Section 2: Finite Element Analysis Theory
1. Method of Weighted Residuals
2. Calculus of Variations
Two distinct ways to develop the underlying
equations of FEA!
FEA Theory
-1-
Section 2: FEA Theory
 Some definitions:
•V = volume of object
•A = surface area
= Au + As
•Au = surface of known
displacements
•As = surface of known stresses
•b = body force
•t = surface stresses (tractions)
FEA Theory
 u  x, y , z  


x   x, y , z  ; u  x    v  x, y , z   .
 w  x, y , z  


-2-
Section 2.1: Weighted Residual Methods
 A group of methods that take governing
equations in the strong form and turn them into
(related) statements in the weak form.
 Applicable to a wide class of problems
(elasticity, heat conduction, mass flow, …).
 A “purely mathematical” concept.
FEA Theory
-3-
2.1: Weighted residual methods (cont.)
 Need to write the equilibrium equations and boundary
conditions in an abstract form as follows:

s x  xy  xz


 bx  0 
x
y
z


 xy s y  yz


 by  0   E  u  x    0,
x
y
z


 xz  yz s z


 bz  0 
x
y
z

u  u 0  0 on Au 
 B  u  x    0.

σ   nˆ   t  0 on As 
FEA Theory
Solve these for u(x)!
-4-
2.1: Weighted residual methods (cont.)
 Let uexact  x  be the exact solution to the problem
(differential equation and boundary conditions)
 E  u exact  x    0 everywhere in V !
B  u exact  x    0 everywhere on A!
 Then, for any choice of vectors W and W’:
W E  u exact  x    0 everywhere in V !
W B  u exact  x    0 everywhere on A!
FEA Theory
-5-
2.1: Weighted residual methods (cont.)
 Integrate these “results” over the entire volume
and surface: W E  u exact  x   dV   W B  u exact  x   dA  0
V
A
 Previous expression is still true if W and W’ are
functions of x (called weighting functions):
 W1  x, y, z  
 W1 x, y , z  






W  x    W2  x, y, z   , W  x    W2  x, y , z  
 W  x, y , z  
 W  x, y , z  
3


 3

  W  x  E  u exact  x   dV   W  x  B  u exact  x   dA  0
V
FEA Theory
A
-6-
2.1: Weighted residual methods (cont.)
 Now, consider an approximate solution to the
same problem:
 uapprox  x, y, z  
 N11  x, y, z  
 N12  x, y , z  






 vapprox  x, y, z    a1 *  N 21  x, y, z    a 2 *  N 22  x, y, z   
 N  x, y , z  
 N  x, y , z  
w

 31

 32

 approx  x, y, z  
 N1n  x, y , z    uexact  x, y , z  

 

 a n *  N 2 n  x, y, z     vexact  x, y, z   .
 N  x, y , z    w  x, y , z  
 3n
  exact

 u approx  x    a k N k  x   u exact  x 
n
k 1
Known functions
 Matrix/vector form of this:
Unknown constants
a 
 uapprox  x, y, z    N11  x, y, z  N12  x, y , z 
N1n  x, y , z    1   uexact  x, y, z  

 


 a
N 2 n  x, y , z    2    vexact  x, y, z   .
 vapprox  x, y, z     N 21  x, y , z  N 22  x, y , z 
 


w
  N  x, y , z  N  x, y , z 

x
,
y
,
z
N
x
,
y
,
z




32
3n
  a   wexact  x, y, z  
 approx
  31
 n
known functions
unknowns
 u approx  x    N  x    a   u exact  x 
FEA Theory
-7-
2.1: Weighted residual methods (cont.)
 Plugging this approximate solution into the
differential equation and boundary conditions
results in some errors, called the residuals.
R E  x, a   E  u approx  x    0 in V !
R B  x, a   B  u approx  x    0 on A!
 Repeating the previous process now gives us an
integral close to but not exactly equal to zero!
I  a    W  x  R E  x, a  dV   W  x  R B  x, a  dA  0 .
V
FEA Theory
A
-8-
2.1: Weighted residual methods (cont.)
 Goal: Find the value of a that makes this integral
as close as possible to zero – “best approximation”.
 Idea: for n different choices of the weighting
functions, derive an equation for a by requiring
that the above integral equal zero:
Equation #1: I1  a    W1  x  R E  x, a  dV   W1 x  R B  x, a  dA  0 .
V
A
Equation #2: I 2  a    W2  x  R E  x, a  dV   W2  x  R B  x, a  dA  0 .
V
A
Equation #n: I n  a    Wn  x  R E  x, a  dV   Wn  x  R B  x, a  dA  0 .
V
 Solve these equations for a!
FEA Theory
A
-9-
2.1: Weighted residual methods (cont.)
 Notes on weighted residual methods:

