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Question 1
ప్రశ్న 1
A straight prismatic bar is subjected to a concentric axial tensile load P, and the stress distribution over a section sufficiently away from the load application is assumed uniform. Which expression gives the average normal stress?
Explanation:
For a concentrically loaded prismatic member with a uniform stress distribution, the average normal stress is equal to the axial force divided by the cross-sectional area. Thus, σ = P/A. The stress acts normal to the cross-section and is tensile when the member is being pulled. The simple expression assumes that local stress concentrations near load application points are not being considered. Therefore P/A is the appropriate average normal-stress relationship for the stated condition.
వివరణ:
Question 2
ప్రశ్న 2
Using the common Strength of Materials sign convention in which tensile normal stress and tensile normal strain are positive, which statement is correct?
Explanation:
A commonly used Strength of Materials convention assigns a positive sign to tensile normal stress and a negative sign to compressive normal stress. Correspondingly, elongating longitudinal strain is positive and shortening longitudinal strain is negative. Stating the convention is important because signs carry information about the physical sense of deformation. Shear stresses require their own directional sign convention and should not be inferred merely from the normal-stress rule. Under the convention specified in the question, Option C is therefore correct.
వివరణ:
Question 3
ప్రశ్న 3
Which statement about the units and dimensions of normal stress and normal strain is correct?
Explanation:
Normal stress is force divided by area, so its dimensions are force per unit area. In SI engineering calculations it is commonly expressed in Pa, MPa or N/mm². Normal strain is the ratio of a change in length to an original length. Because both numerator and denominator have the dimension of length, strain is dimensionless. Therefore Option A correctly distinguishes the dimensions of stress and strain.
వివరణ:
Question 4
ప్రశ్న 4
A steel tie of uniform cross-sectional area 600 mm² carries a concentric tensile load of 120 kN. What is the average tensile stress in the tie?
Explanation:
The average tensile stress is calculated from σ = P/A. The load is 120 kN = 120000 N and the area is 600 mm². Substitution gives σ = 120000/600 = 200 N/mm². Since 1 N/mm² is equal to 1 MPa, the stress is 200 MPa. Therefore Option D is the correct answer.
వివరణ:
Question 5
ప్రశ్న 5
A 1500 mm long bar elongates by 0.75 mm under an axial tensile load. What is its longitudinal strain?
Explanation:
Longitudinal strain is the change in length divided by the original length. Therefore ε = 0.75/1500. This gives ε = 0.00050. Strain has no physical unit because it is a ratio of two lengths expressed in consistent units. Hence Option B is correct.
వివరణ:
Question 6
ప్రశ్న 6
Within its linear elastic range, a material develops a normal stress of 150 MPa at a longitudinal strain of 0.00075. What is its Young's modulus?
Explanation:
Within the linear elastic proportional range, Hooke's law gives σ = Eε. Rearranging gives E = σ/ε. Substitution gives E = 150/0.00075 = 200000 MPa. Since 1000 MPa equals 1 GPa, this is 200 GPa. Therefore Option C is the correct Young's modulus.
వివరణ:
Question 7
ప్రశ్న 7
A uniform steel bar 2 m long and 400 mm² in cross-sectional area carries an axial tensile load of 80 kN. If E = 200 GPa, what is its elastic elongation?
Explanation:
For a uniform prismatic bar under constant axial force, the elastic deformation is δ = PL/(AE). Here P = 80000 N, L = 2000 mm, A = 400 mm² and E = 200000 N/mm². Substitution gives δ = (80000 × 2000)/(400 × 200000). The result is 2.0 mm of elongation. Therefore Option A is correct.
వివరణ:
Question 8
ప్రశ్న 8
Using tension-positive deformation, a 2.5 m long steel member of area 750 mm² carries a concentric compressive load of 150 kN. If E = 200 GPa, what is its axial deformation?
Explanation:
The magnitude of axial deformation is obtained from δ = PL/(AE). Using P = 150000 N, L = 2500 mm, A = 750 mm² and E = 200000 N/mm² gives a magnitude of 2.50 mm. Because the load is compressive, the member shortens rather than elongates. Under the stated tension-positive convention, shortening is represented by a negative deformation. Hence the deformation is -2.50 mm and Option D is correct.
వివరణ:
Question 9
ప్రశ్న 9
A material subjected to uniaxial tension has a longitudinal strain of +0.0012 and a lateral strain of -0.0003. What is its Poisson's ratio?
Explanation:
Poisson's ratio is defined as the negative ratio of lateral strain to longitudinal strain. Thus ν = -(-0.0003)/(0.0012). The result is 0.25. The negative sign in the definition makes Poisson's ratio positive for an ordinary material that contracts laterally when stretched longitudinally. Therefore Option B is correct.
