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7/27/2019 IGCAR FEM , syllabus
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FINITE ELEMENTS ANALYSIS
lectures at IGCAR
Lecture-1
by
Dr.C.Jebaraj
Professor of Mechanical Engineering
AU-FRG Institute for CAD/CAM
CEG Campus, Anna University
Chennai-25
jebaraj@annauniv.edu
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1. Why CAD/CAM/CAE/CAPP.CAx
2. SAFE DESIGN
3. OPTIMUM DESIGN1. In shape
2. In weight
3. In size4. In cost
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Example of an Optimum Design
2
max
2
wLPLM
Z
MyI
MfyI
Mf
R
E
y
f
I
M
maxmaxmax;
Bending formula
BMDQuadratic
Pw/Length
L
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Z-variation
L
d
b
261
2
3121
bdd
bd
y
IZ
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STRAIGHT BAR SUBJECTED TO AN AXIAL LOAD
AE
PLU max
L
P
P= applied load
L= length of rod
E= youngs modulus of material
A= area of cross section
dx
AE
Pdx
du
dx
duE
A
P
E
E
(Elongation ofsmall segment dx)
AE
PLU
AE
PdxxU
L
L
max
AE
Px
)(
0
0
Total elongation
P
dx
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Umax
)()( LINEARAE
PxxU
STRAIGHT BAR SUBJECTED TO AN AXIAL LOAD.
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TAPER ROD SUBJECTED TO AXIAL LOAD
0
log
min
max
0
0
at xU
Lat xU
A(x)
E
P
A(x)EPdxU(x)
L
L
L
P
P
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TAPERED ROD SUBJECTED TO AXIAL LOAD
WITH SELF WEIGHT
f(x)dxE
A(x)
(x)dxP
EA(x)E
(x)dxPU(x)
ht (x)Self weigPAxial Load(x)PTotal load
1
1
NOTE:
i. We may not have explicitly integration formulae
ii.The solution is highly complex than a
logarithmic solution.
P
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Trapezoidal method (Linear)
h)......fff(fff
f(x)dx
n
1432
21
2
NUMERICAL INTEGRATION
ERROR IS SMALL
L0
h 0.05L
f(x)
f1 f2 f3 f4 fn+1 fn
h 0.1 L
f(x)
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SIMPSONS METHOD (QUADRATIC)
Error is small
2...)(2
42
2
9753
86421
h
ffff
.....)fff(fff
f(x)dx
N
h 0.1 L
f(x)
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Summary
1. In the absence of an explicit integration
formulae available, we accept numericalintegration formulation which is an
approximation only
2. Approximation is improved byi) Increasing the No of segments
ie., by decreasing h
ii) Increasing the order of polynomial ofrequirements.
3. Error is minimized and there is convergence
towards exact solution
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Formulation of Problem for Approximation
solution
System L is under equilibriumSubsystem dx is also under
equilibrium
+ d P
AdxAd
AdxAdAFv
0
0
L
dx
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0)(
0)(
Adx
duEA
dx
d
AdxAdx
duEd
AdxAEd
Domain : 0 < x < L
0)()(:..
xA
dx
duxEA
dx
dEDG
BCs : at x=0 , u=0
at x=l , R=P
Pdx
duEA
Boundary
value
problem
P
L
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Approximate Solution
0)()(:..
xAdx
du
xEAdx
d
EDG
Domain : 0 < x < L
BCs : at x=0 , u=0
at x=l , R=P
1
nnxa...........xaxaxaxaa
solutionroximationbe the appuLet
4
43
32
210U*
U*
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3
3
2
210 xaxaxaau*
Assume,
Apply BCs
23120* aau
Substituting 2 into 1 we get
0
):(
)(0)()(*
i
i
da
dR
axR
RESIDUExRxAdx
udxEA
dx
d
Should be minimum
2
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ILLISTRATIVE PROBLEM
Consider the equation
2
2
Xdx
dU
xdx
d
in the domain 1< X < 2
With BCs as U(1) =2 and
21
2 Xdx
dUX
CONSTRUCTION OF TRIAL FUNCTION
)x(..........)x()x()x( 110 nnaaU
2)1(..........)1()1()1( 110 nnaaU
if .....n1,2,3.....i0)1(then2)1(0 i
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2
1..................