It is typical (but not required) to assume that the
known functions satisfy the displacement boundary
conditions exactly on Au. (Essential conditions)
I k  a    Wk  x  R E  x, a  dV   Wk  x  R B  x, a  dA  0 , k  1,2,
V


, n.
As
In some methods, one must integrate the volume
integral by parts to get “appropriate” equations.
Different methods result from different ideas about
how to choose the weighting functions.
FEA Theory
-10-
2.1: Weighted residual methods (cont.)
1. Collocation Method:

Assume only one PDE and one BC to solve!
R E  x, a   RE  x, a  ; R B  x, a   RB  x, a .

Idea: pick n points in object
(at least one in V and one
on A) and require residual
to be zero at each point!
RE  xi , a  =0, i  1, 2,
RB  x j , a  =0, j  1, 2,
, nV .
, nA .
 nV  nA  n.
FEA Theory
-11-
2.1: Weighted residual methods (cont.)
2. Subdomain Method:
Assume only one PDE and one BC to solve!

R E  x, a   RE  x, a  ; R B  x, a   RB  x, a .
Divide object up into n distinct regions (at least one
in V and one on A).
Require integral over
each region to be zero.


I i  a    RE  x, a  dV  0, i  1, 2,
, nV
Vi
I j  a    RB  x, a  dA  0, j  1, 2,
, nA .
Aj
 nV  nA  n.
FEA Theory
-12-
2.1: Weighted residual methods (cont.)
 Notes on collocation and subdomain methods:
 Weighting functions for collocation method are the Dirac
delta functions:
Wi  x     x  xi  , i  1,2,

, nV .Wj  x     x  x j  , j  1,2,
Weighting functions for subdomain method are the
indicator functions:
1 if xi Vi
Wi  x   
, i  1, 2,
 0 if xi Vi


, nA.
1 if x j  Aj
, nV .Wj  x   
, j  1, 2,
 0 if xi  Aj
, nA .
Advantage: Simple to formulate.
Disadvantage: Used mostly for problems with only one
governing equation (axial bar, beam, heat,…).
FEA Theory
-13-
2.1: Weighted residual methods (cont.)
3. Least Squares Method:

Considers magnitude of residual over the object.
I LS  a    R E  x, a  R E  x, a  dV    R B  x, a  R B  x, a  dA  0.
V

Ik a 
FEA Theory
As
Finds minimum by setting derivatives to zero
I LS  a 
R E
R B

x
,
a
R
x
,
a
dV


  E 
 x, a  R B  x, a  dA  0 .

a k
V a k
As a k
 W  x 
 W  x
-14-
2.1: Weighted residual methods (cont.)
4. Galerkin’s Method:

Idea: Project residual of differential equation
onto original approximating functions!
Wk  x  

u approx  x 
a k
 N k  x  , k  1, 2,
, n.
To get W’, must integrate any derivatives in
volume integral by parts!
Let R E  x, a   R E ,deriv  x, a   R E ,noderiv  x, a  ;
R E ,deriv  x, a    R E ,deriv  x, a  dx. (Assume derivative involves x.)
FEA Theory
-15-
2.1: Weighted residual methods (cont.)
  N k  x  R E  x, a  dV   N k  x  R E ,deriv  x, a  nx dA
V
A
Introduces the boundary conditions!
N k  x 

R E ,deriv  x, a  dV
x
V
  N k  x  R E ,noderiv  x, a  dV
V
 I k  a    N k  x  R E  x, a  dV  0 , k  1, 2,
, n.
V
 Must use integrated-by-parts version of I k  a  !
FEA Theory
-16-
2.1: Weighted residual methods (cont.)
 Notes on least square and Galerkin methods:
 More widely used than collocation and subdomain,
since they are truly global methods.
 For least squares method:



For Galerkin’s method:


FEA Theory
Equations to solve for a are always symmetric but tend
to be ill-conditioned.
Approximate solution needs to be very smooth.
Equations to solve for a are usually symmetric but much
more “robust”.
Integrating by parts produces “less smooth” version of
approximate solution; more useful for FEA!
-17-
2.1: Weighted residual methods (cont.)
 Example: 1D Axial Rod “dynamics”

