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ME1401 – INTRODUCTION OF FINITE ELEMENT ANALYSIS
CONTENTS

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Problems (III set)

Advantages of Finite Element Method

Disadvantages of Finite Element Method

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Applications of Finite Element Analysis

UNIT – II

One Dimension Problems

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Syllabus

One Dimensional elements

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Bar, Beam and Truss

Stress, Strain and Displacement
Types of Loading
Finite Element Modeling

Co – Ordinates

Natural Co – Ordinate (ε)
Shape function
Polynomial Shape function

Properties of Stiffness Matrix

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Problem (III set)

UNIT – III

Two Dimension Problems – Scalar variable Problems

Syllabus

Two dimensional elements

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Plane Stress and Plane Strain
Finite Element Modeling

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Constant Strain Triangular (CST) Element
Shape function for the CST element

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Displacement function for the CST element
Strain – Displacement matrix [B] for CST element

Stress – Strain relationship matrix (or) Constitutive matrix [D] for two dimensional
element
Stress – Strain relationship matrix for two dimensional plane stress problems

Stress – Strain relationship matrix for two dimensional plane strain problems
Stiffness matrix equation for two dimensional element (CST element)
Temperature Effects

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Syllabus

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AXISYMMETRIC CONTINUUM

Elasticity Equations

Axisymmetric Elements

Axisymmetric Formulation

Equation of shape function for Axisymmetric element

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Equation of Strain – Displacement Matrix [B] for Axisymmetric element
Equation of Stress – Strain Matrix [D] for Axisymmetric element

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Equation of Stiffness Matrix [K] for Axisymmetric element
Temperature Effects

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Problem (I set)

UNIT – V
ISOPARAMETRIC ELEMENTS FOR TWO DIMENSIONAL CONTINUUM

Isoparametric element
Superparametric element
Subparametric element

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Problem (I set)

ME 1401

INTRODUCTION OF FINITE ELEMENT ANALYSIS
Unit – I

Fundamental Concepts

Syllabus

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Force Method – Internal forces are considered as the unknowns of the problem
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Displacement or stiffness method – Displacements of the nodes are considered
as the unknowns of the problem
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for the unknown displacements > computation of the element strains and stresses
from the nodal displacements > Interpret the results (post processing)
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The boundary conditions on

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displacements to prevail at certain points on the boundary of the body, whereas
the boundary conditions on stresses require that the stresses induced must be in

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equilibrium with the external forces applied at certain points on the boundary of
the body
...


 Rayleigh – Ritz Method (Variational Approach)
It is useful for solving complex structural problems
...
Otherwise, Galerkin’s method of weighted

 Problems (I set)

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1
...
Determine the bending moment and deflection at midspan by using
Rayleigh – Ritz method and compare with exact solutions
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A bar of uniform cross section is clamed at one end and left free at another end
and it is subjected to a uniform axial load P
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Compare
with exact solutions
...
For non –
structural problems, the method of weighted residuals becomes very useful
...
The popular four methods are,


...
Point collocation method,

Residuals are set to zero at n different locations X i, and the weighting function wi

is denoted as (x - xi)
...
Subdomain collocation method,
 1 forxinD1
w1 = 
0 forxnotinD1

3
...


4
...

wi = Ni (x)
 Ni (x) [R (x; a1, a2, a3… an)]2 dx = 0,

i = 1, 2, 3, …n
...
The following differential equation is available for a physical phenomenon
...
Trial function is y = a1x (10-x)
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d

are y (0) = 0 and y (10) = 0
...


2
...
Obtain two term Galerkin solution b using the trial functions:
2
dx
N1(x) = x(x-1); N2(x) = x2(x-1); 0  x 1
...

 Matrix Algebra

are

equal
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are zero
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Unit matrix: All diagonal elements are unity and other elements are zero
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 Gaussian Elimination Method
It is most commonly used for solving simultaneous linear equations
...


