introduction to finite element methods
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Introduction to Finite Element Methods. UNIT I. Numerical Methods – Definition and Advantages. Definition: Methods that seek quantitative approximations to the solutions of mathematical problems Advantages:. What is a Numerical Method – An Example. Example 1: . - PowerPoint PPT PresentationTRANSCRIPT
Introduction to Finite Element Methods
UNIT I
Numerical Methods – Definition and Advantages
Definition: Methods that seek quantitative approximations to the solutions of mathematical problems
Advantages:
What is a Numerical Method – An Example
Example 1: 0)0( uuudtdu
What is a Numerical Method – An Example
Example 1: 0)0( uuudtdu
What is a Numerical Method – An Example
Example 2: 0t- )0( e52 uuu
dtdu
What is a Numerical Method – An Example
Example 2: 1)0( e52 t- uudtdu
What is a Numerical Method – An Example
Example 3: 1)0( 2 utudtdu
What is a Finite Element Method
Discretization
1-D 2-D
3-D?-D Hybrid
Approximation
Numerical Interpolation
Non-exact Boundary Conditions
Applications of Finite Element Methods
Structural & Stress Analysis Thermal Analysis Dynamic Analysis Acoustic Analysis Electro-Magnetic Analysis Manufacturing Processes Fluid Dynamics
Lecture 2
Review
Matrix Algebra • Row and column vectors
• Addition and Subtraction – must have the same dimensions• Multiplication – with scalar, with vector, with matrix
• Transposition –
• Differentiation and Integration
Matrix Algebra • Determinant of a Matrix:
• Matrix inversion -
• Important Matrices• diagonal matrix• identity matrix• zero matrix• eye matrix
Numerical Integration Calculate: dxxfI
b
a
• Newton – Cotes integration
• Trapezoidal rule – 1st order Newton-Cotes integration
• Trapezoidal rule – multiple application
)()()()()()( 1 axab
afbfafxfxf
b
a
b
a
bfafabdxxfdxxfI2
)()()()()( 1
n
n
n x
x
x
x
b
a
x
x
x
xn dxxfdxxfdxxfdxxfdxxfI
1
2
10
1
0
)()()( )()(
1
1
)()(2)(2
n
ii bfxfafhI
Numerical Integration Calculate: dxxfI
b
a
• Newton – Cotes integration
• Simpson 1/3 rule – 2nd order Newton-Cotes integration
)())((
))(()())((
))(()())((
))(()()( 21202
101
2101
200
2010
212 xf
xxxxxxxxxf
xxxxxxxxxf
xxxxxxxxxfxf
b
a
b
a
xfxfxfxxdxxfdxxfI6
)()(4)()()()( 210022
Numerical Integration Calculate: dxxfI
b
a
• Gaussian Quadrature
)(2
)()(2
)(2
)()()(
bfabafab
bfafabI
)()( 1100 xfcxfcI
Trapezoidal Rule: Gaussian Quadrature:
Choose according to certain criteria1010 ,,, xxcc
Numerical Integration Calculate: dxxfI
b
a
• Gaussian Quadrature
• 2pt Gaussian Quadrature
• 3pt Gaussian Quadrature
31
311
1
ffdxxfI
77.055.0089.077.055.01
1
fffdxxfI
111100
1
1
nn xfcxfcxfcdxxfI
abaxx
)(21~Let:
xdxabbafabdxxfb
a
~~)(21)(
21)(
21)(
1
1
Numerical Integration - Example Calculate: dxxeI x
1
0
sin
• Trapezoidal rule
• Simpson 1/3 rule
• 2pt Gaussian quadrature
• Exact solution90933.0
2cossinsin
1
0
1
0
xexedxxeI
xxx
Linear System Solver Solve: bAx
• Gaussian Elimination: forward elimination + back substitution
Example:
0 33322
062
21
321
321
xxxxx
xxx
230
610960
621
3
2
1
xxx
230
031322621
3
2
1
xxx
2330
21500960
621
3
2
1
xxx
Linear System Solver Solve: bAx
• Gaussian Elimination: forward elimination + back substitution
Pseudo code:
Forward elimination: Back substitution:
Do k = 1, n-1Do i = k+1,n
Do j = k+1, nkk
ik
aac
kii
kjijij
cbbb
caaa
Do ii = 1, n-1i = n – iisum = 0Do j = i+1, nsum = sum +
ii
ii a
sumbb
jijba
Finite Element Analysis (F.E.A.) of 1-D Problems
UNIT II
Historical Background
• Hrenikoff, 1941 – “frame work method” • Courant, 1943 – “piecewise polynomial
interpolation” • Turner, 1956 – derived stiffness matrice for
truss, beam, etc• Clough, 1960 – coined the term “finite element”
Key Ideas: - frame work method piecewise polynomial approximation
Axially Loaded Bar Review:
Stress:
Strain:
Deformation:
Stress:
Strain:
Deformation:
Axially Loaded Bar Review:
Stress:
Strain:
Deformation:
Axially Loaded Bar – Governing Equations and Boundary
Conditions• Differential Equation
• Boundary Condition Types • prescribed displacement (essential BC)
• prescribed force/derivative of displacement (natural BC)
LxxfdxduxEA
dxd
0 0)()(
Axially Loaded Bar –Boundary Conditions
• Examples• fixed end
• simple support
• free end
Potential Energy• Elastic Potential Energy (PE)
- Spring case
- Axially loaded bar
- Elastic body
x
Unstretched spring
Stretched bar
0PE
2
21PE kx
undeformed:
deformed:
0PE
L
Adx02
1PE
dvV
Tεσ21PE
Potential Energy• Work Potential (WE)
B
L
uPfdxu 0
WP
Pf
f: distributed force over a lineP: point forceu: displacement
A B
• Total Potential Energy
B
LL
uPfdxuAdx 002
1
• Principle of Minimum Potential Energy For conservative systems, of all the kinematically admissible displacement fields,
those corresponding to equilibrium extremize the total potential energy. If the extremum condition is a minimum, the equilibrium state is stable.
Potential Energy + Rayleigh-Ritz Approach
Pf
A B
Example:
Step 1: assume a displacement field nixau ii
i to1
is shape function / basis functionn is the order of approximation
Step 2: calculate total potential energy
Potential Energy + Rayleigh-Ritz Approach
Pf
A B
Example:
Step 3:select ai so that the total potential energy is minimum
Galerkin’s Method
Pf
A B
Example:
PdxduxEA
xu
xfdxduxEA
dxd
Lx
)(
00
0)()( Seek an approximation so
PdxudxEA
xu
dVxfdxudxEA
dxdw
Lx
Vi
~)(
00~
0)(~
)(
u~
In the Galerkin’s method, the weight function is chosen to be the same as the shape function.
