Introduction to
Finite Element Method
Dr. Santosh Chavan
Assistant Professor
Department of Mechanical Engineering
Scope, Purpose, and Scale
Small-scale, single part
optimization
Large-scale simulation, e.g.
vehicle crash-worthiness
These may both employ the same software,
but with significantly different models!
Large-Scale Simulation, Full Vehicle
Models
Part of the CAE Domain
Such analyses and simulations
likely include multi-physics
and media-structure
interaction
Actual Geometry
Derived
Idealized
Geometry
Eng. Analysis.
Model (FEM)
Simplify
Idealize
De-Feature
Pave
Mesh
Discretize
Analysis Model Creation (FEA)
Amechanicalengineer,astructuralengineer,andapiping
engineermayeachrequiredifferentformsofgeometry
capture.
Geometry Representation
Same Object ...
Multiple/Different Forms of
Geometry Capture
1D
Line
(Curv
e)
3D Solid
(Volume)
2D Surface
(Shell)
Exploded View -Same Geometry,
Different Data Capture
1D Line
(Curve)
3D Solid
(Volume)
2D Surface
(Shell)
A GENERAL PROCEDURE FOR FINITE
ELEMENT ANALYSIS
•Certainstepsinformulatingafiniteelementanalysisofaphysicalproblemarecommon
toallsuchanalyses,whetherstructural,heattransfer,fluidflow,orsomeotherproblem.
•Thesestepsareembodiedincommercialfiniteelementsoftwarepackages.Thestepsare
describedasfollows.
PREPROCESSING
Thepreprocessingstepis,quitegenerally,describedasdefiningthemodeland
includes
Definethegeometricdomainoftheproblem.
Definetheelementtype(s).
Definethematerialpropertiesoftheelements.
Definethegeometricpropertiesoftheelements(length,area,andthelike).
Definetheelementconnectivity's(meshthemodel).
Definethephysicalconstraints(boundaryconditions).
Definetheloadings.
SOLUTION / PROCESSING
•During the solution phase, finite element software assembles the governing algebraic
equations in matrix form and computes the unknown values of the primary field
variable(s).
•The computed values are then used by back substitution to compute additional, derived
variables, such as reaction forces, element stresses, and heat flow.
POSTPROCESSING
Analysisandevaluationofthesolutionresultsisreferredtoaspostprocessing.
Checkequilibrium.
Calculatefactorsofsafety.
Plotdeformedstructuralshape.
Animatedynamicmodelbehavior.
Producecolor-codedtemperatureplots.
Whilesolutiondatacanbemanipulatedmanywaysinpostprocessing,themost
importantobjectiveistoapplysoundengineeringjudgmentindeterminingwhether
thesolutionresultsarephysicallyreasonable.
EXAMPLES OF FINITE ELEMENT ANALYSIS
1. Mathematical Model
Physical
Problems
Mathematical
Model
Solution
Identify control variables
Assumptions (empirical law)
(1) Modeling
(2) Types of solution
Exact Eq. Approx. Eq.
Exact Sol.
◎ ◎
Approx. Sol.
◎ ◎
Eq.
Sol.
2. Finite Element Method
(1) Definition
FEMisanumericalmethodforsolvingasystemof
governingequationsoverthedomainofacontinuous
physicalsystem,whichisdiscretizedintosimplegeometric
shapescalledfiniteelement.
Continuous system
Time-independent PDE
Time-dependent PDE
Discrete system
Linear algebraic eq.
ODE
(2) Discretization
Modeling a body by dividing it into an equivalent system of
finite elements interconnected at a finite number of points on
each element called nodes.
Continuous system Discretization system
System
Discretization
Nodes
Finite Elements
3. Historical Background
4. Analytical Processes of Finite Element Method
(1) Structural stress analysis problem
A.Conditions that solution must satisfy
a. Equilibrium
b. Compatibility
c. Constitutive law
d. Boundary conditions
Above conditions are used to generate a system of equations
representing system behavior.
