c 2003 Tech Science Press
Copyright CMES, vol.4, no.2, pp.329-336, 2003
Variational Formulation and Symmetric Tangent Operator for Shells with Finite
Rotation Field
Yoshitaka Suetake1 , Masashi Iura2 and S. N. Atluri3
Abstract: The objective of this paper is to examine the
symmetry of the tangent operator for nonlinear shell theories with the finite rotation field. As well known, it has
been stated that since the rotation field carries the Lie
group structure, not a vector space one, the tangent operator incorporating the rotation field does not become
symmetric. In this paper, however, it is shown that by
adopting a rotation vector as a variable, the symmetry
can be achieved in the Lagrangean (material) description.
First, we present a general concept for the problem. Next,
we adopt the finitely deformed thick shell problem as an
example. We also present a tensor formula that plays a
key role for the derivation of a symmetric tangent operator.
and rotated 3-D beams is presented by Iura and Atluri
(1989), in which an equilibrium state and a conservative
system are assumed. Sansour and Bufler (1992) suggest
that if we define the configuration space by using a rotation vector instead of a rotation tensor, we can derive
a symmetric tangent operator. In addition, Bufler (1993)
shows that the tangent stiffness of a conservative system
is always symmetric, even if the rotation field is involved.
However, in Bufler (1993), the Gâteaux derivative with
respect to the rotation tensor is defined in a similar manner to the derivative on a vector space. In Simo (1992)
it is proposed to replace the conventional definition of
the second Gâteaux derivative by the covariant derivative,
which is defined by the Levi-Civita connection and which
is symmetric even in a non-equilibrium state. Aforemenkeyword: finite rotation, tangent operator, symmetry, tioned discussions are summarized by Makowski and
shell theory, variational formulation.
Stumpf (1995), in which the question whether the tangent
operator is symmetric or not is controversially discussed.
1 Introduction
In the latest research works, Gotou, Kuwataka and
Iwakuma (2003) derive an unsymmetric tangent stiffness
Many authors have discussed the symmetry and/or unmatrix in which rotational degrees of freedom are exsymmetry of the tangent operator for nonlinear solid mepressed by Eulerian angles. On the other hand, Ijima,
chanics. As well known, when the rotation field is inObiya, Iguchi and Goto (2003) present a symmetric fortroduced as an independent variable, the tangent operator
mulation for space frames by introducing element end
for a nonlinear mechanical system is not always symmetcoordinates. The authors (2003) also present a symmetric
ric. The reason why the symmetry is lost is that a convariational formulation for 3-dimensinal Timoshenko’s
figuration space represented by a rotation tensor is the
beam and discuss the accuracy of the co-rotational forLie group SO(3) and therefore it does not carry a vector
mulation. As for shell analyses, Basar and Kintzel (2003)
space structure.
derive a symmetric tangent equation, in which an asSimo and Vu-Quoc (1986) show that the linearized tan- sumption is introduced for the variation and the linearizagent operator for 3-D finite-strain rods became unsym- tion of rotation fields.
metric at a non-equilibrium configuration. Nour-Omid
In this paper, the symmetry of the tangent operator is
and Rankin (1991) separate the rigid body motion from
reconsidered in the context of the conventional Gâteaux
the total one and also derive an unsymmetric tangent
derivative, while Simo (1992) and Makowski and Stumpf
stiffness matrix for rods and shells including large rota(1995) discuss the problem in terms of the differential
tions. A symmetric tangent stiffness for finitely stretched
manifold and the Lie algebra. Especially, we deal with
the problem in the context of the material (Lagrangean)
1 AIT, Ashikaga, Tochigi, JAPAN
2 TDU, Hikigun, Saitama, JAPAN
and the spatial (Eulerian) descriptions. As a result, a
3 UC
Irvine, Irvine, CA, U.S.A
330
c 2003 Tech Science Press
Copyright CMES, vol.4, no.2, pp.329-336, 2003
symmetric tangent operator with the rotation field can be
derived as far as the Lagrangean description is adopted
∇ φ) = (∇
∇φ) ⊗∇
∇ = φ|i j g i ⊗gg j ,
(5)
and a conservative system is assumed. It is also shown ∇ ⊗ (∇
that the symmetry of the tangent operator holds even at a
where φ|i j is the second covariant derivative of a scalar
non-critical (non-equilibrium) point.
field φ. Note that Eq. (5) is a symmetric.