Galerkin’s Method
Pf
A B
Example:
0)(~
)(
dVxf
dxudxEA
dxdw
Vi 0
~)(
~)(
00 0
L
i
L L
ii
dxudxEAwfdxwdx
dxdw
dxudxEA
1 2 3
1
2
3
Finite Element Method – Piecewise Approximation
x
u
x
u
FEM Formulation of Axially Loaded Bar – Governing Equations
• Differential Equation
• Weighted-Integral Formulation
• Weak Form
LxxfdxduxEA
dxd
0 0)()(
0)()(0
dxxf
dxduxEA
dxdw
L
LL
dxduxEAwdxxwf
dxduxEA
dxdw
00
)()()(0
Approximation Methods – Finite Element Method
Example:
Step 1: Discretization
Step 2: Weak form of one element P2P1x1 x2
0)()()()()(2
1
2
1
x
x
x
x dxduxEAxwdxxfxw
dxduxEA
dxdw
0)()()( 1122
2
1
PxwPxwdxxfxw
dxduxEA
dxdwx
x
Approximation Methods – Finite Element Method
Example (cont):
Step 3: Choosing shape functions - linear shape functions
2211 uuu
lx1 x2
x xx1 x0 x1
lxx
lxx 1
22
1 ;
2
1 ;2
121
xx
11 2
1 ;12 xlxxxl
xx
Approximation Methods – Finite Element Method
Example (cont):
Step 4: Forming element equation
Let , weak form becomes1w 0111211
122
1
2
1
PPdxfdxl
uuEAl
x
x
x
x
1121
2
1
Pdxful
EAul
EA x
x
Let , weak form becomes2w 0112222
122
1
2
1
PPdxfdxl
uuEAl
x
x
x
x
2221
2
1
Pdxful
EAul
EA x
x
2
1
2
1
1
1 1 1 1
2 2 2 22
1 11 1
x
x
x
x
fdxu P f PEAu P f Pl
fdx
E,A are constant
Approximation Methods – Finite Element Method
Example (cont):
Step 5: Assembling to form system equation
Approach 1:
Element 1:
1 1 1
2 2 2
1 1 0 01 1 0 0
0 0 0 0 0 0 00 0 0 0 0 0 0
I I I
I I II I
I
u f Pu f PE A
l
Element 2:1 1 1
2 2 2
0 0 0 0 0 0 00 1 1 00 1 1 00 0 0 0 0 0 0
II II IIII II
II II IIII
u f PE Au f Pl
Element 3:
1 1 1
2 2 2
0 0 00 0 0 00 0 00 0 0 0
0 0 1 10 0 1 1
III III
III III IIIIII
III III III
E Au f Plu f P
Approximation Methods – Finite Element Method
Example (cont):
Step 5: Assembling to form system equation
Assembled System:
1 1 1
2 2 2
3 3 3
4 4 4
0 0
0
0
0 0
I I I I
I I
I I I I II II II II
I I II II
II II II II III III III III
II II III III
III III III III
III III
E A E Al l
u f PE A E A E A E Au f Pl l l lu f PE A E A E A E Au f Pl l l l
E A E Al l
1 1
2 1 2 1
2 1 2 1
2 2
I I
I II I II
II III II III
III III
f Pf f P Pf f P P
f P
Approximation Methods – Finite Element Method
Example (cont):
Step 5: Assembling to form system equation
Approach 2: Element connectivity table Element 1 Element 2 Element 3
1 1 2 3
2 2 3 4
global node index (I,J)
local node (i,j)
eij IJk K
Approximation Methods – Finite Element Method
Example (cont):
Step 6: Imposing boundary conditions and forming condense system
Condensed system:
2 2
3 3
4 4
000
0
I I II II II II
I II II
II II II II III III III III
II II III III
III III III III
III III
E A E A E Al l l u f
E A E A E A E A u fl l l l
u f PE A E A
l l
Approximation Methods – Finite Element Method
Example (cont):
Step 7: solution
Step 8: post calculation
dxdu
dxdu
dxdu 2
21
1 2211 uuu
dxdEu
dxdEuE 2
21
1
Summary - Major Steps in FEM• Discretization • Derivation of element equation• weak form• construct form of approximation solution over
one element• derive finite element model
• Assembling – putting elements together• Imposing boundary conditions• Solving equations• Postcomputation
Exercises – Linear Element
Example 1:E = 100 GPa, A = 1 cm2
Linear Formulation for Bar Element
2
1
2212
1211
2
1
2
1
uu
KKKK
ff
PP
2
1
2
1
, x
xii
x
xji
jiij dxffKdx
dxd
dxdEAKwhere
x=x1 x=x2
1 2 1 1
x
x=x1 x= x2
u1 u2
1P 2Pf(x)
L = x2-x1
ux
Higher Order Formulation for Bar Element
(x)u(x)u(x)u(x)u 332211
)x(u)x(u)x(u)x(u(x)u 44332211
1 3
u1 u3ux
u2
2
1 4
u1 u4
2
ux
u2 u3
3
)x(u)x(u)x(u)x(u)x(u(x)u nn44332211
1 n
u1 un
2
ux
u2 u3
3
u4 ……………
4 ……………
Natural Coordinates and Interpolation Functions
21 ,
21
21
xx
Natural (or Normal) Coordinate:
x=x1 x= x2
x=-1 x=1
xx
0x x l
1xxx 1 2
2/ 2
x xx
lx
1 32
xx=-1 x=1
1 2
xx=-1 x=1
1 42
xx=-1 x=1
3
21 ,11 ,
21
321xxxxxx
1311
1627 ,1
31
31
169
21
xxxxxx
31
311
169 ,1
311
1627
43 xxxxxx
Quadratic Formulation for Bar Element
2
1
1
1
nd , , 1, 2, 32
x
i i ix
la f f dx f d i j x
2
1
1
1
2 x
j ji iij ji
x
d dd dwhere K EA dx EA d Kdx dx d d l
xx x
3
2
1
332313
232212
131211
3
2
1
3
2
1
uuu
KKKKKKKKK
fff
PPP
x=-1 x=0 x=1
31 2
Quadratic Formulation for Bar Elementu1 u3u2f(x)
P3P1
P2
x=-1 x=0 x=11x 2x 3x
21u11u
21u)(u)(u)(u)(u 321332211
xxxxxxxxxx
21 ,11 ,
21
321xxxxxx
1 1 2 2 3 32 2 1 2 4 2 2 1, , d d d d d ddx l d l dx l d l dx l d l x x x
x x x
1 2
2/ 2
x xx
lx
2
l d dxx 2d
dx lx
Exercises – Quadratic Element
Example 2:
E = 100 GPa, A1 = 1 cm2; A1 = 2 cm2
Some Issues
Non-constant cross section:
Interior load point:
Mixed boundary condition:k
Finite Element Analysis (F.E.A.) of I-D Problems – Applications
Plane Truss Problems
Example 1: Find forces inside each member. All members have the same length.
