Sag and Tension Calculation

8/18/2019 Sag and Tension Calculation

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1) Conductors suspended between two equal support A and B

L = Length of span (m)

W = Weight per unit length of conductor (kg)

T = Tension in the conductor (kg)

Let the cordinatioin of P are x, y and assuming that at small length of OP = x

a) The weight (w, kg) wx of conductor acting at a distance x!" from O

#) The working tension T (k$) acting at point O

%oment&s e'uation of a#oe two forces a#out point O, we get

*t support *

x = l!" and y = + (k$ = -./0 kgf)

+ag,

L = .- m "0- m 1-

W = 00- kg!km 02- kg!km --3-0

ultimate strenght = 421 k$ 4-- kg 421

safety factor = 3 "

T = "30 k$ 33- kg "30

"0-.32 kg 3" k$

+ = -"0 m 4/ m 2

The maximum dip (sag) is represented #y the alue of y at either of support * and 5

"

xT y w x⋅ = ⋅ ×

"

"

w x y

T

⋅=

" "( !")

" 2

w l w l S

T T

⋅= =

( )

Ultimated Strength kN T

Saftety factor =


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m

kg!km

kg

kg

k$

m


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2) Conductors suspended between two are not equal support A and B

L = Length of span (m)

W = Weight per unit length of conductor (kg)

T = Tension in the conductor (kg)

h = different in leels #etween two supports

x = 6istance of support at higher leel

x" = 6istance of support at lower leel

Let the cordinatioin of P are x, y and assuming that at small length of OP = x

a) The weight (w, kg) wx of conductor acting at a distance x!" from O

#) The working tension T (k$) acting at point O

from the preious %oment&s e'uation, we get

k$ = -./0 kgf

x 7 x" = l kg = ---!.2 k$

+ag at x,

3) Effect of wind and ice loading

"

"

w x y

T

⋅=

"

,,

2

w xS

T

⋅=

(per u

Ultimated StrengthT

Saftety fa=

"

"

"

2

w xS

T

⋅

=

" "

" , " , " , " ,

" , " ,

" ,

( )( )" "

( )"

"( )

w wS S x x x x x xT T

w l S S h x x

T

T h x x

w l

− = − = − +

⋅− = = −

⋅− =

⋅

,"

l T h x

w l

⋅= −

⋅ " "

l T h x

w l

⋅= +

⋅


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#y neglecting the ice loading, then total weight of conductor per unit length is

8eritcal +ag is defined #y

" "

t ww w w= +

"

2

t w l

S T

⋅

=

S V S Cosθ = ⋅

tan ww

w

θ =


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9xample * distri#ution line of "" k8 with 4 wire of

;onductor 3- mm: weight (kg!m) =

pole span x(m) = 0- x" (m) =

*llowa#le sag for oerhead line +(m) =

Wind elocity (m!s) 4"

diameter of conductor (mm) "-

wind pressure (Pa) = 02-

Weight of conductors #y wind pressure

The maximum tension of conductors is

if the two conductor is alliagned #y T is - deg from T", t

P = T" < Tcos4- = - $

it) conductor area

ctor

×


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3- mm: conductor, has the following data

-00

0-

"- +" (m) = "- max sag = 44

Wt = 1 $

T = .,"30 $

T" = .,"30 $

hen -


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Design Data

Component Wind pressure (Pa)

Cylindrical concrete pole 425

Rectangular concrete pole 900

Square 800

Narrow face to wind 425

Wide face to wind 800

Crossars 800

!nsulators "00

Conductor "80

#ransforer and Siilar ites 900

conductor data $alue

Cross sectional area %&' 240 (85

)$erall outside diater %' 20 (*+5*

Stranding %No,' (9,4 (9,-+5(

Weig.t %/g,/' ""0 508+5

ltiate tensile streng.t %/N' -8+4 29+9


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(50 *0


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Design Condition:

1llowale sag for o$er.ead line 3 5+0

Wind pressure 3 "80 a

Wind $elocity 3 -2 ,s

Cross6sectional area %&' 240

)$erall conductor diaeter %'

ltiate tensile streng.t %7rea/ing load' %/N'

Weig.t of conductor %/g,/'

ole Span #ension on ole deg 45

S( %' S2 %' #( %/N' #2 %/N' %/N'

40 40 0+(98 0+(98 0+0580

45 45 0+22- 0+22- 0+0"52

50 50 0+248 0+248 0+0*25

55 55 0+2*2 0+2*2 0+0*9*

"0 "0 0+29* 0+29* 0+08*0

"5 "5 0+-22 0+-22 0+0942

*0 *0 0+-4* 0+-4* 0+(0(5

*5 *5 0+-*( 0+-*( 0+(08*

80 80 0+-9" 0+-9" 0+(("0

85 85 0+42( 0+42( 0+(2-2

90 90 0+44" 0+44" 0+(-05

95 95 0+4*0 0+4*0 0+(-**

(00 (00 0+495 0+495 0+(450

(05 (05 0+520 0+520 0+(522

((0 ((0 0+545 0+545 0+(595

((5 ((5 0+5"9 0+5"9 0+(""*

(20 (20 0+594 0+594 0+(*40

(25 (25 0+"(9 0+"(9 0+(8(2

(-0 (-0 0+"44 0+"44 0+(885

(-5 (-5 0+""8 0+""8 0+(95*

(40 (40 0+"9- 0+"9- 0+20-0

(45 (45 0+*(8 0+*(8 0+2(02

(50 (50 0+*4- 0+*4- 0+2(*5


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(85 (50 *0 -5

20

-8+4

0+""

Sag and Tension Calculation

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