### Introduction

```EMA 405
Introduction
Syllabus
Textbook: none
 Prerequisites: EMA 214; 303, 304, or 306;
EMA 202 or 221
 Room: 2261 Engineering Hall
 Time: TR 11-12:15
 Course Materials: ecow2.engr.wisc.edu

Instructors
Jake Blanchard, Room 143 ERB,
 phone: 263-0391
 e-mail: [email protected]/* <![CDATA[ */!function(t,e,r,n,c,a,p){try{t=document.currentScript||function(){for(t=document.getElementsByTagName('script'),e=t.length;e--;)if(t[e].getAttribute('data-cfhash'))return t[e]}();if(t&&(c=t.previousSibling)){p=t.parentNode;if(a=c.getAttribute('data-cfemail')){for(e='',r='0x'+a.substr(0,2)|0,n=2;a.length-n;n+=2)e+='%'+('0'+('0x'+a.substr(n,2)^r).toString(16)).slice(-2);p.replaceChild(document.createTextNode(decodeURIComponent(e)),c)}p.removeChild(t)}}catch(u){}}()/* ]]> */
 office hours: TBD

Homeworks – 40%
 Quiz – 20%
 Design Problem – 20%
 Final Project – 20%

Schedule
Topics
Introduction
FEA Theory
Intro to ANSYS
Trusses
Plane Stress/Strain
Axisymmetric
3-D Problems
Beams
Plates
Heat Transfer
Steps
Plasticity
The finite element method
Began in 1940’s to help solve problems in
elasticity and structures
 It has evolved to solve nonlinear, thermal,
structural, and electromagnetic problems
 Key commercial codes are ANSYS,
ABAQUS, Nastran, etc.
 We’ll use ANSYS, but other codes are as
good or better (…a “religious” question)

The Process

Build a model
◦
◦
◦
◦
◦
Geometry
Material Properties
Discretization/mesh
Boundary conditions
Solve
 Postprocessing

Structural Elements
Truss
 Beams
 Planar
 3-D
 Plate

Elements
Truss
 Beam
 Planar
 Shell
 Brick

Finite Element Fundamentals

The building block of FEM is the element
stiffness matrix
3
a
1
2
a
 f1x 
f 
 1 y   k11
 f 2 x  k 21
 
 f2 y   
 f 3 x  k61
 
 f 3 y 
k12
k 22

k62
 u1 
 k16   v1 
 k 26  u 2 
 
   v2 

 k66  u3 
 
 v3 
Now Put Several Together
7
8
9
5
4
6
1
2
3
F  K U 
Global Stiffness
[K] is a composite of the
element stiffness elements
Once K is known, we can
choose forces and calculate
displacements, or choose
displacements and calculate
forces
Boundary conditions are
needed to allow solution
 f 1x 
f 
 1y 
 f 2x 


f
 2y 
 f 3x 


f
 3y 
f 
 4x 
 f 4y 
f 
F    5 x 
 f 5y 
 f 6x 


f
 6y 
 f 7x 


 f 7y 
f 
 8x 
 f 8y 
f 
 9x 
 f 9 y 
 u1 
v 
 1
u 2 
 
v 2 
u 3 
 
v3 
u 
 4
v 4 
u 
U    5 
v 5 
u 6 
 
v 6 
u 
 7
v 7 
u 
 8
v8 
 
u 9 
 v 9 
Element Stiffness
f3y
f3x
3
a
f2y
1
2
f1y
a
f1x
v3
f2x
u3
y
x
v2
v1
u1
u2
How Do We Get Element Stiffness?
