The Modal Method of Analysis

Chapter 6

The Modal Method of Analysis

6.1

Introduction

The method of analysis described in this Chapter, is based on the finite element method, and it is suitable for determining the dynamic response of structures to time-dependent forces. Only simple structures will be considered in this Chapter, but the effects of damping will also be taken into account. The method used here shows that for complex problems involving forced vibrations and damping, their solution can become extremely simple when the modal method of analysis is adopted. 62

The Modal Matrix [Φ]

In this section, we will show how it is possible to uncouple the matrix dynamical equation of motion of an "n" degree of freedom forced vibration system, with the aid of the modal matrix [Φ]. Now the modal matrix [Φ] is a square matrix, whose columns are the "n" eigenmodes of the structure, corresponding to the "n" degrees of freedom of the system, as shown by equation (6.1) [Φ] = [Φ, Φ

2

Φ

Φ J,

3

(6.1)

where [Φι] = the i

a

eigenmode

(6.2)

Xj = the j " relative displacement of the eigenmode "i" Consider a two degree of freedom system, with the modal matrix [Φ] of equation (6.3) [Φ] = [Φ, Φ2] Let [M]

= the mass matrix

and[M]

= [Φ] [Μ][Φ] Τ

(6.3)

(6.4)

T h e Modal Method of Analysis

184

[Μ] [Φ,

Ch.6

(6.5)

φ ] 2

4Ϊ

φ[Μ φ^Μ

[Φ,

Φ1 2

or Τ Φι Μ φ! [Μ]

τ φ, Μ φ

2

(6.6)

= Τ Φ2 Φΐ

τ Φ2 Φς Μ

Μ

but as [Φ,] and [Φ^ are orthogonal, [Φ^ΜΦΛ

= [ Φ / Μ Φ , ] = [0]

(6.7)

Hence, equation (6.6) becomes of the diagonal form shown by equation (6.8)

φ[Μφ, [M]

0 (6.8)

= 0

Mj

ΦίΜφ

2

0 (6.9)

0

Mj

where [MJ

andpvLJ

=

(6.10)

[φίΜφ!

=

[φΐΜφ

(6.11) 2

Sec.

6.2]

The Modal Matrix (Φ)

185

Similarly, it can be shown that [ Κ]

=

[ φ ] [Κ] [φ] τ

Κ,

0

0

Kj

(6.12)

(6.13)

where [Κ,] and[KJ ΝΒ

=

[Φ, ΚΦ,]

=

[Φ/ΚΦ2]

(6.14)

Τ

(6.15)

Both [Μ] and [Κ] are of diagonal form, hence, it will be easier to handle them mathematically; this is taken advantage of by introducing the weighted modal matrix [Φ] which is obtained by dividing [Φ] by the square root of [M], as follows:

[$] = []/vST =

w e i g h t e d m o d a l matrix

[φ"]

[M]

[φ]

[ Μ ] [ Μ ] [φ]

[φΤ

[Κ]

τ

Now =

τ

and

(6.16)

[φ~] 1

= [φ]

τ

= [I] = the identity matrix

(6.17)

RT|

[ Μ ] [Κ] 1

[φ]

but [ Μ ] [Κ] = [λ] 1

.·. [ φ ] [ Μ - ] [Κ] [φ] τ

(6.18) =

[λ]

where

ω [λ]

=

ω,

2

ω*

= the i* radian frequency of vibration.

(6.19)

Ch.6

T h e Modal Method of Analysis

186

Now the matrix dynamical equation of motion of a structure, without damping, is given by: [K] {u,} + [M] (0,}

=

{q,}

(6-20)

where {q,}

= a vector of time-dependent forcing functions.

Let {U,} and

{0,}

= [Φ] { x j

(6.21)

= [Φ] {x,}

(6.22)

Hence, equation (6.20) becomes

[Κ] [Φ] {x } + [Μ] [Φ] {x,} = {q,}

(6.23)

i

Pre-multiplying equation (6.23) by [ Φ ] , we get τ

[ Φ ] [Κ] [Φ] {χ,} + [ Φ ] [Μ] [Φ] {«,} = [ Φ ] {q,}, τ

or

τ

(6.24)

τ

[ Φ ] [ Μ ] [Κ] [Φ] {χ,} + [ Φ ] [ Μ ] [Μ] [Φ] {χ,} = [ Φ ] {q,} τ

1

τ

[λ] {χ,} + [I] {χ.} = {Ρ,},

where {Ρ,}

= [ Φ ] {q,} τ

1

τ

(6.25)

(6.26)

2

ω, 2

α>

2

[λ]

(6.27)

From equation (6.25), it can be seen that the equations are uncoupled and each row equation can be solved independently line by line.