22
11
2
0
2
X
nn
XXXdx
dxa
dx
dxa
dx
dx
dx
UdX
1
if
2
1
2
0
Xdx
d
x
then ....n1,2,3.....i02
X
i
dx
dx
Let
2aaaa(x)U
xaxaxaa(x)U
4321
3
4
2
321
or 4321 aaa2a
2
1)12a4a2(a
dx
UdX 432
2X
2
432 12a4a
4
1a
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Substituting for a1& a2in the expression for , we have
xa)x(a
11xx1xa3x1xa1x4
12xU
22110
2
43
11xx1x23x1x1
1x4
1
20
2
It can be easily seen that the above trail function satisfiesthe conditions imposed on the boundary. Thus the construction of
trial function is over.
xU
when
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WRM APPLICATIONConsider the equation
0
22
xdx
xdUxdx
d
Substituting the trial solution xU for xU , this equation is unlikely to be satisfied
022
xdxxUdx
dxd
This is called as a Residue and is a measure of the error involved.
222
1x2a43x3a1x4
41xR
42
31
aa
aa
i.e., the RHS is a non zero function R(X)
i.e R(x)
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COLLOCATION METHODFor all
ia choose a point ix in the domain and at each such ix
force the residual to be exactly zero
i.e R(x1) = 0R(x2) = 0
R(xn) = 0
The chosen points are called collocation point.
They may be located any where on the boundary or in the domain .
For the present problem we have 2 parameter
Therefore select any two collocation point say1a &. 2a
X1=4/3 & X2=5/3Substituting in the expression for R(x) we have
100
9713
3
8
8
114
3
4
21
21
aa
aa
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1a =2.0993 &
2a = -0.356
11xx1x0.3563x1-x2.09931x412xU 2
therefore
Solving the simultaneous equation
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THE SUB- DOMAIN METHODFor each undetermined parameter
1a choose an intervalx , in the domain.
Then force average of the residual in the each interval to be zero.
x11
0R(x)dxx
1
.
.
.
0R(x)dxx
1
x22
xnn
0R(x)dxx
1
1
X1 i
R(x)dxx
1 iX
x1
x2
x3
x4
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which yields n system of residual equations which can be solved for1
a
The intervalsix are called the sub-domain.
They may chosen in any fashion.
Taking
1
x = 1< X < 1.5,2
x = 1.5 < X
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LEAST SQUARE METHOD
2
1
2 0)( dxxRai for i. 1,2,3n
2
10/ dxaddRXR i i=1,2,3,n
this yield the results
1= 2.3155
2= -0.3816
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THE GALERKIN METHOD
For each parameter 1 we required that a weighted
average of R(x) over the entire domain be zero.
xi
associated with .a i
Weighting functions are the trial function
.
.
.
0dx(x)R(x)
2
1
1
0dx(x)R(x)2
1
i this yields 1= 2.13782 = - 0.3477
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FINITE ELEMENTS ANALYSIS
lectures at IGCAR
Lecture-2&3
by
Dr.C.Jebaraj
Professor of Mechanical Engineering
AU-FRG Institute for CAD/CAM
CEG Campus, Anna University
Chennai-25
jebaraj@annauniv.edu
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Comparison of WRM
Collocation method = R(x) = 0
Ritz method
0.)( FdxWxR
0)(2
dxda
dR
xRi
Sub domain method 0)( dxxR
Least square method
0)(
0.)(
dxxR
FdxWxR
i
Galerkin method
RITZ VARIATIONAL METHOD (weak formulation)
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RITZ VARIATIONAL METHOD (weak formulation)
Staring with equation
in0xfdx
dU
xdx
d
The WR became
0
dxxf
dx
dUx
dx
dxW
Xb
Xa
where W(x)= weighing function
0 dxxWxRie
0)()(
dxxWxfdxxWdxdU
xdx
d
RITZ VARIATIONAL METHOD (contd )
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RITZ VARIATIONAL METHOD (contd ..)
