SlideShare a Scribd company logo
8/29/2023
1
Principles of minimum
potential energy and
Rayleigh-Ritz
So far, structural mechanics using Direct Stiffness approach
Finite element method is used to solve physical problems
Solid Mechanics
Fluid Mechanics
Heat Transfer
Electrostatics
Electromagnetism
….
Physical problems are governed by differential equations which satisfy
Boundary conditions
Initial conditions
One variable: Ordinary differential equation (ODE)
Multiple independent variables: Partial differential equation (PDE)
A systematic technique of solving the differential equations
Differential equations (strong) formulation (today)
Variational (weak) formulation
Approximate the weak form using finite elements
Axially loaded elastic bar
x
y
x=0 x=L
A(x) = cross section at x
b(x) = body force distribution
(force per unit length)
E(x) = Young’s modulus
u(x) = displacement of the bar
at x
x
Differential equation governing the response of the bar
L
x
b
dx
du
AE
dx
d










0
;
0
Second order differential equations
Requires 2 boundary conditions for solution
8/29/2023
2
x
y
x=0 x=L
x
Boundary conditions (examples)
L
x
at
u
x
at
u




1
0
0 Dirichlet/ displacement bc
L
x
at
F
dx
du
EA
x
at
u



 0
0
Neumann/ force bc
Differential equation + Boundary conditions = Strong form
of the “boundary value problem”
Observe:
1. All the cases we considered lead to very similar differential
equations and boundary conditions.
2. In 1D it is easy to analytically solve these equations
3. Not so in 2 and 3D especially when the geometry of the domain is
complex: need to solve approximately
4. We’ll learn how to solve these equations in 1D. The approximation
techniques easily translate to 2 and 3D, no matter how complex the
geometry
A generic problem in 1D
1
1
0
0
1
0
;
0
2
2








x
at
u
x
at
u
x
x
dx
u
d
Approximate solution strategy:
Guess
Where jo(x), j1(x),… are “known” functions and ao, a1, etc are
constants chosen such that the approximate solution
1. Satisfies the boundary conditions
2. Satisfies the differential equation
Too difficult to satisfy for general problems!!
...
)
(
)
(
)
(
)
( 2
2
1
1
0 


 x
a
x
a
x
a
x
u o j
j
j
8/29/2023
3
Potential energy
 
W
loading
of
energy
potential
(U)
energy
Strain 


The potential energy of an elastic body is defined as F
u
F
x
k
k
u
k
1
Hooke’s Law
F = ku
Strain energy of a linear spring
F = Force in the spring
u = deflection of the spring
k = “stiffness” of the spring
Strain energy of a linear spring
F
u u+du
Differential strain energy of the spring
for a small change in displacement
(du) of the spring
Fdu
dU 
For a linear spring
kudu
dU 
The total strain energy of the spring
2
u
0
u
k
2
1
du
u
k
U 
 
dU
Strain energy of a nonlinear spring
F
u u+du
Fdu
dU 
The total strain energy of the spring
curve
nt
dispalceme
force
the
under
Area
du
F
U
u
0


 
dU
8/29/2023
4
Potential energy of the loading (for a single spring as in the figure)
Fu
W 
Potential energy of a linear spring
 
W
loading
of
energy
potential
(U)
energy
Strain 


Fu
ku
2
1
Π 2


F
x
k
k
u
Example of how to obtain the equlibr
Principle of minimum potential energy for a system of springs
For this system of spring, first write down the total potential
energy of the system as:
3x
2
2
3
2
2
2
1 Fd
)
d
d
(
k
2
1
)
d
(
k
2
1










 x
x
x
system
Obtain the equilibrium equations by minimizing the potential energy
1
k F
x
2
k
1x
d 2x
d 3x
d
)
2
(
0
)
d
d
(
k
d
)
1
(
0
)
d
d
(
k
d
k
d
2
3
2
3
2
3
2
2
1
2
Equation
F
Equation
x
x
x
system
x
x
x
x
system














Principle of minimum potential energy for a system of springs
In matrix form, equations 1 and 2 look like






















F
x
x 0
d
d
k
k
k
k
k
3
2
2
2
2
2
1
Does this equation look familiar?
Also look at example problem worked out in class
Axially loaded elastic bar
x
y
x=0 x=L
A(x) = cross section at x
b(x) = body force distribution
(force per unit length)
E(x) = Young’s modulus
u(x) = displacement of the bar
at x
x
F
dx
du
ε 
Axial strain
Axial stress
dx
du
E
Eε 


Strain energy per unit volume of the bar
2
dx
du
E
2
1
σε
2
1
dU 







Strain energy of the bar
Adx
σε
2
1
dV
σε
2
1
dU
U
L
0
x


 


 since dV=Adx
8/29/2023
5
Axially loaded elastic bar
Strain energy of the bar

 







L
L
0
2
0
dx
dx
du
EA
2
1
dx
σεA
2
1
U
Potential energy of the loading
L)
Fu(x
dx
bu
W
0


 
L
Potential energy of the axially loaded bar
L)
Fu(x
dx
bu
dx
dx
du
EA
2
1
0
0
2










 