Given: Axial rod has constant density ρ, area A, length L, and spins at
constant rate ω. It is pinned at x = 0 and has applied force -F at x = L. The
governing equation and boundary conditions for the steady-state rotation of
the rod are:
d 2u
E 2   2 x  0 for 0  x  L;
dx
du
F
u  x  0   0; E  x  L    .
dx
A

Required: Using each of the four weighted residual methods and the
approximate solution u  x   a1 x  a2 x 2, estimate the displacement of the rod.
FEA Theory
-18-
2.1: Weighted residual methods (cont.)
 Some preliminaries:
 Problem has an exact solution given by

u  x   uo *   12  Lx   16 





x 3
L

  x 
 L
FL
 A 2 L2
; uo 

.
EA
F
Approximate solution satisfies essential boundary
condition u(x = 0) = 0.
Two unknown constants → n = 2.
Notation:
V  0  x  L ; Au   x  0 ; As   x  L.
d 2u
du F
2
E  u   E 2   x; B  u   E
 .
dx
dx A
FEA Theory
-19-
2.1: Weighted residual methods (cont.)
 Solution:
1. Collocation Method -


Since n=2, have two collocation points. One must be at
x = L (must have one on As). Assume other at x = L/3.
Equation #1: evaluate residual of E(u) at x = L/3:
d2
RE  x, a   E  u approx   E 2 a1 x  a 2 x 2    2 x  2 Ea 2   2 x.
dx
 Equation #1 is: 2Ea 2  13  2 L  0.
Equation #2: evaluate residual of B(u) at x = L:
RB  x, a   B  u approx   E
FEA Theory
d
F
F
a1 x  a 2 x 2    Ea1  2 Ea 2 x  .
dx
A
A
F
 Equation #1 is: Ea1  2 Ea 2 L   0.
A
-20-
2.1: Weighted residual methods (cont.)
 Solution:
1. Collocation Method -Solve simultaneous equations:
F  2 L2
 2 L
2
a1  
+
; a2  
 uapprox  x   uo *   13  Lx   16  Lx     Lx  .


EA
3E
6E

Plot results:


FEA Theory

-21-
2.1: Weighted residual methods (cont.)
 Solution:
2. Subdomain Method -

Since n=2, have two subdomains. One must be at
x = L (= As). Other must be 0 < x < L (= V).
Equation #1: integrate residual of E(u) over V:
RE  x, a   2 Ea 2   x  I1  a    2 Ea 2   2 x dx  2 ELa 2  12  2 L2 .
L
2
0
 Equation #1 is: 2 ELa 2  12  2 L2  0.

Equation #2: evaluate residual of B(u) at x = L:
RB  x, a   B  u approx   E
FEA Theory
d
F
F
a1 x  a 2 x 2    Ea1  2 Ea 2 x  .
dx
A
A
F
 Equation #1 is: Ea1  2 Ea 2 L   0.
A
-22-
2.1: Weighted residual methods (cont.)
 Solution:
2. Subdomain Method -Solve simultaneous equations:
F  2 L2
 2 L
2
a1  
+
; a2  
 uapprox  x   uo *   12  Lx   14  Lx     Lx  .


EA
2E
4E

Plot results:


FEA Theory

-23-
2.1: Weighted residual methods (cont.)
 Solution:
3. Least Squares Method -For dimensional equality, take   1 L in ILS(a). Once again,
the “integral” over As is just evaluation at x = L.
I LS  a 
R E
1 R B
Ik a 

 x, a  R E  x, a  dV 
 x  L, a  R B  x  L, a  =0.
a k
L a k
V a k

Equation #1: take derivative with respect to a1:

RE

R

F
 x, a   2 Ea 2   2 x  0; B  x, a    Ea1  2 Ea 2 x    E.
a1
a1
a1
a1 
A
E
F
E
F
Equation #1 is:  Ea1  2 Ea 2 x     Ea1  2 Ea 2 L    0.
L
A xL L 
A
FEA Theory
-24-
2.1: Weighted residual methods (cont.)
 Solution:
3. Least Squares Method -
Equation #2: take derivative with respect to a2:
RE

R

F
 x, a   2 Ea 2   2 x  2 E; B  x, a    Ea1  2Ea 2 x    2Ex.
a 2
a 2
a 2
a 2 
A
 2 Ex 
F 
2 EF
2
2
2 2
I 2  a    2 E 2 Ea 2   2 x dx + 
E
a

2
E
a
x


2
E
a

8
E
a
L

E

L

.
 1

2
1
2
L
A
A
0

 x  L

2 EF
So Equation #2 is 2 E 2 a1  8E 2 a 2 L  E  2 L2 
 0.
A
L

Solve equations:


F  2 L2
 2 L
2
a1  
+
; a2  
 uapprox  x   uo *   12  Lx   14  Lx     Lx  .


