CABLE SUPPORTED STRUCTURES

3/22/2005 Prof. dr Stanko Brcic 1

CABLE SUPPORTED STRUCTURES

STATIC AND DYNAMIC ANALYSIS OF CABLES


Cable Supported Structures

Suspension bridgesCable-Stayed BridgesMastsRoof structuresetc


Cable Analysis – Main Assumptions

Cables are flexible material linesIn-extensible or elastic lines

Consequently, the only internal cross - sectional force is the cable tension:

( ) ( )T s T s τ= ⋅


Cable Analysis

Differential equation of equilibrium

If the loading is gravitational (or, constant direction), i.e. if

( ) 0dT q sds

+ =

( ) ( )q s q s e e const= ⋅ =


Cable Analysis

then the cable is a curved line within the plane (of loading):

If is the unit vector within the cable plane perpendicular to the loading direction (horizontal direction in vertical plane – for gravitational loading):

e T const× =h

0h e⋅ =


Cable Analysis

then the component of cable tension in direction of h (i.e. horizontal cable tension for vertical loading) is constant:

h T H const⋅ = =


Cable Analysis – dead load



The cable is loaded by the self-weight q(s) = q = const (in vertical x-yplane)Differential equation of equilibrium is:

2( ) 0 1 0dT q s Hy q yds

′′ ′+ = ⇒ + + =



The general solution is given as

The boundary conditions are:

1

1 2

sinh( )

cosh( )

qxy CH

H qxy C Cq H

′ = − −

= − − +

(0) 0( )

yy l b

==



With notation:

the constants of integration are

sinh( )2 sinhql barH l

λλ λλ

= Φ = +

1

2 cosh( )

CHCq

= Φ

= Φ



So, the final solution is given as:

HYPERBOLIC RELATIONS(deep cable, catenary)

( ) sinh( )

( ) [cosh( ) cosh( )]

qxy xH

H qxy xq H

′ = − −Φ

= Φ − −Φ



The maximum deflection of the cable:

If

then the max deflection is within the span of the cable.

0

0 max

( ) 0

( ) [cosh( ) 1]

Hy x x xq

Hy x yq

′ = ⇒ = = ⋅Φ

= = ⋅ Φ −

00 x l≤ ≤



Extreme values of support denivelation b in order to have are:00 x l≤ ≤

2 2

2max

2 2sinh ( ) sinh ( )

2| | sinh ( )

H Hb orq q

Hb bq

λ λ

λ

− ≤ ≤

≤ =



The cable with horizontal span: b = 0

0 max 1

2

( ) [cosh( ) cosh( )]

( ) sinh( )

, [cosh( ) 1]2 2

qlH

H qxy xq H

qxy xH

H H ql l Hx y fq q H q

λ

λ λ

λ

λ λ

Φ = =

= − −

′ = − −

= = ⋅ = = = −



For horizontal span b = 0:since,

one obtains the max deflection as

2

cosh( ) 12λλ ≅ +

2 2

max 2 8H qlyq H

λ= ⋅ =



The total length of the cable L:

Since

One obtains the length L as:

2

0 0

1x l l

x

dsL dx y dxdx

=

=

′= = +∫ ∫

2 2 21 1 sinh ( ) cosh ( )qx qxyH H

′+ = + −Φ = −Φ

2 sinh( )cosh( )HLq

λ λ= Φ −



For cable with horizontal span, the length L of the cable is:

02 sinh( )

bHLq

λ

λ

= ⇒ Φ =

=



Reactions of supports at A and B:

Vertical component Ya is equal to(after transformation)

0( ) | ( ) |

A A B B

A x B x l

T H Y T H Ydy dyY H Y Hdx dx= =

= + = +

= ⋅ = − ⋅

sinh [ coth( )]2A AqY H Y L b λ= Φ ⇒ = +



Vertical component Yb is obtained from equilibrium condition:

For the cable with horizontal span:

0 [ coth( )]2A B BqY Y qL Y L b λ+ − = ⇒ = −

12A BY Y qL= =



PARABOLIC RELATIONSAssumption: The cable is relatively shallow, orThe gravitational loading is distributed along the horizontal projection of the cable



