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320

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

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