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Ma# 8## S2# R##a! & C#a3# W

1950

Calculations of maximum sag of a transmission line with an ice load on one span

Rodney A. Schaefer

F** a aa* a: ://!*a+#.+.#2/+a#_##Department: Electrical and Computer Engineering

8 8# - O# A!!# 2% 2 $ $## a # a!!# # S2# R##a! & C#a3# W a S!*a' M#. I a ## a!!##

$ !*2 Ma# 8## a a2# a+a $ S!*a' M#. F +# $+a, *#a# !a! #a3#@+.#2.

R#!++## CaS!a#$#, R# A., "Ca*!2*a $ +a+2+ a% $ a a+ *# a !# *a # a" (1950). Masters Teses. Pa#6674.

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L UL TIOns OF

E AXDIUM

S G OF

TR NSMISSION LINE

WITH N

ICE

LO D ON ONE SP N

BY

RODNEY RTHUR

SCH EFER

A

THESIS

submitted to

the

faculty of the

SCHOOL OF

MINES

ND MET LLURGY OF THE

UNIVERSITY

OF

MISSOURI

in p rt l ~ u l i l l m n t

of the work required for the

Degree of

M STER OF S IEN E IN ELE TRI L ENGINEERING

Rolla

Missouri

1950

Approved ~ J : r ¥ :

rofessor of h ~ e t r i l

Engineer rig



CKNOWLEDGEMENT

I wish to

express

my thanks to

Erofessor

Lovett and Dr J

zaborsky of the l e c t r i c ~ Engi-

neering Department for

the i r

encouragement advice

and constructive cri t icism

on

the ideas which are set

forth

in th is

paper

i i



T LE

OF

CONTENTS

Exact

Analysis

•••

Review

of Litera ture

••••

•

• • • • •

•

• • • • • •

• •

• • • •

•

• • • • •

•

• • • • • •

Page

iv

8

• • •

•

• •

•

•

•

• • • • • • • •

• •

• • •

• • • • • • • • • • • • • • • •

• •

• •

•

• •

•

• •

•

• •

•

• •

•

• • • • • • • • • • •

• • •

•

•

• • • •

•

• •

•

• • •

of l lus t r a t ions

Catenary

Cable

Acknowledgement • • • • • • •

Introduction

• • • • • • • • • • •

List

Parabolic Cable

Exact nalysis

3

Catenary

Cable

Approximate nalysis 7

Parabolic Cable Approximate nalysis 2

Conclusions •••

Sununary

•••••••

•

• •

•

•

• • •

•

• • •

•

• • •

•

• • • • • • • •

•

• •

•

• • • • • • • • • • •

3

Bibl iography

36

Vi t

37



iv

LIST

O

ILLUSTR TIONS

] igure

1

]rorce

d iagram

of

an y

i ns ul a t or

••••••••••••••••

P a g e

6

2 Sag

and

i n s u l a t o r swing

in

spans

k l

k and

k

• • • • • • • • 8

3

4

5

6

7

8

Force diagram of k i ns ul a t or

••••••••••••••••••

Force

diagram

of k i ns ul a t or showing horizon-

ta l

and ver t ica l components

of the

maximum

conductor

tensions

• • • • • . • • • • • • • • • • • • • • • • • • • • • •

Sa g and i ns ul a t or swing in

cr i t ica l

spans •••••

Force diagram of i ns ul a t or

] orce d iagram

of

suspension

in su lato r

4 show-

i n g the h o rizo n tal and v e r t i c a l components

of

the maximum

conductor

tensions ••••••••••• . • • ••

Force

diagram

of suspension

i ns ul a t or show-

ing

the

h o rizo n tal and v e r t i c a l components

of

the

maximum conductor tensions

••••••••••••••••

8

3

17

17

2

22



INTRO U TION

important problem concerning

electr ic power

l ines

is the

calculat ion

of the maximum sag occurring

when there

i s

an

ice

load

on

only

one

span

The

object of

this

thesis

is to determine a praot ical method fo r

calculat ing

the

maximum sag of a transmission l ine suspended on

suspension

insula tors with the

aforementioned

loading conditions

This

paper

i s

l imited

to

the

case where the supports

of

the

transmission l ine

are a t

equal

elevations

and a l l

span

lengths

are

equal

he

condition of

ice

on one span of a

transmission

1ine

oocurs

qUite

~ r q u n t y

upon the

melting

of hoar

f rost

ioe

s leet

or snow

from

the l ine I t is extremely impor-

tant to

be able

to calculate

the

maximum

sag

occurring a t

suoh t imes

The

reason for this

i s

the necessity of calcu-

la t ing the

height

of the

supporting

structures

and the

ver-

t i ca l spaoing

of tbe conductors so that the l ine even u Ylder

this worse possible

loading

condition

will

not come

in

contact with the ground thereby causing a l ine to ground

faul t

or

even

come in con ta ct w ith

the other

conductors in

the same

span thereby causing

a

l ine to l ine faul t

addition

to these

conditions i t is

also necessary to be

p ositiv e th at

the

l ine wil l

not

come

close

enough

to

the



2

ground

to

endanger

the l ives of people

in

the

vic ini ty .