L
L
Principle of Minimum Potential Energy
Among all admissible displacements that a body can have, the one
that minimizes the total potential energy of the body satisfies the
strong formulation
Admissible displacements: these are any reasonable displacement
that you can think of that satisfy the displacement boundary
conditions of the original problem (and of course certain minimum
continuity requirements). Example:
Exact solution for the
displacement field uexact(x)
Any other “admissible”
displacement field w(x)
L
0
x
Lets see what this means for an axially loaded elastic bar
A(x) = cross section at x
b(x) = body force distribution
(force per unit length)
E(x) = Young’s modulus
x
y
x=0 x=L
x
F
Potential energy of the axially loaded bar corresponding to the
exact solution uexact(x)
L)
(x
Fu
dx
bu
dx
dx
du
EA
2
1
)
(u exact
0
exact
0
2
exact
exact 









 

L
L
Potential energy of the axially loaded bar corresponding to the
“admissible” displacement w(x)
L)
Fw(x
dx
bw
dx
dx
dw
EA
2
1
(w)
0
0
2










 

L
L
Exact solution for the
displacement field uexact(x)
Any other “admissible”
displacement field w(x)
L
0
x
8/29/2023
6
Example:
L
x
b
dx
u
d
AE 


 0
;
0
2
2
L
x
at
F
dx
du
EA
x
at
u



 0
0
Assume EA=1; b=1; L=1; F=1
Analytical solution is
2
2
2
x
x
uexact 

6
7
1)
(x
u
dx
u
dx
dx
du
2
1
)
(u exact
1
0
exact
1
0
2
exact
exact 











 

Potential energy corresponding to this analytical solution
Now assume an admissible displacement
x
w 
Why is this an “admissible” displacement? This displacement is
quite arbitrary. But, it satisfies the given displacement boundary
condition w(x=0)=0. Also, its first derivate does not blow up.
1
1)
w(x
dx
w
dx
dx
dw
2
1
(w)
1
0
1
0
2












 

Potential energy corresponding to this admissible displacement
Notice
exact
7
1
6
(u ) (w)
  
  
since
Principle of Minimum Potential Energy
Among all admissible displacements that a body can have, the one
that minimizes the total potential energy of the body satisfies the
strong formulation
Mathematical statement: If ‘uexact’ is the exact solution (which
satisfies the differential equation together with the boundary
conditions), and ‘w’ is an admissible displacement (that is quite
arbitrary except for the fact that it satisfies the displacement
boundary conditions and its first derivative does not blow up),
then
(w)
)
(uexact 


unless w=uexact (i.e. the exact solution minimizes the potential
energy)
The Principle of Minimum Potential Energy and the strong
formulation are exactly equivalent statements of the same
problem.
The exact solution (uexact) that satisfies the strong form, renders
the potential energy of the system a minimum.
So, why use the Principle of Minimum Potential Energy?
The short answer is that it is much less demanding than the strong
formulation. The long answer is, it
1. requires only the first derivative to be finite
2. incorporates the force boundary condition automatically. The
admissible displacement (which is the function that you need to
choose) needs to satisfy only the displacement boundary condition
8/29/2023
7
Finite element formulation, takes as its starting point, not the
strong formulation, but the Principle of Minimum Potential
Energy.
Task is to find the function ‘w’ that minimizes the potential energy
of the system
From the Principle of Minimum Potential Energy, that function ‘w’
is the exact solution.
L)
Fw(x
dx
bw
dx
dx
dw
EA
2
1
(w)
0
0
2










 

L
L
Step 1. Assume a solution
...
)
(
)
(
)
(
)
( 2
2
1
1
0 


 x
a
x
a
x
a
x
w o j
j
j
Where jo(x), j1(x),… are “admissible” functions and ao, a1,
etc are constants to be determined from the solution.
Rayleigh-Ritz Principle
The minimization of the potential energy is difficult to perform
exactly.
The Rayleigh-Ritz principle is an approximate way of doing this.
Step 2. Plug the approximate solution into the potential energy
L)
Fw(x
dx
bw
dx
dx
dw
EA
2
1
(w)
0
0
2










 

L
L
Rayleigh-Ritz Principle
 
 
...
)
(
)
(
F
dx
...
b
dx
...
dx
d
dx
d
EA
2
1
,...)
a
,
(a
1
1
0
0
0
1
1
0
0
0
2
1
1
0
0
1
0





















L
x
a
L
x
a
a
a
a
a
L
L
j
j
j
j
j
j
Step 3. Obtain the coefficients ao, a1, etc by setting
,...
2
,
1
,
0
,
0
(w)





i
ai
Rayleigh-Ritz Principle
The approximate solution is
...
)
(
)
(
)
(
)
( 2
2
1
1
0 


 x
a
x
a
x
a
x
u o j
j
j
Where the coefficients have been obtained from step 3
8/29/2023
8
Example of application of Rayleigh Ritz Principle
x
x=0 x=2
x=1
F
E=A=1
F=2






 



