Finite Element Method

1. MAE 5563 Finite Element Methods
Professor J.K. Good
Fall 2019
SR71 1966
Apollo Saturn V
1966
Is the topic of
finite
elements
relevant to
these
amazing
machines?
Computers
 UNIVAC 1 1951
 IBM 360 1964
Theory ~1940
 Hrennikoff
 Courant
Software
 NASTRAN 1968
 ANSYS 1969
Any finite element
calculations on these
marvels were
performed by hand….

2. • Why are we here?
• Most BS level engineers have limited
stress analysis skills.
• B.S. coursework focuses on:
– Beams in bending and shear
– Thin & thick wall vessels
– Truss structures
– Students may apply FE software but cannot
attest to the validity of the results.
• Reality - Today’s structures are often
complex in shape and in material
properties. What are the stresses in the
red plate? If you model these stresses,
when do you know you have correct
results?

3. These problems were solved historically using
the Theory of Elasticity. The Stress Method
requires these steps:
1. Assume functional forms of sx, sy, sz,
txy, tyz, and tzx(x,y,z).
2. Check Body Equilibrium
3. If the functional forms of sx, sy, sy, txy,
tyz, and tzx do not satisfy these equations,
start over!
sx
x

txy
y

txz
z
 fx  0
txy
x

sy
y

tyz
z
 fy  0
txz
x

tyz
y

sz
z
 fz  0
P1
P2

4. 4. Check surface equilibrium
The Tx tractions in the boss regions
integrated over the area should yield
the applied loads P1 and P2.
4. If the assumed forms of stress
don’t satisfy these equations, start
over, assume new forms.
n
x
y
     
x y z
n cos n,x n cos n,y n cos n,z
  
sxnx  txyny  txznz  Tx
txynx  syny  tyznz  Ty
txznx  tyzny  sznz  Tz
P1
P2
0
on free boundaries
x y n
T T T
  

5. 6. Convert stresses to strains using
constitutive equations:
 
x x y z
1
E
 
  s   s  s
 
 
y y x z
1
E
 
  s   s  s
 
 
z z x y
1
E
 
  s   s  s
 
 xy 
E
2 1 
 
txy  yz 
E
2 1 
 
tyz
zx 
E
2 1 
 
tzx

6. 7. Integrate strains to yield
displacements u, v, w(x,y,z):
8. Are these displacements (u, v,
and w) compatible with the
external constraints? Yes: You
are done! No: Start over!
x 
u
x
y 
v
y
z 
w
z
xy 
u
y

v
x
yz 
v
z

w
y
zx 
w
x

u
z
u,v,wwelded boundary 
?
0

7. Theory of Elasticity – A
“Simple” Example
• Let’s consider an end loaded
cantilever beam that we were
familiar with as undergraduates.
• Based on our experience:
• 2D Body Equilibrium is satisfied by
these assumed forms:
x 4 y
2
4
xy 2
d xy (Euler) 0
d
b y (Parabolic Shear Stress)
2
s  s 
t   
xy
x
x
f 0
x y
t
s
  
 
xy y
y
f 0
x y
t s
  
 
c
c
L
x
y
P
1
u
v
Assume the
body intensity
forces fx and fy
are negligible

8. • On the upper and lower boundaries
where y=+c, surface equilibrium
dictates that the shear stress must
vanish:
• On the loaded end (x=0) the shear
stress txy integrated over the area must
be equal to the applied load if surface
equilibrium is to be satisfied:
• Substituting back we have:
• These results appear to be in exact
agreement with what we learned in
strength of materials course work.
  
t      
2
4 2
xy 2 4 2
y c
d 2b
b c =0 d
2 c
c c
2
2
xy 2 2
2
c c
b 3 P
dy b y dy P b
4 c
c
 
 
 t     
   
 
 
s     s 
 
t       
 
 
 
