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Prepared by
Mr. G S D Madhav, Assistant Professor, Dept of Aeronautical Engg
Ms Y Shwetha, Assistant Professor, Dept of Aeronautical Engg
Power Point Presentation
on
Finite element methods III B Tech II Semester
INSTITUTE OF AERONAUTICAL ENGINEERING (Autonomous)
Dundigal, Hyderabad - 500 043
1
UNIT - 1
• Introduction to FEM:
– Stiffness equations for a axial bar element in local co-ordinates using Potential Energy approach and Virtual energy principle
– Finite element analysis of uniform, stepped and tapered bars subjected to mechanical and thermal loads
– Assembly of Global stiffness matrix and load vector
– Quadratic shape functions
– properties of stiffness matrix
2
Axially Loaded Bar Review:
Stress:
Strain:
Deformation:
Stress:
Strain:
Deformation:
3
Axially Loaded Bar Review:
Stress:
Strain:
Deformation:
4
Axially Loaded Bar – Governing Equations and Boundary Conditions
• Differential Equation
• Boundary Condition Types
• prescribed displacement (essential BC)
• prescribed force/derivative of displacement (natural BC)
Lxxfdx
duxEA
dx
d
0 0)()(
5
Axially Loaded Bar –Boundary Conditions
• Examples
• fixed end
• simple support
• free end
6
Potential Energy
• Elastic Potential Energy (PE) - Spring case
- Axially loaded bar
- Elastic body
x
Unstretched spring
Stretched bar
0PE
2
2
1PE kx
undeformed:
deformed:
0PE
L
Adx0
2
1PE
dvV
Tεσ2
1PE
7
Potential Energy
• Work Potential (WE)
B
L
uPfdxu 0
WP
P
f
f: distributed force over a line P: point force u: displacement
A B
• Total Potential Energy
B
LL
uPfdxuAdx 00
2
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.
8
Potential Energy + Rayleigh-Ritz Approach
P
f
A B
Example:
Step 1: assume a displacement field nixau i
i
i to1
is shape function / basis function n is the order of approximation
Step 2: calculate total potential energy
9
Potential Energy + Rayleigh-Ritz Approach
P
f
A B
Example:
Step 3:select ai so that the total potential energy is minimum
10
Galerkin’s Method
P
f
A B
Example:
Pdx
duxEA
xu
xfdx
duxEA
dx
d
Lx
)(
00
0)()(Seek an approximation so
Pdx
udxEA
xu
dVxfdx
udxEA
dx
dw
Lx
V
i
~)(
00~
0)(~
)(
u~
In the Galerkin’s method, the weight function is chosen to be the same as the shape function.
11
Galerkin’s Method
P
f
A B
Example:
0)(~
)(
dVxf
dx
udxEA
dx
dw
V
i 0~
)(~
)(00 0
L
i
L L
ii
dx
udxEAwfdxwdx
dx
dw
dx
udxEA
1 2 3
1
2
3
12
Finite Element Method – Piecewise Approximation
x
u
x
u
13
FEM Formulation of Axially Loaded Bar – Governing Equations
• Differential Equation
• Weighted-Integral Formulation
• Weak Form
Lxxfdx
duxEA
dx
d
0 0)()(
0)()(0
dxxf
dx
duxEA
dx
dw
L
LL
dx
duxEAwdxxwf
dx
duxEA
dx
dw
00
)()()(0
14
Approximation Methods – Finite Element Method
Example:
Step 1: Discretization
Step 2: Weak form of one element P2 P1 x1 x2
0)()()()()(
2
1
2
1
x
x
x
xdx
duxEAxwdxxfxw
dx
duxEA
dx
dw
0)()()( 1122
2
1
PxwPxwdxxfxw
dx
duxEA
dx
dwx
x 15
Approximation Methods – Finite Element Method
Example (cont):
Step 3: Choosing shape functions - linear shape functions
2211 uuu
l x1 x2
x x x1 x0 x1
l
xx
l
xx 12
21 ;
2
1 ;
2
121
x
x
112
1 ;1
2x
lxxx
l
xx
16
Approximation Methods – Finite Element Method
Example (cont):
Step 4: Forming element equation
Let , weak form becomes 1w 01
1121112
2
1
2
1
PPdxfdx
l
uuEA
l
x
x
x
x
1121
2
1
Pdxful
EAu
l
EAx
x
Let , weak form becomes 2w 01
1222212
2
1
2
1
PPdxfdx
l
uuEA
l
x
x
x
x
2221
2
1
Pdxful
EAu
l
EAx
x
2
1
2
1
1
1 1 1 1
2 2 2 2
2
1 1
1 1
x
x
x
x
fdxu P f PEA
u P f Plfdx
E,A are constant
17
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 0
1 1 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
I I I
I I II I
I
u f P
u f PE A
l
Element 2: 1 1 1
2 2 2
0 0 0 0 0 0 0
0 1 1 0
0 1 1 0
0 0 0 0 0 0 0
II II IIII II
II II IIII
u f PE A
u f Pl
Element 3:
1 1 1
2 2 2
0 0 00 0 0 0
0 0 00 0 0 0
0 0 1 1
0 0 1 1
III III
III III IIIIII
III III III
E A
u f Pl
u f P
18
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 A
l lu f PE A E A E A E A
u f Pl l l l
u f PE A E A E A E A
u f Pl l l l
E A E A
l 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 P
f f P P
f f P P
f P
19
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)
e
ij IJk K
20
Approximation Methods – Finite Element Method
Example (cont):
Step 6: Imposing boundary conditions and forming condense system
Condensed system:
2 2
3 3
4 4
0
0
0
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 A
l l lu f
E A E A E A E Au f
l l l lu f P
E A E A
l l
21
Approximation Methods – Finite Element Method
Example (cont):
Step 7: solution
Step 8: post calculation
dx
du
dx
du
dx
du 22
11
2211 uuu dx
dEu
dx
dEuE 2
21
1
22
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 23
Exercises – Linear Element
Example 1:
E = 100 GPa, A = 1 cm2
24
Linear Formulation for Bar Element
2
1
2212
1211
2
1
2
1
u
u
KK
KK
f
f
P
P
2
1
2
1
,
x
x
ii
x
x
ji
jiij dxffKdx
dx
d
dx
dEAKwhere
x=x1 x=x2
1 2 1 1
x
x=x1 x= x2
u1 u2
1P 2Pf(x)
L = x2-x1
u
x
25
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 u3 u
x
u2
2
1 4
u1 u4
2
u
x
u2 u3
3
)x(u)x(u)x(u)x(u)x(u(x)u nn44332211
1 n
u1 un
2
u
x
u2 u3
3
u4 ……………
4 ……………
26
Natural Coordinates and Interpolation Functions
2
1 ,
2
121
x
x
Natural (or Normal) Coordinate:
x=x1 x= x2
x=-1 x=1
x
x
0x x l
1xxx
1 2
2/ 2
x xx
lx
1 3 2
x x=-1 x=1
1 2
x x=-1 x=1
1 4 2
x x=-1 x=1
3
2
1 ,11 ,
2
1321
xxxx
xx
13
11
16
27 ,1
3
1
3
1
16
921
xxxxxx
3
1
3
11
16
9 ,1
3
11
16
2743 xxxxxx
27
Quadratic Formulation for Bar Element
2
1
1
1
nd , , 1, 2, 32
x
i i i
x
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 K
dx dx d d l
x
x x
3
2
1
332313
232212
131211
3
2
1
3
2
1
u
u
u
KKK
KKK
KKK
f
f
f
P
P
P
x=-1 x=0 x=1
3 1 2
28
Quadratic Formulation for Bar Element
u1 u3 u2 f(x) P3 P1
P2
x=-1 x=0 x=1
1x 2x 3x
2
1u11u
2
1u)(u)(u)(u)(u 321332211
xxxx
xxxxxx
2
1 ,11 ,
2
1321
xxxx
xx
1 1 2 2 3 32 2 1 2 4 2 2 1, ,
d d d d d d
dx 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
ld dxx
2d
dx l
x
29
Exercises – Quadratic Element
Example 2:
E = 100 GPa, A1 = 1 cm2; A1 = 2 cm2
30
Some Issues
Non-constant cross section:
Interior load point:
Mixed boundary condition:
k
31
UNIT – 2
Finite Element Analysis of Trusses
32
Plane Truss Problems
Example 1: Find forces inside each member. All members have the same length.
