EMA 405 Introduction
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EMA 405Introduction
SyllabusTextbook: nonePrerequisites: EMA 214; 303, 304,
or 306; EMA 202 or 221Room: 2261 Engineering HallTime: TR 11-12:15Course Materials:
ecow2.engr.wisc.edu
InstructorsJake Blanchard, Room 143 ERB, phone: 263-0391e-mail: [email protected] hours: TBD
GradingHomeworks – 40%Quiz – 20%Design Problem – 20%Final Project – 20%
ScheduleTopicsIntroductionFEA TheoryIntro to ANSYSTrussesPlane Stress/StrainAxisymmetric3-D ProblemsBeamsPlatesHeat TransferMultiple Load StepsPlasticity
The finite element methodBegan 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◦Load
SolvePostprocessing
Structural ElementsTrussBea
msPlana
r3-DPlate
ElementsTrussBea
mPlan
arShellBrick
Finite Element Fundamentals
The building block of FEM is the element stiffness matrix
1
3
2
a
a
3
3
2
2
1
1
666261
262221
161211
3
3
2
2
1
1
v
u
v
u
v
u
kkk
kkk
kkk
f
f
f
f
f
f
y
x
y
x
y
x
Now Put Several Together
45
1
23
6
987
UKF
Global Stiffness
y
x
y
x
y
x
y
x
y
x
y
x
y
x
y
x
y
x
f
f
f
f
f
f
f
f
f
f
f
f
f
f
f
f
f
f
F
9
9
8
8
7
7
6
6
5
5
4
4
3
3
2
2
1
1
9
9
8
8
7
7
6
6
5
5
4
4
3
3
2
2
1
1
v
u
v
u
v
u
v
u
v
u
v
u
v
u
v
u
v
u
U
[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
Element Stiffness
1
3
2
a
a
u3
u2u1
v2v1
v3
f3x
f1x f2x
f2y
f1y
f3y
y
x
How Do We Get Element Stiffness?
accvaccu
accvaccu
cvcu
ayx
yax
yx
ycxccyxv
ycxccyxu
assume
643313
542212
4111
33
22
11
654
321
;
;
;
;0
0;
0;0
),(
),(
3
2
1
3
2
1
1
1
3
2
1
3
2
1
101
011
001
101
011
001
01
01
001
][
u
u
ua
ac
c
c
a
aA
uAc
c
c
c
a
a
u
u
u
cAu
Coordinates of element corners
Substitute coordinates into assumed functions
Rewrite as matrix
equation
Continued…
3
2
1
3
2
1
3
2
1
3
2
1
101
011
00
11
1
101
011
001
u
u
ua
yxa
u
c
c
c
yxu
u
u
ua
ac
c
c
yuuxuuaua
yxu
uu
uu
au
yxa
u
31211
31
21
1
1),(
11
Rewrite assumed functions
Substitute
Multiply
Continued
332211
3
2
1
332211
321
),(
1
),(
1),(
vNvNvNyxv
Similarlya
yN
a
xN
a
y
a
xN
uNuNuNyxu
yuxuuyxaa
yxu
Collect terms
Stress-Strain
332211
332211
3
2
1
),(
),(
1
vNvNvNyxv
uNuNuNyxua
yN
a
xN
a
y
a
xN
a
v
a
v
a
u
a
u
x
v
y
u
a
v
a
v
y
va
u
a
u
x
u
ux
Nu
x
Nu
x
N
x
u
xy
y
x
x
2131
31
21
33
22
11
Stress-Strain
BDBtAdVBDBk
aB
v
u
v
u
v
u
B
TT
V
xy
y
x
011011
100010
0001011
3
3
2
2
1
1
Comes from minimizing total potential energy
(variational principles)
Material Properties
[D] comes from the stress-strain equations
For a linear, elastic, isotropic material
dVDU
ED
D
T
V
xy
y
x
xy
y
x
2
12
100
01
01
1][
][
2
Strain Energy
Final Result for Our Case
200222
011011
011011
200222
211231
211213
14][
2
1
200222
011011
011011
200222
211231
211213
12][
2
2
22
Etk
aA
a
AEtk
or
3
3
2
2
1
1
2
3
3
2
2
1
1
200222
011011
011011
200222
211231
211213
14
v
u
v
u
v
u
Et
f
f
f
f
f
f
y
x
y
x
y
x
Examples
12
3
3
2
2
1
1
2
1
1
2
1
3
14u
Et
f
f
f
f
f
f
y
x
y
x
y
x
u1
Examples
v1
12
3
3
2
2
1
1
2
1
1
2
3
1
14v
Et
f
f
f
f
f
f
y
x
y
x
y
x
Prescribe forces
F
ProcessWhat do we know? – v1=v2=0; f3y=F; all
horizontal forces are 0Remove 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 displacementsBack-solve for reaction forces
Equations
3
3
2
2
3
3
2
22
1
1
202
010
202
140
0
0
0
0
200222
011011
011011
200222
211231
211213
14
0
0
v
u
uEt
F
or
v
u
uEt
F
f
f
f
y
y
x
Solution
1
02
3
3
2
Et
F
v
u
u
Putting 2 Together
a
a
1
2
1 2
3 4
Element 2 Stiffness Matrix
1
3
2
1 (4)
3 (2)
2(3)Rotate
180o
c s 0 0 0 0-s c 0 0 0 00 0 c s 0 00 0 -s c 0 00 0 0 0 c s0 0 0 0 -s c
T =
K’ = TTKTFor 180o rotationK’=KJust rearrange the rows and columns top correspond to global numbering scheme (in red).
4
4
3
3
2
2
1
1
2
4
4
3
3
2
2
1
1
00000000
00000000
00200222
00011011
00011011
00200222
00211231
00211213
14
v
u
vu
v
u
v
u
Et
f
f
ff
f
f
f
f
y
x
y
x
y
x
y
x
4
4
3
3
2
2
1
1
2
4
4
3
3
2
2
1
1
31122100
13122100
11100100
22022000
22022000
11100100
00000000
00000000
14
v
u
vu
v
u
v
u
Et
f
f
ff
f
f
f
f
y
x
y
x
y
x
y
x
Ele
men
t 1
Ele
men
t 2
Element Matrices
4
4
3
3
2
2
1
1
2
4
4
3
3
2
2
1
1
31122100
13122100
11300122
22031011
22013011
11100322
00211231
00211213
14
v
u
vu
v
u
v
u
Et
f
f
ff
f
f
f
f
y
x
y
x
y
x
y
x
Add the element matrices
What if triangles have midside nodes?
3
4
5
26
1
21211
210987
265
24321
),(
),(
ycxycxcycxccyxv
ycxycxcycxccyxu
What about a quadrilateral element?
3 4
21
xycycxccyxv
xycycxccyxu
8765
4321
),(
),(
What about arbitrary shapes?
For most problems, the element shapes are arbitrary, material properties are more general, etc.
Typical solution is to integrate stiffness solution numerically
Typically gaussian quadrature, 4 points
dVBDBkT
V
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