FEM

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: