VIBRATION ANALYSIS OF CRACKED BEAMS (ISOTROPIC AND.pdf

By Sabiju V V

12ME63R05

VIBRATION ANALYSIS OF CRACKED BEAMS (ISOTROPIC AND

FUNCTIONALLY GRADED) AND COMPOSITE PLATES USING FINITE

ELEMENT METHOD

Under the guidance of

Dr Kumar Ray

1

Introduction

Literature review

Objectives

Mathematical formulation

Finite element analysis

Results and Discussions

Conclusions

Scope of Further works

References

CONTENTS

2

Generation and propagation of crack has been crucial in life and operation of structural elements.

Crack propagation is a slow process but fracture is really rapid one.

Vibrating structures behaves differently with the presence of crack.

The parameters (flexibility, stiffness), strain energy will be different for a cracked component.

The change in dynamic response will occur with the initiation and propagation of crack.

It is necessary to study the dynamics of cracked structure to improve the life of it.

INTRODUCTION

3

D.Y. Zheng et al

Analysed a cracked beam with effect of stiffness reduction in

globally rather than local variation

H. Tada et al Explains the correction factors for stress intensity factors for

different conditions of loading.

S. Valiappan et al Proposed a new method for damage analysis in plates and

anisotropic elements.

Z. Friedman et al Explains improved model for analysis of Timoshinko beam

R.D.Mindlin Studies Influence of rotary inertia and shear on flexural motions

of isotropic elastic plates

E. Reissner et al Analysed the effect of transverse shear deformation on the

bending of elastic plates

LITERATURE REVIEW

4

To investigate the change in natural frequencies with the depth and location of crack of an isotropic cantilever beam.

To analyze the forced response of the same with the applied load.

To extend the same analysis for a Functionally graded cracked cantilever beam.

To find free and forced vibration response of a damaged composite plate with varying modulus ratio, stacking sequence, side to thickness ratio and boundary conditions.

OBJECTIVES

5

MATHEMATICAL FORMULATION

Fig.1 isotropic beam with a crack

6

Strain energy will be increased due to the existence of crack.

Where G is the strain energy release rate function and A will be the effective cracked area.

G can be expressed as

Where El = E for plain stress condition and El = E/(1-2 ) for plain strain.

KI1, KI2, KI3 and KII2 are the stress intensity factors due to loads P1, P2 and P3.


c GdA

2 2

1 2 3 2'

1[( ) ]I I I IIG K K K K

E

7

s = depth of crack/ total depth of beam

F1(s) = (1.078s4-3.048s3+3.17s2 1.195s+1.119)/(1-s)3/2

F2(s) = (2.368s4-6.319s3+6.308s23.095s+1.15)/(1-s)3/2

FII(s) = (1.194s4-3.455s3+3.693s22.425s+1.105)/(1-s)3/2

F1, F2 and FII are the correction factors for stress intensity factors[2].


11 1( )I

PK F

bh h

22 22

6( )cI

P LK F

bh h

33 22

6( )I

PK F

bh h

22 ( )II II

PK F

bh h

s = x/h

8

From the total potential energy we have

Elements overall additional flexibility

Cov = Ccrack + Cintact


ci

iP

i = 1,2,3

( 1,2,3)iijj

c i jP

222 31

1 2 22 2

6 6{[ ( ) ( ) ( )]cij

i j

P L PPbc F F F

E P P bh h bh h bh h

2

22

2 2( )}II

PF d

b h h

9

Material properties of Functionally Graded beam is assumed to vary in Z direction only.

The bottom surface is pure metal(Al) and the top surface is pure ceramic

Effect of Poissons ratio on the deformation is much less than that of Youngs modulus. So it is considered as constant through out the analysis.

Power law is utilised for defining the variation in youngs modulus with the depth

k is an index which determines the metal fraction.


2 3Al O

1( ) ( )( )

2

k

t b b

zE z E E E

h

10

Displacement field is based on First order Shear Deformation Theory.

U(x, z, t) =U0 (x, t) z (x, t) W(x, z, t)= W0(x, t)

U0 (x, t) & W0(x, t) are the mid plane displacements.

Strain displacement relation is given by

Using Hamiltons Principle Differential Equation of Motion can be found out.


