Chapter 16

Structural Dynamics 10/14/08

Dynamic Analysis

1

subjected to a time dependent load.

k F(t) m

10/14/08

Dynamic Analysis

2

Free-body diagram of the mass.

T = kx m

10/14/08

F(t)

=

Dynamic Analysis

m

ma = mx

3

F( t ) − k x = m x m x + k x = F( t ) 10/14/08

Dynamic Analysis

4

Solution of D.E. is sum of homogeneous and particular solutions:

Homogeneous :

F( t ) = 0 m x + k x = 0 10/14/08

Dynamic Analysis

5

Let : k ω = m Then : 2

x + ω x = 0 2

ω 2 is the natural circular frequency 10/14/08

Dynamic Analysis

6

2π τ= ω τ is the period (measured in seconds) 10/14/08

Dynamic Analysis

7

Displacement due to simple harmonic motion.

xm

10/14/08

τ

Dynamic Analysis

8

One Dimensional Bar Element

10/14/08

Dynamic Analysis

9

Step 1 - Select Element Type dˆ1x fˆ1ex ( t )

1

2

dˆ 2 x

xˆ

fˆ2ex ( t )

L E - modulus of elasticity A - cross-sectional area ρ - mass density 10/14/08

Dynamic Analysis

10

Step 2 - Select a Displacement Function uˆ = a1 + a 2 xˆ uˆ = N 1dˆ1x + N 2dˆ 2 x xˆ N1 = 1 − L xˆ N2 = L 10/14/08

Dynamic Analysis

11

Step 3 - Define Strain/Displacement and Stress/Strain Relationships

{}

ˆ ∂ u {εx } = ˆ = [ B] dˆ ∂x [ B] = − 1 1   L L ˆ   d dˆ =  1x  ˆ d  2x 

{}

{σ} = [ D]{εx } = [ D][ B] {dˆ} 10/14/08

Dynamic Analysis

12

Step 4 - Derive Element Stiffness and Mass Matrices and Equations With time dependent loading

ˆf ≠ fˆ 1x 2x 10/14/08

Dynamic Analysis

13

Newton’s Second Law

  f = ma

10/14/08

Dynamic Analysis

14

NODAL EQUILIBRIUM EQUATIONS

ˆ ∂ d 1x ˆf e = fˆ + m 1x 1x 1 2 ∂t 2ˆ ∂ d e 2x ˆf = fˆ + m 2x 2x 2 2 ∂t 2

10/14/08

Dynamic Analysis

15

m1 and m2 are obtained by lumping the total mass of the bar equally at the two nodes

ρAL m1 = 2 ρAL m2 = 2 10/14/08

Dynamic Analysis

16

Lumped Mass Model dˆ1x fˆ1ex ( t )

1

m1

2

m2

xˆ

dˆ 2 x fˆ2ex ( t )

L

10/14/08

Dynamic Analysis

17

Equilibrium in Matrix Form fˆ1ex  fˆ1x  m1 ˆ e  = ˆ  +  f 2 x  f 2 x   0

10/14/08

Dynamic Analysis

 ∂ dˆ1x    2 0  ∂ t   ˆ   m 2   ∂ d 2x  2  ∂ t 

18

Equilibrium in Matrix Form

{

} [ ]{ }

{}

  ˆf ( t ) = kˆ dˆ + [ m ˆ ] dˆ e

10/14/08

Dynamic Analysis

19

Defining Terms  1 − 1 AE [k ] =  L − 1 1  ρAL 1 0 [m] = 2 0 1 dˆ 2 ˆ ˆ ∂ d d = ∂ t2

{}

{}

10/14/08

{}

Element Stiffness Matrix Element Lumped Mass Matrix Nodal Displacements Nodal Accelerations

Dynamic Analysis

20

Consistent Mass Matrix

{}

{ }

  X = −ρ uˆ e

{fb } = ∫∫∫[ N] { X} dV T

V

{fb } = −∫∫∫ ρ[ N ]