 Problems (III set)
1
...
Solve by using Gauss – Elimination
method
...
2a+4b+2c = 15, 2a+b+2c = -5, 4a+b-2c = 0
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Gauss – Elimination method
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FEM can handle irregular geometry in a convenient manner
...
Handles general load conditions without difficulty

3
...

4
...

 Disadvantages of Finite Element Method

1
...
It requires longer execution time compared with FEM
...
Output result will vary considerably
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Structural Problems:

1
...
Stress concentration problems typically associated with holes, fillets or
other changes in geometry in a body
...
Buckling Analysis: Example: Connecting rod subjected to axial

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compression
...
Vibration Analysis: Example: A beam subjected to different types of
loading
...
Heat Transfer analysis:
Example: Steady state thermal analysis on composite cylinder
...
Fluid flow analysis:
Example: Fluid flow through pipes
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approach – Galarkin approach – Assembly of stiffness matrix and load vector – Finite
element equations – Quadratic shape functions – Applications to plane trusses
 One Dimensional elements

Bar and beam elements are considered as One Dimensional elements
...

 Bar, Beam and Truss

Bar is a member which resists only axial loads
...
A truss is an assemblage of bars with pin joints and a frame is
an assemblage of beam elements
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the form of vector by the variable x as e x, Displacement is denoted in the form of
vector by the variable x as ux
...
Unit

is Force / Unit volume
...


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(2) Traction (T)
It is a distributed force acting on the surface of the body
...
But for one dimensional problem, unit is Force / Unit length
...

(3) Point load (P)
It is a force acting at a particular point which causes displacement
...

(1) Discretization of structure

 CO – ORDINATES

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(A) Global co – ordinates, (B) Local co – ordinates and (C) Natural co –
ordinates
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Integration of polynomial terms in natural co – ordinates for two dimensional
elements can be performed by using the formula,

L1  L2  L3  dA  !  ! ! X 2 A

     !

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 Shape function

N1N2N3 are usually denoted as shape function
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Shape function need to satisfy the following
(a) First derivatives should be finite within an element; (b) Displacement should
be continuous across the element boundary
...


(2) It is easy to formulate and computerize the finite element equations
...

 Stiffness Matrix [K]

 B DBdv
T

in

Stiffness Matrix [K] =

V

 Properties of Stiffness Matrix


...
It is a symmetric matrix, 2
...
It is an unstable element
...


 Equation of Stiffness Matrix for One dimensional bar element

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[K] =

AE  1  1
l  1 1 



w

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 Finite Element Equation for One dimensional bar element

 F1  AE  1  1 u1 
 
 
F2 
l  1 1  u2 




 The Load (or) Force Vector {F}

F e  Al  

1

 Problem (I set)

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A two noded truss element is shown in figure
...
Calculate the displacement at x = ¼, 1/3 and ½
...
The following
assumptions are made while finding the forces in a truss,


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 Stiffness Matrix [K] for a truss element

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w

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 l2
lm

Ae E e  lm
m2
K  
l e   l 2  lm

2
 lm  m


 lm 
2
 lm  m 
l2
lm 
2 
lm
m 

 l2

 Finite Element Equation for Two noded Truss element

 l2
lm

m2
 lm
  l 2  lm

2
 lm  m


 Problem (II set)

 lm  u1 
 
 lm  m 2  u 2 
 
l2
lm  u 3 

lm
m 2  u 4 
 
 l2

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Consider a three bar truss as shown in figure
...
Calculate (a) Nodal displacement, (b) Stress in each member and
(c) Reactions at the support
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element 2 = 2500 mm2, Area of element 3 = 2500 mm2
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Cantilever beam, 2
...
Over hanging beam, 4
...
Continuous beam
...
Point or Concentrated Load, 2
...
Uniformly

 Problem (III set)

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1
...
Calculate the rotation at Point

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Unit – III Two Dimension Problems – Scalar variable Problems
Syllabus
Finite element modeling – CST & LST elements – Elements equations – Load vectors and

heat transfer
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boundary conditions – Assembly – Applications to scalar variable problems such as torsion,

Two dimensional elements are defined by three or more nodes in a two
dimensional plane (i
...
, x, y plane)
...