For deep cable (catenary) differential equation of equilibrium is

For shallow cable (parabolic relations)

21 0Hy q y′′ ′+ + =

1.0 0ds Hy qdx

′′≈ ⇒ + =



The general solution of diff. eq. is

From boundary conditions y(0)=0, y(l)=b one obtains:

1

21 2

( )

( )2

qy x x CHqy x x C x CH

′ = − +

= − + +

2 10 ( tan )2

b q bC C ll H l

β= = + =



Therefore, the final solution is

The cable shape is parabola.Position of maximum deflection:

2

( ) ( 2 )2

( ) ( )2

q by x l xH l

q by x lx x xH l

′ = − +

= − +

0( ) 02l H by x x

q l′ = ⇒ = +



Therefore, the max deflection is:

Extreme support denivelation to have max deflection within the cable span:

2 2

0 max 2( )8 2 2ql b H by x yH q l

= = + +

2 2 2

max| |2 2 2ql ql qlb or b bH H H

− ≤ ≤ ≤ =



The length of the cable L:

It is obtained:

2 2 2

0

11 1 12

l

L y dx y y′ ′ ′= + + ≈ +∫

2 2

2

1 116 2 2

10 : 16 2

ql bL lH l

qlfor b L lH

⎡ ⎤⎛ ⎞ ⎛ ⎞≅ + +⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

⎡ ⎤⎛ ⎞= ≅ +⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦



Vertical components of support reactions are:

It is obtained:

For horizontal span (b = 0):

0| |A x B x lY Hy Y Hy= =′ ′= = −

2 2A Bql b ql bY H Y H

l l= + = −

2A BqlY Y= =



Hyperbolic or Parabolic relations:the solution depends on H, i.e. on the unknown horizontal cable tensionStatically undetermined problemInitial assumption of H or max cable tension orInitial assumption of the total cable length


Elastic Cables – The Cable Equation

Cable is ideally elastic body (material line) with equivalent

- Modulus of elasticity- Coefficient of thermal expansionTwo loading conditions are considered: initial loading (self-weight) and some additional loading (live load)



For initial loading the cable element is given as:Due to some additional loading the cable moves to the new equilibrium configuration (displacement components u, v). The cable element in the new configuration is

2 2 2ds dx dy= +

2 2 2( ) ( )ds dx du dy dv′ = + + +



Dilatation of a point at the cable element is given as:

If neglecting the small quantities of the 2nd order:

21 ( )2

ds ds dx du dy dv dvds ds ds ds ds ds

ε′ −

= = + +

ds ds dx du dy dvds ds ds ds ds

ε′ −

= = +



Increment of cable tension due to additional load is denoted as τ. The corresponding horizontal component of that increment in tension is h:

cosdx dsh hds dx

τ τ ϕ τ= = ⇒ =



Therefore, the Hooke’s Law is

Consequently, it may be obtained:

t tEAτε α= + ∆

21 ( )2t

dsh dx du dy dv dvdx tEA ds ds ds ds ds

α+ ∆ = + +



After multiplying with the cable equation in differential form is obtained

If neglecting the 2nd order terms, then:

2dsdx

⎛ ⎞⎜ ⎟⎝ ⎠

3

2 2( ) 1( ) ( )

2t

dsh ds du dy dv dvdx tEA dx dx dx dx dx

α+ ∆ = + +

3

2( )

( )t

dsh ds du dy dvdx tEA dx dx dx dx

α+ ∆ = +



Integrating over the span of the cable, one obtains the Integral form of the Cable equation:

Or in the form

2

0 0

1( ) (0) ( )2

l le

t thL dy dv dvtL u l u dx dxAE dx dx dx

α+ ∆ = − + +∫ ∫

0

( ) (0)l

et t

hL dy dvtL u l u dxAE dx dx

α+ ∆ = − + ∫



Where the virtual lengths of the cable are given by

Also, u(l) and u(0) are horizontal displacements of supports A and B

3 2

0 0

( ) ( )l l

e tds dsL dx L dxdx dx

= =∫ ∫



For Hyperbolic relations the virtual lengths may be obtained as:

which may be transformed into

3 2

0 0

cosh ( ) cosh ( )l l

e tqx qxL dx L dxH H

= −Φ = −Φ∫ ∫

3sinh(3 )cosh 3( ) sinh( )cosh( )6 2

sinh(2 )cosh 2( )2 2

e

t

H HLq q

H lLq

λ λ λ λ

λ λ

= Φ − + Φ −

= Φ − +



For the cable with horizontal span (b = 0):

