MAE 5563 Finite Element Methods
Professor J.K. Good
Fall 2019
Is the topic of
finite
elements
relevant to
these
amazing
machines?
Theory ~1940
 Hrennikoff
 Courant
Computers
 UNIVAC 1 1951
 IBM 360 1964
Software
 NASTRAN 1968
 ANSYS 1969
Apollo Saturn V
1966
SR71 1966
Any finite element
calculations on these
marvels were
performed by hand….
• 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?
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).
P1
P2
2. Check Body Equilibrium
s x t xy t xz


 fx  0
x
y
z
t xy s y t yz


 fy  0
x
y
z
t xz t yz s z


 fz  0
x
y
z
3. If the functional forms of sx, sy, sy, txy,
tyz, and tzx do not satisfy these equations,
start over!
4. Check surface equilibrium
y n
Tx  Ty  Tn  0
P1
x
on free boundaries
P2
nx  cos n,x  ny  cos n,y  nz  cos n,z 
s x n x  t xy n y  t xz n z  Tx
t xy n x  s y n y  t yz n z  Ty
t xz n x  t yz n y  s z n z  Tz
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.
6. Convert stresses to strains using
constitutive equations:


1
x 
s x   s y  sz 

E
1
 y  s y    s x  sz  
E


1
z 
sz   s x  s y 

E
E
E
 xy 
t xy  yz 
t yz
21  
21  
E
 zx 
t zx
21  
7. Integrate strains to yield
displacements u, v, w(x,y,z):
u
v
w
x 
y 
z 
x
y
z
u v
v w
 xy  
 yz  
y x
z y
w u
 zx 

x z
?
u, v,w welded boundary  0
8. Are these displacements (u, v,
and w) compatible with the
external constraints? Yes: You
are done! No: Start over!
Theory of Elasticity – A
“Simple” Example
• Let’s consider an end loaded
cantilever beam that we were
familiar with as undergraduates.
L
1
P
v
u
c
c
y
x
• Based on our experience:
s x  d4 xy (Euler)
sy  0
d4 2
t xy  b2 
y (Parabolic Shear Stress)
2
• 2D Body Equilibrium is satisfied by
these assumed forms:
s x t xy

 fx  0
x
y
t xy s y

 fy  0
x
y
Assume the
body intensity
forces fx and fy
are negligible
• On the upper and lower boundaries
where y=+c, surface equilibrium
dictates that the shear stress must
vanish:
 
d4 2
txy
 b2  c =0
y c
2
 d4  
2b2
c2
• 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:
b2 2 
  t xy dy    b2 
y dy  P
2

c
c
c 
c
c 
• Substituting back we have:
3P
Pxy
sx  
xy  
3
2c
I
2 3
(I= c )
3
3P
 b2 
4c
sy  0
3P  y 2 
P 2
VQ
2
t xy    1 


c

y


2

4c  c 
2I
It


• These results appear to be in exact
agreement with what we learned in
strength of materials course work.
• 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
• 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
sx
u 1
Pxy
x 
 sx  s y  

x E
E
EI
s x Pxy
v 1
y 
 s y  s x   

y E
E
EI
v u txy
P
2
2
 xy 



c y
x y
G
2IG


• Integrating the x and y strains yields:
2
2
Pxy
v
 f1  x 
2EI
Px y
u
 f  y
2EI
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:
df1  x 
Px
Py




2EI
dy
2EI
dx
P

c2  y2
2IG
2
df  y 
2


• Let’s now collect the terms which are
functions of x only, y only, and
constants:
Px 2 df1  x 
Fx  

