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