Stress Ribbon and Cable-supported Pedestrian Bridges

ISBN 978-0-7277-4146-2

ICE Publishing: All rights reserved

doi: 10.1680/srcspb.41462.041

Chapter 4
Cable analysis

Analysis of the stress ribbon and cable-supported structure is The equilibrium of forces in the vertical direction can be
based on the understanding of the static and dynamic behaviour defined
of the single cable.
VðxÞ qðxÞ dx ðVðxÞ þ dVðxÞÞ ¼ 0
4.1. Single cable
We assume that a cable of area A and modulus of elasticity E acts VðxÞ qðxÞ dx VðxÞ dVðxÞ ¼ 0 ð4:4Þ
as a perfectly flexible member that is able to resist the normal
force only. Under this assumption, the cable curve will coincide qðxÞ dx ¼ dVðxÞ ¼ Hy00 ðxÞ dx
with the funicular curve of the load applied to the cable and to
the chosen value of the horizontal force H (Figure 4.1). where

Consider a cable supported at two fixed hinges a and b loaded qðxÞ

y00 ðxÞ ¼
by a vertical load q(x). We have: H
1
l ¼ XðbÞ XðaÞ ¼ xðbÞ; y00 ðxÞ ¼ ð4:5Þ
RðxÞ
h ¼ YðbÞ YðaÞ ¼ yðbÞ; ð4:1Þ
H
h qðxÞ ¼
tan ¼ : RðxÞ
l

For the given load q(x) and chosen horizontal force H the cable and
curve is determined by coordinate y(x), sag f(x), the slope of the
tangent y0 (x) ¼ tan ’(x) and radius of the curvature R(x). These qðxÞ ¼ Hy00 ðxÞ
values are derived from the general equilibrium conditions on
QðxÞ
the element ds. See Figure 4.1 for definitions of notation. y0 ðxÞ ¼ þ C1 ð4:6Þ
H
The cable is stressed by a normal force N(x) that has vertical MðxÞ
and horizontal components V(x) and H(x), defined: yðxÞ ¼ þ C 1 x þ C2
H
NðxÞ2 ¼ HðxÞ2 þ VðxÞ2 where Q(x) and M(x) are shear force and bending moment on a
HðxÞ ¼ NðxÞ cos ðxÞ ð4:2Þ simple beam of span l. The constants C1 and C2 are determined
from the boundary conditions:
VðxÞ ¼ NðxÞ sin ðxÞ:
x ¼ 0; y ¼ yðaÞ; yðaÞ ¼ 0 þ C1 0 þ C2 ; C2 ¼ yðaÞ
For a vertical load H ¼ const, we have

dy x ¼ l; y ¼ yðbÞ; yðbÞ ¼ 0 þ C1 l þ C2 ¼ C1 l þ yðaÞ

VðxÞ ¼ H tan ðxÞ ¼ H ¼ Hy0 ðxÞ
dx
hence
dVðxÞ H dy
¼ ¼ Hy00 ðxÞ ð4:3Þ
dx dx dx yðbÞ yðaÞ h
dVðxÞ ¼ Hy00 ðxÞ dx: C1 ¼ ¼ :
l l

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 Stress Ribbon and Cable-supported Pedestrian Bridges

Figure 4.1 Basic characteristics of the single cable We then have

X QðxÞ
q(x) p0 ðxÞ ¼
H
h
pðxÞ ¼ y0 ðxÞ ¼ p0 ðxÞ þ ¼ p0 ðxÞ þ tan
l
H a x
ð4:7Þ
MðxÞ
Y b f ðxÞ ¼
H
q(x) A
N(x) V(x) h MðxÞ h
yðxÞ ¼ þ x ¼ f ðxÞ þ x tan
f(x) H l
f(x) y y(x) b H
H
x) for q(x) constant (Figure 4.2). From Figure 4.2, we have
dy ds R(
H B
dx
f(x) 1 1 q
p0 ðxÞ ¼ ql qx ¼ ðl 2xÞ
H 2 2H
V(x) + dV(x) N(x) + dN(x)
s h
I pðxÞ ¼ p0 ðxÞ þ ¼ p0 ðxÞ þ tan
l
ð4:8Þ
MðxÞ 1 1 1 2 q
f ðxÞ ¼ ¼ qlx qx ¼ xðl xÞ
N H H 2 2 2H
h
yðxÞ ¼ f ðxÞ þ x ¼ f ðxÞ þ x tan
l

