lecture_23 _12-10-2005

7/30/2019 Lecture_23 _12-10-2005

1/6

Lecture Notes of Dr. Ramamurti: FEM for Vibrations Lecture 2312 th Oct 2005

CAE, e-Engineering Solutions, Chennai Center 23- 1

Eigen Value problem

A typical example for Eigen value problem is a turbine, which consists of number of blades

with lacing rods. The schematic view is given below.

In our case let us consider a system, which consist of 4 rods (to simplify the problem

connected with each other at right angles. The pictorial representation is shown below,

Four different modes of deformation would posses the following combinations

1) Tensile / Tensile / Tensile/ Tensile2) Compressive / Compressive/ Compressive / Compressive3) Tensile / Tensile / Compressive / Compressive4) Compressive / Compressive / Tensile / Tensile

Component (before deformation)

Component deformation (mode 1 & mode2)

Tensile deformation

Compressive

deformation

7/30/2019 Lecture_23 _12-10-2005

2/6



Mode 3 Mode 4

Analogy

4 rods forming a square capable of withstanding tension / compression loads.

Each node has 2 dof, u in horizontal and v in vertical direction.

f1h= 21212

vvuu

l

AE

Tensile deformation

Compressive

deformation

u2 u1v1

1

43

2

u4

v4

u3 v4

v2

7/30/2019 Lecture_23 _12-10-2005

3/6



f1v= 41412

vvuul

AE

Net radial force = 4242122 vvuuulAE

Net tangential force = 42142 22

vvvuul

AE

=

y

x

F

F

v

u

v

u

vu

v

u

l

AE

4

1

4

4

3

3

2

2

1

1

00110211

11001120

11001102

2

=m2[I]{}

For mode I: for ei, when =0 e0=1

1(displacement vector in sector 1)= 4421

1

v

u

taking the first two rows

=

1

12

1

1

10

01

)1012()1010(

)1010(1012

2 v

um

v

u

l

AE

=

1

12

2

1

00

04

2 v

um

v

u

l

AE

1

2

142

umul

AE

7/30/2019 Lecture_23 _12-10-2005

4/6



lm

AE22

For mode 2 when =900

ei=cos90+isin90=i

2=i13=i

21=-14=i31=-i1Substituting in basic equation

=

1

12

110

01

)2()(

)(2

2 v

um

iiii

iiii

l

AE

lm

AE

vmvl

AE

umu

l

AE

ivuif

vmviul

AE

umivul

AE

2

42

4

2

)22(2

)22(2

2

1

2

1

1

2

1

11

1

2

11

1

2

11

For mode 3 when 1=1800

3=(cos180+isin180)=-1

lm

AE

vmvl

AE

v

um

v

u

l

AE

A

2

42

10

01

10121010

1010112

2

;

2

1

2

1

1

12

1

1

12413

7/30/2019 Lecture_23 _12-10-2005

5/6



4.1 Simultaneous iterative Scheme for conventional problems

[k]{}=m2{} - Non standard Eigen value problem

[A]{}={} - Standard Eigen value problem

For distributed elasticity the natural frequencies are lower one or closely placed, otherwise

the spacing is quite wide.

For most engineering problem the first 30 Eigen pairs will be below 100cps

Steps involved in SIM (simultaneous Iterative Method)

Choleskey factorization of [k](nxb)[k]=[L][L]T

Iteration I: -

[m]2{}=[k]{} 1

[L][L]T{}=2[m]{} 2

Let [L]T{} ={q}

Then {}=[L]-T{q}

Substitute in 2

[L]{q}=2[m][L]-T{q}

Pre multiply by [L]-1on both sides

[L]-1[L]{q}=2[L]-1 [M][L]-T{q}

If (1/2=)

{q}=[A]{q} where [A]= [L]-1

[M][L]-T

Where is the Eigen value and {q} is the Eigen vector

Iteration logic: complete problem in real domain

Iteration II: Assume {q1}

7/30/2019 Lecture_23 _12-10-2005

6/6



q1 =

1000

01000010

0001

,bn

[q][q]T=[I]

Pre multiply by [A]

[L]-1[m][L]-T{q}

This equation consist of three parts

1. Forward substitution2. Straight forward multiplication3. Back substitution

Let [A]{q}={}(n,m)

IInd iteration

Orthonormalization by [r](m,m)

Let [D]=[L][L]T;Let [r]=[L]-T

Then {q2}={2}(n,m){r}(m,m)

{q2}={2}[L]-T

mnT

mmmnLq

,2,,2

Iteration logic: Complex numbers

Assume {q}n

For computing [A] involves [L]-1[m][L]-T, consider the developed program.

lecture_23 _12-10-2005

Documents