Xb
Xb
Xb
Xa
Xb
Xa
dxdx
dW
dx
dUx
dx
dUxxW
dxdx
dUx
dx
dxW
Integration by parts for first term:
vduuvudv
xUxW
xU
In Ritz method we take
where
is specified, as at the boundary, W(x) = 0.
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0
xAdx
du
XEAdx
d
L
dxxAdx
duxEA
dx
dxw
0
0
The governing equation is
The WR formulation is
where w(x) is the weighting function and
u(x) is the trial solution.
Integrating by parts and re arranging we get
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LPLwPwdxxwxA
dxdx
dw
dx
duXEA
L
L
000
0
Since u(0) = 0 (specified), uw at x = 0 vanishes
LPwdxxwxAdxdx
dw
dx
duXEA
L L
0 0
Integrating by parts and re-arranging we get
)( dvdxxEA Strain energy stored
External
work done
LAW OF CONSERVATION OF ENERGY!!
LHS =
Solution Procedure
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Solution Procedure
,,, ii
(x) wj j
j
a
xaxaxau(x)
a) u(
xaxaxaaLet u
321
3
1
33
221
000
3
3
2
210
Th V i ti l E ti
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The Variational Equation
L(w)B(u,w)
Pw(L)A(x)wdxdxdx
dw
dx
duxEA
1,2,3iw
au
i
3
1j
jj
Now substituting
ii
iii
j
jj
raK
LPdxxAdxdxd
dxdxEAa
)()(..)(3
1
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L
ij
ij dxdx
d
dx
dxEAk
0
)(0
LPdxxxAr i
L
ii
3
3
2
2
1
x
x
x
23
2
1
3
2
1
xdx
d
xdx
d
dx
d
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300
0
1111 dx
dx
d
dx
dxEAk = E(80-0.2x) .1.1 dx
=1.5x104E
dxdxd
dx
dxEAk 2112
= E(80-0.2x) .1. 2x dx
= 3.6x106E
dxdx
d
dx
dxEAk 3113
= E ( 80 - 0.2 x) .1. 3x2dx
= 9.45 x 108 E
Similarly k21=..
k22=
.
.
k33=
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511 103773.12.080)( dxxxdxxAr
7222 104.22.080)( dxxxdxxAr
9333 10598.42.080)( dxxxdxxAr
xPL(L)PP 710311
10x9L)( 9222 PLPP
1027 11333 xPL(L)PP
864 10x9 4510x3 6x1051 a
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119
97
75
3
2
1
14118
1196
10x2710x4.598
10x910x2.4
10x310x.371
10x1.32210x3.8810x9.45
10x2.8810x1.210x3.6
10x9.4510x3.6x105.1
a
a
a
On solving
a1 = 6.6762 x 10-5
a2 = -4.946 x 10-8
a3 = 6.4736 x 10-10
U(x) = a1x + a2x2 + a3x
3
U(X=300) = uL= a1 (300) + a2 (300)2+ a3 (300)
3 = 0.033056 cm
But Exact Value is 0.0378 cm.
CONVERGENCE STUDY fi t
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L
U(x)
0.0378
10Th Order
0
6Th
Order
0.03306
CONVERGENCE STUDY-p refinementincreasing the order of approximate polynomial
NODAL APPROXIMATION METHOD
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NODAL APPROXIMATION METHOD
he
i je
Be(u ,w ) = L e(w )
e
he he
e
e
ee
PiwdxxAdxdx
dw
dx
du
XAE 0 0
Ue= a0+ a1 x
=
1
0
a
a
he
xAAAxA jii
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Ui=
1
0
a
a
Uj=
1
0
a
a
1
0
1
01
a
a
heU
U
j
i
j
i
U
U
hea
a1
1
0
1
01
101
U
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.
1
011
j
i
U
U
hexU
j
i
U
U
he
x
he
x1
2
1j
jjNu
1,2i, iNw
.he
x
N,he
x
N 21 1
;
1
,
1 21
hedx
dN
hedx
dN
hex
AAAxAjii
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Characteristic of shape function
hexN 11
hexN 2
N1 N2
N1= 1 at x=0
N1=0 at x=he
N2= 0 at x=0
N2= 1 at x=he
N1+N2=1
h
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2
11.