2EI
dx
df  y  Py 2 Py 2
G  y 


dy
2EI 2IG
Pc2

2IG
a constant
• The shear strain equation can now be
rewritten as:
F x   G  y  
which means that F(x) must be some
constant we will call d and G(y) must
be some constant e and thus:
2
Pc
de  
2IG
d f1  x  Px 2
from F(x):

d
dx
2EI
d f  y
Py 2 Py 2
from G(y):


e
dy
2EI
2IG
Now we can determine f(y) and f1(x):
3
3
Py
Py
f  y  

 ey  g
6EI 3 6IG
Px
f1  x  
 dx  h
6EI
• If we substitute these expressions back into
our expression for u and v we have:
2
3
3
Px y Py
Py
u


 ey  g
2EI
6EI 6IG
Pxy2 Px3
v

 dx  h
2EI
6EI
• 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:
g0
3
PL
h=  dL
6EI
• The deflection curve for the elastic axis of
the beam can be found by setting y=0 in
v(x):
Px3 PL3

 d L  x
 v y  0 
6EI 6EI
• 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:  v 
0
 
 x x  L,y 0
• Substituting into the derivative of our
deflection equation (v)y=0 yields:
2
PL
d
• Earlier we found: 2EI
Pc2
ed  
so
2IG
PL2 Pc2
e=

2EI 2IG
• We have found all the constants now
and can write u and v:
Px 2 y Py3 Py3  PL2 Pc2 
u




y
2EI
6EI 6IG  2EI 2IG 
Pxy2 Px3 PL2 x PL3
v



2EI
6EI
2EI 3EI
• The deflection equation for the elastic
axis is:
Px3 PL2 x PL3
 v y  0 
6EI


2EI
3EI
• 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:
 u 
0
 
 y  x  L
y0
• Our equation for u was: u x  L,y  0  0
0
Px y Py
Py
u


 ey  g
2EI
6EI 6IG
2
3
3
• If we set x=L we have:
PL2 y Py3 Py3
u


 ey
2EI
6EI 6IG
and taking the derivative wrt y yields:
 u 
PL2
PL2

 e  0 e 
 
2EI
2EI
 y x  L
y0
• Earlier we found:
Pc2
ed  
so
2IG
PL2 Pc2
d=
2EI 2IG
and substituting into our earlier
equation for the deflection curve:
Px3 PL3

 d L  x
 v y  0 
6EI 6EI
we can find the deflection curve for
the elastic axis:
Px3 PL2 x PL3 Pc2



 v y  0 
L  x
6EI 2EI 3EI 2IG
• The slope of the deformation wrt x is:
2
2
2
Px
PL
Pc
 dv 



 
 dx  y  0 2EI 2EI 2IG
and will be non zero at x=L:
2
dv
Pc
 
 y  0  
2IG
 dx 
xL
• At the beam tip where x=0 the
deflection is now:
PL3 Pc2 L

 v y  0 
3EI 2IG
x 0
• 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.
• 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
E = 30*106 psi 10000 lb
 = 0.29
t = 0.5
10000 lb
u (in)
What functions of x and y would produce these
displacements?
v (in)
sx (psi)
What functions of x and y would produce these
stresses?
sy (psi)
• 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.
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:
sx
dU
1
T
dU  s  dV and
2
1
T
U=  s dV
2 Vol
x
for the body.
• 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:
WP due to
inertial forces
WP    u
T
f dV 
V
 u
T
TdS   u i  Pi 
S
WP due to
surface tractions
T
i
WP due to
concentrated loads
• The total potential is thus:
1
 =  sT dV   uT f dV 
2V
V
T
T
 u TdS   u i  Pi 
i
S
• 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
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:
u  a11  a 2 2  a 33
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.
So:
=(a1, a2, a3 )
where these constant coefficients are
three independent unknowns for
which we must solve.
• The variation in the total potential is:



 
a1 
a 2 
a 3  0
a1
a 2
a 3
• Since the forms of variation are
arbitrary:



0
0
0
a1
a2
a3
which generates 3 equations by
which we can solve for our 3
unknowns.
• Refer to Example 1.2.
Piecewise Continuous Functions:
du
u  a1 x 0  x 1
 a1
dx
du
u  a1 2  x 1  x  2
  a1
dx
du 2
11
   EA   dx 
dx 
20
12
2

du
 EA   dx  2 a1
dx 
21

2
  a1  2 a1
 2 a1  2  0
 a1
 a1  1
This yields the exact solution:
du
u 
 x s = E 1 0  x 1
dx
du
u  2  x s = E  1 1  x  2
dx
 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:
My
My
sx  
and  x  
I
EI
 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
2
stiffness:
d v M