 Plane Stress and Plane Strain


...


Plane Stress Analysis:

It is defined to be a state of stress in which the normal stress () and shear

stress () directed perpendicular to the plane are assumed to be zero
...

 Finite Element Modeling
It consists of 1
...
Numbering of nodes
...
Discretization:
The art of subdividing a structure into a convenient number of smaller
components is known as discretization
...
Numbering of nodes:
In one dimensional problem, each node is allowed to move only in  x
direction
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directions i
...
, x and y
...


Nodes
123
234
435
536
637
738
839
931

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It has six unknown displacement degrees of freedom (u 1v1, u2v2, u3v3)
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Shape function N3 = (p3 + q3x + r3y) / 2A
 Displacement function for the CST element

u ( x, y )  N1
Displacement function u = 

 v ( x, y )   0

0

N2

0

N3

N1

0

N2

 u1
 v1 
 
0  u 2
 
 X v 2 
N 3  
u 3 
 
 v3 
 

0

 Strain – Displacement matrix [B] for CST element

in

 q1
1 
0
Strain – Displacement matrix [B] =
2A 
 r1


q2

0

q3

r1

0

r2

0

q1

r2

q2

r3

0
r3 

q3 


r1 = x3 – x2


...


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in



v
0 
(1  v)
E
 v
(1  v)
0 
[D] =

1  v  1  2v 
1  2v 
 0

0
2 


 Stiffness matrix equation for two dimensional element (CST element)


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

(1  v)
v
0 

E
 v
(1  v)
0 
[D] =
1  v  1  2v  
1  2v 
 0

0
2 


Distribution of the change in temperature (ΔT) is known as strain
...

σ = D (Bu - e0)

 Galerkin Approach

Stiffness matrix [K] e = [B]T [D][B] A t
...
It has twelve unknown displacement degrees of freedom
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functions of the element are quadratic instead of linear as in the CST
...
Determine the shape functions N1, N2 and N3 at the interior point P for the

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The two dimensional propped beam shown in figure
...
Determine the nodal displacement and element stresses using plane stress
conditions
...

Take, Thickness (t) = 10mm,

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Young’s modulus (E) = 2x105 N/mm2,

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...
25
...
A thin plate is subjected to surface traction as in figure
...


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By using
these displacement solutions, stresses and strains are calculated for each element
...
For example, in a two dimensional plate, the unknown


...

 Equation of Temperature function (T) for one dimensional heat conduction
Temperature (T) = N1T1 + N2T2

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 Equation of Shape functions (N1 & N2) for one dimensional heat conduction
N1 =

lx
l

N2 =

x
l

 Equation of Stiffness Matrix (K) for one dimensional heat conduction

 1
l  1 1 


K c   Ak 


1

 Finite Element Equations for one dimensional heat conduction
Case (i): One dimensional heat conduction with free end convection

 1  1
0 0  T1 
0
 1 1   hA0 1  T   hT A1



 2 
 

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 Ak

 l

 1  1 hPl 2 1  T1  QAl  PhT l 1

 1 1   6 1 2  T  
2



 2 
1

 Finite element Equation for Torsional Bar element

 1  1 1x 
 1 1   

 2 x 

in

 M 1x  GJ


l
M 2 x 


...
An Aluminium alloy fin of 7 mm thick and 50 mm long protrudes from a wall,

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which is maintained at 120°C
...
The heat
transfer coefficient and thermal conductivity of fin material are 140 W/m2 K and 55

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W/mK respectively
...