V x¼0
ql ð4:9Þ
q(x) max p0 ¼
2H
l
A0 B0
x¼
2
ð4:10Þ
Q ql 2
max f ¼
8H

M The length of the cable (Figure 4.3) is defined

ðs ð l qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
s¼ ds ¼ dx2 þ dy2 ð4:11Þ
0 0

Figure 4.2 Uniformly loaded cable

(a) (b)

q q

H x H x H

h p0max f (x) f max

y y(x)
f (x ) H
p0(x)
p(x )

I /2 I /2
I

42

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 Cable analysis

Figure 4.3 Non-tension length of the cable: (a) initial stage and (b) final stage

(a) (b)

q(x)0 q(x)i

H0 a Hi a
b b
A0 Ai

b H0 b Hi

Ln
0 B0 Bi

Ln
Ds
i

Ds
s

I I

since then

dy ð l qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð l sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2
tan ðxÞ ¼ y0 ðxÞ ¼ and dy ¼ y0 ðxÞ dx: 02 QðxÞ h
dx s¼ 1 þ y ðxÞ dx ¼ 1þ þ dx
0 0 H l
We then have ð l sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 Q2 ðxÞ cos2 2QðxÞh cos2
¼ 2
1þ 2
þ
ds2 ¼ dx2 þ dy2 ¼ dx2 1 þ y02 ðxÞ 0 cos H Hl
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ðl
and 1 Q2 ðxÞ cos2 2QðxÞh cos2
¼ 1þ þ dx: ð4:13Þ
0 cos H2 Hl
ðs ð l qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
s¼ ds ¼ dx2 þ ð1 þ y02 ðxÞÞ
0 0 We can express s as
ð l qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð4:12Þ
ðl
¼ 1 þ y02 ðxÞ dx: 1
0 s¼ ð1 þ BÞ1=2 dx ð4:14Þ
cos 0

Since
where
QðxÞ h QðxÞ
y0 ðxÞ ¼ þ ¼ þ tan Q2 ðxÞ cos2 2QðxÞh cos2
H l H B¼ þ : ð4:15Þ
H2 hl
and
Since
2 1
cos ¼ jBj < 1;
1 þ tan2
h it is possible to use the binomial formula:
tan ¼
l
1 h 2 B B B
¼1þ ; ð1 þ BÞp ¼ 1 þ xþ x2 þ x3 þ . . . ð4:16Þ
cos2 l 1 2 3

43

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 Stress Ribbon and Cable-supported Pedestrian Bridges

Figure 4.4 Initial and final stage of the cable: (a) fixed supports and (b) flexible supports

q(x)i
x
q(x)0

q(x)i
H0, Hi y
a
q(x)0
x Da v0 a
b
A0, Ai Da vi H0 b0 x0
h Hi xi
A0 bi
h0
Ai
hi
b H0, Hi Db v 0 b
y y0
Db vi yi H0
0 0 Hi
B0, Bi
B0
i
i Bi

DaH0 I0 Db H0
I Da Hi Ii DbHi

(a) (b)

If we use only the first two terms, we have and

ðl
p 1 cos2
ð1 þ BÞp ¼ 1 þ B s¼ Q2 ðxÞ dx
lþ
1 cos 2H 2 0
ð
1 l cos l 2
ð1 þ BÞ1=2 ¼ 1 þ B ¼ þ Q ðxÞ dx ð4:18Þ
2 cos 2H 2 0
1 Q2 ðxÞ cos2 2QðxÞh cos2
¼1þ þ The term
2 2H 2 Hl
ðl ðl
and hence D¼ Q2 ðxÞ dx ¼ QðxÞQðxÞ dx ð4:19Þ
0 0
ðl ðl
1 Q2 ðxÞ cos2 is usually determined by a Verescagin rule. The area Q(x) is
s¼ dx þ dx
cos 0 0 2H 2 multiplied by the value of Qt, which occurs at the centre of
ðl the gravity of the area (Figure 4.4).
2QðxÞh cos2
þ dx : ð4:17Þ
0 Hl
For example, for uniform load q(x) ¼ const (Figure 4.5(a)), we
Since have

