PDF Archive

Easily share your PDF documents with your contacts, on the Web and Social Networks.

Share a file Manage my documents Convert Recover PDF Search Help Contact



Chellapilla 3 ICONE10 Double Bellows 2002 .pdf



Original filename: Chellapilla_3_ICONE10-Double Bellows_2002.PDF
Title: Microsoft Word - ICONE10-Double Bellows.doc
Author: CKRao

This PDF 1.3 document has been generated by PScript5.dll Version 5.2 / Acrobat Distiller 5.0 (Windows), and has been sent on pdf-archive.com on 25/06/2011 at 11:18, from IP address 119.235.x.x. The current document download page has been viewed 793 times.
File size: 188 KB (4 pages).
Privacy: public file




Download original PDF file









Document preview


ICONE 10
Tenth International Conference on Nuclear Engineering
April 14-18, 2002
Arlington, Virginia, USA

“DRAFT”

ICONE 10-22092
TRANSVERSE VIBRATIONS OF DOUBLE BELLOWS
EXPANSION JOINT RESTRAINED AGAINST ROTATION
KAMESWARA RAO.C
Bharat Heavy Electricals Limited
Hyderabad, India

Keywords:
Bellow vibration, double bellows, transverse
frequency & elastic constraints

RADHAKRISHNA.M
Indian Institute of Chemical Technology
Hyderabad, India

industry, the elongation/ contraction of the piping
system is quite significant and frequent changes in
temperature or relative mechanical motion of

ABSTRACT

structures with respect to each other. Therefore, the

The paper presents the results of investigation of

bellows are very susceptible to vibrations that can

transverse vibrations of single bellows expansion

be easily excited either structurally through the

joint restrained against rotation on either end. A

fixed ends of the expansion joints or by the fluid

theoretical model is developed based on the

flowing inside the bellows.

Bernoulli-Euler beam theory and includes added

transverse vibration of double bellows expansion

mass of the fluid flowing inside the pipe-bellow-

joint consider the bellows directly welded to pipe

pipe assembly. Neglecting effects of shear and

flange and uses fixed-fixed beam approach for

rotary inertia an exact frequency equation is

investigation of transverse mode shapes and natural

derived for the transverse vibrations of single

frequency [1].

Most studies of

bellows expansion joint including the effects of end
elastic restraints against rotation. Numerical results

However, in real practice the bellow is welded to a

are presented for an example bellow showing the

short pipe nipple. The other end of the nipple has a

effects of variation of elastic restraints and internal

flange and is connected to the pipe flange. Figure 1

pressure on the first four modes of vibration.

depicts the bellow-nipple-flange assembly.

Introduction

The paper presents an analysis of estimating the

Bellows expansion joints are designed to absorb the

frequency and mode shapes of transverse vibrations

axial or transverse displacements in pipeline

of double bellows expansion joint restrained

systems. It is observed that any piping system is

against rotation on either end. The analysis forms

more or less flexible, and so has the natural

the basis of development of a theoretical model for

flexibility to accept small elongation or contraction.

the bellow considering as a Bernoulli-Euler Beam.

However, in many of the applications such as the

The model considers fluid filled bellows and

fossil/nuclear power plants or in aircraft and space

neglects the effect of shear and rotatory inertia.

More precise formulae are developed based on the solution of

Differentiating the above equation (2) and substituting

Bernoulli-Euler differential equation for calculating the natural

into differential equation (1) –

frequency of bellows. Results obtained using the Bisection
method are compared with results presented in thesis [1].

d4 X
[Pπ Rm2 + J ω2] d2X
m tot
X ------- + -------------------- ----- − W2 ---- = 0
dx2
EI
dx2
EI

(4)

√ (Pπ Rm2 + Jω2 )
If c = -----------------------√2 EI

(5)

The Differential Equation
Figure 2 shows the various geometrical dimensions of U-shaped
bellows –where R1 and R2 is the inner and outer radius of bellow,
p is the pitch, h is the convolution height, t is the thickness of
bellows wall and L is the length of bellow.

