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Structural Dynamics 7/29/2005

Dynamic Analysis

1

Spring-Mass system subjected to a time dependent load.

k F(t) m

7/29/2005

Dynamic Analysis

2

Free-body diagram of the mass.

T = kx

F(t) m

7/29/2005

= Dynamic Analysis

m

ma = m&x&

3

F(t ) − k x = m &x& m &x& + k x = F(t ) 7/29/2005

Dynamic Analysis

4

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

Homogeneous :

F(t ) = 0 m &x& + k x = 0 7/29/2005

Dynamic Analysis

5

Let : k ω = m Then : 2

&x& + ω x = 0 2

ω2 is the natural circular frequency 7/29/2005

Dynamic Analysis

6

2π τ= ω τ is the period (measured in seconds) 7/29/2005

Dynamic Analysis

7

Displacement due to simple harmonic motion.

xm

7/29/2005

τ

Dynamic Analysis

8

One Dimensional Bar Element

7/29/2005

Dynamic Analysis

9

Step 1 - Select Element Type dˆ1x ˆf e (t ) 1x

1

2

dˆ 2 x

xˆ

ˆf e (t ) 2x

L E - modulus of elasticity A - cross-sectional area ρ - mass density 7/29/2005

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 7/29/2005

Dynamic Analysis

11

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

{}

ˆ ∂ u {ε x } = ˆ = [B ] dˆ ∂x 1 1 ⎡ ⎤ [B ] = ⎢ − ⎣ L L ⎥⎦ ˆ ⎫ ⎧ d dˆ = ⎨ 1x ⎬ ˆ d ⎩ 2x ⎭

{}

{σ} = [D ]{ε x } = [D ][B ]{dˆ} 7/29/2005

Dynamic Analysis

12

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

ˆf ≠ fˆ 1x 2x 7/29/2005

Dynamic Analysis

13

Newton’s Second Law

7/29/2005

Dynamic Analysis

14

NODAL EQUILIBRIUM EQUATIONS

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

7/29/2005

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 7/29/2005

Dynamic Analysis

16

Lumped Mass Model dˆ1x fˆ1ex (t )

1

m1

2

m2

xˆ

dˆ 2 x fˆ2ex (t )

L

7/29/2005

Dynamic Analysis

17

Equilibrium in Matrix Form e ˆ ⎧⎪f1x ⎫⎪ ⎧⎪fˆ1x ⎫⎪ ⎡m 1 ⎨ˆ e ⎬ = ⎨ˆ ⎬ + ⎢ ⎪⎩f 2 x ⎪⎭ ⎪⎩f 2 x ⎪⎭ ⎣ 0

7/29/2005

Dynamic Analysis

⎧ ∂ dˆ1x ⎫ ⎪ ⎪ 2 0 ⎤⎪ ∂ t ⎪ ⎨ ˆ ⎬ ⎥ m 2 ⎦ ⎪ ∂ d 2x ⎪ 2 ⎪⎩ ∂ t ⎪⎭

18

Equilibrium in Matrix Form

& & ˆf (t ) = kˆ dˆ + [m ˆ ˆ] d e

7/29/2005

Dynamic Analysis

19

Defining Terms ⎡ 1 − 1⎤ AE [k ] = ⎢ ⎥ L ⎣− 1 1 ⎦ ρAL ⎡1 0 ⎤ [m ] = ⎥ ⎢ 2 ⎣0 1 ⎦ dˆ &ˆ& ∂ 2 dˆ d = ∂ t2

{}

{}

7/29/2005

{}

Element Stiffness Matrix Element Lumped Mass Matrix Nodal Displaceme nts Nodal Accelerati ons

Dynamic Analysis

20

Consistent Mass Matrix

{}

{ }

& & X = −ρ uˆ e

{f b } = ∫∫∫ [N ] {X} dV T

V

{f b } = − ∫∫∫ ρ[N ]

T

{}

&uˆ& dV

V

7/29/2005

Dynamic Analysis

21

Consistent Mass Matrix

ˆ {uˆ} = [N ] d &uˆ = [N ] dˆ& & & &uˆ& = [N ] dˆ 7/29/2005

Dynamic Analysis

22

Consistent Mass Matrix

{}

&ˆ& {f b } = − ∫∫∫ ρ[N ] [N ] d dV T

V

{}

&ˆ& {f b } = −[mˆ ] d

[mˆ ] = ∫∫∫ ρ[N ] [N ] dV T

V

7/29/2005

Dynamic Analysis

23

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

V

⎧1 − xˆ ⎫ ⎪ L ⎪ ⎡ xˆ [mˆ ] = ∫∫∫ ρ⎨ xˆ ⎬ ⎢1 − V ⎪ ⎪⎣ L ⎩ L ⎭ xˆ ⎫ L ⎧1 − ⎪ L ⎪ ⎡ xˆ [mˆ ] = ρ A ∫ ⎨ xˆ ⎬ ⎢1 − 0 ⎪ ⎪⎣ L ⎩ L ⎭ 7/29/2005