The Modal Matrix (Φ)

Sec. 6.2]

187

Example 6.1 Determine expressions for the nodal displacements of the two element rod structure of Example 4.1, when it is subjected to periodical axial excitation force of value F(t), at its node 2. Hence, or otherwise, determine the [Φ] matrix, where F(t) = F sin tot. From Example (4.1),

180 [Κ]

=

[KJ

=

(6.28)

1E6 -80

7.598 [M]

=

[M ]

80

1.179 (6.29)

=

n

1.179 [Φ]

-80

22.358

= [Φ,

where 0.492

[Φ,] = 1.0

1 and [ φ ] 2

= -0.215

so that 0.492

1.0 (6.30)

[Φ] = 1.0

-0.215

Now, the equation of motion is: [K] {Uj} + [M] {U,}

= {F(t)>

(6.31)

Let {L\}= [Φ] { x j and {UJ

= [Φ] {x,},

(6.32)

188

T h e Modal Method of Analysis

Ch.6

So that equation (6.31) becomes: [Κ] [Φ] {χ,} + [Μ] [Φ] {x,} = {F(t»

(6.33)

Substituting equations (6.28), (6.29) and (6.30) into equation (6.33), we get

0.492

1.0

180

-80

0.492

1.0

-80

80

1.0

-0.215

1E6

W

1.0

-0.215

10.492

1.0

7.598

1.179

10.492

1.0

10.492

1.0

0

1.0

-0.215

1.179

22.358

1.0

-0.215

1.0

-0.215

K0

(6.34)

Now

0.492

1.0

ΓΚ] = [ψ] [κ][Φ] = τ

1.0

=

180

-80

0.492

1.0

-80

80

1.0

-0.215

1E6 -0.215

0.492

1.0

8.56

197.2

1.0

-0.215

40.64

-97.2

1E6

44.85

0 (6.35)

1E6 0

and

[Μ]

=

[φ]

218.1

τ

[Μ]

[φ]

0.492

1.0

7.598

1.179

0.492

1.0

1.0

-0.215

1.179

22.358

1.0

-0.215

=

The Modal Matrix (Φ)

Sec. 6.2]

0.492

1.0

4.917

7.345

1.0

-0.215

22.938

-3.628

25.36

189

0 (6.36)

0

8.125

and{P,}

=

[φ] {ς,} τ

0.492

1.0

0

1.0

-0.215

[Fsin(tot)

(6.37)

Fsin(a)t) { .}

(6.38)

p

Fsin(cut)

The equation of motion becomes:

[K] { } + [M] XJ

m

Fsin(cot)

(6.39)

Fsin(cot)

or 44.85

0

25.36

0

0

218.1

0

8.125

1E6

F sin(cot) F

(6.40)

190

T h e Modal Method of Analysis

Ch.6

Equation (6.40) can be seen to be uncoupled, with the following two independent equations: 44.85 E6 x + 25.36 x

2

= Fsin((ul)

(6.41)

218.1 E6 x + 8.125 x

3

= F sin(tot)

(6.42)

2

and

3

These two independent equations can both be seen to be of the form k, x, + m χ = F sin(cot)

(6.43)

If we divide through by ην we get k x. + _ i x. m '

=

k but — m.

ω,

f

F _ sincot, m

(6.44)

;

2

=

Hence, equation (6.44) becomes x. + cof χ

=

—

sin((ot)

(6.45)

The complementary function can be found by substituting x, = Ae™ in equation (6.45) ie

(a

x

(

2

+ lOj ) x, 2

=

0

= A cos ω, t + Β sin ω, t

(6.46)

The particular integral is given by F sin(cot) m. ( D

2

_

F sin(cot)

+ ω )

rflj ( - ω

2

2

F sin(cot) nij ω

2

(l - ω /ω ) 2

2

but ω, = k/m,, 2

where D = a differential operator

+ cof)

(6.47)

Sec. 6.2]

The Modal Matrix (Φ)

191

Hence the particular integral becomes _

F sin (tot) (6.48) k, ( l - ω 7 ω ? )

From equations (6.46) and (6.48), the complete solution becomes:

x

A r» · A c o s cat + Β sin ioJt +

=

j

F 8Ϊη(ωΙ) i—ί—

(6.49)

k, (l - coVcof)

and x.

