Prof. A. Meher PrasadProf. A. Meher Prasad

Department of Civil EngineeringDepartment of Civil EngineeringIndian Institute of Technology MadrasIndian Institute of Technology Madras

email: prasadam@iitm.ac.in

Rayleigh’s Method

Background:

Consider that an undamped SDF mass-spring system is in free harmonic motion, then

x = X sin (pt+α)

x = pX cos (pt+α)

The strain energy of the system, V, at any time t is given by

V = ½ kx2 = ½ kX2 sin2(pt+α)

and its kinetic energy, T, is given by

T = ½ mx2 = ½ mp2X2 cos2(pt+α)

The principle of conservation of energy requires that, the sum of V and T be the same.

Note that when V = Vmax, T=0, and when T= Tmax, V =0. Hence

Vmax = Tmax

.

. (E1)

(E2)

(E3)

(E4)

or ½ kX2 = ½ mp2X2

From which we conclude that p2 = k/m

This is identical to the expression obtained from the solution of the governing equation of motion.

As a second example, consider the SDF system shown2

2max 0 0 0

1 a a 1 aV [k y ] y k y

2 L L 2 L

2L2 2 2 2 2

max 0 0 0

0

1 x 1 1T p [ y dx my ] p y [ L m]

2 L 2 3

Equating Vmax and Tmax, we obtain2

2

ak

Lp

1m L

3

μ m

a

L

ky0

(E5)

(E6)

(E7)

Rayleigh Quotient

Consider now a MDF system in free vibration, such that

{x} = {X} sin(pt+α)

{x} = p {X} cos(pt+α)

The maximum strain energy of the system is

½ {F}T[d]{F}

Vmax = ½ {F}T{X} =

½ {X}T[k]{X}

In which {F} are the static modal forces corresponding to the displacement amplitudes {X}, and [d] and [d] are the flexibility and stiffness matrices of the system.

. (E8)

(E9)

The maximum Kinetic energy of the system is given by

Tmax = ½ p2{X}T[m]{X} = p2 Tmax

In which Tmax = ½ {X}T[m]{X}

will be referred to as the maximum pseudo-kinetic energy of the system.

The principle of conservation of energy requires that

Vmax =Tmax = p2 Tmax

where

~

~

~

2 max~

max

Vp

T

The above equation is known as Rayleigh’s quotient.

(E10)

(E11)

(E12)

The Eq.(E12) could also be obtained from the equations of motion of the system as follows:

[m] { x } + [k] { x } = 0

Making use of Eq.(E8) and pre multiplying the resulting equation by {X}T, we obtain,

- p2 {X}T[m]{X} + {X}T[k]{X} = 0

Vmax Tmax

~

where2 max

~

max

Vp

T

..

Properties of Rayleigh quotient

• If and in Eq.(E12) are evaluated for {X} = to the jth mode {Xj}, then the value of p2 will be precisely pj

2.

• If Eq.(E12) is evaluated for a vector {X} which does not correspond to a natural mode, then the resulting value of p2 will not be a natural frequency. Furthermore, to each assumption of {X} there corresponds a different value of p2. In fact, if we recall that (n-1) displacement ratios are necessary to define the configuration of a MDF system, we may conclude that p2 in Eq.(E12) defines a “surface” in a space having (n-1) dimensions.

• It can be shown that

Tmax

~Vmax

(1) The natural frequencies of the system, i.e., the values of p2 obtained when {X} is equal to any of the natural modes, correspond to stationary (maximum, minimum or saddle) points of this surface.

It follows that an error in estimating the mode will produce an error in frequency which is of smaller order (since the surface is ‘flat’ in the neighbourhood of the stationary point).

2. The values of p2 obtained for an arbitrary {X} lies between the lowest and highest natural frequencies of the system(i.e., p1

2 < p2 < pn

2).

It follows that if one assumes an {X} which approximates the fundamental natural mode of the system, then the resulting value of p will be greater than (will represent an upper bound estimate for) p1. Similarly, if one assumes an {X} that approximates the highest natural mode, the resulting value of p will be lower than (will represent a low bound for) the true pn.

Because the fundamental natural mode of a system can normally be estimated with reasonable accuracy, the procedure is ideally suited to the computation of the fundamental natural frequency.Application of the procedure requires the following steps:

1. Estimate the fundamental mode of vibration. This may be done either by assuming directly the displacements of the nodes, or the associated forces and computing the displacements.

