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where the spatial constitutive matrix D is constructed from the components
of the fourth-order tensor c by equating the tensor product δd : c : µ to the
matrix product δdT Dµ to give, after some algebra,

® 
2c 2c 2c +c +c +c
c c c
1111 1122 1133 1112 1121 1113 1131 1123 1132
2c 2c 2212 +c 2213 +c 2223 +c
 
c c c
2222 2233 2221 2231 2232

1
 
2c 3312 +c 3313 +c 3323 +c
c c c
3333 3321 3331 3332
D=  
2 1212 +c 1213 +c 1223 +c
c c c 
1221 1231 1232

sym. 1313 +c 1323 +c
 
c c
1331 1332»
°
2323 +c
c 2332

(7.39)
14 DISCRETIZATION AND SOLUTION



In the particular case of a neo-Hookean material (see Equations (5.28“29)),
D becomes,
® 
» + 2µ » » 0 0 0
» » + 2µ » 0 0 0
 
» » » + 2µ 0 0 0
 
D= ;
0 0 0 µ 0 0
°0 0 0 0 µ 0»
 
0 0 0 0 0 µ

» µ ’ » ln J
»= ; µ= (7.40)
J J
Substituting for δd and µ from Equations (7.26) and (7.37b) respectively
into the right-hand side of Equation (7.38) enables the contribution from
element (e) associated with nodes a and b to emerge as,

DδWc(e) (φ, Na δv a )[Nb ub ] = (Ba δv a )T D(Bb ub ) dv
v (e)

BT DBb dv ub
= δv a · (7.41)
a
v (e)

The term in brackets de¬nes the constitutive component of the tangent
matrix relating node a to node b in element (e) as,

(e)
BT DBb dv
K c,ab = (7.42)
a
(e)
v



7.4.3 INITIAL STRESS COMPONENT
Concentrating attention on the second term in the linearized equilibrium
Equation (7.31), note ¬rst that the gradients of δv and u can be interpolated
from Equations (7.3“4) as,
n
δv = δv a — Na (7.43a)
a=1
n
u= Nb (7.43b)
ub —
b=1

Introducing these two equations into the second term of Equation (7.31)
and noting Equation (2.51b), that is, σ : (u — v) = u · σv for any vectors
u, v, enables the initial stress contribution to the linearized virtual work
15
7.4 LINEARIZED EQUILIBRIUM EQUATIONS



Equation (7.31) for element (e) linking nodes a and b to be found as,

σ : [( ub )T
DδWσ (φ, Na δv a )[Nb ub ] = δv a ] dv
v

= σ : [(δv a · ub ) Nb — Na ] dv
(e)
v

= (δv a · ub ) Na · σ Nb dv (7.44)
v (e)

Observing that the integral in Equation (7.44) is a scalar, and noting that
δv a · ub = δv a · Iub , the expression can be rewritten in matrix form as,
(e)
DδWσ (φ, Na δv a )[Nb ub ] = δv a · K σ,ab ub (7.45a)
(e)
where the components of the so-called initial stress matrix K σ,ab are,

(e)
K σ,ab = ( Na · σ Nb )I dv
(e)
v
3
‚Na ‚Nb
(e)
[K σ,ab ]ij = σkl δij dv; i, j = 1, 2, 3
‚xk ‚xl
v (e) k,l=1



EXAMPLE 7.5: Initial stress component of tangent matrix
[K σ, ab ]
Using the same con¬guration as in examples 7.1“7.4 a typical initial
stress tangent matrix component connecting nodes 1 and 2 can be found
from Equation (7.45) as,
‚N2
σ11 σ12
‚N1 ‚N1 10
‚x1
[K σ,12 ] = dv
01
‚x1 ‚x2 ‚N2
σ21 σ22
v (e) ‚x2