ðl ðl
1 1 l 2 1 q2 l 3
QðxÞ dx ¼ 0 D¼ QðxÞ dx ¼ 2 ql ql ¼ :
0 0 2 2 2 3 2 12

i.e. there is an equilibrium of forces in the vertical direction, For the cable of length l (Figure 4.2(b)), sag f, horizontal force
then H and uniform load q, the length of the cable is defined:
ðl
2QðxÞh cos2 l 1 q2 l 3 q2 l 3
dx ¼ 0 s¼ þ 2
¼lþ
0 Hl 1 2H 12 24H 2

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 Cable analysis

Figure 4.5 Determination of D for: (a) a uniform load and (b) an arbitrary load

(a) (b)
q q(x )

Qi – 1
1/2ql Qi
Q
Q
i–1 i

2/3(1/2ql ) Qi,t
Q
Q

I /2 I /2 hi
I I

for For the vertical load,

ql 2 8f 2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2
ðs ð 1 þ y02 ðxÞ dx
H¼ and s¼lþ : H ds2 H l
8f 3l s¼ ¼
EA 0 dx EA 0 dx
For a general load (Figure 4.5(b)) it is possible to divide the ðl " #
H QðxÞ h 2
length of the girder into the elements of the length h and substi- ¼ 1þ þ dx
EA 0 H l
tute the course of Q(x) by a polygon. Then: ð4:22Þ
ðl !
ðl X
n n ðh
X H cos2
¼ l þ QðxÞ dx
D¼ Q2 ðxÞ dx ¼ Di ¼ Q2 dx EA cos2 H2 0
0 i¼1 i¼1 0
ð
H l 1 l
¼ þ QðxÞ dx :
It is possible to calculate the values of Di for each element: EA cos2 H2 0
Qi 1þ Qi Since
Di ¼ h Qi;t
2
ðl
l cos
where Qi,t is the value of Q(x) at the centre of gravity of the s¼ þ Q2 ðxÞ dx
element. cos 2H 2 0

we have
4.1.1 Elastic elongation of the cable
Since ðl !
H cos2 2
s¼ lþ Q ðxÞ dx ð4:23Þ
NðsÞ ds ds EA cos2 H2 0
¼ ; i:e: NðsÞ ¼ H
H dx dx
and hence
we have
ðl !
ðs ðs 2H l cos2 2
NðsÞ H ds2 s¼ þ Q ðxÞ dx
s¼ ds ¼ ð4:20Þ EA cos cos 2H 2 cos 0
0 EA 0 EA dx !
2H l cos2 ð4:24Þ
and ¼ þ D
EA cos cos 2H 2 cos
ðl
2H l
ds ¼ 1 þ y02 ðxÞ dx: ð4:21Þ ¼ s :
0 EA cos 2 cos

45

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 Stress Ribbon and Cable-supported Pedestrian Bridges

The elastic elongation of the cable (Figure 4.2(b)) is Di

c¼
! ! EA
2H l 2H 8f 2 l H 16f 2
s¼ s ¼ lþ ¼ lþ : and
EA 2 EA 3l 2 EA 3l
cos
4.1.2 Determining the horizontal force Hi d¼ D;
2
For the load q(x)0, horizontal force H0 and temperature t0 the
length of the non-tension cable (Figure 4.4(a)) is defined: we can describe Hi in terms of a cubic equation, i.e.