వివరణ:
Question 10
ప్రశ్న 10
A circular steel rod of diameter 40 mm is subjected to a tensile stress of 120 MPa. If E = 200 GPa and Poisson's ratio ν = 0.30, what is the approximate change in diameter within the linear elastic range?
Explanation:
The longitudinal strain is ε = σ/E = 120/200000 = 0.0006. The lateral strain is -νε = -0.30 × 0.0006 = -0.00018. The diameter change is therefore -0.00018 × 40 = -0.0072 mm. The negative sign indicates lateral contraction under the tensile loading. Hence the diameter decreases by approximately 0.0072 mm and Option C is correct.
వివరణ:
Question 11
ప్రశ్న 11
An isotropic linearly elastic material is subjected to a uniaxial tensile stress of 120 MPa. If E = 200 GPa and ν = 0.30, what is the approximate volumetric strain under the small-strain assumption?
Explanation:
The longitudinal strain is σ/E = 120/200000 = 0.0006. Each lateral strain is -νσ/E = -0.30 × 0.0006 = -0.00018. For small strains, volumetric strain is the algebraic sum of the three mutually perpendicular normal strains. Therefore εv = 0.0006 - 0.00018 - 0.00018 = 0.00024. Option A is the correct volumetric strain.
వివరణ:
Question 12
ప్రశ్న 12
A 100 mm × 100 mm × 100 mm cube of an isotropic linearly elastic material is subjected to uniaxial tensile stress of 100 MPa. If E = 200 GPa and ν = 0.25, what is the approximate increase in volume, neglecting products of strains?
Explanation:
The longitudinal strain is 100/200000 = 0.0005. Each lateral strain is -0.25 × 0.0005 = -0.000125. Hence the first-order volumetric strain is 0.0005 - 0.000125 - 0.000125 = 0.00025. The original cube volume is 100³ = 1000000 mm³, so the volume change is 0.00025 × 1000000 = 250 mm³. Therefore Option D is correct.
వివరణ:
Question 13
ప్రశ్న 13
A stepped steel bar carries a constant tensile force of 60 kN. Segment 1 is 1000 mm long with area 300 mm², and Segment 2 is 1500 mm long with area 600 mm². If E = 200 GPa throughout, what is the total elongation?
Explanation:
The total deformation of a stepped bar is the sum of the axial deformations of its individual segments. For Segment 1, δ1 = 60000 × 1000/(300 × 200000) = 1.00 mm. For Segment 2, δ2 = 60000 × 1500/(600 × 200000) = 0.75 mm. Therefore the total elongation is 1.00 + 0.75 = 1.75 mm. Hence Option B is correct.
వివరణ:
Question 14
ప్రశ్న 14
A 20 mm diameter steel rod of length 3 m has E = 200 GPa. The tensile stress must not exceed 120 MPa and the elongation must not exceed 1.5 mm. What is the maximum permissible axial tensile load considering both limits?
Explanation:
The rod area is π(20²)/4 = approximately 314.16 mm². The stress criterion permits P = 120 × 314.16 = about 37.70 kN. The deformation criterion gives P = δAE/L = 1.5 × 314.16 × 200000/3000 = about 31.42 kN. Both requirements must be satisfied simultaneously, so the smaller permissible load governs. Therefore the maximum allowable load is approximately 31.4 kN and Option C is correct.
వివరణ:
Question 15
ప్రశ్న 15
A circular prismatic bar remains within the linear elastic range while carrying the same axial load P over the same length. If its diameter is doubled without changing the material, what happens to both its average normal stress and its axial deformation?
Explanation:
The cross-sectional area of a circular bar is proportional to the square of its diameter. Doubling the diameter therefore increases its area by a factor of four. Since σ = P/A, the average normal stress becomes one-fourth of its original value. Since δ = PL/(AE) and P, L and E remain unchanged, the deformation also becomes one-fourth. Therefore Option A correctly describes both effects.
వివరణ:
Question 16
ప్రశ్న 16
A stepped axial member carries the same internal tensile force through two adjacent segments, with no external load applied between them. Segment A has twice the cross-sectional area of Segment B. Which statement about their average normal stresses is correct?
Explanation:
With no intermediate external axial load, equilibrium requires the same internal axial force to pass through the two segments. Average normal stress is equal to the internal force divided by cross-sectional area. Segment B has half the area of Segment A, so its stress is twice as large. Young's modulus affects strain and deformation but is not needed to determine the average stress from a known axial force and area. Therefore Option B is correct.
వివరణ:
Question 17
ప్రశ్న 17
Consider the following statements about simple axial stress and strain:
1. Average normal stress under concentric axial loading is P/A.
2. Longitudinal normal strain is dimensionless.
3. Elastic axial deformation of a prismatic bar under a given P, L and A is independent of Young's modulus.
4. Within the linear proportional range, normal stress is proportional to normal strain.
Which statements are correct?