1 1 ’1 ’1 10
= 8 + 4 24t
01
8 8 8 6
’1t 0
=
0 ’1t



7.4.4 EXTERNAL FORCE COMPONENT
As explained in Section 6.5 the body forces are invariably independent of the
motion and consequently do not contribute to the linearized virtual work.
However, for the particular case of enclosed normal pressure discussed in
16 DISCRETIZATION AND SOLUTION



Section 6.5.2, the linearization of the associated virtual work term is given
by Equation (6.23) as,
1 ‚x ‚δv ‚u
p
DδWext (φ, δv)[u] = p — u ’ δv — dξd·
·
2 ‚ξ ‚· ‚·


1 ‚x ‚δv ‚u
’ p — u ’ δv — dξd·
·
2 ‚· ‚ξ ‚ξ

(7.46)
Implicit in the isoparametric volume interpolation is a corresponding surface
representation in terms of ξ and · as (see Figure 7.1),
n
x(ξ, ·) = N a xa (7.47)
a=1

where n is the number of nodes per surface element. Similar surface in-
terpolations apply to both δv and u in Equation (7.46). Considering, as
before, the contribution to the linearized external virtual work term, in
Equation (7.30), from surface element (e) associated with nodes a and b
gives,
p(e)
DδWext (φ, Na δv a )[Nb ub ]
1 ‚x ‚Na ‚Nb
= (δv a — ub ) · p Nb ’ Na dξd·
2 ‚ξ ‚· ‚·


1 ‚x ‚Na ‚Nb
’ (δv a — ub ) · p Nb ’ Na dξd·
2 ‚· ‚ξ ‚ξ


= (δv a — ub ) · kp,ab (7.48)
where the vector of sti¬ness coe¬cients kp,ab is,
1 ‚x ‚Na ‚Nb
kp,ab = p Nb ’ Na dξd·
2 ‚ξ ‚· ‚·


1 ‚x ‚Na ‚Nb
+ p Nb ’ Na dξd· (7.49)
2 ‚· ‚ξ ‚ξ


Equation (7.48) can now be reinterpreted in terms of tangent matrix com-
ponents as,
(e)
p(e)
DδWext (φ, Na δv a )[Nb ub ] = δv a · K p,ab ub (7.50a)
where the external pressure component of the tangent matrix is,
3
(e) (e) (e) (e)
= Ekp,ab ; = Eijk kp,ab k ; i, j = 1, 2, 3
K p,ab K p,ab ij
k=1
17
7.4 LINEARIZED EQUILIBRIUM EQUATIONS



where E is the third-order alternating tensor (Eijk = ±1 or zero, depending
on the parity of the ijk permutation).


EXAMPLE 7.6: External pressure component of tangent ma-
trix K p, ab
Consider the same triangle of Example 7.1 now representing a face
on which pressure p is applied. If the isoparametric coordinates are
renamed ξ, ·, the vectors ‚x/‚ξ and ‚x/‚· in the vector of sti¬ness
coe¬cients given in Equation (7.49) are constant and depend upon
the geometry of the particular surface element, whereas the terms in
parentheses depend only on the element type. Noting that if a = b the
terms in parentheses are zero, the resulting simple integration yields,
N1 = 1 ’ ξ ’ ·; N2 = ξ; N3 = ·
® ®
8 8
‚x ° » ‚x ° »
=0 =6
‚ξ ‚·
0 0
1 ‚x 1 1 ‚x 1
kp,12 = p ’ +p ’
2 ‚ξ 6 2 ‚· 3
1 ‚x 1 1 ‚x 1
kp,13 = p ’ +p ’ ;
2 ‚ξ 3 2 ‚· 6
1 ‚x 1 1 ‚x 1
kp,23 = p ’ +p +
2 ‚ξ 6 2 ‚· 6
Nonzero pressure sti¬ness submatrices are now found from Equation (7.50b)
as,
®  ® 
0 0 ’2 0 0 ’1
p p
K p,12 = ’ ° 0 4»; K p,13 = ’ ° 0 4»;
0 0
2 2
2 ’4 0 1 ’4 0
® 
0 01
p

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