F
33
Arbitrarily Oriented 1-D Bar Element on 2-D Plane
Q2 , v2
11 u ,P
q
22 u ,P
P2 , u2
Q1 , v1
P1 , u1
34
Relationship Between Local Coordinates and Global Coordinates
2
2
1
1
2
2
1
1
cossin00
sincos00
00cossin
00sincos
0
0
v
u
v
u
v
u
v
u
35
Relationship Between Local Coordinates and Global Coordinates
2
2
1
1
2
1
cossin00
sincos00
00cossin
00sincos
0
0
Q
P
Q
P
P
P
36
Stiffness Matrix of 1-D Bar Element on 2-D Plane
2
2
1
1
2
2
1
1
cossin00
sincos00
00cossin
00sincos
cossin00
sincos00
00cossin
00sincos
v
u
v
u
K
Q
P
Q
P
ij
0000
0101
0000
0101
L
AEKij
Q2 , v2
11 , uP
q
22 u ,P
P2 , u2
Q1 , v1
P1 , u1
2
2
1
1
22
22
22
22
2
2
1
1
sincossinsincossin
cossincoscossincos
sincossinsincossin
cossincoscossincos
v
u
v
u
L
AE
Q
P
Q
P
qqqqqq
qqqqqq
qqqqqq
qqqqqq
37
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
w
v
u
w
v
u
000
000
000
000
000
000
0w
0v
u
0w
0v
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
x x
y
z 2
1
- -
-
2
2
2
1
1
1
2
2
2
1
1
1
000
000
000
000
000
000
0
0
0
0
R
Q
P
R
Q
P
R
Q
P
R
Q
P
zzz
yyy
xxx
zzz
yyy
xxx
38
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
w
v
u
w
v
u
L
AE
R
Q
P
R
Q
P
xxxxxxxxxx
xxxxxxxxxx
xxxxxxxxxx
xxxxxxxxxx
xxxxxxxxxx
xxxxxxxxxx
11 u ,Px
22 u ,P
x
x x
y
z 2
1
- -
-
0w
0v
u
0w
0v
u
000000
000000
001001
000000
000000
001001
L
AE
0R
0Q
P
0R
0Q
P
2
2
2
1
1
1
2
2
2
1
1
1
39
Matrix Assembly of Multiple Bar Elements
2
2
1
1
2
2
1
1
0000
0101
0000
0101
v
u
v
u
L
AE
Q
P
Q
P
Element I
Element I I
Element I I I
3
3
2
2
3
3
2
2
3333
3131
3333
3131
4
v
u
v
u
L
AE
Q
P
Q
P
3
3
1
1
3
3
1
1
3333
3131
3333
3131
4
v
u
v
u
L
AE
Q
P
Q
P
40
Matrix Assembly of Multiple Bar Elements
3
3
2
2
1
1
3
3
2
2
1
1
000000
000000
000000
000404
000000
000404
4
v
u
v
u
v
u
L
AE
Q
P
Q
P
Q
P
Element I
3
3
2
2
1
1
3
3
2
2
1
1
333300
313100
333300
313100
000000
000000
4
v
u
v
u
v
u
L
AE
Q
P
Q
P
Q
P
3
3
2
2
1
1
3
3
2
2
1
1
330033
310031
000000
000000
330033
310031
4
v
u
v
u
v
u
L
AE
Q
P
Q
P
Q
P
Element I I
Element I I I
41
Matrix Assembly of Multiple Bar Elements
3
3
2
2
1
1
3
3
2
2
1
1
v
u
v
u
v
u
33333333
33113131
33303000
31301404
33003030
31043014
L4
AE
S
R
S
R
S
R
0v
0u
0v
?u
?v
0u
603333
023131
333300
313504
330033
310435
L4
AE
?S
?R
?S
FR
0S
?R
3
3
2
2
1
1
3
3
2
2
1
1
Apply known boundary conditions
42
Solution Procedures
0v
0u
0v
?u
?v
0u
603333
023131
333300
310435
330033
313504
L4
AE
?S
?R
?S
?R
0S
FR
3
3
2
2
1
1
3
3
2
1
1
2
u2= 4FL/5AE, v1= 0
0v
0u
0vAE5
FL4u
0v
0u
603333
023131
333300
310435
330033
313504
L4
AE
?S
?R
?S
?R
0S
FR
3
3
2
2
1
1
3
3
2
1
1
2
43
Recovery of Axial Forces
05
4
05
4
05
4
0
0
0000
0101
0000
0101
2
2
1
1
2
2
1
1
F
vAE
FLu
v
u
L
AE
Q
P
Q
P
Element I
Element I I
Element I I I
53
51
53
51
0
0
05
4
3333
3131
3333
3131
4
3
3
2
2
3
3
2
2
F
v
u
vAE
FLu
L
AE
Q
P
Q
P
0
0
0
0
0
0
0
0
3333
3131
3333
3131
4
3
3
1
1
3
3
1
1
v
u
v
u
L
AE
Q
P
Q
P
44
Stresses inside members
Element I
Element I I
Element I I I
A
F
5
4
5
41
FP
5
42
FP
FP5
12
FQ5
32
FQ5
33
FP5
13
45
Finite Element Analysis of
Beams
46
Bending Beam
Review
Normal strain:
Pure bending problems:
Normal stress:
Normal stress with bending moment:
Moment-curvature relationship:
Flexure formula:
x
y
yx
Eyx
M M
MydAx
EI
M
1
I
Myx dAyI 2
2
21
dx
ydEIEIM
47
Bending Beam
Review
Deflection:
Sign convention:
Relationship between shear force, bending moment and transverse load:
q(x)
x
y
qdx
dV V
dx
dM
qdx
ydEI
4
4
+ - M M M
+ - V V V
48
Governing Equation and Boundary Condition
• Governing Equation
• Boundary Conditions -----
0at , ? &? &? & ?2
2
2
2
x
dx
vdEI
dx
d
dx
vdEI
dx
dvv
Essential BCs – if v or is specified at the boundary.