T

{}

uˆ dV

V

10/14/08

Dynamic Analysis

21

Consistent Mass Matrix

{}

ˆ { uˆ} = [ N ] d  uˆ = [ N ] dˆ   uˆ = [ N ] dˆ

{} {}

10/14/08

{} {}

Dynamic Analysis

22

Consistent Mass Matrix

{}

ˆ {fb } = −∫∫∫ ρ[ N] [ N] d dV T

V

{}

  ˆ {fb } = −[mˆ] d

[mˆ] = ∫∫∫ ρ[ N] [ N] dV T

V

10/14/08

Dynamic Analysis

23

Consistent Mass Matrix Bar Element [mˆ] = ∫∫∫ ρ[ N] [ N ] dV T

V

1 − xˆ    L [mˆ] = ∫∫∫ ρ xˆ  1 − xˆ V   L  L  xˆ  L 1 −   ˆ x  L ˆ [m] = ρ A ∫  xˆ  1 − 0  L  L  10/14/08

Dynamic Analysis

xˆ  dV L  xˆ  dxˆ L  24

Consistent Mass Matrix Bar Element ˆ ˆ  x x     L 1 − 1 −     L  L  [mˆ] = ρ A ∫  ˆ ˆ x x     0  1 −     L  L  ρ A L 2 [mˆ] =  6 1 10/14/08

Dynamic Analysis

1 − xˆ  xˆ   L  L  dxˆ  ˆ ˆ  x  x    L  L   1  2 25

STEP 5 - Assemble the Global Equations and Apply B.C.’s

  { F( t )} = [ K ] { d} + [ M ] {d}

10/14/08

Dynamic Analysis

26

  { F( t )} = [ K ] { d} + [ M ] {d} Now must solve coupled set of ODE’s instead of set of linear algebraic equations!

10/14/08

Dynamic Analysis

27

Consistent Mass Matrix

[m] = ∫∫∫ ρ[ N] [ N] dV T

V

10/14/08

Dynamic Analysis

28

Beam Element yˆ , vˆ ˆ ,m φ 1 ˆ1

1

xˆ

fˆ1y , dˆ 1y

10/14/08

2 φˆ 2 , m ˆ2 L

Dynamic Analysis

fˆ2y , dˆ 2y

29

Shape Functions

( 2xˆ 3 − 3xˆ 2 L + L3 ) L3 1 3 ( N2 = xˆ L − 2xˆ 2 L2 + xˆL3 ) L3 1 ( N3 = − 2xˆ 3 + 3xˆ 2 L ) L3 1 3 ( N4 = xˆ L − xˆ 2 L2 ) 3 L N1 =

10/14/08

1

Dynamic Analysis

30

Shape Functions 1.000

N1 0.500

N3

N2 L

0.000

N4

0 -0.500 10/14/08

Dynamic Analysis

31

Consistent Mass Matrix

[ m] = ∫∫∫ ρ[ N] [ N] dV T

V

22L  156  22L 2 4 L m [m] =  420  54 13L  2 − 13L − 3L 10/14/08

Dynamic Analysis

− 13L  2  13L − 3L  156 − 22L  2  − 22L 4L  54

32

Lumped Mass Matrix 1  0  m [ m ] = 0 2 0  10/14/08

0 2 αL 210 0

1

0

0

Dynamic Analysis

0 0

0   0   0  2 αL  210  33

Lumped Mass Matrix 2nd and 4th terms account for rotary inertia. α = 0 if this is ignored. α = 17.5 if mass moment of inertia of bar spinning about one end is selected

 m  L     2  2   I= 3 10/14/08

Dynamic Analysis

2

34

Consistent Mass Matrix CST Q 0  [ m] =    0 Q  2 1 1 u1 m [ Q] = 1 2 1 u 2 12 1 1 2 u 3

For each degree of freedom

10/14/08

Dynamic Analysis

35

Consistent Mass Matrix CST 2 0  ρAt 1 [ m] =  12  0 1   0 10/14/08

0 2 0 1 0 1

1 0 2 0 1 0

Dynamic Analysis

0 1 0 2 0 1

1 0 1 0 2 0

0  1 0  1 0  2 36

Lumped Mass Matrix CST Q 0  [ m] =    0 Q 1 0 0 u 1 m [ Q] = 0 0 1 u 2 3 0 0 1 u 3