The remainder of this paper is as follows. Section 2 deals
X0
with preliminaries. In Section 3, it is symbolically shown For shell problems, we define several quantities. Let X
that the second Gâteaux derivative of the total potential be a generic point on the undeformed shell reference surenergy becomes symmetric in the Lagrangean descrip- face S0 . The covariant base vectors at X 0 on S0 are detion. In Section 4, the application of the present con- fined by
cept to the shell problem is presented, in which the functional established by the authors is employed [Suetake,
X0
∂X
1
Aβ ,
A 3 = √ eαβA α ×A
(6)
Iura, and Atluri (1999); Atluri, Iura, and Suetake (2001); A α = α ,
∂ξ
2 A
Atluri, Iura and Vasudevan (2001)]. Concluding remarks
is given in Section 5.
where Greek indices take 1 or 2, e αβ is a permutation
Aα · Aβ ). By the definisymbol, and A = det(A αβ ) = det(A
2 Preliminaries
tion as Eq. (6), the base vector A 3 becomes a unit normal
Throughout this paper, we indicate vectors with an under to S0 and the covariant base vectors in the undeformed
shell domain can be written as
bar like v and tensors with an under tilde like TT .
∼
Let X be a point in the reference configuration B. On the
other hand, let x be a point in the current configuration b. G = A + ξ3 A , ,
G3 = A3.
(7)
α
3 α
α
If we introduce convected coordinates ξ i , base vectors at
points X and x are defined as follows:
The contravariant base vectors A i and G i are defined by
A j = G i ·G
G j = δij where δij is the Kronecker’s delta.
A i ·A
X
∂X
∂xx
Let the vectors X 0 and A 3 be mapped into x 0 and a 3 . The
(1)
Gi = i , gi = i .
∂ξ
∂ξ
base vectors at x 0 are defined by
We refer to G i in the material (Lagrangean) description
and g i in the spatial (Eulerian) description.
(8)
aα = x 0 ,α = A α +uu 0 ,α ,
We define a differential operator ∇ [see, for example, Ogden (1984)]:
where u 0 is a displacement vector at X 0 . If we assume
that a straight fiber normal to S 0 is mapped into another
straight fiber after deformation, the base vector ggi in the
∂
(2) deformed shell domain can be expressed as
∇ = gi i .
∂ξ
By using the operator ∇ , the gradients of a scalar and a
gα = a α + ξ3 a 3 ,α
vector fields, φ and f , are defined as follows:
(9)
At this stage, we introduce an arbitrary rigid rotation RR
∼
(3) and alternative strain measures c i and b α as follows:
∂φ
∇φ = i gi ,
∂ξ
∇ ⊗ f = g i ⊗ f ,i
, g3 = a3.
,
∇ = f ,i ⊗ggi .
f ⊗∇
T
·aai
(4) c i = R
∼
,
b α = R T ·aa3 ,α .
∼
(10)
where ( ),i denotes a partial derivative with respect to ξ i . We also define a stress-resultant vector and a stressTherefore, the gradient of the vector field ∇ φ is given by couple vector for shells:
331
Variational formulation and symmetric tangent operator for shells with finite rotation field
Ni =
Hα =
ξ3
t i µ0 dξ3 ,
ξ3
ξ3t α µ0 dξ3 .
(11) x̃ = x + εhh .
(15)
Note that the variation vector hh does not depend on the
current configuration x. Therefore, the first Gâteaux
(12) derivative of Π is written as
where t i is a stress vector defined by the first PiolaKirchhoff stress tensor as tt i = G i · t and µ0 is defined
∂
x̃
∂ε ε=0 Π̃(x̃ )
x̃ = ( ∂∂ξΠ̃i g i · ∂x̃
∂ε )ε=0
dΠ(xx;hh) =
(16)
∼
= ∇Π ·hh
by using the curvature tensor B α β and the Gaussian curvature B of the undeformed shell reference surface S 0 as
µ0 = 1 − Bρ ρ ξ3 + B(ξ3 )2 . We also introduce alternative where ∇Π = 0 corresponds to a critical point, that is, an
stress-resultant and stress-couple vectors as
equilibrium equation.
