Finite element method (matlab) milan kumar rai

1. Finite Element Method
(FEM)
Milan Kumar Rai
MSc Engineering Geology
Tribhuvan University

2. Course Content
• Introduction, Definition of Finite Element Method,
Differential Equation and Weak Form, Variational Principal,
• Ritz‐Galerkin Method (approximate function, Galerkin Method
and Ritz Method), Finite Element Method (1‐D
• Problem): Construction of approximate function, Element
matrix, Total element matrix and simple example,
• Finite Element Method (2‐D Problem): Construction of
approximate function, Element matrix & total element
• matrix, simple example and Gauss’s method of elimination.

3. Finite Element Method – Introduction
• The Finite Element Method (FEM) is a numerical
method of solving systems of partial differential
equations (PDEs)
• It reduces a PDE system to a system of algebraic
equations that can be solved using traditional linear
algebra techniques.
• In simple terms, FEM is a method for dividing up a
very complicated problem into small elements that can
be solved in relation to each other.

4. • What is finite element analysis, FEM?
• A Brief history of FEM
• What is FEM used for?

5. Lord John William Strutt Rayleigh (late 1800s),
developed a method for predicting the first natural
frequency of simple structures. It assumed a
deformed shape for a structure and then quantified
this shape by minimizing the distributed energy in
the structure.
Ritz then exp Walter ended this into a method, now
known as the Rayleigh-Ritz method, for predicting
the stress and displacement behavior of structures.

6. F.E.M.
• In finite element method, the structure to be
analyzed is subdivided into a mesh of finite-sized
elements of simple shape, and then the whole
structure is solved with quite easiness.
Rectangular Body Circular Plate
Finite Sized Element

7. Finite Sized Elements
• The rectangular panel in the rectangular body and
triangular panel in the circular plate are referred to an
‘element’.
• There’re one-, two- and three-dimensional elements.
• The accuracy of the solution depends upon the
number of the finite elements; the more there’re, the
greater the accuracy.

8. Finite Element of a Bar
• If a uniaxial bar is part of a structure then it’s usually
modeled by a spring element if and only if the bar is
allowed to move freely due to the displacement of
the whole structure. (One dimensional element)
Structure
Uniaxial bar of the
structure
Bar

9. Types of Elements
• Here goes the examples of two- and three-
dimensional finite sized elements.
Triangle
Rectangle
Hexahedron

10. Node
• The points of attachment of the element to other
parts of the structure are called nodes.
• The displacement at any node due to the
deformation of structure is known as the nodal
displacement. Node

11. Why F.E.M.?
Simple trusses can be solved by just using the equilibrium
equations. But for the complex shapes and frameworks
like a circular plate, equilibrium equations can no longer
be applied as the plate is an elastic continuum not the
beams or bars as the case of normal trusses.
Hence, metal plate is divided into finite subdivisions
(elements) and each element is treated as the beam or bar.
And now stress distribution at any part can be determined
accurately.

12. Simple Bar Analysis
• Consider a simple bar made up of uniform material with length
L and the cross-sectional area A. The young modulus of the
material is E.
• Since any bar is modeled as spring in FEM thus we’ve:L
F1 F2x1
x2k

13. Simple Bar Analysis
• Let us suppose that the value of spring constant is k.
Now, we’ll evaluate the value of k in terms of the
properties (length, area, etc.) of the bar:
We know that:
i.e.
Also: i.e.
And i.e.

14. Simple Bar Analysis
• Now substituting the values of x and F is the base
equation of k, we’ll have:
But
Hence, we may write:

15. Simple Bar Analysis
• According to the diagram, the force at node x1 can be
written in the form:
• Where x1 – x2 is actually the nodal displacement between
two nodes. Further:
• Similarly:

16. Simple Bar Analysis
• Now further simplification gives:
• These two equations for F1 and F2 can also be written as,
in Matrix form:
• Or:

17. Simple Bar Analysis
• Here Ke is known as the Stiffness Matrix. So a
uniform material framework of bars, the value of the
stiffness matrix would remain the same for all the
elements of bars in the FEM structure.