2. Compute the values of Vmax and corresponding to the estimated mode.

3. Evaluate p2 from,

The value thus determined is higher than the true fundamental natural frequency of the system, and, unless the assumption regarding the mode was quite poor, it will be close to the actual frequency.

Tmax

~

2 max~

max

Vp

T

If the frequency is computed for several different assumed configurations, the smallest of the computed values will be closest to the exact fundamental frequency, and the associated configuration is closest to the actual fundamental mode.

The details of the procedure are illustrated by a series of examples.

m

k

km

x1

x2

1

1

1

2

1

1.5

(a) (c)(b)

Example #1

Vmax may be evaluated from the deformations of the stories without having to determine first the stiffness matrix of the system.

2 Vmax= k [ (1)2+0] = k

2 Tmax= m [ (1)2+(1)2 ] = 2m

2 k kp 0.5

2m m

Assumption 1: Take x1=x2=1, as shown in Fig. (a)

~

Assumption 2: Take x1=1 and x2= 2, as shown in Fig.(b)

2 Vmax= k [ (1)2+(1)2 ] = 2k

2 Tmax= m [ (1)2+(2)2 ] = 5m

2 2k kp 0.4

5m m

Assumption 3: Take x1=1 and x2= 1.5, as shown in Fig.(c)

2 Vmax= k [ (1)2+(0.5)2 ] = 1.25k

2 Tmax= m [ (1)2+(1.5)2 ] = 3.25m

2 1.25k kp 0.3846

3.25m m

~

~

1. Assumption 3, which leads to the lowest frequency value, is the best of the three approximations considered, and is only slightly off the exact value of,

21

1 k kp = [3- 5] = 0.38197

2 m m

2. That assumptions 1 and 2 would be poor, could have been anticipated by considering the forces necessary to produce the assumed configurations.

The deflection configuration in Fig.(a) is produced by a single concentrated force applied at the first floor level, whereas the configuration in Fig.(b) is produced by a single concentrated force acting at the second floor level.

Clearly, neither of these force distributions is a reasonable approximation to the inertia forces associated with the motion of the system in its fundamental mode.

Discussion

3. That assumptions 1 and 2 The deflection pattern considered in Fig.(c) is produced by lateral forces which are proportional to the weights of the floors. If these forces are denoted by F, the resulting displacements are as shown.

F

F

2F/k

2F/k + F/k

Subject to the justification noted later, this assumption generally leads to an excellent approximation for the fundamental natural frequency of the system.

Fig (2)

4. In the following diagram, the value of p2 determined by application of Rayleigh’s method is plotted as a function of the displacement ratio x2/x1. Note that the two natural frequencies correspond to the maximum and minimum points of the curve, and that in the vicinity of these extremum points the frequency values are insensitive to variations in the displacement ratio x2/x1.

X2/ X1

2p

k m

3

2

1

2.618

-0.6181.618

1-2 -1 0 2 3

0.38197

5. The first step in the solution that has been presented was to estimate the mode of vibration of the system. Alternatively, we could have assumed the distribution of the inertia forces associated with the mode of interest and compute Vmax as the product of these forces and the resulting displacements. For example, for the forces considered in Fig.2,

2 Vmax = F F/k [2+3] = 5 (F2/ k)

2 Tmax = m F/k F/k [22+32] = 13m (F2/ k2)

p2 = 0.3846 (k / m)

which is the same as the answer obtained above.

~

Example # 2

k

k

x1

x2

k x3

m

m

0.5m

2.5

2.5+1.5 = 4

4+0.5 = 4.5

k

k

0.5k

1

1.6

1.8

Assume a deflection configuration equal to that produced by a set of lateral forces equal to the weights of the system.

2Vmax = k [ 2.5+4+ ½(4.5) ] = 8.75 k

( This can also be evaluated from the story deformations as

k [ 2.52 + 1.52 + 0.52 ] = 8.75k )

Tmax

~2 = m [ 2.52 + 42 + ½(4.52) ] = 32.375m

2 8.75k kp 0.27027

32.375m m

The exact value of p2 is 0.2680k/m, and the error in p is only 0.42%.

In the following figure, the value of p2 is plotted as a function of the displacement ratios X2/ X1 and X3/ X1 in the range between –3 and 3. As would be expected, there are three stationary points:

(1) a minimum point of p2=0.268k/m corresponding to the fundamental frequency;

(2) a maximum value of p2=3.732k/m corresponding to the third natural frequency;

(3) a saddle point of p2=2k/m corresponding to the second natural frequency. The values X2/ X1 and X3/ X1 of the associated modes can be read off the figure.