 u  [ A]c
assum e
u ( x, y )  c1  c2 x  c3 y
v( x, y )  c4  c5 x  c6 y
x1  0;
y1  0
x2  a;
y2  0
x3  0;
y3  a
Rewrite as
matrix
equation
c  A1u
Coordinates
of element
corners
u1  c1 ; v1  c4
u2  c1  c2 a; v2  c4  c5 a
u3  c1  c3 a; v3  c4  c6 a
u1  1 0 0   c1 
  
 c 
u

1
a
0
 2 
 2 
u  1 0 a  c 
 3 
 3 
Substitute
coordinates
into assumed
functions
a
A1  1  1
a
 1
 c1 
a
  1
c2    1
c  a   1
 3

0 0
1 0
0 1
0 0 u1 
 

1 0 u2 
0 1 u3 
Continued…
 c1 
 a 0 0  u1 
  1
 u 
c


1
1
0
 2
 2


c  a  1 0 1 u 
 3

 3 
 c1 
Rewrite
 
u  1 x y c2 
assumed
functions
c 
 3
u  1 1
a
x
 a 0 0  u1 
 
y  1 1 0 u 2 
 1 0 1 u3 
u  1 1
a
x
 au1 


y  u1  u2 
 u  u 
 1 3 
Multiply
1
u ( x, y )  au1   u1  u2 x   u1  u3  y
a
Substitute
Continued
1
u ( x, y )  a  x  y u1  xu2  yu3
a
u ( x, y )  N1u1  N 2u 2  N 3u3
x y
N1  1  
a a
x
N2 
a
y
N3 
a
Sim ilarly
v( x, y )  N1v1  N 2 v2  N 3v3
Collect
terms
Stress-Strain
x y
N 3
u N1
N 2
N1  1  
x 

u1 
u2 
u3
a a
x x
x
x
x
u
u1 u 2
N2 
x 
 
a
x
a a
y
v
v1 v3
N3 
y    
a
y
a a
u ( x, y )  N1u1  N 2u 2  N 3u3
u v
u1 u3 v1 v2
 xy       
v ( x, y )  N1v1  N 2 v2  N 3v3
y x
a a a a
Stress-Strain
 u1 
v 
 1
x 
u 2 
 
  y   B  
 
 v2 
 xy 
u3 
 
 v3 
  1 0 1 0 0 0
1
B    0  1 0 0 0 1
a
 1  1 0 1 1 0
k    B  DB dV  tAB T D B 
T
V
Comes from
minimizing total
potential energy
(variational
principles)
Material Properties
[D] comes from
the stress-strain
equations
 For a linear,
elastic, isotropic
material

Strain Energy
 x 
x 
 
 
 y   [ D ]  y 
 
 
 xy 
 xy 


1 
0 

E
 1
[ D] 
0 
2
1  
1  
0 0

2 

T
1
U     D  dV
2 V
Final Result for Our Case
[k ] 
A
AEt
2a 2 1  2


1 
2
 3 
 1 
3 
 2

 2
 2
2

 1    1   0
 1    1   0

2
2
  2
1 2
a
2
[k ] 
Et
4 1  2


1 
2
 3 
 1 
3 
 2

 2
 2
2

 1    1   0
 1    1   0

2
2
  2
 1    1    2 
 1    1    2 
0
0
2 

1 
1 
0 
1 
1 
0 

0
0
2 
 1    1    2 
 1    1    2 
0
0
2 

1 
1 
0 
1 
1 
0 

0
0
2 
or
 f1x 
f 
 1y 
 f 2 x 
Et
 
2
f
4
1


2
y
 
 f3x 
 
 f 3 y 


1 
2
 3 
 1 
3 
 2

 2
 2
2

 1    1   0
 1    1   0

2
2
  2
 1    1    2  u1 
 1    1    2   v1 
0
0
2  u2 
 
1 
1 
0   v2 
1 
1 
0  u3 
 
0
0
2   v3 
Examples
 f1x 
f 
 1y 
 f 2 x 
Et
 
2
f
 2 y  4 1 
 f3x 
 
 f 3 y 

u1

 3  
 1  


  2 

 u1
 1  
 1  


  2 
Examples
 f1x 
f 
 1y 
 f 2 x 
Et
 
2
f
 2 y  4 1 
 f3x 
 
 f 3 y 

v1

 1  
 3  


  2 