Since we adopt g i as base vectors, Eq. (16) is the Eulerian description. However, if we express the base vectors
i
i
R or N i = R · N̂ ,
N̂ = N i ·R
(13)
gi by G i and u = uiG i , Eq. (16) can be regarded as the
∼
∼
Lagrangean description. Consequently, Eq. (16) can be
interpreted as both the Eulerian and the Lagrangean dei
i
i
i
R or H = R · Ĥ .
(14) scriptions. For simplicity, we make the expression of the
Ĥ = H ·R
∼
∼
base vectors remain g i . The definition of the differential
operator ∇ also remains Eq. (2).
We should note that these stress measures are conjugate
In the similar manner to Eq. (16), we can derive the secwith the strain measures defined by Eq. (10).
ond Gâteaux derivative of Π for an infinitesimal transforIn the latter section, a functional for shells will be exmation x ∗ = x + τvv:
pressed by the above alternative stress and strain measures.
3 Gâteaux Derivative of Functional
d 2 Π(xx;hh ,vv) = d[∇Π ·hh](xx;vv)
= ∂τ∂ {∇Π∗ (xx∗ ) ·hh}
τ=0
∂xx ∗
∗
= { ∂τ · (∇ ⊗ ∇Π )} ·hh = v · (∇ ⊗ ∇Π) ·hh.
We discuss here the Gâteaux derivatives of a potential
energy Π = Π(xx), which always exists for a conservative
τ=0
(17)
system. The argument of the potential Π, namely, xx is
expressed by G i in the Lagrangean description, while by
g i in the Eulerian description.
where we consider that the arbitrary vector hh does not
We deal with two mappings: one is represented by a vec- depend on x .
tor field (displacement) u and the other by a rotation field Apparently, the right hand side of Eq. (17) is commutable
R . In the following subsections, both mappings are dis- with respect to h and v . Therefore, we can conclude that
∼
the second Gâteaux derivative of Π is a symmetric bicussed separately.
linear form in both the Lagrangean and the Eulerian de3.1 Vector Field
scriptions. In addition, the above discussion does not depend on whether or not the current configuration point xx
The mapping form X to x shall be represented by a vector
is a critical one (an equilibrium point).
field u such that x = X + u . We introduce a variation of
u ,that is, δuu = εhh, where ε is an infinitesimal parameter 3.2 Rotation Field
and h is an arbitrary vector. Since u belongs to an Euclidean vector space, an infinitesimal transformation x̃x̃ is In this subsection, our discussion is restricted to the rigid
body mapping form X to x such that x = R · X in order
simply given by
∼
c 2003 Tech Science Press
Copyright 332
CMES, vol.4, no.2, pp.329-336, 2003
to clarify the relationship between the rotation field and
the Gâteaux derivative. Since the space of the rotation
field R is SO(3), not a vector space, the second Gâteaux
∼
derivative of the potential Π cannot be simply defined as
Eq. (17). Such concept is similar to the aforementioned
literatures.
Q
In the beginning, we introduce an infinitesimal rotation Q
η
w) = d[∇Π ·η
η ](xx;w
w)
d 2 Π(xx;η
,w
∂
∗ x ∗ η∗
= ∂τ {∇Π (x ) ·η }
τ=0
∗
x
∗ η∗
= { ∂x
∂τ · (∇ ⊗ ∇Π ) ·η
(22)
∂
+∇Π∗ (xx∗ ) · ∂τ
(φφδ ×xx ∗ )}
τ=0
η + ∇Π · (φφδ ×w
w ).
= w · (∇ ⊗ ∇Π) ·η
∼
in order to define an infinitesimal transformation x̃x̃ such
Equation (22) is not commutable with respect to ηη and
that
w . As a result, the second Gâteaux derivative of Π is
not a symmetric bilinear form in the Eulerian description
unless the current configuration point is critical (∇Π =
R ·X
X = Q ·xx.
(18) O ).
x̃ = Q ·R
∼
∼
∼
∼
On the other hand, the situation is changed in the Lagrangean description. As stated by Atluri and Cazzani
RT is represented by
R ·R
(1995), the variation of rotation δR
∼ ∼
a rotation vector ω in the Lagrangean description. Since
the rotation tensor R
R is expressed by the rotation vector
∼
as
ω
(19)
In the Eulerian description, we can represent the infinitesimal rotation Q by
∼
Q = I + εφφ δ × I ,
∼
∼
∼
sinθ
1 − cos θ
where ε is an infinitesimal parameter again, φ δ is an arbi- R = cos θII +
ω,
ω×I +
ω ⊗ω
∼
∼
∼
θ
θ2
trary axial vector, and I is the identity tensor. The second
∼
RT is given by
term in Eq. (19) corresponds to the skew-symmetric ten- the skew-symmetric tensor δR
R ·R
∼
∼
R · R T and means the Eulerian variation of rotation
sor δR
∼ ∼
as stated by Atluri and Cazzani (1995).