Dimensional representation of Rayleigh’s quotient

for a 3 DOF system

**20 squares to the inch

Assumption # 1:

0

4'' 2 2 2

max 40 0

sin

2 ( ) sinl l

o

xy x y

l

nxV EI y dx y EI dx

L L

2 2 2max

22 [[ ( )] 1.5[ ( )] ]3 3L LT m y y p

2 20

3 3[ 1.5 ]4 4

my p 2 20

3 9[ ]4 8

my p 2 20

15

8my p

4 2 42 0

3 2 3 30

8 1 425.98

2 15 15

EIy EI EIp

L my mL mL

The exact value of ,2

325.8463exact

EIp

mL

Example # 3

m 1.5 m

L/3 L/3 L/3

x

Assumption #2

3 3 34 3.5 1.5 9.25

243 243 243

WL WL WL

EI EI EI

3 3 34 1.5 3.5 9.5

243 243 243

WL WL WL

EI EI EI

2 3

max

12 [1.0 9.25 1.5 9.5]

243

W LV

EI

23

2 2

max

max max

23

12 [1.0 9.25 1.5 9.5 ]

243

,

25.8467

W WLT

g EI

V T

EIp

mL

setting

Which is in excellent agreement with the exact value

W 1.5W

Let F1,F2, Fj,…Fn be the inertia forces corresponding to the assumed mode and y1,y2……yj,…..yn be the deflections induced by these forces.

Then,

n

jj =12n

2j

1

F

pW

j

jj

y

gy

m1 m2mj mn

. . . . . . . .

. . . . . . . .F1 F2 Fj Fn

yj

Application to systems with Lumped masses:

As already demonstrated, good accuracy is achieved by taking the forces Fj to be equal to, or proportional to the weights Wj. It should be realized, however that these forces are not the exact inertia forces.Rather they represent the inertia forces associated with a uniform (rigid body) motion of the system.

An improved approximation may be achieved by assuming a configuration for the mode and taking Fj as the inertia forces corresponding to the assumed configuration.It is important to note that y j in Eq. E13 are the deflections produced by the forces Fj, not the deflections assumed for the purpose of estimating Fj

Selection of Fj

Illustration:

•As a guide in the selection of the forces { f }, assume that the fundamental mode varies as a sine curve , as shown in fig (a)

•The inertia forces corresponding to this assumption are shown in fig.(b). For the example considered, these turn out to be the exact forces, and hence the frequency computed from these forces will be exact in this case.

0.5

3/2

1

0.5F

3/2F

0.5F

0.5F

3/2F

0.5F

m

m

0.5m

x1

x2

x3

(a) Assumed Mode (b) Inertia Forces

• If the mode were assumed to increase linearly along the height, the forces and the deflections would be as follows:

F

2F

1.5F

4.5 F/k

8 F/k

9.5 F/k

• In this case

2 2

max

2 2

max

2

2 1*4.5 2*8 1.5*4.5 34.75

2 20.25 64 .5*90.25 129.375

34.750.2686

129.375

(The exact valve is 2- 3 0.26795)

F FV

K K

F FT m m

K KK K

pm m

• Taking the {F} proportional to the masses will not lead to satisfactory results. Directions important

• Take

m1m2

Another example :

F1 F2

Applications to continuous systems

0

max

2" 4 2 2max 04

0 0

max

0

( ) sin

2

2 sin

2

q

a)

)

L

L

xy x y

LV

EI xV EI y dx y dx

L L

V q y dx

Let

b

may be evaluated by either of the following procedures

where is the lateral force

EI, μ, L

associated with the assumed deflection

y(x) = yosin(πx/L)

This force may be determined from the differential equation for beams. Recalling that , 4

4( )

d yEI q x

dx

We obtain, 4

4sino

nxq x EI y

L L

Hence,4

2 2max 4

0

2 sinL

o

nxV EI y

L L

Which is the same as the result found by procedure (a)

2 2 2max

0 0

4 2 24

2 401 4

2 2

0

2

1 2

2 sin

sin

sin

L L

o

L

o

L

o

nxT y dx y dx

L

EI nxy dx

L L EIp

Lnxy dx

L

EIp

L

21 4

1 2

98.82

9.94

EIp

L

EIp

L

where

This is 0.8% too high

This is the exact frequency – Explain why?