2 EI
dx
 The strain energy is:
2
1
1
My
T
U
dVol
 s  dVol 

2 Vol
2 Vol EI2
 M and EI are at best functions of x
and thus:
U
2
1 M 
2
2
    Ey dAdx but  y dA  I
2 x EI A
A
and:
2
2
1 d v 
U    2  EI dx
2 x dx 
 Example: Use of Rayleigh-Ritz
Method on a Beam
y,v
W
x
L
 Pick displacement functions which
satisfy or can be made to satisfy
the kinematic boundary conditions
v(0) = v’(0) = 0. I selected:


 x  
v  a 1  Cos     b x 4  6L2 x 2
 2L  


 Our strain energy expression
yields:

U  EI
5a 2 5  15360abL4  12288b 2 L8
320L3

 The Work Potential results from
the surface traction:
5

9bL
2aL 
T
WP    u Tds  WaL 


5



s
 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.
 10a 5  15360bL4
2 


EI  WL 1 
0
3
 
a
320L
 24576bL8 15360aL4
9WL5

EI 
0
3
b
5
320L
 Now we have two equations with
which we can solve for a and b.
 Solving yields:
a
4
4WL 8  25


W  5120  2560  36 
b
128EI  6  960 
6
EI   960
 Now a and b can be substituted
back into the assumed
displacement function to yield the
solution:
4WL4  8  25  
 x  
v
1  Cos   

 2L  
EI 6  960 




W 5120  2560  36

128EI 6  960

 x4  6L2x 2


 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
 The exact solution from strength
of materials is:

Wx 2
v
6L2  4Lx  x 2
24EI
X(in) RR (in)
0
0.0000
2
0.0036
4
0.0142
6
0.0313
8
0.0538
10
0.0806
12
0.1105
14
0.1420
16
0.1740
18
0.2058
20
0.2372
Exact(in)
0.0000
0.0045
0.0168
0.0352
0.0584
0.0850
0.1140
0.1446
0.1761
0.2080
0.2400

%Error
#DIV/0
-19.7
-15.2
-11.2
-7.9
-5.2
-3.1
-1.9
-1.2
-1.0
-1.2
 Although the % errors are not
appealing, these are small numbers
where % error can be meaningless.
 A chart shows our solution to be
quite acceptable.
v (in) (in)
Deflection
.25
Rayleigh-Ritz
Exact
.20
.15
.10
.05
.00
0
5
10 15
X position (in)
20
 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.
 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:


 x  
v  a 1  Cos     b x 4  6L2 x 2
 2L  


a and b are known and we can solve for the
total potential:
2
2
1 d v
T
=  
 EI dx   u TdS
2
2 x  dx 
S
2 5
W L
TP1  0.0242647
EI
 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:
2 5
W L
TP2  0.025
EI
 A 3rd trial function might be the first
half of the function I chose for the
example:

 x  
v  aa 1  Cos   
 2L  

We must now resolve the problem
for a new constant coefficient aa:
aa 
32    2  WL4
5EI
and the total potential is:
W2L5
TP3  0.0216892
EI
 Finally a 4th trial function might be
the second half of my original trial
function: v  bb x 4  6L2 x 2
We must now resolve the problem
for a new constant coefficient bb:

3W
bb  
128EI

 and the total potential is:
2 5
W L
TP4  0.0210938
EI
 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.
v (in)
0,3
0,2
sol1
sol2
sol3
sol4
0,1
0
0
5
10
x (in)
15
20
 Remember all our solutions satisfy
the kinematic boundary conditions
for the cantilever: v(0)=v’(0)=0.
Which is really “best”?
 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.
Total Potential (W2L5/EI)
TP1
TP2
TP3
-0,019
-0,02
-0,021
-0,022
-0,023
-0,024
-0,025
-0,026
Trial Solutions
TP4
 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.