Chapter 15. Analysis of Sag of Tension

374



Electrical Power Systems



The elastic property of wire is measured by its modulus of elasticity. The modulus of

elasticity is defined as the stress per unit area divided by the deformation per unit length.

Since

s=



T

kg/m2

A



... (15.1)



where

s = stress per unit area (kg/m2)

T = conductor tension (kg)

A = actual cross section of conductor (m2)

Elongation e of the conductor due to the tension is

e=



stress

modulus of elasticity



... (15.2)



Elongation is high if modulus of elasticiy is low. Thus, a small change in the length of

conductor causes large effect on sag and tension of conductor.

Sag and stresses in conductors are dependent on the following things:

1.

2.

3.

4.



Initial tension put on them when they are clamped in place.

Weight of the conductors themselves.

Ice or sleet chinging to them.

Wind pressure.



Stress depends on sag, any span can be used provided the poles or towers are high enough

and strong enough. The matter is merely one of extending the catenary in both directions. Cost

of towers sharply increases with height and loading.



15.2 EFFECT OF TEMPERATURE CHANGE

Sag and stress vary with temperature because of the thermal expansion and contraction of the

conductor. Temperature rise of conductor increase the length of conductor, and hence sag

increases and tension decreases. A temperature fall causes opposite effect. Maximum stress

occurs at the lowest temperature, when the line has contracted and is also possibly covered with

ice and sleet.

If conductor stress is constant while the temperature changes, the change in length of the

conductor is

Dl = l0 aDt

...(15.3)

where

Dt = t1 – t0, Dl = l1 – l0

where

t0 = initial temperature

l0 = conductor length at initial temperature t0

l1 = conductor length at t1.

a = coefficient of linear expansion of conductor per degree centigrade.

Dt = change in temperature in degree centigrade.

Dl = change in conductor length in meter.

If the temperature is constant while the conductor stress changes (i.e., loading), the change

in length of the conductor is



Analysis of Sag and Tension 375



DT

MA

DT = T1 – T0



Dl = l0.



where



...(15.4)



T0 = initial tension of conductor (kg)

DT = change in conductor tension (kg)

M = modulus of elasticiy of conductor (kg-m)

A = actual metal cross section of conductor (m2).



15.3 CALCULATIONS OF LINE SAG AND TENSION

Figure 15.1 shows a conductor suspended freely from two supports, which are at the same lavel

and spaced L meter, takes the form of a catenary curve providing the conductor is perfectly

flexible and conductor weight is uniformly distributed along its length. When sag (d) is very

small in comparison to span L (i.e., the conductor is tightly stretched), the resultant curve can

be considered as parabola. If d < 0.06 L, the error, the error in sag computed by the parabolic

equations is less than 0.5 per cent. If 0.06 L < d < 0.1 L, the error in sag computed by the

parabolic equations is about 2 per cent.



15.3.1



Catenary Method: Supports at Same Level



Figure 15.1 shows a span of conductor with two supports at the same level and separated by a

horizontal distance L. Let O be the lowest point on the catenary curve l be the length of the

conductor between two supports. Let W is the weight of the conductor per unit length (kg/m),

T is the tension of the conductor (kg) at any point P in the direction of the curve, and H is the

tension (kg) at origin O.

Further, s be the length of the curve between points O and P, thus the weight of the portion

s is ws.



Fig. 15.1: Conductor suspended between supports at same level.



Tension T can be resolved into two components, Tx, the horizontal component and Ty, the

vertical component. Then, for equilibrium,



376



Electrical Power Systems



Tx = H



...(15.5)



Ty = ws

Thus the portion OP of the conductor is in equilibrium under the

tension T at P, the weight ws acting vertically downward, and the

horiazontal tension H.

Figure 15.2 shows a triangle, where ds represents a very short

portion of the conductor, in the region of point P. When s is increased by

ds, the corresponding x and y are increased by dx and dy, respectively.

Therefore, we can write,



...(15.6)



tanq =

From Fig. 15.2,



\



dy V

ws

=

=

dx

H

H



...(15.7)

Fig. 15.2



(ds)2 = (dx)2 + (dy)2



FG ds IJ

H dx K



2



=1+



FG dy IJ

H dx K



2



...(15.8)



Using eqns. (15.8) and (15.7), we get



FG ds IJ = 1 + FGH ws IJK

H

H dx K

2



2



ds

\



dx =



1+



FG ws IJ

HHK



2



...(15.9)



Integrating both side of eqn. (15.9),

x=



z



ds

1+



FG ws IJ

HHK



Therefore,

x=



2



FG IJ

H K



ws

H

sinh–1 H + K

w



... (15.10)



where K is constant of integration.