If we had assumed as y (x) the deflection produced by a single concentrated force at the center, we would have found that,

Dunkerley’s Method

x d F

Equation of motion : 0d m x x

Let [m] be the diagonal matrix, 0d m x x

2

11 1 12 2 12

21 1 22 2 22

1 2 2 2

10

1. .

1. .

0. . . . .

. . . . .

1. .

n n

n n

n n nn n

d m I xp

d m d m d mp

d m d m d mp

d d m d mp

set

Det

Given a nth order polynomial equation, (1/p2) 1

11 1 22 22 2

1 1( .... ) .... 0

n n

nn nd m d m d mp p

Sum of the roots of characteristic equation,

11 1 22 22 2 21 2

1 1 1.... ( .... )

1where

nn nn

iiii

i ii

d m d m d mp p p

kk

m d

iilet p

dii is the flexibility coefficient equal to deflection at i resulting from a unit load of i, its reciprocal must be the stiffness coefficient kii, equal to the force per unit deflection at i.

By neglecting these terms(1/p22, …1/pn

2) ,1/p12 is larger than its true

value and there fore p1 is smaller than the exact value of the fundamental frequency

2 2 2 21 11 22

1 1 1 1.......

nnp p p p

The estimate to the fundamental frequency is made by recognizing p2 ,p3 etc are natural frequencies of higher modes and larger than p1.

2 2 2 21 11 22

1 1 1 1.......

nnp p p p

Dunkerley’s Approximation

It provides a lower bound estimate for the fundamental frequency.

Let p = natural frequency of system

pA, pB, pC, …….. pN = exact frequencies of component systems

Then

2 2 2 2 2A B C N

1 1 1 1 1+ + +...........+

p p p p p

or 2 2 2 2 2A B C NT T +T +T +.................+T

The frequency so determined can be shown to be lower than the exact.

k

k

km

m

0.5m

Example # 1

m

m

0.5m

2A

kp =

m2B

kp =

2m2C

kp =

3m/2

2

1 m m= [1+2+1.5]=4.5

p k k

2 k kp =0.2222

4.5m m

If natural modes of component systems A, B, C are close of each other, then the value of p determined by this procedure can be shown to be close to the exact.

Consider the cantilever beam shown for which the component systems A,B,C are indicated .

Since the natural modes of the system are in closer agreement in this case than for the system of the shear beam type considered in the previous example, the natural frequency computed by Dunkereley’s method can be expected to be closer to the exact value than with case before.

m m m/2

m

m

m/2

A

B

C

Example # 2

L/3

2L/3

2L/3

L/3

L

L/3 L/3 L/3

)mL

EI7312.3(

)(mL

EI6.3

mL

54

15mL

6

1

9

1

p

1

mL

6

1

81

8

81

1

p

1

p

1

p

1

p

1

mL

EI6

)2(L

EI3p

mL

EI

8

81

3)2L(

EI3p

mL

EI81

3)L(

EI3p

32

32

33

2

3

2C

2B

2A

2

332C

332B

332A

exactp

boundLowp

EIEI

EI

m

m

m

As expected the agreement is excellent in this case.

Example # 3

Upper bound: Determined by Rayleigh’s method with y(x) = y0 sin(πx/L) is,

Lower bound: Determined by Dunkerley’s approximation

4 32

42 2A B4 3

4 3 4 3

2 4 4

32

4

4 42

4 4

(EI L )

2

48EI;

mL

1 mL L

48EI 48

(EI L )

( )90 48

,

EI EI

3.029 3

pL m

EIp p

L

L mL

p EI EI

where pL m

For m L we find

pL L

m μ

If we consider one mode,

2 2 3 5 32 3 4

0

4 3 4 3

4 4214 4

( )[ 2 ]

3 48 3 5 48

1 1 1[ ]

3 3 2 5 48 90 48

3.117 3

l x l x dx ml l mll l ll

lEI EI lEI EI

l ml l ml

EI EI EI EI

EI EIp

L L

mμ dxx

Consider all modes,

4 321 4 4

1

4 4 321 4

1

48

90 48

n

l mlp

EI n EI

l mlp

EI EI

Limitation of procedures:

• One cannot improve the accuracy of the solution (depends on the deflected shape of structure) in a systematic manner.

• Extension of procedure : Rayleigh - Ritz

• Applicable to systems governed by [H]{X} = {X} ,where [H] is not necessarily symmetric.

• Meaning to a solution : Finding an {X} which when operated by [H] will give a vector is proportional to itself.Then {X}=characteristic vector and = the associated characteristic value.