=

t

- ω . A sin co,t + ω Β cos ω{ + ω (

F cos(tot) k (l - orVojf)

The boundary conditions are that at t = 0; x, = x,(o) and x, = Xj(o). Substituting the first boundary condition, into equation (6.49), we get A = Xi(0)

(6.51)

Substituting the second boundary condition into (6.50), we get

•/ \ x.(o)

=

coF

to Β + π

k, (l - co /co?) 2

or Β

=

*i(°)

coF

ω.

ω k, (l - coVcof)

(6.52)

(

Hence, from equations (6.51) and (6.52),

x,

=

X j

( o ) cos ω t + _ 1

k, ( l - coVco?)

ω.

sin cot '

ω sin tot - —

sin cot

(6.53)

The Modal Method of Analysis

192

Ch.6

The nodal displacements, u and u can be obtained from the expression 2

U

\

2

3

'

(6.54)

• - [Φ] ' U

3

X

3

0.492

1.0

«2

1.0

-0.215

«3

(6.55)

The weighted modal matrix [Φ] can be obtained from the expression:

[φ]

[φ]

(6.56)

[Μ] where

y/25M

0

0

y/JUB

[Μ]

5.036

0

0

2.850

0.492 5.036 [φ]

=

(6.57)

1.0 2.850

10 -0.215 5.036 2.850

0.0977

0.351

0.1986 -0.0754

(6.58)

Sec. 6.3]

63

Damping

193

Damping

The effects of damping on structural vibration is very important and in this section, we will consider three forms of damping, namely, (a)

viscous damping

(b)

Rayleigh damping

(c)

structural damping 63.1

Viscous Damping

The equation of motion for viscous damping is: [K] {U,} + [M] {0,} where, [c] =

+ [c] {(J,} = {q,}

(6.59)

[ N ] [u] [N] d(vol) T

[u] is a matrix of viscous damping terms. It can be seen from equation (6.59), that viscous damping is proportional to a vector of velocities. Let {U,}

= [Φ] {x,}

So that {U,}

= [Φ] {χ,}

{0,}

= [Φ] {χ,}

and

(6.60)

Substituting equations (6.60) into equation (6.59), we get [Κ] [Φ] {χ,} + [Μ] [Φ] {x,} + [c] [Φ] {x,} = {q,}

(6.61)

Pre-multiplying equation (6.61) by [ Φ ] , we get τ

[ Φ ] [Κ] [Φ] {x } + [ Φ ] [Μ] [Φ] {χ,} τ

τ

(

+ [ Φ ] [c] [Φ] {χ,} = [ Φ ] { ς , } , τ

τ

(6.62)

or [Κ] { χ } + [Μ] { } + β (

Xj

{ } = {Ρ,} Xj

(6.63)

where [5] = [ Φ ] [c] [ Φ ] τ

(6.64)

A l t h o u g h [Κ] and [Μ] are of diagonal form, [c] is not, and b e c a u s e of this, equation (6.63) is uncoupled and difficult to solve.

T h e Modal Method of Analysis

6.3.2

Ch.6

Rayleigh D a m p i n g

If in equation (6.63), [c] is related to [K] and [M] in the following form: [c] = α [ Κ ] + β [ Μ ] ,

(6.65)

then equation (6.63) b e c o m e s uncoupled, w h e r e α and β are obtained experimentally. T h i s form of d a m p i n g is called R a y l e i g h d a m p i n g and it is often used in structural dynamics. For Rayleigh d a m p i n g , [Ε] = [ Φ ] ( α [ Κ ] + β [Μ]) [ Φ ] ,

(6.66)

[c] = α [Κ] + β [Μ]

(6.67)

τ

so that,

H e n c e , equation 6.63) b e c o m e s [Κ] {x } + [Μ] { } + ( α [Κ] + β [Μ]) { } f

Xj

=

Xj

{Ρ,}

(6.68)