F
UNIT II
Arbitrarily Oriented 1-D Bar Element on 2-D Plane
Q2 , v2
11 u ,P
q
22 u ,PP2 , u2
Q1 , v1
P1 , u1
Relationship Between Local Coordinates and Global Coordinates
2
2
1
1
2
2
1
1
cossin00sincos00
00cossin00sincos
0
0
vuvu
vu
vu
qqqq
qqqq
Relationship Between Local Coordinates and Global Coordinates
2
2
1
1
2
1
cossin00sincos00
00cossin00sincos
0
0
QPQP
P
P
qqqq
qqqq
Stiffness Matrix of 1-D Bar Element on 2-D Plane
2
2
1
1
2
2
1
1
cossin00sincos00
00cossin00sincos
cossin00sincos0000cossin00sincos
vuvu
K
QPQP
ij
qqqq
qqqq
qqqq
qqqq
0000010100000101
LAEK ij
Q2 , v2
11 , uP
q
22 u ,PP2 , u2
Q1 , v1
P1 , u1
2
2
1
1
22
22
22
22
2
2
1
1
sincossinsincossincossincoscossincos
sincossinsincossincossincoscossincos
vuvu
LAE
QPQP
qqqqqqqqqqqqqqqqqqqqqqqq
Arbitrarily Oriented 1-D Bar Element in 3-D Space
2
2
2
1
1
1
zzz
yyy
xxx
zzz
yyy
xxx
2
2
2
1
1
1
wvuwvu
000000000
000000000
0w0v
u0w0v
u
x, x, x are the Direction Cosines of the bar in the x-y-z coordinate system
- - -
11 u ,Px
22 u ,P
x
xx
y
z2
1
--
-
2
2
2
1
1
1
2
2
2
1
1
1
000000000
000000000
00
00
RQPRQP
RQ
PRQ
P
zzz
yyy
xxx
zzz
yyy
xxx
Stiffness Matrix of 1-D Bar Element in 3-D Space
2
2
2
1
1
1
22
22
22
22
22
22
2
2
2
1
1
1
wvuwvu
LAE
RQPRQP
xxxxxxxxxx
xxxxxxxxxx
xxxxxxxxxx
xxxxxxxxxx
xxxxxxxxxx
xxxxxxxxxx
11 u ,Px
22 u ,P
x
xx
y
z2
1
--
-
0w0v
u0w0v
u
000000000000001001000000000000001001
LAE
0R0Q
P0R0Q
P
2
2
2
1
1
1
2
2
2
1
1
1
Matrix Assembly of Multiple Bar Elements
2
2
1
1
2
2
1
1
0000010100000101
vuvu
LAE
QPQP
Element I
Element II
Element II I
3
3
2
2
3
3
2
2
33333131
33333131
4vuvu
LAE
QPQP
3
3
1
1
3
3
1
1
3333313133333131
4vuvu
LAE
QPQP
Matrix Assembly of Multiple Bar Elements
3
3
2
2
1
1
3
3
2
2
1
1
000000000000000000000404000000000404
4
vuvuvu
LAE
QPQPQP
Element I
3
3
2
2
1
1
3
3
2
2
1
1
333300313100
333300313100
000000000000
4
vuvuvu
LAE
QPQPQP
3
3
2
2
1
1
3
3
2
2
1
1
330033310031
000000000000330033310031
4
vuvuvu
LAE
QPQPQP
Element II
Element II I
Matrix Assembly of Multiple Bar Elements
3
3
2
2
1
1
3
3
2
2
1
1
vuvuvu
3333333333113131
33303000313014043300303031043014
L4AE
SRSRSR
0v0u0v?u?v0u
603333023131333300313504330033310435
L4AE
?S?R?SFR0S?R
3
3
2
2
1
1
3
3
2
2
1
1
Apply known boundary conditions
Solution Procedures
0v0u0v?u?v0u
603333023131333300310435
330033313504
L4AE
?S?R?S?R0SFR
3
3
2
2
1
1
3
3
2
1
1
2
u2= 4FL/5AE, v1= 0
0v0u0vAE5FL4u
0v0u
603333023131333300310435
330033313504
L4AE
?S?R?S?R0SFR
3
3
2
2
1
1
3
3
2
1
1
2
Recovery of Axial Forces
05
40
54
054
00
0000010100000101
2
2
1
1
2
2
1
1
F
vAEFLu
vu
LAE
QPQP
Element I
Element II
Element II I
53
51
535
1
000
54
33333131
33333131
4
3
3
2
2
3
3
2
2
F
vuv
AEFLu
LAE
QPQP
0000
0000
3333313133333131
4
3
3
1
1
3
3
1
1
vuvu
LAE
QPQP
Stresses inside members
Element I
Element II
Element II I
AF
54
54
1FP
54
2FP
FP51
2
FQ53
2
FQ53
3 FP
51
3
Lecture 5
FEM of 1-D Problems: Applications
Torsional Shaft
Review
Assumption: Circular cross section
Shear stress:
Deformation:
Shear strain:
TrJ
TrGJ
GJTL
q
Finite Element Equation for Torsional Shaft
2
1
2
1
2
1
1111
LGJ
TT
ff
Bending Beam
Review
Normal strain:
Pure bending problems:
Normal stress:
Normal stress with bending moment:
Moment-curvature relationship:
Flexure formula:
x
y
y
x
Ey
x
M M
MydAx
EIM
1
IMy
x dAyI 2
2
21dx
ydEIEIM
Bending Beam
Review
Deflection:
Sign convention:
Relationship between shear force, bending moment and transverse load:
q(x)
x
y
qdxdV
Vdx
dM
qdx
ydEI 4
4
+ -M M M
+ -V V V
Governing Equation and Boundary Condition
• Governing Equation
• Boundary Conditions -----
0at , ? &? &? & ? 2
2
2
2
x
dxvdEI
dxd
dxvdEI
dxdvv
Essential BCs – if v or is specified at the boundary.
Natural BCs – if or is specified at the boundary.
dxdv
{Lx
dxvdEI
dxd
dxvdEI
dxdvv
at , ? &? &? & ? 2
2
2
2
2
2
dxvdEI
2
2
dxvdEI
dxd
,0)()(2
2
2
2
xq
dxxvdEI
dxd 0<x<L
Weak Formulation for Beam Element • Governing Equation
,0)()(2
2
2
2
xq
dxxvdEI
dxd
• Weighted-Integral Formulation for one element
2
1
)()()(0 2
2
2
2x
x
dxxqdx
xvdEIdxdxw
• Weak Form from Integration-by-Parts ----- (1st time)2
1
2
1
2
2
2
2
0x
x
x
x dxvdEI
dxdwdxwq
dxvdEI
dxd
dxdw
21 xxx
Weak Formulation• Weak Form from Integration-by-Parts ----- (2nd time)
2
1
2
1
2
1
2
2
2
2
2
2
2
2
0x
x
x
x
x
x dxvdEI
dxdw
dxvdEI
dxdwdxwq
dxvdEI
dxwd
V(x2)
x = x1
M(x2)
q(x)y
x x = x2
V(x1)
M(x1)L = x2-x1
2
1
2
1
2
2
2
2
0x
x
x
x
MdxdwwVdxwq
dxvdEI
dxwd
Weak Formulation• Weak Form
24231211 , , , xMQxVQxMQxVQ
42
21
32112
2
2
2
)()(2
1
QdxdwQ
dxdwQxwQxwdxwq
dxvdEI
dxwdx
x
2
1
2
1
2
2
2
2
0x
x
x
x
MdxdwwVdxwq
dxvdEI
dxwd
Q3
x = x1
Q4
q(x)y(v)
x x = x2
Q1
Q2
L = x2-x1
Ritz Method for Approximation
n
jjj nxuxvLet
1
4 and )()(
42
21
32112
24
12
2
)()(2
1
QdxdwQ
dxdwQxwQxwdxwq
dxd
uEIdx
wdx
x
j
jj
Let w(x)= i (x), i = 1, 2, 3, 4
42
21
32112
24
12
2
)()(2
1
QdxdQ
dxdQxQxdxq
dxd
uEIdxd ii
ii
x
xi
j
jj
i
Q3
x = x1
Q4
q(x)y(v)
x x = x2
Q1
Q2
L = x2-x1
where ; ; ; ;21
423211xxxx dx
dvuxvudxdvuxvu
Ritz Method for Approximation
ij
jijx
ixi
x
ixi quKQ
dxdQQ
dxdQ
4
14321
22
11
2
1
2
1
and where 2
2
2
2 x
xii
x
x
jiij qdxqdx
dxd
dxdEIK
Q3
x = x1
y(v)
x x = x2
Q1
Q2
L = x2-x1
Q4
Ritz Method for Approximation
4
3
2
1
4
3
2
1
44434241
34333231
24232221
14131211
4
3
2
1
44
44
33
33
22
22
11
11
22
11
22
11
22
11
22
11
qqqq
uuuu
KKKKKKKKKKKKKKKK
QQQQ
dxd
dxd
dxd
dxd
dxd
dxd
dxd
dxd
xx
xx
xx
xx
xx
xx
xx
xx
jiij KK where
Q3
x = x1
y(v)
x x = x2
Q1
Q2
L = x2-x1
Q4
Selection of Shape Function
1000010000100001
22
11
22
11
22
11
22
11
44
44
33
33
22
22
11
11
xx
xx
xx
xx
xx
xx
xx
xx
dxd
dxd
dxd
dxd
dxd
dxd
dxd
dxd
4
3
2
1
4
3
2
1
44342414
34332313
24232212
14131211
4
3
2
1
qqqq
uuuu
KKKKKKKKKKKKKKKK
QQQQ
The best situation is -----
Interpolation Properties
Derivation of Shape Function for Beam Element – Local Coordinates
1 1 2 2 3 3 4 4( )v u u u ux
How to select i???
