For a linearly elastic, homogeneous and isotropic material with modulus of rigidity G and modulus of elasticity E, the range of G is:
Try it first: pick an option, then read the explanation.
Asked in SSC JE 07/06/2024 (9 am–11 am)
Detailed Explanation
The elastic constants of an isotropic material are not independent. E and G are linked through Poisson's ratio μ, and μ itself can only take values in a limited range. Put the two limits of μ into the relation and the range of G follows.
\[ E = 2G(1+\mu) \;\Rightarrow\; G = \frac{E}{2(1+\mu)} \]
For an isotropic material \(0 \le \mu \le 0.5\).
\(\mu = 0:\; G = \dfrac{E}{2}\) \(\mu = 0.5:\; G = \dfrac{E}{2(1.5)} = \dfrac{E}{3}\)
Why “E/3 to E/2” Is Correct
G falls as μ rises, so its largest value is E/2 (at μ = 0) and its smallest is E/3 (at μ = 0.5). Hence E/3 ≤ G ≤ E/2, which is option (d).
| Relation | Formula | What it tells you |
|---|---|---|
| E and G | E = 2G(1 + μ) | G = E/2 when μ = 0; G = E/3 when μ = 0.5 |
| E and K | E = 3K(1 − 2μ) | K → ∞ when μ = 0.5 (incompressible) |
| E, G and K | E = 9KG/(3K + G) | Links all three moduli without μ |
| Poisson's ratio | 0 ≤ μ ≤ 0.5 (isotropic) | Steel ≈ 0.3, concrete ≈ 0.15–0.2, rubber ≈ 0.5 |
Why the Other Options Are Wrong
- A. 0 to E/2: G cannot drop to zero; that would need an infinite Poisson's ratio.
- B. 0 to E/3: The upper limit is E/2, not E/3, and the lower limit is not zero.
- C. E/2 to E/4: E/4 is below the true lower limit E/3.
Key Point
E = 2G(1 + μ) and 0 ≤ μ ≤ 0.5 ⇒ E/3 ≤ G ≤ E/2.
Key Concepts for Students
- For steel (μ ≈ 0.3), G ≈ E/2.6 ≈ 0.38E.
- μ = 0.5 means the material is incompressible (no volume change).
- Only two elastic constants are independent for an isotropic material.
Practice more: this is question 1 in Bending Moment and Shear Force MCQs – Part 1. Try the whole set and check every answer instantly.