Mechanical

calcula.t ions

of

elec t r ic

power

l ine

conduc-

to rs a re

mede

on

t he b a s i s

that

the curve t the cell ter

l ine

o f th e

susnended conductor

i s ei ther

a

ca te na ry o r

a

p a r a -

b o l a . hen th e

load

i s assumed

to be uLli forml y dist i -

buted a l ong

the

c e n t e r l ine

of

th e cona.uctor th e conductor

hangs in a cu r v e c a l l e d the catenar y.

hen

the load is

assumed to be

uniformly

d i s t r i buted

along

th e hor i

zon ca.l

th e resu l t ing

curve

is a parRbola. I n a c t u a l p r a c t i c e

w i t h

an

ic e load on the cond.uctor a cona.ition somewhere

between these two extremes exis ts . However is gener ally

agreea. t h a t

th e

assumption of

a

catenar y

curve

gi ve s

more

n e a r l y

accur ate resul t s ,

e s p e c i a l l y

fo r tr ansmission l ines o f

l ong spans i . e . ,

spans g r e a t e r

than 1000 feet .

I t

i s

th e purpose of

th is

paper

to

determine

whether

or

no t

is

p o s s i b l e

to a r r iv e a t

a pract ica l s o l u t i o n of

the

problem

making

c a l c u l a t i o n s ei ther

on

the

b a s i s of the

c a t e n a r y curve

or

th e p a r a b o l i c curve. y a pract ica l s o l u -

t i o n is meant a

form

of

e qua t i ons

s u i t a b l e f o r e ngi ne e ri ng

c a l c u l a t i o n s .



3

REVIEW OF LITER TURE

Practioally a l l the previous

work

attempted

on prob

lems

of th is nature has been done experimentally on spe

ci f ic

eleotr ic

power l ines

to

obtain

data

necessary for

th e p revention of l ine to l ine and l ine-to-ground

faul ts

oocurring

upon the

melting

of

ice from

the

conduotors.

One

of the f i r s t to make a detailed

study

of th is

1

problem

was V

R. Greisser.

In

1913, Greisser

pub-

l ished a

paper

disoussing

in

great

detail a

study

made

upon a l ine of the Washington Water Power Company of

Spokane. Washington, between i t s Lit t le Falls power sta-

t ion and

a

substation near Spokane.

The reason

for the

tes t was the faot that during a few days of fog and frost

conditions short cirCUits

ooourred on the

l ine so fre-

quently

that

the

l ine was practioally

useless.

I t

was

found,

af ter

a thorough inspection, that the

hoar frost

and ice which formed

on the conduotors would

not fa l l

from

a l l spans a t

the

same tim e. Therefore.

the

ioe

loaded

spans

would increase thei r sags. at the same time

deoreas ing the sag

of the adjacent

spans, and deflecting

the suspension insulators unt i l a new oondition

of

sta t io

equilibrium

was established. This oaused

short

oirouits

between the

conduotors.

1

Greisser.

v

Effeots

of

Ice Loading on Trans

mission

Lines, A.I.E.E. Trrols., Vol. 32, Part I I .

pp.

1829-1844,

September 1913.



The experiments were

performed

p lacing equal ly

spaced bags of

rock along the

l ine

to

reproduce

the

ice

loaded condition on the conductors. This i s not

an

exact

representation of the ice loaded

condition,

but app roxi-

mates the

condition,

as i t is

knO\v n

that

ice

or sleet

forming on wires does not do

so

with exact uniformity.

Data of swing

of the

insulators

and sag of

the

suspension

insulators as well

as

the

elast ic

properties of the con

ductors

are

both

ext remely importan t

in

determining the

maximum

sag

of the conductors. Also i t was

proven

that

the

influence of a change

in

temperature was negligible.

2

Somewhat

l a te r

E.

S.

Kealy and

A Wright

published

a

paper describing tes ts

made upon

the Wallen-

paupack-Siegfried

l ine of the Pennsylvania Power and

Light Company The tes ts carried

out

in

this

case

were

pract ical ly

the same as those previously performed by

v

H Greisser on

an

entire ly

different l ine In

this

experiment, i t was also determined. that the swing of the

suspension

insulators, as

well

as

the elas t ic

properties

of

the

conductors,

is extremely

important in

determing

the

maximum

sag

of

the conductors. Therefore,

the

re -

suIts obtained by Healy and Wright

are

simply a ver i f i -

cation of

those

obtained. ear l i e r by Greisser.

u

Healy,

E A and

Wright,

A J Unbalanced Cond uctor

Tensions, Jour. A.I.E.E., Vol. 45, pp. 1064-1070,

September

926



}

In

1935. S.

Noda

and

S Nishiyama

published

a

paper in which they derived

a

method of

aalculating the

maximum

sag

of

an elea t r ia

~ o w e r

l ine having an ice load

on

only

one

span.

However

in arriving

at

a

solution

of

the

problem,

they

made two

incorrect

assumptions which

material ly

affect the

f ina l

answer. Besides these incor-

rect assumptions, two

very serious mistakes

were made in

set t ing

up the

equation of s ta t ic equilibrium.

In the i r ~ p e r Noda and Nishiyama

assumed

in sett ing

up th e requ ired

equations that

both the weight of the

suspension

insulators

and

the elast ic i ty

of

the conductors

could

be

neglected. For extremely long spans, i t

m y be

possible to neglect

the

weight of the

suspension insu

l a to rs but for

short

spans this

weight

has

a

noticeable

affect

upon

the sag.

liowever,

under

n

circumstances

can

the

elas t ic i ty of the conductors

be neglected as

i t

is

c r i t ica l in

a l l length

spans.

In

set t ing

up

the equation of

sta t ic

e ~ u i l i r i u m of

the

insulators ,

a very

serious

mistake

was made

by ne-

glecting

not only

the angles

of

incl inat ion

of

the

vectors

representing the tension in the conductors,

but

also

assuming the d i ~ f e r e n e of

the

conductor tensions to be

perpendicular to th e suspension

insulator .

