17.Introduction to Finite Element Analysis

Intro-FEM

1. Intelligent Structural Systems 17: Introduction to Finite Element Methods
1
11
Instructor: Dr. Song
Dept. of Mechanical Engineering
17. Introduction to Finite Element Method
Foratypicalstructure,ingeneral,anexactanalyticalsolutiondoesnotexist.
TheproblemisovercomebytheuseoftheFiniteElementmethod.The
structureisdividedintoelementswhichareconnectedatafinitenumberof
points,callednodes.Themotionofthepointsintheelementisdefinedinterms
ofnodaldisplacementandusinginterpolationfunctions.Therefore,wefirstfind
thestiffnessandmassmatricesoftheelements.Theelementsareassembledto
determinethestiffnessandmassmatricesofthestructure.

2. Intelligent Structural Systems 17: Introduction to Finite Element Methods
2
22
Instructor: Dr. Song
Dept. of Mechanical Engineering
1 2 3 4 5 6
Element 1
Element 2
Element 3
Element 4
Element 5
Beam
1 3 5 7
642
1
2
3
4
5
Plate
Fig 1 Finite elements

3. Intelligent Structural Systems 17: Introduction to Finite Element Methods
3
33
Instructor: Dr. Song
Dept. of Mechanical Engineering
A Beam Element Under Bending:
Bernoulli–Euler Theory
Fig 2 Beam element
M
1 M
2

1

2
Q
1
Q
2
W
1 W(x)
W
2
y
x
1 2
x
y
z

4. Intelligent Structural Systems 17: Introduction to Finite Element Methods
4
44
Instructor: Dr. Song
Dept. of Mechanical Engineering
Euler-Bernoulli Model
▪No axial load
▪Shear deformation can be neglected
▪Uniform along its length
▪Predicts the correct curvature of actuator and base over the entire
contact area
▪Considers the actuator as a layer
▪Treating bimorphs, the correct neutral axis is used which is the
neutral axis of the layered system

5. Intelligent Structural Systems 17: Introduction to Finite Element Methods
5
55
Instructor: Dr. Song
Dept. of Mechanical Engineering
Let us consider a beam element of uniform cross-section and bending stiffness.
•The bending deformation is measured as a transverse displacement and a
rotation. Hence the degrees of freedom considered per node are a transverse
displacementand a rotation.
•We consider two nodes, 1 and 2, which have transverse displacements: along
the y-axis, represented by W
1and W
2, and rotations along the z-axis
represented by 
1and 
2.
•The nodal forces are shear forces, Q
1and Q
2, and moments, M
1and M
2. The
sign convention is that nodal displacements / rotations and nodal forces are
positive when they are along the positive axis.
Therefore this element,















=
2
2
1
1
}{
W
W
q
Nodal displacement vector (1)

6. Intelligent Structural Systems 17: Introduction to Finite Element Methods
6
66
Instructor: Dr. Song
Dept. of Mechanical Engineering













=
2
2
1
1
}{
M
Q
M
Q
F
Nodal force vector (2)
M
Q Q
M
Fig 3 indicates the sign conventions used in simple beam theory for
positive shear forces Q and bending moment M.
Fig 3 positive shear forces and bending moments

7. Intelligent Structural Systems 17: Introduction to Finite Element Methods
7
77
Instructor: Dr. Song
Dept. of Mechanical Engineering
M
Q
Q+dQ
M+dM
Fig 4 positive shear forces and bending moments
Our objective is to find the stiffness and mass matrices for the element.
The differential equation governing elementary linear-elastic beam
behavior is derived as follows.
Considering a beam subjected to a distributed loading p(x) (force/length).
The differential beam element Is shown in Fig 4
P(x)
dx
1 2

8. Intelligent Structural Systems 17: Introduction to Finite Element Methods
8
88
Instructor: Dr. Song
Dept. of Mechanical Engineering
From force and moment equilibrium of the differential element of the beam.
We have =;0
y
F 0)()( =−+− dxxPdQQQ aroundpoint 2
0;M = 0
2
)()( =+−++−
dx
dxxpQdxdMMM
Then, we havedx
dQ
xp −=)(
(Ignoring the high order term)and dx
dM
Q=
If the distributed loading is we can get 0)(=xp 0
2
2
=
dx
Md
(5)
(3)
(4)

9. Intelligent Structural Systems 17: Introduction to Finite Element Methods
9
99
Instructor: Dr. Song
Dept. of Mechanical Engineering
Also, the curvature kof the beam is related to the moment byEI
M
k ==

1
where is the radius of the deflected curve shown in Fig 5, W(x) is the
transverse displacement in the y direction, Eis the modulus of elasticity,
and Iis the principal moment of inertia about the z axis.  
W(x)
The curvature for small slopes is given bydx
dW
=
(6)
Fig 5 Radius of the deflected curve