Natural BCs – if or is specified at the boundary.
dx
dv
{ Lx
dx
vdEI
dx
d
dx
vdEI
dx
dvv
at , ? &? &? & ?
2
2
2
2
2
2
dx
vdEI
2
2
dx
vdEI
dx
d
,0)()(
2
2
2
2
xq
dx
xvdEI
dx
d 0<x<L
49
Weak Formulation for Beam Element
• Governing Equation
,0)()(
2
2
2
2
xq
dx
xvdEI
dx
d
• Weighted-Integral Formulation for one element
2
1
)()(
)(02
2
2
2x
x
dxxqdx
xvdEI
dx
dxw
• Weak Form from Integration-by-Parts ----- (1st time)
2
1
2
1
2
2
2
2
0
x
x
x
xdx
vdEI
dx
dwdxwq
dx
vdEI
dx
d
dx
dw
21 xxx
50
Weak Formulation
• Weak Form from Integration-by-Parts ----- (2nd time)
2
1
2
1
2
1
2
2
2
2
2
2
2
2
0
x
x
x
x
x
xdx
vdEI
dx
dw
dx
vdEI
dx
dwdxwq
dx
vdEI
dx
wd
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
0
x
x
x
x
Mdx
dwwVdxwq
dx
vdEI
dx
wd
51
Weak Formulation
• Weak Form
24231211 , , , xMQxVQxMQxVQ
4
2
2
1
32112
2
2
2
)()(2
1
Qdx
dwQ
dx
dwQxwQxwdxwq
dx
vdEI
dx
wdx
x
2
1
2
1
2
2
2
2
0
x
x
x
x
Mdx
dwwVdxwq
dx
vdEI
dx
wd
Q3
x = x1
Q4
q(x) y(v)
x x = x2
Q1
Q2
L = x2-x1
52
Ritz Method for Approximation
n
j
jj nxuxvLet1
4 and )()(
4
2
2
1
32112
24
12
2
)()(2
1
Qdx
dwQ
dx
dwQxwQxwdxwq
dx
duEI
dx
wdx
x
j
j
j
Let w(x)= i (x), i = 1, 2, 3, 4
4
2
2
1
32112
24
12
2
)()(2
1
Qdx
dQ
dx
dQxQxdxq
dx
duEI
dx
d iiii
x
x
i
j
j
ji
Q3
x = x1
Q4
q(x) y(v)
x x = x2
Q1
Q2
L = x2-x1
where ; ; ; ;
21
423211
xxxx dx
dvuxvu
dx
dvuxvu
53
Ritz Method for Approximation
i
j
jij
x
i
xi
x
i
xi quKQdx
dQQ
dx
dQ
4
1
4321
2
2
1
1
2
1
2
1
and where2
2
2
2 x
x
ii
x
x
jiij qdxqdx
dx
d
dx
dEIK
Q3
x = x1
y(v)
x x = x2
Q1
Q2
L = x2-x1
Q4
54
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
2
2
1
1
2
2
1
1
2
2
1
1
2
2
1
1
q
q
q
q
u
u
u
u
KKKK
KKKK
KKKK
KKKK
Q
Q
Q
Q
dx
d
dx
d
dx
d
dx
d
dx
d
dx
d
dx
d
dx
d
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
55
Selection of Shape Function
1000
0100
0010
0001
2
2
1
1
2
2
1
1
2
2
1
1
2
2
1
1
44
44
33
33
22
22
11
11
xx
xx
xx
xx
xx
xx
xx
xx
dx
d
dx
d
dx
d
dx
d
dx
d
dx
d
dx
d
dx
d
4
3
2
1
4
3
2
1
44342414
34332313
24232212
14131211
4
3
2
1
q
q
q
q
u
u
u
u
KKKK
KKKK
KKKK
KKKK
Q
Q
Q
Q
The best situation is -----
Interpolation Properties
56
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 u
d d d d d
x
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.
57
Derivation of Shape Function for Beam Element
How to select i???
e.g. Let 3
1
2
1111 xxx dcba
xx 214
1 2
1
Similarly
114
1
214
1
114
1
2
2
2
3
2
2
xx
xx
xx
58
Derivation of Shape Function for Beam Element
In the global coordinates:
)(2
)()(2
)()( 42
3221
11 xdx
dvlxvx
dx
dvlxvxv
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
xx
xx
xx
xxxx
l
xx
xx
xx
xx
xx
xxxx
l
xx
xx
xx
xx
59
Element Equations of 4th Order 1-D Model
u3
x = x1
u4
q(x) y(v)
x x = x2
u1
u2
L = x2-x1
4
3
2
1
44342414
34332313
24232212
14131211
4
3
2
1
4
3
2
1
u
u
u
u
KKKK
KKKK
KKKK
KKKK
q
q
q
q
Q
Q
Q
Q
2
1
2
1
2
2
2
2 x
x
ii
x
x
ji
jiij qdxqandKdx
dx
d
dx
dEIKwhere
x=x2 x=x1
1 1 1
3 2
4
60
Element Equations of 4th Order 1-D Model
u3
x = x1
u4
q(x) y(v)
x x = x2
u1
u2
L = x2-x1
2
1
x
x
ii qdxqwhere
24
23
12
11
22
22
3
4
3
2
1
4
3
2
1
233
3636
323
3636
2
q
q
u
vu
u
vu
LLLL
LL
LLLL
LL
L
EI
q
q
q
q
Q
Q
Q
Q
61
Finite Element Analysis of 1-D Problems - Applications
F
L L L
Example 1.