Ln ¼ s0 s0 d c
bþ þ aHi þ ¼0
l cos H0 l 1 Hi2 Hi ð4:28Þ
¼ þ D0 D
cos 2H02 EA cos2 H0 EA 0 ð4:25Þ aHi3 þ bHi2 þ cHi þ d ¼ 0:
ðl
D0 ¼ Q2x;0 dx: from which the unknown horizontal force Hi can be easily
0
determined.
For the load q(x)i, unknown horizontal force Hi and tempera-
ture ti, the length of the non-tension cable (Figure 4.4(b)) is 4.1.3 Influence of deformation of supports and
defined: elongation of the cable at the anchor blocks
In actual structures it is necessary to include possible deforma-
Lni ¼ si si tions of supports and elongations of the cable at the anchor
blocks.
Lni ¼ Lnð1 þ t ti Þ
ð4:26Þ
ðl Deformations of supports for load 0 and i are defined:
Di ¼ Q2x;i dx
0 V V V V
ai ¼ Ai a a0 ¼ A0 a

where the temperature change ti ¼ ti t0 and is a coefficient H H H H

i ai ¼ Hi a a0 ¼ H0 a
of thermal expansion. Since ð4:29Þ
V V V V
bi ¼ Bi b b0 ¼ B0 b
l cos H H H H
si ¼ þ Di bi ¼ Hi b b0 ¼ H0 b
cos 2Hi2
and depend on values of the reactions and positive unit defor-
and
mations aV , aH , bV and bH . For load 0,
Hi l 1
si ¼ þ D; l0 ¼ Xb Xa H H
EA cos2 Hi EA i a0 b0

V V
h0 ¼ Yb Ya a0 bi0 ð4:30Þ
we have
h
tan 0 ¼ 0:
Lni ¼ si si l0

l cos Hi l Di For load i,

¼ þ Di
cos 2Hi2 EA cos2 EAHi ð4:27Þ
H H
li ¼ Xb Xa ai bi
l cos Hi li Di
Lni Di þ þ ¼0 V V
cos 2Hi2 EA cos2 EAHi hi ¼ Yb Ya ai bi ð4:31Þ
h
If we denote tan i ¼ i:
li
l
a¼ ; The elastic deformations of the cable in the anchor blocks a and
EA cos2
b are defined for load 0:
l
b ¼ Lni ;
cos an;0 ¼ kH0 ð4:32Þ

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 Cable analysis

Figure 4.6 Elastic deformations of the cable at the anchor blocks where
li cos i Hi li Di
H=1 Lni Di þ þ kHi þ ¼0
cos i 2Hi2 EA cos2 i EAHi
ð4:36Þ
If we set
li li Di
a¼ þ k; b ¼ Lni ; c¼
S EA cos2 i cos i EA
and
1
cos i
d¼ Di ;
2
we obtain a cubic equation to determine Hi:
aHi3 þ bHi2 þ cHi þ d ¼ 0: ð4:37Þ
Ik Since the parameters a, b, c and d depend on the span li and
vertical difference hi (which depends on horizontal force Hi),
it is not possible to determine the unknown Hi directly by
solving Equation (4.37); it is therefore necessary to determine
and load i:
Hi by iteration. First, the unknown Hi is determined for zero
deformation of supports and zero elongation of the cable at
an;i ¼ kHi ð4:33Þ
the anchor blocks. For this force, the vertical reactions Ai and
Bi, span length li, vertical difference hi, parameters a, b, c and
where k ¼ ka þ kb (Figure 4.6) expresses the elongation of the
d and new horizontal force Hi are utilised. The computation is
cable at the anchor blocks a and b due to unit horizontal
repeated until the difference between the subsequent solutions
force H ¼ 1. ka and kb are defined:
is smaller than the required accuracy.
ð lka ð lkb
Ska Skb
ka ¼ ds kb ¼ ds ð4:34Þ 4.2. Bending of the cable
0 EA 0 EA
The bending of the cable is derived from the analysis of the
For the load q(x)0, horizontal force H0 and temperature t0, the single cable which is stressed by a known horizontal force H
length of the non-tension cable (Figure 4.4(a)) is defined: (Eibel et al., 1973).