1. Average normal stress under concentric axial loading is P/A.
2. Longitudinal normal strain is dimensionless.
3. Elastic axial deformation of a prismatic bar under a given P, L and A is independent of Young's modulus.
4. Within the linear proportional range, normal stress is proportional to normal strain.
Which statements are correct?
Explanation:
Statement 1 is correct for a concentrically loaded member when average normal stress is being considered. Statement 2 is correct because longitudinal strain is a ratio of two lengths and is therefore dimensionless. Statement 3 is incorrect because δ = PL/(AE), so increasing Young's modulus decreases elastic axial deformation for otherwise fixed quantities. Statement 4 is correct within the linear proportional range described by Hooke's law. Therefore Statements 1, 2 and 4 are correct and Option C is the unique answer.
వివరణ:
Question 18
ప్రశ్న 18
Match the quantity with the appropriate simple relationship for the stated elementary loading conditions:
1. Average normal stress
2. Longitudinal strain
3. Poisson's ratio
4. Elastic elongation of a uniform prismatic bar
P. δ/L
Q. PL/(AE)
R. P/A
S. - lateral strain/longitudinal strain
Which matching is correct?
1. Average normal stress
2. Longitudinal strain
3. Poisson's ratio
4. Elastic elongation of a uniform prismatic bar
P. δ/L
Q. PL/(AE)
R. P/A
S. - lateral strain/longitudinal strain
Which matching is correct?
Explanation:
Average normal stress is P/A, so item 1 matches R. Longitudinal strain is deformation divided by original length, so item 2 matches P. Poisson's ratio is minus the lateral strain divided by the longitudinal strain, so item 3 matches S. Elastic elongation of a uniform prismatic bar under constant axial load is PL/(AE), so item 4 matches Q. Therefore the correct sequence is 1-R, 2-P, 3-S and 4-Q, which is Option D.
వివరణ:
Question 19
ప్రశ్న 19
A 20 mm diameter pin transmits a load of 100 kN through a connection in double shear. Assuming the load is shared equally by the two shear planes, what is the average direct shear stress in the pin?
Explanation:
In double shear there are two resisting shear planes. The area of one shear plane is πd²/4 = π(20²)/4 = about 314.16 mm². The total resisting area is therefore about 628.32 mm². The average direct shear stress is 100000/628.32 = approximately 159.2 N/mm² or MPa. Hence Option B is correct under the equal-load-sharing assumption stated.
వివరణ:
Question 20
ప్రశ్న 20
A 20 mm diameter pin carries a total load of 90 kN equally through three direct shear planes. What is the average shear stress on each resisting plane?
Explanation:
The shear area of one 20 mm diameter circular plane is π(20²)/4 = approximately 314.16 mm². With three equally loaded shear planes, the total resisting area is 3 × 314.16 = approximately 942.48 mm². The average shear stress is therefore 90000/942.48 = approximately 95.5 N/mm². This calculation treats the stress on each plane as an average direct shear stress. Consequently Option A is correct.
వివరణ:
Question 21
ప్రశ్న 21
Which statement most accurately describes the applicability of the elementary form of Hooke's law, σ = Eε, in an axial stress-strain test?
Explanation:
The elementary relation σ = Eε describes a linear proportional relationship between normal stress and normal strain. It is therefore applicable over the range in which Young's modulus can be treated as a constant slope of the stress-strain curve. It should not be extended indiscriminately into nonlinear or plastic deformation. The proportional limit and elastic limit are related concepts but need not be numerically identical for every material. Hence Option C gives the most technically precise description.
వివరణ:
Question 22
ప్రశ్న 22
Two prismatic bars have the same length and carry the same axial tensile load. Bar A has E = 200 GPa and area 300 mm², while Bar B has E = 100 GPa and area 600 mm². Neglecting all other effects, which statement about their elastic elongations is correct?
Explanation:
Axial deformation of a prismatic bar is δ = PL/(AE). Since P and L are the same, the comparison depends on the axial rigidity AE. For Bar A, AE is proportional to 200 × 300 = 60000, while for Bar B it is 100 × 600 = 60000 in consistent relative units. Their axial rigidities are therefore equal. Consequently their elastic elongations are equal and Option D is correct.
వివరణ:
Question 23
ప్రశ్న 23
An aluminium bar is 1500 mm long, has an area of 600 mm² and Young's modulus E = 70 GPa. What axial tensile load will produce an elastic elongation of 1.8 mm?
Explanation:
For a uniform bar, δ = PL/(AE), so P = δAE/L. Substituting δ = 1.8 mm, A = 600 mm², E = 70000 N/mm² and L = 1500 mm gives P = 50400 N. Converting to kilonewtons gives 50.4 kN. The calculation assumes the aluminium remains within its linear elastic range. Therefore Option B is correct.