B. Approach
a. Force (flexibility) method: internal forces as unknowns.
b. Displacement (stiffness) method: nodal disp. As unknowns.
For computational purpose, the displacement method is more
desirable because its formulation is simple. A vast majority of
general purpose FE software's have incorporated the displacement
method for solving structural problems.
(2) Analysis procedures of linear static structural analysis
A. Build up geometric model
a. 1D problem
line
b. 2D problem
surface
c. 3D problem
solid
B. Construct the finite element model
a. Discretize and select the element types
(a) element type
1D line element
2D element
3D brick element
(b) total number of element (mesh)
1D:
2D:
3D:
b. Select a shape function
1D line element: u=ax+b
c. Define the compatibility and constitutive law
d. Form the element stiffness matrix and equations
(a) Direct equilibrium method
(b) Work or energy method
(c) Method of weight Residuals
e. Form the system equation
Assemble the element equations to obtain global system equation
and introduce boundary conditions
C. Solve the system equations
a. elimination method
Gauss’s method (Nastran)
b. iteration method
Gauss Seidel’s method
Displacement field strain field
stress field
D. Interpret the results (postprocessing)
a. deformation plot
b. stress contour
5. Applications of Finite Element Method
Structural Problem Non-structural Problem
Stress Analysis
-truss & frame analysis
-stress concentrated problem
Buckling problem
Vibration Analysis
Impact Problem
Heat Transfer
Fluid Mechanics
Electric or Magnetic Potential
Basic Steps of Solving FEM
1.Deriveanequilibriumequationfromthepotentialenergy
equationintermsofmaterialdisplacement.
2.Selecttheappropriatefiniteelementsandcorresponding
interpolationfunctions.Subdividetheobjectinto
elements.
3.Foreachelement,re-expressthecomponentsofthe
equilibriumequationintermsofinterpolationfunctions
andtheelement’snodedisplacements.
4.Combinethesetofequilibriumequationsforallthe
elementsintoasinglesystemandsolvethesystemforthe
nodedisplacementsforthewholeobject.
5.Usethenodedisplacementsandtheinterpolation
functionsofaparticularelementtocalculate
displacements(orotherquantities)forpointswithinthe
element.
Steps in FEM
1.Discretizeand Select Element Type
2.Select a Displacement Function
3.Define Strain/Displacementand Stress/Strain
Relationships
4.Derive Element Stiffness Matrix & Eqs.
5.AssembleEquations and Introduce B.C’s
6.Solvefor the Unknown Displacements
7.Solvefor Element Stresses and Strains
8.Interpret the Results
24
Typical Application of FEM
•Structural/Stress Analysis
•Fluid Flow
•Heat Transfer
•Electro-Magnetic Fields
•Soil Mechanics
•Acoustics
25
Definitions for this section
For an element, a stiffness matrix
is a matrix such that
where relates local coordinates
and nodal displacements
to local forces of a single element.
Bold denotes vector/matrices.ˆ
k ˆ
f=
ˆ
k
ˆ
d, ˆ
k ˆx,ˆy,ˆz( ) ˆ
d ˆ
f
28
Spring Element
k
1 2
Lˆx ˆ
f
1ˆx,
ˆ
d
1ˆx ˆ
f
2ˆx,
ˆ
d
2ˆx
29
Definitionsˆ
f
1ˆxlocalnodalforce
ˆ
d
1ˆxdisplacement
node
k -spring constantˆxlocal
coordinate
direction ˆ
f
2ˆxlocalnodalforce
ˆ
d
2ˆxdisplacement
node
These are scalar values
30
Stiffness Relationship for a Springˆ
f
1x
ˆ
f
2x
ì
í
ï
î
ï
ü
ý
ï
þ
ï
=
k
11k
12
k
21k
22
é
ë
ê
ê
ù
û
ú
ú
ˆ
d
1x
ˆ
d
2x
ì
í
ï
î
ï
ü
ý
ï
þ
ï
31
Steps in Process
1)Discretize and Select Element Type
2)Select a Displacement Function
3)Define Strain/Displacement and Stress/Strain
Relationships
4)Derive Element Stiffness Matrix & Eqs.