Ln ¼ s0 s0 Figure 4.7 shows a single cable of the area A, moment of inertia

an;0
I and modulus of elasticity E that is fixed to the supports a and
l0 cos 0 H0 l0
¼ þ D0 Figure 4.7 Geometry and internal forces at the cable
cos 0 2H02 EA cos2 0

1 ð4:35Þ
D kH0 N(x) M(x)
H0 EA 0
ð l0 Q(x)
q(x)
D0 ¼ Q2x;0 dx
0 g
Hg , H x
For the load q(x)i, unknown horizontal force Hi and tempera- a b
q(x)
ture ti, the length of the non-tension cable (Figure 4.4(b)) is A h(x)
V(x) h
defined: N(x)
M(x) y(x)
H y, h Hg , H
Lni ¼ Lnð1 þ t ti Þ j(x)
dh b
ds w(x)
Lni ¼ si si an;i M(x) + dM(x) B
dx H
l cos i Hi li 1 j(x)
¼ i þ Di D kHi N(x) + dN(x)
cos i 2Hi2 EA cos2 i Hi EA i V(x) + dV(x)
ð li
Di ¼ Q2x;i dx I
0

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 Stress Ribbon and Cable-supported Pedestrian Bridges

b. The cable is loaded by load g(x) and q(x). Corresponding Since dx2 dx, dx2 can be neglected. We then have
horizontal forces are Hg and H.
VðxÞ dx H d ðxÞ dMðxÞ ¼ 0
It is assumed that erection of the cable is done is such a way that d ðxÞ dMðxÞ
the load g does not cause any bending of the cable. For constant VðxÞ H ¼0
dx dx
g, the shape of the cable given by y(x) is the second-degree
parabola: dVðxÞ d2 ðxÞ d2 MðxÞ
H ¼0
dx dx2 dx2
h g h dVðxÞ
yðxÞ ¼ f ðxÞ þ x ¼ xðl xÞ þ x qðxÞ ¼
l 2Hg l dx ð4:43Þ
g g 2 h d2 wðxÞ
¼ xl x þ x MðxÞ ¼ EI
2Hg 2Hg l dx2
ð4:38Þ !
g g h g l h d2 ðxÞ d d2 w
y0 ðxÞ ¼ l 2 xþ ¼ x þ qðxÞ H EI 2 ¼0
2Hg 2Hg l Hg 2 l dx2 dx2 dx

g 1 d4 wðxÞ d2 ðxÞ
y00 ðxÞ ¼ ¼ : EI H ¼ qðxÞ:
Hg R g dx4 dx2

We then have
The shape of the cable for load q(x) is given by the coordinate
ðxÞ ¼ yðxÞ þ wðxÞ
ðxÞ ¼ yðxÞ þ wðxÞ
ð4:39Þ d4 wðxÞ d2 ðxÞ
d ðxÞ ¼ dyðxÞ þ dwðxÞ EI H ¼ qðxÞ
dx4 dx2
where w(x) is the deformation of the cable due to load d4 wðxÞ d2 yðxÞ d2 wðxÞ
EI H H ¼ qðxÞ
q(x) g(x). dx4 dx2 dx2
d4 wðxÞ d2 wðxÞ d2 y ð4:44Þ
The cable is stressed by normal force N(x), shear force Q(x) and EI H ¼ qðxÞ þ H
dx4 dx2 dx2
by bending moment M(x):
g 1
¼ qðxÞ þ H ¼ qðxÞ þ H
NðxÞ ¼ H cos ðxÞ þ V sin ðxÞ Hg Rg
ð4:40Þ H H
QðxÞ ¼ H sin ðxÞ þ V cos ðxÞ: ¼ qðxÞ þ ¼ qðxÞ g
Rg Hg
These values are derived from the equilibrium conditions on the
element ds. Equation (4.44) can be easily extended to express the elastic
support of the portion of the cable by Winkler’s springs
For vertical load with constant H, (Figure 4.8):

V¼0 V¼0
VðxÞ qðxÞ dx þ kwðxÞ dx ðV þ dVÞ ¼ 0
VðxÞ qðxÞ dx ðV þ dVÞ ¼ 0
qðxÞ dx þ kwðxÞ ¼ dV ð4:45Þ
ð4:41Þ
qðxÞ dx ¼ dV
dV
dV qðxÞ þ kwðxÞ ¼ :
qðxÞ ¼ dx
dx
The characteristic of the spring k(x) is a ‘stress’ that corresponds
and to its unit deformation.