1[ sinh(3 ) 3sinh( )]2 3

sinh(2 ) ,2 2 2

e

t

HLq

H l qlLq H

λ λ

λ λ λ

= +

= + = Φ =



For the Parabolic relations, the virtual cable lengths are obtained as

After integration, it is obtained

2 3/ 2 2

0 0

(1 ) (1 )l l

e tL y dx L y dx′ ′= + = +∫ ∫

2 4 22 2

2 22

96 3 11 8 1 8 tan tan5 2 4

161 tan , tan ,3 8 4

e

t

f f fL ll l l

f ql b fL l fl H l l

β β

λβ β

⎧ ⎫⎡ ⎤⎪ ⎪⎛ ⎞ ⎛ ⎞ ⎛ ⎞≈ + + + + +⎢ ⎥⎨ ⎬⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎢ ⎥⎪ ⎪⎣ ⎦⎩ ⎭

⎡ ⎤⎛ ⎞≈ + + = = =⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦



By the partial integration one obtains

For the boundary conditions v(0) = 0 and v(l) = 0, and for the parabolic relations, since

2

0 20 0

|l l

ldy dv dy d ydx v vdxdx dx dx dx

= −∫ ∫

2

2

d y q constdx H

= − =



One obtains

Therefore, the Integral form of the Cable equation becomes

0 0

l ldy dv qdx vdxdx dx H

=∫ ∫

2

0 0

1( ) (0) ( )2

l le

t thL q dvtL u l u vdx dxAE H dx

α+ ∆ = − + +∫ ∫



or, if neglecting the small terms of the 2nd order,

If the supports are fixed, thenu(l) = u(0) = 0

0

( ) (0)l

et t

hL qtL u l u vdxAE H

α+ ∆ = − + ∫


Live Load and Temperature Effects upon the Cable (Parabolic relations)

The cable is loaded by the self-weight constantly distributed along the spanDead Load: q(x) = q = constDifferential equation of equilibrium

Additional (static) loadingp(x) = p = const

0Hy q′′ + =



Due to additional load p, increment in horizontal component of cable tension is denoted as h, so differential equation of equilibrium is given as

Due to initial equilibrium equation ⇒( ) ( ) ( ) 0H h y v q p′′+ ⋅ + + + =

( ) ( )qH h v h p aH

′′+ − = −



It is possible to neglect the productas relatively small, so one obtains

Equation (a) or (b), together with the cable equation, say in the form

h v′′⋅

( )qHv h p bH

′′ − = −

0

( )l

et t

hL qtL vdx cAE H

α+ ∆ = ∫



define the problem of additional load (live load)Consider eqs. (b) and (c). From (b) ⇒

If v(0) = v(l) = 0, then

* 2

hq pv A constH H

′′ = = − =

2*

1( ) ( )2

v x A x lx= −



Therefore:

If this integral is introduced into the cable equation (c), one obtains the linear equation for unknown tension increment h

3*

0

1( )12

l

v x dx A l= −∫

3 2 3

2 3

1 112 12

et t

hL ql q ltL p hAE H H

α+ ∆ = −



With notation (catenary, or cable parameter)

horizontal tension increment h is obtained as

2 3

* 3e

q l AEH L

λ =

*

* *

1212 12

L tt

e

Lp AEh H tq L

λα

λ λ= ⋅ ⋅ − ∆ ⋅

+ +



The total cable tension is equal to

Or

The cable parameter is the measure of elastic properties of the cable:for the cable is in-extensible

1H H h= +

*1

* *

12H (1 )12 12

tt

e

Lp AEH tq L

λ αλ λ

= + ⋅ − ∆ ⋅+ +

*λ →∞



If eq. (a) is considered:

Then

So, for the boundary conditions v(0)=v(l)=0, the solution is:

( ) ( )qH h v h p aH

′′+ − = −

* ( )hq pv A const

H H h H h′′ = = − =

+ +

2*

1( ) ( )2

v x A x lx= −



The cable equation, for ∆t=0 and for u(0)=u(l)=0, is given as:

Therefore,

2

0 0

12

l lehL q vdx v dx

AE H′= +∫ ∫

3 2 2 3* *

0 0

1 112 12

l l

vdx A l v dx A l′= − =∫ ∫



With notation for non-dimensional increment of cable tension:

and the catenary (cable) parameter

hH

α =

2 3

* 3e

q l EAH L

λ =



the Cable equation may be written in the non-dimensional form of:

i.e. as the cubic equation in non-dimensional increment of cable tension h

3 2 2* * *(2 ) (1 ) [2( ) ( ) ]24 12 24

p pq q

λ λ λα α α+ + ⋅ + + ⋅ = +



The catenary parameter may be also written as:

If is a large number, then the cable is inextensible: If is a small number, then the cable behaves as the taut string:

3*

8( )e

f EAl qL

λ = ⋅

*λEA→∞

*λ

0fl→



Obtaining solution of the cubic cable equation, tension increment is given. The final horizontal cable tension, due to additional load p, is given as:

Also, new deflection shape is given as1 (1 )H H h Hα= + = +

2*

1( ) ( )2

v x A x lx= −



Non-dimensional cable equation

represents the change in cable tension due to additional load.Non-linear dependence on additional load pNon-linearity is greater for smaller values of catenary parameter

3 2 2* * *(2 ) (1 ) [2( ) ( ) ]24 12 24

p pq q

λ λ λα α α+ + ⋅ + + ⋅ = +

*λ



For > 5000 non-linearity is relatively small

Limiting value of non-dimensional increment of cable tension is:

*λ

*

lim pqλ

α→∞

=



In the case of the cable equation where the second-order term is neglected, i.e.

and diff. equation of equilibrium (a), is used, the corresponding quadratic equation for increment of cable tension h is obtained

0

( )l

ehL q v x dxAE H

= ∫



In the case of simultaneous additionalload p and temperature load ∆t is considered, the cubic equation for incremental cable tension is obtained

3 2* *

2*

(2 ) (1 2 )24 12

[2( ) ( ) ]24

t tt t

e e

tt

e

L LAE AEt tL H L H

Lp p AEtq q L H

λ λα α α α α

λ α

+ + + ∆ ⋅ + + + ∆ ⋅ =

= + − ∆



New cable deflection after additional load p is given as:

New deflection may be transformed into the form similar to initial deflection due to dead load:

1( ) ( ) ( )y x y x v x= +

21 *( ) ( )

2q by x B lx x xH l

= ⋅ − +



where B is the non-dimensional deflection amplification:

Coefficient B is always > 1.0, sinceOnly for in-extensible cable

*

*

1 (1 ) 01

1 0

pB for hvq

pB for hvq

α

α

′′= ⋅ + ≠+

′′= + − =

pq

α>

* * 1.0EA Bλ →∞ ⇒ →∞ ⇒ →



Also, it may be concluded that the amplification of deflection B, due to additional loading, is greater for greater relative additional loading p/q, while the cable parameter λ is smaller.


Free Vibrations in the Cable Plane

Cable is in the equilibrium configuration due to its self-weightParabolic approximation: dead load is uniform over the horizontal projection of cable: q(x) = q = constStatic equilibrium configuration: y(x)Free vibration in cable plane –dominant vertical motion v=v(x,t)



Free vibrations of a cable around its static equilibrium position.Position of a cable during free vibrations is given by:

1( , ) ( ) ( , )y x t y x v x t= +



According to D’Alembert’s Principle, the total load acting upon cable is:

(active and inertial forces), where m is cable mass per unit length:

1( , ) ( ) ( , )q x t q x mv x t= −

2

( )( ) 9.81q x mm x gg s

=



Horizontal component of cable tension is equal to

where H is due to dead load q, while h(t) is the cable tension increment due to inertial forces