2
...
The fin is rectangular in shape and is 120 mm long, 40mm
wide and 10mm thick
...
Use two elements
...
3W/mm°C, h = 1 x 10-3 W/ mm2 °C, T=20°C
...
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– Body forces and temperature effects – Stress calculations – Boundary conditions –
Applications to cylinders under internal or external pressures – Rotating discs

 Elasticity Equations

Elasticity equations are used for solving structural mechanics problems
...
The types of elasticity equations are
1
...

z y

in

 yz 

ex – Strain in X direction, ey – Strain in Y direction
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 xy - Shear Strain in XY plane,  xz - Shear Strain in XZ plane,

 yz - Shear Strain in YZ plane

2
...

3
...
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 y  yz  xy
 x  xy  xz


 By  0


 Bx  0 ;
y
z
x
x
y
z
 z  xz  yz


 Bz  0
z
x
y

σ – Stress, τ – Shear Stress, Bx - Body force at X direction,

B y - Body force at Y direction, Bz - Body force at Z direction
...
Compatibility equations

There are six independent compatibility equations, one of which is
2
2
 2 e x  e y   xy


2
2
xy
y
x


...

 Axisymmetric Elements


...

Those types of problems are solved by a special two dimensional element called

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w

w

as axisymmetric element
...
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u 
u r , z    
w

The stress σ is given by

 r 
 
 
Stress,      
 z 
 rz 
 
The strain e is given by


...
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in

v
v
0 
1  v
 v 1 v
v
0 
E


D 
 v
v 1 v
0 
1  v 1  2v  
1  2v 
0
0
0

2 



 Equation of Stiffness Matrix [K] for Axisymmetric element

K   2  rABT DB

r1  r 2  r 3
; A = (½) bxh
3

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r

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 Temperature Effects

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 3z
r

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 f t

 F1u 
 F w
 1 


F u
 2 
 F2 w
 F3u 


 F3 w



1
...
Take E=200GPa and

in

υ= 0
...



...
For the figure, determine the element stresses
...
1x105N/mm2 and
υ= 0
...
The co – ordinates are in mm
...
05mm,

w

w

w

w1=0
...
02mm, w2=0
...
0mm, w3=0
...


3
...
By using two
elements on the 15mm length, calculate the displacements at the inner radius
...
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The four node quadrilateral – Shape functions – Element stiffness matrix and force
vector – Numerical integration - Stiffness integration – Stress calculations – Four node
quadrilateral for axisymmetric problems
...
A large number of elements may be used to obtain reasonable

resemblance between original body and the assemblage
...
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drawback, isoparametric elements are used
...


 Superparametric element
If the number of nodes used for defining the geometry is more than number of
nodes used for defining the displacements, then it is known as superparametric
element
...
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 Subparametric element

If the number of nodes used for defining the geometry is less than number of

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w


...


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nodes used for defining the displacements, then it is known as subparametric

 Equation of Shape function for 4 noded rectangular parent element

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 Equation of Stiffness Matrix for 4 noded isoparametric quadrilateral element
1

1

T

K   t   B DB J 
1 1

J 11


...

0
 0

2 


in

 Equation of element force vector

 Fx 
 [N ]   ;
Fy 
T


...


w

 Numerical Integration (Gaussian Quadrature)
The Gauss quadrature is one of the numerical integration methods to calculate the
definite integrals
...
In

w

w

N 4

N 4

0

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i

0


0 

0 

N 4 

 
N 4 
 


Number of

Location

Corresponding Weights

Points

xi

wi

1

2

3

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000

x1, x2 = 

x1, x3  

1
 0
...
000

3
 0
...
555555
9
8
 0
...
000

2
...
3478548451
0
...
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x1, x4=  0
...
3399810436

4

 Problem (I set)

x

1

w

1
...

2
...
Compare with exact solution
...
For the isoparametric quadrilateral element shown in figure, determine the
local co –ordinates of the point P which has Cartesian co-ordinates (7, 4)
...
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4
...
Determine (i) Jacobian
matrix,

(ii) Strain – Displacement matrix and (iii) Element Stresses
...
25, u=[0,0,0
...
004,0
...
004,0,0] T, Ɛ= 0, ɳ=0
...
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in

Assume plane stress condition

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