RT = (T
T · δω
ω) × I ,
R ·R
δR
∼ ∼
∼
∼
Substitution of Eq. (19) into Eq. (18) leads to the following expression:
where
η ; η = φ δ ×xx .
x̃ = (II + εφφδ × I ) ·xx = x + εη
∼
∼
(20)
Since the variation with the rotation field has a vector
space structure locally, the first Gâteaux derivative of Π
can be defined in the same way as Eq. (16):
η) = ∇Π ·η
η.
dΠ(xx;η
T=
∼
sin θ
I 1−cosθ ω × I
θ ∼ + θ2
∼
ω ⊗ω
ω
+ θ12 (1 − sinθ
θ )ω
(24)
,
(25)
[See Atluri and Cazzani (1995)]. In Eqs. (23) to (25),
the rotation vector ω is written as ω = θee where θ is a
rotation angle and e is a unit rotation axis.
ω with
If we replace the variation of the rotation vector δω
Q can
ω, instead of Eq. (19), the infinitesimal rotation Q
εω̂
∼
(21) be written by
The second Gâteaux derivative of Π, however, cannot be
defined as Eq. (17), since the variation vector ηη = φ δ ×xx
apparently depends on the current configuration xx. For
w , in which w
an infinitesimal transformation xx∗ = x + τw
w = φ δ ∗ ×xx , the
also involves the rotation field such that w
second Gâteaux derivative of Π is defined as follows:
(23)
R ·R
RT = I + ε(T
T · ω̂
ω) × I ,
Q = I + δR
∼
∼
∼ ∼
∼
∼
∼
(26)
ω is an arbitrary vector. Substitution of Eq. (26)
where ω̂
into Eq. (18) leads to the following infinitesimal transformation:
Variational formulation and symmetric tangent operator for shells with finite rotation field
T · ω̂
ω) × I } ·xx = x + εη ; η
x̃x = {II + ε(T
∼
∼
∼
T · ω̂
ω) ×xx
= (T
∇Π ·
.
∼
(28) Eq. (25) and x ∗ = x +τw
, we can straightforwardly derive
the following expressions:
The second Gâteaux derivative of Π also can be defined
as Eq. (22) for an infinitesimal transformation xx ∗ = x +
τw :
∗
∗
∂
ω · ω̇
ω)(R
R −T
T)
= θ12 (ω
∂τ τ=0 T
∼
∼
∼
1−cosθ
1
ω − θ2 (ω
ω · ω̇
ω)ω
ω} × I
+ θ2 {ω̇
∼
1
sinθ
ω ⊗ω
ω +ω
ω ⊗ ω̇
ω − θ22 (ω
ω · ω̇
ω )ω
ω ⊗ω
ω}
+ θ2 (1 − θ ){ω̇
(31)
d 2 Π(xx ;η , w ) = d[∇Π · η ](xx ; w )
∗
∂
= ∂τ
{∇Π∗ (xx∗ ) · η }
=
∼
ω , instead of ω , in
Since T ∗ is evaluated for ω ∗ = ω + τω̇
dΠ(xx; η ) = ∇Π · η .
τ=0
(30)
where we consider tensor formulae cc ·aa ×bb = a ·bb ×cc and
AT .
A ·aa = a ·A
∼
x∗
[ ∂x
∂τ
T · ω̂
ω) × ( ∂τ∂ +(T
x ∗ )}
∼
τ=0
T ∗ )T · (xx × ∇Π)
τ=0 ∼
T T · {( ∂τ∂ x ∗ ) × ∇Π}].
+T
∼
τ=0
∂
ω · [( ∂τ
= ω̂
At the same time, we obtain an infinitesimal transformaω.