1 2 3 41 2 3 4
( )dv d d d du u u ud d d d dx x x x x x
and
where 1 21 1 2 3 2 4 dv dvu v u u v u
d dx x
Let 32 xxx iiiii dcba
Find coefficients to satisfy the interpolation properties.
Derivation of Shape Function for Beam Element
How to select i???
e.g. Let 31
21111 xxx dcba
xx 2141 2
1
Similarly
1141
2141
1141
22
23
22
xx
xx
xx
Derivation of Shape Function for Beam Element
In the global coordinates:
)(2
)()(2
)()( 42
3221
11 xdxdvlxvx
dxdvlxvxv
12
1
2
12
11
3
12
1
2
12
1
2
12
11
3
12
1
2
12
1
4
3
2
1
2
23
12
231
xxxx
xxxxxx
l
xxxx
xxxx
xxxxxx
l
xxxx
xxxx
Element Equations of 4th Order 1-D Modelu3
x = x1
u4
q(x)y(v)
xx = x2
u1
u2
L = x2-x1
4
3
2
1
44342414
34332313
24232212
14131211
4
3
2
1
4
3
2
1
uuuu
KKKKKKKKKKKKKKKK
qqqq
QQQQ
2
1
2
1
2
2
2
2 x
xii
x
xji
jiij qdxqandKdx
dxd
dxdEIKwhere
x=x2 x=x1
1 1 1
3 2
4
Element Equations of 4th Order 1-D Modelu3
x = x1
u4
q(x)y(v)
x x = x2
u1
u2
L = x2-x1
2
1
x
xii qdxqwhere
24
23
12
11
22
22
3
4
3
2
1
4
3
2
1
2333636
3233636
2
q
q
uvu
uvu
LLLLLL
LLLLLL
LEI
qqqq
QQQQ
Finite Element Analysis of 1-D Problems - Applications
F
LLL
Example 1.
Governing equation:
Lxxqdx
vdEIdxd
0 0)(2
2
2
2
Weak form for one element
04322112
2
2
2
21
2
1
Q
dxdwQxwQ
dxdwQxwdxwq
dxvd
dxwdEI
xx
x
x
where )( 11 xVQ )( 12 xMQ )( 23 xVQ )( 24 xMQ
Finite Element Analysis of 1-D Problems Example 1.Approximation function:
3 2
1 4
x=x1
x=x2
12
1
2
12
11
3
12
1
2
12
1
2
12
11
3
12
1
2
12
1
4
3
2
1
2
23
12
231
xxxx
xxxxxx
l
xxxx
xxxx
xxxxxx
l
xxxx
xxxx
)(2
)()(2
)()( 42
3221
11 xdxdvlxvx
dxdvlxvxv
Finite Element Analysis of 1-D Problems Example 1.
Finite element model:
2
2
1
1
22
22
3
4
3
2
1
2333636
3233636
2
q
qv
v
LLLLLL
LLLLLL
LEI
QQQQ
P1 , v1 P2 , v2 P3 , v3 P4 , v4
M1 , q1 M2 , q2 M3 , q3 M4 , q4
I II III
Discretization:
Matrix Assembly of Multiple Beam Elements
Element I
Element II
112 2
12
232 2
243
3
3
4
4
6 3 6 3 0 0 0 03 2 3 0 0 0 0
6 3 6 3 0 0 0 03 3 2 0 0 0 020 0 0 0 0 0 0 000 0 0 0 0 0 0 000 0 0 0 0 0 0 000 0 0 0 0 0 0 00
I
I
I
I
vL LQL L L LQ
vL LQL L L LQ EI
vL
v
q
q
q
q
1
1
212 2
223
332 2
34
4
4
0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0
0 0 6 3 6 3 0 00 0 3 2 3 0 020 0 6 3 6 3 0 00 0 3 3 2 0 0
0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0
II
II
II
II
v
vQ L LQ L L L LEI
vQ L LLQ L L L L
v
q
q
q
q
Matrix Assembly of Multiple Beam Elements
Element II I
1
1
2
23
1 32 2
2 3
3 42 2
4 4
0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 02
0 0 0 0 6 3 6 30 0 0 0 3 2 30 0 0 0 6 3 6 30 0 0 0 3 3 2
III
III
III
III
v
v
EIQ vL LLQ L L L LQ vL LQ L L L L
q
q
q
q
4
4
3
3
2
2
1
1
22
2222
2222
22
3
4
4
3
3
2
2
1
1
233000036360000
322333003633663600003223330036336636000032300003636
2
q
q
q
q
v
v
v
v
LLLLLL
LLLLLLLLLLLL
LLLLLLLLLLLL
LLLLLL
LEI
MPMP
MPMP
Solution Procedures
???0?000
233000036360000
340300360123600003403003601236000032300003636
2
0
0?0???
4
4
3
3
2
2
1
1
22
222
222
22
3
4
4
3
3
2
2
1
1
q
q
q
q
v
v
v
v
LLLLLL
LLLLLLL
LLLLLLL
LLLLLL
LEI
MFP
MP
MPMP
Apply known boundary conditions
???0?000
360123600003601236000032300003636
233000036360000
340300003403
2
????0
00
4
4
3
3
2
2
1
1
22
22
222
222
3
3
2
1
1
4
4
3
2
q
q
q
q
v
v
v
v
LLLL
LLLLLL
LLLLLL
LLLLLLLLLL
LEI
PPMP
MFP
MM
Solution Procedures
????
2303630
34004
2
0
00
4
4
3
2
22
222
22
3
4
4
3
2
q
vLLLLL
LLLLLL
LEI
MFP
MM
4
4
3
22
3
3
2
1
1
360300300000003
2
????
