Advanced Methods of Structural Analysis

5.2 Cable with Neglected Self-Weight 115

Increasing of the thrustHleads to decreasing of the angle ̨ 0 and ̨ 1 ; as a result, the
sag of the cable is decreasing and tension in both parts of the cable is increasing. In
fact, we used, here, the obvious physical statement: in case of concentrated forces
the cable presents a set of straight portions.
The tensions in the left and right portions of the cable can be determined using
expression (5.6).
Shear forces areQA 1 DRAD 12 kN;Q 1 BDRAPDRBD 8 kN.
Therefore

NA 1 D

q

Q^2 A 1 CH^2 D

p

122 C 242 D26:83kNI

N 1 BD

q

Q^21 BCH^2 D

p

.8/^2 C 242 D25:30kN:

Now let us consider the same problem using the concept of the reference beam
(Fig.5.3c). The bending moment at pointCequalsMCDPab=lD.20 10 15/=
25 D 120 kNm:According to (5.4)thesagf at pointC isfCDMC^0 =H D
120=24 D 5 m:Obtained result presents thedistancebetween chordABand
cable and does not related to adoptedx–ycoordinate system.

5.2.1.2 Inverse Problem

Determine a thrust of the cable, if total lengthLof the cable is given.
Length of the cable shown in Fig.5.3a equals

LD

a

cos ̨ 0

C

la

cos ̨ 1

Da

p

1 Ctan^2 ̨ 0 C.la/

p

1 Ctan^2 ̨ 1 :

Length of the cable in terms of active forcePand thrustHmay be presented as
follows

LDa

s

1 C

P^2

H^2



la

l

 2

C.la/

r

1 C

P^2

H^2

a^2

l^2

:

Solving this equation with respect toHleads to following expression for thrust in
terms ofP; L,anda

HDP

2l 0 ı.1ı/

q

l 04 2l 02 .2ı^2 2ıC1/C4ı^2 4ıC 1

;l 0 D

L

l

;ıD

a

l

:

LetıD0:4,LD1:2lD 30 m, sol 0 D1:2. In this case the thrust equals

HDP

2 1:20:4.10:4/

p

1:2^4  2 1:2^2 .20:4^2  2 0:4C1/C 4 0:4^2  4 0:4C 1

D0:7339P