λ=

(mtot W2 )1/4
--------------(EI)1/4

(6)

Fig. 1 : Flange-Pipe Nipple-Bellow assembly

d(X4)
d X2
------+ 2c2 ---- − λ4 X = 0
dx2
dx4

A

(7)

The general solution of the equation is given byX = A sinh αx + B cosh αx + C sin βx + D cosβx

(8)

The first three derivatives of equation (8) are given as
follows–

5

Fig. 2 : Geometry of Bellows

The general form of the differential equation of vibration of
bellows for single or double bellows expansion joint is given by –
∂4W
∂2W
J * ∂4ω
∂2
2
EI -------- + Pπ Rm ------ − ----------- + m tot ------- = 0
∂x2
∂x2 ∂t2
∂t2
∂x4

(1)

d(X)
---- = Aα cosh αx + Bα sinh αx + Cβ cos β Dβ sinβx
dx

(9)

d2(X)
----- = Aα2 sinh αx + Bα2 cosh αx - Cβ2 sin βx - Dβ2cos βx
dx2
(10)
d3(X)
----- = Aα3 cosh αx + Bα3 sinh αx - Cβ3 cos βx + Dβ3 sinβx
dx3
(11)

Where EI is the bending stiffness, P-internal pressure, J-mass
moment of inertia per unit length, m tot –total mass of bellows per

Where the roots of the equation are α & β and their

unit length includes bellows material mass and fluid mass, x –

values are given by -

axial coordinate, R m –is the mean radius of bellow, W-deflection

α = √ -c2 + √ (c4 + λ4)

(12)

β = √ c2 + √ (c4 + λ4)

(13)

and t-time [2].
Using the technique of Separation of variables, the lateral
deflection of the bellows axis ‘w’ can be expressed asW (x, t) = X (x) T (t)

(2)

A, B, C, D are arbitrary constants

Exact Solution of Double Bellows Expansion Joint

As the end B does not rotate the third boundary

Lateral Mode Natural Frequency

condition, slope is zero and given by -

A case of vibration of bellows in lateral mode is shown in

∂W (L, t)
------------ = 0
∂x

figure 2. It is seen that the pipe has pure translational motion

(18)

due to the geometry and physical symmetry of the system

The fourth boundary condition at end ‘B’ is the shear

provided the coriolis component of force acting on the

force Q (L, t) of the bellows and is given by-

bellows from the fluid flowing inside is neglected.
As a mathematical approximation one half of the system is
considered with its left end having a rotational stiffness ‘R’
and right end fixed to the vertical rollers and is shown in
figure 2.

ks
d3 X(L) −W2{Ms + (m p + mf3) a}
----------= ------------------------------- * X (L) + ---- (L) (20)
EI
EI
d x3
−W 2 {Ms + (m p + mf3) a} k s X(L)
If b = --------------------------------- − -------EI
EI

Š

B

5
$

∂3W(L, t) {Ms + (m p + mf3) a} ∂2W(L, t) k s W(L, t)
-------- = ------------------------ ---------- + --------(19)
EI
∂t2
EI
∂x 3

Ms is the equivalent lateral support mass, k

(21)

s

is the

equivalent spring stiffness of the lateral support, m p is

?

the mass per unit length of the connecting pipe of

Fig. 3 : Mathematical model of universal expansion
joint for lateral mode

length ‘a’ and m

f3

is the mass of the fluid per unit

length in the connecting pipe.
Substitution of the general solution (8) and its
derivatives into boundary condition expressions (14),
(15), (16) and (17) will give a set of linear equations
Boundary Conditions

with respect to the constants A, B, C and D.