Procedure:

1. Assume an {X}.

2. Compute [H]{X}

3. If step 2 results in a vector which is proportional to {x}, then {X} is a characteristic vector, and the factor of proportionality is the associated characteristic value.

Stodola Method – (Method of Iteration)

5. It can further be shown that if, at the end of a cycle, we compute the values of which will make the elements or components of the derived and assumed vectors equal, the highest characteristic value lies between the largest and smallest of these values.

4. In general, the vector computed in 2 will not be proportional to {X}. Now if we take as our next assumption the result of step 2 and repeat the process, the procedure will converge to the characteristic vector associated with the largest characteristic value .

2

-1 2

2

[ ]{ } [ ]{ } 0

- [ ]{ } [ ]{ } 0

[ ] [ ]{ } { }

[ ]{ } { }

m x k x

p m X k X

m k X p X

G X p X

2

2

2

0- [ ][ ]{ } { }

1[ ][ ]{ } { }

1[ ]{ } { }

p d m X x

d m X Xp

H X Xp

Stiffness Matrix Flexibility Matrix

)sin( ptXx

[ ][ ]{ } { } 0d m x x

Converges to highest natural frequency and mode

Converges to fundamental natural frequency and mode

2 -1 0

-1 2 1

0 -1 1

k k

1 1 11

1 2 2

1 2 3

dk

1 0 0

[ ] 0 1 0

0 0 0.5

m m

2

1 1 1 1 0 01

1 2 2 0 1 0

1 2 3 0 0 0.5

mX X

k p

Using Flexibility formulation first,we obtain

Example Frame

k

k

x1

x2

k x3

m

m

0.5m

2

2

1 1 1 2

1 2 1

1 2 1 2

1 3.5 3.5

1.6

1 2.5

1 4.0 2.5 4.5

1 4.5

1

1

6.0 3.75 3.5 3.83

1.8 6.9 3.83

.71

kX X

mp

kLet

mp

3.7 3.7

4 6.399 3.75 3.70 3.75

1.914 7.385 3.83

1 3.729 3.729

1.729 6.454 3.733 3.727 3.733

1.996 7.452 3.733

11.0

1.732 3

1.999 2

exact is

2

21 0 0 2 1 0

0 1 0 -1 2 -1

0 0 2 0 -1 1

2 -1 0

-1 2 -1

0 - 2 2

p mX

p mX X

XK

K

ξ

1 10 3 3.33

4 3 5 3.75 3.33 4

4 3 16 3 4

1 3.5 3.5

3 2 -5.6 3.73

8 5 6.2 3.8

1 1

1 0

1 0

1 3

-1 -4 3 4

1 4

8

Obviously it is diverging from the fundamental mode

1 3.6 3.6

1.6 -5.971 3.732 3.6 3.8

3.6 - 5.971- 6.7423.7321

1- 2.60 -1.771

7

1.771 6

3.7

.742 3.87

321exact

2 [ ]{ }

{ } [ ]{ }

T

T

X k Xp

X m X Rayleigh Quotient

Combining iteration with Rayleigh Method

2

1

1[ ]{ } { }

[ ] [ ][ ]

[ ] { } { }j j

H X Xp

Where H d m

Let H X V

1 121

1 1

[ ]{ }

{ } [ ]{ }

T

j j

Tj j

V k Vp

V m V

Stodola Method:

From Rayleigh’s Quotient,

If convergence is incomplete, the Rayleigh Quotient gives the better approximation.Any error in the first mode frequency computed by the Rayleigh Quotient is always on high side

1

121

1 1

121

1 1

{ } [ ]{ } [ ] [ ] { }

[ ][ ][ ]{ }. .

{ } [ ]{ }

[ ]{ }

{ } [ ]{ }

j j j

T

j j

Tj j

T

j j

Tj j

But V H X d m X

V k d m Xi e p

V m V

V m Xp

V m V

Stodola convergence

1 1 1 2 2 ....... n nX

1 1 2 2H ........... n nX H H H

H X X

2 1 1 1 2 2 2 ....... n n nX

nλ

2 2 23 1 1 1 2 2 2 ....... n n nX

Assumed mode can be expressed as,

Highest characteristic value

On Iteration,

121

1 1

1i

n

T

S ST

S S

for i n

X m Xp

X m X

After ‘s’ Iterations,

1 21 1 2 2 ..............

s s

ss n n n

n n

X

nnX .......2211

Stodola process for the second mode

1 1 1 1 2 1 2 1

11

1 1

[ ] ........

. .