q
vLL
LLL
LEI
PPMP
????0000
360312600003061236000032300030636
230300036306000
340300004303
2
????0
00
4
4
3
2
3
2
1
1
22
22
222
222
3
3
2
1
1
4
4
3
2
q
q
v
vv
v
LLLL
LLLLLL
LLLLLL
LLLLLLLLLL
LEI
PPMP
MFP
MM
Shear Resultant & Bending Moment Diagram
FL72
FL71
FL
F79
F73
2P
F
Plane Flame
Frame: combination of bar and beam
E, A, I, LQ1 , v1 Q3 , v2
Q2 , q1
P1 , u1
Q4 , q2
P2 , u2
2
2
2
1
1
1
22
2323
22
2323
4
3
2
2
1
1
460260
61206120
0000
260460
61206120
0000
q
q
vu
vu
LEI
LEI
LEI
LEI
LEI
LEI
LEI
LEI
LAE
LAE
LEI
LEI
LEI
LEI
LEI
LEI
LEI
LEI
LAE
LAE
QQPQQP
Finite Element Model of an Arbitrarily Oriented Frame
q x
y
q x
y
Finite Element Model of an Arbitrarily Oriented Frame
2
2
2
1
1
1
22
2323
22
2323
4
3
2
2
1
1
406206
0000
60126012
206406
0000
60126012
q
q
vu
vu
LEI
LEI
LEI
LEI
LAE
LAE
LEI
LEI
LEI
LEI
LEI
LEI
LEI
LEI
LAE
LAE
LEI
LEI
LEI
LEI
QQPQQP
local
global
Plane Frame Analysis - Example
Rigid Joint Hinge Joint
Beam II
Beam I Beam
Bar
F F F F
Plane Frame Analysis
2
2
2
1
1
1
22
2323
22
2323
4
3
2
2
1
1
406206
0000
60126012
206406
0000
60126012
q
q
vu
vu
LEI
LEI
LEI
LEI
LAE
LAE
LEI
LEI
LEI
LEI
LEI
LEI
LEI
LEI
LAE
LAE
LEI
LEI
LEI
LEI
QQPQQP
I
I
P1 , u1
P2 , u2
Q2 , q1
Q4 , q2
Q1 , v1
Q3 , v2
Plane Frame Analysis
P1 , u2
Q3 , v3
Q2 , q2Q4 , q3
Q1 , v2
P2 , u3
4
3
3
2
2
2
22
2323
22
2323
4
3
2
2
2
1
460260
61206120
0000
260460
61206120
0000
q
q
vu
vu
LEI
LEI
LEI
LEI
LEI
LEI
LEI
LEI
LAE
LAE
LEI
LEI
LEI
LEI
LEI
LEI
LEI
LEI
LAE
LAE
QQPQQP II
Plane Frame Analysis
2
2
2
1
1
1
22
2323
22
2323
4
3
2
2
1
1
406206
0000
60126012
206406
0000
60126012
q
q
vu
vu
LEI
LEI
LEI
LEI
LAE
LAE
LEI
LEI
LEI
LEI
LEI
LEI
LEI
LEI
LAE
LAE
LEI
LEI
LEI
LEI
QQPQQP
I
I
4
3
3
2
2
2
22
2323
22
2323
4
3
2
2
2
1
460260
61206120
0000
260460
61206120
0000
q
q
vu
vu
LEI
LEI
LEI
LEI
LEI
LEI
LEI
LEI
LAE
LAE
LEI
LEI
LEI
LEI
LEI
LEI
LEI
LEI
LAE
LAE
QQPQQP II
Plane Frame Analysis
FL81
F21
F21
FL83
Finite Element Analysis (F.E.A.) of 1-D Problems – Heat Conduction
UNIT IV
Heat Transfer Mechanisms Conduction – heat transfer by molecular
agitation within a material without any motion of the material as a whole.
Convection – heat transfer by motion of a fluid.
Radiation – the exchange of thermal radiation between two or more bodies. Thermal radiation is the energy emitted from hot surfaces as electromagnetic waves.
Heat Conduction in 1-DHeat flux q: heat transferred per unit area per unit time (W/m2)
dTq kdx
T TA AQ CAx x t
Q: heat generated per unit volume per unit time C: mass heat capacity
Governing equation:
Steady state equation:
0d dTA AQdx dx
: thermal conductivity
Thermal Convection
( )sq h T T
Newton’s Law of Cooling
2: convective heat transfer coefficient ( )oh W m C
Thermal Conduction in 1-D
Dirichlet BC:
Boundary conditions:
Natural BC:
Mixed BC:
Weak Formulation of 1-D Heat Conduction(Steady State Analysis)
• Governing Equation of 1-D Heat Conduction -----
( )( ) ( ) ( ) 0d dT xx A x AQ xdx dx
0<x<L
• Weighted Integral Formulation -----
• Weak Form from Integration-by-Parts -----
0
( )0 ( ) ( ) ( ) ( )L d dT xw x x A x AQ x dx
dx dx
0 0
0LL dw dT dTA wAQ dx w A
dx dx dx
Formulation for 1-D Linear Element
Let 1 1 2 2(x) (x) (x)T T T
f2
x1 x2
1 2
T1
x
T2
f1
2 11 2( ) , ( )x x x xx x
l l
x2 x1
1T1
2T2
22
11 )( ,)(
xTAxf
xTAxf
Formulation for 1-D Linear Element
Let w(x)= i (x), i = 1, 2
2 2
1 1
2
2 2 1 11
0 ( ) ( )x x
jij i i i
j x x
ddT A dx AQ dx x f x fdx dx
1122
2
1
)()( fxfxQTK iiij
jij
2
1
2212
1211
2
1
2
1
TT
KKKK
ff
2 2
1 21 1
1 2 , Q , f , fx x
jiij i i
x xx x
dd dT dTwhere K A dx AQ dx A Adx dx dx dx
Element Equations of 1-D Linear Element
2
1
2
1
2
1
1111
TT
LA
ff
2
1 21
1 2 Q , f , fx
i ix x x xx
dT dTwhere AQ dx A Adx dx
f2
x1 x2
1 2
T1
x
T2
f1
1-D Heat Conduction - Example
A composite wall consists of three materials, as shown in the figure below. The inside wall temperature is 200oC and the outside air temperature is 50oCwith a convection coefficient of h = 10 W(m2.K). Find the temperature alongthe composite wall.
t1 t2 t3
0 200oT C 50oT C
x
1 2 3
1 2 3
1 2 3
70 , 40 , 20
2 , 2.5 , 4
W m K W m K W m K
t cm t cm t cm
Thermal Conduction and Convection- Fin
Objective: to enhance heat transfer
dx
t
x
w 2 ( )2 ( ) 2 ( )
lossc c
h T T w th T T dx w h T T dx tQA dx A
Governing equation for 1-D heat transfer in thin fin
0c cd dTA A Qdx dx
0c cd dTA Ph T T A Qdx dx
2P w t where
Fin - Weak Formulation(Steady State Analysis)
• Governing Equation of 1-D Heat Conduction -----
( )( ) ( ) 0d dT xx A x Ph T T AQdx dx
0<x<L
• Weighted Integral Formulation -----
• Weak Form from Integration-by-Parts -----
0
( )0 ( ) ( ) ( ) ( ) ( )L d dT xw x x A x Ph T T AQ x dx
dx dx
0 0
0 ( )LL dw dT dTA wPh T T wAQ dx w A
dx dx dx
Formulation for 1-D Linear Element
Let w(x)= i (x), i = 1, 2
2 2
1 1
2
1
2 2 1 1
0
( ) ( )
x xji
j i j ij x x
i i
ddT A Ph dx AQ PhT dxdx dx
x f x f
1122
2
1
)()( fxfxQTK iiij
jij
2
1
2212
1211
2
1
2
1
TT
KKKK
ff
2 2
1 1
1 2
1 2
, Q ,
,
x xji
ij i j i ix x
x x x x
ddwhere K A Ph dx AQ PhT dxdx dx
dT dTf A f Adx dx
Element Equations of 1-D Linear Element
1 1 1
2 2 2
1 1 2 11 1 1 26
f Q TA Phlf Q TL
f2
x=0 x=L1 2
T1
x
T2
f1
2
1 21
1 2 Q , , x
i ix x x xx
dT dTwhere AQ PhT dx f A f Adx dx
Lecture 7
Finite Element Analysis of 2-D Problems
2-D Discretization
Common 2-D elements:
2-D Model Problem with Scalar Function- Heat Conduction
• Governing Equation
0),(),(),(
yxQ
yyxT
yxyxT
x in W
• Boundary Conditions
Dirichlet BC:
Natural BC:
Mixed BC:
Weak Formulation of 2-D Model Problem
• Weighted - Integral of 2-D Problem -----
• Weak Form from Integration-by-Parts -----
0 ( , )w T w T wQ x y dxdyx x y y
W
T Tw dxdy w dxdyx x y y
W W
( , ) ( , ) ( , ) 0T x y T x yw Q x y dAx x y y
W
Weak Formulation of 2-D Model Problem• Green-Gauss Theorem -----
where nx and ny are the components of a unit vector, which is normal to the boundary G.
xT Tw dxdy w n ds
x x x
W G
sinjcosijninn yx
yT Tw dxdy w n ds
y y y
W G
Weak Formulation of 2-D Model Problem
• Weak Form of 2-D Model Problem -----
0 ( , )w T w T wQ x y dxdyx x y y
W
x yT Tw n n dsx y
G
EBC: Specify T(x,y) on G
NBC: Specify on Gx yT Tn nx y
where is the normal
outward flux on the boundary G at the segment ds.