When x = 0 , s = 0, and hence K = 0



\

\



FG ws IJ

HHK

FG wx IJ

H

s=

sinh H H K



x=



H

sinh–1

w



w



...(15.11)



Analysis of Sag and Tension 377



when

\



\

or we can write



x=



FG IJ

H K

2H

F wL IJ

l=

sinh G

H 2H K

w



l

H

wL

=

sinh

2H

2

w



...(15.12)



LM 1 wL + 1 FG wL IJ +...OP

MN1! 2H 3 ! H 2H K PQ

F w L IJ

l = L G1 +

H 24H K

l=



or approximately,



L

l

,s= ,

2

2



3



2H

w



2 2



2



From eqns. (15.7) and (15.11), we get,



...(15.13)



FG IJ

H K



dy ws

wx

=

= sinh

H

dx

H

\



dy = sinh



FG wx IJ dx

HHK



...(15.14)



Integrating both sides of eqn. (15.14), we get,



\



z



FG wx IJ dx

HHK

H

F wx IJ +K

y=

cos G

HHK

w

y=



sinh



...(15.15)



1



If the lowest point of the curve is taken as the origin, when x = 0, y = 0, then K1 =

by the series, cosh(0) = 1.

Therefore,

y=



LM FG IJ OP

N H K Q



H cosh wx - 1

H

w



-H

, since

w



...(15.16)



The curve of the eqn. (15.16) is called a catenary. Eqation (15.16) can also be written as



H

y=

w



LMR1 + 1 F wx I +...U - 1OP

|

|

MNS 2 ! GH H JK V PQ

|

|

W

T

2



378



Electrical Power Systems



or in approximate form,

y=



wx 2

2H



...(15.17)



From Fig. 15.1,

T=



\



H2 +V 2



T = H 1+



From eqns. (15.18) and (15.7), we get,

T = H 1+

From eqn. (15.16), we get



FG V IJ

HHK



2



FG dy IJ

H dx K



...(15.18)



2



...(15.19)



FG IJ

H K



wx

dy

= sinh

H

dx



From eqns. (15.19) and (15.20), we get



T = Hcosh



...(15.20)



FG wx IJ

HHK



...(15.21)



FG

H



IJ

K



L

whereas the total tension in the conductor at the support at x = 2 is



T = Hcosh

or



FG wL IJ

H 2H K



L 1 F wL IJ

T = H M1 + G

MN 2 ! H 2H K



...(15.22)



2



+



FG IJ +...OP

H K PQ



1 wL

4 ! 2H



4



...(15.23)



The sag or deflection of the conductor for a span of length L between supports on the same

L

level is [at x =

, y = d , from eqn. (15.16)]

2

d=

or



LM FG IJ OP

N H K Q



H cosh wL - 1

2H

w



LM FG IJ FG IJ

MN H K H K



L 1 . wL + 1 wL

d=

4 ! 2H

2 2 2H



...(15.24)



3



+



FG IJ +...OP

H K PQ



1 wL

6 ! 2H



5



Analysis of Sag and Tension 379



or approximately,

d=



wL2

8H



...(15.26)



Fig. 15.3: Parameters of catenary.



The safety code gives the minimum (required) clearance height for the line above ground

and if this is added to the sag, the minimum height of the insulator support points can be found.

From Fig. 15.3, c is the ordinate of the lowest point of the curve with respect to the directrix

and y is the ordinate of the point of tangency with respect to the directrix.

From Fig. 15.3,

Ty =

or if



s=



l

w

2

l

, then

2



...(15.27)



380



Electrical Power Systems



Ty = ws



and



...(15.28)



...(15.29)

Tx = wc

where Tx can be defined as the mass of some unknown length c of the conductor and similarly

T and Ty also can be defined. Then at equilibrium

Tx = H

...(15.30)

...(15.31)

Ty = V

where

H = horizontal tension in conductor

V = weight of conductor per meter of span times distance from point of maximum

sag to support.

Thus, from the triangle of forces (Fig. 15.1)

T=



H2 +V 2

Using eqns. (15.32), (15.31), (15.30), (15.29) and (15.28) we get,



bwcg + bwsg

F

I

T = H c + s Kw

2



T=



\



2



...(15.32)



2



2



...(15.33)



From eqns. (15.29) and (15.30), we get,



H

w

From eqns. (15.11) and (15.34), we have

c=



s = c sinh



...(15.34)



FG x IJ

H cK



...(15.35)



From eqns. (15.15) and (15.34), we get,



LM FG IJ OP + K

N H KQ



x

y = c cosh c



1



...(15.36)



FG L IJ . From Fig. 15.3, when x = 0 , y = c,

H 2K

c = c cosh b0g + K



where x is half of the span length

\

\



1



K1 = 0 and therefore,



LM FG IJ OP

N H KQ



x

y = c cosh c

Squaring eqn. (15.35). we get.