Governing equation:
Lxxqdx
vdEI
dx
d
0 0)(
2
2
2
2
Weak form for one element
04322112
2
2
2
21
2
1
Q
dx
dwQxwQ
dx
dwQxwdxwq
dx
vd
dx
wdEI
xx
x
x
where )( 11 xVQ )( 12 xMQ )( 23 xVQ )( 24 xMQ
62
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
xx
xx
xx
xxxx
l
xx
xx
xx
xx
xx
xxxx
l
xx
xx
xx
xx
)(2
)()(2
)()( 42
3221
11 xdx
dvlxvx
dx
dvlxvxv
63
Finite Element Analysis of 1-D Problems Example 1.
Finite element model:
2
2
1
1
22
22
3
4
3
2
1
233
3636
323
3636
2
q
q
v
v
LLLL
LL
LLLL
LL
L
EI
Q
Q
Q
Q
P1 , v1 P2 , v2 P3 , v3 P4 , v4
M1 , q1 M2 , q2 M3 , q3 M4 , q4
I II III
Discretization:
64
Matrix Assembly of Multiple Beam Elements
Element I
Element I I
11
2 2
12
23
2 2
24
3
3
3
4
4
6 3 6 3 0 0 0 0
3 2 3 0 0 0 0
6 3 6 3 0 0 0 0
3 3 2 0 0 0 02
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 00
I
I
I
I
vL LQ
L L L LQ
vL LQ
L L L LQ EI
vL
v
q
q
q
q
1
1
21
2 2
22
3
33
2 2
34
4
4
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 6 3 6 3 0 0
0 0 3 2 3 0 02
0 0 6 3 6 3 0 0
0 0 3 3 2 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
II
II
II
II
v
vQ L L
Q L L L LEI
vQ L LL
Q L L L L
v
q
q
q
q
65
Matrix Assembly of Multiple Beam Elements
Element I I I
1
1
2
2
3
1 3
2 2
2 3
3 4
2 2
4 4
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 02
0 0 0 0 6 3 6 3
0 0 0 0 3 2 3
0 0 0 0 6 3 6 3
0 0 0 0 3 3 2
III
III
III
III
v
v
EI
Q vL LL
Q L L L L
Q vL L
Q 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
2330000
36360000
32233300
3633663600
00322333
0036336636
0000323
00003636
2
q
q
q
q
v
v
v
v
LLLL
LL
LLLLLLLL
LLLL
LLLLLLLL
LLLL
LLLL
LL
L
EI
M
P
M
P
M
P
M
P
66
Solution Procedures
?
?
?
0
?
0
0
0
2330000
36360000
340300
360123600
003403
003601236
0000323
00003636
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
LLLL
LL
LLLLL
LL
LLLLL
LL
LLLL
LL
L
EI
M
FP
M
P
M
P
M
P
Apply known boundary conditions
?
?
?
0
?
0
0
0
360123600
003601236
0000323
00003636
2330000
36360000
340300
003403
2
?
?
?
?
0
0
0
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
LL
LL
LLLL
LL
LLLL
LL
LLLLL
LLLLL
L
EI
P
P
M
P
M
FP
M
M
67
Solution Procedures
?
?
?
?
230
3630
34
004
2
0
0
0
4
4
3
2
22
222
22
3
4
4
3
2
q
q
q
v
LLL
LL
LLLL
LL
L
EI
M
FP
M
M
4
4
3
2
2
3
3
2
1
1
3603
0030
000
0003
2
?
?
?
?
q
q
q
v
LL
L
L
L
L
EI
P
P
M
P
?
?
?
?
0
0
0
0
360312600
003061236
0000323
00030636
2303000
36306000
340300
004303
2
?
?
?
?
0
0
0
4
4
3
2
3
2
1
1
22
22
222
222
3
3
2
1
1
4
4
3
2
q
q
q
q
v
v
v
v
LL
LL
LLLL
LL
LLLL
LL
LLLLL
LLLLL
L
EI
P
P
M
P
M
FP
M
M
68
Shear Resultant & Bending Moment Diagram
FL7
2
FL7
1
FL
F7
9
F7
3
2P
F
69
Plane Flame
Frame: combination of bar and beam
E, A, I, L Q1 , 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
460
260
6120
6120
0000
260
460
6120
6120
0000
q
q
v
u
v
u
L
EI
L
EI
L
EI
L
EIL
EI
L
EI
L
EI
L
EIL
AE
L
AEL
EI
L
EI
L
EI
L
EIL
EI
L
EI
L
EI
L
EIL
AE
L
AE
Q
Q
P
Q
Q
P
70
Finite Element Model of an Arbitrarily Oriented Frame
q x
y
q x
y
71
Finite Element Model of an Arbitrarily Oriented Frame
2
2
2
1
1
1
22
2323
22
2323
4
3
2
2
1
1
40
620
6
0000
60
1260
12
20
640
6
0000
60
1260
12
q
q
v
u
v
u
L
EI
L
EI
L
EI
L
EIL
AE
L
AEL
EI
L
EI
L
EI
L
EIL
EI
L
EI
L
EI
L
EIL
AE
L
AEL
EI
L
EI
L
EI
L
EI
Q
Q
P
Q
Q
P
local
global
72
Plane Frame Analysis - Example
Rigid Joint Hinge Joint
Beam II
Beam I Beam
Bar
F F F F
73
Plane Frame Analysis
2
2
2
1
1
1
22
2323
22
2323
4
3
2
2
1
1
40
620
6
0000
60
1260
12
20
640
6
0000
60
1260
12
q
q
v
u
v
u
L
EI
L
EI
L
EI
L
EIL
AE
L
AEL
EI
L
EI
L
EI
L
EIL
EI
L
EI
L
EI
L
EIL
AE
L
AEL
EI
L
EI
L
EI
L
EI
Q
Q
P
Q
Q
P
I
I
P1 , u1
P2 , u2
Q2 , q1
Q4 , q2
Q1 , v1
Q3 , v2
74
Plane Frame Analysis
P1 , u2
Q3 , v3
Q2 , q2 Q4 , q3
Q1 , v2
P2 , u3
4
3
3
2
2
2
22
2323
22
2323
4
3
2
2
2
1
460
260
6120
6120
0000
260
460
6120
6120
0000
q
q
v
u
v
u
L
EI
L
EI
L
EI
L
EIL
EI
L
EI
L
EI
L
EIL
AE
L
AEL
EI
L
EI
L
EI
L
EIL
EI
L
EI
L
EI
L
EIL
AE
L
AE
Q
Q
P
Q
Q
PII
75
Plane Frame Analysis
2
2
2
1
1
1
22
2323
22
2323
4
3
2
2
1
1
40
620
6
0000
60
1260
12
20
640
6
0000
60
1260
12
q
q
v
u
v
u
L
EI
L
EI
L
EI
L
EIL
AE
L
AEL
EI
L
EI
L
EI
L
EIL
EI
L
EI
L
EI
L
EIL
AE
L
AEL
EI
L
EI
L
EI
L
EI
Q
Q