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 7/29/2005

Dynamic Analysis

⎛⎜1 − xˆ ⎞⎟⎛⎜ xˆ ⎞⎟ ⎤ ⎝ L ⎠⎝ L ⎠ ⎥ dxˆ ⎥ ⎛⎜ xˆ ⎞⎟⎛⎜ xˆ ⎞⎟ ⎥ ⎝ L ⎠⎝ L ⎠ ⎦ 1⎤ ⎥ 2⎦ 25

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

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

7/29/2005

Dynamic Analysis

26

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

7/29/2005

Dynamic Analysis

27

Consistent Mass Matrix

[m ] = ∫∫∫ ρ[N ] [N ] dV T

V

7/29/2005

Dynamic Analysis

28

Beam Element yˆ , vˆ ˆ1 φˆ 1 , m

1

xˆ

fˆ1y , dˆ 1y

7/29/2005

2 φˆ , m 2 ˆ 2 L

Dynamic Analysis

fˆ2y , dˆ 2y

29

Shape Functions

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

7/29/2005

1

3 ˆ

2 ˆ

Dynamic Analysis

3

30

Shape Functions 1.000

N1 0.500

N3

N2 L

0.000

N4

0 -0.500 7/29/2005

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 7/29/2005

Dynamic Analysis

− 13L ⎤ 2 ⎥ − 3L ⎥ 156 − 22L ⎥ 2 ⎥ − 22L 4L ⎦ 54 13L

32

Lumped Mass Matrix ⎡1 ⎢ 0 ⎢ m [m] = ⎢0 2⎢ ⎢0 ⎢⎣ 7/29/2005

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 7/29/2005

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

7/29/2005

Dynamic Analysis

35

Consistent Mass Matrix - CST ⎡2 ⎢0 ⎢ ρAt ⎢1 [m] = ⎢ 12 ⎢ 0 ⎢1 ⎢ ⎢⎣ 0 7/29/2005

0 1 0 1 0⎤ ⎥ 2 0 1 0 1⎥ 0 2 0 1 0⎥ ⎥ 1 0 2 0 1⎥ 0 1 0 2 0⎥ ⎥ 1 0 1 0 2⎥⎦ Dynamic Analysis

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

7/29/2005

Dynamic Analysis

37

Lumped Mass Matrix - CST ⎡1 ⎢0 ⎢ ρAt ⎢0 [m] = ⎢ 3 ⎢0 ⎢0 ⎢ ⎢⎣0 7/29/2005

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 m = ρAt

7/29/2005

Dynamic Analysis

1 2⎤ ⎥ 2 1⎥ 4 2⎥ ⎥ 2 4⎦

39

Consistent Mass Matrix - Quad ⎡4 ⎢0 ⎢ ⎢2 ⎢ m ⎢0 [m] = ⎢ 36 1 ⎢ ⎢0 ⎢2 ⎢ ⎢⎣0 7/29/2005

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!

7/29/2005

Dynamic Analysis

41

HRZ Lumping 1. Hinton, Rock, and Zienkiewicz 2. Compute the diagonal terms of consistent

mass matrix. 3. Compute total mass of element, m 4. Compute s by adding diagonal coefficients associated with translational D-O-F that are in same direction. 5. Scale all diagonal coefficients by multiplying by m/s 7/29/2005

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⎦ 7/29/2005

Dynamic Analysis

43

HRZ - Beam Element 22L 54 − 13L ⎤ ⎡ 156 ⎢ 22L 2 2 ⎥ 4 13 3 − L L L m ⎥ [m] = ⎢ 420 ⎢ 54 13L 156 − 22L ⎥ ⎢ 2 2 ⎥ ⎣− 13L − 3L − 22L 4L ⎦ m = ρAL ρAL s = 312 × 420 m = 420 s 312 7/29/2005

Dynamic Analysis

44

HRZ - Beam Element ⎡ 420 ⎢ 312 × 156 ⎢ ⎢ 0 m ⎢ [mˆ ] = 420 ⎢ 0 ⎢ ⎢ ⎢ 0 ⎢⎣

7/29/2005

0

0

420 × 4 L2 312

0

0

420 × 156 312

0

0

Dynamic Analysis

⎤ 0 ⎥ ⎢ 39 ⎥ ⎥ ⎢ ⎥ 2 ⎥ 0 L ⎢ ⎥ m ⎥= ⎢ ⎥ ⎥ 78 ⎢ 39 ⎥ 0 ⎥ ⎢ ⎥ ⎥ 2 ⎢ L 420 ⎣ ⎥⎦ 2 ⎥ × 4L ⎥⎦ 312

