192
14 On the Plasticity Conditions of an Isotropic Body
which coincides with formulas (1.16) and (1.17). By subtracting the mean stress
from the three first Eqs. (14.25), we obtain
σ x − σ 0 = 2G(ε x − ε 0 ),
σ y − σ 0 = 2G(ε y − ε 0 ),
σ z − σ 0 = 2G(ε z − ε 0 );
τ xy = 2Gε xy ,
τ yz = 2Gε yz ,
τ zx = 2Gε zx ,
(14.26)
where
ε ij =
1
2
γ ij , (i, j ∼ x, y, z).
In the tensor form, dependencies (14.26) are represented by formulas (1.16)
D σ = 2GD ε , σ 0 = Kε 0 .
Let us designate deviator guides of the stress and strain tensor using D σ and D ε ,
respectively.
D σ =
1
τ 0
D σ , D ε =
2
γ 0
D ε .
(14.27)
Taking into account that in the elastic stage of strain
τ 0 = Gγ 0 ,
(14.28)
the law (1.16), taking into account formulas (14.27) and (14.28), can finally be
represented as follows:
D σ = D ε .
(14.29)
In this manner, Hooke’s law formulation is represented by the following three
dispositions.
1. Linear invariants of the stress (T σ ) and strain (T ε ) tensors are proportional; or
otherwise, changes in the body element volume are proportional to the mean
nominal stress:
σ 0 = Kε 0 .
(14.30)
2. Guides of the stress and strain tensors coincide
D σ = D ε .
(14.31)