Lektion 1

FEM
M5
Introduktion til FEM
Søren Heide Lambertsen

Hvad anvender man FEA til.

Elementtyper
1D
1 frihedsgrad

Elementtyper
2D
2-3 frihedsgrader

Elementtyper
2D

Elementtyper
3D
6 frihedsgrader(Bjælke) 3 frihedsgrader for Solid

Elementtyper
3D

Eksempel 1
K 1
D 1
D 2

Kx=F
[k]{d}={R}
K
[Stivhedsmatrice] {Flytninger}= {kraftvektor}
F
x

Eksempel 1
R 1 R 2
K 1
[k]{d}={R}
D 1
D 2

Løsning eksempel 1

Eksempel 2
k 1 k 2
U 1
U 2
U 3

Matrice eksempel 2
R 1 R 2 R 3
k 1 k 2
U 1
L 1 L
U 2
2
U 3

• Løsning eksempel 2

• Løsning eksempel 2

Pause
18

Diskretisering af model
Diskretisere
19

Elementtyper
20

Mesh
21

Convergense
22

Bjælkeelement.
DOF =
23

Stivhedsmatricen
6X6 matrice
∙
=
24

Stivhedsmatricen
∙
u 1
v 1
θ 1
u = 2
v 2
θ 2
R u1
R v1
R θ1
R u2
R v2
R θ2
25

Stivhedsmatricen
∙
u 1
v 1
θ 1
u = 2
v 2
θ 2
R u1
R v1
R θ1
R u2
R v2
R θ2
26

Stivhedsmatricen
q 1 q 2
u 1 u 2
v 1
v 2
EA/L 0 0
∙
u 1
v 1
θ 1
u = 2
v 2
θ 2
R u1
R v1
R θ1
R u2
R v2
R θ2
27

Stivhedsmatricen
q 1 q 2
u 1 u 2
v 1
v 2
EA/L 0 0
−EA/L 0 0
∙
u 1
v 1
θ 1
u = 2
v 2
θ 2
R u1
R v1
R θ1
R u2
R v2
R θ2
28

Stivhedsmatricen
http://people.civil.aau.dk/~lda/Notes/
Deformationsmetoden for Rammekonstruktioner
Side 72
29

Stivhedsmatricen
30

Stivhedsmatricen
q 1 q 2
u 1 u 2
v 1
v 2
EA/L 0 0
−EA/L 0 0
∙
u 1
v 1
θ 1
u = 2
v 2
θ 2
R u1
R v1
R θ1
R u2
R v2
R θ2
31

Stivhedsmatricen
32

Stivhedsmatricen
q 1 q 2
u 1 u 2
v 1
v 2
EA/L 0 0
0 12EI/L 3
−EA/L 0 0
∙
u 1
v 1
θ 1
u = 2
v 2
θ 2
R u1
R v1
R θ1
R u2
R v2
R θ2
33

Stivhedsmatricen
q 1 q 2
u 1 u 2
v 1
v 2
EA/L 0 0
0 12EI/L 3
−EA/L 0 0
∙
u 1
v 1
θ 1
u = 2
v 2
θ 2
R u1
R v1
R θ1
R u2
R v2
R θ2
34

Stivhedsmatricen
35

Stivhedsmatricen
q 1 q 2
u 1 u 2
v 1
v 2
EA/L 0 0
0 12EI/L 3 6EA/L
−EA/L 0 0
∙
u 1
v 1
θ 1
u = 2
v 2
θ 2
R u1
R v1
R θ1
R u2
R v2
R θ2
36

Stivhedsmatricen
q 1 q 2
u 1 u 2
v 1
v 2
EA/L 0 0
0 12EI/L 3 6EA/L
0
−EA/L 0 0
∙
u 1
v 1
θ 1
u = 2
v 2
θ 2
R u1
R v1
R θ1
R u2
R v2
R θ2
37

Stivhedsmatricen
38

Stivhedsmatricen
q 1 q 2
u 1 u 2
v 1
v 2
EA/L 0 0
0 12EI/L 3 6EA/L
0 −6EA/L 2
−EA/L 0 0
∙
u 1
v 1
θ 1
u = 2
v 2
θ 2
R u1
R v1
R θ1
R u2
R v2
R θ2
39

Stivhedsmatricen
40

Stivhedsmatricen
q 1 q 2
N 1 N 2
V 1
V 2
EA/L 0 0
0 12EI/L 3 6EA/L
0 6EA/L 2 4EA/L
−EA/L 0 0
∙
u 1
v 1
θ 1
u = 2
v 2
θ 2
R u1
R v1
R θ1
R u2
R v2
R θ2
41

Stivhedsmatricen
u 1
q 1 q 4
u q
u 3
4 3
q 2
u 2
q 5
u 5
V 1
V 4
V 3
V 2
V 5
Frihedsgrader og matrice
42

Stivhedsmatricen
u 1
q 1 q 4
u q
u 3
4 3
q 2
u 2
q 5
u 5
V 1
V 4
V 3
V 2
V 5
3 frihedsgrader i hvert punkt
15X15 matrice
43