10/14/08

Dynamic Analysis

37

Lumped Mass Matrix CST 1 0  ρAt 0 [m] =  3 0 0  0 10/14/08

0 0 0 0 0  1 0 0 0 0 0 1 0 0 0  0 0 1 0 0 0 0 0 1 0  0 0 0 0 1

Dynamic Analysis

38

Consistent Mass Matrix Quad Q 0  [m] =  0 Q    4 2 2 4 [ Q] = m  36 1 2  2 1

1 2 4 2

2 1  2  4

m = ρAt

10/14/08

Dynamic Analysis

39

Consistent Mass Matrix Quad 4 0  2  m 0 [ m] =  36 1  0 2  0 10/14/08

0 2 0 1 0 2 0  4 0 2 0 1 0 2 0 4 0 2 0 1 0  2 0 4 0 2 0 1 0 2 0 4 0 2 0  1 0 2 0 4 0 2 0 1 0 2 0 4 0  2 0 1 0 2 0 4 Dynamic Analysis

40

Hybrid Methods Attempts have been made to combine consistent and lumped mass approaches to achieve some of the benefits of each!

10/14/08

Dynamic Analysis

41

HRZ Lumping  Hinton, Rock, and Zienkiewicz  Compute the diagonal terms of  

 10/14/08

consistent mass matrix. Compute total mass of element, m Compute s by adding diagonal coefficients associated with translational D-O-F that are in same direction. Scale all diagonal coefficients by multiplying by m/s Dynamic Analysis

42

HRZ - Bar Element ρ A L 2 1  [ mˆ] =   6 1 2  m = ρAL ρAL s = 4× 6 m 3 = s 2 ρ A L 3 0 [ mˆ] = 6 0 3 10/14/08

Dynamic Analysis

43

HRZ - Beam Element 22L 54 − 13L   156  22L 2 2  4 L 13 L − 3 L  [m] = m  420  54 13L 156 − 22L   2 2  − 13L − 3L − 22L 4L  m = ρAL ρAL s = 312 × 420 m = 420 s 312 10/14/08

Dynamic Analysis

44

HRZ - Beam Element  420  312 × 156   0 m  [ mˆ ] = 420  0    0 

10/14/08

0

0

420 × 4 L2 312

0

0

420 × 156 312

0

0

Dynamic Analysis

   39     2  0 L   m =    78  39  0     2   L 420  2  × 4L  312 0

45

HRZ – Quadratic Serendipity 1 36

3 76

8 36

16 76 3x3 Gauss Rule

10/14/08

2x2 Gauss Rule

Dynamic Analysis

46

HRZ – Quadratic Lagrangian 1 36

1 36 4 36

16 36

16 36

3x3 Gauss Rule

10/14/08

4 36 2x2 Gauss Rule

Dynamic Analysis

47

% error in natural frequencies of a thick simply-supported plate. Half of the plate modeled with 8-noded 24 d-o-f elements Mode m n 1 1 2 1 2 2 3 1 3 2 3 3 4 2

10/14/08

Type of Mass Matrix Consistent (%) HRz Lumping (%) Ad Hoc Lumping (%) 0.11 0.32 0.32 0.4 0.45 0.45 0.35 2.75 4.12 5.18 0.05 5.75 4.68 2.96 10.15 13.78 5.18 19.42 16.88 1.53 31.7

Dynamic Analysis

48

Optimal Lumping  Only translational d-o-f  Based on consistent mass matrix  Use appropriate quadrature rule  Chose integration points to coincide with nodal locations  [m] will be diagonal

10/14/08

Dynamic Analysis

49

 Let p be the highest order complete polynomial in shape function N  let m be the highest order derivative in strain energy (m = 1 elasticity, m = 2 bending)  Chose quadrature rule with degree of precision 2(p-m) 10/14/08