EA
2E
4E
Same as subdomain method!
FEA Theory
-25-
2.1: Weighted residual methods (cont.)
 Solution:
4. Galerkin’s Method -
Weighting functions are

Integrate general expression for volume integral
by parts first:
W1  x   N1  x   x; W2  x   N 2  x   x 2 .
  x    2 x dx
 N k  x  R E  x, a  dV   N k  x  * Euapprox
L
V
0
xL
  x  
  N k  x  Euapprox
x 0
  x  dx
  N k  x  * Euapprox
L
0
Set this equal to
zero for k = 1,2!
+  N k  x  *   2 x dx.
L
FEA Theory
0
-26-
2.1: Weighted residual methods (cont.)
 Solution:
4. Galerkin’s Method -
Equation #1 uses N1(x)=x in previous:
  x 
I1  a    x * Euapprox

xL
x 0
  1* E  a1  2a 2 x  dx   x *  2 x dx  0.
L
L
0
0
 Equation #1 is  Ea1L  Ea 2 L2  13  2 L3 

FL
 0.
A
Equation #2 uses N2(x)=x2 in previous:
  x 
I 2  a    x * Euapprox

2
xL
x 0
  2 x * E  a1  2a 2 x  dx   x 2 *  2 x dx  0.
L
L
0
0
FL2
 Equation #2 is  Ea1 L  Ea 2 L   L 
 0.
A
2
FEA Theory
4
3
3
1
4
2
4
-27-
2.1: Weighted residual methods (cont.)
 Solution:
4. Galerkin’s Method -Solve simultaneous equations:
F 7  2 L2
 2 L
2
7
x
1 x

a1  

; a2  
 uapprox  x   uo *  12  L   4  L     Lx  .


EA
12E
4E

Plot results:


FEA Theory

-28-
Section 2: Finite Element Analysis Theory
1. Method of Weighted Residuals
2. Calculus of Variations
Two distinct ways to develop the underlying
equations of FEA!
FEA Theory
-29-
Section 2.2: Calculus of Variations
 A formal technique for associating minimum or
maximum principles with weak form equations
that can be solved approximately.
 A more physically motivated approach than
weighted residuals.
 Not all problems amenable to this technique.
FEA Theory
-30-
2.2: Calculus of Variations (cont.)
 Minimum/Maximum Principles (Variational
Principles) involve the following:
A set of equations E  u  x    0 and boundary
conditions B  u  x    0 to solve for u  x  .
 A scalar quantity J  u  x   “related” to E and B (called
a functional).
 A variational principle states that solving E  u  x    0
and B  u  x    0 is equivalent to finding the function that
gives J  u  x   a maximum or minimum value.
 Requires  J  u  x    0 -- “First variation of J  u  x  
must be zero (stationarity)”.

FEA Theory
-31-
2.2: Calculus of Variations (cont.)
 What is a functional?
 A function takes a point in space as input and returns
a scalar number as output.
E.g., u  x, y, z   x  2 y  3 z.
(Vector-valued function u  x  gives vector as output.)
 A functional takes a function as input and returns a
scalar number as output.
a
E.g.,  f  x     1 f  x 2 dx
0


Arc-length of f(x) from
x=0 to x=a.
FEA Theory
-32-
2.2: Calculus of Variations (cont.)
 A few examples:
 a, b  =  4,2  ; f1  x  =
 a, b  =  4,2  ; f 2  x  =
1
2
x
1
2
 f1 x  
x 6 
2
4

2
  1 1 dx  4.4721.
2
0
 f2  x   
4
0
1 x  dx  9.2936.
2
 Recall that straight line is shortest distance
between two points. How do we prove that?
Let  = scalar #, g  x   any function such that g  0   0  g  4  .
 Should have
FEA Theory
 f1 x  g  x    f1 x  for all  and g  x  .
-33-
2.2: Calculus of Variations (cont.)
 For a given function g  x , consider a Taylor series
expansion of arc-length formula in terms of α:
 f1  x    g  x    f1  x     *  dd

 d
 d
1 2  d2

 f1  x    g  x   2  *  d 2
  0

 d

 f1  x    g  x     d
  0 
4

0

 f  x    g  x  
1
  0

2


1   f1  x    g  x  dx 
  0
4




f
x


g
x
g
x








1
 
dx 
2


0 1   f  x    g   x 
1

 0
4
f1 x  g   x 
0
1   f1 x 

FEA Theory
2
dx
= some number β;
Assume β > 0.
-34-

2.2: Calculus of Variations (cont.)
 Suppose that α is small and negative:
 f1  x    g  x     f1  x     *  dd