M¼0 M¼0
1 2 1 2
VðxÞ dx H d ðxÞ þ MðxÞ 2 qðxÞ dx
ð4:42Þ VðxÞ dx H d ðxÞ þ MðxÞ 2 qðxÞ dx ð4:46Þ
2
ðMðxÞ þ dMðxÞÞ ¼ 0: þ kwðxÞ 12 dx ðMðxÞ þ dMðxÞÞ ¼ 0

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 Cable analysis

Figure 4.8 Geometry and internal forces at flexibly supported cable Figure 4.9 Bending moments: (a) at support and (b) under point
load
M(x)
N(x)
(a) q
Q(x) q(x)
g
g
Hg , H x Hg , H x Hg , H
a b g
q(x) A h(x)
V(x) h Df(x) q
N(x)
M(x) y(x)
H y, h Hg , H
f(x)
dh b M
ds w(x)
M(x) + dM(x) B
k(x) H f(x)
N(x) + dN(x) k(x)
V(x) + dV(x)
dx F
I
q
g
(b)
Since dx2 dx, dx2 can be neglected and we then have: –x x g
? ?
4 2
d wðxÞ d wðxÞ H g, F
EI H þ kwðxÞ ¼ qðxÞ þ
dx4 dx 2 R g
ð4:47Þ
H
¼ qðxÞ g
Hg M

The solution of Equation (4.45) can be written as:

wðxÞ ¼ wh ðxÞ þ wp ðxÞ ð4:48Þ

where the particular solution wp(x) corresponds to deformation

of the cable without bending stiffness; the homogenous solution
can be written:
x x x¼0
wh ðxÞ ¼ Ae þ Be þ C þ Dx
rffiffiffiffiffiffi ql gl ð4:50Þ
H
ð4:49Þ w0p ð0Þ ¼ ¼ q ð0Þ g ð0Þ ¼
¼ : 2H 2Hg
EI
A direct solution is possible only for special cases (Figure 4.9), The homogenous solution is:
for example the course of the bending moment M(x) in the x
vicinity of the support of the cable loaded by uniform loads g For x ¼ 1, M ¼ 0, w ¼ 0, e ! 1, and therefore A ¼ 0 and
and q and corresponding horizontal forces are Hg and H is D ¼ 0. Thus
solved for an infinitively long cable. x
wh ðxÞ ¼ B e þC
The particular solution is: w0h ðxÞ ¼ Be x

q g
wp ðxÞ ¼ ðxÞ yðxÞ ¼ ðl xÞx ðl xÞx w00h ðxÞ ¼ 2
Be x
2H 2Hg
w0 ðxÞ ¼ w0h ðxÞ þ w0p ðxÞ
qlx qx2 glx gx2
¼ þ
2H 2H 2Hg 2Hg For x ¼ 0
ql qx gl gx
w0p ðxÞ ¼ þ
2H H 2Hg Hg w0 ðxÞ ¼ 0 ¼ B e0 þ
q g ð4:51Þ
w00p ðxÞ ¼ þ ¼ q B¼
H Hg

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 Stress Ribbon and Cable-supported Pedestrian Bridges