1( ) ( ) ( )H t H h t H const= + =



Differential equation of “equilibrium”, according to D’Alembert’s Principle, is

Substituting, one obtains

or

1 1 1 0H y q′′+ =

( ) ( ) 0H h y v q mv′′+ ⋅ + + − =

0Hy Hv hy hv q mv′′ ′′ ′′ ′′+ + + + − =



Due to differential equation of static equilibrium: and neglecting the term (as the product of small increments), one obtains dif. equation of free vibrations of a cable:

0Hy q′′ + =

hv′′

0 ( )qmv Hv h aH

′′− + =



Equation (a), together with the Cable equation, for ∆t = 0:

defines free vibrations of a cable in its plane. As may be seen, in the Cable equation the influence of the 2nd

order term is neglected too.

0

( ) ( , ) ( )l

eh t L q v x t dx bAE H

= ∫

2v′



For harmonic free vibrations:

So, equations (a) and (b) become( , ) ( ) ( )i t i tv x t v x e h t h eω ω= ⋅ = ⋅

2

0

( .1)

( ) ( .2)l

e

qm v Hv h AH

hL q v x dx AAE H

ω ′′+ =

= ∫


Non-Symmetric Free Vibrations in the Cable Plane

Non-symmetric free vibrations are defined by non-symmetric function v(x,t) along the span of the cable:

In that case, it follows from (A.2) that also

0

( , ) 0l

v x t dx =∫

( ) 0h t =



Therefore, differential eq. of free non-symmetric vibrations is given as

The general solution is given by

where

2 0m v Hvω ′′+ =

1 2( ) sin( ) cos( )v x C kx C kx= +

mkH

ω=



Integration constants are determined from the boundary conditions:

The 2nd conditions is due to non-symmetry of free vibrations. Of course, the condition v(l) = 0 must be also fulfilled

(0) 0 ( ) 02lv v= =



From the boundary conditions one obtains:

The frequency equation is, therefore,

2 10 sin( ) 02klC C= =

sin( ) 0 , 1,2,3,2 2kl kl n nπ= ⇒ = = …



Since

the natural circular frequencies of free non-symmetric vibrations in the cable plane are obtained as:

mkH

ω=

2 1,2,3,...nn H n

l mπω = =



The corresponding natural shapes are given as:

It may also be seen that the solution satisfies the condition v(l) = 0

2( ) sin( ) 1,2,3,...n nnxv x C nlπ

= =


Symmetric Free Vibrations in the Cable Plane

Symmetric free vibrations of the cable in its plane are given by equations (A):

2

0

( )l

e

qm v Hv hH

hL q v x dxAE H

ω ′′+ =

= ∫



The general solution of differential equation is given as:

where the general solution of homogeneous equation is

while is any particular solution of non-homogeneous equation.

( ) ( ) ( )h pv x v x v x= +

1 2( ) sin( ) cos( )hv x C kx C kx= +

( )pv x



The particular solution is

so the general solution is given as

2( )pqhv x

m Hω=

1 2 2( ) sin( ) cos( )hqhv x C kx C kx

m Hω= + +



Integration constants are obtained from the boundary conditions:

It may be obtained:

(0) 0 ( ) 0v v l= =

2 12 2

(1 cos ),sin

qh qh klC Cm H m H klω ω

−= − = − ⋅



Therefore, the general solution of diff. equation is obtained as:

This general solution is inserted into the integral of the 2nd equation (the cable equation).

