5.4 Cable with Self-Weight 127
Formula (5.22) for tension at any point can be rewritten as follows
N.x/DH
r
1 Csinh^2
q 0
H
xDHcosh
q 0
H
x: (5.26)
The minimum tensionNDHoccurs atxD 0 , to say in the lowest point.
In order to obtain equation of the catenary, we need to expressyin terms ofx
dyDdxtanD
W
H
dxD
q 0 s
aq 0
dxDsinh
x
a
dx:
After integrating this equation from pointCto pointD,weget
Zy
a
dyD
Zx
0
sinh
x
a
dx!yjyaDacosh
x
a
ˇ
ˇ
ˇ
x
0
:
The equation of the curve, assumed by the cable, and corresponding slope at any
point are
y.x/Dacosh
x
a
D
H
q 0
cosh
q 0
H
x
tanD
dy
dx
Dsinh
q 0
H
x: (5.27)
Below is presented analysis of typicalcases of cables carrying a uniformly dis-
tributed load along the cable itself:
1.Supports located at the same level
2.Supports located at different elevation and saddle point within the span
3.Supports located on the different elevation and cable has not a saddle point within
the span
5.4.2 Cable with Supports Located at the Same Level
In this case a cable has the axis of symmetry. It is pertinent here to derive the simple
and useful expression for maximum tension for the cable in terms ofq 0 ,f,andH.
Maximum tensionNmaxoccurs at the supports.xD ̇l=2/,then
NmaxDH
r
1 Csinh^2
q 0 l
2H
!sinh^2
q 0 l
2H
D
Nmax^2
H^2
1: (5.28)