Simple Stress and Strain MCQ with answer and solution

Concept

Material symmetry planes reducing elastic constants

Formula

C_ijkl with symmetry planes

Step-by-step solution

1. In 3D generalized Hooke's law, 81 elastic stiffness coefficients relate stress to strain tensors. 2. Stress tensor symmetry (σ_ij = σ_ji) and strain tensor symmetry (ε_kl = ε_lk) reduce the constants to 36. 3. Hyperelastic energy conservation (strain energy function) further reduces the matrix to 21 constants. 4. The subsequent reduction from 21 to 13 (monoclinic), 9 (orthotropic), 5 (transversely isotropic), and 2 (isotropic) is due to material symmetry. Answer: B

Explanation

A generalized 3D anisotropic elastic material requires 21 independent elastic stiffness coefficients after accounting for stress/strain tensor symmetries and strain energy conservation. Any further reduction in the number of independent constants (e.g. 13 for monoclinic, 9 for orthotropic, 2 for isotropic) is achieved through material symmetry planes.

Why the other options are wrong

A) Isotropy is the extreme end-member case with full symmetry, not the general cause of reduction across all anisotropic classes. C) Strain compatibility ensures continuous deformation, not constant reduction. D) Stress/strain tensor symmetry establishes the 36-constant symmetric matrix.

Shortcut

Reduction below 21 constants = Planes and axes of material symmetry.

Common mistake

Confusing tensor symmetry (which yields 21 constants) with material symmetry (which reduces below 21).