...(15.37)



Analysis of Sag and Tension 381



LM

N



FG x IJ OP

H c KQ



...(15.38)



LM

N



FG x IJ OP

H c KQ



...(15.39)



2

s2 = c2 sinh



Squaring eqn. (15.37), we get,



y2 = c2 cosh 2



Subtracting eqn. (15.38) from eqn. (15.39),



LM

N



FG IJ

HK



FG

H



2 x

2 x

y2 – s2 = c2 cosh c - sinh c



\

\



IJ OP

KQ



y2 – s2 = c2

y=



c2 + s 2



From eqns. (15.33) and (15.40),

Tmax = wy

Also

Tmax = w c 2 + s 2



...(15.40)

...(15.41)

...(15.42)



According to eqn. (15.41), maximum tension T occurs at the supports where the conductor

V

s

is at an angle to the horizontal whose tangent is

or , since V = ws and H = wc, at supports,

H

c

y=c+d

...(15.43)

From eqns. (15.40) and (15.43), we get

c+d=



\



c=



c2 + s2

s2 - d2

2d



From eqns. (15.41) and (15.43), we can write,

Tmax = w(c + d)

Substituting eqn. (15.44) into eqn. (15.45),

Tmax =



w 2

(s + d2)

2d



...(15.44)

...(15.45)



...(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.



382



Electrical Power Systems



Therefore,

Tmin = H

But

H = wc,

\

Tmin = wc

From eqns. (15.48) and (15.44),

Tmin = w

From Fig. 15.3,



...(15.47)

... (15.48)



FG s - d IJ

H 2d K

2



2



c=y–d



...(15.49)



The conductor length is

l = 2s

From eqns. (15.50) and (15.35),



...(15.50)



FG IJ

HK



x

l = 2c sinh c



...(15.51)



From eqns. (15.45) and (15.48), we get,

Tmax = Tmin + wd



...(15.52)



15.3.2



Parabolic Method



In the case of short span between supports, the sag is small and the curve can be considered as

parabola. For the sake of simplicity, the following assumptions are made:

1. Throughout the span, tension is considered uniform

2. The change in conductor length due to elastic stretch or temperature expansion is equal

to the change of length of conductor equal in length to the horizontal distance between

the supports.

Let P be any point on the parabolic curve as shown in Fig. 15.4, such that are OP is equal

to x. The portion OP is in equilibrium.



Fig. 15.4: Parameters of parabola.



Analysis of Sag and Tension 383



under the action of T, H and wx.

For equilibrium,

Tx = H and Ty = wx

Taking moments about P,

H.y = wx



FG x IJ

H2K



wx 2

...(15.53)

2H

For short span with small sag, Tmax – Tmin can be considered as small. Therefore Tmax »

Tmin = H. or T = Tmax = Tmin = H.

Therefore, eqn. (15.53) can be written as:

\



y=



y=



wx 2

2T



when



x=



L

,y=d

2



\



d=



wL2

8T



Since T = H,

Also



...(15.54)



...(15.55)



wL2

8H

From eqns. (15.13) and (15.56), we get

d=



...(15.56)



F 8d IJ

l = L G1 +

H 3L K

2



2



...(15.57)



Example 15.1: A transmission line conductor has been suspended freely from two towers and

has taken the form of a catenary that has c = 487.68 m. The span between the two towers is

152 m, and the weight of the conductor is 1160 kg/km. Calculate the following:

(a) Length of the conductor (b) Sag

(c) Maximum and minimum value of conductor tension using catenary method.

(d) Approximate value of tension by using parabolic method.

Solution:

(a) From eqn. (15.12),



FG IJ

H K



l=



since



2H

wL

sinh

w

2H



c=



H

w



384



Electrical Power Systems



LM FG IJ OP

N H KQ



L

l = 2c sinh 2c



c = 487.68 m, L = 152 m



FG

H



\



152

l = 2 × 487.68 sinh 2 ´ 487.68



\



IJ

K



l = 152.576 m.



Using eqn. (15.13)



FG w L IJ

H 24H K

F L IJ

l = L G1 +

H 24c K

FG1 + b152g IJ

l = 152 G

H 24 ´ b487.68g JK m

2 2



l = L 1+



2



2



\



2



2



\



2



\



l = 152.615 m.



(b) Using eqn. (15.24),



LM FG IJ OP

N H K Q

L FLI O

d = c Mcosh G 2 c J - 1P

N H K Q

d=



H cosh wL - 1

2H

w



since



c=



H

w



\



152

d = 487.68 cosh 2 ´ 487.68 - 1



\



d = 5.934 m



\



LM FG

N H



IJ OP

K Q



(c) Using eqn. (15.45)

Tmax = w(c + d)

w = 1160 kg/km = 1.16 kg/m

c = 487.68 m, d = 5.934 m

\



Tmax = 1.16 × (487.68 + 5.934) kg