P
Q
Q
P
I
I
4
3
3
2
2
2
22
2323
22
2323
4
3
2
2
2
1
460
260
6120
6120
0000
260
460
6120
6120
0000
q
q
v
u
v
u
L
EI
L
EI
L
EI
L
EIL
EI
L
EI
L
EI
L
EIL
AE
L
AEL
EI
L
EI
L
EI
L
EIL
EI
L
EI
L
EI
L
EIL
AE
L
AE
Q
Q
P
Q
Q
PII
76
Plane Frame Analysis
FL8
1
F2
1
F2
1
FL8
3
77
UNIT- 3 Finite element analysis constant
strain triangle
78
Finite element formulation for 2D: Step 1: Divide the body into finite elements connected to each other through special points (“nodes”)
x
y
Su
ST u
v
x
px
py
Element ‘e’
3
2
1
4
y
x v
u
1
2
3
4
u1
u2
u3
u4
v4
v3
v2
v1
4
4
3
3
2
2
1
1
v
u
v
u
v
u
v
u
d
79
44332211
44332211
vy)(x,N vy)(x,N vy)(x,N vy)(x,Ny)(x,v
u y)(x,Nu y)(x,Nu y)(x,Nu y)(x,Ny)(x,u
4
4
3
3
2
2
1
1
4321
4321
v
u
v
u
v
u
v
u
N0N0N0N0
0N0N0N0N
y)(x,v
y)(x,uu
dNu
80
TASK 2: APPROXIMATE THE STRAIN and STRESS WITHIN EACH ELEMENT
...... vy)(x,N
uy)(x,Ny)(x,vy)(x,u
vy)(x,N
vy)(x,N
v y)(x,N
vy)(x,Ny)(x,v
u y)(x,N
u y)(x,N
u y)(x,N
u y)(x,Ny)(x,u
11
11
xy
44
3
3
22
11
y
44
3
3
22
11
x
xyxy
yyyyy
xxxxx
Approximation of the strain in element ‘e’
81
4
4
3
3
2
2
1
1
B
44332211
4321
4321
xy
v
u
v
u
v
u
v
u
y)(x,Ny)(x,Ny)(x,Ny)(x,N y)(x,N y)(x,Ny)(x,Ny)(x,N
y)(x,N0
y)(x,N0
y)(x,N0
y)(x,N0
0y)(x,N
0y)(x,N
0 y)(x,N
0y)(x,N
xyxyxyxy
yyyy
xxxx
y
x
dBε
82
Displacement approximation in terms of shape functions
Strain approximation in terms of strain-displacement matrix
Stress approximation
Summary: For each element
Element stiffness matrix
Element nodal load vector
dNu
dBD
dBε
eV
k dVBDBT
S
eT
b
e
f
SS
T
f
V
TdSTdVXf NN
83
Constant Strain Triangle (CST) : Simplest 2D finite element
• 3 nodes per element • 2 dofs per node (each node can move in x- and y- directions) • Hence 6 dofs per element
x
y
u3
v3
v1
u1
u2
v2
2
3
1
(x,y)
v u
(x1,y1)
(x2,y2)
(x3,y3)
84
166212 dNu
3
3
2
2
1
1
321
321
v
u
v
u
v
u
N0N0N0
0N0N0N
y)(x,v
y)(x,uu
The displacement approximation in terms of shape functions is
321
321
N0N0N0
0N0N0NN
1 1 2 2 3 3u (x,y) u u uN N N
1 1 2 2 3 3v(x,y) v v vN N N
85
Formula for the shape functions are
A
ycxbaN
A
ycxbaN
A
ycxbaN
2
2
2
333
3
2222
1111
12321312213
31213231132
23132123321
33
22
11
x1
x1
x1
det2
1
xxcyybyxyxa
xxcyybyxyxa
xxcyybyxyxa
y
y
y
triangleofareaA
where
x
y
u3
v3
v1
u1
u2
v2
2
3
1
(x,y)
v u
(x1,y1)
(x2,y2)
(x3,y3)
86
Properties of the shape functions:
1. The shape functions N1, N2 and N3 are linear functions of x and y
x
y
2
3
1
1
N1
2
3
1
N2
1 2
3
1
1
N3
nodesotherat
inodeat
0
''1N i
87
2. At every point in the domain
yy
xx
i
i
3
1i
i
3
1i
i
3
1i
i
N
N
1N
88
3. Geometric interpretation of the shape functions At any point P(x,y) that the shape functions are evaluated,
x
y
2
3
1 P (x,y)
A1
A3
A2
A
A
A
A
A
A
3
3
22
11
N
N
N
89
Approximation of the strains
xy
u
v
u v
x
y
x
Bdy
y x
332211
321
321
332211
321
321
000
000
2
1
y)(x,Ny)(x,N y)(x,N y)(x,Ny)(x,Ny)(x,N
y)(x,N0
y)(x,N0
y)(x,N0
0y)(x,N
0 y)(x,N
0y)(x,N
bcbcbc
ccc
bbb
A
xyxyxy
yyy
xxx
B
90
Element stresses (constant inside each element)
dBD
Inside each element, all components of strain are constant: hence the name Constant Strain Triangle
91
IMPORTANT NOTE: 1. The displacement field is continuous across element boundaries 2. The strains and stresses are NOT continuous across element boundaries
92
Element stiffness matrix
eV
k dVBDBT
AtkeV
BDBdVBDBTT
t=thickness of the element A=surface area of the element
Since B is constant
t
A
93
S
eT
b
e
f
SS
T
f
V
TdSTdVXf NN
Element nodal load vector
94
Element nodal load vector due to body forces
ee A
T
V
T
bdAXtdVXf NN
e
e
e
e
e
e
Ab
Aa
Ab
Aa
Ab
Aa
yb
xb
yb
xb
yb
xb
b
dAXNt
dAXNt
dAXNt
dAXNt
dAXNt
dAXNt
f
f
f
f
f
f
f
3
3
2
2
1
1
3
3
2
2
1
1
x
y
fb3x
fb3y
fb1y
fb1x
fb2x
fb2y
2
3
1
(x,y)
Xb Xa
95
EXAMPLE: If Xa=1 and Xb=0
03
03
03
0
0
0
3
2
1
3
3
2
2
1
1
3
3
2
2
1
1
tA
tA
tA
dANt
dANt
dANt
dAXNt
dAXNt
dAXNt
dAXNt
dAXNt
dAXNt
f
f
f
f
f
f
f
e
e
e
e
e
e
e
e
e
A
A
A
Ab
Aa
Ab
Aa
Ab
Aa
yb
xb
yb
xb
yb
xb
b
96
Element nodal load vector due to traction
e
TSS
T
SdSTf N
EXAMPLE:
x
y
fS3x
fS3y
fS1y
fS1x
2
3
1
e
lS
along
T
SdSTtf
31 31N
97
Element nodal load vector due to traction
EXAMPLE:
x
y
fS3x
2
3 1
e