w00 ðxÞ ¼ w00h ðxÞ þ w00p ðxÞ ¼ 2

e x
q d2 wðxÞ 1
¼ ðwi 1 2wi þ wi þ 1 Þ
dx2 h
ð4:54Þ
MðxÞ ¼ EJw00 ðxÞ d4 wðxÞ 1
rffiffiffiffiffiffi ð4:52Þ 4
¼ 4 ðwi 2 4wi 1 þ 6wi 4wi þ 1 þ wi þ 2 Þ
H dx h
¼
EI In this way, a solution of Equation (4.47) was substituted by a
pffiffiffiffiffiffiffiffiffiffiffiffiffiffi x
MðxÞ ¼ H EI e þ qEI solution of a system of linear equations. From the discrete
values of the deflections, the bending moments and shear
The bending moment at an infinitively long cable that is loaded forces were determined.
by point load F and by uniform load q (Figure 4.9(b)) can be
derived similarly. For uniform load g that does not cause the From the discrete values of deformations, the bending moment
bending, the cable is stressed by horizontal force Hg; for and shear forces were determined from:
uniform load q and point load F the cable is stressed by hori-
zontal force H. The course of bending moment is defined: EIi
Mi ¼ ðwi 1 2wi þ wi þ 1 Þ
h2
F ð4:55Þ
MðxÞ ¼ e x
þ EI q M Mi 1
2 Ti ¼ i þ 1 :
2h
rffiffiffiffiffiffi
H
¼ ð4:53Þ The described analysis of the bending of the cable was verified
EI
during the loading tests of the cables that were developed for
q g
q¼ the Elbe Bridge (Figure 2.10). The cable was formed from 18
H Hg strands of 15.5 mm diameter that were grouted in steel tubes.
The tubes and mortar are composite with the strands and
Rather then solving the equations for different loading condi-
contribute to the resistance of the strands to the live load
tions, the author developed a program in which the deformation
(Figure 2.42). The tested cable was loaded with a point load
and corresponding shear forces and bending moments were
situated at its mid-span (Figure 4.11). During the test, the
solved using the finite difference method (Tomishenko and
strain in the steel tubes at the mid-span and support sections
Goodier, 1970). This approach enables us to describe a local
was carefully measured. The bending moment was calculated
stiffening of the cable and supporting of portion of the cable
from these strains. Figure 4.12, which depicts the arrangement
by Winkler springs.
and results of the test, highlights that a good agreement of
results has been achieved.
In the analysis, the cable was divided into short elements of
length h (in the actual structure, a cable of length 100 m was
During the analysis, the cable is initially analysed as a perfectly
divided into 10 000 elements of length of 0.010 m). The elements
flexible member. The initial state is chosen as the state in which
can have different stiffness values given by EI and can be
erection guarantees that the cable is without bending. To
supported by springs of different stiffness.
analyse loading, the unknown horizontal force is determined
and then used to determine the deformation, shear forces and
The derivations were substituted by well-known formulae
bending moments.
(Figure 4.10)
Figure 4.11 Loading test of the stay cable
Figure 4.10 Finite difference method

h h h h

Wi – 2 Wi – 1 Wi Wi + 1 Wi + 2

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 Cable analysis

Figure 4.12 Deformation and bending moments of the tested In actual cables loaded by a uniform load, the bending moment
cable is nearly zero along the length of the cable. Significant values of
bending moments originate only close to the supports and
P = 31.5 kN
1 1 under the point load (Figure 4.9); their course is exponential,
2 3 2 which must be taken into consideration when we analyse the
H 0.500 H = 1730 kN structures by modern non-linear programs by finite element
methods. To cover the concentration of stresses, a very fine
0.300 1.050 16.640 1.050 0.300
mesh of elements has to be used close to supports and point
19.340
loads.
part 1 part 2 part 3

To understand the difference between the behaviour of the beam

and of the cable, a stress ribbon and beam structure of span
I = 2.39 ¥ 105 m4 I = 1.03 ¥ 105 m4 I = 0.69 ¥ 105 m4 L ¼ 33, 66 and 99 m loaded by uniform load and by vertical
M (kN m) –18 deflection of supports are examined here. The stress ribbon was
modelled as a cable and analysed by the above process.
–9
The structure has area A ¼ 1.25 m2, moment of inertia
0 I ¼ 0.0065104 m4 and modulus of elasticity E ¼ 36 000 MPa.
measurement The stress ribbon structures have sag at mid-span fL/2 ¼
9 calculation 0.02L, corresponding horizontal force Hg ¼ gL2/8fL/2 ¼
6.25(gL) ¼ 195.3125L. The dead load g ¼ 31.25 kN/m does not
y (mm) 0 cause bending of the structure.
20
Figure 4.13(a) and Table 4.1 show the deflection and bending
40 moment in the beam and cable for load p ¼ 20 kN/m. It is
evident that the deflections and bending moments in the stress
60
ribbon are very low.