3
x y
3
2
2 2
xy 2
3 P Pxy 2
xy (I= c ) 0
2 I 3
c
3P y P VQ
1 c y
4c 2I It
c

9. • If in fact we look closer this may or
may not be the case. To be exact the
load P would have to be applied
parabolically to the tip of the beam
(x=0) and be reacted parabolically at
the beam root (x=L). If they are not
our stress solution may be in error at
the tip and root but by St.Venant’s
principle we can state that they
become accurate away from these
surfaces.
• St. Venant’s Principle:
If the loading distribution on a small section
of an elastic body is replaced by another
loading which has the same resultant force
and moment as the original loading, then no
appreciable changes will occur in the stresses
in the body except in the region near the
surface where the loading is altered. Rivello

10. • Now we must determine the
deformations which result from
the stress functions that have
satisfied body and surface
equilibrium. First we convert
the stress functions to strain
functions.
• Constitutive Relationships
x
x x y
u 1 Pxy
x E E EI
s

 
   s  s   
 

x
y y x
v 1 Pxy
y E E EI
s
 
 
   s  s   
 

 
xy 2 2
xy
v u P
c y
x y G 2IG
t
 
      
 

11. • Integrating the x and y strains yields:
where f(y) and f1(x) are as yet
unknown functions of y and x only.
• If we substitute u and v into the shear
strain equation we have:
• Let’s now collect the terms which are
functions of x only, y only, and
constants:
 
2
Px y
u f y
2EI
    
2
1
Pxy
v f x
2EI

 
   
 
2 2
1
2 2
df y df x
Px Py
2EI dy 2EI dx
P
c y
2IG

   
  
 
 
2
1
df x
Px
F x
2EI dx
  
 
  2 2
df y Py Py
G y
dy 2EI 2IG

  
2
Pc
2IG
a constant
  

12. • The shear strain equation can now be
rewritten as:
which means that F(x) must be some
constant we will call d and G(y) must
be some constant e and thus:
from F(x):
from G(y):
Now we can determine f(y) and f1(x):
   
F x G y
  
2
Pc
d e
2IG
  
  2
1
d f x Px
d
dx 2EI
 
  2 2
d f y Py Py
e
dy 2EI 2IG

   
 
3 3
Py Py
f y ey g
6EI 6IG

    
 
3
1
Px
f x dx h
6EI
  

13. • If we substitute these expressions back into
our expression for u and v we have:
• Now we must determine the constants e, g,
d, and h using kinematic (displacement
related) boundary conditions. First we know
at the root of the cantilever (x=L, y=0) that
u and v are zero. From our equations above:
• The deflection curve for the elastic axis of
the beam can be found by setting y=0 in
v(x):
2 3 3
Px y Py Py
u ey g
2EI 6EI 6IG

     
2 3
Pxy Px
v dx h
2EI 6EI

   
3
PL
g 0 h= - dL
6EI
 
   
3 3
y 0
Px PL
v d L x
6EI 6EI

   

14. • We now have more than one route by
which we can proceed. Our first route
will be to assume the slope of the
deflection of the root of the cantilever
will be zero:
• Substituting into the derivative of our
deflection equation (v)y=0 yields:
• Earlier we found:
• We have found all the constants now
and can write u and v:
x L,y 0
v
0
x  

 

 

 
2
PL
d
2EI
 
2 2 2
Pc PL Pc
e d so e=
2IG 2EI 2IG
   
2 3 3 2 2
Px y Py Py PL Pc
u y
2EI 6EI 6IG 2EI 2IG
 

     
 
 
 

15. • The deflection equation for the elastic
axis is:
• At x=L this yields the deflection
PL3/3EI, the value we found before in
strength of materials.
• The 2nd route we could have chosen
would be to set the slope at the root to
zero using the expression:
• Our equation for u was:
2 3 2 3
Pxy Px PL x PL
v
2EI 6EI 2EI 3EI

   
 
3 2 3
y 0
Px PL x PL
v
6EI 2EI 3EI

  
x L
y 0
u
0
y 

 


 

