I have also thought about this question and have not seen an answer in the literature, but I will give you some thoughts. Unfortunately, the link you give is behind a pay wall.
First, strains do not transform as a 4-vector under rotations, boosts, or strains. Instead, they transform as part of a 2nd rank tensor $\Theta^{ab}$. The strains are group parameters like rotation angles and are measured in radians. We do a strain to an object. If we do a strain transformation to a vector (x,y,z), then the components get fractionally squashed/expanded or parallelopipeded as we have learned in a materials engineering class (where strains belonging to a group is never mentioned). In this answer to another question, I showed how rotations and strains are all the transformations of GL(3,R) and how to do them. Now, let's identify all the invertible 4x4 matrix transformations of a 4-vector (x,y,z,t), which is the group GL(4,R).
We can describe what these transformations do by just talking about the matrices $M$ that are very close to the identity matrix, where all elements in the matrix $\Theta$ are <<1. All these elements are in radians.
$$
M=I+\Theta
$$
$$
\Theta = \begin{bmatrix}
0 & \theta^{12} &-\theta^{13} & \theta^{14} \\
-\theta^{12} & 0 & \theta^{23} & \theta^{24} \\
\theta^{13} &-\theta^{23} & 0 & \theta^{34} \\
-\theta^{14} &-\theta^{24} &-\theta^{34} & 0 \\
\end{bmatrix}_{Antisymmetric}
\
+ \begin{bmatrix}
\epsilon^{11} & \epsilon^{12} & \epsilon^{13} & \lambda^1 \\
\epsilon^{12} & \epsilon^{22} & \epsilon^{23} & \lambda^2 \\
\epsilon^{13} & \epsilon^{23} & \epsilon^{33} & \lambda^3 \\
\lambda^1 & \lambda^2 & \lambda^3 & \epsilon^{44} \\
\end{bmatrix}_{Symmetric}
$$
By taking products of these matrices we build all the matrices $M$ in the group for any size elements in $\Theta$.
$$
M=e^{\Theta}=I+\Theta+\dfrac{\Theta^2}{2!}+ \dfrac{\Theta^3}{3!}+ \dfrac{\Theta^4}{4!} +…
$$
The $\theta$ are antisymmetric, make M orthogonal ($M^T =M^{-1}$), and leave the length ${x_1}^2+{x_2}^2+{x_3}^2+{x_4}^2$ invariant. Because lengths are invariant, the transformations are called rotations. The $\epsilon$ do not leave lengths invariant and are called strains. The $\lambda^1,\lambda^2,\lambda^3$ are the standard Lorentz Boost parameters which are space-time strains, though usually this is not pointed out.
Now the old (eg: for a,b=1,2,3 only) Hooke's law is that stress is proportional to strain for small strains
$$
T_{Symmetric}^{ab}=C^{ab}_{cd}\Theta_{Symmetric}^{cd}
$$
Presumably (guessing) this generalizes to the GL(4,R) covariant relation for a,b=1,2,3,4.
$$
T^{ab}=C^{ab}_{cd}\Theta^{cd}
$$
which may be the answer to you question. Notice that the stress energy tensor no longer has just symmetric components. This is because once we include strains, we can look at $T_{Symmetric}$ from a strained frame and we will find it has picked up antisymmetric components! This is because the commutator of two strains is a rotation (which is the reason a free falling cat can rotate itself by a sequence of strains to its body). $T_{Antisymmetric}$ can be identified as torque densities.
Masses in GR do strains. For example, the Schwarzschild metric is simultaneous stretch of t and squash of x (assuming x points toward the mass) by equal amounts ($\epsilon^{11}=-\epsilon^{44}=\frac{GM}{r}$). A gravity wave going in the z-direction does the squash/stretch strain ($\epsilon^{11}=-\epsilon^{22}=h_+$) or the parralelopiped strain which is called the other polarization ($\epsilon^{12}=\epsilon^{21}=h_X$).
These thoughts end here, uncompleted. Many things remain to be explained. For example, how do you do the space-time rotations ($\theta^{14},\theta^{24},\theta^{34}$)? Physically, what are Hooke's constants $C^{ab}_{cd}$ for mixed symmetries or when any of the indices are =4 ? Physically, what are the antisymmetric components of T ?
