dynamic characteristics of a generalised suspension system

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International Journal of Mechanical Sciences 50 (2008) 30–42

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Dynamic characteristics of a generalised suspension system

Ding Zhoua, Tianjian Jib,�

aCollege of Civil Engineering, Nanjing University of Technology, Nanjing 210009, People’s Republic of ChinabSchool of Mechanical, Aerospace and Civil Engineering, The University of Manchester, Manchester, UK

Received 9 August 2006; received in revised form 24 May 2007; accepted 27 May 2007

Available online 13 June 2007

Abstract

The paper studies analytically the free vibration of a generalised suspension system. The system is composed of a rigid body with the

symmetric shape about its centroid and two symmetrically inclined massless strings with infinite axial stiffness. The two hanging points

have horizontal and vertical elastic supports. The rigid body is eccentrically acted by horizontal, vertical and rotational springs. A typical

application of the model is to represent a section of cable suspended bridges involving the lateral vibration of small amplitude. The

governing differential equations with three degrees-of-freedom are derived where the effect of gravitational potential is considered. The

static equilibrium position is determined simultaneously. When the elastic constraints are applied at the centroid of the rigid body,

the vibration modes of the system can be divided into symmetric and antisymmetric ones which are studied independently. The analytical

expressions of natural frequencies for some special cases are also provided. The effects of various parameters, especially the length and

inclined angle of the strings, on the natural frequencies of the system are examined. Some useful conclusions are obtained. Simple model

demonstration and verification are conducted to verify some findings in the paper.

r 2007 Elsevier Ltd. All rights reserved.

Keywords: Suspension system; Pendulum system; Cable-supported bridge; Inclined string; Lateral vibration; Dynamic characteristic

1. Introduction

The lateral vibration of the London Millennium Foot-bridge has drawn particular attention in design andresearch communities around the world [1–9]. The lateralvibration of bridges, such as suspended footbridges, archbridges and truss bridges, were also observed in Japan[5,10], Canada and New Zealand [2]. For explaining thecauses of the unexpected lateral vibration of the Millen-nium Footbridge, Roberts [7,8] modelled the bridge as abeam then converted it to a single degree-of-freedom(SDOF) system; Blekherman [1] used a pendulum systemto simulate the vibration; and Strogaze et al. [9] also treatedthe system as a SDOF system. Using a section model,McRobie et al. [3] investigated experimentally on thephenomenon of human-structure lock in. Ellis [4] examinedthe influence of crowd size on floor vibration induced bywalking. Pachi and Ji [6] examined statistically the

e front matter r 2007 Elsevier Ltd. All rights reserved.

ecsci.2007.05.007

ing author.

ess: [email protected] (T. Ji).

frequency and velocity of people walking on footbridgesand shopping floors.For investigating the lateral vibration of a suspended

footbridge, it may be appropriate that a slice/section is cutfrom the bridge and the actions on the slice from itsneighbouring parts are represented by springs. If a modernpedestrian bridge deck is modelled, such as the LondonMillennium Footbridge, each segment is restrained in six-degrees-of-freedom (three translational motions and threerotational motions). However, if the attention is paid to thelateral vibration, the effect of the longitudinal vibrationscould be ignored without significant errors. In such case,the suspended rigid body is confined in planar motion. Thissimplified model may capture the nature of the lateralvibration of the footbridge. In other words, the lateralvibration of a suspended bridge is modelled as a suspendedsystem consisting of a rigid body (the deck of the bridge)and two strings (the hangers of the bridge). The verticaland lateral springs on the top ends of the strings representthe actions of the suspended cables. The vertical, lateraland rotational springs are added to the rigid body to

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Nomenclature

xA; yA dynamic displacements at the left end of therigid body

xB; yB dynamic displacements at the right end of therigid body

xC ; yC dynamic displacements of the left hanging pointxD; yD dynamic displacements of the right hanging

pointx0C ; y

0C static displacements of the left hanging point

x0D; y0D static displacements of the right hanging point

xM ; yM dynamic displacements at the centroid of therigid body

xe; ye dynamic displacements at the elastic constraintson the rigid body

e eccentric distance of the elastic constraints onthe rigid body

a dynamic rotational angle of the rigid bodys0 original distance between two hanging pointskx; ky stiffness coefficients of the hanging points in the

x and y directionsq;j dynamic angle displacements of the left and

right stringsb; h length and thickness of the rigid bodyl length of the stringsy static inclined angle of the stringsM ; I mass and rotary inertia of the rigid bodyk0x; k0y; k0r stiffness coefficients of the elastic con-

straints on the rigid body in the horizontal,vertical and rotational directions

g gravity accelerationo circular frequencyT ;U kinetic energy and potential energy of the

system

l l

q

x

y

kykx ky kx

kox

C D

koy

O1

e

��

�

D. Zhou, T. Ji / International Journal of Mechanical Sciences 50 (2008) 30–42 31

simulate the actions from the neighbouring decks. Fewpublications on modelling such a system have been foundand studied after searching widely. Thus these constitutethe basis for the current study. It is clear that the planarmotion of the suspended system is nonlinear in nature if thebridge makes large lateral vibration. In the present study,the dynamic characteristics of the suspended system havebeen analysed when the bridge makes vibration of smallamplitude.

In the present study, the model abstracted from thepractical problems is extended to a more general mechanicsproblem. The strings can be inclined either inward oroutward and the elastic constraints on the rigid body canbe applied asymmetrically. Therefore, the results obtainedfrom this study can also be applied to other systems whichcan be represented by such a model, for example thevibrating griddles to separate ores and the lateral vibrationof cable-cars. This paper studies the dynamic character-istics of the system in detail, which will provide the basisfor investigating the response of such a system subjected toexternal forces such as human actions.

Section 2 gives the basic assumptions and description ofthe model. In Section 3, the geometric relationshipsbetween 11 displacement variables are derived and threeindependent displacement variables are selected. The freevibration of the system is studied in Section 4. Section 5studies the dynamic characteristics of the system when theelastic constraints are applied at the centroid of the rigidbody. Simple experimental demonstration and verificationare also provided. Section 6 summarises the conclusionsobtained from this study.

A BM, I

b

kor

h

O2

Fig. 1. A generalised suspension system.