2 2 2

(1 cos )( ) sin cossin

qh kl qh qhv x kx kxm H kl m H m Hω ω ω

−= − ⋅ ⋅ − ⋅ +



It is obtained

Inserting this into the cable equation, and dividing by it is obtained

0 2

2 2 20

(1 cos )( ) sinsin

qh kl qh qhv x dx kl lm Hk kl m Hk m Hω ω ω

−= ⋅ − ⋅ + ⋅∫

0h ≠

2 2 2 2

3 2 2 2 2 2 2

(1 cos ) sin 0sin

eL q kl q qkl lAEl m H k kl m H k m Hω ω ω

−+ ⋅ + ⋅ − ⋅ =



Since

the last equation may be transformed as

2 3 Hk km

ω =

2 2

3 3 3

(1 cos ) sin 0( ) sin

eL q kl kl klAEl H kl kl

⎡ ⎤−+ + − =⎢ ⎥

⎣ ⎦



With notation

it may be obtained

from which

2 3

* 3e

m q l AEkl lH H L

ω ω λ= = =

*3 3

2(1 cos )1 0sin

eLAEl

λ ω ωω ω

⎧ ⎫−⎡ ⎤+ − =⎨ ⎬⎢ ⎥⎣ ⎦⎩ ⎭



the frequency equation is obtained:

If the cable parameter is relatively large (cable tends to be in-extensible, which is NOT the case of cable-stayed bridges), then (c) becomes

3

*

4tan( ) ( ) ( ) ( )2 2 2

cω ω ωλ

= −

*λ

tan( ) ( )2 2

dω ω=



Roots of the frequency equation (d) for in-extensible cable are

For n = 1 (i.e. the lowest root) is obtained as

2 2

4(2 1) 1 1,2,3,...(2 1)n n n

nω π

π⎡ ⎤

= + − =⎢ ⎥+⎣ ⎦

1 2.86ω π=



As may be seen, the roots of the frequency equation (c) depend on the cable parameter (which depends upon the loading, cross section, span, modulus of elasticity, cable tension).

*λ

3

*

4tan( ) ( ) ( ) ( )2 2 2

cω ω ωλ

= −



If the cable parameter is relatively small, the cable behaves as the taut string, which IS usually the case of cable-stayed bridges, the frequency equation (c) may be approximated as

or

tan( ) (2 1) , 1, 2,3,...2 n n nω ω π→∞ ⇒ = − =

,3 ,5 ,...ω π π π=



Considering the notation for the natural circular frequencies of free symmetric vibrations in plane of the cable are:

where are solutions of the frequency equation

ω

1,2,3,...nn

H nl mωω = ⋅ =

nω



Symmetric natural modes of free vibrations of the cable are given by

where is an arbitrary constant

2

1 cos( ) 1 sin cossin

n nn n n

n n

qhv x k x k xm H

ωω ω

⎡ ⎤−= − −⎢ ⎥

⎣ ⎦

nh


Orthogonality of Natural Modes –Non-Symmetric Vibrations

Two different natural modes of non-symmetric free vibrations are considered:

The first one is multiplied by and integrated over the span:

2

2

0 ( )

0 ( )p p p

q q q

m v Hv A

m v Hv p q

ω

ω

′′+ =

′′+ = ≠

2 0l l

p p q p qm v v dx H v v dxω ′′0 0

+ =∫ ∫

( )qv x



Partial integration of the 2nd integral:

Due to the boundary conditions, v(0)=v(l)=0, boundary terms are zero, while, due to the 2nd of eqs. (A),

00 0

[ ]l l

lp q p q p q p qv v dx v v v v v v dx′′ ′ ′ ′′= − +∫ ∫

2

0 0

l l

p q q p qH v v dx m v v dxω′′ = −∫ ∫



Consequently, it may be obtained

Since the natural frequencies are different, it follows that

i.e. natural modes are orthogonal.

2 2

0

( ) 0l

p q p qm v v dxω ω− ⋅ =∫

0

0l

p qv v dx =∫



Natural modes of non-symmetric free vibrations are given by

Since

for integration constants

2( ) sin( ) , 1, 2,3,...n n n nnv x C x nlπα α= = =

2 2

0 2

l

n nlv dx C= ⋅∫

2 ( 1,2,3,...)nC nl

= =



natural modes of free non-symmetric vibrations are orthonormal:

0

01

2 2( ) sin , 1, 2,3,...

l

n m

n n n

n mv v dx

n m

nv x x nl l

πα α

≠⎧= ⎨ =⎩

= ⋅ = =

∫


Orthogonality of Natural Modes –Symmetric Vibrations

Symmetric mode number p:

Symmetric mode number q:

