mech593 finite element methods finite element analysis (f.e.a.) of 1-d problems dr. wenjing ye
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MECH593 Finite Element Methods
Finite Element Analysis (F.E.A.) of 1-D Problems
Dr. Wenjing Ye
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
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 dx
duxEAxwdxxfxw
dx
duxEA
dx
dw
0)()()( 1122
2
1
PxwPxwdxxfxw
dx
duxEA
dx
dwx
x
Approximation Methods – Finite Element Method
Example (cont):
Step 3: Choosing shape functions - linear shape functions
2211 uuu
lx1 x2
x
l
xx
l
xx 12
21 ;
2
1 ;
2
121
11 2
1 ;1
2x
lxxx
l
Approximation Methods – Finite Element Method
Example (cont):
Step 4: Forming element equation
Let , 1w 2 2
1 1
2 11 1 2 2 1 1 1
10
x x
x x
u uEA dx f dx x P x P
l l
1121
2
1
Pdxful
EAu
l
EAx
x
Let , 2w 2 2
1 1
2 12 2 2 2 2 1 1
10
x x
x x
u uEA dx f dx x P x P
l l
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 PEAu P f Pl
fdx
E,A are constant
0)()()( 1122
2
1
PxwPxwdxxfxw
dx
duxEA
dx
dwx
x
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 Au f Pl
u 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 A
l lu 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 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
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
00
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 l u fE 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
dx
du
dx
du
dx
du 22
11
2211 uuu dx
dEu
dx
dEuE 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 4.1:E = 100 GPa, A = 1 cm2
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
jiji
ij dxffKdxdx
d
dx
dEAKwhere
x=x1 x=x2
2 1
x
x=x1 x= x2
u1 u2
1P 2Pf(x)
L = x2-x1
u
x
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 u3u
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
u2u3
3
u4 ……………
4 ……………
Natural Coordinates and Interpolation Functions
2
1 ,
2
121
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
2
1 ,11 ,
2
1321
13
11
16
27 ,1
3
1
3
1
16
921
3
1
3
11
16
9 ,1
3
11
16
2743
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
2
1
1
1
2
xj ji i
ij ji
x
d dd dwhere K EA dx EA d K
dx dx d d l
3
2
1
332313
232212
131211
3
2
1
3
2
1
u
u
u
KKK
KKK
KKK
f
f
f
P
P
P
=-1 =0 =1
Quadratic Formulation for Bar Element
u1 u3u2f(x)P3
P1
P2
=-1 =0 =11x 2x 3x
2
1u11u
2
1u)(u)(u)(u)(u 321332211
2
1 ,11 ,
2
1321
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
1 2
2/ 2
x xx
l
2
ld dx
2d
dx l
Quadratic Formulation for Bar Element
u1 u3u2f(x)P3
P1
P2
=-1 =0 =11x 2x 3x
7 8 1
8 16 83
1 8 7
EAK
l
Exercises – Quadratic Element
Example 4.2:
E = 100 GPa, A1 = 1 cm2; A1 = 2 cm2
Some Issues
Non-constant cross section:
Interior load point:
Mixed boundary condition:k
Application - Plane Truss Problems
Example 4.3: Find forces inside each member. All members have the same length.
F
Arbitrarily Oriented 1-D Bar Element on 2-D Plane
Q2 , v2
11 u ,P
22 u ,P
P2 , u2
Q1 , v1
P1 , u1
2
2
1
1
2
2
1
1
cossin00
sincos00
00cossin
00sincos
0
0
v
u
v
u
v
u
v
u
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
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
AEK ij
Q2 , v2
11 , uP
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
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 xy
z2
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
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 xy
z2
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
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 II
Element II 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
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 II
Element II I
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
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
Recovery of Axial Forces
05
40
54
05
40
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 II
Element II I
53
51
535
1
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
Stresses inside members
Element I
Element II
Element II I
A
F
5
4
5
41
FP
5
42
FP
FP5
12
FQ5
32
FQ5
33
FP5
13