0

0 0

x

xx

xzx

u zx

xw

x

2

1[ ] 0

t

e e et

T U W dt

11

In composite plates also FSDT is used to define the displacement field.

The linear strain-displacement relation is as follows

=

=

=

+

=

+

=

+

Constitutive relation for the each lamina is as follows

=

11 =11

11221 22 =

22

11221 12 =

2111

11221 Q44 = G23, Q55 = G13, Q66 = G12


0

0

0

( , , , ) ( , , )( , , , )

( , , , ) ( , , , ) ( , , )

( , , , ) ( , , , ) 0

x

y

u x y z t x y tu x y z t

v x y z t v x y z t z x y t

w x y z t w x y z t

12

Constitutive equation for the damage materials[8]

Fig. 2 Illustration of damage material

AI-A1: A1, = the area of damaged material with unit normal ni , Ai* : = the effective

area of A1. A2-A2,: A2 = the area of damaged material with unit normal n, A* 2 = the

effective area of A2


=

13

Thus the stress tensor can be expressed as follows

Depends on the variation of i the state of stress as well as the stiffness at the desired element changes. It can vary from 0.1 to 0.9.


11 12 21 22

=

111 1

121 2

211 1

221 2

14

The boundary conditions are as follows

Fig. 3 SS-1 Boundary condition is used for cross ply stacking sequence and SS2 for angle ply stacking sequence.


15

A typical cracked beam subjected to axial force, shear force and bending moment

Fig. 4 finite element model of beam.

The relationship between displacement and force can be expressed as

FINITE ELEMENT ANALYSIS

uj(Uj)

(Qj) (Qi)

ui(Ui)

Lc vj(Vj) vi(Vi)

a

Le

i j

j i j

j i e i ovl j

j i j

u u U

v v L C V

16

u, v, are deflections and U,V,Q are corresponding loads.

is the overall flexibility matrix of the cracked beam.

Cintact is the flexibility matrix for the intact beam.

Total flexibility will be the sum of the above two.

The above expression gives rise to the stiffness matrix of the cracked beam element, where L is a transformation matrix that is obtained using equilibrium conditions

Ctot = Cintact + Covl


ovlC

1 T

c totK LC L

17

Formulation for plate element

The strain-displacement relation for plate is given as

where


0 0

1 1 1

0 0

2 2 2

0 0

6 6 6

0

4 4

0

5 5

xx

yy

xy

yz

zx

uzk

x

vzk

x

u vzk

y x

w v

y z

u w

z x

0 01

u

x

0 02

v

x

0 0 06

u v

y x

0 04

w

y

0x

w

x

0

5

0

1xk

x

0

2

yk

y

0

6

yxky x

18

Mid plane strain vector is defined as follows

Also

Where

[L] is the operational matrix.


0 0 0 0 0 0 0 01 2 6 1 2 6 4 5T

k k k

8 1 5 18 5X XX

L

0 0 0 T

x yu v w

19

Potential Energy=Strain Energy


1

2

T

U dV

20

Potential energy can be written as

Where

Eight noded isoparametric element is considered here

Fig. 5 Eight noded isoparametric element in natural coordinates with node number.


Q

1

2

T

U D dA

1 1

[ ] [ ][ ]zkN

T

k zk

D T Q T dz

1(-1,-1) 2(1,-1)

3(1,1) 4(-1,1)

5(-0,-1)

6(1,0)

7(0,1)

21

Element bending stiffness matrix is defined as

where

Keij is computed numerically as

J is Jacobian matrix given by

J =


( ) ( ) ( ) ( )e e T e eK B DB dA ( ) ( )[ ] [ ][ ]e eB L N

1 1

( )

1 1

dete Tij i jK B DB Jd d

x y

x y

22

The following characteristics were studied for the beams

Free vibration characteristics.

cracked/intact Vs dcrack/d

cracked/intact Vs Lcrack/L

cracked = natural frequency of cracked beam

intact = natural frequency of intact beam

dcrack = depth of crack Lcrack = location of crack

d = depth of beam L = length of beam

Forced vibration characteristics

Deflection Vs applied frequency (for various crack depths)

RESULTS AND DISCUSSIONS

23

The stress correction factors for stress intensity factors are found out from the correlation results from stress concentration data book[2]

Polynomial expressions are made by curve fitting using MS Excel.

Free vibration response

cracked / intact Vs depth ratio

cracked / intact Vs location ratio

depth ratio = depth of crack/depth of beam

location ratio = location of crack from fixed end/length of beam


24

Fig. 6 natural frequency ratio Vs depth ratio


y1 = -5.726x6 + 11.09x5 + 0.891x4 - 16.38x3 + 14.84x2 - 5.763x + 1.042

y2 = 9.4395x6 - 38.99x5 + 64.972x4 - 56.061x3 + 26.797x2 - 7.1942x + 1.0352

y3 = 16.129x6 - 62.007x5 + 96.132x4 - 76.999x3 + 33.957x2 - 8.2487x + 1.0364

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

0 0.2 0.4 0.6 0.8 1 1.2

cra

ck

ed

na

tura

l fr

eq

ue

nc

y r

atio

depth ratio

Series1

Series2

Series3

25

Fig. 7 natural frequency ratio Vs location ratio


y1 = -0.033x2 - 0.043x + 1.000

y2 = -0.1156x2 - 0.0618x + 1.0039

y3 = 0.1501x3 - 0.3758x2 + 0.0121x + 1

0

0.2

0.4

0.6

0.8

1

1.2

0 0.2 0.4 0.6 0.8 1 1.2

cra

ck

ed

na

tura

l fr

eq

ue

nc

y r

atio

location ratio

Series1

Series2

Series3

26

Figure 6 and 7 depicts the relation between the first three natural frequencies with the depth and location of crack respectively.

It is clear that frequencies decrease gradually with the crack propagation.

From the figures (6,7) it is clear that variation of frequencies with depth is more predominant than that with the location

Approximate polynomial expressions are found out by curve fitting using MS Excel


27

Natural frequency ratio to the depth ratio

1 = -5.726s6 + 11.09s 5 + 0.891s 4 - 16.38s 3 + 14.84s 2 - 5.763s + 1.042

2= 9.439s 6 - 38.99s 5 + 64.97s 4 - 56.06s 3 + 26.79s 2 - 7.194s + 1.035

3 = 16.12s 6 - 62.00s 5 + 96.13s 4 - 76.99s 3 + 33.95s 2 - 8.248s + 1.036

Natural frequency ratio to the location ratio

1 = -0.033 2 - 0.043 + 1.000

2 = -0.115 2 - 0.061 + 1.003 a = location ratio

3 = 0.150 3 - 0.375 2 + 0.012 + 1


28

Previous equations gives us natural frequencies readily after supplying depth ratio or location ratio. (Modulus of elasticity, density of the materials, length, breadth, depth, shear correction factor, Poisson's ratio remain constants)

Depth and location of crack can be found out if natural frequencies are measured/supplied. Thus it can be decided whether the crack is severe or mild.

Depth of crack

s = -2.154 16 +2.012 1

5 + 8.373 14 -17.86 1

3+4.05 12 -5.418 1+1.009

Location of crack

= -72.51 12+126.7 1-54.24


29

Fig. 8 natural frequency ratio Vs depth ratio for an FGM beam


y 1= 29.56x6 - 108.1x5 + 157.9x4 - 117.3x3 + 46.94x2 - 9.940x + 1.031

y2 = -28.39x6 + 87.54x5 - 100.3x4 + 50.05x3 - 7.274x2 - 2.578x + 1.008

y 3= -40.68x6 + 118.1x5 - 122.1x4 + 48.32x3 + 0.061x2 - 4.800x + 1.049

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

0 0.2 0.4 0.6 0.8 1 1.2

Series1

Series2

Series3

Poly. (Series1)

Poly. (Series2)

Poly. (Series3)

30


y1 = -15.23x6 + 52.83x5 - 71.62x4 + 46.90x3 - 13.71x2 + 0.059x + 1.000

y2 = 0.361x3 - 0.509x2 - 0.408x + 1.006

y3 = 25.38x6 - 75.02x5 + 78.68x4 - 31.98x3 + 2.067x2 + 0.207x + 0.998

0

0.2

0.4

0.6

0.8

1

1.2

0 0.2 0.4 0.6 0.8 1 1.2

_c

rac

ke

d/

_in

tac

t

Series1

Series2

Series3

Poly. (Series1)

Poly. (Series2)

Poly. (Series3)

Fig. 9 natural frequency ratio Vs location ratio for an FGM beam

31

Forced response

Applied force is sinusoidal (F0cost) in nature.

Deflection Vs forcing frequency curves are plotted for different crack depths.


0 100 200 300 400 500 600 700 800 900 100010

-10

10-8

10-6

10-4

10-2

100

102

Frequency (rad/s)

Response (

mm

)

0 100 200 300 400 500 600 700 800 900 100010

-10

10-8

10-6

10-4

10-2

100

102

Frequency (rad/s)

Response (

m)

Fig 10 forced response of Intact beam Fig 11 forced response of Beam with 5 mm crack 32


0 100 200 300 400 500 600 700 800 900 100010

-10

10-8

10-6

10-4

10-2

100

102

Frequency (rad/s)

Response (

m)

Fig 12 Forced response of beam with 50 mm

crack

Fig 13 Forced response of beam with depth

as crack depth 33

Fig. 10 to 13 represent the forced characteristic curve (deflection Vs applied frequency)

The response changes drastically as the crack advances.

As the crack depth become depth of beam it separates, with the application of load the separated piece will deflect all of a sudden and then no further movement. (Shown clearly in Fig. 13)

Free and forced response of a composite plate are shown ahead

Nondimensional natural frequency Vs % of damage

Nondimensional central deflection Vs % of damage has been studied

for different E1/E2, stacking order, side/depth ratios


34


y1 = -4E-09x6 + 5E-07x5 - 2E-05x4 + 0.0006x3 - 0.0077x2 + 0.0638x + 0.4491

y2 = -3E-09x6 + 3E-07x5 - 2E-05x4 + 0.0004x3 - 0.005x2 + 0.0416x + 0.2927

y3 = -8E-09x6 + 1E-06x5 - 5E-05x4 + 0.0011x3 - 0.016x2 + 0.1329x + 0.9331

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

0 5 10 15 20 25 30 35 40 45

no

nd

ime

nsi

on

al c

en

tra

l d

efle

ctio

n

% of damage

central deflections Vs % of damage (cross ply)

W_max1

W_max2

W_max3

E1/E2=10

E1/E2 = 25

E1/E2=40

Fig. 14 Central deflection Vs % of damage of cross ply composite plate with different E1/E2 ratios 35


y1 = 5E-08x6 - 6E-06x5 + 0.0003x4 - 0.0079x3 + 0.1102x2 - 0.8581x + 14.915

y2 = 6E-08x6 - 8E-06x5 + 0.0004x4 - 0.0098x3 + 0.1358x2 - 1.0609x + 18.476

y3 = 3E-08x6 - 4E-06x5 + 0.0002x4 - 0.0055x3 + 0.0777x2 - 0.5992x + 10.349

0

2

4

6

8

10

12

14

16

18

20

0 5 10 15 20 25 30 35 40 45

no

nd

ime

nsi

on

al n

atu

ral fr

eq

ue

nc

y

% of damage

natural frequencies Vs % of damage (cross ply)

Omega_natural1

Omega_natural2

Omega_natural3E1/E2 = 10

E1/E2 = 25

E1/E2 = 40

Fig. 15 Natural frequency Vs % of damage of cross ply composite plate with different E1/E2 ratios 36


y1 = -3E-09x6 + 3E-07x5 - 2E-05x4 + 0.0004x3 - 0.0048x2 + 0.0362x + 0.2478

y2 = -2E-09x6 + 2E-07x5 - 1E-05x4 + 0.0002x3 - 0.003x2 + 0.0232x + 0.1577

y3 = -6E-09x6 + 7E-07x5 - 3E-05x4 + 0.0008x3 - 0.011x2 + 0.0821x + 0.5753

0

0.2

0.4

0.6

0.8

1

1.2

0 5 10 15 20 25 30 35 40 45

no

nd

ime

nsi

on

al c

en

tra

l d

efle

ctio

n

% of damage

Central deflection Vs % of damage (angle ply)