Dynamic Analysis

50

Three noded bar element p=2 m=1 2(p-m) = 2 Three point quadrature rule. Newton -Cotes has points at the nodes. (Simpson’ Rule) 1  4 b+a 1 ∫a f (x)dx = ( b − a )  6 f ( a ) + 6 f  2  + 6 f ( b )  b

10/14/08

Dynamic Analysis

51

1

m ij = ρA ∫ N i N jdx = ∫ N i N j J dξ −1

L J= 2 L ( 1 − ( − 1) )  1 N i ( − 1) N j ( − 1) + 4 N i ( 0) N j ( 0) + 1 N i ( 1) N j ( 1)  2 6 6 6  i ≠ j m ij = 0

m ij = ρA

1 0 0 ρAL   [ m] = 0 1 0  6  0 0 4 10/14/08

1

Dynamic Analysis

3

2

52

Serendipity 1 − 12 1 3

10/14/08

Dynamic Analysis

53

Lagrangian 1 36

4 9

10/14/08

Dynamic Analysis

1 9

54

Mass Matrices  Product [m]{a} must yield the correct total force on an element (F = ma) when {a} represents a rigid-body translational acceleration.  Consistent mass matrices, [m] and [M] are positive definite.  Lumped mass matrix is positive semi-definite when zero terms appear on main diagonal.  Lumped mass matrix is indefinite when negative terms appear on main diagonal. 10/14/08

Dynamic Analysis

55

Mass Matrices  Special treatment may be needed to handle the last two cases.

10/14/08

Dynamic Analysis

56

Best Type ? 1. Consistent matrices usually more accurate for flexural problems. 2. Consistent matrices give upper bounds on natural frequencies.

10/14/08

Dynamic Analysis

57

Best Type ? 1. Lumped matrices usually give natural 2. 3. 4. 5.

frequencies less than exact values. Simpler to form. Occupy less storage. Require less computational effort. Usually more important in time-history than in vibration problems.

10/14/08

Dynamic Analysis

58

Damping 1. Structural damping is not viscous. 2. Due to mechanisms such as hysteresis 3. 4. 5. 6.

and slip in connections. Mechanisms not well understood. Awkward to incorporate into structural dynamic equations. Makes equations computationally difficult. Effects usually approximated by viscous damping.

10/14/08

Dynamic Analysis

59

Types of Damping Models  Phenomenological Damping

Methods (models actual dissipative mechanisms) ➲ ➲ ➲

Elastic-Plastic Hysteresis Loss Structural Joint Friction Material Micro-cracking

 Spectral Damping Methods ➲ ➲ 10/14/08

Introduce Viscous Damping Relies on Fraction of Critical Damping Dynamic Analysis

60

Critical Damping ξ

Fraction of Critical Damping ξ = 1 Critical Damping

Critical Damping marks the transition between oscillatory and non- oscillatory response of a structure

10/14/08

Dynamic Analysis

61

Critical Damping Ratio 0.5% ≤ ξ ≤ 5% 2% ≤ ξ ≤ 15% 2% ≤ ξ ≤ 15%

Steel Piping Bolted or riveted steel structures Reinforced or Prestresses Concrete

Actual value may depend on stress level. 10/14/08

Dynamic Analysis

62

Rayleigh or Proportional Damping Damping matrix is a linear combination of stiffness and mass matrices:

[ C] = α [ K ] + β [ M ] 10/14/08

Dynamic Analysis

63

Rayleigh or Proportional Damping [C] is orthogonal damping matrix. Modes may be uncoupled by eigenvectors associated with undamped problem.