10. Intelligent Structural Systems 17: Introduction to Finite Element Methods
10
1010
Instructor: Dr. Song
Dept. of Mechanical Engineering
Combining Eqns. (5), (6), and (7), the differential equation for the
displacement W(x) is4
4
()
0, 0
d W x
EI x h
dx
=  
(8)
Integrating Eq. (8) four times, we have43
2
2
3
1
2
1
6
1
)( CxCxCxCxW +++=
(9)2
2
dx
Wd
k=
(7)
where C
i( i = 1,2,3,4 ) are constants of integration and can be
determined in terms of nodal displacements as follows:

11. Intelligent Structural Systems 17: Introduction to Finite Element Methods
11
1111
Instructor: Dr. Song
Dept. of Mechanical Engineering
The boundary displacements and rotations are:22101
|
)(
;)(;|
)(
;)0( ====
== hxx
dx
xdW
WhW
dx
xdW
WW
(10)
Introducing Eq. (9) into Eq. (10), we get232
2
1
243
2
2
3
1
13014
2
1
|
)(
2
1
6
1
)(
|
)(
;)0(
=++=
=+++=
====
=
=
ChChC
dx
xdW
WChChChChW
C
dx
xdW
WCW
hx
x
(11)

12. Intelligent Structural Systems 17: Introduction to Finite Element Methods
12
1212
Instructor: Dr. Song
Dept. of Mechanical Engineering
From Eqs. (10), we get the solution1413
221132
221131
;
)323(
2
)22(
6
WCC
hWhW
h
C
hWhW
h
C
==
−+−−=
+−+=
(12)
Therefore the expression for bending displacement is2
32
2
32
1
32
1
32
23
2231)(















+





−+














−





+















+





−





+














+





−=
h
h
x
h
x
W
h
x
h
x
h
h
x
h
x
h
x
W
h
x
h
x
xW
(13)

13. Intelligent Structural Systems 17: Introduction to Finite Element Methods
13
1313
Instructor: Dr. Song
Dept. of Mechanical Engineering
The bending displacement is related to the nodal forces as follows:
Positive moment and shear forces
M Q Q M22
2
23
3
102
2
103
3
|;|
|;|
M
dx
Wd
EIQ
dx
Wd
EI
M
dx
Wd
EIQ
dx
Wd
EI
hxhx
xx
=−=
−==
==
==
(14)2
2
dx
Wd
EIM=
and3
3
dx
Wd
EIQ=
From Eqns. (4), (6), and (7), we have
Then,

14. Intelligent Structural Systems 17: Introduction to Finite Element Methods
14
1414
Instructor: Dr. Song
Dept. of Mechanical Engineering
Substituting Eq. (13) into Eq. (14), we get)4626(
)612612(
)2646(
)612612(
221122
221132
221121
221131
+−+=
−+−−=
+−+=
+−+=
hWhW
h
EI
M
hWhW
h
EI
Q
hWhW
h
EI
M
hWhW
h
EI
Q
(15)
Eq. (15) has the form [k]{q}={F}}{}]{[ Fqk=
(16)

15. Intelligent Structural Systems 17: Introduction to Finite Element Methods
15
1515
Instructor: Dr. Song
Dept. of Mechanical Engineering
where













=
2
2
1
1
}{
M
Q
M
Q
F ,}{
2
2
1
1
















=
W
W
q
and [k], the stiffness matrix, is given by











−
−
−
=
2
22
3
4)(
612
264
612612
][
hsym
h
hhh
hh
h
EI
k
(17)
W(x) can be expressed as 24231211
)()()()()( +++= xLWxLxLWxLxW

16. Intelligent Structural Systems 17: Introduction to Finite Element Methods
16
1616
Instructor: Dr. Song
Dept. of Mechanical Engineering
where the interpolation function L’s are defined as32
4
32
3
32
2
32
1
)(
23)(
2)(
231)(






+





−=






−





=






+





−=






+





−=
h
x
h
h
x
hxL
h
x
h
x
xL
h
x
h
h
x
hxxL
h
x
h
x
xL
(18)

17. Intelligent Structural Systems 17: Introduction to Finite Element Methods
17
1717
Instructor: Dr. Song
Dept. of Mechanical Engineering
L
1(x)
h
L
2(x)
h
L
3(x)
h
L
4(x)
h

18. Intelligent Structural Systems 17: Introduction to Finite Element Methods
18
1818
Instructor: Dr. Song
Dept. of Mechanical Engineering
Use of Kinetic Energy and Potential
Energy
Stiffness:
The bending displacement is expressed as}{)}({
)()()()(
24231211
qxL
xLWxLxLWxLW
T
=
+++=
(19)
The potential energy can be written as )}(]{[)}({
2
1
),(
)(
2
1
)(
0
2
2
2
tqktq
dx
x
txW
xEItV
T
h
=










=
(20)

19. Intelligent Structural Systems 17: Introduction to Finite Element Methods
19
1919
Instructor: Dr. Song
Dept. of Mechanical Engineering
where











