Question

The number of independent elastic constants that a fully anisotropic linear elastic material can have is .

• A. 36
• B. 21
• C. 10
• D. 2

Hint:

Use the symmetry of the stiffness matrix from strain energy to reduce the 36 raw entries of a 6 by 6 matrix.

Answer 1

The Correct Option is B

Step 1: Set up the generalized Hooke's law.
For a linear elastic material, stress is related to strain through the stiffness matrix \(C\):
\[ \sigma_{ij} = C_{ijkl}\,\varepsilon_{kl} \]
Written in the compact 6 by 6 matrix (Voigt) notation, stress and strain each have 6 independent components (3 normal + 3 shear), so \(C\) is a 6 by 6 matrix, which has \(6 \times 6 = 36\) entries.

Step 2: Reduce 36 using the strain energy argument.
For an elastic material, the strain energy density \(U\) is a scalar function of strain, and stress is obtained as \(\sigma_i = \partial U/\partial \varepsilon_i\).
Because mixed partial derivatives of \(U\) can be taken in either order, \(C_{ij} = C_{ji}\), so the stiffness matrix must be symmetric.

Step 3: Count the independent terms of a symmetric 6 by 6 matrix.
A symmetric \(6 \times 6\) matrix has 6 diagonal terms plus \(\frac{6 \times 5}{2} = 15\) off-diagonal terms (the terms below the diagonal equal the ones above it), giving
\[ 6 + 15 = 21 \]
independent elastic constants.

Step 4: Note why the other numbers are wrong.
36 (option A) is the raw entry count of \(C\) before using symmetry. 10 and 2 (options C, D) are smaller counts that come from extra material symmetry (such as orthotropic or isotropic materials), not from the most general, fully anisotropic (triclinic) case, which is what the question asks about.

Final Answer:
A fully anisotropic linear elastic material has 21 independent elastic constants. \[ \boxed{21} \]