T T T Tn n

T

T

m X m m m

m Xi e

m

11 1 1

1 1

1 1-

1 1

Tm X

X XTm

Tm

ITm

X

The vector, 11 X has a zero first mode component

Any Vector,

1 1S I -1

1 1

Tm

Tm

Then premultiplying any arbitrary vector {X} by the sweeping

matrix [S1] removes the first mode component

2 1m H S

[H] 2{X}=λ2 [X]

Define sweeping matrix [S1] to be,

Matrix iteration is carried out for,

2

1 1 1 1 0 01

1 2 2 0 1 0

1 2 3 0 0 0.5

11 1

21

1 2 1

31 2

2

11 1

21 2 1

31 2

2

d m mk

md m x

k p

H

0.5m

m

m

k

k

k

x3

x2

x1

2

1

1 1

1 1

1 1

1.0

1.732

1.999

1 1.732 0.9995

5.997825

0.1667 0.2888 0.1666

0.2888 0.5002 0.2886

0.3333 0.5773 0.3331

T

T

T

T

k

mp

m

m

m

m

1 1

11 1

1 1

0.8333 0.2888 0.1666

0.2888 0.4999 0.2886

0.3333 0.5773 0.6669

0.37795 0.0776 0.12185

0.0776 0.1337 0.070

0.2442 0.1550 0.2564

T

T

mS I

m

H H S

21H x x

Iteration

1 0.4221

1 0.1331

1 0.6536

1 0.5709

0.3154 0.0354

1.554 0.6353

1 0.5085

0.062 0.0163

1.113 0.5394

X

22

22

1 0.5045

0.032 0.0083

1.061 0.5214

1

0.016

-1.034

(0.5045)(1) (0.032)( )(0.0083) ( 0.061) ( 0.5214)1 2

(0.5045) (0.00

mm m

pm

2 283) (0.5214)2

0.78137 2.000

0.39052

mm

Execute the first few natural frequencies and the associated modes of the beam shown , and study the rate of convergence of the results as a function of the stiffness of the spacing ,i.e. or an appropriate dimensionless measure of it.

Application of Rayleigh Ritz Procedure

L/4 3L/4

1 2 3

2 3( ) sin sin sin ...........

x x xy x a a a

L L L

The coefficients ‘a’ in this expression must be such that the value of

p2,determined from

When expressed in terms of the dimensionless distancex

L

The expression for y(x) becomes,

1 2 3( ) sin sin 2 sin 3 ...........y a a a

2 max

max

Vp

T

2max max

j j

V Tp

a a

is stationary.

This requires that,

Consider first only Two Terms in the Series,

22 2

max 20

1 22 2

max 3 20

212 2 2

1 2 1 230

122 2

1 1 230

1 1 ( ) [ ( / 4)]2 2

1 1 ( ) [ ( / 4)]2 2

1 1 [ sin 4 sin 2 ] sin sin

2 2 4 2

1 [ sin 8

2

L yV EI x k y L

x

EI yV k y L

L

EIa a k a a

L

EIa a a

L

2 22

2 2 2 21 1 2 2

sin sin 2 16 a sin 2 ]

1 sin 2 sin sin sin

2 4 4 2 2K a a a a

1 1 12 2 2 22

1 1 2 2

3 0 0 0

2 21 1 2 2

22 2 2 2

1 2 0 1 2 1 23

sin 8 sin sin 2 16 sin 2 ]1

2

1 2

2 2

1 116 2

2 2MAX

a a a aEI

L

K a a a a

EIV a a a a a a

L

where

3

0 2

12 2

0 0

1 1( ) ( )

2 2

L

MAX

KL

EI

T y x x L y

21

max 1 2

0

2 2 2 2max 1 1 2 2

2 21 2

1[ sin sin 2 ]

2

1[ sin 2 sin sin 2 sin 2

21 1 1

[ 0 ]2 2 2

T L a a

T L a a a a

L a a

41 1 2[ ( 2 ] [ ]1 0 1 2 132 2

41 1 2[16 (2 2 ] [ ]2 0 2 1 232 2

EIa a a p L a

L

EIa a a p L a

L

Application of the above Equation for aj =a1 and aj =a2 leads to ,

or,

Canceling the factor ½ on the two sides of these equations , and Introducing the dimensionless frequency parameter

2 2

0 240

2

p pEIpL

We obtain after rearrangement of term:

0 0 11

02

20 0

(1 ) 2

2 (16 2

a a

aa

0 0 0

0 0 0

(1 - ) 20

2 (16 2 )