1 2  d2

 f1  x    g  x    2  *  d 2
  0


 f  x    g  x  
1
  0
Negligible!

 f  x    g  x     f  x     *    f  x  !
1
1
1
Can’t happen!!!
 Same problem if β < 0 and α small and positive.
So, must have β = 0!
4
f1 x  g   x 
0
1   f1 x 

FEA Theory
2
dx  0.
-35-

2.2: Calculus of Variations (cont.)
 Integrate by parts:
4

0
x4




4


f
x
f
x




d


1
1
dx  
* g  x 
 
 g  x dx  0.
2
2
2


dx
0



1   f1 x 
 1   f1 x 
 x 0
 1   f1  x  
f1 x  g   x 
 But g  x  0   0 and g  x  4   0.

  Must equal zero!!!

f1  x 
d 


dx    
 g  x  dx  0 for any choice of g  x  .
2
2
dx  1  f  x 
0 1   f  x 
0
 1   
1

f1 x 

 constant, or f1 x   some other constant.
2
1   f1 x 
4
f1 x  g   x 
4
 f1  x   mx  b and must pass through  0,0  and  4,2   f1  x   12 x !
FEA Theory
-36-
2.2: Calculus of Variations (cont.)
 Key ideas in this “proof”:



Considered an arbitrary increment of the input
function.
Derivative of the functional forced to be zero.
This implies a certain equation must equal zero.
 Calculus of Variations gives you a “direct” way of
performing these calculations!
FEA Theory
-37-
2.2: Calculus of Variations (cont.)
 Some definitions:
 General form of a functional is


u
u
u
 2u
 nu
J  u  x     E  x, u  x  ,  x  ,  x  ,  x  , 2  x  , , n  x   dV
x
y
z
x
z


V


u
u
u
 2u
 mu
+  B  x, u  x  ,  x  ,  x  ,  x  , 2  x  , , m  x   dA.
x
y
z
x
z


A

A variation of u  x  is  u  x    v  x  ,

Note: if u  x  must satisfy some boundary conditions,
so must u  x    u  x .
FEA Theory
1.
-38-
2.2: Calculus of Variations (cont.)
FEA Theory
-39-
2.2: Calculus of Variations (cont.)
 Some properties of the variation of u  x  :
 Derivatives and variations can interchange.
 u  x  
v  x 


 u  x     v  x    
 

.
z
z
z
 z 
 Integrals and variations can interchange.
  u  x  dV    u  x  dV .
V
V

Variation of sum is sum of variations.
 u  x   u  x    u  x    u  x .

Variation of product obeys “product rule”.
 u  x  u  x   u  x   u  x    u  x  u  x .
FEA Theory
-40-
2.2: Calculus of Variations (cont.)
 Some properties of the variation of u  x  :
 “Chain rule” applies to dependent variables only!

u
 E  x, u  x  ,  x  ,
x

 E 
 nu
u
, n x  
x
,
u
x
,
  x,

z
x
 u 
E 
u
+ u  x, u  x  ,  x  ,
  x  
x
+
  u 
 nu
, n x    
z
  x 
E 
u
+
x
,
u
x
,
  x,
 nu 
x
 z n 
   nu 
 nu
, n  x     n .
z
  z 
 
FEA Theory

 nu
, n x   u
z

-41-
2.2: Calculus of Variations (cont.)
 Let’s go back to arc-length example:
 d
f
x


*
 1  
 d
 f1  x    g  x  
=

 f1  x 
4
 f  x    
1
=


 f  x    
1
0
f1  x  g   x 

1  f1  x 
Just like before!
FEA Theory

  0
 f  x .

1
 


*    f   x  

 dx
1   f   x 
2
0
2
Thus, we see that

4
1
1  f1  x  dx    1  f1  x  dx
0
4
2
 f  x    g  x 
2
1
4 2
dx

0
1
2
1
4 1
2

0
=
* 2 f1  x  *  f1  x 
dx
2
1  f1  x 


 g x
-42-
2.2: Calculus of Variations (cont.)
 Minimum/Maximum Principles (Variational
Principles) involve the following:


A set of equations E  u  x    0 and boundary
conditions B  u  x    0 to solve for u  x  .
A scalar quantity J  u  x   “related” to E and B (called
a functional).
What is the relation?
FEA Theory
-43-
2.2: Calculus of Variations (cont.)
 Let’s consider a 1D version of this:
xb
u  x   u  x  ; E  u  x    E  u  x   ; J  u  x     E  x, u , u , u   dx.
xa
 Want to minimize J(u), so require δJ(u) = 0:
xb
 J  u  x      E  x, u , u , u   dx
xa
 E

E
E


   u 
u 
 u  dx
u
u 
u 

xa 
xb
 E

E d
E d 2
   u 
u 
 u   dx  0.