( )n x yT Tq s i j n i n jx y
FEM Implementation of 2-D Heat Conduction – Shape Functions
Step 1: Discretization – linear triangular element
T1
T2
T3
332211 TTTT
Derivation of linear triangular shape functions:
Let ycxcc 2101
Interpolation properties
1 at node0 at other nodes
i
i
ith
0 1 1 2 1
0 1 2 2 2
0 1 3 2 3
100
c c x c yc c x c yc c x c y
10 1 1
1 2 2
2 3 3
1 11 01 0
c x yc x yc x y
1
1 1 2 3 3 2
1 2 2 2 3
3 3 3 2
1 11
1 1 02
1 0 e
x y x y x yx y
x y x y y yA
x y x x
Same 3 1 1 3
2 3 1
1 3
12 e
x y x yx y
y yA
x x
1 2 2 1
3 1 2
2 1
12 e
x y x yx y
y yA
x x
FEM Implementation of 2-D Heat Conduction – Shape Functions
linear triangular element – area coordinates
T1
T2
T3
2 3 3 21
1 2 3
3 2
12 e e
x y x yx y Ay yA A
x x
3 1 1 32
2 3 1
1 3
12 e e
x y x yx y Ay yA A
x x
1 2 2 13
3 1 2
2 1
12 e e
x y x yx y Ay yA A
x x
A3
A1
A2
12
3
Interpolation Function - Requirements
• Interpolation condition• Take a unit value at node i, and is zero at all other nodes
• Local support condition• i is zero at an edge that doesn’t contain node i.
• Interelement compatibility condition• Satisfies continuity condition between adjacent elements
over any element boundary that includes node i
• Completeness condition• The interpolation is able to represent exactly any
displacement field which is polynomial in x and y with the order of the interpolation function
Formulation of 2-D 4-Node Rectangular Element – Bi-linear Element
1 1 2 2 3 3 4 4( , )u u u u ux
ba1
ba
b1
ab1
a1
43
21
xx
xx
1 2 3
4
Note: The local node numbers should be arranged in a counter-clockwise sense. Otherwise, the area Of the element would be negative and the stiffness matrix can not be formed.
Let
• Weak Form of 2-D Model Problem -----
Assume approximation:
and let w(x,y)=i(x,y) as before, then1
( , ) ( , )n
j jj
u x y u x y
1 1
0e
n ni i
j j j j i i nj j
T T Q dxdy q dsx x y y
W G
0 ( , )e e
nw T w T wQ x y dxdy wq dsx x y y
W G
1e e
n
ij j i i nj
K T Qdxdy q ds W G
e
j ji iijK dxdy
x x y y
W
where
FEM Implementation of 2-D Heat Conduction – Element Equation
1e e
n
ij j i i nj
K T Qdxdy q ds W G
223 23 31 23 12
223 31 31 31 12
223 12 31 12 12
4 e
l l l l l
K l l l l lA
l l l l l
FEM Implementation of 2-D Heat Conduction – Element Equation
1 1
2 2
3 3
Q qF Q q
Q q
e
e
i i
i i n
Q Qdxdy
q q ds
W
G
Assembly of Stiffness Matrices
)2(35
)2(24
)2(4
)1(33
)2(1
)1(22
)1(11 uU,uU,uuU,uuU,uU
( ) ( )
( ) ( ) ( )
1
e
e e
ne e e
i i i n ij jj
F Q dxdy q ds K u W G
Imposing Boundary Conditions
The meaning of qi:
(1) (1) (1)1 12 23 31
(1) (1)12 23
(1) (1) (1) (1) (1) (1) (1) (1) (1)2 2 2 2 2
(1) (1) (1) (1)2 2
n n n nh h h
n nh h
q q ds q ds q ds q ds
q ds q ds
G
(1) (1) (1)1 12 23 31
(1) (1)23 31
(1) (1) (1) (1) (1) (1) (1) (1) (1)3 3 3 3 3
(1) (1) (1) (1)3 3
n n n nh h h
n nh h
q q ds q ds q ds q ds
q ds q ds
G
(1) (1) (1)1 12 23 31
(1) (1)12 31
(1) (1) (1) (1) (1) (1) (1) (1) (1)1 1 1 1 1
(1) (1) (1) (1)1 1
n n n nh h h
n nh h
q q ds q ds q ds q ds
q ds q ds
G
1
2
3
112
3
1
2
3
112
3
Imposing Boundary Conditions
(1) ( 2)23 41
(1) (2)n nh h
q qEquilibrium of flux:
(1) ( 2) (1) ( 2)23 41 23 41
(1) (1) (2) (2) (1) (1) (2) (2)2 1 3 4; n n n n
h h h h
q ds q ds q ds q ds
FEM implementation:
(1) (2)2 2 1q q q
(1) (1)12 23
(1) (1) (1) (1) (1)2 2 2n n
h h
q q ds q ds
Consider
( 2) ( 2)12 41
(2) (2) (2) (2) (2)1 1 1n n
h h
q q ds q ds
(1) ( 2 )12 12
(1) (1) (2) (2)2 2 1n n
h h
q q ds q ds (1) ( 2)31 34
(1) (1) (2) (2)3 3 4n n
h h
q q ds q ds
(1) (2)3 3 4q q q
(1) (1)23 31
(1) (1) (1) (1) (1)3 3 3n n
h h
q q ds q ds ( 2) ( 2)34 41
(2) (2) (2) (2) (2)4 4 4n n
h h
q q ds q ds
Calculating the q Vector
Example:
293T K0nq
1nq
2-D Steady-State Heat Conduction - Example
0nq
0.6 m
0.4 m
A
B C
DAB and BC:
CD: convection CmWh o
250 CT o25
DA: CT o180
x
y
Finite Element Analysis of Plane Elasticity
Review of Linear ElasticityLinear Elasticity: A theory to predict mechanical response of an elastic body under a general loading condition.
