Advanced Methods of Structural Analysis

5.2 Cable with Neglected Self-Weight 117

Constant of integration are obtained from boundary conditions: atxD 0 (supportA)

yD 0 and atxDl(supportB)yD 0. These conditions applied to equationy.x/

lead to following constants of integration:C 2 D 0 andC 1 Dql=2H.Nowthe

shape of a cable and slope in terms of loadqand thrustHis described by equations

y.x/D

ql^2

2H



x

l



x^2

l^2



(5.9a)

tanD

dy

dx

D

ql

2H



2

x

l

 1



: (5.9b)

Equationy.x/presents symmetrical parabola. AtxDl=2amaximumy-coordinate

equals to

ymaxD

ql^2

8H

: (5.10)

The tensionNat any sectionxin terms of thrustHis

N.x/D

H

cos

DH

p

1 Ctan^2 DH

r

1 C

q^2 l^2

4H^2



2

x

l

 1

 2

: (5.11)

This equation may be obtained from (5.6), where shearQDRAqx.

At the lowest point (xDl=2) a tensionNDH. The maximum tension occurs

at supports

NmaxDH

r

1 C

q^2 l^2

4H^2

: (5.12)

2.The concept of the reference beam leads to the following procedure. The bend-
ing moment for reference beam at any sectionxisMx^0 D .ql=2/xqx^2 =2.
According to (5.4) and taking into account the direction of they-axis (Fig.5.4),
we immediately get the expression (5.9a) fory.x/, and as result, the formulas
(5.9b–5.12) for slope,ymax,tensionN.x/, and maximum tensionNmax.

5.2.2.2 Inverse Problem

Expression for total lengthLof the cable is

LD

Zl

0

s

1 C



dy

dx

 2

dx: (5.13)

Since the sag of the cable isfDql^2 =8H, then expression (5.9b) for slope at anyx

in terms of sagfmay be presented as

dy

dx

D

ql

2H



2x

l

 1



D

4f

l



2x

l

 1



: (5.14)