5] Noda S and Nishiyama, S Mechanical

Characterist ics

of

fransmission Lines, Byojun College of Engineering

Ivtemoirs

105-138,

1935.



T

Figure 1 . l orce diagram

of

any

insulator.

l4

~ r l =

Lwb

k

a

i s

the

equation of s ta t ic equilibrium used by Hocla

and

Nishiyama in the solution

of

th is

problem.

Where

~

=

maximum tension

in

conductor of span k

T:k

1

=

maximum tension in conductor

of

span k ., . 1

L

= length of

span

w

=

weight

o f oonducto r

per unit length

=

horizontal

displacement of the

end of the

suspension insulator

a

length of

the suspension

insulators.

AB

can

readily

be seen, the

equation

of s ta t ic

equil i -

brium fo r th is

conductor

should be

aos =

I k

Va

2

h

k

2

where tl and 1 k-t l are

the

angles, the oond.uctors of

spans

k and k+ 1 respectively, make with the

horizontal

a t the suspension insula tors .

4 )

Ibid., p 1 1



Also

in

expressing the

equi libr ium equat ion in

terms

of sag, an

incorrect expression

was

used.

Noda and

Nishiyama used the ~ u t i o n

where

d is the maximum sag. while the correct eQ.uation is

. l6

= L

w

S r

where

is

the

horizontal

component

of

the

maximum

con-

ductor

tensions.

Because of the

mistakes and incorrect

s s u n ~ t i o n s

made by Noda and ~ i s h i y m in the i r paper, the result ing

s olu tion th at

t hey ob tained does

not

fa l l with in eng i-

neering accuracy; and,

therefore, is

of no practical

engineering value.

5 Noda and Nishiyama

op c i t

p.

105.

l6 ) Brown Jr. L., Engineering

mechanics John

filley

~ Sons,

196, 194 1.



C TEN RY

C LE

EUCT

ANALYSIS

There

are

several

methods y which a solution of

this

problem may be obtained.. The solution

that

would

give

the

most accurate resul ts is an exact

analysis

with

the

assump

t io n th at the center

l ine

a tf the eond ucto r

i s

a catenary

curve.

Therefore attemptiIlg to

arrive

t

a solution of

the

problem,

the

f1rS t method should

and wil l

be

the one

that

wil l

give the

most

aceura.te

results

k

k 1

k

Jr f

f

L

Figa.re

2. Sag and insulator swing

in

spans k-l k , and

k+l

T { f

Y1gure 3.

Foree diagram of

k insula

to r

From the force diagram of k insulator ,

i t

i s possible

to obtain a

relationship between

the conductor

tensions and



l l

9

the

horizontal

swing of

the

suspension

insulator by

taking

the

summation of moments of the

forces

about pQint o

Therefore.

= Wsbk

T

k

1 CoSP

k

1

a

2

- b

k

2

j

t

T

k

T 1

2

Sin J k-t- 1

bk} -

T

k

Cos ¢k

Va

2

- b

k

2

Sin

¢k

b

k

- Q

where

W

s

=

weight

of

the suspension

insulator

a -

length

of

the

suspension

insulator

b

k

- horizontal swing of the suspension insulator

T

k

and Tkt- l =

maximum

eonductor

tensions

in

spans k

and k 1- 1 , respectively

¢k and

¢k

= angle conductors of spans k and k t

respectively

make with the horizontal

a t

the suspension

insulator.

l Tow

Similarly

Sin

~ t

=

2

3

4

5

where Su

-

uns tretched leng th

of the conductor in

one span

w =weight of conductor per

unit

length.



6

J.O

Subs ti .tu ting equa t ions

2 .

4 ;. and

5} in equat ion

J.

and

s implt lyiDg

W s ~ t V r T k ~ t = ~ W S u } 2

V

a

_ ~ 2

.

t

w ub lt

V c

-

lWS

u

V 2 - k

2

T

w S u ~ = a

SoJ.ving e ~ u a t i o n t6 f o r ~

(lk2

- CWSu)2 - V

k

J.2 - WS

u

2J

~

, - - - - - - : : - r ~ ~ = = = ~ - _ r : : : = = : : : : : : : : : = = : ; _ ; 9 - - I ( 7

l ~

T

2wS

l

a

t

[V k

-l_B,,/ -

V

k

i

-l_B

a

I

2

] 2

Kow to b t ~

a

solution 10r i . t

wiJ.J.

be necessa ry

to

t ind

T

k

and

T

k

1 l.

in

terms

of known quanti t ies .

The

equa.t1Qn to r

the

unatretched 1.ength

of

a

conductor

sup

ported

at equal.

elevations i s

L :

span. l.ength

l ~ T

----------1

-----------_.: ]

l8}

Tm =maximum ccmductor tension

beto.re ice load on

span 0;:

which i s

the

ice l .oade d span.

as

shown

on

:page 1.7

E

=

modulus

of

elas t ic i ty

of

condactor

=

eross-sectional.

area. of the

conductor

Now

in

th is probl.em... the conductor i s not sUI>ported

a t equal e l.evations . Irowever. the difference

i n

el.evation

of the

supports

is due

only

to the

swing

of the

suspension



insul tors

Therefore,

the difference

in

the elevation

i s extremely small and

can

be

neglected.

Since the unstretched

length

of the conductor before

and f ter the ice

load

o span is

equal, the following

equation can

be writ ten

fo r

span

lO



12

To

be

abl.e

to arr ive t

a general. sOJ.tn;iOll

01

1Jhe

problem by th i s

method,

i s necessary to

solve

eOl1stioHs

9 )

anQ 10)

Tor

Tk anQ Tk respectivel .y.

