Electrical Power Systems

Squaring eqn. (15.37), we get,
s 2 = c 2 sinh 2 x
Analysis of Sag and Tension 381
L IO NM HG cKJ
QP ...(15.38)
L IO NM HG cKJ
QP ...(15.39)
y 2 = c 2 cosh 2 x
Subtracting eqn. (15.38) from eqn. (15.39),
\ y 2 – s 2 = c 2
L
NM
I - HG KJ
y2 – s2 x
2 = c cosh sinh
c
\ y = 2 2
c + s
2 2
IO HG KJ QP x
c
...(15.40)
rom eqns. (15.33) and (15.40),
T max = wy ...(15.41)
Also
Tmax = 2 2
w c + s
...(15.42)
According to eqn. (15.41), maximum tension T occurs at the supports where the conductor
is at an angle to the horizontal whose tangent is V s
or , since V = ws and H = wc, at supports,
H c
y = c + d ...(15.43)
rom eqns. (15.40) and (15.43), we get
c + d = 2 2
c + s
2 2
\
s
c =
- d
2d
...(15.44)
rom eqns. (15.41) and (15.43), we can write,
Tmax = w(c + d)
Substituting eqn. (15.44) into eqn. (15.45),
...(15.45)
Tmax = w
2 d
(s2 + d 2 ) ...(15.46)
which gives the maximum value of the conductor tension.
A line tangent to the conductor is horizontal at the point (0), where sag is maximum and has
greatest angle from the horizontal at the supports. Supports are at the same level, thus, the
weight of the conductor in one half span on each side is supported at each tower.
At the point of maximum sag (midspan), the vertical component of tension is zero. Thus,
minimum tension occurs at the point of maximum sag. The tension at this point (at y = c) acts
in a horizontal direction and is equal to the horizontal component of tension.