86
3 Continuum Mechanics and Nonlinear Elasticity
are strain invariants, which are independent of coordinates. To write these invariants
in terms of components defined by E = E ij e i e j , we need to compute
E
2
= E · E = (E ij e i e j ) · (E kl e k e l )
= E ij E kl (e j · e k )e i e l = E ij E kl δ jk e i e l
= E ij E jl e i e l = E ik E kj e i e j ,
and Eq. (2.21) yields
tr E = E ii
tr E
2
= E ik E kj e i · e j = E ik E kj δ ij = E ik E ki = E ik E ik .
(3.67)
The strain invariants become
¯
I 1 = E ii
¯
I 2 =
1
2 (E ii E jj − E ij E ij )
¯
I 3 = det[E ij ].
(3.68)
Clearly, any combination of ¯
I 1 , ¯
I 2 , and ¯
I 3 is also invariant. One particularly useful
form is given by
I 1 = 3 + 2 ¯
I 1 = 3 + 2E ii
I 2 = 3 + 4( ¯
I 1 + ¯
I 2 ) = 3 + 4E ii + 2(E ii E jj − E ij E ij )
I 3 = 1 + 2 ¯
I 1 + 4 ¯
I 2 + 8 ¯
I 3 = det
δ ij + 2E ij
.
(3.69)
For some problems, it is possible to choose a coordinate system where the base
vectors align with the principal directions at every point. Examples include uniaxial
stretching of a bar, biaxial stretching of a membrane, and inflation of an isotropic
circular cylinder. In these cases, we can solve the problem without dealing directly
with shear, and the governing equations become much simpler.
For example, the dyadic representation of the strain tensor in principal coordinates is
E = E 1 N 1 N 1 + E 2 N 2 N 2 + E 3 N 3 N 3 ,
(3.70)
where the E i are principal strains, and the invariants of (3.69) become
I 1 = 3 + 2(E 1 + E 2 + E 3 )
I 2 = 3 + 4(E 1 + E 2 + E 3 ) + 4(E 1 E 2 + E 2 E 3 + E 3 E 1 )
I 3 = (1 + 2E 1 )(1 + 2E 2 )(1 + 2E 3 ).
(3.71)
3 Continuum Mechanics and Nonlinear Elasticity
are strain invariants, which are independent of coordinates. To write these invariants
in terms of components defined by E = E ij e i e j , we need to compute
E
2
= E · E = (E ij e i e j ) · (E kl e k e l )
= E ij E kl (e j · e k )e i e l = E ij E kl δ jk e i e l
= E ij E jl e i e l = E ik E kj e i e j ,
and Eq. (2.21) yields
tr E = E ii
tr E
2
= E ik E kj e i · e j = E ik E kj δ ij = E ik E ki = E ik E ik .
(3.67)
The strain invariants become
¯
I 1 = E ii
¯
I 2 =
1
2 (E ii E jj − E ij E ij )
¯
I 3 = det[E ij ].
(3.68)
Clearly, any combination of ¯
I 1 , ¯
I 2 , and ¯
I 3 is also invariant. One particularly useful
form is given by
I 1 = 3 + 2 ¯
I 1 = 3 + 2E ii
I 2 = 3 + 4( ¯
I 1 + ¯
I 2 ) = 3 + 4E ii + 2(E ii E jj − E ij E ij )
I 3 = 1 + 2 ¯
I 1 + 4 ¯
I 2 + 8 ¯
I 3 = det
δ ij + 2E ij
.
(3.69)
For some problems, it is possible to choose a coordinate system where the base
vectors align with the principal directions at every point. Examples include uniaxial
stretching of a bar, biaxial stretching of a membrane, and inflation of an isotropic
circular cylinder. In these cases, we can solve the problem without dealing directly
with shear, and the governing equations become much simpler.
For example, the dyadic representation of the strain tensor in principal coordinates is
E = E 1 N 1 N 1 + E 2 N 2 N 2 + E 3 N 3 N 3 ,
(3.70)
where the E i are principal strains, and the invariants of (3.69) become
I 1 = 3 + 2(E 1 + E 2 + E 3 )
I 2 = 3 + 4(E 1 + E 2 + E 3 ) + 4(E 1 E 2 + E 2 E 3 + E 3 E 1 )
I 3 = (1 + 2E 1 )(1 + 2E 2 )(1 + 2E 3 ).
(3.71)