 v1



1




 1  


  2 
Prescribe forces
F
Process
What do we know? – v1=v2=0; f3y=F; all
horizontal forces are 0
 Remove rigid body motion – arbitrarily set
u1=0 to remove horizontal translation;
hence, f1x is a reaction
 Reduce matrix to essential elements for
calculating unknown displacements – cross
out rows with unknown reactions and
columns with displacements that are 0
 Solve for displacements
 Back-solve for reaction forces

Equations
1 
2
 3 
 1 
3 
 2

 2
 2
2

 1    1   0
 1    1   0

2
2
  2
 f1x 
f 
 1y 
 0 
Et
 
2
f
4
1


 2y 
 0 
 
 F 
or

0
Et
 
0

 
2
4
1


F 
 
0
2
 0 1 

2
0



2  u2 
 
0  u3 
2   v3 
 1    1    2   0 
 1    1    2   0 
0
0
2  u2 
 
1 
1 
0  0 
1 
1 
0  u3 
 
0
0
2   v3 
Solution
u 2 
  
  2F  
0
u3  
 
Et
v 
 1 
 3
Putting 2 Together
3
4
2
a
1
1
2
a
Element 2 Stiffness Matrix
2
(3)
3
1
T=
1 (4)
Rotate 180o
2
c s 0 0 0 0
-s c 0 0 0 0
0 0 c s 0 0
0 0 -s c 0 0
0 0 0 0 c s
0 0 0 0 -s c
3 (2)
K’ = TTKT
For 180o rotation
K’=K
Just rearrange the rows and columns top correspond to
global numbering scheme (in red).
Element Matrices
 f 1x 
f 
 1y 
 f 2x 


Et
 f2y 


2
 f 3x  4 1  
 f 3y 


f
4
x


 f4y 



 f 1x 
f 
 1y 
 f 2x 


Et
 f2y 


2
 f 3x  4 1  
 f 3y 


 f 4x 
 f4y 




1 
2
 3 
 1 
3 
 2

 2
 2
2

0
  1     1   
  1     1   
0

2
2
  2
 0
0
0

0
0
 0

0
0

0

0
0

0
0

0
0
0
0
0
0
0
1 
0
0
0
0
2
0
0
2
0
1 
0
0  1     2
0  1     2
 1     1     2
 1     1     2
0
0
2
1 
1 
0
1 
0
0
0
1 
0
0
0
0
0
0
2
0
0
1 
0
2
0
2
 2
0
1 
 1   
 1   
0
2
0
0
0
0
0
0
0
0
0
0
0  u1 
0  v1 
 

0 u 2 

0 v 2 
 
0  u 3 

0  v 3 
 
0 u 4 

0 v 4 
0
0   u1 
0
0   v1 
 
 1     1    u 2 

 2
 2  v 2 
 
2
 2  u 3 

 1     1    v 3 
 
3 
1    u 4 

1 
3    v 4 
 f 1x 
f 
 1y 
 f 2x 


Et
 f2y 


2
 f 3x  4 1  
 f 3y 


f
4
x


 f4y 




1 
2
 1     1   
 2
0
0   u1 
 3 
 1 
3 
 2
 1     1   
2
0
0   v1 

 
 2

 2
3 
0
0
1 
 1     1    u 2 

 





1



1


0
3


1


0

2


2

 v 2 
 
  1     1   
0
1 
3 
0
2
 2  u 3 


2
1 
0
0
3 
 1     1    v 3 
  2
 
 0
0
 1   
 2
2
 1    3  
1    u 4 


0
 1   
2
 2
 1    1  
3    v 4 
 0
What if triangles have midside nodes?
u ( x, y)  c1  c2 x  c3 y  c4 x  c5 xy  c6 y
2
3
2
v( x, y)  c7  c8 x  c9 y  c10 x  c11 xy  c12 y 2
2
4
5
1
2
6
u( x, y)  c1  c2 x  c3 y  c4 xy
v( x, y)  c5  c6 x  c7 y  c8 xy
3
4
1
2