18. Further Extension
• Similarly for two different materials bars joined together, we may write:
;
F1 F2
x1 x2
k1
x3
F3
k2

19. Importance
• FEM has become very familiar in subdivision of
continuum. It gives reliable and accurate results if the
number of elements are kept greater.
• Modern computer technology had helped this
analysis to be very easy and less time consuming.
• Large structures under loadings are now easily solved
and stresses on each and every part are now being
determined.

20. Development of Theory
• Rayleigh-Ritz Method
• Total potential energy equation
• Galerkin’s Method

21. Galerkin method
• Galerkin suggested that the residue should be
multiplied by a weighting function that is a part of
the suggested solution then the integration is
performed over the whole domain!!!
• Actually, it turned out to be a VERY GOOD idea

22. 1D Rod Elements• To understand and solve 2D and 3D problems we must
understand basic of 1D problems.
• Analysis of 1D rod elements can be done using Rayleigh-Ritz
and Galerkin’s method.
• To solve FEA problems same are modified in the Potential-
Energy approach and Galerkin’s approach

23. 1D Rod Elements
• Loading consists of three types : body force f , traction force T,
point load Pi
• Body force: distributed force , acting on every elemental volume
of body i.e. self weight of body.
• Traction force: distributed force , acting on surface of body i.e.
frictional resistance, viscous drag and surface shear
• Point load: a force acting on any single point of element

24. 1D Rod Elements• Element strain energy
• Element stiffness matrix
• Load vectors
• Element body load vector
• Element traction-force vector
qkqU eT
e

][
2
1










11
11
][
e
eee
l
AE
k







1
1
2
flA
f eee








1
1
2
ee Tl
T

Element -1 Element-2

25. Bar application
   

n
i
ii xaxu
1

  02
2



xF
x
u
EA
     xRxF
dx
xd
aEA
n
i
i
i 1
2
2

Applying Galerkin method
         
 Domain
j
n
i Domain
i
ji dxxFxdx
dx
xd
xaEA 


1
2
2

26. In Matrix Form
         

















 Domain
ji
Domain
i
j dxxFxadx
dx
xd
xEA 

 2
2
Solve the above system for the “generalized
coordinates” ai to get the solution for u(x)

27. Same conditions on the functions are
applied
• They should be at least twice differentiable!
• They should satisfy all boundary conditions!
• Let’s use the same function as in the collocation
method:
  






l
x
Sinx
2



28. Substituting with the approximate
solution:
         
 Domain
j
n
i Domain
i
ji dxxFxdx
dx
xd
xaEA 


1
2
2
 
























ll
fdx
l
x
Sindx
l
x
Sin
l
x
Sina
l
EA
00
1
2
2222


 ll
a
l
EA
2
22
1
2







EA
fll
EA
f
a
2
3
2
1 52.0
16



29. Substituting with the approximate
solution: (Int. by Parts)
         
 Domain
j
n
i Domain
i
ji dxxFxdx
dx
xd
xaEA 


1
2
2

 ll
a
l
EA
2
22
1
2







EA
fll
EA
f
a
2
3
2
1 52.0
16


   
       



Domain
ij
l
i
j
Domain
i
j
dx
dx
xd
dx
xd
dx
xd
x
dx
dx
xd
x




0
2
2
Zero!

30. What did we gain?
• The functions are required to be less differentiable
• Not all boundary conditions need to be satisfied
• The matrix became symmetric!

31. The Finite Element
Method
2nd order DE’s in 1-D

32. Objectives
• Understand the basic steps of the finite element
analysis
• Apply the finite element method to second order
differential equations in 1-D

33. The Mathematical Model
• Solve:
• Subject to:
Lx
fcu
dx
du
a
dx
d








0
0
  00 ,0 Q
dx
du
auu
Lx









34. Step #1: Discretization
• At this step, we divide the
domain into elements.
• The elements are
connected at nodes.
• All properties of the
domain are defined at
those nodes.