2

0

lp e

p p p p p

h Lq qm v Hv h v dxH AE H

ω ′′+ = = ∫

2

0

lq e

q q q q q

h Lq qm v Hv h v dxH AE H

ω ′′+ = = ∫



It may be obtained (considering diff. equations of motion):

Considering both cable equations, it may be obtained:

2 2

0 0 0

( )l l l

p q p q p q q pqm v v dx h v dx h v dxH

ω ω⎛ ⎞

− = −⎜ ⎟⎝ ⎠

∫ ∫ ∫

0 0

0l l

ep q q p

L h v dx h v dxAE

⎛ ⎞− =⎜ ⎟

⎝ ⎠∫ ∫



Therefore, the natural modes of symmetric vibrations are orthogonal:

Also, the following relations are valid:

0

0l

p qv v dx =∫

0 0

l l

p q q ph v dx h v dx=∫ ∫



The natural modes of free symmetric vibrations may be written as

where

( ) 1 tan( )sin( ) cos( )2n n n nv x A k x k xω⎛ ⎞= − −⎜ ⎟

⎝ ⎠

2 2 2n n

nn n

qh qhAm H k Hω

= =



It is possible to determine constants An in order to have orthonormalnatural modes of symmetric free vibrations

0

01

l

n m

n mv v dx

n m≠⎧

= ⎨ =⎩∫


Free Vibrations of Cables outside of the Cable Plane

Free vibrations without the change of cable shape – PENDULUM ModeShallow cable with horizontal slope (b=0)Dead load of the cable q(x) = mg = constParabolic relations


Free Vibrations of Cables outside of the Cable Plane – Pendulum Mode



In the pendulum mode the cable retains its shape and the plane of cable rotates about horizontal axis x.Generalized coordinate: angle ϕbetween x-y (initial vertical) plane and current cable plane during pendulum motion.



Element of the cable arc:- length ds- mass dm = m ds = q/g ds (g=9.81)

Differential equation of motion of the cable element:

2 2

sinx

x

J M dJ qds

dJ dm mds

ε ϕ η ϕ

η η

= ⇒ = − ⋅

= =



Substituting, one obtains:

Integrating over the length of the cable, one obtains:

or

2 2sin ( 9.81 / )m ds mg ds g m sη ϕ η ϕ= − ⋅ =

2 sins s

m ds mg dsϕ η ϕ η= −∫ ∫

* sin 0l gϕ ϕ+ =



where the equivalent pendulum length is given by:

Since

2

*s

s

dsl

ds

η

η=∫

∫

2 21 ( 1 )ds d ds y dxη ξ′ ′= + = +



so, the equivalent pendulum length is given as

With approximation:

2 2

0*

2

0

1

1

l

l

dl

d

η η ξ

η η ξ

′+=

′+

∫

∫

2 211 12

y y′ ′+ ≈ +



And also, for parabolic relations and horizontal span:

It may be obtained

22 2

2

4( ) ( )2 8q f qll l fH l H

η ξ ξ ξ ξ= − = − =

2

*2

81 ( )4 7 0.8085 1 ( )5

fll f ffl

+= ⋅ ≈ ⋅

+



Therefore, the equivalent pendulum length is about 80% of max cable deflectionFor relatively small angles of inclination ϕ: so, differential equation of motion becomes

sinϕ ϕ≈

2

*

0 gl

ϕ ω ϕ ω+ = =



and represents the pendulum mode oscillation of the cable about horizontal axis in direction of span.For initial conditionsthe solution is given as

0 0(0) , (0)ϕ ϕ ϕ ϕ= =

00( ) cos sint t tϕϕ ϕ ω ω

ω= +



If the angle ϕ is not small enough (but it is still small!), then one may assume approximation

Differential equation of motion is

31sin6

ϕ ϕ ϕ≈ −

2 2 2

*

1(1 ) 06

gl

ϕ ω ϕ ϕ ω+ − = =



For the initial conditions

the solution may be obtained as:0(0) , (0) 0ϕ ϕ ϕ= =

30 0

1( ) cos (cos cos3 )192

t t t tϕ ϕ ω ϕ ω ω= + −

CABLE SUPPORTED STRUCTURES

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