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A bar produces a lateral strain of magnitude 60 x 10−5 when subjected to a tensile stress of magnitude 300 MPa along the axial direction. The elastic modulus of the material if the Poisson’s ratio is 0.3 is
- a)200 GPa
- b)150 GPa
- c)125 GPa
- d)100 GPa
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
A bar produces a lateral strain of magnitude 60 x 10−5 when subjected...
Lateral strain = Axial strain x poisson′ s ratio ∈y=∈x x μ
= 150 GPa
Community Answer
A bar produces a lateral strain of magnitude 60 x 10−5 when subjected...
Given:
- Lateral strain (ε) = 60 x 10^(-5)
- Tensile stress (σ) = 300 MPa
- Poisson's ratio (ν) = 0.3
To find:
Elastic modulus (E)
Formula:
Poisson's ratio is defined as the ratio of lateral strain to axial strain.
ν = -ε_lat / ε_axial
Elastic modulus is defined as the ratio of stress to strain.
E = σ / ε_axial
Solution:
Step 1: Calculate the axial strain (ε_axial) using Poisson's ratio and lateral strain.
ε_axial = -ε_lat / ν
ε_axial = -60 x 10^(-5) / 0.3
ε_axial = -2 x 10^(-4)
Step 2: Calculate the elastic modulus (E) using stress and axial strain.
E = σ / ε_axial
E = 300 x 10^6 / (-2 x 10^(-4))
E = -1500 x 10^9
E = 150 GPa
Answer:
The elastic modulus of the material, when the Poisson's ratio is 0.3, is 150 GPa. Therefore, the correct answer is option B.
- Lateral strain (ε) = 60 x 10^(-5)
- Tensile stress (σ) = 300 MPa
- Poisson's ratio (ν) = 0.3
To find:
Elastic modulus (E)
Formula:
Poisson's ratio is defined as the ratio of lateral strain to axial strain.
ν = -ε_lat / ε_axial
Elastic modulus is defined as the ratio of stress to strain.
E = σ / ε_axial
Solution:
Step 1: Calculate the axial strain (ε_axial) using Poisson's ratio and lateral strain.
ε_axial = -ε_lat / ν
ε_axial = -60 x 10^(-5) / 0.3
ε_axial = -2 x 10^(-4)
Step 2: Calculate the elastic modulus (E) using stress and axial strain.
E = σ / ε_axial
E = 300 x 10^6 / (-2 x 10^(-4))
E = -1500 x 10^9
E = 150 GPa
Answer:
The elastic modulus of the material, when the Poisson's ratio is 0.3, is 150 GPa. Therefore, the correct answer is option B.
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A bar produces a lateral strain of magnitude 60 x 10−5 when subjected to a tensile stress of magnitude 300 MPa along the axial direction. The elastic modulus of the material if the Poisson’s ratio is 0.3 isa) 200 GPab) 150 GPac) 125 GPad) 100 GPaCorrect answer is option 'B'. Can you explain this answer?
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A bar produces a lateral strain of magnitude 60 x 10−5 when subjected to a tensile stress of magnitude 300 MPa along the axial direction. The elastic modulus of the material if the Poisson’s ratio is 0.3 isa) 200 GPab) 150 GPac) 125 GPad) 100 GPaCorrect answer is option 'B'. Can you explain this answer? for Mechanical Engineering 2025 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A bar produces a lateral strain of magnitude 60 x 10−5 when subjected to a tensile stress of magnitude 300 MPa along the axial direction. The elastic modulus of the material if the Poisson’s ratio is 0.3 isa) 200 GPab) 150 GPac) 125 GPad) 100 GPaCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A bar produces a lateral strain of magnitude 60 x 10−5 when subjected to a tensile stress of magnitude 300 MPa along the axial direction. The elastic modulus of the material if the Poisson’s ratio is 0.3 isa) 200 GPab) 150 GPac) 125 GPad) 100 GPaCorrect answer is option 'B'. Can you explain this answer?.
A bar produces a lateral strain of magnitude 60 x 10−5 when subjected to a tensile stress of magnitude 300 MPa along the axial direction. The elastic modulus of the material if the Poisson’s ratio is 0.3 isa) 200 GPab) 150 GPac) 125 GPad) 100 GPaCorrect answer is option 'B'. Can you explain this answer? for Mechanical Engineering 2025 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A bar produces a lateral strain of magnitude 60 x 10−5 when subjected to a tensile stress of magnitude 300 MPa along the axial direction. The elastic modulus of the material if the Poisson’s ratio is 0.3 isa) 200 GPab) 150 GPac) 125 GPad) 100 GPaCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A bar produces a lateral strain of magnitude 60 x 10−5 when subjected to a tensile stress of magnitude 300 MPa along the axial direction. The elastic modulus of the material if the Poisson’s ratio is 0.3 isa) 200 GPab) 150 GPac) 125 GPad) 100 GPaCorrect answer is option 'B'. Can you explain this answer?.
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