Equation (6.68) is uncoupled and it consists of a n u m b e r of independent e q u a t i o n s , each of the following form: X j

+ 2 ζ ω, Xj + ω ,

= Ρ,

2 X j

(6.69)

where ξ and 2 ζ ω,

= the modal d a m p i n g ratio = ω α + β

(6.70)

2

T h e two u n k n o w n s α and β can be obtained in terms of ζ „ ω , and from the first t w o equations of (6.70). From the first two equations of (6.70), w e get, β + ω, α 2

and β + ω

2 2

α

= 2 ζ, ω,

(6.71)

= 2 ^ ω

(6.72)

2

Solution of equations (6.71) and (6.72) gives α and β

= 2 ( ω ζ, - ω, ζ,) / ( ω 2

= 2 ω, ω ( ω ζ, - ω, 2

2

2 2

- ω, )

(6.73)

2

/ (ω

2 2

- ω, ) 2

(6.74)

Damping

Sex. 6.3]

195

and from equation (6.70), the d a m p i n g ratios for the other m o d e s is given b y :

f,

-

^

+

2

JL

2ω,

(6.75)

T h e separate effects of mass-proportional and stiffness-proportional d a m p i n g are s h o w n in Figure 6 . 1 .

Figure 6.1 Mass-Proportional and Stiffness-Proportional Damping. F r o m Figure 6 . 1 , it can be seen that the effects of mass-proportional d a m p i n g o n ζ , are larger for smaller values of to, and vice-versa for stiffness-proportional d a m p i n g . Additionally, it can be seen that for stiffness-proportional d a m p i n g , ξ is linearly proportional to ω,. T o obtain the complimentary function of equation (6.70), it can b e written in the form:

the solution of which, according to Case et al [3], is given b y :

+ Be

{-μ/Σοι, - γΌι/Σηψ - (k/ng) I 2

196

T h e Modal Method of Analysis

Ch.6

N o w (k/nij) is usually very m u c h larger than (μ/2 m ) , hence, the expression for x, can b e simplified to the form 2

{

cos

C

rr

11

"H

J

C a s e et al [3] also s h o w that if equation (6.70) is in the form: Γη Xj +

μ

χ, + kj = Ρ sin ωΐ,

the particular integral is given b y : *i

=

Ρ nij D

2

sin

at

+ μ ϋ + kj

where D = a differential operator. C a s e et al [3] g i v e the following expression for the particular integral: Ρ [(k, - ω X

'

2

(k, -

Ρ

m ^ s i n o>t - μ ω coscat]

«Vf * μ

2

ω

2

ί i\ sinut - μ ω cosoat 1

\2

1

- ϋ 2 ω,

+ μ

2

ω

2

T h e m a x i m u m amplitude of this forced vibration, namely, x ^ , , , is given b y :

Damping

6.3]

6.3.3

197

Structural D a m p i n g

For structural d a m p i n g , Petyt [7] s h o w s that the equation of m o t i o n is given b y : ( F] • m

) {U,} • |M] (0,}

=

{q,}

(6.76)

where j = /^T Petyt explains that this form of d a m p i n g can only b e used w h e n the excitations is h a r m o n i c , and that the c o m p l e x stiffness matrix ([K] + j [ H ] ) can b e obtained b y letting

the C o m p l e x Y o u n g ' s m o d u l u s

= E ( l + jt|),

where Ε and η

(6.77)

= Y o u n g ' s m o d u l u s of elasticity

= the material loss factor = 2 χ 10" for p u r e a l u m i n i u m 5

=

1.0 for hard rubber.

H e n c e , from equation (6.77) [Η]

= η[Κ],

(6.78)

S o that equation (6.76) b e c o m e s p q f l + j T u W

+ rMHu,}

=

{q,}

(6.79)

Letting {UJ

= [ Φ ] { x j in equation (6.79), w e get

Μ σ + ί η Η Φ Η χ , , + ΙΜΠΦΠ*} = {q,}

(6.80)

Pre-multiplying equation (6.80) by [ Φ ] , w e get τ

[ K ] ( l - Η η ) {*} + [M] {x } = {P,} (

(6.81)

Equation (6.81) is uncoupled, and each horizontal line of equations can be solved quite independently of any other horizontal line of equations.

For further reading o n this topic, see Petyt [7] and T h o m s o n [8].