2. The model

The idealised model used in this study is shown in Fig. 1.The following assumptions are made for establishing

the model:

�
The lateral and torsional vibration of a suspended foot-bridge can be represented by a slice/section of the bridgewith appropriate constraints from its neighbouring parts. � The slice of the bridge deck is considered as a rigid body.
As observed in practice, the elastic deformation of thesection is much smaller than its movements.
� The hangers of the bridge are considered as strings with infi-
nite axial stiffness but without mass and transverse stiffness.
� The actions of cables on the top ends of the hangers are
represented by horizontal and vertical springs withstiffnesses of kx and ky, respectively.
� The actions of the neighbouring decks on the slice are
represented by horizontal, vertical and rotationalsprings with stiffnesses of k0x, k0y and k0r, respectively.
� The suspended system is symmetric to the axis O1–O2
but the elastic constraints can be applied at a distance e

from or at the centroid of the rigid body.
� Small amplitude vibrations of the system are considered.

ARTICLE IN PRESSD. Zhou, T. Ji / International Journal of Mechanical Sciences 50 (2008) 30–4232

The rigid body has a length b, mass M and rotary inertiaI around its centroid and is suspended by two symme-
trically inclined strings with the same length l. The lowerend of each string has a pinned connection to one end ofthe rigid body and the upper end to an elastic hangingpoint. The two inclined angles of the strings under gravityat static equilibrium state are the same and defined by y.The changes of the angles allow the system to be outwardinclined (y40Þ, inward inclined (yo0Þ or vertical (y ¼ 0Þ.The initial distance between the two elastic hanging pointsis s0 before considering the gravity of the rigid body. Theseform the basis of the derivation and study.
As the system experiences small amplitude vibration, thelinearization of the movements of the system is reasonable.It is significant to consider the following in the derivation:

�
The second order small quantities of the displacementshould be considered when the gravitational potential isconcerned. � The small amplitude vibration of the system should be
based on its static equilibrium state.
� The swaying angle of the left string can be different from
that of the right string during vibration.

3. Geometrical relations

3.1. Coordinates and displacements

The system has three degrees-of-freedom, namely thereare three independent displacement coordinates. For thesimplicity in derivation, two displacements perpendiculareach other at each of the control points, A, B, C, D andthree rotational angles of the strings C–A, D–B and therigid body A–B are introduced in the analysis. Therelationships between these 11 displacement variables canbe identified through the geometric relationships of thesystem at the deformed position as shown in Fig. 1.

The coordinates of points A and B at the deformedposition can be expressed in two different manners, whichform the equalities or relationships between the concerneddisplacement variables as follows:

xA þ l sin y ¼ xC þ l sinðyþ qÞ, (1)

xB þ l sin yþ b ¼ xD þ 2l sin yþ b� l sinðy� jÞ, (2)

yA þ l cos y ¼ yC þ l cosðyþ qÞ, (3)

yB þ l cos y ¼ yD þ l cosðy� jÞ, (4)

As A–B is a rigid body, the displacements xA, xB, yA, yB arerelated to each other. The following relationships are true:

xB ¼ xA � bð1� cos aÞ, (5)

yB ¼ yA þ b sin a. (6)

As strings C–A and D–B has no transverse stiffness, thesupport forces at the hanging points C and D must be

along the axes of the two strings in both static and dynamicequilibrium positions. This leads to the following relation-ships:

kxðxC þ x0CÞ

kyðyC þ y0CÞ¼ tanðyþ qÞ; �

kxðxD þ x0DÞ

kyðyD þ y0DÞ¼ tanðy� jÞ,

(7)

kxx0Ckyy0C

¼ tan y; x0D ¼ �x0C ; y0C ¼ y0D. (8)

3.2. Second order approximations

Considering the dynamic angles q and j are the smallquantities, there are the following approximate formulaeup to the second order small quantities:

sinðyþ qÞ � sin yð1� q2=2Þ þ cos yq, (9a)

cosðyþ qÞ � cos yð1� q2=2Þ � sin yq, (9b)

sinðy� jÞ � sin yð1� j2=2Þ � cos yj, (9c)

cosðy� jÞ � cos yð1� j2=2Þ þ sin yj, (9d)

tanðyþ qÞ ¼ tan yþ1

cos2 yqþ

tan ycos2 y

q2, (9e)

tanðy� jÞ ¼ tan y�1

cos2 yjþ

tan ycos2 y

j2 (9f)

1

tanðyþ qÞ¼

1

tan y�

1

sin2 yqþ

cos y

sin3yq2 (9g)

3.3. Relationships between displacements

From Eqs. (5) and (6), one has

ðxB � xA þ bÞ2 þ ðyB � yAÞ2¼ b2. (10)

Substituting Eqs. (9a)–(9d) into Eqs. (1)–(4) gives

xA ¼ xC þ l sinðyþ qÞ � l sin y ¼ xC þ l cos yq

� l sin yq2=2, ð11Þ

xB ¼ xD þ l sin y� l sinðy� jÞ ¼ xD þ l cos yj

þ l sin yj2=2, ð12Þ

yA ¼ yC þ l cosðyþ qÞ � l cos y ¼ yC � l sin yq

� l cos yq2=2, ð13Þ

yB ¼ yD þ l cosðy� jÞ � l cos y ¼ yD þ l sin yj

� l cos yj2=2. ð14Þ

It is well known that

xM ¼ ðxA þ xBÞ=2 ¼ ½xC þ xD þ l cos yðqþ jÞ

� l sin yðq2 � j2Þ=2�=2, ð15Þ

ARTICLE IN PRESSD. Zhou, T. Ji / International Journal of Mechanical Sciences 50 (2008) 30–42 33

yM ¼ ðyA þ yBÞ=2 ¼ ½yC þ yD � l sin yðq� jÞ

� l cos yðq2 þ j2Þ=2�=2, ð16Þ

xe ¼ xM þ e cos a, (17)

ye ¼ yM þ e sin a, (18)

a ¼ ðyB � yAÞ=b ¼ yD � yC þ l sin yðqþ jÞ

þ l cos yðq2 þ j2Þ=2. ð19Þ

Substituting Eqs. (11)–(14) into Eq. (10) gives

½xC � xD � g� sin yðq2 þ j2Þ=2þ cos yðq� jÞ�2

þ ½ZC � ZD � cos yðq2 � j2Þ=2� sin yðqþ jÞ�2 ¼ g2,

ð20Þ

where xC ¼ xC=l, xD ¼ xD=l, ZC ¼ yC=l, ZD ¼ yD=l,g ¼ b=l.