This

is

impossibl.e, in a pr t i l

form aHa therefore,

a mathem2tically

or re t

solu tian 1

th is :orOb Lem i ll a orm SUiGaDJ.e t·or 1 rC? c-

t i l

cal .culations

cannot

be

oDtainea.

i1

the ceuter

li11e

of

the conductor

is

assumed

to be a catenary

curve.



13

PARABOLIC CABLE

EX T

A1\jALYSIS

Since

1 t was

1mpossib.le

to

a r r ive a t

a ma them8 GiceJ..ly

c o r re c t

so lu t ion

of

the prOb.lem witih resuJ. ts iLl a pract ica l

orm

by

assumiLlg the ceLl ter . liLle 1

the

couO.uc tor

to

be a

ca tenary an attemp t l' i.l.l be mane o

ar r ive

a t a sO.lutiiou by

assuming the

cen te r

l i ne

1

the cono.uc Gor to be a

parabO.la.

~ ~ t ~ H k t l

~

v

Figure

4

Force Qiagram of insu. la tor Showil1g horizo l ta.l

ana. v e r t i c a l componen Gs of

the

maximum

conQuctor tensious.

From Figure 4, i s 1)OSSible to obtain

re.lB.tioIlShlp

between the hor izon ta l componen ts

of lihe

maximum COllQUC tor

tens ions

and

the hor i ZOrl ta.l swing 1 the 8uspensioLl iIlSU.l-

a t o r by tak ing a summation 1 moments of the 'orces about

o



Therefore,

lIvL

o

=

Wsb

k

lH

k t

- H

k

a

-

b

k

2

Vb

k

::

a

2

where = V

k

Vk-tl

= to ta l

ver t ica l component of

the

maximum conductor

tensions.

14

l l l }

an d Hk l - horizontal components of

the

maximum

conductor

tensions

in spans k

and

kt-l, r e s p e c t i v e l y .

Since the span length of a l l

spans

is

e ~ u l

th e

ver t ica l

components

of

the

maximum

conductor

tensions

are e 9.ual.

Theref

or e

t

[ =

1[k

t V k t l =

wS

u

SUbstituting equation l 3 j in

e ~ t l t i o n

t I l}

, I}

2 2

W

s

b

k

t

l:B:kt-l - l kl

a -

b

k

WSU

k

=Q

:

Solving

equation

l14)

fo r

b

k

2

13)

l14

l15 j

Now

to obtain

a

s o l u t i o n

fo r

b

k

i t

is

necessary

to

f i n d Hk and H

k t

i n terms of known q u a n t i t i e s .

Since

th e

unstretched l e n g t h of the cond uctor s

for a l l

spans i s not

only equ al but also known an equation

for

i k and Bk tl in

terms

of th e u n st re tc h ed l en g th can

be o b t i n ~ d



15

In general

d

=

wL

2

mr

2

~

_

3 L

l16

}

17 }

where both equations 16

and \17

J are

Ior

conductors

supported at equal elevations. however, as mentioned

before,

the

elevation of the s u ~ r t s

is

not

equal

but

wil l

be

t reated

as

such

as

this

difference

in

elevation

i s

extremely

small.

SUbstituting equation

15

in

l16}

and. simplifying

l IS

Solving equation

l18J

for

t EAl u - 1 liZ

r

ow

:t

or spa.n k

II.

_ W

2

L

2

EA = 0

24

l IS

Similarly fo r

E pan

k

T

1

20 }

Now i t

i s

necessary to solve equations

20

and

1

hr 2_

w

2

L

T

b

k

b

k

1 }2

n

k l

.

=

0

24 21

t2 l

fo r l\: and R

k

l respectively.

This

can

be done

but will

yield

suCh an u n w l ~ equation that

i t is impossible to

l7}

Noda and

Nishiyama.

op.ci t . 1 5

t8l

Ib id .



16

simplify

enough

so

as

to

be

able

t

con tiulJ.e

with

a

pr c t i c l

Bolu-GioB of the

prOblem. Thererore

,

i t is also

impract ica l to

rr ive

t

a

mathematically

correc t

sollJ.tion

or this uroblem in a form slJ.itabLe Tor ur ct ic l

calcu12tiolls

by

assuming

the

cen ter l ine

o t the COilCluctor

to

be a

pctra-

bol ic

curve.

Since i t i s im r c t ic l to

rr ive e.t a

mathematically

cor rec t

sollJ.tion

of

th is problem assuming

the

cerl ter l ine

of the COIlo.uctor

to

be

e i ther

a

catsIlary or

a

parabola

the next step i s

to

t ry to derive

a methoQ

of

calculat ing

the

maximum

sag th t wil l give

an

approximate solut ion.



17

CATENARY

CABLE

APPROXIMATE ANALYSIS

s-

l

6

Figure ·

sag

and

inSula tor

swing

in

c r i t i ca l

spans.

t i ~ t attempt to arrive

a t

a . olution

of

the

problem

that

gives approximate r es ults w ill be

made

by

S8tl I I l ng the center l ine of the

conductor

to be a cate-

n

attempt.ing

to

arrive

at a

solution

of

this

problem

tha t gives

approximate

results several asaumpt ions and

simplif icat ions

wil l have to

be

made. One

fact

that

i s

known and has been

shown

in pract ica l

tes ts

i s that the

:fifth

insula.tor from

the

ice loaded. span has an extremely

9

sme]]

swing. Therefore i s possible to

assume the

swing

of the

f i f th

insula tor from the ice loaded span to

be zero Without material ly ~ t i n g the resu l ts

2 3 4

2

J

Figu re6 Force

d 1a gra m

o 1nsu.lator 5.