As shown in figure 3, end ’A’ of the bellow is connected to

B+D=0

(22)

{B (αL) 2 − D (βL) 2 – T (A αL + C βL)} = 0

(23)

a pipe nipple, and considered to have a rotational stiffness
of ‘R’. Now the first two boundary conditions are written as
follows -

Aα cosh αL + Bα sinh αL + Cβ cos βL - Dβ sinβL=0
(24)

Deflection is zero at W (0, t) = 0 and
R ∂ X (0)
EI ∂2 X (0)
---- ------------ = ---- ------L2
∂x2
L
∂x
∂2 X (0)
∂ X (0)
---------- = T ----------∂x2
∂x

(14)

[α3 cosh αL + b sinh αL] A + [α3 sinh αL + b cosh
αL] B − [β3 cos βL − b sin βL] C + [β3 sin βL + b

(15)

cos βL] D = 0

(25)

Substituting c1, c2, c3, c4 for the terms in the brackets in

According to [1], the maximum allowable pressure in

equation (25) we get -

bellows is –

c1 = α cosh αL + b sin hαL

(26)

c2 = α3 sinh αL + b cosh αL

(27)

c3 = β3 cos βL − b sinβL

(28)

c4 = β3 sinβL + b cos βL

(29)

3

Pmax = π k* P
--------6.666L2

(32)

Where k is the equivalent axial stiffness of bellows -

For a non-trivial solution the determinant formed by the

Substitution of the above numerical values into the
expressions 4 & 5, we get –

coefficients of the system of algebraic equations above

c = √ 616.49 + 0.0001135ω2

(33)

must be equal to zero-

λ = 1.0029√ω

(34)

The expansion of the above determinant results in the

b = 0.10235ω2

(35)

frequency equation for the system shown in figure 2.
0

1

+ T (α)

(α)

2

0

1

+T(β)

A

+(β)

2

B

αcosh(αL) αsinh(αL) β cosβL -βsinβL
c1

c2

c3

=0

(30)

C

c4

The study involves the effect of varying the equivalent
support stiffness and mass on the frequency of vibration.
Using expressions in (12), (13), (32), (33), (34) and (35)

D

the frequency equation (31) is solved by applying
bisection method. The results are presented in Tables 1,2
2

2

T {b [(α −β ) sinβL sinh αL + 2αβ (1- cos βL cosh αL]

and 3 and the same are also graphically represented in

− αβ (α2+ β2) (α sinh αL cos βL+ β cosh βL sin βL) + b

figures 4 and 5 respectively.

(α2 + β2) (α sinβL cosh αL - β cos βL sinh αL -αβ (α2
+β2) cos βL cosh βL

(31)

Tables 1 and 2 presents the frequencies obtained by
varying the support stiffness ks for mode numbers N=1 and

Fig 4: Transverse Vibrations-Double Bellows
(Lateral Mode) & for N=1 and N=2 & Ms =0

N=1
N=2

2. It is seen that as stiffness increases from a value of 102

16000

to 1015, frequency increases from 331.40rad/s to

14000

3400.1597rad/s, a steep rise by almost 90%. The

Lateral Frequency, w rad/s

12000

frequency attains almost a constant value from 108

10000

8000

onwards.

6000

4000

The effect of varying the support mass on the frequency of

2000

0
1

10

100

1000

10000

100000

1000000

10000000 100000000

1E+09

1E+10

Support Stiffness, ks (kg/m)

vibration is also studied and the results presented in Table
3. It is seen that there is a steep drop of 87% in frequency

Results and Discussion

as the support mass increases from 1.0 to 100 for N=1 and

Let us take a double bellows expansion joint having the

by a small drop of 1% for N=2.

following geometrical and physical parameters -bellow
length L=0.0693m, mass moment of inertia per unit
length J=0.001153kgm, EI =5.078Nm2, total bellows
mass mtot = 5.138kg/m, total connecting pipe mass,
mp+mf3 = 5.0kg/m. and assuming a= L.


Chellapilla_3_ICONE10-Double Bellows_2002.PDF - page 1/4
Chellapilla_3_ICONE10-Double Bellows_2002.PDF - page 2/4
Chellapilla_3_ICONE10-Double Bellows_2002.PDF - page 3/4
Chellapilla_3_ICONE10-Double Bellows_2002.PDF - page 4/4

Related documents


chellapilla 2 vibrations of bellows using fea vetomac 2 2002
chellapilla 3 icone10 double bellows 2002
chellapilla 4 icone 11 vibrations of bellows 2003
chellapilla 7 icone 13 2005
chellapilla 5 axial vib bellows tws 2004
chellapilla 11 vibrations of circular plates cbitconf 2009


Related keywords