Applying Eqs. (8), (9e) and (9f), Eq. (7) can be written as

xC ¼ Z0Cc1qþ c2ZC þ c1qZC þ Z0Cc3q2, (21)

xD ¼ Z0Cc1j� c2ZD þ c1jZD � Z0Cc3j2, (22)

in which, x0C ¼ x0C=l, Z0C ¼ y0C=l and

c1 ¼ky

kx cos2 y; c2 ¼

ky

kx

tan y; c3 ¼ky tan ykx cos2 y

. (23)

Substituting Eqs. (21) and (22) into Eq. (20) and ignoringthe small quantities higher than second order, one has

ð1þ c22ÞðZC þ ZDÞ2þ ½sin2 yþ gðsin y� 2Z0Cc3Þ

þ ðZ0Cc1 þ cos yÞ2�q2

þ 2½c2ðZ0Cc1 þ cos yÞ � sin y� gc1�qZC

þ 2½c2ðZ0Cc1 þ cos yÞ þ sin y�qZD

� 2gc2ðZC þ ZDÞ � 4ZCZD � 2gðZ0Cc1 þ cos yÞq

¼ �½sin2 yþ gðsin y� 2Z0Cc3Þ þ ðZ0Cc1 þ cos yÞ2�j2

þ 2½ðZ0Cc1 þ cos yÞ2 � sin2 y�qj

þ 2½c2ðZ0Cc1 þ cos yÞ þ sin y�jZC

þ 2½c2ðZ0Cc1 þ cos yÞ � sin y� gc1�jZD

� 2gðZ0Cc1 þ cos yÞj. ð24Þ

In the following analysis, q, ZC and ZD are selected as theindependent generalised coordinates. In such a case, withinthe scope of the second order small quantity, j is different,however, not independent, from q. It can be assumed to be

j ¼ a0qþ a1ZC þ a2ZD þ b0q2 þ b1Z2C þ b2Z2D þ b3ZCZD

þ b4qZC þ b5qZD. ð25Þ

Substituting Eq. (25) into Eq. (24), ignoring the smallquantities higher than second order and comparing thecoefficients of the same terms in the two sides of Eq. (24)gives

a0 ¼ 1; a1 ¼ a2 ¼ c2=l,

b0 ¼ �½2 sin2 yþ gðsin y� 2Z0Cc3Þ�=ðglÞ,

b1 ¼ f2½c2lþ sin y�a1 � 1� c22

� ½sin2 yþ gðsin y� 2Z0Cc3Þ þ l2�a21g=ð2glÞ,

b2 ¼ f2½c2l� sin y� gc1�a1 � 1� c22

� ½sin2 yþ gðsin y� 2Z0Cc3Þ þ l2�a21g=ð2glÞ,

b3 ¼ f½2c2l� gc1�a1 þ 1� c22

� ½sin2 yþ gðsin y� 2Z0Cc3Þ þ l2�a21g=ðglÞ,

b4 ¼ f�½2 sin2 yþ gðsin y� 2Z0Cc3Þ�a1 þ 2 sin yþ gc1g=ðglÞ,

b5 ¼ f�½2 sin2 yþ gðsin y� 2Z0Cc3Þ�a1 � 2 sin y� gc1g=ðglÞ,

(26)

where

l ¼ Z0Cc1 þ cos y. (27)

Substituting Eq. (25) into Eq. (16) gives

yM ¼l2fð1þ sin ya1ÞZC þ ð1þ sin ya1ÞZD þ ðsin yb0 � cos yÞq2

þ ðsin yb1 � cos ya21=2ÞZ2C þ ðsin yb2 � cos ya21=2ÞZ

2D

þ ðsin yb3 � cos ya21ÞZCZD

þ ðsin yb4 � cos ya1ÞqZC þ ðsin yb5 � cos ya1ÞqZDg ð28Þ

3.4. Linearised displacements and velocities

Substituting Eq. (25) into Eqs. (15)–(19), the lineariseddisplacements are

xM ¼ lðlqþ c2ZCÞ, (29)

yM ¼l

2ð1þ sin ya1ÞðZC þ ZDÞ, (30)

a ¼ 2sin yg

q�1

gð1� sin ya1ÞZC þ

1

gð1þ sin ya1ÞZD, (31)

xe ¼ xM , (32)

ye ¼ yM þ ea. (33)

Differentiating Eqs. (29)–(31) once in respect to time t, thelinearised velocities are given as

_xM ¼ lðl _qþ c2 _ZCÞ, (34)

_yM ¼l

2ð1þ sin ya1Þð_ZC þ _ZDÞ, (35)

_a ¼ 2sin yg

_q�1

gð1� sin ya1Þ_ZC þ

1

gð1þ sin ya1Þ_ZD. (36)

4. Free vibration

4.1. Governing differential equations

The kinetic energy T of the system is

T ¼ 12

M _x2M þ

12

M _y2M þ

12

I _a2. (37)


The potential energy U of the system is

U ¼ �MgyM þ12

kxðxC þ x0CÞ2� 1

2kxx0C

2þ 1

2kyðyC þ y0CÞ

2

� kyy0C2

þ 12

kxðxD þ x0DÞ2� 1

2kxx0D

2þ 1

2kyðyD þ y0DÞ

2� kyy0D

2

þ 12 k0xx2

M þ12 k0yðyM þ eaÞ2 þ 1

2 k0r a2. ð38Þ

Using the following Lagrangian equations:d

dt

qT

q _qþ

qU

qq¼ 0;

d

dt

qT

q_ZC

þqU

qZC

¼ 0;d

dt

qT

q_ZD

þqU

qZD

¼ 0

(39)

and ignoring the small quantities higher than second order,one obtains the static equilibrium equation and threedynamic equations, respectively, as follows:

�Mgl

2ð1þ sin ya1Þ þ kyl2Z0C þ kxl2x0Cðc2 � Z0Cc1a1Þ ¼ 0,

(40)

m11 m12 m13

m21 m22 m23

m31 m32 m33

2664

3775

€q

€ZC

€ZD

8>><>>:

9>>=>>;þ

k11 k12 k13

k21 k22 k23

k31 k32 k33

2664

3775

q

ZC

ZD

8>><>>:

9>>=>>;

¼

0

0

0

8>><>>:

9>>=>>;, ð41Þ

where

m11 ¼Ml2l2 þ 4Isin2 yg2

,

m12 ¼Ml2c2l� 2Isin yg2ð1� sin ya1Þ,

m13 ¼ 2Isin yg2ð1þ sin ya1Þ; m21 ¼ m12,

m22 ¼Ml2 c22 þ14ð1þ sin ya1Þ

2� �

þI

g2ð1� sin ya1Þ

2,

m23 ¼Ml2

4ð1þ sin ya1Þ

2�

I

g2ð1� sin2 ya2

1Þ; m31 ¼ m13,

m32 ¼ m23; m33 ¼Ml2

4þ

I

g2

� �ð1þ sin ya1Þ

2,

k11 ¼ �Mglðsin yb0 � cos yÞ þ k0xl2l2 þ 4ðk0r þ k0ye2Þsin2 yg2

þ 2kxl2½ðZ0Cc1Þ2� x0CZ

0Cðc1b0 � 2c3Þ�,

k12 ¼ �Mgl

2ðsin yb4 � cos ya1Þ þ k0xl2c2l

� 2ðk0r þ k0ye2Þsin yg2ð1� sin ya1Þ

þ kxl2½Z0Cc1ðZ0Cc1a1 þ c2Þ � x0CZ0Cðc1b4 � 2c3a1Þ

þ x0Cc1� þ k0yelsin ygð1þ sin ya1Þ,

k13 ¼ �Mgl

2ðsin yb5 � cos ya1Þ þ 2ðk0r þ k0ye2Þ

�sin yg2ð1þ sin ya1Þ þ kxl2½Z0Cc1ðZ0Cc1a1 � c2Þ

� x0CðZ0Cc1b5 þ c1 � 2Z0Cc3a1Þ�

þ k0yelsin ygð1þ sin ya1Þ,

k21 ¼ k12; k22 ¼ �Mglðsin yb1 � cos ya21=2Þ

þ kyl2 þ k0xl2c22 þk0yl2

4ð1þ sin ya1Þ

2

þ1

g2ðk0r þ k0ye2Þð1� sin ya1Þ

2

þ kxl2½c22 þ ðZ0Cc1Þ

2a21 � 2x0CZ

0Cðc1b1 � c3a

21Þ�

�k0y

2gelð1� sin2 ya2

1Þ,

k23 ¼ �Mgl

2ðsin yb3 � cos ya2

1Þ þk0yl2

4ð1þ sin ya1Þ

2

�1

g2ðk0r þ k0ye2Þð1� sin2 ya2

1Þ

þ kxl2½Z0Cc1a1ðZ0Cc1a1 � c2Þ

� x0CðZ0Cc1b3 þ c1a1 � 2Z0Cc3a2

1Þ�

þk0y

gel sin ya1ð1þ sin ya1Þ,

k31 ¼ k13; k32 ¼ k23,

k33 ¼ �Mglðsin yb2 � cos ya21=2Þ

þ kyl2 þk0yl2

4þ

1

g2ðk0r þ k0ye2Þ

" #ð1þ sin ya1Þ

2

þ kxl2½ðZ0Cc1a1 � c2Þ2� 2x0CðZ

0Cc1b2 þ c1a1

� Z0Cc3a21Þ� þ

k0y

2gelð1þ sin ya1Þ

2. ð42Þ

4.2. Static solution and frequency equations

Substituting Eq. (8) into Eq. (40) gives the staticdisplacements at the static equilibrium position:

Z0C ¼ Z0D ¼Mg

2kyl; x0C ¼ �x

0D ¼

Mg

2kxltan y. (43)

From the compatibility relation, the static inclined angle yshould satisfy the following equation:

s0 � b

l¼ 2 sin yþ

Mg

kxltan y ¼ 2ðsin yþ tan yo2

g=o2xÞ, (44)

where o2g ¼ g=l, o2

x ¼ 2kx=M, s0 is the original distancebetween two hanging points before the rigid body is placedinto position.

ARTICLE IN PRESS

Table 1

The effect of inclined angle and eccentric distance of the elastic constraints

on the natural frequencies

y (degree) e (m) o1 (1/s) o2 (1/s) o3 (1/s)

�10 0.0 2.3395 3.4798 4.4439 (sy)

0.5 2.3777 3.4806 4.4465

1.0 2.4920 3.4841 4.4608

1.5 2.6650 3.4921 4.4899

2.0 2.8703 3.5095 4.5398

2.5 3.0706 3.5482 4.6212

0 0.0 2.5020 (ro) 3.4116 (sw) 4.4721 (sy)

0.5 2.5424 3.4116 (sw) 4.4773

1.0 2.6583 3.4116 (sw) 4.4937

1.5 2.8354 3.4116 (sw) 4.5244

2.0 3.0534 3.4116 (sw) 4.5757

2.5 3.2878 3.4116 (sw) 4.6583

10 0 2.5116 3.4840 4.4439 (sy)

0.5 2.5461 3.4851 4.4469

1.0 2.6498 3.4897 4.4623

1.5 2.8061 3.5004 4.4937

2.0 2.9881 3.5239 4.5478

2.5 3.1555 3.5756 4.6363

20 0.0 2.4283 3.6529 4.3626 (sy)

0.5 2.4531 3.6524 4.3588

1.0 2.5398 3.6587 4.3709

1.5 2.6716 3.6747 4.4028

2.0 2.8225 3.7054 4.4625

2.5 2.9619 3.7555 4.5626


The solutions of Eq. (41) are assumed to be

q ¼ q0e�jot; ZC ¼ ZC0e

�jot; ZD ¼ ZD0e�jot, (45)

where q0, ZC0 and ZD0 are the unknown constants, o is thenatural frequency of the system and j ¼

ffiffiffiffiffiffiffi�1p

. SubstitutingEq. (45) into Eq. (41) gives the following frequencyequations:

k11 k12 k13

k21 k22 k23

k31 k32 k33

2664

3775� o2

m11 m12 m13

m21 m22 m23

m31 m32 m33

2664

3775

8>><>>:

9>>=>>;

q0

ZC0

ZD0

8>><>>:

9>>=>>;

¼

0

0

0

8>><>>:

9>>=>>;. ð46Þ

Solving Eq. (46), three natural frequencies and theircorresponding eigenvectors can be obtained.

4.3. Examples

Consider a suspension system with the physical parameters:k0x ¼ k0y ¼ 1000N=m, k0r ¼ 10Nm, kx ¼ ky ¼ 500N=m,l ¼ b ¼ 5m, M ¼ 100 kg, I ¼ 1000 kgm2. Table 1 gives thenatural frequencies varying with the inclined angle y and theeccentric distance e. It is shown from Table 1 that:

Note: ro, rotational mode; sw, swaying mode; sy, symmetric mode.