2S)

18

Since it

i s assumed t ha t

there i s no swing

of suspen-

s ion

i n su l a to r

S,

the

maximum conductor

t ens ion in

span 6

a f t e r

ic e

load

i n span 0 i s equal to the tens ion before

th e

ic e

load in

span

From

Figure 6 , the follow ing equation can be obtained

by

tak ing

the summation of the forces along

the

horizonta l

2F

x

• T

6

Cos ¢6

-

T

S

Cos ¢S =0 . 22)

where

T

S

and

T

6

•

maximum

conductor

t eps ion

in

spans Sand

6,

respect ive ly

¢S and

¢6 =a ng le c onducto rs

of spans Sand 6,

respec-

t i ve ly make w ith the hor izonta l a t the

suspension i n su l a to r s

Cos ¢5

=

VS

2

- wS

u

)2 23)

TS

Cos

¢

V

6

2

-

wSu)2

24)

T

SUbst i tu t ing equat ions 23) and

24)

in to equation 22)

and s impl i fy ing

wSu)2 - YTs

2

- wSu)2 = 0

o r ;;62 _ wSu)2 =

V

_ wSu)2 26)

Squaring

both

s ides of

equat ion

26)

. T6

2

- wSu. 2 :: T

s

2

- WSu)2

27)

Since the u ns tre tc he d le ng th s o f

the

conductors

in

a l l

spans are

equal

28



19

Now

the unst;re tched l eng th

01 the COuQuc:t:;or

il l spall

5

bel 'ore alia. a ' ter the

ice

10aet on Spa.i:l 0 i s nOli ouly

the

same but; i s a.lSo a kn01'1 Ll qU8.LftiliY. There1'ore

Su = (L b4)

[JT(\;O >4l]

2

[

241 [W(L

+b4) +

- - - - - - - - - - - - - -

44,512 T5

w(L b4 2 . 2T5 _ l

2EA

w(L

b4)

12

(29)

where b4 i s the

horizon tal

swing 01

suspension

in su la to r

4.

To obtai l l a solu t ion 01' t h i s

pro

Olem

by

as

suming

'the

cen t e r l i ne

o f

the o n u ~ o r to be a parabola ,

i s neces-

sa ry

to

so lve

equet ion

(29)

fo r

b4- Since

t h i s i s impossible

in a form su i tab le fo r

pre .c t ica l

ca lcula t ions a prac t ica l

s o lu t i on

01'

th e

problem

caIlIlot be

obta ined

by

t h i s

methOd.



20

PARABOLIC CABLE

APPROXINillTE

ANALYSIS

Another possible

way

to arr iv e a t

a solution of

the

problem i s

by

assuming

the

center l ine of the conductor

to be

a

parabol ic curve.

Previously was proven

tha t

the maximum conductor

tenaion

in spans

5 and 6 i s

equal on

the assumption

that

the horizontal swing or

insulator

5

i s

zero. Therefore

the ang les ¢ and ¢6

are

equal and

the

horizontal com-

ponents

or

the

maximum

conductor

tensions

in

these

spans

are eq ua l.

Figure

30

V=

c

Force diagram

or suspension insulator

4

showing the hor izonta l and ver t ica l com-

ponents or the

maximum

em ductor tensions.

Since HS

i s

a known

quanti ty i s

po ssible to obtain

an equation

ro r

H rrom

Figure

7 by taking

the

summation

or moments of the forces about

the

point

4.

M 4 HsVa2-b42t-wSub4t-Ws

b4

H 4 ~ _ b 4 0

or

wSu

b

4

31



Similarly

from

the foree

diagrams

of

suspension

insulators

3

2.

and

respectively

o -

f

4

t

wS

u

T

sl

r

b

3

Yaz - b3

2

wS

u

sf

=

r

be

Va

Z

_ bz

2

wSu t

W

s

-

zT

-

2

b

l

Va

Z

_ bIZ

where

32}

o3}

l34J

b4 ba

bZ

and

bl =

horizontal

swing of suspension

insulators

4 3

2

and

1

respectively.

R

4

z

nd

1

-

horizontal component of the

maximum

conductor tension

in

spans 4 3 2

and 1 respec-

t ively.

w

=

Su

-

a

_

W

s

-

weight

per

foot

of the conductor

unst re tched length of the conductor

length of

the

suspension insulators

weight of

the suspension insulators



22

b

v w; AI

Figure 8.

Force

diagram

of suspension insulator

0

showing the

horizontal and

vert ical components

of

the maximum conductor tensions.

An equation fo r o can be obtained by

taking

the

summation of moments of the forces about point O

or

where

s

{w

w}

Su

: ; : : : : : = = 2 ~ ; ;= _

b

0

Va

2

_ b

o

2

6

w

=

-

-

horizontal swing

of

suspension

insulator

a

horizontal component of the maximum conductor

tension in span

0

load per unit length

on

the conductor

of span

0

includes

the weight

of

the conductor

and

the

weight of

the

ice load.

The unst re tched length of span 5 before and f t r

the

ice load on span

0

i s

the

same.. Therefore

371



23

I t was previously proven that

E::

5

= Jr . Therefore,

L.,..wZL3 _

=

L r

b

4t w2{L-r

b

4)3

_

B: L-tb

4 }

24H

2

XI

24H

2

EX l3B)

MUltiplying

both sides

of

e ~ u a t i o n t38} by ~

~ L T wZIt3u - 241f

3

L =

24lI2u.{L

7 b

4

}

w 2 { L b 4 J ~ A

- 2 ~ L b 4 } 39)

Collecting terms

Z.m2EA.b4 -

w ~ 3b

4

L

2

-t 3b4

b4

3

24JFb4 = a 40)

3

w b

4

is

ext remely small compared to the res t

of

the

b

4

terms,

and

therefore,

can

be

neglected.

