# What are the elastic constants for tetragonal zirconia polycrystal (TZP)?

I have six known elastic constants of TZP. What is the relationship between the known and the unknown constants?

Known constants (GPa):

$c_{11}=327\\ c_{12}=100\\ c_{13}=62\\ c_{33}=264\\ c_{44}=59\\ c_{66}=64$

How can I find the others? Is $c_{22}=c_{11}$? Does TZP exibhit transverse isotropy?

Are the relationships between the tensor components and the coordinate axes given by: $c_{11}$ for x-direction, $c_{22}$ for y and $c_{33}$ for z; $c_{23}$ (4) for yz, $c_{31}$ (5) for zx, and $c_{12}$ (6) for xy?

• Can you give us some context for how you were presented with these constants? The 6 you have are enough to define an isotropic material, and while I can't verify that TZP is isotropic, that may be what is being implied here. – Trevor Archibald Apr 17 '15 at 22:11
• Well, TZP doesn't seem like isotropy ..... You may just search " elastic constants for tetragonal zirconia " in google , you will find a context giving the constants. I give a link here: onlinelibrary.wiley.com/doi/10.1111/j.1151-2916.1998.tb02533.x/… – Jimmy Apr 18 '15 at 12:38

Generalized Linear Elasticity Tensor

In general, stress and strain can each be modeled by a 2nd-rank tensor with $3\times 3=9$ elements. A full linear relationship between the stress and strain tensors requires each element to be paired with every other element, giving a 4th-rank tensor with $9\times 9=81$ elements. By symmetry of the stress tensor, i.e. $\sigma_{ij}=\sigma_{ji}$, every element $C_{ijkl}=C_{jikl}$ in the full elasticity tensor. The number of unique elements is now 54, since one of every three elements has a symmetric partner. Likewise by symmetry of the strain tensor, i.e. $\varepsilon_{ij}=\varepsilon_{ji}$, every element $C_{ijkl}=C_{ijlk}$ in the full elasticity tensor. The number of unique elements is now 36, again a reduction of one-third. There is an additional strain-energy symmetry because the strain energy contribution of one elastic constant depends on $C_{ijkl}\varepsilon_{ij}\varepsilon_{kl}$, and reversing the order to $C_{klij}\varepsilon_{kl}\varepsilon{ij}$ should have no effect on the total strain energy, giving $C_{ijkl}=C_{klij}$. The number of unique elements is now 21.

Voigt Reduced Elasticity Tensor

Using Voigt notation, the 21 elements may be expressed in a symmetric $6\times 6$ 2nd-rank tensor or matrix, which relates a $6$-element 1st-rank stress tensor or vector to a similar strain vector. The elements of the stress and strain vectors are $\sigma_{ii}\rightarrow\sigma_{i}$ and $\varepsilon_{ii}\rightarrow\varepsilon_{i}$ for axial stresses and strains, and $\sigma_{23}\rightarrow\sigma_{4}$, $\sigma_{31}\rightarrow\sigma_{5}$ and $\sigma_{12}\rightarrow\sigma_{6}$ for shear stresses, with corresponding doubled shear strains $2\varepsilon_{23}\rightarrow\varepsilon_{4}$, $2\varepsilon_{31}\rightarrow\varepsilon_{5}$ and $2\varepsilon_{12}\rightarrow\varepsilon_{6}$. I will refer to all shear stresses as $\tau_{i}$ and strains as $\gamma_{i}$. The coefficients then go from 4 subindices to 2 subindices, each ranging from 1 to 6, and I will write with lower case as $c_{ij}$, as the OP has.

Hookean Elasticity

Assuming Hookean linear elasticity, 9 of the 21 elements of the Voigt tensor immediately vanish, specifically those relating shear stresses ($\tau_{ij}$) to axial strains ($\varepsilon_{kl}$) and axial stresses ($\sigma_{ij}$) to shear strains ($\gamma_{kl}$), leaving 12 elements. The elements thus eliminated are 14, 15, 16, 24, 25, 26, 34, 35, and 36.

Three of the remaining 12 relate shear stress in one sense to shear strain in another sense (e.g. $\tau_{12}$ to $\gamma_{23}$), and thus also vanish under Hookean linear elasticity. The elements thus eliminated are 45, 56, and 46.

Tetragonal Symmetry in Elasticity

Only 9 elements remain, in three categories:

Axial: $c_{11},c_{22},c_{33}$

Axial due to Poisson effect: $c_{12},c_{13},c_{23}$

Shear: $c_{44},c_{55},c_{66}$

By symmetry of the tetragonal lattice, lattice parameters $a_{1}=a_{2}\neq a_{3}$:

$c_{11} = c_{22}$ because the elastic response must be the same in direction $1$ and $2$ due to the lattice symmetry.

$c_{13} = c_{23}$ because the Poisson effect is similarly symmetric.

$c_{44} = c_{55}$ because the elastic shear response must be the same in the sense of direction pairs $1,3$ ($c_{44}$) and $2,3$ ($c_{55}$), since directions $1$ and $2$ are symmetric.

$c_{33}$ and $c_{66}$ do not share any symmetries.

This leaves 6 unique elements, which are the ones you have, i.e. $c_{11},c_{12},c_{13},c_{33},c_{44},c_{66}$.