�
Increasing the eccentric distance always results in theincrease of the natural frequencies except for y ¼ 0� atwhich the second natural frequency is independent ofthe eccentric distance and for y ¼ 20� and e ¼ 0:5 atwhich the third natural frequency is slightly smaller thanthat when y ¼ 20� and e ¼ 0:0. � When y40�, the second natural frequency increases
with the increase of the inclined angle. However, thethird natural frequency decreases with the increase ofthe inclined angle.
� The effect of the eccentric distance on the second and
third natural frequencies is smaller than that on the firstnatural frequency.
� The mode corresponding to the first natural frequency
mainly reflects the vibration of the rigid body in therotational direction. The mode corresponding to thesecond natural frequency mainly reflects the vibration ofthe rigid body in the horizontal direction and the modecorresponding to the third natural frequency mainlyreflects the vibration of the rigid body in the verticaldirection.

5. Free vibration when e ¼ 0

When e ¼ 0, the structure is symmetric about the axisO1–O2. In such a case, there are three vibration modes: onesymmetric and two antisymmetric. It is obvious that thesymmetric and antisymmetric vibrations can be studiedseparately, which greatly simplifies the derivation.

5.1. Symmetric vibration

5.1.1. Displacements and velocities

For the symmetric vibration, it is obvious that

q ¼ �j; xC ¼ �xD; yC ¼ yD,

xA ¼ xB ¼ xM ¼ 0; yA ¼ yB ¼ yM ; a ¼ 0 ð47Þ

From Eq. (11), one has

xC ¼ l sin y� l sinðyþ qÞ ¼ l sin yq2=2� l cos yq. (48)

Substituting Eq. (7) into the above equation gives

1

2l sin yq2 � l cos yq ¼

ky

kx

tanðyþ qÞðyC þ y0CÞ � x0C . (49)

Considering Eqs. (9g) and (8), Eq. (49) can be given as

yC ¼ �ld1qþ ld2q2, (50)

where

d1 ¼kx

ky

cos2 ysin y

þx0C

sin2 y

� �,

d2 ¼kx

ky

cos yðsin2 yþ 2Þ

2 sin2 yþ

cos y

sin3 yx0C

� �. ð51Þ

ARTICLE IN PRESS

0

1

2

3

4

5

6

7

8

9

20 30 40 50 60 70 80

α g,

α x

αgαxΔ

θ

Fig. 2. Functions ag and ax vary with the inclined angle y.

D. Zhou, T. Ji / International Journal of Mechanical Sciences 50 (2008) 30–4236

From Eq. (16), one has

yM ¼ yC � l sin yq� l cos yq2=2. (52)

Substituting Eq. (51) into Eq. (52) gives

yM ¼ �ðsin yþ d1Þlqþ d2 �1

2cos y

� �lq2. (53)

From Eq. (53), the following linearised velocity can beobtained:

_yM ¼ �ðsin yþ d1Þl _q. (54)

5.1.2. Governing differential equation

The kinetic energy T of symmetric vibration is

T ¼ 12

M _y2M ¼

12

Mðsin yþ d1Þ2l2 _q2. (55)

The potential energy U of symmetric vibration is

U ¼ �MgyM þ kxðxC þ x0CÞ2� kxx0C

2þ kyðyC þ y0CÞ

2

� kyy2C0 þ

12k0yy2

M

¼ ½Mgðsin yþ d1Þ � 2kxl cos yx0C � 2kylZ0Cd1�lq

þ �Mg

ld2 �

12cos y

�þ kxðcos

2 yþ sin yx0CÞ þ kyðd21 þ 2Z0Cd2Þ

þ12

k0yðsin yþ d1Þ2

�l2q2. ð56Þ

Using the following Lagrangian equation:

d

dt

qT

q _qþ

qU

qq¼ 0 (57)

one obtains the static equation and the governing dynamicequation, respectively

Mgðsin yþ d1Þ � 2kxl cos yx0C � 2kylZ0Cd1 ¼ 0, (58)

Mðsin yþ d1Þ2 €qþ 2 �

Mg

ld2 �

1

2cos y

� ��þ kxðcos

2 yþ sin yx0CÞ þ kyðd21 þ 2Z0Cd2Þ

þ12k0yðsin yþ d1Þ

2

�q ¼ 0. ð59Þ

5.1.3. Solution

From Eq. (58), the static solution which is the same asEq. (43) can be obtained once again.

The natural frequency of the symmetric vibration of thesystem can be obtained from Eq. (59) and is

o2 ¼cos y

ðsin yþ d1Þ2

g

lþ

k0y

M

þ 21

ðsin yþ d1Þ2

cos2 y �kx

Mþ d2

1

ky

M

� �. ð60Þ

It is clear that the above frequency should be one of thethree natural frequencies obtained from Eq. (46).

5.1.4. Solutions for special cases

From Eq. (60), the natural frequency for several specialcases can be easily given as follows:

y ¼ 0 or kx ¼ 1; o2 ¼ o20y þ o2

y ¼k0y þ 2ky

M. (61)

This is the natural frequency of a SDOF systemconsisting a mass M and two parallel springs withstiffnesses k0y and 2ky.

ky ¼ 1; o2 ¼ o20y þ ago2

g þ axo2x (62)

in which o20y ¼ k0y=M, o2

y ¼ 2ky=M and

ag ¼cos y

sin2 y; ax ¼

1

tan2 y(63)

Fig. 2 plots ag and ax with respect to the inclined angle y inthe interval ½20�; 80��. From Eqs. (61) to (63), it is knownthat:

�
When y ¼ 0, i.e. the strings are vertical, the horizontalstiffness at the hanging points has no effect on thesymmetric vertical vibration of the rigid body. � When kx ¼ 1, the inclined angle has no effect on the
symmetric vibration.
� When ky ¼ 1, not only the horizontal stiffness kx at the
hanging points but also the gravity affect the naturalfrequency of the symmetric vibration.
� When ky ¼ 1, the natural frequency monotonically
increases with the decrease of the inclined angle andcloses to infinite when y is close to zero.
� When ky ¼ 1, the natural frequency is the same for �y
and y.


