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MECh300H Introduction to Finite Element Methods. Finite Element Analysis (F.E.A.) of 1-D Problems. Historical Background . Hrenikoff, 1941 – “frame work method” Courant, 1943 – “piecewise polynomial interpolation” Turner, 1956 – derived stiffness matrice for truss, beam, etc - PowerPoint PPT Presentation
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MECh300H Introduction to Finite Element Methods
Finite Element Analysis (F.E.A.) of 1-D Problems
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
lxx
lxx 1
22
1 ;
2
1 ;2
121
11 2
1 ;12 xlxxxl
Approximation Methods – Finite Element Method
Example (cont):
Step 4: Forming element equation
Let , weak form becomes1w 0111211
122
1
2
1
PPdxfdxluuEA
l
x
x
x
x
1121
2
1
PdxfulEAu
lEA x
x
Let , weak form becomes2w 0112222
122
1
2
1
PPdxfdxluuEA
l
x
x
x
x
2221
2
1
PdxfulEAu
lEA 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 Pf 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 fE A E A E A E A u fl l l l
u f PE A E Al 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
2 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
Natural (or Normal) Coordinate:
x=x1 x= x2
=-1 =1
x
0x x l
1xxx 1 2
2/ 2
x xx
l
1 32
=-1 =1
1 2
=-1 =1
1 42
=-1 =1
3
21 ,11 ,
21
321
1311
1627 ,1
31
31
169
21
31
311
169 ,1
311
1627
43
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
2
1
1
1
2 x
j ji iij ji
x
d dd dwhere K EA dx EA d Kdx dx d d l
3
2
1
332313
232212
131211
3
2
1
3
2
1
uuu
KKKKKKKKK
fff
PPP
=-1 =0 =1
Quadratic Formulation for Bar Elementu1 u3u2f(x)
P3P1
P2
=-1 =0 =11x 2x 3x
21u11u
21u)(u)(u)(u)(u 321332211
21 ,11 ,
21
321
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
1 2
2/ 2
x xx
l
2
l d dx 2d
dx l
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