Therefore,

Zw4EALb42t [3W

2

EJ.L2 t a ~ E A _ R ) ] b

4

=

a

l41.}

or

b

4

[ aw2EALb4

i

3w2m L

2

-t ZuZ{EA K>l

= 0 \42)

Therefore,

b

a.m2ta; EA}

aw

2

EAL

2

4 :: 3i EAL 43)

or

Again the

unst re tched leng th of span

4

before and

af ter the

ice load on span a

i s

the same. therefore,

L-t

2 . ~

=

Ltb b4T

wZlL-tb

3

b

4

)3

-

~ L + b 3 - b 4

. 24K

4

2

EI 45)

w

lLt b

3

-b

4

}3 =

3 2 2

L 3L {b3-

b

4}+ 3L b3-

b

4} -

3 3

T b

3

b

4

l46 }



4

Since

b

3

and

b

4

are both small

as compared to L the

l a s t three terms of eCluation l46J are

negligible

compared

to the

res t of the

terms

in

the equa tion. Because

of this

fac t ,

the

l a s t

three terms of equation l46}

wil l be

ne

glected.

~ h e r e f o r e rewriting equation l46}

L- tb

3

b

4

}3 = L 3 ~ 3 L 2 b 3 - b 4 } 3 L t b 3 - b 4 } 2

47}

Substituting equation

l47 into

equation 45

w

2

L

3

H L

w

2

L

3

w

2

L

2

b w2L

2

b 2 2

24ftZ - f iT h4.-

b

3

-

3 t 4 w L

b

3

8

R4

2

48

Collecting

terms

of equation l48

: ~ 2 b

3

2

r

1

1-+ ~ 2 L

2b

41 - ~ J b

3

- [ I T : ~ ~ 2 L - b 4 _ R 4 l

b

0;0.4

l 4 . EA 4

E j

4

2 2

_ wZL

3

{ f.i 1 L

4

- n (1I4. - II} =0 l49

24R

2

li4

2

The coeff icients of the b

3

and b

4

terms

in equation

49

are pract ical ly the

same

varying

only

by

one minus

b

4

in

the second term of each coefficient. ~ i n e

b

4

i s

extremely small when compared with

L, the

two o e f f i i ~ n t s

are

pract ical ly

the

same

and can

be

written

as fo llows:

L5



25

There

is

very l i t t l difference between the

small

Quantities

b3

and

b4 •

Therefore,

b3 - b4

is

extremely

small. Also, the coefficient of

t h i s

term in equation L50

i s

r e l a t i v e l y

small.

fo

f u r t h e r simplify t h i s equation,

the 03

b

4

terms w i l l be neglected, an d equation LBOJ

can

be

wr i t t e n

as follow s:

Dividing equation l51}

b 2

3

=L

2

:EI

4

2

-

2

3R

2

by

W

2

L

~ ; ; ; ;

and

simplifying

H

4

2

_

8II

4

2

Ut

4

X}

wZEA

{52}

or

L

2

{:Ei

4

2

-

2

_

SH

4

2

lH

4

- H

Z

wen

53

Similarly,

i t i s

possible to obtain b2 b l , an d b

o

by

s e t t i n g

the u n st re tc h ed l en g th of the conductors before the

ic e

load

on span Q equal

to

the u n st re tc h ed l e ng t h of the

conductors f t r the ice

load on

span

0

fo r spans 3,

an d

1 ,

r e s p e c t i v e l y .

t h e r e f o r e ,

LEeR

3

2 -

HZ

_

H

3

2 H

3

-

fl }

3liZ w

2

EA

l54}

L2 liz

2

_ liZ}

i

8HZ {HZ H}

wZE

55)



56

To

be able to calcu la te the

maximum

sag in any span

of a power

transmission l ine is necessary to

know

the

horizontal component of the maximum conductor tension in

tha t span

and

the

effec t ive span length. The same

is true

when attempting to c alc ula te

the

maximum sag in

the

only

ice

loaded span

of

a

power

transmission

l i ne

Therefore,

to be able to calculate the maximum sag

in

the only ice

loaded span,

i s

necessary

to calculate not only H

o

but

also b

o

• Since i s impossible

to

make these calculations

by subs t i tu t ing

known

values

in one

equation,

a series of

calculat ions has to be made

to determine

H

o

and b

o

• The

method

to

use in calcula t ing

these

values is to calculate

b

4

by using equation 44 fo r

which

a l l terms are known

Then

calcula te H

4

from equation 31 . Using th is

value

for

B4

calculate b

3

by

using equation 53 . After a l te r -

nat el y c al cu la ti ng the

horizontal swing

of

the suspension

insu la tors

and

the

horizontal component

of

the maximum

conductor

tension

for

spans 4, 3,

2,

1,

and

0,

the

maximQm

sag in

span

0, the ice loaded span, can be calculated by

subst i tu t ing

in the

fol lowing equat ion:

2

dO

=

W L

2b

>

SH

o

57



27

In order to determine whether or

not

the above

approximate

solut ion

gives

sat isfactory

resul ts a

cal

cula t ion

using reasonable values i s made.