5)Assemble Equations and Introduce B.C.’s
6)Solve for the Unknowns (Displacements)
7)Solve for Element Stresses and Strains
8)Interpret the Results
32
Step 1 -Select the Element Type
L
T
k
1 2ˆx Tˆ
d
1ˆx ˆ
d
2ˆx
33
Step 2 -Select a Displacement Function
Assume a displacement function
Assume a linear function.
Number of coefficients = number of d-o-f
Write in matrix form.ˆu ˆu=a
1
+a
2
ˆx ˆu=1ˆx[]
a
1
a
2
ì
í
î
ü
ý
þ
34
ˆu(0)=a
1
+a
2
(0)=
ˆ
d
1x
=a
1
ˆu(L)=a
1
+a
2
(L)=
ˆ
d
2
=a
2
L+
ˆ
d
1x ˆu Express as function of and ˆ
d
1x ˆ
d
2x a
2=
ˆ
d
2x
-
ˆ
d
1x
L
Solve fora
2:
35
ˆu=a
1
+a
2
ˆx Substituting back into:
Yields:ˆu=
ˆ
d
2x-
ˆ
d
1x
L
æ
è
ç
ç
ö
ø
÷
÷
ˆx+
ˆ
d
1x
36
In matrix form:ˆu=1-
ˆx
L
ˆx
L
é
ë
ê
ù
û
ú
ˆ
d
1x
ˆ
d
2x
ì
í
ï
îï
ü
ý
ï
þï
ˆu=N
1
N
2[ ]
ˆ
d
1x
ˆ
d
2x
ì
í
ï
îï
ü
ý
ï
þï
Where:
N
1=1-
ˆx
L
andN
2=
ˆx
L
37
Shape Functions
N
1and N
2are called Shape Functions or
Interpolation Functions. They express the
shape of the assumed displacements.
N
1=1 N
2=0at node 1
N
1=0 N
2=1at node 2
N
1+ N
2=1
Recall Fourier Transform, in which the basis
functions are sinusoidal functions.
Here, the bases are linear functions!
38
1 2
N
1
L
39
1 2
N
2
L
40
2
1
N
1 N
2
L
41
Step 3 -Define Strain/Displacement and
Stress/Strain RelationshipsT=kd
d=ˆu(L)-ˆu(0)
d=
ˆ
d
2x-
ˆ
d
1x
T -tensile force-total elongation
Here is where
physics comes into
play!
42
Step 4 -Derive the Element
Stiffness Matrix and Equationsˆ
f
1x=-T
ˆ
f
2x
=T
T=-
ˆ
f
1x
=k
ˆ
d
2x
-
ˆ
d
1x( )
T=
ˆ
f
2x=k
ˆ
d
2x-
ˆ
d
1x( )
ˆ
f
1x=k
ˆ
d
1x-
ˆ
d
2x( )
ˆ
f
2x=k
ˆ
d
2x-
ˆ
d
1x( )
43
ˆ
f
1x
ˆ
f
2x
ì
í
ï
î
ï
ü
ý
ï
þ
ï
=
k-k
-kk
é
ë
ê
ù
û
ú
ˆ
d
1x
ˆ
d
2x
ì
í
ï
î
ï
ü
ý
ï
þ
ï
ˆ
k
é
ë
ù
û
º
k-k
-kk
é
ë
ê
ù
û
ú Stiffness Matrix
44
Step 5 -Assemble the Element Equations to Obtain the
Global Equations and Introduce the B.C.K[]=
ˆ
k
(e)é
ë
ù
û
e=1
N
å
F{}=
ˆ
f
(e)
{}
e=1
N
å
Note: not simple addition!
An example later.
(e) indicates
“element” index
45
Step 6 -Solve for Nodal
DisplacementsSolve:
K[]d{}=F{}
46
Step 7 -Solve for Element Forces
Oncedisplacementsateach
nodeareknown,thensubstitute
backintoelementstiffnessequations
toobtainelementnodalforces.
47
Two Spring Assembly
k
1
1
2
k
2
1
2
3
x
F
3x F
2x
48
Forelement1:
ˆ
f
1x
ˆ
f
3x
ì
í
ï
îï
ü
ý
ï
þï
=
k
1
-k
1
-k
1
k
1
é
ë
ê
ù
û
ú
ˆ
d
1x
ˆ
d
3x
ì
í
ï
îï
ü
ý
ï
þï
Forelement2:
ˆ
f
3x
ˆ
f
2x
ì
í
ï
îï
ü
ý
ï
þï
=
k
2-k
2
-k
2k
2
é
ë
ê
ù
û
ú
ˆ
d
3x
ˆ
d
2x
ì
í
ï
îï
ü
ý
ï
þï 49
Assemble Global force matrix
F
3x
=
ˆ
f
3x
(1)
+
ˆ
f
3x
(2)
F
2x
=
ˆ
f
2x
(2)
F
1x
=
ˆ
f
1x
(1) This is just adding.
51
F
3x=-k
1d
1x+k
1d
3x+k
2d
3x-k
2d
2x
F
2x
=-k
2
d
3x
+k
2
d
2x
F
1x=k
1d
1x-k
1d
3x
In the matrix form:
F
1x
F
2x
F
3x
ì
í
ï
î
ï
ü
ý
ï
þ
ï
=
k
1
0 -k
1
0k
2-k
2
-k
1-k
2k
1+k
2
é
ë
ê
ê
ê
ù
û
ú
ú
ú
d
1x
d
2x
d
3x
ì
í
ï
î
ï
ü
ý
ï
þ
ï
or
F[]=K[]d{} Substitution will give
you these equations
52
Global Force Matrix: Global Displacement Matrix:
F
1x
F
2x
F
3x
ì
í
ï
î
ï
ü
ý
ï
þ
ï
d
1x
d
2x
d
3x
ì
í
ï
î
ï
ü
ý
ï
þ
ï
Global Stiffness Matrix:
k
10-k
1
0k
2-k
2
-k
1
-k
2
k
1
+k
2
é
ë
ê
ê
ê
ù
û
ú
ú
ú But, this way is really cumbersome!
Computers can’t do this either!
53
Assembly of [K] -
An Alternative Method
k
1
1
2
k
2
1
2
3
x
F
3x F
2x
54
Forelement1:
ˆ
f
1x
ˆ
f
3x
ì
í
ï
îï
ü
ý
ï
þï
=
k
1
-k
1
-k
1
k
1
é
ë
ê
ù
û
ú
ˆ
d
1x
ˆ
d
3x
ì
í
ï
îï
ü
ý
ï
þï
Forelement2:
ˆ
f
3x
ˆ
f
2x
ì
í
ï
îï
ü
ý
ï
þï
=
k
2-k
2
-k
2k
2
é
ë
ê
ù
û
ú
ˆ
d
3x
ˆ
d
2x
ì
í
ï
îï
ü
ý
ï
þï Recall that
55
Assembly of [K] -An
Alternative Methodnode 1 3
[k
(1)
]=
k
1
-k
1
-k
1k
1
é
ë
ê
ê
ù
û
ú
ú
node 3 2
[k
(2)
]=
k
2
-k
2
-k
2
k
2
é
ë
ê
ê
ù
û
ú
ú
Insert row and
column 2 with zeros
Flip row, flip
columns, and insert
row 1 with zeros
1
3
3
2
56
Expand Local [k] matrices to
Global Sizek
1
10-1
000
-101
é
ë
ê
ê
ê
ù
û
ú
ú
ú
ˆ
d
1x
(1)
ˆ
d
2x
(1)
ˆ
d
3x
(1)
ì
í
ï
ï
î
ï
ï
ü
ý
ï
ï
þ
ï
ï
=
f
1x
(1)
f
2x
(1)
f
3x
(1)
ì
í
ï
î
ï
ü
ý
ï
þ
ï
k
2
000
01-1
0-11
é
ë
ê
ê
ê
ù
û
ú
ú
ú
ˆ
d
1x
(2)
ˆ
d
2x
(2)
ˆ
d
3x
(2)
ì
í
ï
ï
î
ï
ï
ü
ý
ï
ï
þ
ï
ï
=
f
1x
(2)
f
2x
(2)
f
3x
(2)
ì
í
ï
î
ï
ü
ý
ï
þ
ï
Computers
can do this!
