Sag and Tension with Supports at Equal Levels Fig. 41.62 shows a span of a wire with the two supports at the same elevation and separated by a
41.42. Sag and Tension with Supports at Equal Levels Fig. 41.62 shows a span of a wire with the two supports at the same elevation and separated by a
horizontal distance 2l. It can be proved that the conductor AB forms a catenary with the lowest point O forming the mid-point (where the curve is straight).
A.C. Transmission and Distribution
Let W be the weight of the wire per unit length and let point O be chosen as the reference point for measuring the co-ordinates of different points on the wire. Consider a point P having co-ordinates of x and y. The tension T at point P can be resolved into two rectangular components. T x
2 and T 2 y so that T = T x + T y . If S is the length of the arc OP,
then its weight is WS which acts vertically downward through the centre of gravity of OP. There are four forces acting on OP —two vertical and two
horizontal. Since OP is in Fig. 41.62 equilibrium, the net force is zero. Equating the horizontal and vertical components, we have,
T 0 =T x and T y =W S .
It may be noted that the horizontal component of tension is constant throughout the length of the wire :
Since line PT is tangential to the curve OB at point P, tan θ =
It is also seen from the elementary piece PP ′ of the line that tan θ = dy/dx
dy/dx = tan θ = T or dy/dx = WS/T
0 ... ( i)
If PP ′ = dS, then dS =
2 2 () 2 dx + () dy = dx 1(/) + dy dx
Integrating both sides, we have x =
0 sinh
where C is the constant of integration. Now, when
x = 0, S = 0. Putting these values above, we find that C = 0.
− 1 WS
sinh Wx
x = sinh
or S = 0
... ( ii)
T 0 Substituting this value of S in Eq. ( i) ,we get
dy
= sinh or dy sinh Wx = dx dx T
Wx
1662 Electrical Technology
0 Wx ∫
Ex
T
y = sinh
T dx = cosh
T 0
where D is also the constant of integration. At the origin point O, x = 0 and y = 0. Hence, the above equation becomes
T
+ D ∴ D =− 0 W W W Substituting this value of D in the above equation, we get
++ D
0 cosh 0 T 0 T
This is the equation of the curve known as catenary. Hence, when a wire is hung between two supports, it forms a catenary ( a) The tension at point P (x, y) is given by
2 2 2 sinh 2 Wx T
=T x 2 +T y 2 =T 02 +W S = T 02 + T 02 T
—from Eq. (iii)
=− T 2 cosh
T T cosh
T 0 ... ( iv)
( b) Tension at points A and B where x = ± l is given by T = T 0 cosh (W l /T 0 )
... ( v)
( c) The maximum sag is represented by the value of y at either of the two points A and B for which x = + l and x = − l respectively. Writing y = d max and putting x = ± l in Eq. (iii), we get,
Wl
d max 0 = cosh − 1
( vi)
( d) The length of the wire or conductor in a half span is as seen from Eq. (ii) above,
sinh Wl
Approximate Formulae
The hyperbolic sine and cosine functions can be expanded into the following series
sinh z = z + z + z + z + ... and cosh z =+ 1 z + z + z + 3! ... 5!7! 2! 4!6!
Using the above, the approximate values of T d and S points at A and B may be found as follows :
Wl 22
T =T 0 cosh T = 0
2 0 —neglecting higher powers
0 —if is negligible
Wl
0 2 T 0 ∴ T =T 0 — i.e. tension at the supports is very approximately equal to the horizontal tension acting at any point on the wire.
= Wl = Wl (approx)
∴ d = Wl
It should be noted that W and T should be in the same units i.e. kg-wt or newton.
A.C. Transmission and Distribution
( iii) S Wl Wl = Wl 0 sinh + 0 =
—neglecting higher terms. W
3 + ... =+ l Wl
Wl 22
The total length of the wire along the curve is L = 2S = 2 l +
This length consists of the unstertched length L u and the stretch or extension. ∴ unstretched length L u = L —extension ∆ L
Substituting the value of T from the relation d = Wl 2 /2T, we have
— the equation of a parabola
It shows that the catenary curve formed by the sagging is very approximately like a parabola in shape.
The above formulae are sufficiently accurate for all practical purposes provided sag is less than 10% of the span.
Overhead cables are economical and easy to maintain, which are usually made of bare copper or aluminum
wires