lS
along
T
SdSTtf
32 32N
fS3y
fS2x
fS2y
(2,0)
(2,2)
(0,0)
0
1ST
tt
dyNtfex l alongS
122
1
)1(32
2 322
0
0
3
3
2
y
x
y
S
S
S
f
tf
f
Similarly, compute
1
2
98
Recommendations for use of CST 1. Use in areas where strain gradients are small 2. Use in mesh transition areas (fine mesh to coarse mesh) 3. Avoid CST in critical areas of structures (e.g., stress concentrations, edges of holes, corners) 4. In general CSTs are not recommended for general analysis purposes as a very large number of these elements are required for reasonable accuracy.
99
Example
x
y
El 1
El 2
1
2 3
4
300 psi 1000 lb
3 in
2 in Thickness (t) = 0.5 in E= 30×106 psi n=0.25
(a) Compute the unknown nodal displacements. (b) Compute the stresses in the two elements.
100
Realize that this is a plane stress problem and therefore we need to use
psiE
D 7
210
2.100
02.38.0
08.02.3
2
100
01
01
1
nn
n
n
Step 1: Node-element connectivity chart
ELEMENT Node 1 Node 2 Node 3 Area
(sqin)
1 1 2 4 3
2 3 4 2 3
Node x y
1 3 0
2 3 2
3 0 2
4 0 0
Nodal coordinates
101
Step 2: Compute strain-displacement matrices for the elements
332211
321
321
000
000
2
1
bcbcbc
ccc
bbb
AB
Recall
123312231
213132321
xxcxxcxxc
yybyybyyb
with
For Element #1:
1(1)
2(2)
4(3)
(local numbers within brackets)
0;3;3
0;2;0
321
321
xxx
yyy
Hence
033
202
321
321
ccc
bbb
200323
003030
020002
6
1)1(B
Therefore
For Element #2:
200323
003030
020002
6
1)2(B
102
Step 3: Compute element stiffness matrices
7
)1(T)1()1(T)1()1(
10
2.0
05333.0
02.02.1
3.00045.0
2.02.02.13.04.1
3.05333.02.045.05.09833.0
BDB)5.0)(3(BDB
Atk
u1 u2 u4 v4 v2 v1
103
7
)2(T)2()2(T)2()2(
10
2.0
05333.0
02.02.1
3.00045.0
2.02.02.13.04.1
3.05333.02.045.05.09833.0
BDB)5.0)(3(BDB
Atk
u3 u4 u2 v2 v4 v3
104
Step 4: Assemble the global stiffness matrix corresponding to the nonzero degrees of freedom
014433 vvuvuNotice that
Hence we need to calculate only a small (3x3) stiffness matrix
710
4.102.0
0983.045.0
2.045.0983.0
K
u1
u2 v2
u1 u2
v2
105
Step 5: Compute consistent nodal loads
y
x
x
f
f
f
f
2
2
1
yf2
0
0
ySy ff2
10002
The consistent nodal load due to traction on the edge 3-2
lbx
dxx
dxN
tdxNf
x
x
xS y
2252
950
250
3150
)5.0)(300(
)300(
3
0
2
3
0
3
0 233
3
0 2332
3 2
3232
xN
106
lb
ffySy
1225
100022
Hence
Step 6: Solve the system equations to obtain the unknown nodal loads
fdK
1225
0
0
4.102.0
0983.045.0
2.045.0983.0
10
2
2
1
7
v
u
u
Solve to get
in
in
in
v
u
u
4
4
4
2
2
1
109084.0
101069.0
102337.0
107
Step 7: Compute the stresses in the elements
)1()1()1(dBD
With
00109084.0101069.00102337.0
d
444
442211
)1(
vuvuvuT
Calculate
psi
1.76
1.1391
1.114)1(
In Element #1
108
)2()2()2(dBD
With
44
224433
)2(
109084.0101069.00000
d
vuvuvuT
Calculate
psi
35.363
52.28
1.114)2(
In Element #2
Notice that the stresses are constant in each element
109
UNIT - 4
Heat transfer analysis
110
Thermal Convection
( )sq h T T
Newton’s Law of Cooling
2: convective heat transfer coefficient ( )oh W m C
111
Thermal Conduction in 1-D
Dirichlet BC:
Boundary conditions:
Natural BC:
Mixed BC:
112
Weak Formulation of 1-D Heat Conduction (Steady State Analysis)
• Governing Equation of 1-D Heat Conduction -----
( )( ) ( ) ( ) 0
d dT xx A x AQ x
dx dx
0<x<L
• Weighted Integral Formulation -----
• Weak Form from Integration-by-Parts -----
0
( )0 ( ) ( ) ( ) ( )
Ld dT x
w x x A x AQ x dxdx dx
0 0
0
LLdw dT dT
A wAQ dx w Adx dx dx
113
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
2
2
1
1 )( ,)(x
TAxf
x
TAxf
114
Formulation for 1-D Linear Element
Let w(x)= i (x), i = 1, 2
2 2
1 1
2
2 2 1 1
1
0 ( ) ( )
x x
jij i i i
j x x
ddT A dx AQ dx x f x f
dx dx
1122
2
1
)()( fxfxQTK iii
j
jij
2
1
2212
1211
2
1
2
1
T
T
KK
KK
Q
Q
f
f
2 2
1 21 1
1 2 , Q , f , f
x x
jiij i i
x xx x
dd dT dTwhere K A dx AQ dx A A
dx dx dx dx
115
Element Equations of 1-D Linear Element
2
1
2
1
2
1
11
11
T
T
L
A
Q
Q
f
f
2
1 21
1 2 Q , f , f
x
i i
x x x xx
dT dTwhere AQ dx A A
dx dx
f2
x1 x2
1 2
T1
x
T2
f1
116
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 50oC with a convection coefficient of h = 10 W(m2.K). Find the temperature along the composite wall.