57
Net Forcef
1x
(1)
0
f
3x
(1)
ì
í
ï
î
ï
ü
ý
ï
þ
ï
+
0
f
2x
(2)
f
3x
(2)
ì
í
ï
î
ï
ü
ý
ï
þ
ï
=
F
1x
F
2x
F
3x
ì
í
ï
î
ï
ü
ý
ï
þ
ï
58
k
1
10-1
000
-101
é
ë
ê
ê
ê
ù
û
ú
ú
ú
ˆ
d
1x
(1)
ˆ
d
2x
(1)
ˆ
d
3x
(1)
ì
í
ï
ï
î
ï
ï
ü
ý
ï
ï
þ
ï
ï
+k
2
000
01-1
0-11
é
ë
ê
ê
ê
ù
û
ú
ú
ú
ˆ
d
1x
(2)
ˆ
d
2x
(2)
ˆ
d
3x
(2)
ì
í
ï
ï
î
ï
ï
ü
ý
ï
ï
þ
ï
ï
=
F
1x
F
2x
F
3x
ì
í
ï
î
ï
ü
ý
ï
þ
ï 59
ˆ
d
1x
(1)
ˆ
d
2x
(1)
ˆ
d
3x
(1)
ì
í
ï
ï
î
ï
ï
ü
ý
ï
ï
þ
ï
ï
=
ˆ
d
1x
(2)
ˆ
d
2x
(2)
ˆ
d
3x
(2)
ì
í
ï
ï
î
ï
ï
ü
ý
ï
ï
þ
ï
ï
=
d
1x
d
2x
d
3x
ì
í
ï
î
ï
ü
ý
ï
þ
ï Compatibility
Displacements of the shared nodes are equal.
60
k
1
0 -k
1
0k
2 -k
2
-k
1
-k
2
k
1
+k
2
é
ë
ê
ê
ê
ù
û
ú
ú
ú
d
1x
d
2x
d
3x
ì
í
ï
î
ï
ü
ý
ï
þ
ï
=
F
1x
F
2x
F
3x
ì
í
ï
î
ï
ü
ý
ï
þ
ï 61
Assembling the Components
Boundary Conditions
•Must Specify B.C’s to prohibit rigid body
motion.
•Two type of B.C’s
–Homogeneous -displacements = 0
–Nonhomogeneous -displacements = nonzero
value
62
Homogeneous B.C’s
•Delete row and column corresponding to
B.C.
•Solve for unknown displacements.
•Compute unknown forces (reactions)
from original (unmodified) stiffness
matrix.
63
k
1
1
2
k
2
1
2
3
x
F
3x F
2x
64
k
10 -k
1
0k
2 -k
2
-k
1-k
2k
1+k
2
é
ë
ê
ê
ê
ù
û
ú
ú
ú
0
d
2x
d
3x
ì
í
ï
î
ï
ü
ý
ï
þ
ï
=
F
1x
F
2x
F
3x
ì
í
ï
î
ï
ü
ý
ï
þ
ï
k
2
-k
2
-k
2
k
1
+k
2
é
ë
ê
ù
û
ú
d
2x
d
3x
ì
í
î
ü
ý
þ
=
F
2x
F
3x
ì
í
î
ü
ý
þ
F
1x=-k
1d
3x Example: Homogeneous BC, d
1x=0
65