-Q, 2-S, 3-P and 4-R, making Option A correct.
వివరణ:
Question 20
ప్రశ్న 20
Two isotropic linear-elastic materials have the same Young's modulus E. Material B has a higher Poisson's ratio than Material A, with both ratios lying between 0 and 0.5. Which comparison is correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
At fixed E, the shear modulus is G = E/[2(1 + ν)]. Increasing ν therefore increases the denominator and causes G to decrease. The bulk modulus is K = E/[3(1 - 2ν)]. Increasing ν toward 0.5 decreases the denominator of this expression and therefore increases K. Thus Material B has lower shear modulus but higher bulk modulus. Option B is the correct comparison.
వివరణ:
Question 21
ప్రశ్న 21
An isotropic linear-elastic material has K = 120 GPa and G = 45 GPa. Which pair approximately represents E and ν?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The relation E = 9KG/(3K + G) can be used when K and G are known. Substitution gives E = 9 × 120 × 45/(360 + 45) = 48600/405 = 120 GPa. Poisson's ratio is ν = (3K - 2G)/[2(3K + G)]. This gives ν = (360 - 90)/(810) = 270/810 = 1/3, or approximately 0.333. Therefore Option C is correct.
వివరణ:
Question 22
ప్రశ్న 22
For an isotropic linear-elastic material, G = 75 GPa and ν = 0.20. What are the corresponding values of E and K?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Young's modulus is obtained from E = 2G(1 + ν). Therefore E = 2 × 75 × 1.20 = 180 GPa. Bulk modulus is K = E/[3(1 - 2ν)] = 180/[3 × 0.60]. This gives K = 180/1.80 = 100 GPa. Hence Option B gives the correct pair.
వివరణ:
Question 23
ప్రశ్న 23
Under uniaxial tensile stress, an isotropic linear-elastic material develops a longitudinal strain of 0.00050 and a first-order volumetric strain of 0.00020. What is Poisson's ratio?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Under uniaxial stress, the two lateral strains are each equal to -νεlongitudinal. The first-order volumetric strain is therefore εv = εlongitudinal(1 - 2ν). Substitution gives 0.00020 = 0.00050(1 - 2ν). Hence 1 - 2ν = 0.40, which gives ν = 0.30. Therefore Option C is correct.
వివరణ:
Question 24
ప్రశ్న 24
An isotropic linear-elastic material has E = 150 GPa and G = 60 GPa. What are its Poisson's ratio and bulk modulus?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
From E = 2G(1 + ν), Poisson's ratio is ν = E/(2G) - 1. Substitution gives ν = 150/120 - 1 = 0.25. The bulk modulus is then K = E/[3(1 - 2ν)] = 150/[3 × 0.50]. This gives K = 100 GPa. Therefore Option B correctly provides both elastic constants.
వివరణ:
Question 25
ప్రశ్న 25
For an isotropic linear-elastic material having Poisson's ratio ν = 0, which relationship among E, G and K is correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Setting ν = 0 in G = E/[2(1 + ν)] gives G = E/2. Setting ν = 0 in K = E/[3(1 - 2ν)] gives K = E/3. These results are mutually consistent with the general relationship E = 9KG/(3K + G). A zero Poisson's ratio means that uniaxial longitudinal deformation produces no lateral strain in the idealized isotropic linear-elastic model. Therefore Option B is correct.
వివరణ:
Question 26
ప్రశ్న 26
Consider the following statements about elastic constants:
1. A three-dimensional isotropic linear-elastic material can be characterized by two independent elastic constants.
2. The relationship E = 2G(1 + ν) may be applied without qualification to any anisotropic material.
3. Requiring both K and G to be positive leads to -1 < ν < 0.5 for an isotropic linear-elastic material.
4. For fixed positive E, K increases without bound as ν approaches 0.5 from below.
Which statements are correct?
1. A three-dimensional isotropic linear-elastic material can be characterized by two independent elastic constants.
2. The relationship E = 2G(1 + ν) may be applied without qualification to any anisotropic material.
3. Requiring both K and G to be positive leads to -1 < ν < 0.5 for an isotropic linear-elastic material.
4. For fixed positive E, K increases without bound as ν approaches 0.5 from below.
Which statements are correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Statement 1 is correct because isotropic linear elasticity contains only two independent elastic constants. Statement 2 is incorrect because E = 2G(1 + ν) is an isotropic relationship and cannot generally be imposed on anisotropic elasticity. Statement 3 is correct because positivity of the bulk and shear moduli produces the stability interval -1 < ν < 0.5. Statement 4 is correct because K = E/[3(1 - 2ν)], whose denominator approaches zero as ν approaches 0.5 from below. Therefore Statements 1, 3 and 4 are correct and Option B is the unique answer.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: B. Two సమాధానం: B.
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Question 2 ప్రశ్న 2Answer: B. E = 2G(1 + ν) సమాధానం: B.
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Question 3 ప్రశ్న 3Answer: C. 80 GPa సమాధానం: C.
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Question 4 ప్రశ్న 4Answer: C. 175 GPa సమాధానం: C.
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Question 5 ప్రశ్న 5Answer: B. 0.25 సమాధానం: B.
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Question 6 ప్రశ్న 6Answer: C. 0.30 సమాధానం: C.
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Question 7 ప్రశ్న 7Answer: B. E = 150 GPa, ν = 0.25 సమాధానం: B.
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Question 8 ప్రశ్న 8Answer: D. 5/3 సమాధానం: D.
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Question 9 ప్రశ్న 9Answer: A. E, G and K have dimensions of stress, while Poisson's ratio is dimensionless. సమాధానం: A.
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Question 10 ప్రశ్న 10Answer: C. Ratio of hydrostatic stress magnitude to the corresponding volumetric strain magnitude సమాధానం: C.
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Question 11 ప్రశ్న 11Answer: B. 0.00050 సమాధానం: B.
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Question 12 ప్రశ్న 12Answer: C. 375 cm³ సమాధానం: C.
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Question 13 ప్రశ్న 13Answer: B. G = τ/γ సమాధానం: B.
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Question 14 ప్రశ్న 14Answer: A. G = 80.77 GPa and K = 175 GPa సమాధానం: A.
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Question 15 ప్రశ్న 15Answer: B. 1, 2 and 3 only సమాధానం: B.
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Question 16 ప్రశ్న 16Answer: B. +0.00020 సమాధానం: B.
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Question 17 ప్రశ్న 17Answer: B. Bulk modulus tends to infinity and E approaches 3G. సమాధానం: B.
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Question 18 ప్రశ్న 18Answer: C. They imply ν ≈ 0.667, which is incompatible with a stable isotropic linear-elastic material having positive bulk and shear moduli. సమాధానం: C.
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Question 19 ప్రశ్న 19Answer: A. 1-Q, 2-S, 3-P, 4-R సమాధానం: A.
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Question 20 ప్రశ్న 20Answer: B. Material B has lower G but higher K. సమాధానం: B.
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Question 21 ప్రశ్న 21Answer: C. E = 120 GPa, ν = 0.333 సమాధానం: C.
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Question 22 ప్రశ్న 22Answer: B. E = 180 GPa and K = 100 GPa సమాధానం: B.
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Question 23 ప్రశ్న 23Answer: C. 0.30 సమాధానం: C.
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Question 24 ప్రశ్న 24Answer: B. ν = 0.25 and K = 100 GPa సమాధానం: B.
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Question 25 ప్రశ్న 25Answer: B. G = E/2 and K = E/3 సమాధానం: B.
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Question 26 ప్రశ్న 26Answer: B. 1, 3 and 4 only సమాధానం: B.