t1 t2 t3
0 200oT C 50oT C
x
12 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
117
Thermal Conduction and Convection- Fin
Objective: to enhance heat transfer
dx
t
x
w
2 ( )2 ( ) 2 ( )loss
c c
h T T w th T T dx w h T T dx tQ
A dx A
Governing equation for 1-D heat transfer in thin fin
0c c
d dTA A Q
dx dx
0c c
d dTA Ph T T A Q
dx dx
2P w t where
118
Fin - Weak Formulation (Steady State Analysis)
• Governing Equation of 1-D Heat Conduction -----
( )
( ) ( ) 0d dT x
x A x Ph T T AQdx dx
0<x<L
• Weighted Integral Formulation -----
• Weak Form from Integration-by-Parts -----
0
( )0 ( ) ( ) ( ) ( ) ( )
Ld dT x
w x x A x Ph T T AQ x dxdx dx
0 0
0 ( )
LLdw dT dT
A wPh T T wAQ dx w Adx dx dx
119
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 x
jij i j i
j x x
i i
ddT A Ph dx AQ PhT dx
dx dx
x f x f
1122
2
1
)()( fxfxQTK iii
j
jij
2
1
2212
1211
2
1
2
1
T
T
KK
KK
Q
Q
f
f
2 2
1 1
1 2
1 2
, Q ,
,
x x
jiij i j i i
x x
x x x x
ddwhere K A Ph dx AQ PhT dx
dx dx
dT dTf A f A
dx dx
120
Element Equations of 1-D Linear Element
1 1 1
2 2 2
1 1 2 1
1 1 1 26
f Q TA Phl
f Q TL
f2
x=0 x=L
1 2
T1
x
T2
f1
2
1 21
1 2 Q , ,
x
i i
x x x xx
dT dTwhere AQ PhT dx f A f A
dx dx
121
Time-Dependent Problems
122
Time-Dependent Problems
In general,
Key question: How to choose approximate functions?
Two approaches:
txutxu jj ,,
txu ,
xtutxu jj ,
123
Model Problem I – Transient Heat Conduction
Weak form:
txfx
ua
xt
uc ,
)()(0 2211
2
1
xwQxwQdxwft
ucw
x
u
x
wa
x
x
;
21
21
xx dx
duaQ
dx
duaQ
124
Transient Heat Conduction
let:
)()(0 2211
2
1
xwQxwQdxwft
ucw
x
u
x
wa
x
x
n
j
jj xtutxu1
, and xw i
FuMuK
2
1
x
x
jiij dx
xxaK
2
1
x
x
jiij dxcM
i
x
x
ii QfdxF 2
1
ODE!
125
Time Approximation – First Order ODE
tfbudt
dua Tt 0 00 uu
Forward difference approximation - explicit
Backward difference approximation - implicit
1k k k k
tu u f bu
a
1k k k k
tu u f bu
a b t
126
Stability Requirment
max21
2
critt
QuMK where
Note: One must use the same discretization for solving the eigenvalue problem.
127
Transient Heat Conduction - Example
02
2
x
u
t
u10 x
0,0 tu 0,1
t
t
u
0.10, xu
0t
128
Transient Heat Conduction - Example
129
Transient Heat Conduction - Example
130
Transient Heat Conduction - Example
131
Transient Heat Conduction - Example
132
Transient Heat Conduction - Example
133
UNIT – 5
Dynamic Analysis
134
Axi-symmetric Analysis
Cylindrical coordinates:
zzryrx ;sin ;cos qq
zr , , q
• quantities depend on r and z only • 3-D problem 2-D problem
135
Axi-symmetric Analysis
136
Axi-symmetric Analysis – Single-Variable Problem
0),(),(),(1
002211
zrfua
z
zrua
zr
zrura
rr
Weak form:
dswq
rdrdzzrwfwuaz
ua
z
w
r
ua
r
w
e
e
n
),(0 002211
where zrn n
z
zruan
r
zruaq
),(),(2211
137
Finite Element Model – Single-Variable Problem
j j
j
u u
Ritz method:
e
i
e
i
e
j
n
j
e
ij 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
where
e
e
i if frdrdz
e
e
i i nQ q ds
138
Single-Variable Problem – Heat Transfer
Heat Transfer:
Weak form
1 ( , ) ( , )( , ) 0
T r z T r zrk k f r z
r r r z z
0 ( , )
e
e
n
w T w Tk k wf r z rdrdz
r r z z
wq ds
( , ) ( , )n r z
T r z T r zq k n k n
r z
where
139
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
1
2 e
r z r zr z
z zA
r r
3 1 1 3
2 3 1
1 3
1
2 e
r z r zr z
z zA
r r
1 2 2 1
3 1 2
2 1
1
2 e
r z r zr z
z zA
r r
140
4-Node Axi-symmetric Element
1 1 2 2 3 3 4 4( , )T r z T T T T
1 2
3 4
a
b
h
x
r
z
1 2
3 4
1 1 1
1
a b a b
a b a b
x h x h
x h x h
141
Single-Variable Problem – Example
Step 1: Discretization
Step 2: Element equation
e
j je i iijK rdrdz
r r z z
e
e
i if frdrdz
e
e
i i nQ q ds
142
Review of CST Element
• Constant Strain Triangle (CST) - easiest and simplest finite element – Displacement field in terms of generalized coordinates
– Resulting strain field is
– Strains do not vary within the element. Hence, the name constant strain triangle (CST) • Other elements are not so lucky. • Can also be called linear triangle because displacement field is
linear in x and y - sides remain straight.
143
Constant Strain Triangle
• The strain field from the shape functions looks like:
– Where, xi and yi are nodal coordinates (i=1, 2, 3) – xij = xi - xj and yij=yi - yj – 2A is twice the area of the triangle, 2A = x21y31-x31y21
• Node numbering is arbitrary except that the sequence 123 must go clockwise around the element if A is to be positive.
