A note about this meta post:
I just have one example right now, but I will have three in the final.
I thought it would be interesting to have a problem about something I know a bit about. Right now it seems a bit mathy, but I wanted to ground the problem on something real. It feels more 'real-life' if you need to understand the spec in addition to golfing. The problem is I don't want it to seem like homework. Another problem is the actual computation that needs to take place isn't actually that hard once you understand the simplifications of the problem.
Let me know what you think.
\$\def\tensor#1{\smash{\underline{\underline{\smash{#1}}}}}\$
Challenge
Calculate the strain tensor and volume percent change of a cube given its material properties and stress tensor.
Background
Common Terms

Strain: ε, The amount of elongation per unit length, Units: \$\frac{in}{in}\$
Normal Stress: σ, The amount of force per unit area perpendicular to the cross section, Units: \$\frac{lbs}{in^2} = psi\$
Shear Stress: τ, The amount of force per unit area parallel to the cross section, Units: \$\frac{lbs}{in^2} = psi\$
Young's Modulus: E, The relationship between stress and strain: \$σ = Eε\$, Units: psi

Poisson's Ratio: ν, The relationship between strain in different directions.
For a uniaxial bar: \$ε_{22} = -ν ε_{11}\$, Units: unitless
Index Notation: A short form for tensors written with subscripts \$i,j,k,l\$ to denote which element within the tensor. The number of subscripts the tensor has indicates what order it is. \$σ_{ij} \equiv \tensor{σ}\$ (Second Order)
Kroniker Delta: \$δ_{ij}\$, has the value of 1 if i=j, otherwise its value is 0. Index Notation for the Identity Tensor.
$$
δ_{ij} =
\left[\begin{array}{ccc}
δ_{11} & δ_{12} & δ_{13}\\
δ_{21} & δ_{22} & δ_{23}\\
δ_{31} & δ_{32} & δ_{33}\\
\end{array}\right] =
\left[\begin{array}{ccc}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1\\
\end{array}\right] = I
$$
Tensors
A tensor in this context can be thought of as three directional components for each of the three positive faces of the cube.

Although a Stress Tensor is not a matrix, it can be represented in matrix form by a 3x3 or other matrices:
$${\tensor{σ}} =
\left[\begin{array}{ccc}
σ_{xx} & σ_{xy} & σ_{xz}\\
σ_{yx} & σ_{yy} & σ_{yz}\\
σ_{zx} & σ_{zy} & σ_{zz}\\
\end{array}\right] =
\left[\begin{array}{ccc}
σ_{11} & σ_{12} & σ_{13}\\
σ_{21} & σ_{22} & σ_{23}\\
σ_{31} & σ_{32} & σ_{33}\\
\end{array}\right] =
\left[\begin{array}{c}
σ_{11}\\
σ_{12}\\
\vdots \\
σ_{33}\\
\end{array}\right] =
\left[\begin{array}{cccc}
σ_{11} & σ_{12} & \dots & σ_{33}\\
\end{array}\right]
$$
If a Tensor and a Kroniker Delta share the same indices, they are combined.
$$
σ_{ij} δ_{kj} = σ_{ik}
$$
If a Tensor has repeating indices, then it is taken as a zero order tensor and the indices are summed.
$$
σ_{ij} δ_{ij} = σ_{ii} = \sum_{i=1}^{3} σ_{ii} \equiv tr(\tensor{σ}) = σ_{11} + σ_{22} + σ_{33}
$$
Stress Strain Relationship
The relationship in one dimension is \$σ=Eε\$. For three dimensions we can use Hook's law to find the relationship between the strain tensor and the stress tensor as follows:
$$
σ_{ij} = C_{ijkl} ε_{kl}
$$
This general case would need \$3^4 = 81\$ independent material properties to calculate the strain tensor. If we assume the cube has a symetric \$\tensor{σ}\$, symetric \$\tensor{ε}\$, is Elastic, Isotropic, Linear, and Homogeneous, then we only need two independent material properties: (Young's modulus: E, Poisson's Ratio: ν) or (Lamé modulus: λ, Shear modulus: μ). We can use either pair of values, but for this example it is much easier to calculate the strain tensor from the stress tensor using Young's Modulus and Poisson's Ratio.
And so we can simplify the stress tensor by what we know.
$${\tensor{σ}} =
\left[\begin{array}{ccc}
σ_x & σ_{xy} & σ_{xz}\\
& σ_y & σ_{yz}\\
Sym & & σ_z\\
\end{array}\right]
$$
And our new relationship is:
$$
ε_{ij} = \frac{1+ν}{E} σ_{ij} - \frac{ν}{E} σ_{kk} δ_{ij}
$$
Calculating the Dilation

The Volumetric Strain can be found by calculating the trace of the strain tensor for very small values of strain. This is because for small values \$ε^3 \ll ε^2 \ll ε\$.
$$
{\frac {ΔV}{V_0}} \approx tr(\tensor{ε})
$$
Putting it all together
Therefore, in summary we can calculate the strain tensor with the following:
Using Index Notation:
$$
ε_{ij} = \frac{1+ν}{E} σ_{ij} - \frac{ν}{E} σ_{kk} δ_{ij}
$$
Using Matrix Notation:
$$
ε_{ij} = \frac{1+ν}{E}
\left[\begin{array}{ccc}
σ_{11} & σ_{12} & σ_{13}\\
σ_{12} & σ_{22} & σ_{23}\\
σ_{13} & σ_{23} & σ_{33}\\
\end{array}\right]
- \frac{ν}{E} tr(\tensor{σ})
\left[\begin{array}{c}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1\\
\end{array}\right]
$$
Then calculate the volumetric change:
Index Notation
$$
{\frac {ΔV}{V_0}} = ε_{ij} δ_{ij} = ε_{ii}
$$
Matrix Notation
$$
{\frac {ΔV}{V_0}} = tr(\tensor{ε})
$$
Input
One positive long: Young's Modulus (Usually in the range 100 psi - 100,000,000 psi)
One positive decimal: Poissons's Ratio (Usually in the range 0.01 - 0.5)
Array of signed decimals for Stress Tensor (9 values)
Output
Array of Strain Tensor, to at least five significant figures (same format as input)
Percent volume change of cube, to at least five significant figures
Examples
(simple example for now)
Input:
Young's Modulus: 29,000,000 psi
Poisson's Ratio: 0.30
Stress Tensor: [[50000, 0, 0]
[ 0, -10000, 0]
[ 0, 0, 25000]] psi
Output:
Strain Tensor: [[0.00157, 0, 0]
[ 0, -0.00112, 0]
[ 0, 0, 0.00045]] in/in
Dilation: 0.08966%
Rules
IO is flexible
This is code-golf, least number of bytes for each language wins