Stivhedsmatricen
u 1
q 1 q 4
u q
u 3
4 3
q 2
u 2
q 5
u 5
V 1
V 4
V 3
V 2
V 5
44

Stivhedsmatricen
u 1
q 1 q 2
u q
u 3
2 3
q 4
u 4
q 5
u 5
V 1
V 2
V 3
V 4
V 5
X
45

Stivhedsmatricen
u 1
q 1 q 2
u q
u 3
2 3
q 4
u 4
q 5
u 5
V 1
V 2
V 3
V 4
V 5
X X
X XX X
X XX X
X XX X
X X
46

Stivhedsmatricen
u 1
q 1 q 4
u q
u 3
4 3
q 2
u 2
q 5
u 5
V 1
V 4
V 3
V 2
V 5
Nyt nummersystem.
47

Stivhedsmatricen
u 1
q 1 q 4
u q
u 3
4 3
q 2
u 2
q 5
u 5
V 1
V 4
V 3
V 2
V 5
X
X
X
X
48

Stivhedsmatricen
u 1
q 1 q 4
u q
u 3
4 3
q 2
u 2
q 5
u 5
V 1
V 4
V 3
V 2
V 5
X
X
X
X
49

Stivhedsmatricen
u 1
q 1 q 4
u q
u 3
4 3
q 2
u 2
q 5
u 5
V 1
V 4
V 3
V 2
V 5
X X
X X
50

Stivhedsmatricen
u 1
q 1 q 4
u q
u 3
4 3
q 2
u 2
q 5
u 5
V 1
V 4
V 3
V 2
V 5
X
X
X
X
51

Stivhedsmatricen
u 1
q 1 q 4
u q
u 3
4 3
q 2
u 2
q 5
Nu 5
V 1
V 4
V 3
V 2
V 5
X
X
X
X
52

Stivhedsmatricen
u 1
q 1 q 4
u q
u 3
4 3
q 2
u 2
q 5
u 5
V 1
V 4
V 3
V 2
V 5
X
X
X
X
X
X
XX
53

Stivhedsmatricen
u 1
q 1 q 4
u q
u 3
4 3
q 2
u 2
q 5
u 5
V 1
V 4
V 3
V 2
V 5
X
X
X
X X
X XX X
X XX
54

Stivhedsmatricen
u 1
q 1 q 4
u q
u 3
4 3
q 2
u 2
q 5
u 5
V 1
V 4
V 3
V 2
V 5
X
X
X
XX X X
X XX X
X XX
X
X
55

Eksempel
P
1 2 3
P
56

Eksempel
1 2 3
P
0
D =
0
0
0
57

Eksempel
1 2 3
P
R =
R 1N
R 1V
0
0
−P
0
R 3N
R 3V
0
58

Eksempel
59
X
X
X
X
X
X
0
0
0
X
X
X
X
X
X
0
0
0
X
X
X
X
X
X
0
0
0
X
X
X
XX
XX
XX
X
X
X
X
X
X
XX
XX
XX
X
X
X
X
X
X
XX
XX
XX
X
X
X
0
0
0
X
X
X
X
X
X
0
0
0
X
X
X
X
X
X
0
0
0
X
X
X
X
X
X
∙
0
0
0
0
=
R 1N
R 1V
0
0
−P
0
R 3N
R 3V
0

Eksempel
60
X
X
X
X
0
X
X
0
0
X
X
X
X
0
X
X
0
0
X
X
X
X
0
X
X
0
0
XX
XX
XX
X
XX
X
X
X
X
XX
XX
XX
X
XX
X
X
X
X
XX
XX
XX
X
XX
X
X
X
X
X
X
X
0
X
0
0
X
X
X
X
X
0
X
0
0
X
X
X
X
X
0
X
0
0
X
X
∙
0
0
0
0
=
0
−P
0 0
0
R 1N
R 1V
R 3N
R 3V

Eksempel
61
X
X
X
X
0
X
X
0
0
X
X
X
X
0
X
X
0
0
X
X
X
X
0
X
X
0
0
XX
XX
XX
X
XX
X
X
X
X
XX
XX
XX
X
XX
X
X
X
X
XX
XX
XX
X
XX
X
X
X
X
X
X
X
0
X
0
0
X
X
X
X
X
0
X
0
0
X
X
X
X
X
0
X
0
0
X
X
∙
0
0
0
0
=
0
−P
0 0
0
R 1N
R 1V
R 3N
R 3V
k reduceret
k 12

Eksempel
k reduceret D = R
D = R k
−1
reduceret
Reaktioner bliver da:
R reaktioner = k 12 D
62

FEM
BM5 kursusgang
Introduktion til FEM (Modal analyse)
Søren Heide Lambertsen

I dag
1. Hvad er modal analyse
2. Egenfrekvens,Dæmpning
3. Beregning´s grundlaget for modal analyse
4. Eksempel 1
5. Opsætning af modal analyse i Ansys workbench
6. Opsætning af Prestress modal analyse i Ansys Workbench

Modal analyse
Beregningsmetoden giver mulighed for at bestemme egenfrekvenser
for en struktur.