5.2. Antisymmetric vibration

5.2.1. Displacements and velocities

From the antisymmetric vibration, there is ZD ¼ �ZC ,therefore,

j ¼ qþ b0q2 � b6Z2C þ b7qZC , (64)

xC ¼ Z0Cc1qþ c2ZC þ c1qZC þ Z0Cc3q2, (65)

xD ¼ Z0Cc1qþ c2ZC þ ðZ0Cc1b7 � c1ÞqZC

þ Z0Cðc1b0 � c3Þq2 � Z0Cc1b6Z2c , ð66Þ

yM ¼l

2½ðsin yb0 � cos yÞq2 � sin yb6Z2C þ sin yb7qZC �, (67)

_xM ¼ lðl _qþ c2 _ZCÞ; _yM ¼ 0; _a ¼ 2sin yg

_q�2

g_ZC , (68)

where

b6 ¼2

gl; b7 ¼

4 sin yþ 2gc1

gl. (69)

5.2.2. Governing differential equation

The kinetic energy T of antisymmetric vibration is

T ¼ 12

Ml2ðl _qþ c2 _ZCÞ2þ 2

I

g2ðsin y _q� _ZCÞ

2. (70)

The potential energy U of antisymmetric vibration is

U ¼ � 12

Mgl½ðsin yb0 � cos yÞq2 � sin yb6Z2C þ sin yb7qZC �

þ kxl2fðZ0Cc1qþ c2ZCÞ2� x0C ½ðZ

0Cc1b7 � 2c1ÞqZC

þ Z0Cðc1b0 � 2c3Þq2 � Z0Cc1b6Z2C �g

þ 12

k0xl2ðlqþ c2ZCÞ2þ 2

k0r

g2ðsin yq� ZCÞ

2. ð71Þ

Using the following Lagrangian equations:

d

dt

qT

q _qþ

qU

qq¼ 0;

d

dt

qT

q_ZC

þqU

qZC

¼ 0 (72)

one obtains the dynamic equations:

ðMl2l2 þ 4I

g2sin2 yÞ €qþ ðMl2lc2 � 4

I

g2sin yÞ€ZC

þ �Mglðsin yb0 � cos yÞ�þ 2kxl2Z0CðZ

0Cc21 þ 2c3x

0C � c1b0x

0CÞ þ k0xl2l2

þ4k0r

g2sin2 y

�q

þ ½�12Mgl sin yb7 þ kxl2ð2Z0Cc1c2 þ 2c1x

0C � c1b7x

0CZ0CÞ

þ k0xl2lc2 � 4k0r

g2sin y�ZC ¼ 0,

Ml2lc2 � 4I

g2sin y

� �€qþ ðMl2c22 þ 4

I

g2Þ€ZC

þ �1

2Mgl sin yb7 þ kxl2ð2Z0Cc1c2 þ 2c1x

0C � c1b7x

0CZ0CÞ

�

þk0xl2lc2 � 4k0r

g2sin y

�q

þ Mgl sin yb6 þ 2kxl2ðc22 þ x0CZ0Cc1b6Þ þ 2kyl2

�

þk0xl2c22 þ 4k0r

g2

�ZC ¼ 0. ð73Þ

5.2.3. Solution

In Eq. (73), assuming q ¼ q0e�jot and ZC ¼ ZC0e

�jot oneobtains

ðb11 � a11o2Þq0 þ ðb12 � a12o2ÞZC0 ¼ 0,

ðb12 � a12o2Þq0 þ ðb22 � a22o2ÞZC0 ¼ 0, (74)

where

a11 ¼Ml2l2 þ 4I

g2sin2 y; a12 ¼Ml2lc2 � 4

I

g2sin y,

b11 ¼ �Mglðsin yb0 � cos yÞ þ 2kxl2Z0C

�ðZ0Cc21 þ 2c3x0C � c1b0x

0CÞ þ k0xl2l2 þ 4

k0r

g2sin2 y,

b12 ¼ �12Mgl sin yb7 þ kxl2ð2Z0Cc1c2 þ 2c1x

0C � c1b7x

0CZ0CÞ

þ k0xl2lc2 � 4k0r

g2sin y,

a22 ¼Ml2c22 þ 4I

g2,

b22 ¼Mgl sin yb6 þ 2kxl2ðc22 þ x0CZ0Cc1b6Þ

þ 2kyl2 þ k0xl2c22 þ 4k0r

g2. ð75Þ

The solution of Eq. (75) is

o21;2 ¼

�BffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiB2 � 4ACp

2A(76)

in which,

A ¼ a11a22 � a212; B ¼ 2b12a12 � b11a22 � a11b22,

C ¼ b11b22 � b212. ð77Þ

It is clear that the above natural frequencies should be twoof the three natural frequencies obtained from Eq. (46).

5.2.4. Solutions for special cases

(1) y ¼ 0.

o2 ¼b2ky

2Iþ o2

0r, (78)

o2 ¼ o20x þ

o2go

2x

o2x þ o2

g

, (79)


where o20r ¼ k0r=I . Eqs. (78) and (79) provide the natural

frequencies of the rotational and swaying vibrations of therigid body, respectively.

It can be seen from Eqs. (78) and (79) that:

�

Ta

Th

o0x

0

0

1

10

100

The rotational vibration of the rigid body is independentof the horizontal stiffnesses kx, k0x and the verticalstiffness k0y.
� The swaying vibration is independent of the vertical
stiffnesses ky, k0y and the rotational stiffness k0r.

Eq. (79) can be rewritten as

o ¼ o0xb; b2 ¼ 1þ1

ðo0x=ogÞ2þ k0x=ð2kxÞ

, (80)

where o0x ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffik0x=M

pis the natural frequency of the rigid

body without the actions of the strings. b is themagnification factor to consider the suspension effect. Itis seen from Eq. (80) that the lateral natural frequency ofthe system with suspension is always larger than that of thesystem without suspension. This indicates that the lateralnatural frequency of a suspended bridge estimatedusing the theory of beams will be smaller than it shouldbe. Table 2 gives the magnification factors when o0x=og

varies between 0.01 and 100 and k0x=kx changes between 1and 100. The results indicate that the larger the o0x=og

and/or the larger the k0x=kx, the smaller the suspensioneffect, vice versa.