Expanding,we obtain the following quadratic equation in λ0

λ0²-(17+3ρ0)λ0+16+18ρ0=0

λ0=1/2 {17+3ρ0 ±2

0 0225 30 9 }

The modes are defined by the ratio a2/a1 this is given by,

a2 / a1 = - 0 0

0

1

2

2 0

1 0 0

2

16 2

a

a

Consider Three Terms in Series

1

24max 1 2 33

0

2

1 2 3

42 2 2 2 2 2

1 2 33

2 2 21 2 3

2 sin 4 sin 2 9 sin 3

3 K a sin sin sin

4 2 4

sin 16 sin 2 81 sin 3 0 0 0

1 K a *1 *

2

EIV a a a d

L

a a

EIa d a d a d

L

a a

1 2 1 3 2 3

1 1 1 1 12 2 2

2 2 2 2 2

a a a a a a

42 2 2 2 2 2

max 1 2 3 0 1 2 3 1 2 1 3 2 33

1 1 1 1 12 a 16* 81* ( a 2 2

2 2 2 2 2

EIV a a a a a a a a a a

L

or,

2 2 2 2 2 2max 1 2 3

2 2 2max 1 2 3

2 sin sin 2 sin 3 0 0 0

1 1 12

2 2 2

T L a d a d a d

T L a a a

or,

Application of the Equation for aj =a1 and aj =a2 leads to,

0 0 1 1

0 0 0 2 0 2

3 30 0 0

(1 ) 2

2 (16 2 ) 2

2 81

o a a

a a

a a

1 2

2 16 2 2 0

2 81

o o o o

o o o o

o o o o

Expanding we obtain the following cubic equation in :3 2(98 4 ) (1393 278 ) (1296 1474 ) 0o o o o o

Modes: These are defined by the ratios a2 / a1 and a3 / a1 .

Considering the first two equations (13) and eliminating a3 we obtain:

2

1

2(1 )

16o

o

a

a

Considering the first and third of equation (13) , and eliminating a2 we obtain:

3

1

(1 )

81o

o

a

a

(16)

(17)

(15)

This leads to the determinantal equation,

Note that Equations (16) and (17) are independent of o. However o

enters in these equations indirectly through o .The equations are valid irrespective of the order of o considered (i.e. for all three modes)

In considering the second mode ,it is more convenient to express it in terms of the ratios a1/ a2 and a3 / a2 (i.e normalize it with respect to a2).These ratios are given by,

1

2

16

2(1 )o

o

a

a

and

3

2

(16 )

2 (81 )o

o

a

a

(18)

(19)

Convergence of natural frequencies and modes

No of terms used

Frequency coefficient Fundamental mode Second mode Third mode

1 1 1 a1 a2 a3 a1 a2 a3 a1 a2 a3

(a) For o=1

1 2 1

2 1.8760 18.124 1 -0.0877 0.0877 1

3 1.8665 18.089 82.044 1 -0.0867 -0.0109 0.0864 1 -0.0235 0.0129 0.0224 1

(b) For o=5

1 6 1

2 3.7526 28.247 1 -0.3178 0.3178 1

3 3.6661 27.097 87.237 1 -0.3050 -0.0344 0.3007 1 -0.1456 0.0723 0.1238 1

Mass Condensation or Guyan Reduction

• Extensively used to reduce the number of D.O.F for eigen value extraction.

• Unless properly used it is detrimental to accuracy• This method is never used when optimal damping is used for mass

matrix

Let 'm' represent those to be restrained

Let 's' represent those to be condensed

. .

0

0

. .

mmm ms mm ms

ssm ss sm ss

m

s

k m u

uk k m m

u

Master d o f

Slav

k k m m

u

eu d o f

• Assumption: Slave d.o.f do not have masses – only elastic forces are important

1

1

Guass Elimination Scheme

0

1 1 ---

,

ms sm ss

T

s ss ms m

m

m T

s ss ms

T T

r r

m m m

u k k u

s s s s m m

IuT u T

u k k

k T k T m T m T

• Choice of Slave d.o.f

– All rotational d.o.f

– Find ratio, neglect those having large values for this ratio

– If [ Mss ] = 0, diagonal, [Kr] = same as static condensation then there is no loss of accuracy

Reduced eigen problem

master d.o.f

Slave d.o

1

f

1

.

r m r mk u m u

m m m m m m m m

ii

ii

k

m

1 T Ts ss i ss ms i ms m ii iiu k m k m u

Subspace Iteration Method

• Most powerful method for obtaining first few Eigen values/Eigen vectors

• Minimum storage is necessary as the subroutine can be implemented as out-of core solver

