It is customary, and convenient, to gather all the nodal displacements uxei and uyei

in one column, indicated by ue , according to

∼

⎡ ¤

uxe1

⎢ node 1 ⎥

⎢ uye1 ⎥

⎢ ⎥

⎢ ⎥

⎢ ⎥

⎢ ⎥

⎢ uxe2 ⎥

ue = ⎢ u node 2 ⎥ . (18.48)

⎢ ye2 ⎥

∼

⎢ ⎥

⎢ ⎥

.

⎢ ⎥

.

.

⎢ ⎥

⎢ ⎥

⎣ uxen ¦

node n

uyen

Using this de¬nition, the strain column for an element e can be rewritten as

µ = B ue , (18.49)

∼

∼

with B the so-called strain displacement matrix:

⎡ ¤

‚N1 ‚N2 ‚Nn

···

0 0 0

‚x ‚x ‚x

⎢ ⎥

⎢ ⎥

⎢ ⎥

‚N1 ‚N2 ‚Nn

···

B=⎢ ⎥.

0 0 0 (18.50)

‚y ‚y ‚y

⎢ ⎥

⎣ ¦

‚N1 ‚N1 ‚N2 ‚N2 ‚Nn ‚Nn

···

‚y ‚x ‚y ‚x ‚y ‚x

Clearly, a similar expression holds for µw . So, patching everything together, the

∼

double inner product µ w : σ may be written as:

µ w : σ = ( µw )T σ

∼

∼

= ( µ w )T H µ

∼ ∼

= wT BT H B ue , (18.51)

∼e ∼

where we stores the components of the weighting vector w structured in the

∼

same way as ue . This result can be exploited to elaborate the left-hand side of

∼

Eq. (18.24):

(∇ w)T : σ d = wT BT H B d ue . (18.52)

∼e ∼

e e

The element coef¬cient matrix, or stiffness matrix K e is de¬ned as

Ke = BT H B d . (18.53)

e

321 18.4 Galerkin discretization

Step 6 Writing the force per unit volume vector f as

f = fx ex + fy ey , (18.54)

and the weighting function w within element as

e

= N T wxe ex + N T wye ey ,

w| (18.55)

∼∼ ∼∼

e

the second integral on the right-hand side of Eq. (18.24) can be written as

w·f d = ( N T wxe fx + N T wye fy ) d

∼∼ ∼∼

e e

= wT f v , (18.56)

∼e e ∼

where

⎡ ¤

N1 fex

⎢ N1 fey ⎥

⎢ ⎥

⎢ ⎥

⎢ ⎥

⎢ ⎥

⎢ N2 fex ⎥

⎢ ⎥

fv = ⎢ N2 fey ⎥ d . (18.57)

∼e

e⎢ ⎥

⎢.⎥ .⎥

⎢.

⎢ ⎥

⎣ Nn fex ¦

Nn fey

The ¬rst integral on the right-hand side of Eq. (18.24) may, if applicable, be

elaborated in exactly the same manner. This results in

w·( σ · n) d = wT f p , (18.58)

∼e e ∼

e

where is structured analogously to f v . The contribution of the right-hand side

fp

∼e ∼e

triggers the abbreviation:

f = f v + f p, (18.59)

∼e e e

∼ ∼

often referred to as the element load contribution.

Step 7 Putting all the pieces together, the discrete weak formulation of

Eq. (18.24) is written as:

Nel Nel

wT K e ue = wT f e . (18.60)

∼e ∼e

∼ ∼

e=1 e=1

Following an equivalent assembling procedure as outlined in Chapter 14 the

following result may be obtained:

wT K u = wT f , (18.61)

∼

∼ ∼ ∼

Infinitesimal strain elasticity problems

322

where w and u contain the global nodal weighting factors and displacements,

∼

∼

respectively, and K is the global stiffness matrix. This equation should hold for

all weighting factors, and thus

K u = f. (18.62)

∼ ∼

18.5 Solution

As outlined in Chapter 14 the nodal displacements u may be partitioned into two

∼

groups. The ¬rst displacement group consists of components of u that are pre-

∼

scribed: up . The remaining nodal displacements, which are initially unknown, are

∼

gathered in uu . Hence:

∼

uu

u= ∼

. (18.63)

∼

up

∼

In a similar fashion the stiffness matrix K and the load vector f are partitioned. As

∼

a result, Eq. (18.62) can be written as:

f

K uu K up uu

=

∼ ∼u . (18.64)

fp

up

K pu K pp ∼ ∼

The force column f is split into two parts: f u and f p . The column f u is known,

∼ ∼ ∼ ∼

since it stores the external loads applied to the body. The column f p , on the other

∼

hand, is not known, since no external load may be applied to points at which the

displacement is prescribed. In Eq. (18.64) f u is known and f p is unknown. The

∼ ∼

following set of equations results:

K uu uu = f u ’ K up up , (18.65)

∼ ∼

∼

f = K pu uu + K pp up . (18.66)

∼ ∼

∼p

The ¬rst equation is used to calculate the unknown displacements uu . The result

∼

is substituted into the second equation to calculate the unknown forces f p .

∼

18.6 Example

Consider the bending of a beam subjected to a concentrated force (Fig. 18.2). Let

the beam be clamped at x = 0 and the point load F be applied at x = L. It is inter-

esting to investigate the response of the bar for different kinds of elements using

a similar element distribution. Four different elements are tested: the linear and

the quadratic triangular element, and the bi-linear and bi-quadratic quadrilateral

323 18.6 Example

Table 18.1 Comparison of the relative accuracy of

different element types for the beam bending case.

uc /uL

Element type

Linear triangle 0.231

Bi-linear quadrilateral 0.697

Quadratic triangle 0.999

Bi-quadratic quadrilateral 1.003

F

h

L

x=0 x=L

Figure 18.2

A beam that is clamped on one side and loaded with a vertical concentrated force at the other

side.

element. The meshes for the linear triangular and the bi-linear quadrilateral

element are shown in Fig. 18.3. The displacement at x = L can be computed

using standard beam theory, giving

FL3

uL = , (18.67)

3EI

with

bh3