2 
u
u  dx
u  dx

xa 
-44xb
FEA Theory
2.2: Calculus of Variations (cont.)
 Integrate 2nd term by parts:
x  xb
xb
 d  E  
 E

E d
x u dx  u  dx   u * u   x  dx  u   u  dx

x  xa
a
a 
x  xb
 E

  *  u  involves the boundary conditions!
 u
 x x
xb
a


Essential BC’s: E.g., u  x  xa   0 or  u  x  xb   0.
E
Natural BC’s: E.g, E
x  xa   0 or


x  xb   0.


u
u
 Other BC’s: E.g., E
x  xa   some number.

u 
FEA Theory
-45-
2.2: Calculus of Variations (cont.)
 Integrate 3rd term by parts twice:
x x
x
x
2
 d  E  d

 E d

E d
x u dx 2  u  dx   u * dx  u   x  dx  u  * dx  u  dx


x x
b
b
b
a
a
a
x  xb
x  xb
xb
 d  E 

 d 2  E 

 E d


*  u  
 
* u 
  2 
*  u  dx.


 u  dx
 x xa  dx  u  
 x xa xa  dx  u  

x  xb
x  xb
 d  E 

 E d

and  
*  u  involve BC's!

 u * dx  u  

 x xa
 dx  u  
 x xa
FEA Theory
-46-
2.2: Calculus of Variations (cont.)
 Pull all of this together:
x x
x
x
 d  E 

 E 
 E

 J  u  x    0     u  dx   *  u     
*  u  dx

u 
 u 
 x x x  dx  u  
x 

b
b
b
a
a
a
x  xb
x  xb
xb
 d  E 

 d 2  E 

 E d


*  u  
 
* u 
 2 
*  u  dx.


 u  dx
 x xa  dx  u  
 x xa xa  dx  u  

 E d  E  d 2  E  
  J u  x   0    u *   
 2
 dx


xa
 u dx  u  dx  u  
+ boundary condition terms.
xb
FEA Theory
-47-
2.2: Calculus of Variations (cont.)
 Assuming all boundary conditions are either essential
or natural, end up with:
 E d  E  d 2  E  
0  u*  
 2
 dx for any choice of  u


xa
 u dx  u  dx  u   
xb
E d  E  d 2  E 

 
 2
 0!


u dx  u   dx  u  
The Euler equation for J  u  x  
FEA Theory
-48-
2.2: Calculus of Variations (cont.)
 The “relation” between J  u  x   being minimum and
E  u  x    0 is as follows:
If you can find an operator E  x, u, u, u  such that
E d  E  d 2  E 
E u  x  
 
 2
 0,


u dx  u   dx  u  
xb
then solving  J  u  x      E  x, u , u , u   dx  0 is the
xa
same as solving E  u  x    0 .
FEA Theory
-49-
2.2: Calculus of Variations (cont.)
 Some notes:

If you have boundary conditions that neither essential nor natural,
then must explicitly include a “boundary term” in the functional.
xb
J  u  x     E  x, u, u, u  dx   B  x, u , u , u , u   
xa
x  xb
x  xa
.
(See Slide #10 for general statement of this idea.)

As number of dependent variables increases (e.g., 2D), one
functional will produce multiple Euler equations:
J  u  x, y  , v  x, y   
 

E x, y, u , v, ux , xv , uy , yv dA
Area
E   E

  u
u x   x
FEA Theory
   E
   u
 y   y

E   E    E 
  v    v   0
  0 and
v x   x  y   y 

-50-
2.2: Calculus of Variations (cont.)
 Notes:
 There are no general procedures for finding the operator
E for a given set of equations E  u  x    0.
 However, E is known for many of the more common finite
element analysis problems.
 Special case for which E can always be found:
E  u  x    M  x   u  x   b  0;
M  x   = matrix of derivative operators such that M  x   satisfies
 u  x  M  x  u  x  dV   u  x  M  x  u  x  dV  BC's
1
V
2
2
1
V
for all possible choices of u1  x  and u 2  x  . “Self-adjoint” equations
FEA Theory
-51-
2.2: Calculus of Variations (cont.)
 Notes:
 For self-adjoint equations, E and J  u  can be shown
to be:
1
E  u  x  M  x   u  x   u  x  b;
2
1