1
2
0
2
1)
Fu(x
dx
dx
du
2
1
(u)










 
x
at
applied
F
load
of
Energy
Potential
Energy
Strain
The potential energy of this bar (of length 2) is
Let us assume a polynomial “admissible” displacement field
2
2
1
0
u x
a
x
a
a 


Note that this is NOT the analytical solution for this problem.
Example of application of Rayleigh Ritz Principle
For this “admissible” displacement to satisfy the displacement
boundary conditions the following conditions must be satisfied:
0
4
2
2)
u(x
0
0)
u(x
2
1
0
0








a
a
a
a
Hence, we obtain
2
1
0
2
0
a
a
a



Hence, the “admissible” displacement simplifies to
 
2
2
2
2
1
0
2
u
x
x
a
x
a
x
a
a






Now we apply Rayleigh Ritz principle, which says that if I plug
this approximation into the expression for the potential energy , I
can obtain the unknown (in this case a2) by minimizing 
 
   
 
4
3
0
2
3
8
0
2
3
4
2
F
dx
2
dx
d
2
1
1)
Fu(x
dx
dx
du
2
1
(u)
2
2
2
2
2
2
1
2
2
2
0
2
2
2
2
0
2





































a
a
a
a
a
x
x
a
x
x
a x
at
evaluated
 
 
2
2
2
2
2
1
0
2
4
3
2
u
x
x
x
x
a
x
a
x
a
a










Hence the approximate solution to this problem, using the
Rayleigh-Ritz principle is
Notice that the exact answer to this problem (can you prove this?) is









2
1
2
1
0
uexact
x
for
x
x
for
x
8/29/2023
9
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
0
0.2
0.4
0.6
0.8
1
x
Exact solution
Approximate
solution
The displacement solution :
How can you improve the approximation?
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
-1.5
-1
-0.5
0
0.5
1
1.5
x
Stress
Approximate
stress
Exact Stress
The stress within the bar:

More Related Content

Similar to Rayleigh Ritz Method 1 D Bar Problem.pdf

Calculus of variations
Calculus of variationsCalculus of variations
Calculus of variations
Solo Hermelin
 
Selected ch25
Selected ch25Selected ch25
Selected ch25
xajo2013
 
What are free particles in quantum mechanics
What are free particles in quantum mechanicsWhat are free particles in quantum mechanics
What are free particles in quantum mechanics
bhaskar chatterjee
 
Elastic beams
Elastic beams Elastic beams
Elastic beams
Bhavik A Shah
 
Unit23
Unit23Unit23
Unit23
sayed30
 
Curso de Analisis por elementos finitos
Curso de Analisis por elementos finitosCurso de Analisis por elementos finitos
Curso de Analisis por elementos finitos
Enrique C.
 
Cs jog
Cs jogCs jog
Chapter 3 wave_optics
Chapter 3 wave_opticsChapter 3 wave_optics
Chapter 3 wave_optics
Gabriel O'Brien
 
Solution to schrodinger equation with dirac comb potential
Solution to schrodinger equation with dirac comb potential Solution to schrodinger equation with dirac comb potential
Solution to schrodinger equation with dirac comb potential
slides
 
Shape1 d
Shape1 dShape1 d
Shape1 d
Manoj Shukla
 
Mit2 092 f09_lec06
Mit2 092 f09_lec06Mit2 092 f09_lec06
Mit2 092 f09_lec06
Rahman Hakim
 
unit8.pdf
unit8.pdfunit8.pdf
Physics formulas list
Physics formulas listPhysics formulas list
Physics formulas list
hannagrauser1
 
Welcome to the presentation.pptx
Welcome to the presentation.pptxWelcome to the presentation.pptx
Welcome to the presentation.pptx
TayebaTakbirOrnila
 
APPLICATION OF HIGHER ORDER DIFFERENTIAL EQUATIONS
APPLICATION OF HIGHER ORDER DIFFERENTIAL EQUATIONSAPPLICATION OF HIGHER ORDER DIFFERENTIAL EQUATIONS
APPLICATION OF HIGHER ORDER DIFFERENTIAL EQUATIONS
AYESHA JAVED
 
Quantum theory of dispersion of light ppt
Quantum theory of dispersion of light pptQuantum theory of dispersion of light ppt
Quantum theory of dispersion of light ppt
tedoado
 
4 b5lecture62008
4 b5lecture620084 b5lecture62008
4 b5lecture62008
Prudhvi Bade
 
Lecture_3.pdf
Lecture_3.pdfLecture_3.pdf
Lecture_3.pdf
BudiAgung19
 
Mathematical formulation of inverse scattering and korteweg de vries equation
Mathematical formulation of inverse scattering and korteweg de vries equationMathematical formulation of inverse scattering and korteweg de vries equation
Mathematical formulation of inverse scattering and korteweg de vries equation
Alexander Decker
 
Ph 101-9 QUANTUM MACHANICS
Ph 101-9 QUANTUM MACHANICSPh 101-9 QUANTUM MACHANICS
Ph 101-9 QUANTUM MACHANICS
Chandan Singh
 

Similar to Rayleigh Ritz Method 1 D Bar Problem.pdf (20)

