CHAPTER 14 TRANSVERSE VIBRATION OF EULER BEAM It was recognized by the early researchers that the bending effect is the single most important

factor in a transversely vibrating beam. The Euler-Bernoulli model includes the strain energy due

to the bending and the kinetic energy due to the lateral displacement. The Euler-Bernoulli model

dates back to the 18th century. Jacob Bernoulli (1654-1705) first discovered that the curvature of

an elastic beam at any point is proportional to the bending moment at that point. Daniel Bernoulli

(1700-1782), nephew of Jacob, was the first one who formulated the differential equation of

motion of a vibrating beam. Later, Jacob Bernoulli's theory was accepted by Leonhard Euler

(1707}1783) in his investigation of the shape of elastic beams under various loading conditions.

Many advances on the elastic curves were made by Euler. The Euler-Bernoulli beam theory,

sometimes called the classical beam theory, Euler beam theory, Bernoulli beam theory, or

Bernoulli-Euler beam theory, is the most commonly used because it is simple and provides

reasonable engineering approximations for many problems. However, the Euler-Bernoulli model

tends to slightly overestimate the natural frequencies. This problem is exacerbated for the natural

frequencies of the higher modes.

Mathematical formulation:

w

xdx

Figure 1: A beam under transverse vibration

Consider a long slender beam as shown in figure 1 subjected to transverse vibration. The free

body diagram of an element of the beam is shown in the figure 2. Here, is the bending

moment, is the shear force, and is the external force per unit length of the beam.

Since the inertia force acting on the element of the beam is

),( txM

),( txV ),( txf

2

2 ),()(t

txwdxxA

247

Figure 2: Freebody diagram of a section of a beam under transverse vibration

balancing the forces in direction gives z

2

2 ),()(),()(t

txwdxxAVdxtxfdVV =+++ ,

where is the mass density and is the cross-sectional area of the beam. The moment equation about the

)(xA

y axis leads to

02

),()()( =+++ MdxdxtxfdxdVVdMM

By writing

dxxVdV = and dx

xMdM =

and disregarding terms involving second powers in , the above equations can be written as dx

2

2 ),()(),(),(t

txwxAtxfx

txV

=+

0),(),( = txV

xtxM

By using the relation x

MV = from above two equations

2

2

2

2 ),()(),(),(t

txwxAtxfx

txM

=+

248

From the elementary theory of bending of beam, the relationship between bending moment and

deflection can be expressed as

2

2 ),()(),(x

txwxEItxM =

where E is the Youngs modulus and is the moment of inertia of the beam cross section

about the axis. Inserting above two equations, we obtain the equation of the motion for the

forced transverse vibration of a non-uniform beam:

)(xI

y

),(),()(),()( 22

2

2

2

2

txft

txwxAx

txwxEIx

=+

For a uniform beam above equation reduces to

),(),()(),()( 22

4

4

txft

txwxAx

txwxEI =+

For free vibration, , and so the equation of motion becomes 0),( =txf

0),(),( 22

4

42 =

+

ttxw

xtxwc

where

AEIc =

Initial Conditions:

Since the equation of the motion involves a second order derivative with respect to time and a

fourth order derivative with respect to , two initial equations and four boundary conditions are

needed for finding a unique solution for . Usually, the values of transverse displacement

and velocity are specified as and at

x

),( txw

)(0 xw )(0.

xw 0=t , so that the initial conditions become:

)()0,( 0 xwtxw ==

)()0,( 0.

xwttxw ==

Free Vibration:

The free vibration solution can be found using the method of separation of variables as

)()(),( tTxWtxw =

249

Substituting this equation in the final equation of motion and rearranging leads to

22

2

4

42 )()(

1)()(

=== adt

tTdtTdx

xWdxW

c

where is a positive constant. Above equation can be written as two equations: 2=a0)()( 44

4

= xWdx

xWd

0)()( 222

=+ tTdt

tTd where,

EIA

c

2

2

24 ==

The solution to time dependent equation can be expressed as

tBtAtT sincos)( += where, A and B are constant that can be found from the initial conditions. For the solution of

displacement dependent equation we assume, sxCexW =)(

where C and are constant, and derive the auxiliary equation as: s

=2,1s , is =2,1 Hence the solution of the equation becomes:

xixixx eCeCeCeCxW +++= 4321)( where , , and are constant. Above equation can also be expressed as: 1C 2C 3C 4C

xCxCxCxCxW sinhcoshsincos)( 4321 +++= or,

( ) ( ) ( ) ( xxCxxCxxCxxCxW ) sinhsinsinhsincoshcoscoshcos)( 4321 ++++++= The constants , , and can be found from boundary conditions. The natural frequencies

of the beam are computed from:

1C 2C 3C 4C

( ) 422 AlEIl

AEI

==

250

The function is known as normal mode or characteristic function of the beam and )(xW is called the natural frequency of vibration. For any beam, there will be infinite number of normal

modes with one natural frequency associated with each normal mode. The unknown

constant , , and and value of 1C 2C 3C 4C can be determined from boundary conditions of the beam.

Boundary Conditions:

Hinged, M(0, t) =0, D(0, t) =0; Free, M(0, t)=0, Q(0, t)=0

1. Free End:

Bending Moment = 022

=

xwEI ,

Shear Force = 022

=

xwEI

x.

2. Simply supported (pinned) end:

Bending Moment = 022

=

xwEI ,

Deflection = 0=w .

The value of nL is unknown and is determined using the boundary condition of the beam given. For different boundary conditions we get different equations.

While the case of simply supported beam admits a closed form solution, the other equation has to

be solved numerically. All these equations have a number of solutions each corresponding to

different modes of vibration. The first, second and third positive solutions were determined for

251

each case and the solutions obtained are used to determine the frequencies of vibration for three

modes.

1. Free End:

Frequency Equation: 0sin =ln , Mode Shape: xCw nnn sin= .

32

3

2

1

===

lll

2. Simply supported (pinned) end:

Frequency Equation: 1cos.cos =ll nn

Mode Shape: ( )

+

++= xx

xxxxxxCw nn

nn

nnnnnn

coshcoscoscosh

sinhsinsinhsin

137165.14995608.10

853205.7730041.4

4

3

2

1

====

llll

We now consider the effect of different masses at different locations. A lower bound

approximation method Dunkerleys Equation is used to determine the frequency with lumped

masses attached to the beam.

The Dunkerleys formulae is given as

252

2 2 2 21 2

1 1 1 1 .......

sys beam

sys

beam

i

wherefundamental frequency of the total systemfundamental frequency of the beam alone

fundamental frequency of the ith mass mounted on the beam alone

= + + +

==

=

The natural frequency of the mass alone is determined from the static deflection value obtained

by using the strength of materials theory

2

( )/i i i

F K xK m

= =

Rigorously speaking, all real system are continuous system A continuous system for analysis purpose can be reduces to a finite number of discrete

models. Each discrete model can be reduced to an eigenvalue problem.

In can of continuous systems the solution yield infinite number of eignvalus and eigenfunctions where as in discrete system the number of eignvalues and vector are finite.

The concept of orthogonality is applicable to both discrete and continuous systems. The eigen value problem in case of discrete system takes the from of algebraic equations

while in continuous systems differential equations and some times integral equations are

obtained. Eigenvectors of the discrete system becomes eigenfunction of the continuous

system.

Geometric boundary conditions (or essential or imposed

boundary conditions)

resulting from conditions of purely geometric compatibility

Ex

Boundary conditions Here, deflections and slope at

both ends are zero constitute geometric b.c s

Natural boundary conditions (additional or dynamic b.cs)

resulting from the balance of moments or forces.

253

Bending moment is zero at both the ends.

So in case of a simply supported beam, the natural boundary conditions are bending moment is

zero at both the ends and in geometric boundary condition is deflections are are zero at both the

ends. For a cantilever beam, while the fixed end has geometric bcs (deflection and slope zero)

and free end has natural bcs (shear force and bending moment zero)

Approximate methods

In exact method difficulties arises in

Solving roots of the characteristic equation. Except for very simple boundary conditions, one has to go for numerical solution.

In determining the normal modes of the system Determination of steady state response

So quick for determination of the natural frequencies of a system, when a very accurate result is not of much importance one should go for an approximate method.

Modeling

.

Approximation

Series solution

Approximate method where approximation error should be within acceptable limits one may

assume a series solution as

=

=

0)()(),(

nnn xtftxy (1)

where ( )n x is the normal modes and ( )nf t is the time function which depends upon initial conditions and forcing function. There are certain difficulties that limit the application of

classical analysis of continua to a very simple geometries only.

The infinite series sometimes converge very slowly and it is difficult to estimate how many terms are needed for engineering accuracy.

The formulation and computation efforts are prohibitive for systems of engineering complexity.

254

The special methods (as approximate methods) treat the continuous systems, for vibration

analysis purpose, as discrete systems. This can be done with one of the following methods.

Retaining n natural modes only and considering them as generalized coordinates and computing the n weighing functions fn(t) to best fit the initial conditions or the

forcing functions.

Substituting for the n modes an equal number of known functions n(x) that satisfy the geometric conditions of the problem and then compute the functions

fn(t) to best fit the differential equation, the remaining boundary conditions, and

the intitial conditions of the forcing functions.

Taking as generalized coordinates the n physical conditions of a certain number of points of the system q, (q,"qn) considering them as functions of time, and computing them to fit the differential equation; and the initial and boundary

conditions.

The main advantage of all theses methods is that instead of dealing with one or more

partial differential equations, one deals with a larger number of ordinary differential

equations (usually liner with constant coefficients) which are particularly suitable for

solution in fast computing machines.

Rayleighs Method

(Lord Rayleigh (1842-1919) 1868 fellow of Trinity collage. His first experimental investigation

was in electricity, but soon he turned to Acoustics & Vibration. He started writing Theory of

Sound on a boat trip up the Nile in 1872 and the book was published in 1877. 1904-Got Nobel

Prize in physics.)

-Rayleigh method gives a fast and rather accurate computation of the fundamental frequency of

the system.

-It applies for both discrete and continuous systems.

Consider a discrete, conservative system describe of by way of Matrix equation

(2) 0=+ KxMx

255

The equation above is satisfied by a set of n eigenvalues and normalized eigenvectors ,

which satisfy the equation

2i i

i =1, 2----n (3) ii MK 22=Multiplying both sides of (3) by and dividing by a scalar , which is a quadratic form,

we have

iT ,iT M

iiTiiT

i MK =2 (4)

If we know the eigenvector , we can obtain the corresponding eigen value by eqi 2i n (4). However in general, we donot know the eigenvectors is advance. Suppose that we consider an

arbitrary vector Z in eqn(3)

MzzKzzzR T

T

== )(2 (5) where R(z) depends on the vector z and is called Rayleighs quotient . When the vector z

coincides with an eigenvector , Rayleighs quotient coincides with the corresponding

eigenvalues.

i

From vector algebra it is known that we can express any vector z by a linear combination of then

linearly independent vectors (expansion theorem):

(6) =

==n

j

i ZCcizz1

)(

where Z is a square modal matrix [z(1)z(2)] and C = {c1 c2.cn}

If the vector z(i) have been normalized so that

,IMZZ T = then =KZZ T diag (7) Pwww n =]........,[ 22221using (6) in (6) and using orthogonal Property

256

=

==== ni

n

iT

T

TT

TT

ICCPCC

MZCZCKZCZCzR

1

2

1

22

)(

(8)

Example: Using Rayleigh quotient method, find the fundamental frequency for a cantilever

beam assuming the approximate function as the static deflection curve.

x

Static deflection of a cantilever beam can be found using bending equation as follows.

2

2

2.

dxydEIxwxM ==

or, 13

321 Cwx

dxdyEI +=

214

461 CxCwxEIy ++=

BCs lx =

==

0

0

dxdyy

31 61 wlC =

lxxCwxC =

+=

1

4

2 461

8624

44

4 wlwlwl +=

=

257

( )434 3424

lxlxEI

wy += (deflection from free end)

To measure x from fixed end

One may substitute '( )x l x= in the above equation.

[ ]4'34' 3)(4)(24

lxllxlETwy +=

( )[ ]32233 43324

llxxlxlEI

w += replacing x by x

[ ]43224322334 333333324

llxxlxxlxlxllxlEI

w +++++=

[ ]2234 6424

xllxxEI

w +=

Taking static deflection curvee ( )2234 6424

)( xllxxEI

wx +=

)12124(24

)( 223 xllxxEI

wdx

xd += w = weight/unit length

)122412(24

)( 222

2

llxxEI

wdx

xd += )2(2

22 llxxEIw +=

potential energy dxx

yEIl

2

0

2

221

=

= lo dxyxmwEK 22 )(21.

)(24

)64(21

)()(2

.21

2222

222234

14

02

IIE

dxwxllxxm

IdxlxEIwEI

wl

o

l

+

=

=

2

2

2

2

2

2

xyEI

xtum

( ) dttEI

EIwIo

l

41 22

1 = otlx

ltxdxdttxt

==+==

==,0

( )20102

1522

1 555 wllowtwEo

l

==

=

258

+++= l dxlxxllxxlxlxEImwI 0 65274462822

2 )1248423616()24(21

22726388

54729

)24(712

684

88

536

716

921

EIwxlxllxxlxlxm

l

o

+////

/++=

22

999999

)24(7128

88

536

716

921

EIwllllllm

+/

/++=

+++= 2

29

)24(71281

536

1716

91

21

EIwml

229

)24(296657.0

EIwml =

5 2

2 9

2 (24 )20 0.96657

l EIm t

/

52)(591963.59

mlEI= Ans

Uniformly loaded shaft:

4 2

4 2 0d y d yEIdx dt

+ = Euler Equation

General Solution

cosh sinh cos sin = + + +y a x b x c x d x .(A) 2

4

2 24( )

=

= =n n nEI

EI ElPL

I

n depends on the bcs

259

Beam Conf 21( ) fundamentall 22( ) second model 23( ) thirdmodel

Simply supported 9.87 39.5 88.9

Cantilever 3.52 22.0 61.7

Free-free 22.4 61.7 121.0

Clamped-clamped 22.4 61.7 121.0

Clamped-hinged 15.4 50.0 104.0

Hinged-free 0 50.0 4 2

4 2 0 + =d md y

dx EIdt

( ), ( ) = y x g t 4 2

4 2( ) . 0+ =d y my d gg tdx EI dt

or, 4 2

4 2 ( ) ( = =

d y my d g g t consdx EI dt

4 2

4 0=> =d y my

dx EI

4

44 0 =d y ydx Roots i ,

For simply supported

Bcs y = 0, x = 0 & L

BM = 0 2

2=> d ydx at x = 0 & L y = 0 eqn. (A)

at x = 0 y = 0 = a + c, =>2

2

d ydl

= a c = 0

sinh sin => = +y b x d x

15.4 2)t

=> a = c =0

260

x = l, y = 0 sinh sin 0 => + =b l d l sinh 0 as sinh 0 => =b l l , b=02

2 0 sinh sin = => =d y b l d ldx 0

Also, sin 0 =d l As 0, sin 0d =l if (d = 0, y = 0, leads to trivial) sin 0 = l Frequency Equation

sin sin => =l n , n = 1,2,3 => =l n , n = 1,2,3 0 l

For n = 1, first mode =l 2 2

4( ) =nEInL

44 => =n

EInL

for simply supported beam carrying has maximum deflection at midpoint. 35

384 = WL

EI W = total weight

45 / 384= gL EI

Hence, 2 2 2 5 384

= ngn 4

5384 =

EI gL

For Cantilever Beam 4

=384 gl

EI

Shaft carrying several loads:

Dunkerleys Method (Semi empirical) approximate solution

According to Dunkerleys empirical formula

261

1 2

2 2 2

1 1 1 1.......... 2 = + + +sn n n n

1 2

2 2 2

1 1 1 1..........= + + + 2sn n n n

f f f f

Dunkerleys lowerbound approximation.

Let W1, W2,.Wn be the concentrated loa sses m1, m2, . mn and 1, 2, 3 are the static deflections of the s n the load acts alone as shaft. Also let the shaft carry a uniformly di it length over its whole

span and static deflection at the mid span du

Let

n = Frequency of tranverse vibration of thens = Frequency with distributed load actingn1, n2...... = Frequency of tranverse vibrati

Energy Method (Rayleighs upper bound

Max P.E.(Extreme Position)=Max K.E (Me

1 1 2 2 3 31 1 1 ....2 2 2

= + + +W Y W Y W Y

ds on the sharft due to ma

haft under each load. Whe

stributed mass of m per une to the load of this mass be s. Also

whole system.

alone

on when each of W1, W2,W3. act alone.

approximation):

an Position)

262

Neglects mass of the beam difficult to apply in the presence of many masses.

Max P.E. 1 1 1 1( ......)2= + +g m y m y

2= g my Max K.E. 2 2 21 1 2 2 3 3

1 1 1 ......2 2 2

= + + +m v m v m v

2 2 21 1 2 2 3 3

1 1 1( ) ( ) ( ) ......2 2 2

= + +m y m y m y +

2 212= my

2 212 2

=> = g my my 2

2=> = ng my

my

Unlike Dunker leys formula, which is valid for lateral vibration of shafts only, Rayleighs

method is valid for a system performing oscillatory motion in any manner i.e., bending, torsional

or longitudinal motions.

Example:

Consider a shaft carrying three discs as shown in the figure. The influence coeffiecients are,

3

113

256= la

EI,

3

22 48= la

EI,

3

333

256= la

EI

263

Influence Coefficients: ija deflection at station i due to

unit load at station j

Using Dunkerleys formula 3

21

1 3256 =

mlEI

, 3

22

1 248 =

mlEI

, 3

23

1 3256 =

mlEI

3 3

2

1 3 2 3256 48 256 = + +n

ml ml mlEI EI EI

3

2

3

3

15.36 Dunkerly

16.199EI = Exact

nEIml

ml

= 3(3+10.66+3)ml=

256EI

23 3

256 15.3616.66

= =n EI Eml mlI

33.9191 =n EIml

Rayleigh method: Flexibility influence coefficient displacement at i due to unit load at j with all other forces equal to zero.

3

119

768= la

EI,

3

1211

768= la ,

EI

3

137

768= la

EI

3

2216

768= la

EI,

3

2311

768= la ,

EI

3

339

768= la

EI

By Maxwells reciprocal theorem the remaining influence coefficients can easily be determined.

The static deflections are therefore given by,

1 1 11 2 12 3 13= + +X m ga m ga m ga 2 1 21 2 22 3= + + 23X m ga m ga m ga 3 1 31 2 32 3 33= + +X m ga m ga m ga 1 3 2 and 2= = =m m m m m

3

138768

= ml gXEI

, 3

254768

= ml gXEI

, 3

338768

= ml gXEI

264

2 1

1

=

=

=> =

n

i ii

n n

i ii

g m X

m X3

16.2055EI=ml

( which is slightly higher than the exact value 3EI16.199ml

)

Example:

A steam turbine blade of length l, can be considered as a uniform cantilever beam, mass

m per unit length with a tip mass M. The flexural rigidity of the blades is EI. Determine

fundamental bending frequency.(Use Rayleigh Method)

23 3

.256 2563 3

n

n

g

g g EI EImgl mgl

= => = = =

Sol:

Assuming Y(x,t)=Y(x)cos t xY(x)=A(1-cos )

2l

xY( , ) Y(x)sin t=- A(1-cos )sin2l = x t t

2

0

1. .2

= = lK E T m y dx 2 2

0

1 (1 cos )2 2

= l xm Al 2dx 2 2

2

0

(1 cos )2 2 = lmA x dxl

2 2

02

2 =

lmA x

2 2

00

00

2sin sin 2( )2 22 2 2( )

2 2

+

l l

ll

x xmA xl lx

l

2 2 3 2 .2(1 0) (0 0)2 2

=

mA l ll

265

2 22 23 4 3 2( )

2 2 4

= = mA ll mA l

The K.E. of tipmass { }2 2 21 1 1= ( ( ))2 2 2

= =M y l M A M A

The K.E. of the system = 2 2 2 23 2 1 mA4 2

+ l M A

22 42

2 30

1P.E. V=2 6

= l d y EI

4EI dx

dx lA

Strain Energy

Equating max P.E. with max K.E.,

23

3.0382( 0.232 )

EIM ml l

= +

Exercise Problems

1. A shaft 40 mm diameter and 2.5 m long has a mass of 15 kg per m length. It is fixed at

both the ends and carries three masses 90 kg, 140 kg and 60 kg at 0.8 m, 1.5 m and 2 m

respectively from the left support. Taking E = 200 GN/m2, find the frequency of the

transverse vibrations. (Hinds: 1 L = 4.730). 2. A rotor has a mass of 12 kg and is mounted midway on a 24 mm diameter horizontal

shaft supported at the ends by two bearings. The mass of the shaft is 2 kg and bearings

are 1 m apart. The shaft rotates at 2400 rpm. If the center of mass of the rotor is 0.1 mm

away from the geometric center of the rotor due to a certain manufacturing defect, find

the amplitude of steady state vibration and the dynamic force transmitted to the bearing.

Take = 0.01 and E = 200 GN/ m . 23. A rotor has a mass of 15 kg and is mounted midway on a 24 mm diameter horizontal

shaft supported at the ends by two bearings. The mass of the shaft is 2 kg and bearings

are 1 m apart. Find the first two natural frequency using energy principle. E = 200 GN/

m . 2

266

CHAPTER 14 TRANSVERSE VIBRATION OF EULER BEAMApproximate methods

Rayleighs MethodStatic deflection of a cantilever beam can be found using beUniformly loaded shaft:General Solution

Dunkerleys Method (Semi empirical) approximate solutionUsing Dunkerleys formulaExercise Problems