9. chapter 9 - fe in gobal coordinates _a4l

8/11/2019 9. Chapter 9 - Fe in Gobal Coordinates _a4l

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______________________Basics of the Finite Element Method Applied in Civil Engineering

89

CHAPTER

ELEMENT SHAPE FUNCTIONS IN GLOBAL COORDINATES

9.1THE GENERAL PROCEDURE

The general procedure for defining the shape functions is presented for a 2D

quadrilateral element, with the following assumptions: C0 continuity and

linear polynomial approximation. The 3D expansion is immediate. Note that

in the following example an element with one DOF/node is considered (i.e.

d- a generalized displacement) in order to simplify writing. The expressions

are available for all components (u, v, w) for a vector-type unknown

function.

The displacement of an interior pointM(x,y) lying inside the elements area

(see figure 9.1) is written using four parameters as

xyayaxaayxd 4321),( +++= (9.1)

Fig. 9.1 The displacement filed as a relationship of nodal displacements

yM (x,y)

12

3

4

12

34


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Chapter 9 Element Shape Functions in Global Coordinates_________________________

90

The ai parameters can be calculated by applying the condition that in thenodal points, the displacement function must have the nodal values:

+++=

+++=

+++=

+++=

444434214

334333213

224232212

114131211

yxayaxaa

yxayaxaa

yxayaxaa

yxayaxaa

(9.2)

In matrix form:

aCM=

=

4

3

2

1

4444

3333

2222

1111

4

3

2

1

1

1

1

1

a

a

a

a

yxyx

yxyx

yxyx

yxyx

where CMis the coordinates matrix, or

aC M= (9.3)

By solving the matrix equations the polynomial parameters are found:

Ca 1M= (9.4)

If the linear approximation is written in a matrix form:

=

4

3

2

1

]1[),(

a

a

a

a

xyyxyxd (9.5)

and the polynomial coefficients are replaced

CM1

]1[),( = xyyxyxd (9.6)


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91

the shape functions yields:

[ ] 11),( = MCN xyyxyx (9.7)

The same procedure can be followed for a quadratic approximation,

assuming a C0class of continuity. In this case, the displacement field is

2

8

2

7

2

6

2

54321),( yaxyayxaxaxyayaxaayxd +++++++= =

= [ ]

8

2

1

2222

...1

a

a

a

yxyyxxxyyx (9.8)

A complete 2D quadratic polynomial is described by 8 coefficients and

consequently the conditions regarding the displacement functions on each

node i,di= i, must be applied for 8 nodes. This time, CMwill be an 8 8matrix.

9.2 EXAMPLE 1 The 2D spar (link or elastic spring) element

The finite element shown in figure 9.2 is defined by two nodes, 1 and 2, in

the (x,y) plane, with known cross section areaAand Young modulusE. The

length and orientation are defined by the nodal coordinates.

12 xxL

x =

12 yyL

y = (9.9)

22

yx LLL +=

The displacement along the element axissis approximated by a first degree

polynomial:

[ ]

=+=2

1

211)(

a

assaasd (9.10)

The nodal displacements are d1 and d2with their projections in the globalcoordinate system (u1, v1) and (u2, v2). The relationship between dand the

projections uand vis


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vLLu

LLd yx += (9.11)

The polynomial coefficients are withdrawn from the subsequent equalities:

11)0( add ==

LaaLdd212

)( +==

Fig. 9.2 The 2D spar (link) element

The displacement function yields

=+=

2

1

1211)()(

d

d

L

s

L

sdd

L

sdsd (9.12)

emphasizing the shape functions matrix N,

=

L

s

L

s1N (9.13)

The longitudinal strain is

B=

=

=

2

11)(

d

d

L

s

L

s

ss

sds (9.14)

x

1

x1

y1

x2

y2

2

u1

v1

u2

v2 )d(s

(9.19)


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Due to the transformation relationship (9.10) the nodal displacements can bewritten as

=

1

1

1v

u

L

L

L

Ld

yx

(9.15)

=

2

2

2v

u

L

L

L

Ld

yx

Substituting in the strain relationship, the following relationship outcomes

[ ]

=

3

2

1

1

2

1

u

u

v

u

LLLLL

yxyxs , (9.16)

emphasizing the Bmatrix of derivatives in the global coordinate system

[ ]yxyx LLLLL

=2

1B . (9.17)

In the stiffness matrix expression =eV

T dVEBBk the Ematrix reduces to a

single term E, while the Bmatrix is a constant, so that both are withdrawn

outside the integral sign. The elements volume is AL. Thus, the elemental

stiffness matrix yields

[ ]yxyx

y

x

y

x

LLLL

L

L

L

L

L

AE

=3

k (9.18)


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9.3 EXAMPLE 2 Triangular element in 2D stress/strain state

For the triangular element shown in figure 9.3, the displacement field is

uniquely defined by [ ]Tvu=d and the stress and strain vectorscomponents are [ ]

xyyx

T = and [ ]xyyxT = .

Note: for the plane strainhypothesis z, which is normal to the (x,y) plane,

is not zero, but because z= 0 it has no contribution to the internal work.

Fig. 9.3 The 2D triangular solid element

The general expression of the displacement field is

[ ]

=

3

2

1

1),(

a

a

a

yxyxd (9.19)

The unknown parameters a are determined by replacing the nodal

displacement (assumed known) and the nodal coordinates (xi,yi)i=1,3

1

2

u1

v1

u2

v2

u3

v3

3

M (x,y)


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95

=

3

2

1

33

22

11

3

2

1

1

11

a

aa

yx

yxyx

(9.20)

thus

a= CM-1

e (9.21)

with

=

123123

211332

122131122332

1

2

1

xxxxxx

yyyyyy

yxyxyxyxyxyx

AMC (9.22)

and

=

33

22

11

1

1

1

det2

yx

yx

yx

A = 2 the triangle area.

In matrix form, the displacement field yields,

[ ] eyx Cd M11 = (9.23)

The element strains are:

[ ] [ ]

==

=

3

2

1

211332

1

2

1010

u

uu

yyyyyyAx

ux uCM

[ ] [ ]

==

=

3

2

1

123123

1

2

1100

v

v

v

xxxxxxAy

vy vCM (9.24)

[ ] [ ]

+

=

+

=

3

2

1

1

3

2

1

1010100

v

v

v

u

u

u

x

v

y

uxy MM CC

in matrix form:


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96

=

=

3

3

2

2

1

1

211213313223

123123

211332

000

000

v

u

v

u

vu

yyxxyyxxyyxx

xxxxxx

yyyyyy

xy

y

x

= B e (9.25)

Note: the matrix Bis constant throughout the element, due to the first degree

polynomial used as shape function.

The elasticity matrix, for plane stress isotropic state is

=

2

100

01

01

1 2

EE (9.26)

while for plane strain isotropic state

( )( )( )

( )

+=

12

2100

011

01

1

211

1EE (9.27)

The stiffness matrix is defined by the general relationship =eV

TdVEBBk

where dV = t dx dy, with t the thickness of the element (t = 1 for plane

strain). The product is constant, hence tATEBBk= , with Athe area of theelement.

The nodal forces due to initial strain are


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== VTT

AtdV 000 EBEBr , (9.28)

and the nodal loads due to body forces

= VT

f dVfNr with dxdyf

f

r

ri

y

x

fy

fx

i

i

=

N (9.29)

The body forces are distributed to the element nodes in equal parts

3

At

f

fr

y

x

fi

= . (9.30)

9. chapter 9 - fe in gobal coordinates _a4l

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