 
2 3 3
Px y Py Py
u ey g
2EI 6EI 6IG

     
0
x L,y 0
u 0
  

16. • If we set x=L we have:
and taking the derivative wrt y yields:
• Earlier we found:
and substituting into our earlier
equation for the deflection curve:
we can find the deflection curve for
the elastic axis:
2 3 3
PL y Py Py
u ey
2EI 6EI 6IG

    
2 2
x L
y 0
u PL PL
e 0 e
y 2EI 2EI


 

     
 

 
2 2 2
Pc PL Pc
e d so d=-
2IG 2EI 2IG
   
   
3 3
y 0
Px PL
v d L x
6EI 6EI

   
   
3 2 3 2
y 0
Px PL x PL Pc
v L x
6EI 2EI 3EI 2IG

    

17. • The slope of the deformation wrt x is:
and will be non zero at x=L:
• At the beam tip where x=0 the
deflection is now:
• The 1st term is the deformation due to
bending strain. The 2nd term is the
deformation due to shear strain. For
large L (L>20c) you will find the 1st
term is dominant. For small L (L<20c)
you find the 2nd term becomes
important.
2 2 2
y 0
dv Px PL Pc
dx 2EI 2EI 2IG

 
  
 
 
2
y 0
x L
dv Pc
dx 2IG


 
 
 
 
 
3 2
y 0
x 0
PL Pc L
v
3EI 2IG


 

18. • Theory of Elasticity
– Benefits: Closed form solutions for
stresses, strains, and deformations
as functions of x,y and z are
obtained. This can be the best
format for providing solutions to
problems to other users.
– Disadvantages: A good deal of
mathematical prowess is required to
solve non-trivial problems. It may
require much iteration and time to
come to a solution which satisfies all
the stress (kinetic) and deformation
(kinematic) boundary conditions.
• A Finite Element Solution
10000 lb
10000 lb
E = 30*106 psi
 = 0.29
t = 0.5

19. u (in)
v (in)
What functions of x and y would produce these
displacements?

20. sx (psi)
What functions of x and y would produce these
stresses?
sy (psi)

21. • This is why we are here. The
theory of elasticity is very
difficult to apply for the complex
problems we must solve.
• We must resort to other
methods which may not give us
closed form functions for the
displacements, strains and
stresses. Discrete results are
better than none!
• Anyone can execute a finite
element code. My objective is to
explain the science of the
method to you and how you
should know when your output
results are correct.

22. Potential Energy &
Equilibrium
• Definition: The total potential
energy  of an elastic body is:
= Strain Energy+Work Potential
(U) (WP)
• Definition: U is the strain energy
due to stresses and strains within
the elastic body. For linear elastic
problems at a point in space the
strain energy is:
   
   
T
T
Vol
1
dU dV and
2
1
U= dV
2
 s 
s 

dU
sx
x for the body.

23. • The Work Potential (WP) is the ability
of external and internal forces to
perform work on the body as these
forces move through deformations
which resulted from the stresses and
strains. The WP is always defined to
be negative in expressions although
calculation may result in a positive
contribution to the total potential:
   
       
T
V
T T
i i
i
S
WP u f dV
u T dS u P
  

 

WP due to
inertial forces
WP due to
surface tractions
WP due to
concentrated loads

24. • The total potential is thus:
• Principle of Minimum Potential
Energy: “Among all compatible
states of deformation of an elastic
body in stable equilibrium, the true
deformations are those for which
the total potential is a minimum.” -
Rivello, Theory and Analysis of
Aircraft Structure, McGraw-Hill,
1969.
• Refer to Example 1.1
 =
1
2
s
 T 
 