Calculus of variations
Calculus of variationsCalculus of variations
Calculus of variations
 
Selected ch25
Selected ch25Selected ch25
Selected ch25
 
What are free particles in quantum mechanics
What are free particles in quantum mechanicsWhat are free particles in quantum mechanics
What are free particles in quantum mechanics
 
Elastic beams
Elastic beams Elastic beams
Elastic beams
 
Unit23
Unit23Unit23
Unit23
 
Curso de Analisis por elementos finitos
Curso de Analisis por elementos finitosCurso de Analisis por elementos finitos
Curso de Analisis por elementos finitos
 
Cs jog
Cs jogCs jog
Cs jog
 
Chapter 3 wave_optics
Chapter 3 wave_opticsChapter 3 wave_optics
Chapter 3 wave_optics
 
Solution to schrodinger equation with dirac comb potential
Solution to schrodinger equation with dirac comb potential Solution to schrodinger equation with dirac comb potential
Solution to schrodinger equation with dirac comb potential
 
Shape1 d
Shape1 dShape1 d
Shape1 d
 
Mit2 092 f09_lec06
Mit2 092 f09_lec06Mit2 092 f09_lec06
Mit2 092 f09_lec06
 
unit8.pdf
unit8.pdfunit8.pdf
unit8.pdf
 
Physics formulas list
Physics formulas listPhysics formulas list
Physics formulas list
 
Welcome to the presentation.pptx
Welcome to the presentation.pptxWelcome to the presentation.pptx
Welcome to the presentation.pptx
 
APPLICATION OF HIGHER ORDER DIFFERENTIAL EQUATIONS
APPLICATION OF HIGHER ORDER DIFFERENTIAL EQUATIONSAPPLICATION OF HIGHER ORDER DIFFERENTIAL EQUATIONS
APPLICATION OF HIGHER ORDER DIFFERENTIAL EQUATIONS
 
Quantum theory of dispersion of light ppt
Quantum theory of dispersion of light pptQuantum theory of dispersion of light ppt
Quantum theory of dispersion of light ppt
 
4 b5lecture62008
4 b5lecture620084 b5lecture62008
4 b5lecture62008
 
Lecture_3.pdf
Lecture_3.pdfLecture_3.pdf
Lecture_3.pdf
 
Mathematical formulation of inverse scattering and korteweg de vries equation
Mathematical formulation of inverse scattering and korteweg de vries equationMathematical formulation of inverse scattering and korteweg de vries equation
Mathematical formulation of inverse scattering and korteweg de vries equation
 
Ph 101-9 QUANTUM MACHANICS
Ph 101-9 QUANTUM MACHANICSPh 101-9 QUANTUM MACHANICS
Ph 101-9 QUANTUM MACHANICS
 

Recently uploaded

AI + Data Community Tour - Build the Next Generation of Apps with the Einstei...
AI + Data Community Tour - Build the Next Generation of Apps with the Einstei...AI + Data Community Tour - Build the Next Generation of Apps with the Einstei...
AI + Data Community Tour - Build the Next Generation of Apps with the Einstei...
Paris Salesforce Developer Group
 
ELS: 2.4.1 POWER ELECTRONICS Course objectives: This course will enable stude...
ELS: 2.4.1 POWER ELECTRONICS Course objectives: This course will enable stude...ELS: 2.4.1 POWER ELECTRONICS Course objectives: This course will enable stude...
ELS: 2.4.1 POWER ELECTRONICS Course objectives: This course will enable stude...
Kuvempu University
 
Sri Guru Hargobind Ji - Bandi Chor Guru.pdf
Sri Guru Hargobind Ji - Bandi Chor Guru.pdfSri Guru Hargobind Ji - Bandi Chor Guru.pdf
Sri Guru Hargobind Ji - Bandi Chor Guru.pdf
Balvir Singh
 
原版制作(Humboldt毕业证书)柏林大学毕业证学位证一模一样
原版制作(Humboldt毕业证书)柏林大学毕业证学位证一模一样原版制作(Humboldt毕业证书)柏林大学毕业证学位证一模一样
原版制作(Humboldt毕业证书)柏林大学毕业证学位证一模一样
ydzowc
 
Applications of artificial Intelligence in Mechanical Engineering.pdf
Applications of artificial Intelligence in Mechanical Engineering.pdfApplications of artificial Intelligence in Mechanical Engineering.pdf
Applications of artificial Intelligence in Mechanical Engineering.pdf
Atif Razi
 
FULL STACK PROGRAMMING - Both Front End and Back End
FULL STACK PROGRAMMING - Both Front End and Back EndFULL STACK PROGRAMMING - Both Front End and Back End
FULL STACK PROGRAMMING - Both Front End and Back End
PreethaV16
 
Tools & Techniques for Commissioning and Maintaining PV Systems W-Animations ...
Tools & Techniques for Commissioning and Maintaining PV Systems W-Animations ...Tools & Techniques for Commissioning and Maintaining PV Systems W-Animations ...
Tools & Techniques for Commissioning and Maintaining PV Systems W-Animations ...
Transcat
 