144
Constant Strain Triangle
• Stiffness matrix for element k =BTEB tA
• The CST gives good results in regions of the FE model where there is little strain gradient
– Otherwise it does not work well.
145
Linear Strain Triangle
• Changes the shape functions and results in quadratic displacement distributions and linear strain distributions within the element.
146
Linear Strain Triangle
• Will this element work better for the problem?
147
Example Problem
• Consider the problem we were looking at:
5 in.
1 in.
0.1 in.
I 0.113 /12 0.008333 in4
M c
I1 0.5
0.008333 60 ksi
E 0.00207
ML2
2EI
25
2 29000 0.008333 0.0517 in.
1k
1k
148
Bilinear Quadratic
• The Q4 element is a quadrilateral element that has four nodes. In terms of generalized coordinates, its displacement field is:
149
Bilinear Quadratic • Shape functions and strain-displacement
matrix
150
Example Problem
• Consider the problem we were looking at:
5 in.
1 in.
0.1 in.
.in0345.0008333.0290003
1252.0
EI3
PL
00207.0E
ksi60008333.0
5.01
I
cM
in008333.012/11.0I
3
43
0.1k
0.1k
151
Quadratic Quadrilateral Element
• The 8 noded quadratic quadrilateral element uses quadratic functions for the displacements
152
Quadratic Quadrilateral Element
• Shape function examples:
• Strain distribution within the element
153
Quadratic Quadrilateral Element
• Should we try to use this element to solve our problem?
• Or try fixing the Q4 element for our purposes.
– Hmm… tough choice.
154
Isoparametric Elements and Solution
• Biggest breakthrough in the implementation of the finite element method is the development of an isoparametric element with capabilities to model structure (problem) geometries of any shape and size.
• The whole idea works on mapping. – The element in the real structure is mapped to an
‘imaginary’ element in an ideal coordinate system – The solution to the stress analysis problem is easy and
known for the ‘imaginary’ element – These solutions are mapped back to the element in the
real structure. – All the loads and boundary conditions are also mapped
from the real to the ‘imaginary’ element in this approach
155
Isoparametric Element
1 2
3 4
(x1, y1) (x2, y2)
(x3, y3) (x4, y4)
X, u
Y,v
(-1, 1)
x
h
2 (1, -1)
1 (-1, -1)
4 3 (1, 1)
156
Isoparametric element
• The mapping functions are quite simple:
X
Y
N1 N2 N3 N4 0 0 0 0
0 0 0 0 N1 N2 N3 N4
x1
x 2
x3
x4
y1
y2
y3
y4
N1 1
4(1x )(1h)
N2 1
4(1 x )(1h)
N3 1
4(1 x )(1 h)
N4 1
4(1x )(1 h)
Basically, the x and y coordinates of any point in the element are interpolations of the nodal (corner) coordinates.
From the Q4 element, the bilinear shape functions are borrowed to be used as the interpolation functions. They readily satisfy the boundary values too. 157
Isoparametric element
• Nodal shape functions for displacements
u
v
N1 N2 N3 N4 0 0 0 0
0 0 0 0 N1 N2 N3 N4
u1
u2
u3
u4
v1
v2
v3
v4
N1 1
4(1x )(1h)
N2 1
4(1 x )(1h)
N3 1
4(1 x )(1 h)
N4 1
4(1x )(1 h) 158
• The displacement strain relationships:
x u
X
u
x
x
X
u
h
h
X
y v
Y
v
x
x
Y
v
h
h
Y
x
yxy
u
Xv
Yu
Y
v
X
x
X
h
X0 0
0 0x
Y
h
Yx
Y
h
Y
x
X
h
X
u
xu
hv
xv
h
But,it is too difficult to obtainx
Xand
h
X159
Isoparametric Element
u
x
u
X
X
x
u
Y
Y
x
u
h
u
X
X
h
u
Y
Y
h
u
xu
h
X
x
Y
xX
h
Y
h
u
Xu
Y
It is easier to obtainX
xand
Y
x
J
X
x
Y
xX
h
Y
h
Jacobian
defines coordinate transformation
Hence we will do it another way
X
x
N ixX i
Y
x
N ixYi
X
h
N ihX i
Y
h
N ihYi
u
Xu
Y
J 1
u
xu
h
160
Gauss Quadrature
• The mapping approach requires us to be able to evaluate the integrations within the domain (-1…1) of the functions shown.
• Integration can be done analytically by using closed-form formulas from a table of integrals (Nah..) – Or numerical integration can be performed
• Gauss quadrature is the more common form of numerical integration - better suited for numerical analysis and finite element method.
• It evaluated the integral of a function as a sum of a finite number of terms
I dx becomes I Wiii1
n
1
1
161
Gauss Quadrature
• Wi is the ‘weight’ and i is the value of f(x=i)
162
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
bfafabdxxfdxxfI
2
)()()()()( 1
n
n
n x
x
x
x
b
a
x
x
x
x
n dxxfdxxfdxxfdxxfdxxfI
1
2
10
1
0
)()()( )()(
1
1
)()(2)(2
n
i
i bfxfafh
I
163
Numerical Integration Calculate: dxxfI
b
a
• Newton – Cotes integration
• Simpson 1/3 rule – 2nd order Newton-Cotes integration
)())((
))(()(
))((
))(()(
))((
))(()()( 2
1202
101
2101
200
2010
212 xf
xxxx
xxxxxf
xxxx
xxxxxf
xxxx
xxxxxfxf
b
a
b
a
xfxfxfxxdxxfdxxfI
6
)()(4)()()()( 210
022
164
Numerical Integration Calculate: dxxfI
b
a
• Gaussian Quadrature
)(2
)()(
2
)(
2
)()()(
bfab
afab
bfafabI
)()( 1100 xfcxfcI
Trapezoidal Rule: Gaussian Quadrature:
Choose according to certain criteria 1010 ,,, xxcc
165
Numerical Integration Calculate: dxxfI
b
a
• Gaussian Quadrature
• 2pt Gaussian Quadrature
•3pt Gaussian Quadrature
3
1
3
11
1
ffdxxfI
77.055.0089.077.055.0
1
1
fffdxxfI
111100
1
1
nn xfcxfcxfcdxxfI
ab
axx
)(21~Let:
xdxabbafabdxxf
b
a
~~)(2
1)(
2
1)(
2
1)(
1
1
166
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