ω = ω + εω̂
tion for the rotation vector as ω̃
ω ) ×xx
T ∗ · ω̂
τ=0 ∼
∼
The first Gâteaux derivative of the potential Π has the
same form as Eq. (21):
∂
T ∗ · ω̂
ω) ×xx ∗ }
∂τ τ=0 {(T
∼
= ∇Π · {( ∂τ∂ (27)
333
∗
x∗
and
T · ω̂
ω) ×xx }
· (∇ ⊗ ∇Π ) · {(T
∼
∂
T ∗ · ω̂
ω) ×xx ∗ }]
+∇Π (xx ) · ∂τ {(T
∼
τ=0
∂
T ∗ · ω̂
ω) ×xx ∗ }.
= w · (∇ ⊗ ∇Π) · η + ∇Π · ∂τ
{(T
∼
τ=0
∗
x∗
∂ T · ω̇
ω) ×xx .
x ∗ = w = (T
∼
∂τ τ=0
(32)
(29) Substituting Eqs. (31) and (32) into the square bracket
in Eq. (30), after careful and straightforward calculation,
we can finally obtain the following formula:
In Eq. (29), both T ∗ and x ∗ correspond to an infinites∼
ω .
ω ∗ = ω + τω̇
imal transformation for the rotation as ω
Especially, likewise Eq. (27), the vector w is given by
∂
R ·X
X ) × ∇Π}
( ∂τ
T ∗ )T .{(R
T · ω̇
ω) ×xx .
τ=0 ∼
w = (T
∼
(33)
∼
X ) × ∇Π} = Γ (X
X , ∇Π).ω̇
ω,
T T · {( ∂τ∂ R∗ ·X
+T
∼
∼
The first term of the right hand side of Eq. (29) is comτ=0 ∼
mutable with respect to η and w and therefore symmet· X and the tensor Γ (aa,bb) is
ric bilinear form. The second term, however, does not where we consider x = R
∼
∼
seem to be symmetric. On the other hand, since the ro- given by
tation vector ω makes a vector space, it is expected that
the second Gâteaux derivative also becomes symmetric
(aa,bb) = 1θ [C1 I + 1−cosθ
(aa ⊗bb +bb ⊗aa)
as suggested by Sansour and Bufler (1992). In what fol- Γ
θ
∼
∼
lows, we examine whether the second term of Eq. (29) is + 1 (cosθ − sinθ ){(aa ×bb) ⊗ω
ω +ω
ω ⊗ (aa ×bb)}
θ
θ
2(1−cosθ)
1
symmetric or not.
(34)
ω ·bb)(aa ⊗ω
ω
}{(ω
+ θ2 {sinθ −
θ
The second term of the right hand side of Eq. (29) is
ω ⊗aa) + (ω
ω ·aa)(bb ⊗ω
ω +ω
ω ⊗bb)}
+ω
1
rewritten as
ω ],
+ θ2 C2ω ⊗ω
334
c 2003 Tech Science Press
Copyright CMES, vol.4, no.2, pp.329-336, 2003
where
ω ·aa ×bb )
C1 = − sinθ(aa ·bb) + 1θ (cosθ − sinθ
θ )(ω
2(1−cosθ)
1
ω ·bb)(ω
ω ·bb)
}(ω
+ θ2 {sinθ −
θ
α
(35)
and
α
3
ω,cc3, N̂ ,Ĥ ,N̂ )
F(uu0 ,ω
α
α 3
= S0 [ −WC (N̂ ,Ĥ ,N̂ )
α
α
3
Aα +uu 0 ,α ) + Ĥ ·cc3 ,α + N̂ ·cc3
R · N̂ · (A
+R
∼
√
m · (R
R ·cc3 −A
A 3 )] Adξ1 dξ2 ,
−pp ·uu0 −m
(38)
∼
C2 = (sinθ − θ cos θ)(aa ·bb)
ω a b
−{sinθ + 3θ (cosθ − sinθ
θ )}(ω ·a ×b )
8(1−cosθ)
1
5 sinθ
ω ·bb)(ω
ω ·bb).
}(ω
+ θ {sinθ − θ +
θ2
where WC is a complementary energy density, p is an ex(36) ternal force, is m an external moment, and the rotation
ω by Eq.
tensor R is represented by the rotation vector ω
∼
(23). In this section we also assume a conservative sysNote that Γ (aa,bb) is a symmetric tensor. Therefore, the
tem.
∼
ω.