45

HRZ – Quadratic Serendipity 1 36

3 76

8 36

16 76 3x3 Gauss Rule

7/29/2005

2x2 Gauss Rule

Dynamic Analysis

46

HRZ – Quadratic Lagrangian 1 36

1 36 4 36

16 36

16 36

3x3 Gauss Rule

7/29/2005

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

7/29/2005

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

7/29/2005

Dynamic Analysis

49

Š Let p be the highest order complete

polynomial in shape function N Š let m be the highest order derivative in strain energy (m = 1 elasticity, m = 2 bending) Š Chose quadrature rule with degree of precision 2(p-m)

7/29/2005

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) b

⎡1 ⎤ 4 ⎛b+a⎞ 1 ∫a f (x)dx = (b − a )⎢⎣ 6 f (a ) + 6 f ⎜⎝ 2 ⎟⎠ + 6 f (b )⎥⎦ 7/29/2005

Dynamic Analysis

51

1

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

L J= 2 L 4 1 ⎡1 ⎤ m ij = ρA (1 − (− 1))⎢ N i (− 1)N j (− 1) + N i (0)N j (0 ) + N i (1)N j (1)⎥ 2 6 6 ⎣6 ⎦ i ≠ j m ij = 0

⎡1 0 0⎤ ρAL ⎢ ⎥ [m] = 0 1 0⎥ ⎢ 6 ⎢⎣0 0 4⎥⎦ 7/29/2005

1

Dynamic Analysis

3

2

52

Serendipity 1 − 12 1 3

7/29/2005

Dynamic Analysis

53

Lagrangian 1 36

4 9

7/29/2005

Dynamic Analysis

1 9

54

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

7/29/2005

Dynamic Analysis

55

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

7/29/2005

Dynamic Analysis

56

Best Type ? 1. Consistent matrices usually

more accurate for flexural problems. 2. Consistent matrices give upper bounds on natural frequencies.

7/29/2005

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.

7/29/2005

Dynamic Analysis

58

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

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

7/29/2005

Dynamic Analysis

59

Types of Damping Models Š Phenomenological Damping Methods

(models actual dissipative mechanisms) Ø Ø Ø

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

Š Spectral Damping Methods Ø Ø

7/29/2005

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

7/29/2005

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. 7/29/2005

Dynamic Analysis

62

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

[C] = α [K ] + β [M ] 7/29/2005

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⎠ 7/29/2005

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 7/29/2005

)

(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 7/29/2005

( Hz ) Dynamic Analysis

66

Results in generalized eigenproblem

([K ] − λ [M ]){D } = {0} λ=ω

7/29/2005

2

Dynamic Analysis

67

Trivial Solution:

[K ] − λ [M ]

≠0

{D } = 0

7/29/2005

Dynamic Analysis

68

Nontrivial Solution:

[K ] − λ[M ]

{D} ≠ 0

7/29/2005

Dynamic Analysis

=0

69

λi

{D }

i

7/29/2005

≡ Roots of Characteristic Polynomial (eigenvalues) ≡ Associated

Dynamic Analysis

Eigenvectors

70

ω i Natural Frequencies

{D }

i

7/29/2005

Normal Modes

Dynamic Analysis

71

Natural Frequencies Š [K] and [M] n x n then there are n eigenvalues and n eigenvectors Š [K] and [M] positive definite then eigenvalues are

all positive Š Mii = 0 infinite eigenvalue Š Mii < 0 negative eigenvalue - imaginary frequency Š Use condensation to remove ith equation if Mii = 0

7/29/2005

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 eigenvalue th

7/29/2005

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

7/29/2005

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 )} 7/29/2005

}

given as initial conditions Dynamic Analysis

75

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

{D } [M ]{D } = 0 {D } [K ]{D } = 0 {D } [C ]{D } = 0 T i

j

T i

j

T i

j

i≠j 7/29/2005

Dynamic Analysis

76

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

i

T i

T i

7/29/2005

2 i

i

i

Dynamic Analysis

i

i

77

Eigenvectors are linearly independent

[φ] = matrix of {Z} 7/29/2005

eigenvectors (mode shapes) {D} = [φ]{Z} modal amplitudes Dynamic Analysis

78

Substitute into:

{

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

7/29/2005

Dynamic Analysis

}

79

{

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

[φ ]{Z ( 0 ) } = {D (0 )}

7/29/2005

Dynamic Analysis

given

80

}

Mode Displacement Method

[ ]

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

7/29/2005

Dynamic Analysis

81

Mode Displacement Method Pre-multiply by [φ]T T & & [M ][φ ]{Z }+ [φ ] [C ][φ ]{Z& }

[φ ] T T ext + [φ ] [K ][φ ]{Z } = [φ ] {R } [φ ]{Z& ( 0 ) } = {D& (0 )} [φ ]{Z ( 0 ) } = {D (0 )} given T

7/29/2005

Dynamic Analysis

82

Mode Displacement Method

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

7/29/2005

Dynamic Analysis

83

Mode Displacement Method

[ ]

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

7/29/2005

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

7/29/2005

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 )} 7/29/2005

Dynamic Analysis

86

Reduce size of problem:

m