J  u     u  x  M  x   u  x   u  x  b  dV .
2

V 
(Depending on problem details, may be necessary to
integrate J  u  by parts before taking variation.)
FEA Theory
-52-
2.2: Calculus of Variations (cont.)
 Example: axial deformation of fixed rod with axial load –

f  x
d
du
 0;   x   .
 x 
dx
E
dx
u  x  0  0  u  x  L .
Can re-write governing equations as:
 0
 d
 dx
M 
FEA Theory
  u  x    f E x    0 
   .



1    x  0   0
d
dx
u x 
b
-53-
2.2: Calculus of Variations (cont.)
 Example:
 Functional is then calculated as follows:
1 u 
E  
2  
 0
 d
 dx
  u   u   f E x   1 d 1
  
 = 2 u dx  2 



1      0 
d
dx
L
J u   

1
2
u ddx  12 
du
dx
 12  2 
uf  x 
E
du
dx
 12  2 
uf  x 
E
 dx.
0

FEA Theory
Euler equations for this functional:
E d  E 
f x
  du   0  12 ddx + E   dxd   12    0, or
u dx   dx 
E d  E
  d
 dx   dx
d
dx
+
f  x
E

d 1
du
1 du

0




u

0,
or





2 dx
dx 2
dx  0.

 0.
-54-
;
2.2: Calculus of Variations (cont.)
 So, what’s all of this have to do with finite elements?
 Have a set of equations E  u  x    0 and boundary
conditions B  u  x    0 to solve for u  x  .
 Have a functional J  u  x     E x, u, ux , uy , uz dV


V
related to E and B via the Euler equations on E .
 Finite element analysis attempts to find the best
approximate solution to
 J  u  x      E  x, u, ux , uy , uz  dV  0.
V
Weak form of governing equations!
FEA Theory
-55-
2.2: Calculus of Variations (cont.)
 Look more closely at 1D version:
xb
u*
xa

E  x ,u ,u 
u
 dxd

E  x ,u ,u 
  u 
 dx  boundary terms  0.
 Suppose we make “usual” approximation –
n
u  x   uapprox  x    a k N k  x 
k 1
n
  u  x    uapprox  x     a k N k  x .
k 1
A “space” of trial functions
Must belong to same “space”
E.g., if uapprox  x   a 0  a1 x  a 2 x 2 , then  uapprox  x    a 0   a1 x   a 2 x 2 .
FEA Theory
-56-
2.2: Calculus of Variations (cont.)
 Plug in approximations (ignoring boundary terms for
now) –
xb n
  a k N k  x  *
xa k 1
xb
n

or   a k *  N k  x  *
k 1
xa


E x ,u uapprox ,uuapprox
u





E x ,u uapprox ,uuapprox
u
d
dx





E x ,u uapprox ,u uapprox
d
dx

 u 


dx  0,

dx  0.

E x ,u uapprox ,u uapprox
  u 
 Since each ak is arbitrary, best approximation comes
from
xb
 Nk  x  *
xa
FEA Theory



E x ,u uapprox ,uuapprox
u


d
dx



E x ,u uapprox ,uuapprox
  u 

dx  0, k  1, 2,
Function of a1, a2, …, an  Get n equations for n constants!
-57-
,n
2.2: Calculus of Variations (cont.)
 Notice the following:
E d  E 
If
 
 E  u  x   , then

u dx  u  


E x ,u uapprox ,uuapprox
u


d
dx



E x ,u uapprox ,uuapprox
 u 

  E u  u
xb
  N k  x  * RE  x, a  dx  0, k  1, 2,
, n.
approx
  R  x, a  .
E
Galerkin’s
method!
xa
Galerkin’s Method and Calculus of Variations give
same equations when “proper” E is used!
FEA Theory
-58-
2.2: Calculus of Variations (cont.)
 Notice something else:
  a   J  u  uapprox   J  u  uexact 
xb

  E  x, u  uapprox , u   uapprox
 dx  J u  uexact .
xa
 a k 
xb
  x, u  u
E
a k
approx

