lecture 4 jan 25

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LECTURE 4

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THE FINITE ELEMENT METHOD

or NODAL APPROXIMATION METHOD:

The basic concept behind the Finite

element method is going from part to whole

Name FINITE ELEMENTcoined by

Clough

Fitting of a number of piecewise continuous

polynomials to approximate the variation of

the field variable over the entire domain

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STEPS INVOLVED IN THE FINITE ELEMENT

METHOD:

Discretisation of the structure

Selection of suitable displacement model

Derivation of elemental matrices and loadvectors

Assembly of elemental equations to obtainoverall stiffness matrix

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STEPS INVOLVED IN THE FINITE ELEMENT

METHOD:contd

Imposition of boundary conditions

Solutions for the unknown nodal

displacements

Computation of elemental strains and

stresses

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1

2

3

2

1

10 kN

A 1= 2sq.cmA 2= 1sq.cm

L 1= 10 cm

L 2= 10cm

E= 2x107N/cm2 BC:

U1= 0

Pl= 10kN

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u1

u2

u3

N1u1+ N2u2

N1u2+ N2u3

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u(x) = a1+ a2x

u(x) = N1u1+ N2u2

Here Nis are called Shape functions or

Interpolation functionsShape functions are used to interpolate

the field variable over the element in

terms of nodal values of the field variable

1=N+N

1=)(N0.=)0(Nx/=(x)N

0=)(N1.=)0(Nx/-1=(x)

21

121

111

N

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It can be verified that

= 0 i j= 1 i = j

=

(Kronecker Delta Function)

.

N xi j( )

ij

1 2 1 2

u1

u2N1 N2

1

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To provide for the possibility of a constant

or uniform field when u is constant at all

points in the domainWe have

u1= u2= . = un= c

(x)=u(x)==u(x)n

1j=

i

n

1j=

ii NcNc

1=(x)N

NN

i

n

1j=

21

or

ccc

The above properties are very important

properties of shape functions.

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In FEA, we use the nodal approximation to

specify the unknown function in terms of its

values at selected nodal po ints, through a

Nodal App rox imation

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Now let us consider the numerical example

of the tapered beam whose area of crosssection varies uniformly from A1to A2at the

free end and subjected to its own self

weight and a point load at the end.

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A(x) = A1(A1- A2) x/l

ie.A(x) = 80(80-20)x/300

= (800.2x)

Specific weight = 0.075 N/cm3Young's Modulus E = 2 x 107N/cm2

Example

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The governing equation is

in 0 < x < L

With B.Cs i) u(0) = 0

and

ii)At x=lP]

dx

du[EA(x)

0=A(x)+]dx

du[EA(x)

dx

d

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Weak form is given by

Substituting in the weak formu(x) = N1u1+N2u2

And w(x) as N1first and then N2we get a

system of two equations in two unknownsnamely u

1and u

2

w(0)P(0)-))w(P(dxw(x)A(x)=dxdxdw

dxdu)(

00

llxEA

ll

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w(0)P(0)-))w(P(dxNA(x)

=dx

dx

dN

dx

)d(N)(

1

0

12211

0

ll

uNuxEA

l

l


=dxdx

dN

dx

)d(N)(

2

0

22211

0

ll

uNuxEA

l

l

----1

----2

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=dx

dx

dN

dx

)d()(dx

dx

dN

dx

)d(N)(

1

0

212

0

111

0

ll

uN

xEAuxEA

l

ll


=dxdx

dN

dx

)d()(dx

dx

dN

dx

)d(N)(

2

0

222

0

121

0

ll

uN

xEAuxEA

l

ll

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ee r=u][ eK

dxdx

dN

dx

dNEA(x)=

ji

0

l

e

ijK

dxN)(A= j0

xrl

e

j

These 2 equations can be written in matrix form a

2

1

2

1

2221

1211

r

r

u

u

KK

KK

Where

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1N

l

xN =

2l

x

= 1 -

dN

dx

1

l

1 dN

dx

2

l

1

== -

We know that the shape functions for a

two noded element are given by

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K11 dxdx

dN

dx

dNEA(x) 11

0

l

=

xA-A

-A 2110

lE

l

dx1

2

l

=

l

AAE

l

E

2

)()