−
−
−
=
















+−
−
+−
+−
















+−
−
+−
+−
=
=


2
22
3
0
4
0
""
4)(
612
264
612612
3
63
32
63
3
63
32
63
4
)}()}{(){(][
hsym
h
hhh
hh
h
EI
dx
xh
h
x
xh
h
x
xh
h
x
xh
h
x
h
EI
dxxLxLxEIk
T
h
h
T
(21)
(22)
The expression for [k] is the same as in Eq. (17)

20. Intelligent Structural Systems 17: Introduction to Finite Element Methods
20
2020
Instructor: Dr. Song
Dept. of Mechanical Engineering
Mass Matrices:
Lump Mass
The simplest way of generating the element mass matrix is by
lumping the mass at nodes, a method used in early days of the
finite element method. It has several drawbacks:
1. Lumping is an arbitrary process, resulting in a lack of control
over error
2. It could lead to singular mass matrices, although by definition
the mass matrix should be positive and definite. This arises in
bending problems because lumped masses are generally
regarded as point masses, so that mass coefficients
corresponding to the rotational coordinate is zero.

21. Intelligent Structural Systems 17: Introduction to Finite Element Methods
21
2121
Instructor: Dr. Song
Dept. of Mechanical Engineering
ConsistentMass Matrices:
Let m(x) be the mass per unit length of the beam.
From Eq. (19), W= { L(x) }
T
{ q }
The element’s kinetic energy has the form











=








=
••

qmq
dx
t
txW
xmtT
T
h
][
2
1
),(
)(
2
1
)(
0
2
(23, 24)

22. Intelligent Structural Systems 17: Introduction to Finite Element Methods
22
2222
Instructor: Dr. Song
Dept. of Mechanical Engineering
where the mass matrix [m] is given by











−
−
−
=
































+





−






−











+





−






+





−
































+





−






−











+





−






+





−
=
=


2
22
32
32
32
32
0
32
32
32
32
0
4)(
22156
3134
135422156
420
23
2
231
23
2
231
)}()}{(){(][
hsym
h
hhh
hh
mh
dx
h
x
h
h
x
h
h
x
h
x
h
x
h
h
x
hx
h
x
h
x
h
x
h
h
x
h
h
x
h
x
h
x
h
h
x
hx
h
x
h
x
m
dxxLxLxmm
T
h
h
T
(25)

23. Intelligent Structural Systems 17: Introduction to Finite Element Methods
23
2323
Instructor: Dr. Song
Dept. of Mechanical Engineering
Therefore, the mass matrix for the element, [m], is given by











−
−
−
=
2
22
4)(
22156
3134
135422156
420
][
hsym
h
hhh
hh
mh
m
(26)

24. Intelligent Structural Systems 17: Introduction to Finite Element Methods
24
2424
Instructor: Dr. Song
Dept. of Mechanical Engineering
Stiffness Matrix Assemblage
Let us consider a simple two-spring system as shown in the figure
k
a k
b

25. Intelligent Structural Systems 17: Introduction to Finite Element Methods
25
2525
Instructor: Dr. Song
Dept. of Mechanical Engineering
Let us break the system into two spring elements
k
a
F
1 F
2
u
1 u
2
F
3 F
4
u
3 u
4k
b





=












−
−






=












−
−
4
3
4
3
2
1
2
1
;
F
F
u
u
kk
kk
F
F
u
u
kk
kk
bb
bb
aa
aa
(1)
If we just combine the two matrices,











−
−
−
−
=
bb
bb
aa
aa
kk
kk
kk
kk
k
00
00
00
00
(2)

26. Intelligent Structural Systems 17: Introduction to Finite Element Methods
26
2626
Instructor: Dr. Song
Dept. of Mechanical Engineering
Constraints:
In our example,32
uu=
Therefore, we represent our system by coordinates u
1, u
2 and u
4.
They are related to u
1, u
2 , u
3 and u
4as follows:





















=












4
2
1
4
3
2
1
100
010
010
001
u
u
u
u
u
u
u
(3)
(4)q q R

27. Intelligent Structural Systems 17: Introduction to Finite Element Methods
27
2727
Instructor: Dr. Song
Dept. of Mechanical Engineering
The new stiffness matrix will be









−
−+−
−
=
























−
−
−
−










=
bb
bbaa
aa
bb
bb
aa
aa
kk
kkkk
kk
kk
kk
kk
kk
k
0
0
100
010
010
001
00
00
00
00
1000
0110
0001
][
(5)
The second approach is direct superposition.









−
−=










−
−
=
bb
bbbaa
aa
a
kk
kkkkk
kk
k
0
0
000
,
000
0
0
(6)

28. Intelligent Structural Systems 17: Introduction to Finite Element Methods
28
2828
Instructor: Dr. Song
Dept. of Mechanical Engineering
Since u
2= u
3, by superposition,









−
−+−
−
=
bb
bbaa
aa
kk
kkkk
kk
k
0
0
(7)
If u
1= 0, we eliminate the corresponding row and column, so that





−
−+
=
bb
bba
kk
kkk
k
(8)