(2) ky ¼ 1, kx ¼ 1, ya0.In such a case, only the swaying vibration occurs and

j ¼ q� 1þ 2sin yg

� �tan yq2. (81)

The governing differential equation becomes

ðMl2 cos2 yþ 4I sin2 y=g2Þ €q

þ fMgl cos y½1þ tan2 yð1þ 2 sin y=gÞ� þ k0xl2 cos2 y

þ 4k0r sin2 y=g2gq ¼ 0. ð82Þ

The solution of Eq. (82) is

o2 ¼ fMgl cos y½1þ tan2 yð1þ 2 sin y=gÞ� þ k0xl2 cos2 y

þ 4k0r sin2 y=g2g=ðMl2 cos2 yþ 4I sin2 y=g2Þ. ð83Þ

ble 2

e value of b versus o0x=og and k0x=kx

=og k0x=kx ¼ 1 k0x=kx ¼ 5 k0x=kx ¼ 25 k0x=kx ¼ 100

.01 1.732 1.183 1.039 1.010

.1 1.721 1.183 1.039 1.010

.0 1.291 1.134 1.036 1.010

.0 1.005 1.005 1.004 1.003

.0 1.0 1.0 1.0 1.0

It is clear that the natural frequency exists only when

Mgl cos y½1þ tan2 yð1þ 2 sin y=gÞ� þ k0xl2 cos2 y

þ 4k0r sin2 y=g240. ð84Þ

Considering a uniform rectangular rigid body, its rotary inertiaaround the centroid of the body is I ¼Mðb2

þ h2Þ=12. If

k0x ¼ k0r ¼ 0, Eqs. (83) and (84) can be simplified as

o ¼ ogm; m ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ tan2 yð1þ 2 sin y=gÞcos y½1þ tan2 yð1þ dÞ=3�

s, (85)

1þ tan2 yð1þ 2 sin y=gÞ40, (86)

where m is the magnification factor to consider the inclinedangle and d ¼ h=b is the thickness-length ratio of the rigidbody. m varies with the inclined y from �60� to 60� aregiven in Fig. 3 for different g ¼ b=l. Fig. 4 gives the criticalinclined angle with respect to g when 1þ tan2 yð1þ2 sin y=gÞ ¼ 0 is held. It is shown from Figs. 3 and 4 that:

�
When y40 the natural frequency monotonically in-creases with the increase of g. However, when yo0 thenatural frequency monotonically decreases with thedecrease of g. � When y40 the natural frequency monotonically in-
creases with the increase of y. However, when yo0 andgX0:5 the natural frequency monotonically decreaseswith the decrease of y.
� The natural frequency is independent of the mass of the
rigid body. Increasing the thickness-length ratio d alwaysresults in the decrease of the natural frequency. However,for the small inclined angles such as yo40o, the effect of don the natural frequency is very small and negligible.ffiffiffiffiffiffiffip
� The critical inclined angle is y ¼ � arcsinð g=23 Þ which
is independent of the rotary inertia of the rigid body.Only when the inclined angle is lager than the criticalangle, the system oscillates. If the inclined angle issmaller than the critical angle, the system is in anunstable equilibrium state.
� The critical angle can only be negative. When 0ogo1,
there is always a critical angle at which the naturalfrequency of the system becomes zero, however, whengX2 (bX2lÞ, Eq. (86) is always held. Therefore, thenecessary condition of an unstable equilibrium state isthat the two strings are cross each other.

5.3. Examples

Consider a symmetric suspension system with the physicalparameters: kx ¼ 1000N=m, ky ¼ 10 000N=m, e ¼ 0, k0x ¼

k0y ¼ k0r ¼ 0, M ¼ 1000 kg, I ¼ 10 000 kgm2, b ¼ 5m. Theeffect of string length and inclined angle on the initial distancebetween two hanging points and on the natural frequencies ofthe system is studied, as given in Figs. 5–8, respectively. It is

ARTICLE IN PRESS

0

1

2

3

4

5

-40 20 40 60

μ

γ=0.25, δ=0.5

γ=0.5, δ=0.25

γ=0.5, δ=0.5

γ=1.0, δ=0.0

γ=1.0, δ=0.25

γ=1.0, δ=0.5

γ=2.0, δ=0.25

δ=2.0, δ=0.5

γ=0.25, δ=0.0

γ=0.25, δ=0.25

γ=0.5, δ=0.0

γ=2.0, δ=0.0

-60 -20 0

θ

0.5

1.5

2.5

3.5

4.5

Fig. 3. The magnification factor m varies with the inclined angle y for different length ratios of the rigid body to the strings g and thickness-length ratio of

the rigid body d.

0

0 0.5 1 1.5 2

Stable equilibrium state

-10

-20

-30

-40

-50

-60

-70

-80

-90

Unstable equilibrium state

θ

γ

Fig. 4. The critical angle with respect to the length ratio g. The dashed line

is the angle at which two hanging points intersect.

0

5

10

15

20

25

30

35

-30 -20 -10 10 20 30 40 50 60

s0

θ

-5

0

-10

l = 2

l = 5

l = 7

l = 10

Fig. 5. The initial distance between two hanging points.


shown from Fig. 5 that the initial distance between two hang-ing points monotonically increases with the increase of theinclined angle. However, when yo0 (inward inclined strings),the initial distance between two hanging points could be nega-tive with the further decrease of the inclined angle. This meansthat in such a case, the two strings are cross each other beforethe rigid body is placed. The longer the strings, the larger theinclined angle at which the intersection of two strings occurs.

Fig. 6 gives the natural frequency of the symmetricmode. It is shown that the natural frequency is independentof the sign of the inclined angle. The natural frequency ofthe symmetric mode monotonically decreases with theincrease of the absolute value of the inclined angle.The longer the strings, the lower the natural frequency ofthe symmetric mode.

ARTICLE IN PRESS

0

1

2

3

4

4.5

5

-20 10 20 30 40 50 60

ωs

l = 2

l = 5

l = 7

l = 10

2.5

3.5

1.5

0.5

0-10-30

θ

Fig. 6. Natural frequency of the symmetric mode when no elastic

constraints applied on the rigid body.

0

1

-20 10 20 30 40 50 60

ωa2

l = 2

l = 5

l = 7

l = 10

θ

-30 -10 0

0.2

0.4

0.6

0.8

1.2

1.4

Fig. 7. Natural frequency of the first antisymmetric mode when no elastic

constraints applied on the rigid body.

0

1

3

3.5

4

-20 10 20 30 40 50 60

l = 2

l = 5

l = 7

l = 10

θ

-30 -10 0

0.5

1.5

ωa1

2

2.5

Fig. 8. Natural frequency of the second antisymmetric mode when no

elastic constraints applied on the rigid body.