.)(
21
211
he
0
1111
AAheE
dxhehehe
AAAE
dxdx
dN
dx
dNxEAk
2
11.
21
21
12112
AA
he
E
dxhehehe
AAAEkk
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xAAhe
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1
2
0
2
36P
heA
heA
Pdxhe
x
he
AAAr
ji
ji
i
On substituting the value
7070
7070
100
1 Ek
0
6
2006
220
1001 Rr
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For element 2
50505050
100
2
Ek
O
Or
6
1406
160
1002
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For element 3
3030
3030
100
3 Ek
P
O
r680
6
100
100
3
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On assembling
3
2
1
k
k
k
4
3
2
1
u
u
u
u
r1
r2
r3
=
17070 U
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4
3
2
1
303030305050
50507070
7070
100UU
U
U
EX
6
806
1006
1406
160
6
2006
220
100
P
R
0
0
17070 U
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4
3
2
1
3030308050
5012070
7070
100UU
U
U
EX
6
806
240
6
3606
220
100
P
R
0
0
A li Gl b l b d diti U 0
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Appling Global boundary condition U1= 0
P
OO
U
UUE
80
240360
6
100
3030
30805050120
1004
3
2
U4= 0.0355,
U3= 0.0188,U2= 0.0088.
Umaxexact = 0.0378
CONVERGENCE STUDY h improvement
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CONVERGENCE STUDY- h improvement
increasing No of element
L0.0378
0
0.0355
3 element
6 element
10 element
U(x)
C i f fi
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L
U(x)
0.0378
10Th Order
0
6Th
Order
0.03306
Comparison of refinement
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L0.0378
EXACT
0
0.0355
3 element
6 element
10 element
U(x)
Comparison of refinement (contd)
FINITE ELEMENTS ANALYSIS
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FINITE ELEMENTS ANALYSIS
lectures at IGCAR
Lecture-4
by
Dr.C.Jebaraj
Professor of Mechanical EngineeringAU-FRG Institute for CAD/CAM
CEG Campus, Anna University
Chennai-25
jebaraj@annauniv.edu
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HEAT TRANSFER BYCONDUCTION & CONVECTION
Heat transfer Problems using FEM
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Heat transfer Problems using FEM
K=conductivity
coeff..
h=convection coeff
T= Ambient Temp
P= Perimeter
A=area of cross
section
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Heat in = Heat out
)()( TThPdxAdqqqA
f luxheatdx
dTKq
0)()(
)(0
TTdxhPdx
dTKAd
TTdxhPdqA
dxby 0)()( TThPdx
dTKA
dx
d
q + dqq
dx
MATHEMATICAL FORUMULATION
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MATHEMATICAL FORUMULATION
0)()(
TThPdx
dTKA
dx
d
0 < X < L
@ x = 0, T=TO
@ x=L,)(
TThAdx
dTKA
BCs:
If the end is open toatmosphere
0
dx
dTKA If the end is insulated
GDE:
DOMAIN:
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Approximate Solution Procedure
Let the App solution be, T* = a0+ a1x + a2x2
+ a3x3
Apply B.Cs and get 23120* aaT
Substituting in to GDE*T
RESIDUExRTThP
dx
TdKA
dx
d )(0)()(
*
*
Minimisation of residue by RITZ METHOD
RITZ FORMULATION
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RITZ FORMULATION
0.)( dxFWxR
Let (x)be the weighting function
0)()()( dxxTThP
dx
dTKA
dx
d
0)()()()( dxxhPTdxxhPTdxxdxdT
KAdx
d
Integration by parts vduuvudvdv
u
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ddT
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0
).().()(
)(
xLx dxdTKAx
dxdTKAxdxxhPT
dxxhPTdxdx
d
dx
dTKA
)(),( LTB
3
3
2
20 1 xaxaxaaT*
Q)(
)(
dxxhPT
dxxhPTdxdx
d
dx
dTKA
A
3
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321 ,,j j On applying eqn B into eqn A we get
e
j
ee
he
i
ee
hejiee
i
QdxATh
dxxTPhdxdx
d
dx
dAKa
)(00
;aTi
ii
*
1 B
ii raK
Where i=1,2,3 & j=1,2,3 and we can calculate Kij
and similarly ri
RITZ E ti f Fi it l t
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RITZ Equation for a Finite element
O heI Je
)(),( ee LTB
he
eee
he
