A completely general way of describing the constitutive equation of a linear elastic material (Hooke's law) is by the following equation:
T = L . E or Tij = Lijkl . Ekl | ||
Where: | T | is the stress tensor |
L | is the elasticity tensor with a total of 81 indices | |
E | is the strain tensor |
The assumptions implicit in this equation is that the reference configuration, i.e. E = 0 also is stress free, i.e. T = 0. For most materials the above equation is valid only for small strains and rotations.
Note that the strain tensor is also described by the differential equation:
Eij = (ui,j + uj,i) e.g. E12 = (dx/dy + dy/dx)
Before starting to simplify the strain tensor we take a look at the elasticity tensor complete with the indices:
The first step in reducing the elasticity tensor is to check equilibrium as this demands that the stress tensor is symmetric. Mathematically this can be written:
Tij = Tji => Lijkl = Ljikl
From the equation above we see that this implies that Ljikl= Lijkl and the number of indices in the elasticity tensor can be reduced by 27 to 54.
Likewise the strain tensor is symmetric:
LijklEkl = ½ LijklEkl + ½ LijlkEkl => Elk = Ekl
From the equation above we see that this implies that Ljikl= Ljilk and the number of indices can be reduced by 18 to36.
Thus the elasticity tensor for linearly elastic anisotropic materials include 36 independent elements:
The indices of the elasticity matrix can be simplified if we introduce the following equalities:
T1=T11 | T2=T22 | T3=T33 | T4=T23 | T5=T32 | T6=T12 |
E1=E11 | E2=E22 | E3=E33 | E4=2E23 | E5=2E32 | E6=2E12 |
We can then simply formulate the relationship for an anisotropic linear elastic material as:
TK = LKLEL
In this equation the indices K and L run from 1 to 6, and L is thus a 6x6 matrix with the 36 independent elasticities as described above (Voigt matrix). Note that the new modified elasticity matrix L is no longer a tensor.