V
 dV  u
 T f
 
V
 dV 
u
 T T
 
S
 dS  ui
 T
Pi
 
i


26. The Rayleigh-Ritz Method
• We have seen how the PMPE applies
to linearly elastic discrete problems.
Now we will apply it to continuums.
• A set of deformations are assumed.
The form of deformation may be
polynomial functions but need not be:
where a1, a2, and a3 are constant
coefficients for which we shall solve.
1, 2, and 3 may be polynomials,
transcendentals, etc.
• The deformation function (u) must
satisfy kinematic boundary
conditions.
• We then develop the total potential as
a function of the constant coefficients.
u  a11  a22  a33

27. So: =(a1, a2, a3 )
where these constant coefficients are
three independent unknowns for
which we must solve.
• The variation in the total potential is:
• Since the forms of variation are
arbitrary:
which generates 3 equations by
which we can solve for our 3
unknowns.
• Refer to Example 1.2.
 

a1
a1 

a2
a2 

a3
a3  0
1 2 3
0 0 0
a a a
  
  
  

30. u  1
a x 0  x 1
 
du
dx
 1
a
u  1
a 2  x
  1 x  2
 
du
dx
  1
a
 
1
2
EA
2
du
dx





 dx
0
1
 
1
2
EA
2
du
dx





 dx  2 1
a
1
2

Piecewise Continuous Functions:
  1
2
a  2 1
a

 1
a
 2 1
a  2  0

 1
a 1
u  x s = E
du
dx
1 0  x 1
 
u  2  x s = E
du
dx
 1 1 x  2
 
This yields the exact solution:

31.  The expression for the Total
Potential can be applied to
different types of structure other
than the rod in Example 1.2.
 Beam structures are examples.
Classic beams are defined as
structures whose length dimension
exceeds any cross sectional
dimension by a factor of ten.
 In such cases the only substantial
stresses and strains are calculated
from Euler’s expressions:
sx  
My
I
and x  
My
EI

32.  Also recall that the curvature of a
beam (2nd derivative of the
displacement) at some location x is
equal to the bending moment at that
location divided by the bending
stiffness:
 The strain energy is:
 M and EI are at best functions of x
and thus:
and:
d2
v
dx2 
M
EI
U 
1
2
M
EI




2
Ey2
A

x
 dAdx but y2
dA  I
A

   
2
T
2
Vol Vol
1 1 My
U dVol dVol
2 2 EI
 s  
 
U 
1
2
d2
v
dx2






2
EI
x
 dx

33.  Example: Use of Rayleigh-Ritz
Method on a Beam
 Pick displacement functions which
satisfy or can be made to satisfy
the kinematic boundary conditions
v(0) = v’(0) = 0. I selected:
 Our strain energy expression
yields:
 
4 2 2
x
v a 1 Cos b x 6L x
2L
 

 
   
 
 
 
 
W
x
y,v
L
 
2 5 4 2 8
3
5a 15360abL 12288 b L
U EI
320 L
   



34.  The Work Potential results from
the surface traction:
 The Total Potential  is the sum of
the strain energy and the work
potential. Now we minimize the
Total Potential with respect to the
unknown coefficients a and b.
 Now we have two equations with
which we can solve for a and b.
WP   uT
s
 Tds  W aL 
9bL5
5

2aL








a

10a5
15360bL4
320L3 EI  WL 1
2





 0

b

24576bL8
15360aL4
320L3 EI 
9WL5
5
 0

35.  Solving yields:
 Now a and b can be substituted
back into the assumed
displacement function to yield the
solution:
a 
4WL4
8  25
 
EI 6  960
 
 
 
6
6
W 5120 2560 3
b
128EI 960
   
 
 
 
 
 
 
 
4
6
6
4 2 2
6
4 WL 8 25 x
v 1 Cos
2L
EI 960
W 5120 2560 3
x 6L x
128EI 960
   

 
 
 
 
 
 
 
  
 
 

36.  How accurate is the solution?
 Without comparison you cannot be
sure. You can expect that the
values of a and b are the best
possible solutions for the
displacement functions chosen
because the Principle of Minimum
Potential Energy has been
employed.
 We can substitute numerical values
for the inputs and compare:
 W=10 lb/in
 E = 10,000,000 psi
 I = 1/12 in4
 L = 20 in