eeeehe
ee
QdxxTAh
dxxTphdxdx
d
dx
dTAK
0
00
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T should be expressed in term of e
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J
I
T
T
T should be expressed in term of
nodal variables
eJI
0 he
heaahe, Tx
aa, T@ x
J
I
10
10
00
Putting it in matrix form
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Putting it in matrix form
1
0
1
01
a
a
heT
T
J
I
J
I
T
T
hea
a1
1
0
1
01
matrix form cont*
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J
I
T
T
he
T
1
*
1
01 x1
J
I
T
T
he
x
he
x1
J
I
JI
T
TNN
xxaaT* 110
1
0
a
a
2 T
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321
2
1
,,iNxfunctionweighting
T
Tre q wheqNT
i
J
I
j
j
jj
*
II
he
eee
heeee
eheee
QdxxTAh
dxxTphdxdx
d
dx
dTAK
0
0
*
0
I
Repeating the RITZ Eqn. I have
On substituting II in I
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On substituting II in I
he
i
ee
he
ij
eeijhe ee
j
j
dxNTPh
dxNNphdxdx
dN
dx
dNAKq
0
00
2
1
;;12121 21hex N
hex-; N,; j,i
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heKA
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322
hehP
he
KAK
21
12
611
11 hehP
he
KAKe
dxNhPTrly 11lll
21he
hPTdxhe
x
hPT
hehPTdxNhPTr
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2 22 hPTdxNhPTr
11
2hehPTre
eJ
Ie rT
TK
1
1
221
12
611
11 hehPT
T
ThehP
he
KA
J
I
On Appling the element stiffness matrix and load
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vectors and introducing the continuity condition
we get
3
2
1
kk
k
4
3
2
1
T
T
T
T
2
1
r
r
2
1
r
r
2
1
r
r=
On applying the BCs and solving for the
remaining system of equations we get the nodal
values
ILLUSTRATION BROBLEM
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ILLUSTRATION BROBLEM80OC
8 cm
1 cm4cm
K = 3 w/cm 0c
h = 0.1 w/cm20cT= 20
0c
51 2 3 4
421 3
Let the element be of equal length he = 2 cm
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The element matrices are
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hA
hehP
he
KA
K
e
0
00
21
12
6
11
11
ThA
ThehPf e
0
1
1
2
The element matrices for ELEMENT 1, 2 & 3 are
20
20;
666.6667.5
667.5666.6ee fK
The element matrices for ELEMENT 4 is
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28
20
;066.7667.5
667.5666.6ee
fK
The assemblies of the element matrices are
28
40
40
40
20
006.7667.5000
667.5337.13667.500
0667.533.13667.50
00667.5333.13667.5
000667.5667.6
5
4
3
2
1
T
T
T
T
T
Appling the boundary condition T1= 80oC
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1
28
4040
80667.540
006.7667.500
667.5333.13667.500667.5333.13667.5
00667.5337.13
5
4
3
2
T
TT
T
3.308.329.399.5380T
Exact solution is
6.302.332.403.5480T
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Description Solidmechanic
Equation
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mechanic
= Variable U=axial
displacement
(x) = property
associated with
the DERIVATIVE
of the variable.
E=youngs
modulus
= Property
associated with
the variable0
f(x) = Load not
at all associated
with the variablerA(x)
0xrAdx
dyxEA
dx
d:GDE
Description Fluidmechanic
Equation
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mechanic
0Qdx
dK
dx
d:GDE XX
f(x) = Load not
at all associated
with the variable
Q=volume of flow
rate
= Property
associated with
the variable
h = convec co-eff
(x) = property
associated with
the DERIVATIVE
of the variable.
Kxx= Permeability
Co-eff
= Variable=fluid head
Description Electromagnetic
Equation
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magnetic
0:
dx
dv
dx
dGDE
= Variable V=Electric
potential
= Property
associated with
the variable
0
(x) = property
associated with
the DERIVATIVE
of the variable. permissivity
f(x) = Load not
at all associated
with the variable
=Volume
charge
Density
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