For a material with Young’s modulus E = 200 GN/m² and modulus of rigidity G = 80 GN/m², what is the value of Poisson’s ratio (μ)?
Mechanical Engineering
Strength of Materials
Difficulty: Medium
Choose an option
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A0.15
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B0.20
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C0.25
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D0.30
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E0.40
Answer
Correct Answer: 0.25
Explanation
Introduction / Context:Poisson’s ratio is a key elastic constant that relates lateral strain to longitudinal strain in materials. It can be derived from the relationship between Young’s modulus (E) and shear modulus (G).
Given Data / Assumptions:
- E = 200 GN/m².
- G = 80 GN/m².
- Isotropic, homogeneous, linearly elastic material.
Concept / Approach:The relation between E, G, and μ is:E = 2 * G * (1 + μ) Rearranging: μ = (E / (2 * G)) - 1
Step-by-Step Solution:μ = (200 / (2 * 80)) - 1= (200 / 160) - 1= 1.25 - 1= 0.25
Verification / Alternative check:Cross-check using bulk modulus relation: E = 3K(1 - 2μ). Substituting μ = 0.25 maintains consistency for isotropic elasticity.
Why Other Options Are Wrong:
- 0.15, 0.20: underestimate value.
- 0.30, 0.40: overestimate value.
Common Pitfalls:
- Using wrong formula (mixing with bulk modulus relation).
- Incorrect unit handling (ensure consistency in GN/m²).
Final Answer:0.25