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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
w(0)P(0)-))w(P(dxNA(x)
=dxdx
dN
dx
)d(N)(
2
0
22211
0
ll
uNuxEA
l
l
----1
----2
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w(0)P(0)-))w(P(dxNA(x)
=dx
dx
dN
dx
)d()(dx
dx
dN
dx
)d(N)(
1
0
212
0
111
0
ll
uN
xEAuxEA
l
ll
w(0)P(0)-))w(P(dxNA(x)
=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