Stress: measurement of force intensity
zzzyzx
yzyyyx
xzxyxx
zxxz
zyyz
yxxy
with
xx xy
yx yy
2-D
Review of Linear ElasticityTraction (surface force) :
Equilibrium – Newton’s Law
0
0
Static
xyxxx
yx yyy
fx y
fx y
0F
x xx x xy y
y xy x yy y
t n n
t n n
t
Dynamic
xyxxx x
yx yyy y
f ux y
f ux y
Review of Linear Elasticity
Strain: measurement of intensity of deformation
1 1 2 2
y yx xxx xy xy yy
u uu ux y x y
Generalized Hooke’s Law
yyxx zzxx
yyxx zzyy
yyxx zzzz
E E E
E E E
E E E
zxzxyzyzxyxy GGG
12EG
zzyyxx
zzzz
yyyy
xxxx
eGe
GeGe
2
22
1 1 2E
Plane Stress and Plane Strain
Plane Stress - Thin Plate:
xy
y
x
22
22
xy
y
x
12E00
01
E1
E
01
E1
E
xy
y
x
33
2212
1211
xy
y
x
C000CC0CC
Plane Stress and Plane Strain
Plane Strain - Thick Plate:
xy
y
x
xy
y
x
12E00
0211
E1211
E
0211
E211
E1
xy
y
x
33
2212
1211
xy
y
x
C000CC0CC
Plane Stress: Plane Strain:
Replace E by and by21 E
1
Equations of Plane ElasticityGoverning Equations(Static Equilibrium)
Constitutive Relation (Linear Elasticity)
Strain-Deformation(Small Deformation)
xy
y
x
33
2212
1211
xy
y
x
C000CC0CC
0yxxyx
0yx
yxy
xu
x
yv
y
yu
xv
xy
0yvC
xuC
yxvC
yuC
x
0xvC
yuC
yyvC
xuC
x
22123333
33331211
Specification of Boundary Conditions
EBC: Specify u(x,y) and/or v(x,y) on G
NBC: Specify tx and/or ty on G
where
is the traction on the boundary G at the segment ds.
yyyxyxyyxyxxxxyx nntnntjtitsT ; ;)(
UNIT V
Weak Formulation for Plane Elasticity
dxdyyvC
xuC
yxvC
yuC
xw0
dxdyxvC
yuC
yyvC
xuC
xw0
221233332
333312111
W
W
GW
GW
dstwdxdyyvC
xuC
yw
xv
yuC
xw0
dstwdxdyxv
yuC
yw
yvC
xuC
xw0
y222122
332
x1331
12111
where
y2212x33y
y33x1211x
nyvC
xuCn
xv
yuCt
nxv
yuCn
yvC
xuCt
are components of traction on the boundary G
Finite Element Formulation for Plane Elasticity
n
1jj
22ij
n
1jj
21ij
2i
n
1jj
12ij
n
1jj
11ij
1i
vKuKF
vKuKF Let
n
1jjj
n
1jjj
v)y,x()y,x(v
u)y,x()y,x(u
dxdyyy
Cxx
CK
Kdxdyxy
Cyx
CK
dxdyyy
Cxx
CK
ji22
ji33
22ij
21ji
ji33
ji12
12ij
ji33
ji11
11ij
W
W
W
G W
G W
dxdyfdstF
dxdyfdstF
yiyi2
i
xixi1
i
where
and
Constant-Strain Triangular (CST) Element for Plane Stress Analysis
Let1 2 3 1 1 2 2 3 3
5 6 7 1 1 2 2 3 3
( , )( , )
u x y c c x c y u u uv x y c c x c y v v v
1 1, xu F
1 1, yv F
2 2, xu F
3 3, xu F
2 2, yv F
3 3, yv F
2 3 3 2
1 2 3
3 2
12 e
x y x yx y
y yA
x x
3 1 1 3
2 3 1
1 3
12 e
x y x yx y
y yA
x x
1 2 2 1
3 1 2
2 1
12 e
x y x yx y
y yA
x x
Constant-Strain Triangular (CST) Element for Plane Stress Analysis
111 12 13 14 15 16 1
121 22 23 24 25 26 1
231 32 33 34 35 36 2
2241 42 43 44 45 46
3351 52 53 54 55 56
3361 62 63 64 65 66
14
x
y
x
ye
x
y
Fk k k k k k uFk k k k k k vFk k k k k k uFvk k k k k kAFuk k k k k kFvk k k k k k
2 2 2 2 211 11 2 3 33 3 2 21 12 2 3 3 2 33 2 3 22 22 3 2 33 2 3
2 231 11 3 1 2 3 33 1 3 3 2 32 12 3 1 3 2 33 1 3 3 2 33 11 3 1 33 1 3
41 12 2 3 33 1 3
; ;
; ;
k c y y c x x k c y y x x c y y k c x x c y y
k c y y y y c x x x x k c y y x x c x x x x k c y y c x x
k c y y c x x x
23 2 42 22 1 3 3 2 33 2 3 3 1 43 12 1 3 3 1 33 1 3
2 244 22 1 3 33 3 1 51 11 1 2 2 3 33 2 1 3 2 52 12 1 2 33 2 1 3 2
53 11 1 2 3 1 33 2 1 1 3 54
; ;
; ;
;
x k c x x x x c y y y y k c x x y y c x x
k c x x c y y k c y y y y c x x x x k c y y c x x x x
k c y y y y c x x x x k c
2 212 1 2 1 3 33 2 1 1 3 55 11 1 2 33 2 1
61 12 2 3 33 2 1 3 2 62 22 2 1 3 2 33 1 2 2 3 63 12 3 1 33 2 1 1 3
64 22 1 3 2 1 33 1 2 3 1 65 12 1 2 2 1
;
y y x x c x x x x k c y y c x x
k c y y c x x x x k c x x x x c y y y y k c y y c x x x x
k c x x x x c y y y y k c y y x x c
2 2 233 2 1 66 22 2 1 33 1 2 x x k c x x c y y
4-Node Rectangular Element for Plane Stress Analysis
Let
443322118765
443322114321
vvvvxycycxcc)y,x(vuuuuxycycxcc)y,x(u
by
ax1
by
ax
by1
ax
by1
ax1
43
21
4-Node Rectangular Element for Plane Stress Analysis
For Plane Strain Analysis:21
EE
1
and
Loading Conditions for Plane Stress Analysis
n
1jj
22ij
n
1jj
21ij
2i
n
1jj
12ij
n
1jj
11ij
1i
vKuKF
vKuKF
G W
G W
dxdyfdstF
dxdyfdstF
yiyi2
i
xixi1
i
Evaluation of Applied Nodal Forces
G
dstF xi1
i
tdy16y1
by1
axdstFF
b
0
2
ox2)A(1
2)A(
x2
G
3.383dy168y
16y
8y1100dy1.0
16y11000
8y1
88F
8
0 2
3
2
28
0
2)A(
x2
tdy16y1
by
axdstFF
b
0
2
ox3)A(1
3)A(
x3
G
350dy168y
8y100dy1.0
16y11000
8y
88F
8
0 2
38
0
2)A(
x3
Evaluation of Applied Nodal Forces
tdy16
8y1by1
axdstFF
b
0
2
ox2)B(1
2)B(
x2
G
7.216dy168y
16y
32y5
43100dy1.0
168y11000
8y1
88F
8
0 2
3
2
28
0
2)B(
x2
tdy16
8y1by
axdstFF
b
0
2
ox3)B(1
3)B(
x3
G
7.116dy168y
16y2
32y3100dy1.0
168y11000
8y
88F
8
0 2
3
2
28
0
2)B(
x3
Element Assembly for Plane Elasticity
4
4
3
3
2
2
1
1)A()A(
y
x
y
x
y
x
y
x
vuvuvuvu
FFFFFFFF
3
3
4
4
2
2
1
1
��������A
B
1 2
3 4
3 4
65
6
6
5
5
4
4
3
3)B()B(
y
x
y
x
y
x
y
x
vuvuvuvu
FFFFFFFF
3
3
4
4
2
2
1
1
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Element Assembly for Plane Elasticity
1 2
3 4
65
A
B
6
6
5
5
4
4
3
3
2
2
1
1
)B(y
)B(x
)B(y
)B(x
)B(y
)A(y
)B(x
)A(x
)B(y
)A(y
)B(x
)A(x
)A(y
)A(x
)A(y
)A(x
vuvuvuvuvuvu
0000000000000000
0000000000000000
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FFFFFFFF
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3
3
4
4
23
23
14
14
2
2
1
1
Comparison of Applied Nodal Forces
Discussion on Boundary Conditions
• Must have sufficient EBCs to suppress rigid body translation and rotation• For higher order elements, the mid side nodes cannot be skipped while applying EBCs/NBCs
Plane Stress – Example 2
Plane Stress – Example 3
Evaluation of Strains
44332211
44332211
vvvv)y,x(vuuuu)y,x(u
by
ax1
by
ax
by1
ax
by1
ax1
43
21
4
1jj
jj
j
4
1jj
j
4
1jj
j
xy
y
x
vx
uy
vy
ux
xv
yu
yvxu
Evaluation of Stresses
4
4
3
3
2
2
1
1
xy
y
x
vuvuvuvu
aby
ax1
b1
aby
abx
by1
a1
abx
by1
a1
ax1
b1
ax1
b10
abx0
abx0
ax1
b10
0aby0
aby0
by1
a10
by1
a1
Plane Stress Analysis Plane Strain Analysis
xy
y
x
22
22
xy
y
x
12E00
01
E1
E
01
E1
E
xy
y
x
xy
y
x
12E00
0211
E1211
E
0211
E211
E1
Finite Element Analysis of 2-D Problems – Axi-symmetric Problems
Axi-symmetric ProblemsDefinition:
A problem in which geometry, loadings, boundaryconditions and materials are symmetric about one axis.
