Lektion 1

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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
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FEM M5 Introduktion til FEM Søren
Page 4 and 5:
Elementtyper 1D 1 frihedsgrad
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Elementtyper 2D
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Elementtyper 3D
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Kx=F [k]{d}={R} K [Stivhedsmatrice]
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Løsning eksempel 1
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Matrice eksempel 2 R 1 R 2 R 3 k 1
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• Løsning eksempel 2
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Diskretisering af model Diskretiser
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Mesh 21
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Bjælkeelement. DOF = 23
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Stivhedsmatricen ∙ u 1 v 1 θ 1 u
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Stivhedsmatricen q 1 q 2 u 1 u 2 v
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Stivhedsmatricen http://people.civi
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Page 35 and 36:
Stivhedsmatricen 35
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Stivhedsmatricen q 1 q 2 N 1 N 2 V
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Stivhedsmatricen u 1 q 1 q 4 u q u
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Page 51 and 52: Stivhedsmatricen u 1 q 1 q 4 u q u
Page 57 and 58: Eksempel 1 2 3 P 0 D = 0 0 0 5
Page 59 and 60: Eksempel 59 X X X X X X 0 0 0 X X X
Page 61 and 62: Eksempel 61 X X X X 0 X X 0 0 X X X
Page 63 and 64: FEM BM5 kursusgang Introduktion til
Page 65 and 66: Modal analyse Beregningsmetoden giv
Page 67 and 68: Fjeder/masse system u = u t K u = d
Page 69 and 70: Fjeder/masse system u = u sin (ωt)
Page 71 and 72: Fjeder/masse system u = u sin (ωt)
Page 73 and 74: Fjeder/masse system f = ma f = mu K
Page 75 and 76: Fjeder/masse system f = ma f = mu K
Page 77 and 78: Fjeder/masse system c = 0 f = 0 K f
Page 79 and 80: Fjeder/masse system 0 = mu + ku K m
Page 81 and 82: Fjeder/masse system 0 = mu + ku K u
Page 83 and 84: Fjeder/masse system 0 = −muω 2 s
Page 85 and 86: Fjeder/masse system 0 = −muω 2 s
Page 87 and 88: Dæmpning c = 0 C cr = 2m k m c < c
Page 89 and 90: Dæmpning c = 0 ξ < 0,15
Page 91 and 92: Dæmpning c = 0 ξ < 0,15 ω d = ω
Page 93 and 94: Dæmpning c = 0 ξ < 0,15 ω d = 30
Page 95 and 96: Matrise system D = D sin (ωt) D =
Page 101: Mass Consistent mass metrix 0 L [N]
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Lektion 1

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