37.  The exact solution from strength
of materials is:
 
2
2 2
Wx
v 6L 4Lx x
24EI
  
X(in) RR (in) Exact(in) %Error
0 0.0000 0.0000 #DIV/0
2 0.0036 0.0045 -19.7
4 0.0142 0.0168 -15.2
6 0.0313 0.0352 -11.2
8 0.0538 0.0584 -7.9
10 0.0806 0.0850 -5.2
12 0.1105 0.1140 -3.1
14 0.1420 0.1446 -1.9
16 0.1740 0.1761 -1.2
18 0.2058 0.2080 -1.0
20 0.2372 0.2400 -1.2

38.  Although the % errors are not
appealing, these are small numbers
where % error can be meaningless.
 A chart shows our solution to be
quite acceptable.
.00
.05
.10
.15
.20
.25
0 5 10 15 20
X position (in)
Deflection
(in)
Rayleigh-Ritz
Exact
v
(in)

39.  In general we wish to solve
problems where exact answers are
not known. The trial functions we
choose for the deformation may be
good or bad choices for
representing the true deformations.
 How do we decide we have
obtained a good solution?
 We must select several trial
functions, use the theory of minimum
total potential to evaluate unknown
coefficients, and compare the total
potentials that resulted from all trial
functions.
 Whichever trial function yields the
lowest total potential will be the best
solution by our theory.
 If several of our trial functions yield
comparable total potentials, we gain
confidence that our solution(s) are
good.

40.  In our example of the uniformly loaded
cantilever we might propose several
solutions.
 The 1st solution might be the trial function
proposed in the example:
a and b are known and we can solve for the
total potential:
 Let’s pretend we do not know the exact
solution is exact and that it is just another
trial solution candidate. It’s total potential
is:
 
4 2 2
x
v a 1 Cos b x 6L x
2L
 

 
   
 
 
 
 
2 5
W L
TP1 0.0242647
EI
 
2 5
W L
TP2 0.025
EI
 
   
2
2
T
2
x S
1 d v
= EI dx u T dS
2 dx
 
 
 
 
 
 

41.  A 3rd trial function might be the first
half of the function I chose for the
example:
We must now resolve the problem
for a new constant coefficient aa:
and the total potential is:
 Finally a 4th trial function might be
the second half of my original trial
function:
We must now resolve the problem
for a new constant coefficient bb:
x
v aa 1 Cos
2L
 

 
 
 
 
 
 
  4
5
32 2 WL
aa
EI



2 5
W L
TP3 0.0216892
EI
 
 
4 2 2
v bb x 6L x
 
3W
bb
128EI
 

42.  and the total potential is:
 If we produce a chart of our
solutions we can see how much
they differ. We see solutions 1 and
2 are similar. We see solutions 3
and 4 are similar.
 Remember all our solutions satisfy
the kinematic boundary conditions
for the cantilever: v(0)=v’(0)=0.
Which is really “best”?
2 5
W L
TP4 0.0210938
EI
 
0
0.1
0.2
0.3
0 5 10 15 20
v
(in)
x (in)
sol1 sol2
sol3 sol4

43.  How do we evaluate what the best
solution is? Our 2nd solution had
the minimum total potential. By the
Theory of Minimum Total Potential
it is the best solution.
-0.026
-0.025
-0.024
-0.023
-0.022
-0.021
-0.02
-0.019
TP1 TP2 TP3 TP4
Total
Potential
(W
2
L
5
/EI)
Trial Solutions

44.  Read Chapter 1 in
Chandrupatla and Belegundu.
This chapter discusses
principles we will use to
develop and understand finite
element theory.
 We will ensure we understand
these theories by first applying
them to continuum problems.
 I will assign homework next
lecture. Have you ever done
symbolic mathematics using
codes like Mathematica,
Mathcad or Matlab? These are
all available to you through
OSU. I suggest you download
one and be prepared to use it.