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is a vector equation, the weighting function is chosen to be a vector ¬eld as
well: w. Taking the inner product of the weight function w with the momentum
Eq. (18.1) yields a scalar expression, which upon integration over the domain
yields

w·(∇ · σ + f) d = 0. (18.14)

This integral equation must hold for all weighting functions w. By application of
the product rule, it can be shown that
∇ · ( σ · w) = ( ∇ · σ ) ·w + ( ∇ w)T : σ . (18.15)
Infinitesimal strain elasticity problems
316

In Eq. (18.15) the double dot product ( ∇ w)T : σ is used. The de¬nition of the
double dot product of two tensors A and B is

A : B = tr( A · B) = Aij Bji , (18.16)

where the Einstein summation convention has been used. With respect to a Carte-
sian basis and using index notation it is straightforward to prove Eq. (18.15). First
of all notice that

σ · w = σij wj ei . (18.17)

Consequently

∇·( σ · w) = ( σij wj ) . (18.18)
‚xi
Application of the product rule of differentiation yields

∇·( σ · w) = ( σij wj )
‚xi
‚σij ‚wj
= wj + σij . (18.19)
‚xi ‚xi
With the identi¬cations
‚σij
( ∇ · σ ) ·w = wj , (18.20)
‚xi
and
‚wj
( ∇ w)T : σ = σij , (18.21)
‚xi
the product rule according to Eq. (18.15) is obtained.
Use of this result in Eq. (18.14) yields

∇·( σ · w) d ’ ( ∇ w)T : σ d + w·f d = 0. (18.22)

The ¬rst integral may be transformed using the divergence theorem Eq. (16.5).
This yields

( ∇ w)T : σ d = w·( σ · n) d + w·f d . (18.23)




18.4 Galerkin discretization

Within the context of the ¬nite element method, the domain is split into a num-
ber of non-overlapping subdomains (elements) e , such that this integral equation
is rewritten as
317 18.4 Galerkin discretization

Nel Nel
( ∇ w) : σ d = w·( σ · n) d + w·f d
T
. (18.24)
e e e
e=1 e=1

Clearly, the above e ( .) d denotes the integral along those element boundaries
that coincide with the domain boundary ( e = e © ). As an example of how
to proceed based on Eq. (18.24) a plane strain problem is considered.

Step 1 The integrand of the integral in the left-hand side of Eq. (18.24) is
elaborated ¬rst. If the double inner product of a symmetric tensor A and a
skew-symmetric tensor B is taken, then
A : B = 0 if AT = A and BT = ’B . (18.25)
This can easily be proved. Using the properties of the double inner product it
follows that
A : B = AT : BT = ’A : B . (18.26)
This equality can only hold if A : B = 0.
The dyadic product ( ∇ w)T appearing in ( ∇ w)T : σ may be split into a
symmetric and a skew-symmetric part:
1 1
( ∇ w)T =( ∇ w) + ( ∇ w)T ’ ( ∇ w) ’ ( ∇ w)T . (18.27)
2 2
Because of the symmetry of the Cauchy stress tensor it follows that
1
( ∇ w)T : σ = ( ∇ w) + ( ∇ w)T : σ . (18.28)
2
Notice that the expression 1 ( ∇ w) + ( ∇ w)T has a similar form to the in¬nites-
2
imal strain tensor µ de¬ned as
1
µ= ( ∇u) + ( ∇u)T , (18.29)
2
where u denotes the displacement ¬eld. This motivates the abbreviation:
1
µw = ( ∇ w) + ( ∇ w)T . (18.30)
2
Consequently, the symmetry of the Cauchy stress tensor allows the double inner
product ( ∇ w)T : σ to be rewritten as

( ∇ w)T : σ = µw : σ . (18.31)

Step 2 To elaborate further, it is convenient to introduce a vector basis. Here, a
Cartesian vector basis {ex , ey , ez } is chosen. In the plane strain case it is assumed
that µzz = µxz = µyz = 0 while the displacement ¬eld is written as
u = ux ( x, y) ex + uy ( x, y) ey . (18.32)
Infinitesimal strain elasticity problems
318

Likewise, the weighting function w is written as
w = wx ( x, y) ex + wy ( x, y) ey . (18.33)
In the plane strain case, the matrices associated with the tensors µ and µ w are
given by
⎡ ¤ ⎡ ¤
µxx µxy 0
µxx µxy 0 w w
⎢ ⎥ ⎢w ⎥
µ w = ⎣ µxy µyy 0 ¦ .
µ = ⎣ µxy µyy 0 ¦ , w (18.34)
0 00 0 00
Consequently, the inner product µw : σ equals
µw : σ = µxx σxx + µyy σyy + 2µxy σxy .
w w w
(18.35)
Notice the factor 2 in front of the last product on the right-hand side due to the
symmetry of both µ w and σ . It is convenient to gather the relevant components of
µ w , µ (for future purposes) and σ in a column:

( µw )T = µxx µyy
w w w
2µxy





µT = µxx µyy 2µxy , (18.36)



(notice the 2 in front of µxy and µxy ) and
w


σ T = σxx σyy σxy . (18.37)



This allows the inner product µ w : σ to be written as
µw : σ = ( µw )T σ . (18.38)





Step 3 The constitutive equation according to Eq. (18.4) may be recast in the
form
σ = H µ. (18.39)
∼ ∼

Dealing with the isotropic Hooke™s law and plane strain conditions and after
introduction of Eqs. (18.12) and (18.13) into Eq. (18.4), the matrix H can be
written as
⎡ ¤ ⎡ ¤
4 ’2 0
110
⎢ ⎥ G⎢ ⎥
H = K ⎣ 1 1 0 ¦ + ⎣ ’2 4 0 ¦ . (18.40)
3
000 0 03
Consequently
µ w : σ = ( µw )T H µ . (18.41)
∼ ∼
319 18.4 Galerkin discretization

Step 4 With u = ux ex + uy ey and x = xex + yey the strain components may be
written as
¤⎡ ¤
⎡ ‚ux
µxx ‚x
⎥⎢ ⎥

⎥⎢ ⎥
⎢ ⎢ ⎥
⎢ ⎥⎢ ‚uy
⎥.
µ = ⎢ µyy ⎥ = ⎢ (18.42)

‚y
⎢ ⎥⎢


⎣ ¦⎣ ¦
‚uy
‚ux
‚y + ‚x
2µxy

This is frequently rewritten as

ˆ
µ = B u, (18.43)



ˆ
with B an operator de¬ned by
⎡ ¤

0
‚x
⎢ ⎥
⎢ ⎥
ˆ⎢ ⎥

B=⎢ 0 ⎥, (18.44)
‚y
⎢ ⎥
⎣ ¦
‚ ‚
‚y ‚x

while u represents the displacement ¬eld:



ux
u= . (18.45)

uy


Step 5 Within each element the displacement ¬eld is interpolated according to
n
ux | = Ni ( x, y) uxei = N T uxe
∼∼
e
i=1
n
= Ni ( x, y) uyei = N T uye ,
uy (18.46)
∼∼
e
i=1

where uxe and uye denote the nodal displacements of element e in the x- and
∼ ∼
y-direction, respectively. Using this discretization the strain within an element can
be written as
⎡ ¤
‚Ni
uxei
‚x
⎢ ⎥
n⎢ ⎥
⎢ ⎥
‚Ni
µ= ⎢ ⎥.
uyei (18.47)
‚y
⎢ ⎥

i=1 ⎣ ¦
‚Ni ‚Ni
‚y uxei + ‚x uyei
Infinitesimal strain elasticity problems

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