వివరణ:
Question 24
ప్రశ్న 24
A two-segment steel bar has E = 200 GPa throughout. Segment AB is 800 mm long with area 500 mm² and carries 100 kN tension, while Segment BC is 1200 mm long with area 400 mm² and carries 40 kN compression. Taking elongation as positive, what is the algebraic change in the total length?
Explanation:
The deformation of Segment AB is positive because its internal axial force is tensile. Its elongation is 100000 × 800/(500 × 200000) = +0.80 mm. Segment BC is in compression, so its deformation is 40000 × 1200/(400 × 200000) = 0.60 mm in magnitude and is taken as -0.60 mm. The algebraic total is therefore +0.80 - 0.60 = +0.20 mm. Hence the complete bar undergoes a net elongation of 0.20 mm and Option A is correct.
వివరణ:
Question 25
ప్రశ్న 25
Consider the following statements for an isotropic material under small, uniaxial, linear-elastic deformation using tension-positive normal strain:
1. Tensile longitudinal strain is positive.
2. For an ordinary material with positive Poisson's ratio, lateral strain under uniaxial tension has the same sign as longitudinal strain.
3. To first order, volumetric strain is the sum of three mutually perpendicular normal strains.
4. For a uniform axial bar with fixed P, L and A, doubling Young's modulus halves its elastic deformation.
Which statements are correct?
1. Tensile longitudinal strain is positive.
2. For an ordinary material with positive Poisson's ratio, lateral strain under uniaxial tension has the same sign as longitudinal strain.
3. To first order, volumetric strain is the sum of three mutually perpendicular normal strains.
4. For a uniform axial bar with fixed P, L and A, doubling Young's modulus halves its elastic deformation.
Which statements are correct?
Explanation:
Statement 1 is correct under the explicitly stated tension-positive strain convention. Statement 2 is incorrect because an ordinary material with positive Poisson's ratio contracts laterally under longitudinal tension, giving lateral strain the opposite sign. Statement 3 is correct for infinitesimal strains because the first-order volumetric strain is εx + εy + εz. Statement 4 is correct because δ = PL/(AE), so doubling E while all other stated quantities remain fixed halves the deformation. Therefore Statements 1, 3 and 4 are correct and Option C is the unique answer.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: B. P/A సమాధానం: B.
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Question 2 ప్రశ్న 2Answer: C. Tensile stress is positive and compressive stress is negative. సమాధానం: C.
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Question 3 ప్రశ్న 3Answer: A. Stress has dimensions of force per area, while strain is dimensionless. సమాధానం: A.
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Question 4 ప్రశ్న 4Answer: D. 200 MPa సమాధానం: D.
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Question 5 ప్రశ్న 5Answer: B. 0.00050 సమాధానం: B.
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Question 6 ప్రశ్న 6Answer: C. 200 GPa సమాధానం: C.
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Question 7 ప్రశ్న 7Answer: A. 2.0 mm సమాధానం: A.
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Question 8 ప్రశ్న 8Answer: D. -2.50 mm సమాధానం: D.
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Question 9 ప్రశ్న 9Answer: B. 0.25 సమాధానం: B.
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Question 10 ప్రశ్న 10Answer: C. Decrease of 0.0072 mm సమాధానం: C.
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Question 11 ప్రశ్న 11Answer: A. 0.00024 సమాధానం: A.
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Question 12 ప్రశ్న 12Answer: D. 250 mm³ సమాధానం: D.
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Question 13 ప్రశ్న 13Answer: B. 1.75 mm సమాధానం: B.
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Question 14 ప్రశ్న 14Answer: C. 31.4 kN సమాధానం: C.
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Question 15 ప్రశ్న 15Answer: A. Both become one-fourth of their original values. సమాధానం: A.
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Question 16 ప్రశ్న 16Answer: B. The average stress in Segment B is twice that in Segment A. సమాధానం: B.
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Question 17 ప్రశ్న 17Answer: C. 1, 2 and 4 only సమాధానం: C.
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Question 18 ప్రశ్న 18Answer: D. 1-R, 2-P, 3-S, 4-Q సమాధానం: D.
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Question 19 ప్రశ్న 19Answer: B. 159.2 MPa సమాధానం: B.
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Question 20 ప్రశ్న 20Answer: A. 95.5 MPa సమాధానం: A.
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Question 21 ప్రశ్న 21Answer: C. It represents the linear proportional stress-strain relationship within the material's proportional range. సమాధానం: C.
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Question 22 ప్రశ్న 22Answer: D. The two bars have equal elongation. సమాధానం: D.
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Question 23 ప్రశ్న 23Answer: B. 50.4 kN సమాధానం: B.
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Question 24 ప్రశ్న 24Answer: A. +0.20 mm సమాధానం: A.
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Question 25 ప్రశ్న 25Answer: C. 1, 3 and 4 only సమాధానం: C.
