CONTINUUM MECHANICS for ENGINEERS

( q)
( q)
σ = (λδijεkk + 2µε ijn j
ij) n j
q
= λ ε + 2µε
( q)
ij n j
( q)
But n j and ε (q) satisfy the fundamental equation for the eigenvalue problem,
namely, εij = ε (q)δij , so that now
n ( q)
j n ( q)
j
( q)
σ ijn j
q
= n +
= [ λε + ]
kk 2µε q
and because εkk = ε (1) + ε (2) + ε (3) is the first invariant of strain, it is constant
for all ( q)
so that
n i
( q)
σ ijn j
( )
=
n ( )
i kk
( )
λεkk i
2µε ( q)
( n q) i
( q)
( ) ni λε [ ( ε ε 2µε
1) + ( 2) + ( 3) ] ( ) +
{ q } ni
q
This indicates that n (q = 1, 2, 3) are principal directions of stress also,
i
with principal stress values
= (q = 1, 2, 3)
σ ( q)
λε [ ( ε ε 2µε
1) + ( 2) + ( 2) ] ( ) + q
We may easily invert Eq 6.2-2 to express the strain components in terms
of the stresses. To this end, we first determine ε ii in terms of σ ii from Eq 6.2-2
by setting i = j to yield
σ ii = 3λε kk + 2µε ii = (3λ + 2µ)ε ii
(6.2-3)
Now, by solving Eq 6.2-2 for ε ij and substituting from Eq 6.2-3, we obtain the
inverse form of the isotropic constitutive equation,
1
2µ 3 2
σ
λ
λ µ δσ
⎛
⎞
⎝ + ⎠
εij = ⎜
(6.2-4)
ij +
ij kk⎟
By a formal — although admittedly not obvious — rearrangement of this
equation, we may write
εij = ⎨⎢
+ ⎥ −
⎬ (6.2-5)
ij ij kk
( q)
λ µ
λ
µ λ µ λ µ σ
λ
λ µ δσ
+ ⎧⎪
⎡ ⎤
⎫⎪
1
( 3 + 2 ) ⎩⎪ ⎣ 2( + ) ⎦ 2(
+ ) ⎭⎪