CONTINUUM MECHANICS for ENGINEERS

for which all directions are principal directions as explained in Section 3.6.
The classical physical example for this is the stress in a fluid at rest which
is termed hydrostatic stress, and for which σ M = –p 0 , the static pressure.
Every state of stress σ ij may be decomposed into a spherical portion and
a portion S ij known as the deviator stress in accordance with the equation
σ = S + δ σ = S + δ σ
1
3
ij ij ij M ij ij kk
(3.10-3)
where δ ij is the Kronecker delta. This equation may be solved for S ij, which
then appears in the symmetric matrix form
⎡S11
S12 S13⎤
⎡σ11
− σ M σ12 σ13
⎤
⎢
⎥ ⎢
⎥
⎢
S12 S22 S23⎥
=
⎢ σ12 σ22 − σ M σ23
⎥
⎣
⎢S13
S23 S33⎦
⎥
⎣
⎢ σ13 σ23 σ33 − σ M ⎦
⎥
(3.10-4)
Also from Eq 3.10-3, we notice immediately that the first invariant of the
deviator stress is
Sii σii δiiσkk (3.10-5)
(since δ ii = 3), so that the characteristic equation for the deviator stress
(analogous to Eq 3.6-6 for σ ij) is
for which the deviator stress invariants are
S 3 + II SS – III S = 0 (3.10-6)
1
II S = − = S IS II + S IIS III + S IIIS I (3.10-7a)
2 SS ij ji
III = S ε Sijk S1i S2j 3k
1
= − = 0
3
= S S ISII
III (3.10-7b)
( q)
Finally, consider a principal direction n j of σij such that the eigenvalue
[ ] =
[ ] =
equation σij − σ δ
( q) ij
q
n
( )
j 0 is satisfied. Then, from the definition of Sij, we
have Sij + σMδij −σ
δ
( q) ij
q
n
( )
j 0 , or
( )
⎡
q
Sij − σ −σ
n
q M δ
⎤ ( )
ij j = 0
⎣⎢ ( ) ⎦⎥
(3.10-8)