Properties of Fluids, Fluid Statics, Pressure Measurement
Properties of Fluids, Fluid Statics, Pressure MeasurementProperties of Fluids, Fluid Statics, Pressure Measurement
Properties of Fluids, Fluid Statics, Pressure Measurement
Indrajeet sahu
 
A high-Speed Communication System is based on the Design of a Bi-NoC Router, ...
A high-Speed Communication System is based on the Design of a Bi-NoC Router, ...A high-Speed Communication System is based on the Design of a Bi-NoC Router, ...
A high-Speed Communication System is based on the Design of a Bi-NoC Router, ...
DharmaBanothu
 
ITSM Integration with MuleSoft.pptx
ITSM  Integration with MuleSoft.pptxITSM  Integration with MuleSoft.pptx
ITSM Integration with MuleSoft.pptx
VANDANAMOHANGOUDA
 
Call For Paper -3rd International Conference on Artificial Intelligence Advan...
Call For Paper -3rd International Conference on Artificial Intelligence Advan...Call For Paper -3rd International Conference on Artificial Intelligence Advan...
Call For Paper -3rd International Conference on Artificial Intelligence Advan...
ijseajournal
 
UNIT-III- DATA CONVERTERS ANALOG TO DIGITAL CONVERTER
UNIT-III- DATA CONVERTERS ANALOG TO DIGITAL CONVERTERUNIT-III- DATA CONVERTERS ANALOG TO DIGITAL CONVERTER
UNIT-III- DATA CONVERTERS ANALOG TO DIGITAL CONVERTER
vmspraneeth
 
3rd International Conference on Artificial Intelligence Advances (AIAD 2024)
3rd International Conference on Artificial Intelligence Advances (AIAD 2024)3rd International Conference on Artificial Intelligence Advances (AIAD 2024)
3rd International Conference on Artificial Intelligence Advances (AIAD 2024)
GiselleginaGloria
 
一比一原版(uoft毕业证书)加拿大多伦多大学毕业证如何办理
一比一原版(uoft毕业证书)加拿大多伦多大学毕业证如何办理一比一原版(uoft毕业证书)加拿大多伦多大学毕业证如何办理
一比一原版(uoft毕业证书)加拿大多伦多大学毕业证如何办理
sydezfe
 
Introduction to Computer Networks & OSI MODEL.ppt
Introduction to Computer Networks & OSI MODEL.pptIntroduction to Computer Networks & OSI MODEL.ppt
Introduction to Computer Networks & OSI MODEL.ppt
Dwarkadas J Sanghvi College of Engineering
 
paper relate Chozhavendhan et al. 2020.pdf
paper relate Chozhavendhan et al. 2020.pdfpaper relate Chozhavendhan et al. 2020.pdf
paper relate Chozhavendhan et al. 2020.pdf
ShurooqTaib
 
Digital Twins Computer Networking Paper Presentation.pptx
Digital Twins Computer Networking Paper Presentation.pptxDigital Twins Computer Networking Paper Presentation.pptx
Digital Twins Computer Networking Paper Presentation.pptx
aryanpankaj78
 
Particle Swarm Optimization–Long Short-Term Memory based Channel Estimation w...
Particle Swarm Optimization–Long Short-Term Memory based Channel Estimation w...Particle Swarm Optimization–Long Short-Term Memory based Channel Estimation w...
Particle Swarm Optimization–Long Short-Term Memory based Channel Estimation w...
IJCNCJournal
 
一比一原版(USF毕业证)旧金山大学毕业证如何办理
一比一原版(USF毕业证)旧金山大学毕业证如何办理一比一原版(USF毕业证)旧金山大学毕业证如何办理
一比一原版(USF毕业证)旧金山大学毕业证如何办理
uqyfuc
 
Accident detection system project report.pdf
Accident detection system project report.pdfAccident detection system project report.pdf
Accident detection system project report.pdf
Kamal Acharya
 

Recently uploaded (20)

AI + Data Community Tour - Build the Next Generation of Apps with the Einstei...
AI + Data Community Tour - Build the Next Generation of Apps with the Einstei...AI + Data Community Tour - Build the Next Generation of Apps with the Einstei...
AI + Data Community Tour - Build the Next Generation of Apps with the Einstei...
 
ELS: 2.4.1 POWER ELECTRONICS Course objectives: This course will enable stude...
ELS: 2.4.1 POWER ELECTRONICS Course objectives: This course will enable stude...ELS: 2.4.1 POWER ELECTRONICS Course objectives: This course will enable stude...
ELS: 2.4.1 POWER ELECTRONICS Course objectives: This course will enable stude...
 