• Basic Steps

– Establish p starting vectors, where p is the number of Eigen values/vectors required P<<n

– Use simultaneous inverse iteration on ‘p’ vectors and Ritz analysis to extract best Eigen values/vectors

– After iteration converges, use STRUM sequence check to verify on missing Eigen values

Tk m L D L

• Method is called “Subspace” iteration because it is equivalent to iterating on whole of ‘p’ dimension (rather that n) and not as simultaneous iteration of “p’ individual vectors

• Starting vectors

• Strum sequence property

For better convergence of initial lower eigen values ,it is better if subspace is increased to q > p such that,

q = min( 2p , p+8)

Smallest eigen value is best approximated than largest value in subspace q.

Starting Vectors

(1) When some masses are zero, for non zero d.o.f have one as vector entry.

0 0 0

2 1 0, { }

0 0 0

1 0 1

m X

(2) Take ratio .The element that has minimum value will have 1 and rest zero in the starting vector.

3 2

2 0,

4 4

8 1

Diagonal k m

/u uk m

• Starting vectors can be generated by Lanczos algorithm- converges fast.

• In dynamic optimisation , where structure is modified previous vectors could be good starting values.

Eigen value problem

0 1

0 0{ }

1 0

0 0

X

/ 3/ 2, ,1,8u uk m

[ ][ ] [ ][ ][ ]

[ ] , [ ]

[ ] [ ][ ] [ ]

[ ] [ ][ ] [ ]

n p n p

Tp p

T

k m

k

k

m I

(1)

(2)

(3)

Eqn. 2 are not true. Eigen values unless P = n

If [] satisfies (2) and (3),they cannot be said that they are true

Eigen vectors. If [] satisfies (1),then they are true Eigen vectors.

Since we have reduced the space from n to p. It is only necessary that subspace of ‘P’ as a whole converge and not individual vectors.

Algorithm:

Pick starting vector XR of size n x p

For k=1,2,…..1

1 1 1

1 1 1

1 1 1 1 1

1 1 1

[ ][ ] [ ]{ }

[ ] { } [ ]{ }

[ ] { } [ ]{ }

[ ] { } [ ] { }[ ]

[ ] { } [ ]

k k

Tk k k

Tk k k

k k k k k

k k k

k X m X

k X k X

m X m X

k Q m Q

X X Q

k+1 { X }k+1 - k

static

p x p

p x pSmaller eigen value problem, Jacobi

Factorization

Subspace Iteration

1

1

1

1 1 1

1 1 1 1 1

1 1 1

[ ] [ ][ ][ ]

[ ][ ] [ ]

[ ] [ ] [ ]

[ ] [ ] [ ]

[ ] [ ] [ ] [ ] [ ]

[ ] [ ] [ ]

k

T

k k

Tk k

Tk k k

k k k k k

k k k

k L D L

k X Y

k X Y

M X Y

k Q M Q

Y Y Q

Sturm sequence check

1 1 1

1

[ ] [ ] [ ]

[ ] [ ][ ][ ]

[ ][ ] [ ][ ]

[ ][ ]

T

k k ki i i

ki

k k M

k L D L

k M

k

(1/2)nm2 + (3/2)nm

nq(2m+1)

(nq/2)(q+1)

(nq/2)(q+1)

n(m+1)

(1/2)nm2 + (3/2)nm

4nm + 5n

nq2

Total for p lowest vector.

@ 10 iteration with nm2 + nm(4+4p)+5np

q = min(2p , p+8) is 20np(2m+q+3/2)

This factor increases as that iteration increases.

N = 70000,b = 1000, p = 100, q = 108 Time = 17 hours

Use the subspace Iteration to calculate the eigen pairs (1,1) and (2,2) of the problem K = M ,where

2 1 0 0 0

1 2 1 0 2;

0 1 2 1 0

0 0 1 1 1

K M

2

2 1 0 0 0 0

1 2 1 0 2 0

0 1 2 1 0 0

0 0 1 1 0 1

X

2

2 2

2 1

4 2

4 3

4 4

2 1 6 44 ; 8

1 1 4 3

a

M

ndX

K

Example

2 2

2

1 2 1 102 4 8 4 2 4 2 8

;1 11 2

04 4 2 4 2 42 4

1 1

4 41 1

2 2

1 2 1 2

4 4

2 2

2 2

Q

and X