ω and ω̇
second term of Eq. (29) is commutable with ω̂
For simplicity, we omit several terms that concern boundThe formula (33) and the tensor Γ (aa,bb) play a key role
ary conditions and therefore only the principal part of the
∼
for the derivation of the symmetric tangent operator in functional is presented in Eq. (38). In addition, for the
shell problems.
application to numerical formulation, the rotation field RR
Consequently, the second Gâteaux derivative, Eq. (29),
can be rewritten as
4.2 Gâteaux Derivatives
d 2 Π(xx; η , w )
ω) ×xx } · (∇ ⊗ ∇Π) · {(T
T · ω̇
ω) ×xx }
T · ω̂
= {(T
∼
Γ(X
X , ∇Π).ω̇
ω.
ω .Γ
+ω̂
∼
∼
or ω is assumed to be constant within a sub-region, i.e.,
within a finite element.
(37)
∼
In this subsection, we calculate the first and the second
Gâteaux derivatives of the functional F. Here, we express the first Gâteaux derivative simply as an operator δ
and the second one as ∆.
ω. The symmetry
ω and ω̇
which is also commutable with ω̂
From Eq. (38), we obtain the following expression as the
of Eq. (37) also does not depend on whether or not the
first Gâteaux derivative of F:
current configuration is critical.
4 Variational Formulation for Shells
In the previous section, it is confirmed that the second
Gâteaux derivative of the total potential energy becomes
symmetric in the Lagrangean description even when involving the rotation field. This section deals with the application of the above discussion to shell problems. As an
example of the nonlinear solid mechanics, we consider a
finitely deformed thick shell model that incorporates finite rotations.
4.1 Mixed Variational Principle
α ∂WC
α ∂WC
S0 [ − δN̂ · ∂N̂ α − δĤ · ∂Ĥ α
3
α
R · N̂ · (A
Aα +uu0 ,α )
−δN̂ · ∂WC3 + δR
∼
∂N̂
α
α
RT · (A
Aα +uu0 ,α ) + δuu0 ,α ·R
R · N̂
+δN̂ ·R
∼
∼
α
α
3
3
c 3 · N̂
+δĤ ·cc3,α + δcc 3 ,α · Ĥ + δN̂ ·cc√
3 + δc
R ·cc3 + δcc 3 ·R
RT ) ·m
m] Adξ1 dξ2 .
−δuu0 · p − (δR
∼
∼
δF =
(39)
For linear elastic material, the complementary energy
density WC is given by a bilinear form of the stress measures, for example,
A mixed type variational principle for shells employed
α
α 3
α
β
1
C (s) · N̂
here has been already established by Suetake, Iura, and WC (N̂ ,Ĥ ,N̂ ) = 2 (N̂ ·C
∼ αβ
(40)
α
β
3
3
Atluri (1999). The variational principle involves the rotaC (b) · Ĥ + N̂ ·C
C (d) · N̂ ),
+Ĥ ·C
∼ αβ
∼
tion tensor R as an independent variable. The functional
∼
for the mixed variational principle is written by using the
(s)
, C (b) , and C (d) are constant symmetric comstress and the strain measures defined in Section 2 as fol- where C
∼ αβ ∼ αβ
∼
pliance tensors.
lows:
335
Variational formulation and symmetric tangent operator for shells with finite rotation field
In what follows, we discuss only the Lagrangean description. Therefore, it is assumed that the variation of rotaR · RT , can be expressed as Eq. (24). Considering
tion, δR
∼ ∼
Eq. (40) and the vector formula a ×bb ·cc = a ·bb ×cc again,
we obtain the following expression from Eq. (39):
∆δF
α
α
R · N̂ ) ×T
T · ∆ω
ω +R
R · ∆N̂ }
= S0 [δuu0 ,α · { − (R
α
∼
∼
α
∼
T T · (R
R · N̂ ) × ∆uu0 ,α + δN̂ ·R
RT · ∆uu0 ,α
ω ·T
+δω
∼
∼
α
∼
m)} · ∆ω
ω ·Γ
Γ (N̂ ,A
Aα +uu0 ,α ) −Γ
Γ (cc3 ,m
ω
+δω
∼
∼
α
ω ·T
T T · {m
R · ∆cc3 − (A
Aα +uu0 ,α ) ×R
R · ∆N̂ }
m ×R
+δω
∼
∼
α
3
α
α
∼
(42)
R ·cc3 ) ×m
T · {(∆R
R ·cc 3 ) ×m
m} −T
m}
T · {(R
−∆T
T
∼
∼
α
β
∼
(44)
R · N̂ ) × (A
Aα +uu0 ,α )}
T T · {(R
∆T
∼
∼
α
3
∼ αβ
Next we calculate the second Gâteaux derivative of F. In
the calculation of the second Gâteaux derivative, we must
pay attention to the underlined terms in Eq. (41), because
it includes ∆TT . The Gâteaux derivative of the underlined
∼
terms in Eq. (41) involves the following part:
R · N̂ ) × (A
Aα +uu0 ,α )}
+ T T · {(∆R
∼
C (s) · N̂
+δcc 3 · ∆N̂ + δN̂ · ∆cc3 − δN̂ ·C
∼ αβ
α
β
3
3 √
(b)
(d)
Ĥ
C
Ĥ
N̂
C
N̂
−δ ·C ·
− δ ·C · )] Adξ1 dξ2 .