Fig. 7 gives the natural frequency of the first antisym-metric mode which mainly reflects the swaying vibration ofthe rigid body. It is seen that the natural frequencygenerally increases with the increase of the inclined angle,especially for the long strings. However, when the absolutevalue of the inclined angle is small, such as yj jo10o, theeffect of the inclined angle on the natural frequency is

negligible. The longer the strings, the lower the naturalfrequency of the first antisymmetric mode.Fig. 8 gives the natural frequency of the second

antisymmetric mode which mainly reflects the rotationalvibration of the rigid body. It is seen that when yo0�, thenatural frequency monotonically decreases with the de-crease of the inclined angle. When 0�oyo45�, the naturalfrequency decreases with the increase of the inclined angle.However, when y445�, the natural frequency increaseswith the increase of the inclined angle. The longer thestrings, the lower the natural frequency of the secondantisymmetric mode.Table 3 studies the effect of stiffness coefficients of the

hanging points and the elastic constraints on naturalfrequencies when e ¼ 0 and k0x ¼ 10k0y, k0r ¼ 0:1m2 � k0y,kx ¼ kl sin y, ky ¼ kl cos y, y ¼ 15

�

, l ¼ b ¼ 5m, M ¼

1000 kg, I ¼ 10 000 kgm2. The first antisymmetric modemainly reflects the rotary vibration of the rigid body andthe second antisymmetric mode mainly reflects the swayingvibration of the rigid body. It is seen that all the naturalfrequencies monotonically increase with the increases ofthe stiffnesses of the hanging points and the elasticconstraints on the rigid body. However, the effect of thestiffnesses of the elastic constraints of the rigid body on thenatural frequency of the first antisymmetric mode issmaller than that on the other two natural frequencies.

5.4. Verification

Simple model demonstration and tests are conducted toverify the theoretical predictions. A uniform hollowaluminium bar with square section treated as a rigid bodyand suspended by two symmetric strings is used for tests,


which allows to creating different inclined angles. Threetypical forms are shown in Fig. 9 including vertical,outward inclined and inward inclined suspension systems.Fig. 9d shows an unstable equilibrium state of the systemwhere two strings are cross each other. An additional

Table 3

The effect of stiffness coefficients of the hanging points and the elastic

constraints on the natural frequencies when e ¼ 0

kl (N/m) k0y (N/m) os (1/s) oa1 (1/s) oa2 (1/s)

100 5000 2.2786 0.90910 7.0746

10 000 3.1925 0.93620 10.003

20 000 4.4936 0.98816 14.144

40 000 6.3397 1.0847 20.001

200 5000 2.3199 0.97214 7.0781

10 000 3.2221 0.99752 10.005

20 000 4.5146 1.0464 14.146

40 000 6.3547 1.1380 20.002

400 5000 2.3990 1.0856 7.0845

10 000 3.2795 1.1084 10.009

20 000 4.5558 1.1526 14.149

40 000 6.3840 1.2363 20.005

800 5000 2.5460 1.2778 7.0958

10 000 3.3885 1.2972 10.018

20 000 4.6349 1.3352 14.155

40000 6.4406 1.4081 20.009

Fig. 9. Typical forms of a suspension system used for demonstration and test

(c) inward inclined suspension system; (d) unstable equilibrium state.

horizontal string is applied to prevent the out planemovement of the system. When a small and allowablemovement is applied on the system, the aluminium bar willmove to and balance at the new position withoutoscillation. The bar has a length of b ¼ 0:45m, a totalmass of M ¼ 0:094 kg and the rotary inertiaI ¼ 0:0016 kgm2. In the tests, an initial lateral displace-ment is applied to the bar and the bar is suddenly releasedto generate free vibration. The number of oscillations iscounted and a stopwatch is used to record the time ofvibration. The swaying natural frequencies of nine differentforms of suspension are measured and listed in Table 4together with the theoretical predictions. It can be seen thatthere are good agreement between the measured andcalculated lateral natural frequencies. The maximum erroris about 2.4%. It is shown that when y ¼ �52:7� andl ¼ 0:584m, the system is in an unstable equilibrium state.

6. Conclusions

The dynamic characteristics of a generalised suspensionsystem are studied analytically in this paper. It is shownthat the system has three vibration modes, which mainlyreflects the vertical vibration, rotational vibration andhorizontal (swaying) vibrations of the system. When theeccentric distance of the elastic constraints is zero (i.e. the

s: (a) vertical suspension system; (b) outward inclined suspension system;

ARTICLE IN PRESS

Table 4

Comparison of the measured and predicted lateral natural frequencies

Parameters Experiment (Hz) Theory (Hz) Error (%)

y ¼ 0�, l ¼ 0:415m 0.778 0.774 �0.5

y ¼ �10:15�, l ¼ 0:431m 0.773 0.769 �0.5

y ¼ �22:28�, l ¼ 0:825m 0.543 0.536 �1.3

y ¼ �24:22�, l ¼ 0:431m 0.786 0.785 �0.1

y ¼ �30:69�, l ¼ 0:485m 0.718 0.717 �0.1

y ¼ �52:7�, l ¼ 0:584m UES UES –

y ¼ 24:66�, l ¼ 0:405m 0.930 0.929 �0.1

y ¼ 35:95�, l ¼ 0:420m 1.16 1.14 �1.7

y ¼ 48:4�, l ¼ 0:324m 1.67 1.71 2.4

UES, Unstable equilibrium state.


system is symmetric) the conclusions obtained from thepresent study are summarised as follows:

�
The vibration of the system can be divided into onesymmetric mode and two antisymmetric modes. In-creasing the string length will result in the decrease ofnatural frequencies. The natural frequency of thesymmetric mode is independent of the sign of theinclined angle y. � The natural frequency of the symmetric mode mono-
tonically decreases with the increase of the absolutevalue of the angle y. However, the natural frequencymainly reflecting the swaying vibration increases withthe increase of the angle y.
� When y ¼ 0, the lateral natural frequency of a
suspended bridge estimated using the theory of beamsis always smaller than it should be. The suspensioneffect is a function of two ratios, o0x=og and k0x=kx.The effect of the spring stiffnesses at the hanging pointson the natural frequency of the swaying mode isgenerally small.
� When the spring stiffnesses at the hanging points are
infinite and there are no elastic constraints on the rigidbody, the swaying natural frequency of the system isindependent of the mass of the rigid body. The rotaryinertia of the rigid body always decreases the naturalfrequency. When y40, the natural frequency increaseswith the increase of the inclined angle y and the lengthratio of the rigid body to the strings g. When yo0, the

natural frequency decreases with the decrease of g. Thecritical angle of an unstable equilibrium state of thesystem is only depends on g. The necessary condition ofan unstable equilibrium state of the system is that thetwo stings are cross each other.
� The effect of the thickness of the rigid body on the
natural frequencies is generally negligible.

Acknowledgment

The work reported in this paper has been conducted aspart of the project, Prediction of floor vibration induced bywalking loads and verification using available measure-ments, funded by the UK Engineering and PhysicalSciences Research Council (EPSRC grant No GR/S74089), whose support is gratefully acknowledged.

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[3] McRobie A, Morgenthal G, Lasenby J, Ringer M. Section model

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