1 β ξ = αω−  2 ω 10/14/08

Dynamic Analysis

64

If critical damping ratio is known at two frequencies then: β 1 ξ = αω−  2 ω ( ξ 2 ω 2 − ξ 1 ω1 ) α=2 2 2 ω 2 − ω1

(

β = 2 ω1 ω 2 10/14/08

)

( ξ 1 ω 2 − ξ 2 ω1 )

(ω

2 2

Dynamic Analysis

−ω

2 1

)

65

Natural Frequencies and Mode Shapes Undamped, Unforced Response

{ D} = { D } sin ωt

{ D } = ω{ D } cos ωt { D } = − ω { D } sin ωt 2

{ D}

amplitudes of nodal d - o - f

ω circular frequency ω f = 2π 10/14/08

( Hz ) Dynamic Analysis

66

Results in generalized eigenproblem

( [ K ] − λ [ M ] ) { D } = { 0} λ=ω

10/14/08

2

Dynamic Analysis

67

Trivial Solution:

[ K ] − λ[ M]

{ D} = 0

10/14/08

Dynamic Analysis

≠0

68

Nontrivial Solution:

[ K ] − λ[ M ]

{ D} ≠ 0

10/14/08

=0

Dynamic Analysis

69

λi

{ D} 10/14/08

≡ Roots of Characteristic Polynomial (eigenvalues)

i

≡ Associated

Dynamic Analysis

Eigenvectors

70

ω i Natural Frequencies

{D}

10/14/08

i

Normal Modes

Dynamic Analysis

71

Natural Frequencies  [K] and [M] n x n then there are n

eigenvalues and n eigenvectors  [K] and [M] positive definite then eigenvalues are all positive  Mii = 0 infinite eigenvalue  Mii < 0 negative eigenvalue - imaginary frequency  Use condensation to remove ith equation if Mii = 0 10/14/08

Dynamic Analysis

72

Rayleigh Quotient

{ D} [ K ]{ D} λ= { D} [ M ]{ D} T

T

{ D}

[ K ] symmetric [ M ] positive definite th

approximat ion to i eigenvector

λ approximat ion to i th eigenvalue 10/14/08

Dynamic Analysis

73

Rayleigh Quotient

λ min

{ v} [ K ]{ v} ≤ ≤ λ max T { v} [ M ]{ v}

λ min

smallest eigenvalue

λ max

l arg est eigenvalue

T

10/14/08

Dynamic Analysis

74

Modal Methods When [K], [C], [M] are known and time independent the problem is linear.

{

ext    [ M ] { D} + [ C ] { D} + [ K ] { D} = R

{ D ( 0 ) } , { D ( 0 ) } 10/14/08

}

given as initial conditions Dynamic Analysis

75

Modal Methods Assume orthogonal damping, such as Rayleigh Damping. Modes can be uncoupled:

{ D} [ M ] { D} { D} [ K ] { D} { D} [ C] { D} T i

j

=0

T i

j

=0

T i

j

=0

i≠j 10/14/08

Dynamic Analysis

76

{ D} [ M ] { D} = 1 { D} [ K ] { D} = ω { D} [ C] { D} = 2ξ ω T i

i

T i

T i

10/14/08

2 i

i

i

Dynamic Analysis

i

i

77

Eigenvectors are linearly independent

[ φ] = matrix of

eigenvectors (mode shapes)

{ Z} 10/14/08

{ D} = [ φ]{ Z}

modal amplitudes Dynamic Analysis

78

Substitute into:

{

ext    [ M ] { D } + [ C ] { D } + [ K ] { D} = R { D ( 0 ) } , { D ( 0 ) } given

10/14/08

Dynamic Analysis

}

79

{

ext    [ M ] [ φ] { Z} + [ C ] [ φ] { Z} + [ K ] [ φ] { Z} = R [ φ ] { Z ( 0 ) } = { D ( 0 ) }

[ φ ] { Z ( 0 )} = { D ( 0 ) }

10/14/08

Dynamic Analysis

given

80

}

Mode Displacement Method

[ ]

2    [ I ] { Z } + [ ξ ] { Z } + ω { Z} = { p}

10/14/08

Dynamic Analysis

81

Mode Displacement Method Pre-multiply by [φ]T

[ φ]

T

T   [ M ] [ φ] { Z } + [ φ] [ C ] [ φ] { Z }

[ K ] [ φ] { Z} = [ φ] { R } [ φ] { Z ( 0 ) } = { D ( 0 ) } [ φ] { Z ( 0 ) } = { D ( 0 ) } given