2

A+

2

A( 2121

=

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=

K12dx

dx

dN

dx

dNA(x) 21

0

E

l

=

= x

l

E

l

A-A

-A 2110

)2

A+

2

A( 21

l

E

K12 K21=

1

l

dx1

l

l

AAE

2

)( 21

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K22

[ ]Ke

11-

1-1

2

A+A 21

l

E

Therefore the element stiffness matrix will be

=

= dxdx

dN

dx

dNA(x) 22

0

E

l

=

=

A-A

-A 2110

lE

l

)2

A+

2

A( 21

l

E

dx1

2

l

l

AAE

2

)( 21

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Similarly the element nodal load vector will be

dxN)( 1

0

1 l

xAr

dxAl

])l

x-(1

l

)A-(A-[ 211

0

=

=

=

=

ll 6

A

3

A21

dxN)( 20

2 l

xAr

dxAl

])l

x(l

)A-(A-[ 2110

ll

3

A

6

A 21

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Therefore the assembled load vector will be

Case - I: Discretize the Tapered Bar into 3elements.

The length of each element = 100 cm.''l

=er

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7070

7070

10011-

1-1

2

A+A 21

1

1 E

l

EK

5050

5050

10011-

1-1

2

A+A 32

2

2 E

l

EK

3030

3030

10011-

1-1

2

A+A

43

3

3 E

l

E

K

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[ ]

]

]

K1

[K

[K

2

3

=[K]

The global stiffness matrix will become

100

E

3030-

30-30+5050-

50-50+707070-70

100

E

3030-00

30-8050-0

050-12070

0070-70

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12

21

2 2

2

61

AA

AAl

r

r

6

2006

220

100x

680

6

100

100x

6

1406

160

100x

1r

2

r 3r

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Similarly the assembled global load vector

will become

[R] = +

|r|

|r|

||

3

2

1r

3

2

1

P

P

P

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The global load vector is

[R] = +

6

80

6

100

6

140

6

160

6

200

6

220

100x

R

O

O

P

P

O

O

R

+

80

240

360

220

6

100x=

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Now the total system of equation will be

E

100

- 70

-70 120 - 50

50 80 - 30

- 30 30

70

u

u

u

u

1

2

3

4

P

O

O

R

80

240

360

220

6

100x

=

Now applying the Boundary conditions i.e. u1 = 0 ..

Delete the first row and first column of elements and

the system of equation will reduce to

120 - 50

- 50 80 - 30

30 30

u

u

u

2

3

4

P

O

O

80

240

360

6

100x=

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The data are E = 2 x 107N/cm2= 0.075 N/ccand P = 1 x 105N.

On solving the above equation we get

u4 = 0.035501997 cm

u3 = 0.018818567 cmu2 = 0.008778557 cm

The deflection at mid section of the bar by

interpolation is

= 0.01379856 cm2

u+u=U

32

50x

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Example 2 Let us consider the discretization

with 2 elements

h = 150 cm

The assembled stiffness matrix will be

[K] =E

150

65 - 65

65 + 35

-35 35

65

Similarly the assembled load vector will be

[R] = +x 150

210

6

+120

6

90

6

180

6

R

O

P

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After applying the B.Cs the global system of

equation will become

=

On solving the above set of simultaneous

equations we get

u3=0.033068406 cm (Tip displacement)

u2=0.011607692 cm (Mid section

displacement)

E

u150

3

100 - 35

-35 35

u2

x 150

240

6

80

6

O

P

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[ ]Ke 11-1-1

2A+A 21

lE

[ ]Ke

11-

1-1

l

EA

For a bar of constant cross section A1= A2

=

1

1

2

Aler

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Example 3

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WEAK FORM OF GOVERNING

EQUATION FOR THERMAL

PROBLEMS

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where

k = Thermal conductivity coefficient

h = Thermal convection coefficientA = Area of cross section subjected to

CONDUCTION

p= Perimeter is the area exposed to

CONVECTION

T = Atmospheric Temp. , T = Variable

Q = Heat Source

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(q + dq) q + hp dx(T - T)=0

by dx we getdq + hp(T - T) = 0

dx

d(-kA(x) dT ) + hp(T - T)=0

dx dx

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Boundary conditions:

i) At x= 0 T = To

ii) At the free end any one of the following

three possible boundary conditions could

be specified

1. If free end is insulated _ kA dT/dx = 0

2. If free end is open to atmosphere_ kA dT/dx|=l = hA(T- T)

3. Specified temperature T(l) = Tl

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0)(

TThp

dx

dTKA

dx

d

0)()( dxxRxwThe weak form can be obtained by

The governing equation for heat transfer in

a one dimensional problem is given by

For a bar of length l with wall temperature T

the weak form of the governing equation

becomes

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l

dxTThpdx

dTKA

dx

dxw

0

0)()(

l l

dxTThpxwdxdx

dTKA

dx

dxw

0 0

0)()()(

1

dxdx

dTKA

dx

dxwI

l

0

1 )(

)(xwu dwdu

dxdx

dTKA

dx

ddv

dx

dTKAv

Let

and

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vduuvI1

dxdxdw

dxdTKA

dxdTKAxwI

ll

001 )(

0)()()(000

dxTThpxwdxdx

dw

dx

dTKA

dx

dTKAxw

lll

Substituting the above term in equation 1,

we get

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0)()()()(0000

dxTxhpwdxxTxhpwdxdx

dw

dx

dTKA

dx

dTKAxw

llll

)()()()()(000

TThAxwdxTxhpwdxxTxhpwdxdxdwdxdTKA L

lll

Boundary term B1(T,w) B2(T,w) l(w)

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Substituting in the weak form

T(x) = N1T1+N2T2

And w(x) as N1first and then N2we get a

system of two equations in two unknownsnamely T1and T2 which can be written as

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2

1

2

1

2221

1211

2

1

2221

1211

q

q

T

T

KK

KK

T

T

KK

KK

convcond

Where dxdx

dN

dxdNkA(x) =K ji

l

eijcond 0

dxNhpT =q je

j

l

0

dxNNhp(x) =K ji

l

conv

e

ij 0

L t th l t b f l l th l

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Let the elements be of equal length

The element matrices are

l

hA

+hP l

+-

-

l

KA] =[Ke

0

00

21

12

611

11

hA T

+hPl T] =[fe 011

2

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Boundary conditions:

at x = 0, T(0) = T

at x = L,

conduction = convection loss

For a typical linear element

)- T= hA (Tdx

dTKA ll

(x/l) =N- (x/l) =N

J

I 1

Let the elements be of equal length l 2

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Let the elements be of equal length

The element matrices are

cml = 2

hA

+hp l

+-

-

l

kA] =[Ke

0

00

21

12

611

11

hA T

+Thpl

] =[qe0

1

1

2

Th l t t i f ELEMENT (1)

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The element matrices for ELEMENT (1),

(2) & (3) are

20

20

66666675

66756666} =; {q

..-

. -.=][K e

e

therm

20

20

66703330

33306670} =; {q

..

..=][K econv

e

20

20

66

66 } =; {q

=][K econde

Th l t t i f ELEMENT (4) i

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The element matrix for ELEMENT (4) is

28

20

06676675

66756666} =; {q

..-

. -.=][K etherm

e

=][K conde

66

66

400

00

66703330

33306670

.

..

..=][K conv

e

8

0

20

20} ={qe

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On assembly we get

6.667 - 5.667 0 0 0

-5.667 13.33 - 5.667 0 0

0 - 5.667 13.33 - 5.667 0

0 0 - 5.667 13.33 - 5.667 0 0 0 - 5.667 7.066

*

T1

T2

T3

T4T5

=

20

20+20

20+20

20+20 28

B l i B d diti t

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By applying Boundary condition at

at x = 0 T = T0 = 80

By solving we get

13.33 - 5.667 0 0

-5.667 13.33 - 5.667 0

0 - 5.667 13.33 - 5.667

0 0 - 5.667 7.066

*

T2

T3

T4

T5

=

40 + 5.667*80

40

40

28

C;.=T

0

2 9553C;.=T 0

3

8839

C;.=T 04 8232 C;.=T0

5 2930

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Boundary condition: Free end insulated

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h = 10 W/cm2 oC

K = 70 W /cm oC

T0= 140oC

T = 40oC

= 5 cmRadius r = 1 cm

Area A = r2 = cm2

Perimeter p = 2r = 2

lecture 4 jan 25

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