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This presentation is about finite element method for bar elements to graduate schools!

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17 views22 pages

Lec 03

This presentation is about finite element method for bar elements to graduate schools!

Uploaded by

abrha.eyassu
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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2 Node Line Elements (Bars)

Abrham E.
May, 2024
Sub Topics
Bar Elements

Governing Equation

Total Potential Energy

Ritz-Method

Galerkin’s Method
Bar Elements
(1)The longitudinal dimension or axial
dimension is much larger that the
transverse dimension(s). The
intersection of a plane normal to the
longitudinal dimension and the bar
defines the cross sections.
(2)The bar resists an internal axial force
along its longitudinal dimension.
Bar Elements cont.
Bar Elements cont.
Bar Elements cont.

• It must be in equilibrium.
•It must satisfy the elastic stress–strain
law (Hooke’s law)
•The displacement field must be
compatible.
•It must satisfy the strain–displacement
equation.
Governing Equation
The governing differential equation of
the bar element is given by
d  du 
 AE q  0
dx  dx 
boundar yconditions
u 0
x0

 A E du 
  P
 dx  xL
Approximate Solution
Admissible displacement function is
continuous over the length and satisfies
any boundary condition:

Principles of Minimum Potential Energy – Of all


kinematically admissible displacement equations,
those corresponding to equilibrium extermize the
TPE. If the extremum is a minimum, the equilibrium
state is stable.
Kinematically admissible
Displacement Functions
those that satisfy the single-valued nature of
displacements (compatibility) and the
boundary conditions
Usually Polynomials
Continuous within element.
Inter-element compatibility. Prevent
overlap or gaps.
Allow for rigid body displacement and
constant strain.
Total Potential Energy (TPE)

 p  U W
dU   x  x  y  z  d  x
Strain d U   x d  x d V
Energy
 ε x 
(Internal U 
work)
    xd x dV
0 
V

1
U    x x d V
2 V
Total Potential Energy

 p work
External U Wof loads

M
ˆf dˆ
W   Xˆ b uˆ dV   x
Tˆ u dS 
i1
i i
V S

TPE
1
p    x  AE  x dx  W
2 L
Total Potential Energy
1
2

Strain energy 1
2L  T
 Adx

External energy Pu 2

1
Total potential energy Π    T  Adx  Pu 2
2 L
Ritz-Method
Ritz-Method
• Using the Ritz-method, approximate
displacement function is obtained by:
1. Assume arbitrary displacement
  a 1 f 1  a 2 f 2  ...  a n f n
2. Introduce this into the TPE functional
3. Performing differentiation and integration
to obtain a function
4. Minimizing the resulting function
d
 0 f o r i  1,2,..., n
dai
Ritz-Method
• Consider the linear elastic one-dimensional
rod with a body force shown below

• The potential energy of this system is:


2
2  du 
Π  1
2 
0
EA   dx  2u 1
 dx 
Ritz-Method
• Consider the polynomial function:

• To be kinematically admissible u must


satisfy the boundary conditions u = 0 at
both (x = 0) and (x = 2)
&

• Thus:
Ritz-Method
• TPE of this system becomes:
Π  4
3
a  2a3
2
3

• Minimizing the TPE: Π 8


 a3  2  0
a 3 3
 a3  3
4

• Thus, an approximate u is given by:

• Rayleigh-Ritz method assumes trial


functions over entire structure
Galerkin-Method
For the one-dimensional rod considered in the
pervious example, the governing equation is:

The Galerkin method aims at setting the


residual relative to a weighting function Wi,
to zero. The weighting functions, Wi, are
chosen from the basis functions used for
constructing û (approximate displacement
function).
Galerkin-Method
• Using the Galerkin-method, approximate
displacement function is obtained by:
1. The governing DE is written in residual form
d  du 
RES  A E 
dx  dx 
2. Multiply this by weight function f introduce
into the TPE, and equate to zero
3. Performing differentiation and integration to
obtain a function L f R d x  0
4. Minimizing the resulting function, to obtain
the approximate u
Galerkin-Method
• Consider the rod shown below

• Multiplying by Φ and integrating gives (parts):


Galerkin-Method
• The function Φ is zero at (x = 0) and (x = 2)
and EA(du/dx) is the tension in the rod,
which equals 2 at (x = 1). Thus:

• Using the same polynomial function for u and


Φ and if u1 and Φ1 are the values at (x = 1):

• Setting these and


E=A=1 in the integral:
Galerkin-Method
• Equating the value in the bracket to zero
and performing the integral:
u1  3
4
u  0.75  2x  x 
2

• In elasticity problems Galerkin’s method turns


out to be the principle of virtual work: (A
deformable body is in equilibrium when the
total work done by external forces is equal to
the total work done by internal forces).

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