Sri Guru Hargobind Ji - Bandi Chor Guru.pdf
Sri Guru Hargobind Ji - Bandi Chor Guru.pdfSri Guru Hargobind Ji - Bandi Chor Guru.pdf
Sri Guru Hargobind Ji - Bandi Chor Guru.pdf
 
原版制作(Humboldt毕业证书)柏林大学毕业证学位证一模一样
原版制作(Humboldt毕业证书)柏林大学毕业证学位证一模一样原版制作(Humboldt毕业证书)柏林大学毕业证学位证一模一样
原版制作(Humboldt毕业证书)柏林大学毕业证学位证一模一样
 
Applications of artificial Intelligence in Mechanical Engineering.pdf
Applications of artificial Intelligence in Mechanical Engineering.pdfApplications of artificial Intelligence in Mechanical Engineering.pdf
Applications of artificial Intelligence in Mechanical Engineering.pdf
 
FULL STACK PROGRAMMING - Both Front End and Back End
FULL STACK PROGRAMMING - Both Front End and Back EndFULL STACK PROGRAMMING - Both Front End and Back End
FULL STACK PROGRAMMING - Both Front End and Back End
 
Tools & Techniques for Commissioning and Maintaining PV Systems W-Animations ...
Tools & Techniques for Commissioning and Maintaining PV Systems W-Animations ...Tools & Techniques for Commissioning and Maintaining PV Systems W-Animations ...
Tools & Techniques for Commissioning and Maintaining PV Systems W-Animations ...
 
Properties of Fluids, Fluid Statics, Pressure Measurement
Properties of Fluids, Fluid Statics, Pressure MeasurementProperties of Fluids, Fluid Statics, Pressure Measurement
Properties of Fluids, Fluid Statics, Pressure Measurement
 
A high-Speed Communication System is based on the Design of a Bi-NoC Router, ...
A high-Speed Communication System is based on the Design of a Bi-NoC Router, ...A high-Speed Communication System is based on the Design of a Bi-NoC Router, ...
A high-Speed Communication System is based on the Design of a Bi-NoC Router, ...
 
ITSM Integration with MuleSoft.pptx
ITSM  Integration with MuleSoft.pptxITSM  Integration with MuleSoft.pptx
ITSM Integration with MuleSoft.pptx
 
Call For Paper -3rd International Conference on Artificial Intelligence Advan...
Call For Paper -3rd International Conference on Artificial Intelligence Advan...Call For Paper -3rd International Conference on Artificial Intelligence Advan...
Call For Paper -3rd International Conference on Artificial Intelligence Advan...
 
UNIT-III- DATA CONVERTERS ANALOG TO DIGITAL CONVERTER
UNIT-III- DATA CONVERTERS ANALOG TO DIGITAL CONVERTERUNIT-III- DATA CONVERTERS ANALOG TO DIGITAL CONVERTER
UNIT-III- DATA CONVERTERS ANALOG TO DIGITAL CONVERTER
 
3rd International Conference on Artificial Intelligence Advances (AIAD 2024)
3rd International Conference on Artificial Intelligence Advances (AIAD 2024)3rd International Conference on Artificial Intelligence Advances (AIAD 2024)
3rd International Conference on Artificial Intelligence Advances (AIAD 2024)
 
一比一原版(uoft毕业证书)加拿大多伦多大学毕业证如何办理
一比一原版(uoft毕业证书)加拿大多伦多大学毕业证如何办理一比一原版(uoft毕业证书)加拿大多伦多大学毕业证如何办理
一比一原版(uoft毕业证书)加拿大多伦多大学毕业证如何办理
 
Introduction to Computer Networks & OSI MODEL.ppt
Introduction to Computer Networks & OSI MODEL.pptIntroduction to Computer Networks & OSI MODEL.ppt
Introduction to Computer Networks & OSI MODEL.ppt
 
paper relate Chozhavendhan et al. 2020.pdf
paper relate Chozhavendhan et al. 2020.pdfpaper relate Chozhavendhan et al. 2020.pdf
paper relate Chozhavendhan et al. 2020.pdf
 
Digital Twins Computer Networking Paper Presentation.pptx
Digital Twins Computer Networking Paper Presentation.pptxDigital Twins Computer Networking Paper Presentation.pptx
Digital Twins Computer Networking Paper Presentation.pptx
 
Particle Swarm Optimization–Long Short-Term Memory based Channel Estimation w...
Particle Swarm Optimization–Long Short-Term Memory based Channel Estimation w...Particle Swarm Optimization–Long Short-Term Memory based Channel Estimation w...
Particle Swarm Optimization–Long Short-Term Memory based Channel Estimation w...
 
一比一原版(USF毕业证)旧金山大学毕业证如何办理
一比一原版(USF毕业证)旧金山大学毕业证如何办理一比一原版(USF毕业证)旧金山大学毕业证如何办理
一比一原版(USF毕业证)旧金山大学毕业证如何办理
 
Accident detection system project report.pdf
Accident detection system project report.pdfAccident detection system project report.pdf
Accident detection system project report.pdf
 