Examples:
Axi-symmetric AnalysisCylindrical coordinates:
zzryrx ;sin ;cos qq
zr , , q
• quantities depend on r and z only• 3-D problem 2-D problem
Axi-symmetric Analysis
Axi-symmetric Analysis – Single-Variable Problem
0),(),(),(1002211
zrfuaz
zruazr
zrurarr
Weak form:
dswq
rdrdzzrwfwuazua
zw
rua
rw
e
e
n
G
W
),(0 002211
where zrn nz
zruanr
zruaq
),(),(
2211
Finite Element Model – Single-Variable Problem
j jj
u u
Ritz method:
ei
ei
ej
n
j
eij QfuK
1
where ( , ) ( , )j jr z x y
iw
Weak form
11 22 00
e
j je i iij i jK a a a rdrdz
r r z z
W
where
e
ei if frdrdz
W
e
ei i nQ q ds
G
Single-Variable Problem – Heat Transfer
Heat Transfer:
Weak form
1 ( , ) ( , ) ( , ) 0T r z T r zrk k f r zr r r z z
0 ( , )
e
e
n
w T w Tk k wf r z rdrdzr r z z
wq ds
W
G
( , ) ( , )n r z
T r z T r zq k n k nr z
where
3-Node Axi-symmetric Element
1 1 2 2 3 3( , )T r z T T T
1
2
3
2 3 3 2
1 2 3
3 2
12 e
r z r zr z
z zA
r r
3 1 1 3
2 3 1
1 3
12 e
r z r zr z
z zA
r r
1 2 2 1
3 1 2
2 1
12 e
r z r zr z
z zA
r r
4-Node Axi-symmetric Element
1 1 2 2 3 3 4 4( , )T r z T T T T
1 2
34
a
b
x
r
z1 2
3 4
1 1 1
1
a b a b
a b a b
x x
x x
Single-Variable Problem – Example
Step 1: Discretization
Step 2: Element equation
e
j je i iijK rdrdz
r r z z
W
e
ei if frdrdz
W
e
ei i nQ q ds
G
Time-Dependent Problems
Time-Dependent Problems
In general,
Key question: How to choose approximate functions?
Two approaches:
txutxu jj ,,
txu ,
xtutxu jj ,
Model Problem I – Transient Heat Conduction
Weak form:
txfxua
xtuc ,
)()(0 2211
2
1
xwQxwQdxwftucw
xu
xwa
x
x
;21
21xx dx
duaQdxduaQ
Transient Heat Conduction
let:
)()(0 2211
2
1
xwQxwQdxwftucw
xu
xwa
x
x
n
jjj xtutxu
1
, and xw i
FuMuK
2
1
x
x
jiij dx
xxaK
2
1
x
xjiij dxcM
i
x
xii QfdxF
2
1
ODE!
Time Approximation – First Order ODE
tfbudtdua Tt 0 00 uu
Forward difference approximation - explicit
Backward difference approximation - implicit
1k k k ktu u f bu
a
1k k k ktu u f bu
a b t
Time Approximation – First Order ODE
tfbudtdua Tt 0 00 uu
- family formula:
1 11k k k ku u t u u
Equation
11
1 1k k kk
a tbu t f fu
a tb
Time Approximation – First Order ODE
tfbudtdua Tt 0 00 uu
Finite Element Approximation
11
223 3 3 3
k kk k
f ftb tba u a u t
Stability of – Family Approximation
Stability
Example
11 1
a tbA
a tb
FEA of Transient Heat Conduction
FuMuK
- family formula for vector:
1 11k k k ku u t u u
1
1 11 1k k k ku M K t M K t u t f t f
Stability Requirment
max212
critt
QuMK where
Note: One must use the same discretization for solvingthe eigenvalue problem.
Transient Heat Conduction - Example
02
2
xu
tu 10 x
0,0 tu 0,1 t
tu
0.10, xu
0t
Transient Heat Conduction - Example
Transient Heat Conduction - Example
Transient Heat Conduction - Example
Transient Heat Conduction - Example
Transient Heat Conduction - Example
Transient Heat Conduction - Example
Model Problem II – Transverse Motion of Euler-Bernoulli Beam
Weak form:
txftuA
xuEI
x,2
2
2
2
2
2
21
2
1
423211
2
2
2
2
2
2
)()(
0
xx
x
x
xwQxwQ
xwQxwQ
dxwftuAw
xu
xwEI
22
11
2
2
42
2
3
2
2
22
2
1
xx
xx
xuEIQ
xuEI
xQ
xuEIQ
xuEI
xQ
Where:
Transverse Motion of Euler-Bernoulli Beam
let:
n
jjj xtutxu
1
, and xw i
FuMuK
21
2
1
423211
2
2
2
2
2
2
)()(
0
xx
x
x
xwQxwQ
xwQxwQ
dxwftuAw
xu
xwEI
Transverse Motion of Euler-Bernoulli Beam
FuMuK
2
1
2
2
2
2x
x
jiij dx
xxEIK
2
1
x
xjiij dxAM
i
x
xii QfdxF
2
1
ODE Solver – Newmark’s Scheme
tuuu
ututuu
sss
ssss
1
21 2
1
11 sss uuu qqqwhere
Stability requirement:
21
2max2
1
critt
FuMK 2where
ODE Solver – Newmark’s Scheme
212 ,
21
Constant-average acceleration method (stable)
312 ,
21
02 ,21
582 ,
23
22 ,23
Linear acceleration method (conditional stable)
Central difference method (conditional stable)
Galerkin method (stable)
Backward difference method (stable)
Fully Discretized Finite Element Equations
Transverse Motion of Euler-Bernoulli Beam
04
4
2
2
xw
tw 10 x
0,0 tw 0,1 t
tw
xxxxw 1sin0,
0,1 tw 0,0 t
tw
00, x
tw
Transverse Motion of Euler-Bernoulli Beam
Transverse Motion of Euler-Bernoulli Beam
Transverse Motion of Euler-Bernoulli Beam