Therefore,

assume

the

following

representa t ive values for

an

electr ic power

l ine

a = 2.5 fee t • length of suspension insulators

L •

700

fee t • span length

d =20

fee t

•

sag of

a l l spans

before

an

ice

load on

span

0

A =0.166 sq. in =cross-sect ional

area

of conductor

E •

16

x 10

6

psi modulus

of

elas t ic i ty of

conductor

w • 0.641

lbs

per f t weight

of

the conductor

per

foot .

W • 1 3 lbs

per f t

= load per foot on

ice

loaded

span. Includes the

weight

of

the

conductor

and

one-half

iNs

=

60 Ibs

2

H a H

=

: -

inch of ice

weight of the suspension insulators lO

=1962 Ibs .

10

b4 •

8

1962 3

0.641 2 16xlO

O

0.166 700

- 8 1962 2 - 700

0.641 2

700

Knowlton

A.E., standard Handbook for Electr ical

n g i n e e r ~

7th

Ed. , N.Y., MCGraw-Hill, 1941,

PI>:.

1160-61.



b

4

= 106,208 fee t

Of

course,

th is re su lt

i s

impossible.

The

only

other

ob

ta inable r esu l t for b

4

i s zero as obtained

from

equation

42 . Since th i s also i s

not

a

correct

resu l t the above

derived

method, wil l not give a reasonable

solution

of the

problem.

The above

resul ts indicate

tha t

the assumptions

and approx ima tions

made

in the above analysis

in

order to

make

possible

a solut ion in a form that can be used for

pract ica l

calculat ions

lead to

impractical

resul ts

Simplifying

assumptions were

previously made on

equa-

t ions

40 , 45 , and 50 . These assumptions tend to give

impossible r esu l t s

I f

these assumptions are not made

an

approximate

solut ion

of

the problem may be obtained. How

ever , wil l not be a prac t ica l solution

as

i t involves

the

solu t ion

of

cubic

equations.

The horizonta l compm

en t

of

the conductor

tensions in

spans

5, 4

3, 2,

and

0

can be calculated

by

using pre-

viously derived equat ions 31 , 32 ,

33 ,

34 , md 36

as no simplif icat ions were made on these

equations.

To

determine

the

horizontal swing of suspension insula-

to r 4 i s possible

to

use

equation

40 without making

any

simplif icat ions

• Therefore,

H ~ b w 2 E A 3 b 4 L 2 + 3 b 4 2 L + b 4 3 ) 2 4 H ~ 4

0

58

Collect ing terms

w

2

EAb

4

3

t 3w

2

LEAb

4

2

3w

2

L

2

EA

4H EA 4H

3

b

4

• 0 59



29

Dividing

by w

2

and simplifying

b

4

3

T

3Lb

4

2

t

[3L

2

24H2 1

. L)]b

4

:

60)

or

b

4

b4 T

3Lb

4

[

3L

2

24H

2

1 T

] =

61

Solving equation 61) for b

4

b

4

=

62)

and

63)

Of course b

4

cannot be

zero

and,

therefore,

can

be deter-

mined

from equ atio n 6 3).

The

horizonta l

svnng of

suspension insula tor

3 can

be

det ermined , w ithout any

s implif ica t ions , from equation

45).

Therefore,

Collect ing

terms

65



30

66

67

68

In d e t e r l n i n i n g b 3 b

b

l

and b

O

t i s n e c e s s a r y to

solve c u b i c equa t ions

65) ,

66) , 6 7 ) , a nd

6 8 ) ,

r e s p e c -

t i v e l y .

To be ab le to ca lcu la te

th e

maximum sa g

in th e ice

loaded s p a n o f a t r ansmiss ion

l i n e , t

i s

n e c e s s a r y

to

ca l cu l a t e

the

hor izon ta l component

of

th e c o n d u c t o r t ens ion

an d th e ho r i zon t a l s w i n g

of

the

s u s p e n s i o n insu la

to r in th2.t

s p a n .

t i s imposs ib le

to

make these

ca l cu l a t i ons

d i r e c t l y ,

an d t he r e fo re , t i s

necessa ry to make

a se r i e s of

ca lcu-

l a t i ons by a l t e r n a t e l y

ca l cu l a t i ng

th e hor i zon ta l swi ng o f



31

th e su sp en sio n i nsu la tors and the horizontal components

or

the

conductor

ten sion s fo r

spans

5, 4,

3,

2,

1,

and

After

determining b

o

and

H

o

the maximum sag

in

the

ice

loaded span can be determined by sUbstituting these values

in to equation

57 .

In order to determine

whether

or not the above approx-

imate solut ion

gives

sa t i s fac tory

resu l t s ,

a

calculat ion

using

r ea sonabl e v alu es

fo r the l ine constants i s

made.

Thererore,

assume the representat ive values

l i s ted

on

page

27.

700 ] 3

700

2

16xl0

6

0.166

3 700

2

b

4

=

13 ,972 i t

Of course, th i s resu l t i s impossible . The only other

obtainable

re su l t fo r b

4

i s zero

as

obtained from equation

62 .

Since t h i s a lso i s

not

a correct

resu l t ,

the above

derived

method,

wil l

not

give

a

reasonable

solution

of

the

problem. The

above

resu l t s indicate that the

assumption

tha t

the

swing of the f i f t h

insulator from the

ice loaded

span i s zero leads to impract ical resu l t s .



32

CONCLUSIONS

In

enueavoring

to

determine

a

simple method

for

cal

culating the maximum sag of a transmission l ine

suspended

on suspension insulators, with an

ice

load on only one

span,

the

center

l ine of the

conductor

was assmued to be a cate-

nary

and

also a

parabola.

With both

of

the

aforementioned

assumptions,

i t was impossible to arrive

at either

a

mathe-

matically

correct

solution or

an

approximate

solution

suit

able

fo r pract ical

calculations.

