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