Figure 4.13 Deformation and bending moments at beam and stress ribbon: (a) uniform load and (b) vertical deflection of support

(a) (b)
q=g+p
p
g g

y
1.00 m

Hg

Hg , HD Hg , Hq Hg , HD 1.00 m
HD

L L

M
stress ribbon
beam

stress ribbon
beam

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 Stress Ribbon and Cable-supported Pedestrian Bridges

Table 4.1 Comparison of the static effects Figure 4.14 Natural modes

L: m
f(1)
33.00 66.00 99.00

Mp Beam (MNm) 1.815 7.262 16.335

ML/2 Beam (MNm) 0.908 3.631 8.168
yL/2 Beam (m) 0.5271 8.4332 42.6933 f(2)
Hq Stress ribbon (MN) 10.180 19.768 28.939
Mp Stress ribbon (MNm) 0.106 0.339 0.594
ML/2 Stress ribbon (MNm) 0.034 0.039 0.036
yL/2 Stress ribbon (m) 0.115 0.0726 0.1679
f(3)

Figure 4.13(b) and Table 4.2 show the deflection and bending
moment in the beam and stress ribbon stressed by a vertical
deflection of support ¼ 1 m. With the increasing span
length, the bending moments in the beams are reduced propor- f(4)
tionally to the square of their length.

On the contrary, the bending moments in the stress ribbon have sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
significant values that in the longer spans are even higher that in 1 1 Hn2 EI 2 n2
the beam. It is therefore necessary to carefully analyse the fðnÞ ¼ þ ð4:57Þ
2 l2 l4
bending of cables (or prestressed bands) in structures where
significant vertical deformations can occur (e.g. in the cables
where H is horizontal force, is mass of the cable per unit
of a cable-supported structure).
length, f is sag of the cable, E is the modulus of elasticity, A is
area and I is moment of inertia.
Note: the above-described analysis is hypothetical since a beam
of the above dimensions would fail.
The term
4.3. Natural modes and frequencies
EAf 2 2
The natural modes and frequencies of a single cable (Figure
2l 4
4.14) can be determined according the following formules
(Strasky and Pirner, 1986): in Equation (4.56) describes the normal stiffness of the cable
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi that has to elongate when vibrating in the first mode. This is
1 1 H EAf 2 2 EI 2 the reason why, in some cases, the first mode is higher than
fð1Þ ¼ þ þ 4 ð4:56Þ
2 l2 2l 4 l the second mode.

The term

Table 4.2 EI 2 2
n
l4
L: m
in Equation (4.57) describes the bending stiffness of the cable
33.00 66.00 99.00 which, in engineering calculations, is insignificant.

Mp,a Beam (MNm) 1.291 0.323 0.143 Natural modes of vibrations are defined as:
Mp,b Beam (MNm) 1.291 0.323 0.143
Hq Stress ribbon (MN) 8.266 13.542 19.730 l
Mp,a Stress ribbon (MNm) 1.429 0.787 0.637 X
n i x
2
Mp,b Stress ribbon (MNm) 2.485 1.166 0.839 wðx; tÞ ¼ Ai cosð2 fi Þt þ Bi sinð2 fi Þt sin
i¼1
l

52

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 Cable analysis

where Ai, Bi are determined by the progression of the right-hand Strasky J and Pirner M (1986) DS-L Stress ribbon footbridges.
side of equations: Dopravni stavby, Olomouc, Czechoslovakia.
Timoshenko SP and Goodier JN (1970) Theory of Elasticity.
wðx; 0Þ ¼ gðxÞ McGraw-Hill, New York.
w0 ðx; 0Þ ¼ hðxÞ:

REFERENCES
Eibel J, Pelle K and Nehse H (1973) Zur Berechnung von
Spannbandbrücken – Flache Hängebänder. Verner, Verlang,
Düsseldorf.

53

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