Calculate the bulk modulus of an alloy whose modulus of elasticity is 180 GPa and Poisson's ratio is 0.25.
Try it first: pick an option, then read the explanation.
Correct answer: A. 120 GPa
Asked in SSC JE 06/11/2024 (Morning)
Detailed Explanation
E, K and μ are linked by E = 3K(1 − 2μ).
\[ K = \frac{E}{3(1 - 2\mu)} = \frac{180}{3(1 - 0.5)} = \frac{180}{1.5} = 120\ \text{GPa} \]
Why “120 GPa” Is Correct
Substituting gives K = 120 GPa.
| Relation | Formula |
|---|---|
| Young's modulus and rigidity | E = 2G(1 + μ) |
| Young's modulus and bulk modulus | E = 3K(1 − 2μ) |
| All three moduli | E = 9KG/(3K + G) |
| Poisson's ratio | μ = (3K − 2G)/(6K + 2G) |
Only two elastic constants are independent for a homogeneous, isotropic, linearly elastic material.
Why the Other Options Are Wrong
- B. 140 GPa: 140 GPa does not satisfy E = 3K(1 − 2μ).
- C. 100 GPa: 100 GPa does not satisfy the relation.
- D. 160 GPa: 160 GPa does not satisfy the relation.
Key Point
K = E/[3(1 − 2μ)]; when μ = 0.25, K = 2E/3.
Key Concepts for Students
- When μ = 1/3, K = E.
- As μ → 0.5, K → ∞ (incompressible).
- G = E/[2(1 + μ)] = 72 GPa for this alloy.
Practice more: this is question 3 in Stress and Strain MCQs – Part 1. Try the whole set and check every answer instantly.