Strength of Materials & Elasticity

### **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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