35. Step #2: Element Equations• Let’s concentrate our
attention to a single
element.
• The same DE applies on
the element level, hence,
we may follow the
procedure for weighted
residual methods on the
element level!
21
0
xxx
fcu
dx
du
a
dx
d








   
21
2211
21
,
,,
Q
dx
du
aQ
dx
du
a
uxuuxu
xxxx















36. 2D Truss• 2 DOF
• Transformations
• Modified Stiffness Matrix
• Methods of Solving

37. 2D Truss
• Transformation Matrix
• Direction Cosines







ml
ml
L
00
00
][
   2
12
2
12 yyxxle 
el
xx
l 12
cos

 
el
yy
m 12
sin

 

38. 2D Truss
• Element Stiffness Matrix



















22
22
22
22
][
mlmmlm
lmllml
mlmmlm
lmllml
l
AE
k
e
ee
e

39. Methods of Solving
• Elimination Approach
• Eliminate Constraints
• Penalty Approach

40. Elimination Method• Set defection at the constraint to equal zero

41. Elimination Method• Modified Equation
• DOF’s 1,2,4,7,8 equal to zero

42. 2D Truss
• Element Stresses
• Element Reaction Forces
 qmlml
l
E
e
e 

 QKR



43. 2D Truss
• Development of Tables
• Coordinate Table
• Connectivity Table
• Direction Cosines Table

44. 2D Truss• Coordinate Table
• E.g;

45. 2D Truss• Connectivity Table
• E.g;

46. 2D Truss
   2
12
2
12 yyxxle 
el
xx
l 12
cos

 
el
yy
m 12
sin

 

47. Example 1D Rod ElementsExample 1
Problem statement: (Problem 3.1 from Chandrupatla and Belegunda’s
book)
Consider the bar in Fig.1, determine the following by hand calculation:
1) Displacement at point P 2) Strain and stress
3) Element stiffness matrix 4) strain energy in element
2
1.2eA in
6
30 10E psi  1 0.02q in
2 0.025q in
Given:

48. Solution:
1) Displacement (q) at point P
We have
1
2 1
2
( ) 1
( )
2
(20 15) 1 0.25
(23 15)
x x
x x


  

   

Now linear shape functions N1( ) and N2( ) are given by
1
1
( ) 0.375
2
N



  And 2
1
( ) 0.625
2
N



 

51. Trusses
• A truss is a set of bars that are connected at
frictionless joints.
• The Truss bars are generally oriented in the plain.

52. Trusses
• Now, the problem lies in the transformation of the
local displacements of the bar, which are always in
the direction of the bar, to the global degrees of
freedom that are generally oriented in the plain.

53. Equation of Motion











































0
0
0000
0101
0000
0101
2
1
2
2
1
1
F
F
v
u
v
u
h
EA

54. Transformation Matrix
   
   
   
    DOF
dTransforme
DOFLocal
v
u
v
u
CosSin
SinCos
CosSin
SinCos
v
u
v
u











































2
2
1
1
2
2
1
1
00
00
00
00




    DOF
dTransforme
DOFLocal T 

55. The Equation of Motion Becomes• Substituting into the FEM:
• Transforming the forces:
• Finally:
     FTK 
         FTTKT
TT

    FK 

56. Projects
• Heat transfer in a 2-D heat sink
• 2-D flow around a blunt body in a wind tunnel
• Vibration characteristics of a pipe with internal fluid
flow
• Panel flutter of a beam
• Rotating Timoshenko beam/blade

57. Heat transfer in a 2-D heat sink
• The heat sink will have heat flowing from one side
• Convection transfer on the surfaces
• Different boundary conditions on the other three
sides
• Plot contours of temperature distribution with
different boundary conditions

58. 2-D flow around a blunt body in a
wind tunnel
• Potential flow in a duct
• Rectangular body with different Dimensions
• Study the effect of the body size on the flow speed
on both sides
• Plot contours of potential function, pressure, and
velocity potential

59. Vibration characteristics of a pipe
with internal fluid flow
• Study the change of the natural frequencies with the
flow speed under different boundary conditions and
fluid density
• Indicate the flow speeds at which instabilities occur

60. Panel flutter of a beam
• A fixed-fixed beam is subjected to flow over its
surface
• Plot the effect of the flow speed on the natural
frequencies of the beam
• Indicate the speed at which instability occurs

61. Rotating Timoshenko beam/blade
• Rotating beams undergo centrifugal tension that
results in the change of its natural frequencies
• Study the effect of rotation speed on the beam
natural frequencies and frequency response to
excitations at the root

62. Example 2D Truss

68. Conclusion
• Good at Hand Calculations, Powerful when applied to
computers
• Only limitations are the computer limitations