+ [ φ]

10/14/08

T

T

Dynamic Analysis

ext

82

Mode Displacement Method

[ φ] [ M ] [ φ] = [ I ] T [ φ] [ C] [ φ] = [ ξ ] T 2 [ φ] [ K ] [ φ] = [ ω ] T

10/14/08

Dynamic Analysis

83

Mode Displacement Method

[ ]

2    [ I ] { Z } + [ ξ ] { Z } + ω { Z} = { p}

10/14/08

Dynamic Analysis

84

Modes Uncouple:

[ ]

2    [ I ] { Z} + [ ξ ] { Z} + ω { Z} = { Z} Z + 2 ξ ω Z  + ω 2 Z = p i = 1, n i

10/14/08

i

i

i

i

Dynamic Analysis

i

85

[ φ ] { Z ( 0 ) } = { D ( 0 ) } [ φ ] { Z ( 0 )} = { D ( 0 ) } T T  [ φ ] [ M ] [ φ ] { Z ( 0 ) } = [ φ ] [ M ] { D ( 0 ) } T  [ I ] { Z ( 0 ) } = [ φ ] [ M ] { D ( 0 ) } T  { Z ( 0 ) } = [ φ ] [ M ] { D ( 0 ) } T { Z ( 0 )} = [ φ ] [ M ] { D ( 0 ) } 10/14/08

Dynamic Analysis

86

Reduce size of problem:

m << n eq m

{ D } = ∑ { φ} i Z i i =1

10/14/08

Dynamic Analysis

87

Error Estimate:

{ R } − [ M ] { D } − [ C ] { D } − [ K ] { D} {R } ext

e( t ) =

ext

For an accurate analysis : e ( t ) ≤ 1%

10/14/08

Dynamic Analysis

88

In many structural dynamics problems, more modes participate in the quasi-static response than in the dynamic response. For a small m value, the mode displacement method may have difficulty in predicting the quasi-static response. 10/14/08

Dynamic Analysis

89

Mode Acceleration method Method

[ ]

2    [ I ] { Z } + [ ξ ] { Z } + ω { Z} = { p}

10/14/08

Dynamic Analysis

90

Modal transformation only on inertial and damping terms

{

ext    [ M ] [ φ ] { Z } + [ C ] [ φ ] { Z } + [ K ] { D} = R [ φ ] { Z ( 0 ) } = { D ( 0 ) }

[ φ ] { Z ( 0 )} = { D ( 0 ) }

10/14/08

Dynamic Analysis

given

91

}

 } − [C][ φ]{Z } [K ] {D} = {R ext } − [M ][φ]{Z  } − [C][ φ]{Z  }) {D} = [K ] {R } − [K ] ([M ][φ]{Z −1

10/14/08

ext

−1

Dynamic Analysis

92

[ φ] [ M ] [ φ] = [ I ] −T [ M ] [ φ] = [ φ] T

10/14/08

Dynamic Analysis

93

{ D} = [ K ] { R } − [ K ] −1

10/14/08

ext

−1

( [ φ ] { Z } − [ C ] [ φ ] { Z } )

Dynamic Analysis

−T

94

[ φ] [ K ] [ φ] = [ ω ] T 2 −1 [ φ] [ K ] [ φ] [ ω ] = [ I ] −T −T 2 −1 [ K ] [ φ] [ ω ] = [ φ] [ I ] = [ φ] −T 2 −1 −1 [ φ] [ ω ] = [ K ] [ φ] T

10/14/08

2

Dynamic Analysis

95

{ D} = [ K ] {R } − [ K ] [ φ] −1 ext 2 −1  −1 { D} = [ K ] {R } − [ φ] [ω ] {Z} + [ K ] [ C][ φ]{Z } −1