T T · ∆ω
ω) × I when ∆R
R·
T T · R = −(T
where we consider ∆T
∼
∼
∼
∼
∼
T · ∆ω
ω) × I and a vector product between a vector
R T = (T
∼
∼
∼
α
RT · (A
Aα +uu0 ,α ) ×T
T · ∆ω
ω
+δN̂ ·R
+δcc 3 ,α · ∆Ĥ + δĤ · ∆cc3,α
(41)
∼
∼
RT ·m
T · ∆ω
ω
m ×T
−δcc 3 ·R
α
u0 ,α ·R
R · N̂
S0 [δu
∼
α
m}
T T · {(R
R · N̂ ) × (A
Aα +uu 0 ,α ) − (R
R ·cc3 ) ×m
+δω ·T
∼
∼
∼
α
3
m)
RT ·m
+δcc 3 ,α · Ĥ + δcc 3 · (N̂ −R
∼
α
β
RT · (A
Aα +uu0 ,α ) −C
C (s) · N̂ }
+δN̂ · {R
∼
∼ αβ
α
β
(b)
C · Ĥ )
+δĤ · (cc3 ,α −C
∼ αβ
√
3
3
(d)
C · N̂ ) − δuu0 · p ] Adξ1 dξ2 .
+δN̂ · (cc3 −C
∼
δF =
∼
∼
∼
A simply means a × I ) ·
and a second-order tensor as a ×A
∼
∼
A . In Eq. (44), we present only the part concerning the
∼
tangent operator, that is, we omit the part concerning the
external force.
It should be noted that since the tensor ΓΓ (aa,bb)
∼
is symmetric, Eq. (44) is apparently commutable
α
α
3
ω , δcc3, δN̂ ,δĤ ,δN̂ ) and
with respect to (δuu0 , δω
α
α
3
ω, ∆cc3 , ∆N̂ ,∆Ĥ ,∆N̂ ). Thus, we can conclude
(∆uu0 , ∆ω
that the tangent operator for shells with the finite rotation
field becomes symmetric in the Lagrangean description
even at non-equilibrium point.
T
∼
∼
5 Concluding Remarks
We discuss the symmetry of the tangent operator for nonlinear mechanical systems that are conservative. In parIn view of the formula (33), it is easily understood
ticular, we deal with the problem in both the Lagrangean
that the expression (42) is rewritten by using the tensor
and the Eulerian descriptions.
Γ (aa,bb) as:
∼
If independent variables are only vector fields, the tangent operator, that is, the second Gâteaux derivative of
the functional becomes symmetric in both descriptions.
In addition, the symmetry of the tangent operator does
α
Aα +uu0 ,α ) −Γ
Γ (cc3 ,m
ω
m)} · ∆ω
Γ(N̂ ,A
(43) not depend on whether or not the current configuration
{Γ
∼
∼
point is at an equilibrium state.
Finally, we obtain the second Gâteaux derivative of F as
follows:
On the other hand, when the functional for the mechanical system incorporates the finite rotation field, we cannot simply reveal the symmetry problem because of the
336
c 2003 Tech Science Press
Copyright special orthogonality of the rotation group. In the Eulerian description the tangent operator does not become
symmetric unless the current state satisfies the equilibrium condition. In the Lagrangean description, however,
the tangent operator always becomes symmetric even at
a non-critical point.
As an example, we adopt the finitely deformed thick shell
problem that involves the rotation field. We evaluate the
second Gâteaux derivative of the well-established mixed
type functional and confirm that the obtained tangent operator has a symmetric bilinear form.
CMES, vol.4, no.2, pp.329-336, 2003
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