, u   uapprox
 dx
xa



E x ,u uapprox ,uuapprox


E x ,u uapprox ,uuapprox
xb


u
*
uapprox
a k



E x ,u uapprox ,uuapprox
u

xa

xb
FEA Theory xa

u

* Nk  x  

 dx

* N   x  dx
*

E x ,u uapprox ,uuapprox
u

uapprox
a k
k
-59-
2.2: Calculus of Variations (cont.)
 Integrate 2nd term by parts (and ignore boundary
terms again):

a k

xb

xa



E x ,u uapprox ,uuapprox
u
xb
  Nk  x  *
xa

xb

a k

* Nk  x  


E x ,u uapprox ,uuapprox
 0   Nk  x  *
xa


u



d
dx
d
dx




E x ,u uapprox ,u uapprox
u

E x ,u uapprox ,uuapprox
u


E x ,u uapprox ,u uapprox
u


d
dx


 * N  x dx

dx.

dx  0.

k

E x ,u uapprox ,uuapprox
u
Rayleigh-Ritz Method on  gives same equations as
J = 0 !
FEA Theory
-60-
2.2: Calculus of Variations (cont.)
 Example: 1D Axial Rod “dynamics”

Given: Axial rod has constant density ρ, area A, length L, and spins at
constant rate ω. It is pinned at x = 0 and has applied force -F at x = L. The
governing equation and boundary conditions for the steady-state rotation of
the rod are:
d 2u
E 2   2 x  0 for 0  x  L;
dx
du
F
u  x  0   0; E  x  L    .
dx
A

Required: Using the calculus of variations on an appropriate variational
principle along with the approximate solution u  x   a1 x  a2 x 2, estimate the
displacement of the rod.
FEA Theory
-61-
2.2: Calculus of Variations (cont.)
 Solution:
 Find appropriate variational principle:
2
d 2u
d
E  u   E 2   2 x is self-adjoint, with  M   E 2 and b   2 x.
dx
dx
L
E  u*E
1
2
2
d u
dx 2
 u  x  J  
2

1
2
u*E
d 2u
dx 2

 u  2 x dx.
0

Problem: there is a nonzero boundary condition –
B   12 u  x  L  * FA (Work done by applied force.)
L
 J   12 u  x  L  * FA  

1
2

u * E ddxu2  u  2 x dx.
2
0
Needs to be integrated by parts!
FEA Theory
-62-
2.2: Calculus of Variations (cont.)
 Solution:
 Doing this gives:
L
x

L
J   12 u  x  L  * FA   12 u * E du
dx x 0  

1
2
E
  dx   u  2 x dx
L
du 2
dx
0
L
=  u  x  L  * FA  

1
2
E
  dx   u  2 x dx.
L
du 2
dx
0

0
0
Require the first variation to equal zero:
L


d  u 
 J   u  x  L  * FA    u *  2 x  E du
*
dx  0.
dx
dx
0
FEA Theory
-63-
2.2: Calculus of Variations (cont.)
 Solution:
 Using the given approximate function:
u  x   a1 x  a 2 x 2   u  x    a1 x   a 2 x 2 .
 J    a1 L   a 2 L2  * FA
L


   a1 x   a 2 x 2  *  2 x  E  a1  2a 2 x  *  a1  2 a 2 x  dx  0.
0

After some integrating, result is:

FEA Theory
=0
 J   13  L  FLA  ELa1  EL a 2   a1
2 3

1
4
 L 
2 4
2
FL2
A

=0
 EL a1  EL a 2  a 2  0.
2
4
3
3
-64-
2.2: Calculus of Variations (cont.)
 Solution:
 Solve the two equations to get:


F 7  2 L2
 2 L
2
7
x
1 x

a1  

; a2  
 uapprox  x   uo *  12  L   4  L     Lx  .


EA
12E
4E
Same as Galerkin’s method solution!
FEA Theory
-65-
2.2: Calculus of Variations (cont.)
 What if we had forgotten about the BC?
 Functional becomes:
L
L
x

L
2
2
du
1
J   12 u * E du

E
dx

u

x dx




dx x 0
 2 dx



0
0
L
=  12 u  x  L  * FA  

1
2
E
  dx   u  2 x dx.
L
du 2
dx
0

So the first variation becomes:
L

0

d  u 
 J   u  x  L  * 2FA    u *  2 x  E du
*
dx  0.
dx
dx
0
Force is cut in half!
FEA Theory
-66-
2.2: Calculus of Variations (cont.)
 Solution:
 Solution becomes:


F
7  2 L2
 2 L
2
a1  

; a2  
 uapprox  x   uo *   127  Lx   14  Lx     2xL  .


2EA
12E
4E
FEA Theory
-67-