10/14/08

ext

−1

−T

−1   {Z} + [ K ] [C][ φ]{Z }

Dynamic Analysis

96

{ D} = [ K ] {R } − [ K ] [ φ] {Z } + [ K ] −1[ C][ φ]{Z } −T T −1 ext 2 −1  −1 { D} = [ K ] {R } − [ φ] [ω ] {Z} + [ K ] [ φ] [ φ] [ C][ φ]{Z } −1 ext 2 −1  2 −1 { D} = [ K ] {R } − [ φ] [ω ] {Z} + [ φ] [ω ] [ ξ]{Z } −1

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ext

−1

−T

Dynamic Analysis

97

{ D} = [ K ] { R } − [ φ ] [ ω −1

ext

{ D} = [ K ] { R } −1

ext

] ( { Z } + [ ξ ] { Z } )

2 −1

 1  2ξ i   − ∑ { φ} i  2 Z i + Z i  ωi i =1  ωi  m

First term on RHS represents quasi-static response, the second term represents corrections for inertia and viscous effects. 10/14/08

Dynamic Analysis

98

Solve for Z terms as before:

Z + 2 ξ ω Z  + ω2 Z = p i i i i i i

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Dynamic Analysis

i = 1, n

99

Mass Condensation  Reduces number of d-o-f.  Reduces expense of computing eigenvalues.  Detrimental to accuracy.  Not used with optimal lumping.

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Dynamic Analysis

100

Guyan Reduction   K mm   KT   ms

K ms   M mm  − λ K ss   M Tms

M ms    D m   0  =     M ss    D s   0 

m - master degree of freedom s - slave degree of freedom 10/14/08

Dynamic Analysis

101

Guyan Reduction m - master degree of freedom s - slave degree of freedom Basic Assumption: For lowest frequency modes the inertial forces on slave d-o-f are less important than elastic forces transmitted by master d-o-f. Thus we ignore all mass except Mmm 10/14/08

Dynamic Analysis

102

Guyan Reduction   K mm   KT   ms

0    D m   0  =     0    D s   0 

K ms   M mm  − λ K ss   0

{ D } = −[ K ] [ K ] { D } −1

s

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ss

T

ms

Dynamic Analysis

m

103

Guyan Reduction  D m    = [ T]{ Dm }  D s  I   [ T ] =  −1 T   − K ss K ms 

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Dynamic Analysis

104

Guyan Reduction

( [ K r ] − λ [ M r ] ) { D m } = { 0} T [ K r ] = [ T] [ K ] [ T] T [ Mr ] = [ T] [ M] [ T] Both [Kr] and [Mr] are generally full. [Mr] contains both mass and stiffness terms 10/14/08

Dynamic Analysis

105

Guyan Reduction

[ Cr ] = [ T] [ C] [ T] T ext ext {R r } = [ T ] {R } ext    [ M r ]{D m } + [ Cr ]{D m } + [ K r ]{ D m } = {R r } T

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Dynamic Analysis

106

Compute Slaves

{D } s

i

= − [ K ss − λ i M ss ]

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−1

[K

Dynamic Analysis

T ms

− λiM

T ms

]{ D } m

107

i

Choosing Master D-O-F  Choose d-o-f where inertia is most important  These are characterized by large mass to stiffness ratios.  Each d-o-f that has a time varying applied load should be chosen.  Master d-o-f should not be clustered.  Process can be automated 10/14/08

Dynamic Analysis

108

Process for Choosing Master D-OF  Scan diagonal coefficients of [K] and [M].  Choose d-o-f for which Kii/Mii is largest.

 This becomes first slave.  Condense [K] and [M] by one order.  Repeat process using condensed matrices till a user specified number of d-o-f remain.  These are Master d-o-f chosen in near 10/14/08 109 optimal way. Dynamic Analysis

Number of Master D-O-F  Choose cut-off frequency ωc  Take this to be about three times the highest frequency of interest.  Terminate selection of master d-o-f when Kii/Mii < ωc2

 Can combine manual and automatic selection (i.e. Choose each d-o-f that has a time varying applied load and then automatically select others.)  Number of Master d-o-f may be 10% - 20% of total d-o-f. 10/14/08 Dynamic Analysis 110