-
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