Fjeder/masse system
u = u t
K
m
u
F

Fjeder/masse system
u = u t
K
u = du
dt
m
u
F

Fjeder/masse system
u = u t
K
u = du
dt
m
u
a = u = du2
d 2 t
F

Fjeder/masse system
u = u sin (ωt)
K
m
u
F

Fjeder/masse system
u = u sin (ωt)
K
u = uω cos(ωt)
m
u
F

Fjeder/masse system
u = u sin (ωt)
K
u = uω cos(ωt)
a = u = −uω 2 sin (ωt)
m
u
F

Fjeder/masse system
f = ma
f = mu
K
m
u
F

Fjeder/masse system
f = ma
f = mu
K
f = mu
f − ku − cu = mu
m
u
F

Fjeder/masse system
f = ma
f = mu
K
f = mu
f − ku − cu = mu
f = mu + cu + ku
m
u
F

Fjeder/masse system
f = ma
f = mu
K
f = mu
f − ku − cu = mu
f = mu + cu + ku
m
u
F

Fjeder/masse system
c = 0 f = 0
K
m
u
F

Fjeder/masse system
c = 0 f = 0
K
f = mu + cu + ku
m
u
F

Fjeder/masse system
c = 0 f = 0
K
f = mu + cu + ku
0 = mu + ku
m
u
F

Fjeder/masse system
0 = mu + ku
K
m
u
F

Fjeder/masse system
0 = mu + ku
K
u = u sin (ωt)
m
u
F

Fjeder/masse system
0 = mu + ku
K
u = u sin (ωt)
u = −uω 2 sin (ωt)
m
u
F

Fjeder/masse system
0 = mu + ku
K
u = u sin (ωt)
u = −uω 2 sin (ωt)
m
u
0 = −muω 2 sin (ωt) + ku sin (ωt)
F

Fjeder/masse system
0 = −muω 2 sin (ωt) + ku sin (ωt)
K
m
u
F

Fjeder/masse system
0 = −muω 2 sin (ωt) + ku sin (ωt)
K
0 = −mω 2 + k
m
u
F

Fjeder/masse system
0 = −muω 2 sin (ωt) + ku sin (ωt)
K
0 = −mω 2 + k
ω =
k m
m
u
F

Dæmpning

Dæmpning
c = 0
C cr = 2m
k m
c < c cr

Dæmpning
c = 0
C cr = 2m
k m
c < c cr
ω d = ω 1 − ξ 2 ξ = C
C cr

Dæmpning
c = 0
ξ < 0,15

Dæmpning
c = 0
ξ < 0,15
ω d = ω 1 − ξ 2

Dæmpning
c = 0
ξ < 0,15
ω d = ω 1 − ξ 2
ω d = 30
1 − 0,15 2 = 29,66 rad/s

Dæmpning
c = 0
ξ < 0,15
ω d = 30
ω d = ω 1 − ξ 2
1 − 0,15 2 = 29,66 rad/s
ω d = 30
2π = 4,77Hz
ω = 29,66
2π
= 4,72Hz

Dæmpning
c = 0
ξ < 0,15
ω d = 30
ω d = ω 1 − ξ 2
1 − 0,15 2 = 29,66 rad/s
ω d = 30
2π = 4,77Hz
ω = 29,66
2π
ω d ≈ ω
= 4,72Hz

Matrise system
D = D sin (ωt)

Matrise system
D = D sin (ωt)
D
= − D ω 2 sin (ωt)

Matrise system
D = D sin (ωt)
D
= − D ω 2 sin (ωt)
0 = [m]{D} + [k]{D}

Matrise system
D = D sin (ωt)
D
= − D ω 2 sin (ωt)
0 = [m]{D} + [k]{D}
0 = − m D ω 2 sin ωt + [k] D sin (ωt)

Matrise system
D = D sin (ωt)
D
= − D ω 2 sin (ωt)
0 = [m]{D} + [k]{D}
0 = − m D ω 2 sin ωt + [k] D sin (ωt)

Matrise system
D = D sin (ωt)
D
= − D ω 2 sin (ωt)
0 = [m]{D} + [k]{D}
0 = − m D ω 2 sin ωt + [k] D sin (ωt)
0 = − m D ω 2 + [k] D 0 = (−[m]ω 2 + [k]) D

Mass
Mass lumping
m = m 2
1 0
0 1

Mass
Consistent mass metrix
0
L
[N] T N ρA dx =
Beam:
m
420
156 22L 54 −13L
22L 4L 2 13L −3L 2
54 13L 156 −22L
−13L −3L 2 −22L 4L 2

Eksempel
EI 12 −6L m
L 3 −6L −4L2 − ω2
420
156 −22L
−22L −4L 2 v 2
θ 2
= 0 0
ω 1 = 3,533
ω 2 = 34,81
1/2
EI
mL 3
1/2
EI
mL 3
102

Eksempel
ω 1 = 3,533
ω 2 = 34,81
1/2
EI
mL 3
1/2
EI
mL 3
103