Question Statement:
The number of independent elastic constants for a homogeneous isotropic linear elastic material is _________.
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2019 Q.32 / AE Q.22)**
**1. Intuitive Physical Setup & Given Parameters**:
- Material symmetry classification: Homogeneous and Isotropic linear elastic material.
**2. Step-by-Step Mathematical Derivation**:
1. **General Anisotropic 3D Material**:
- The general 4th-rank stiffness tensor $C_{ijkl}$ possesses 81 components.
- Minor symmetries of stress and strain tensors ($\sigma_{ij} = \sigma_{ji}, \epsilon_{kl} = \epsilon_{lk}$) reduce 81 to 36.
- Major thermodynamic potential symmetry ($C_{ijkl} = C_{klij}$) reduces 36 to **21 independent constants**.
2. **Orthotropic Material**:
- Possesses 3 mutually orthogonal planes of symmetry $\implies$ **9 independent constants**.
3. **Transversely Isotropic Material**:
- Possesses one axis of rotational symmetry $\implies$ **5 independent constants**.
4. **Isotropic Material**:
- Invariant under arbitrary spatial rotations and reflections.
- Any two independent constants (such as Young's modulus $E$ and Poisson's ratio $\nu$, or Lamé constants $\lambda$ and $\mu$) completely specify all other elastic moduli ($G, K$) via:
$G = \frac{E}{2(1+\nu)}, \quad K = \frac{E}{3(1-2\nu)}$
- Number of independent elastic constants = **2**.
**3. Comprehensive Options & Distractor Analysis**:
- **Question Type**: Numerical Answer Type (NAT).
- **Exact Value**: **2** (Acceptable Range: [2, 2]).
**4. Official Master Key & Exam Takeaway**:
- **Official Answer**: **2.0**
- **Exam Summary Hierarchy**: Anisotropic = 21, Orthotropic = 9, Transversely Isotropic = 5, Isotropic = **2**.
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