W_max1

W_max2

W_max3

E1/E2=10

E1/E2=40

E1/E2=25

Fig. 16 central deflection Vs % of damage of angle ply composite plate with different E1/E2 ratios 37


y1 = 8E-08x6 - 1E-05x5 + 0.0005x4 - 0.0117x3 + 0.1586x2 - 1.1708x + 20.078

y2 = 1E-07x6 - 1E-05x5 + 0.0006x4 - 0.0147x3 + 0.1986x2 - 1.4742x + 25.169

y3 = 5E-08x6 - 6E-06x5 + 0.0003x4 - 0.0077x3 + 0.1042x2 - 0.7558x + 13.179

0

5

10

15

20

25

30

0 5 10 15 20 25 30 35 40 45

no

nd

ime

nsi

on

al n

atu

ral fr

eq

ue

nc

y

% of damage

natural frequency Vs % of damage (angle ply)

Omega_natural

Omega_natural2

Omega_natural3

E1/E2=10

E1/E2=25

E1/E2=40

Fig. 17 natural frequency Vs % of damage of cross ply composite plate with different E1/E2 ratios 38


y = -4E-09x6 + 5E-07x5 - 2E-05x4 + 0.0006x3 -

0.0078x2 + 0.0646x + 0.4544

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 10 20 30 40 50

no

nd

ime

nsi

on

al c

en

tra

l d

efle

ctio

n

% of damage

central deflection Vs % of damage

central deflection

y = 5E-08x6 - 6E-06x5 + 0.0003x4 - 0.0079x3 +

0.1097x2 - 0.8525x + 14.825

0

2

4

6

8

10

12

14

16

0 10 20 30 40 50

no

nd

ime

nsi

on

al n

atu

ral fr

eq

ue

nc

y

% of damage

natural frequency Vs % of damage

natural frequency

Fig. 18 central deflection Vs % of damage Fig. 19 natural frequency Vs % of damage

Moderately

thick plate

39


y = -6E-09x6 + 7E-07x5 - 4E-05x4 + 0.0009x3 -

0.0114x2 + 0.0882x + 0.621

0

0.2

0.4

0.6

0.8

1

1.2

1.4

0 10 20 30 40 50

no

nd

ime

nsi

on

al c

en

tra

l d

efle

ctio

n

% of damage

central deflection Vs % of damage

central deflection

y = 4E-08x6 - 6E-06x5 + 0.0003x4 - 0.007x3 +

0.0961x2 - 0.7202x + 12.629

0

2

4

6

8

10

12

14

0 10 20 30 40 50

no

nd

ime

nsi

on

al n

atu

ral fr

eq

ue

nc

y

% of damage

natural frequency Vs % of damage

natural frequency

Fig. 20 central deflection Vs % of damage Fig. 21 natural frequency Vs % of damage

Thick plate

40

Natural frequency reduces with the advance of crack in beams.

Central deflection goes on increase with the increase in damage.

Natural frequency goes on decrease with the increase in damage.

With increase in E1/E2, both central deflection and natural frequency increases.

FSDT is accurate for thin plates only so the analysis of thick and moderately thick plate may result in error.

CONCLUSIONS

41

It includes

Non-linear free and forced vibration analysis of Functionally Graded cracked beam with different boundary conditions.

Buckling analysis of composite beam and plate with crack

Free and forced vibration analysis of composite shells with crack.

SCOPE OF FURTHER WORK

42

1. Free vibration analysis of a cracked beam by finite element method, D.Y. Zheng et al 2003 Journal of Sound and Vibration, 457475

2. The Stress Analysis of Cracks Handbook, H. Tada et al, ASME Press, New York, 2000.

3. An improved two-node Timoshenko beam finite element, Z. Friedman et al1992, computers & structures, 473-481,

4. Simplified models for the location of cracks in beam structures using measured vibration data, J.K. Sinha, M.I. Friswell and S.Edwards 2001 Journal of Sound and Vibration, 13-38.

5. The effect of transverse shear deformation on the bending of elastic plates, E. Reissner, ASME. Appl.Mech, 12[2].69-77, 1945

6. Influence of rotary inertia and shear on flexural motions of isotropic elastic plates, R.D.Mindlin, ASME.Appl.Mech, 18, 31-38, 1951.

7. .Mechanics of Composite Materials, Arthur Kaw, CRC Press, Taylor & Francis Group, 2006.

8. Finite element analysis of anisotropic damage mechanics problems, S. Valiappan, V. Murti and Zhang Wohua, Engineering Fracture Mechanics, 1990

REFERENCES

43

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VIBRATION ANALYSIS OF CRACKED BEAMS (ISOTROPIC AND.pdf