Rayleigh Ritz Method 1 D Bar Problem.pdf

  • 1. 8/29/2023 1 Principles of minimum potential energy and Rayleigh-Ritz So far, structural mechanics using Direct Stiffness approach Finite element method is used to solve physical problems Solid Mechanics Fluid Mechanics Heat Transfer Electrostatics Electromagnetism …. Physical problems are governed by differential equations which satisfy Boundary conditions Initial conditions One variable: Ordinary differential equation (ODE) Multiple independent variables: Partial differential equation (PDE) A systematic technique of solving the differential equations Differential equations (strong) formulation (today) Variational (weak) formulation Approximate the weak form using finite elements Axially loaded elastic bar x y x=0 x=L A(x) = cross section at x b(x) = body force distribution (force per unit length) E(x) = Young’s modulus u(x) = displacement of the bar at x x Differential equation governing the response of the bar L x b dx du AE dx d           0 ; 0 Second order differential equations Requires 2 boundary conditions for solution
  • 2. 8/29/2023 2 x y x=0 x=L x Boundary conditions (examples) L x at u x at u     1 0 0 Dirichlet/ displacement bc L x at F dx du EA x at u     0 0 Neumann/ force bc Differential equation + Boundary conditions = Strong form of the “boundary value problem” Observe: 1. All the cases we considered lead to very similar differential equations and boundary conditions. 2. In 1D it is easy to analytically solve these equations 3. Not so in 2 and 3D especially when the geometry of the domain is complex: need to solve approximately 4. We’ll learn how to solve these equations in 1D. The approximation techniques easily translate to 2 and 3D, no matter how complex the geometry A generic problem in 1D 1 1 0 0 1 0 ; 0 2 2         x at u x at u x x dx u d Approximate solution strategy: Guess Where jo(x), j1(x),… are “known” functions and ao, a1, etc are constants chosen such that the approximate solution 1. Satisfies the boundary conditions 2. Satisfies the differential equation Too difficult to satisfy for general problems!! ... ) ( ) ( ) ( ) ( 2 2 1 1 0     x a x a x a x u o j j j
  • 3. 8/29/2023 3 Potential energy   W loading of energy potential (U) energy Strain    The potential energy of an elastic body is defined as F u F x k k u k 1 Hooke’s Law F = ku Strain energy of a linear spring F = Force in the spring u = deflection of the spring k = “stiffness” of the spring Strain energy of a linear spring F u u+du Differential strain energy of the spring for a small change in displacement (du) of the spring Fdu dU  For a linear spring kudu dU  The total strain energy of the spring 2 u 0 u k 2 1 du u k U    dU Strain energy of a nonlinear spring F u u+du Fdu dU  The total strain energy of the spring curve nt dispalceme force the under Area du F U u 0     dU
  • 4. 8/29/2023 4 Potential energy of the loading (for a single spring as in the figure) Fu W  Potential energy of a linear spring   W loading of energy potential (U) energy Strain    Fu ku 2 1 Π 2   F x k k u Example of how to obtain the equlibr Principle of minimum potential energy for a system of springs For this system of spring, first write down the total potential energy of the system as: 3x 2 2 3 2 2 2 1 Fd ) d d ( k 2 1 ) d ( k 2 1            x x x system Obtain the equilibrium equations by minimizing the potential energy 1 k F x 2 k 1x d 2x d 3x d ) 2 ( 0 ) d d ( k d ) 1 ( 0 ) d d ( k d k d 2 3 2 3 2 3 2 2 1 2 Equation F Equation x x x system x x x x system               Principle of minimum potential energy for a system of springs In matrix form, equations 1 and 2 look like                       F x x 0 d d k k k k k 3 2 2 2 2 2 1 Does this equation look familiar? Also look at example problem worked out in class Axially loaded elastic bar x y x=0 x=L A(x) = cross section at x b(x) = body force distribution (force per unit length) E(x) = Young’s modulus u(x) = displacement of the bar at x x F dx du ε  Axial strain Axial stress dx du E Eε    Strain energy per unit volume of the bar 2 dx du E 2 1 σε 2 1 dU         Strain energy of the bar Adx σε 2 1 dV σε 2 1 dU U L 0 x        since dV=Adx
  • 5. 8/29/2023 5 Axially loaded elastic bar Strain energy of the bar           L L 0 2 0 dx dx du EA 2 1 dx σεA 2 1 U Potential energy of the loading L) Fu(x dx bu W 0     L Potential energy of the axially loaded bar L) Fu(x dx bu dx dx du EA 2 1 0 0 2              L L Principle of Minimum Potential Energy Among all admissible displacements that a body can have, the one that minimizes the total potential energy of the body satisfies the strong formulation Admissible displacements: these are any reasonable displacement that you can think of that satisfy the displacement boundary conditions of the original problem (and of course certain minimum continuity requirements). Example: Exact solution for the displacement field uexact(x) Any other “admissible” displacement field w(x) L 0 x Lets see what this means for an axially loaded elastic bar A(x) = cross section at x b(x) = body force distribution (force per unit length) E(x) = Young’s modulus x y x=0 x=L x F Potential energy of the axially loaded bar corresponding to the exact solution uexact(x) L) (x Fu dx bu dx dx du EA 2 1 ) (u exact 0 exact 0 2 exact exact              L L Potential energy of the axially loaded bar corresponding to the “admissible” displacement w(x) L) Fw(x dx bw dx dx dw EA 2 1 (w) 0 0 2              L L Exact solution for the displacement field uexact(x) Any other “admissible” displacement field w(x) L 0 x