The reason that i t was impossible to obtain either

a mathematically cor rect solu tion or n approximate

solution

by assuming the center l ine

of

the conductor to be a cate-

nary

i s

that

the equation

fo r the tLi1stretched length of

the conductor

is a

transcendental

equation of

the type

Y

=

osh

X

CX

T ] Sinh E

Y

where

A B 0,

D and

E

are constants. In

this

problem,

i t was necessary to

obtain

a general solution

of the t r a n s ~

cendental

equation for

X.

Since

this

is

impractical,

there

can be no pract ical solution

of

the problem when the

center

l ine

of

th e conductor i s assumed to be a catenary.

t

was

also

impractical

to

obtain an exact solution

of

the problem by assuming the center l ine of

the

conductor

to be a

parabola.

The

reason for this

was that

the solu

t ions

of

the cubin equations

20}

and 21 fo r and ~ l



33

respectively were extremely

complicated

Since the

un-

w l ~

equations

for

R

and

could not

be

simplified

k k-t-l

enough

so

as to continue with the solution

of

the problem

i t was impractical to obtain a

mathematically

correct

solu-

t ion

of the problem by assuming the

center l ine

of the

conductor

to be a

parabola

Since no mathematically correct solution of the

prob-

lem

was

possible the only

other

alternative

was to

attempt

to derive a method of

determining the

maximum sag

that

gives approximate resul ts To do th is the

horizontal

swing of the f i f th insulator from the ice loaded span was

assumed to be z ro s preViously discussed i t was found

impossible

to

determine

an approximate solution of the

problem by assuming the center l ine of t he conduc tor to be

a

catenary

When the center l ine

of

the conductor was assumed

to

be a parabola a solution was

obtained based

upon certain

assumptions previously specif ied However upon assuming

representative

values for the l ine constants

i t

was found

that

a reasonable solution of the problem was

not

obtained

ome

of the

reasons

for

not

obtaining

reasonable

results

are that the

l s t

three t ~ s

of

equation

l46}

and the

{b

3

- b

4

} term of equation 50} are not negligible as

assumed This must be true even

though

these terms

are

small

as compared

to the other terms

of these equat ions



34

f

these terms are not neglected then

the

solution

of

the

problem is

so

unwieldy as

to

be

of

no practical value

Therefore since

i t

i s

impracti ca l to arrive

at

a

solution of the problem

by

the

previous methods

i t

can

be

said th at there i s no simple solution

of

the problem i f

the

el s t ic i ty of

the

conductor is

to

be

taken into account

then

the f ina l resul t

will

not be

ei ther m t h e m t i l ~

correct

or

even

an

approximate

solution

and

therefore

will

have

no pract ica l

value

he

results of

this

investigation

show that by the

methods used in th is thesis

a solution in a form

suitable

for

engineering calculations is

not possible similar

s olu tio n t o this problem

has

not

been

previously developed

in any

k n o v ~

l i ter ture

on the

subject Thus the results

of this thesis while in a negative form y be

of

value

in

further

consideration of the

subject

by other forms of

analysis



SU v M RY

In

attempting

to

derive

pr c t ic l

method

of calcu

Ilat ing

the m ximum sag of power transmission l ine with

ice load

on

one

span

w s proven

that

i s

impossi-

ble

to

obtain mathematically

correct

solut ion in

form

sui t ble for

pr c t ic l

calculations assuming the center l ine

of the conductor to be e i ther catenary or parabola.

Also

i s

impossible

to

rr ive t

pr c t ic l

solution

of

the

problem

th t w il l give

approximate resul t s by assuming

the

center l ine of the

conductor to be

ei ther

catenary

or parabola and making cer t in assumptions which m de

possible form

of the

equations th t could e used for

pr c t i c l c lcul t ions



36

BIBLIOGRAP1rY

1.. Brown .. J 'l.

L.

Engineering

Mechanics.

2nd

Ed.

N. Y.

1942.

PI .

194 205 .

2.

Greisser,

V H. Effects of

Ice

Loading on Transmission

Lines.

A.I.E.E.

Trans.,

Vol.

32,

pp .

lS29-1844. p t 2.

1913.

3. Healy, E. S .,

and

Wright, A.

J Unbalanced

Conductor

Tensions. Trans.

A.I.E.E.,

Vol.

45, 1'p.

1064-107 0

l1926)

4:. Knowlton, A.

E. Standard

liandbook fo r

E l ect r i cal

Engineers.

7th

Ed . N. Y., MCGraW-Hill,

1941.

]; p . 1160-1.161..

5 .

Martin,

J sa g Calculations The

Use

of rJArtin s

Tables.

Co];)perweld

S t e e l

Co.

l1.931}.

Nash,

J

F

and

Nash. F

... J r

Sag an d

Tensian

Calculations For

Cable

an d Wire Slland Using Catenary

] ormulas. E l e c t r i o a l lmgineering, 1[01. 64 p p. 685

692

1945}.

7 . Noda, S .,

and

Nishiyama, S. Mechanical

Characteristios

of Transmission Lines. RyoJun

College

of Engineering

Memoirs. Vol. 8 llll-

105-138

l193Sh



V T

Rodney Arthur Schaefer was born in t Charles

~ s s o u r i

on July

20

1926 Elementary and

secondary

schooling

was

received

in

t

Charles and a Bachelor

of Science in E l ec tr ic al Engineering was obtained in

~ u n e 1947

trom

the

Missouri

School of

~ l n s and

~ e t a l l u r g y

University

of

Missouri

at

Rolla

~ s s o u r 1

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