  • 6. 8/29/2023 6 Example: L x b dx u d AE     0 ; 0 2 2 L x at F dx du EA x at u     0 0 Assume EA=1; b=1; L=1; F=1 Analytical solution is 2 2 2 x x uexact   6 7 1) (x u dx u dx dx du 2 1 ) (u exact 1 0 exact 1 0 2 exact exact                Potential energy corresponding to this analytical solution Now assume an admissible displacement x w  Why is this an “admissible” displacement? This displacement is quite arbitrary. But, it satisfies the given displacement boundary condition w(x=0)=0. Also, its first derivate does not blow up. 1 1) w(x dx w dx dx dw 2 1 (w) 1 0 1 0 2                Potential energy corresponding to this admissible displacement Notice exact 7 1 6 (u ) (w)       since Principle of Minimum Potential Energy Among all admissible displacements that a body can have, the one that minimizes the total potential energy of the body satisfies the strong formulation Mathematical statement: If ‘uexact’ is the exact solution (which satisfies the differential equation together with the boundary conditions), and ‘w’ is an admissible displacement (that is quite arbitrary except for the fact that it satisfies the displacement boundary conditions and its first derivative does not blow up), then (w) ) (uexact    unless w=uexact (i.e. the exact solution minimizes the potential energy) The Principle of Minimum Potential Energy and the strong formulation are exactly equivalent statements of the same problem. The exact solution (uexact) that satisfies the strong form, renders the potential energy of the system a minimum. So, why use the Principle of Minimum Potential Energy? The short answer is that it is much less demanding than the strong formulation. The long answer is, it 1. requires only the first derivative to be finite 2. incorporates the force boundary condition automatically. The admissible displacement (which is the function that you need to choose) needs to satisfy only the displacement boundary condition
  • 7. 8/29/2023 7 Finite element formulation, takes as its starting point, not the strong formulation, but the Principle of Minimum Potential Energy. Task is to find the function ‘w’ that minimizes the potential energy of the system From the Principle of Minimum Potential Energy, that function ‘w’ is the exact solution. L) Fw(x dx bw dx dx dw EA 2 1 (w) 0 0 2              L L Step 1. Assume a solution ... ) ( ) ( ) ( ) ( 2 2 1 1 0     x a x a x a x w o j j j Where jo(x), j1(x),… are “admissible” functions and ao, a1, etc are constants to be determined from the solution. Rayleigh-Ritz Principle The minimization of the potential energy is difficult to perform exactly. The Rayleigh-Ritz principle is an approximate way of doing this. Step 2. Plug the approximate solution into the potential energy L) Fw(x dx bw dx dx dw EA 2 1 (w) 0 0 2              L L Rayleigh-Ritz Principle     ... ) ( ) ( F dx ... b dx ... dx d dx d EA 2 1 ,...) a , (a 1 1 0 0 0 1 1 0 0 0 2 1 1 0 0 1 0                      L x a L x a a a a a L L j j j j j j Step 3. Obtain the coefficients ao, a1, etc by setting ,... 2 , 1 , 0 , 0 (w)      i ai Rayleigh-Ritz Principle The approximate solution is ... ) ( ) ( ) ( ) ( 2 2 1 1 0     x a x a x a x u o j j j Where the coefficients have been obtained from step 3
  • 8. 8/29/2023 8 Example of application of Rayleigh Ritz Principle x x=0 x=2 x=1 F E=A=1 F=2            1 2 0 2 1) Fu(x dx dx du 2 1 (u)             x at applied F load of Energy Potential Energy Strain The potential energy of this bar (of length 2) is Let us assume a polynomial “admissible” displacement field 2 2 1 0 u x a x a a    Note that this is NOT the analytical solution for this problem. Example of application of Rayleigh Ritz Principle For this “admissible” displacement to satisfy the displacement boundary conditions the following conditions must be satisfied: 0 4 2 2) u(x 0 0) u(x 2 1 0 0         a a a a Hence, we obtain 2 1 0 2 0 a a a    Hence, the “admissible” displacement simplifies to   2 2 2 2 1 0 2 u x x a x a x a a       Now we apply Rayleigh Ritz principle, which says that if I plug this approximation into the expression for the potential energy , I can obtain the unknown (in this case a2) by minimizing          4 3 0 2 3 8 0 2 3 4 2 F dx 2 dx d 2 1 1) Fu(x dx dx du 2 1 (u) 2 2 2 2 2 2 1 2 2 2 0 2 2 2 2 0 2                                      a a a a a x x a x x a x at evaluated     2 2 2 2 2 1 0 2 4 3 2 u x x x x a x a x a a           Hence the approximate solution to this problem, using the Rayleigh-Ritz principle is Notice that the exact answer to this problem (can you prove this?) is          2 1 2 1 0 uexact x for x x for x
  • 9. 8/29/2023 9 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0 0.2 0.4 0.6 0.8 1 x Exact solution Approximate solution The displacement solution : How can you improve the approximation? 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 -1.5 -1 -0.5 0 0.5 1 1.5 x Stress Approximate stress Exact Stress The stress within the bar: