# Chapter 13 : Sag and Tension Analysis¶

## Example 13.1 Page No : 690¶

In [1]:
import math

# GIVEN DATA
c = 1600. ; # Length of conductor in feet
L = 500. ; # span b/w conductors in ft
w1 = 4122. ; # Weight of conductor in lb/mi

# CALCULATIONS
# For case (a)
l = 2 * c *( math.sinh(L/(2*c)) ) ; # Length of conductor in ft umath.sing eq 13.6
l_1 = L * (1 + (L**2)/(24*c**2) ) ; # Length of conductor in ft umath.sing eq 13.8

# For case (b)
d = c*( math.cosh( L/(2*c) ) - 1 ) ; # sag in ft

# For case (c)
w = w1/5280 ; # Weight of conductor in lb/ft . [1 mile = 5280 feet]
T_max = w * (c + d) ; # Max conductor tension in lb
T_min = w * c ; # Min conductor tension in lb

# For case (d)
T = w * (L**2)/(8*d) ; # Appr value of tension in lb umath.sing parabolic method

# DISPLAY RESULTS
print ("EXAMPLE : 13.1 : SOLUTION :-") ;
print " a) Length of conductor using eq 13.6 , l = %.3f ft "%(l) ;
print "  &  Length of conductor using eq 13.8 , l = %.4f ft "%(l_1) ;
print " b) Sag , d = %.1f ft "%(d) ;
print " c) Maximum value of conductor tension using catenary method , T_max = %.1f lb "%(T_max) ;
print "     Minimum value of conductor tension using catenary method , T_min = %.1f lb "%(T_min) ;
print " d) Approximate value of tension using parabolic method , T = %.2f lb "%(T) ;

EXAMPLE : 13.1 : SOLUTION :-
a) Length of conductor using eq 13.6 , l = 502.037 ft
&  Length of conductor using eq 13.8 , l = 502.0345 ft
b) Sag , d = 19.6 ft
c) Maximum value of conductor tension using catenary method , T_max = 1264.4 lb
Minimum value of conductor tension using catenary method , T_min = 1249.1 lb
d) Approximate value of tension using parabolic method , T = 1246.55 lb


## Example 13.2 Page No : 697¶

In [2]:
import math

# GIVEN DATA
L = 500. ; # span b/w conductors in ft
p = 4. ; # Horizontal wind pressure in lb/sq ft
t_i = 0.50 ; # Radial thickness of ice in inches
d_c = 1.093 ; # outside diameter of ACSR conductor in inches
w1 = 5399. ; # weight of conductor in lb/mi
s = 28500. ; # ultimate strength in lb

# CALCULATIONS
# For case (a)
w_i = 1.25 * t_i * (d_c + t_i) ; # Weight of ice in pounds per feet

# For case (b)
w = w1/5280 ; # weight of conductor in lb/ft . [1 mile = 5280 feet]
W_T = w + w_i ; # Total vertical load on conductor in pounds per feet

# For case (c)
P = ( (d_c + 2*t_i)/(12) )*p ; # Horizontal wind force in lb/ft

# For case (d)
w_e = math.sqrt( P**2 + (w + w_i)**2 ) ; # Effective load on conductor in lb/ft

# For case (e)
T = s/2 ;
d = w_e * L**2/(8*T) ; # sag in feet

# For case (f)
d_v = d * W_T/w_e ; # vertical sag in feet

# DISPLAY RESULTS
print ("EXAMPLE :13.2 : SOLUTION :-") ;
print " a) Weight of ice in pounds per feet , w_i = %.4f lb/ft "%(w_i) ;
print " b) Total vertical load on conductor in pounds per feet , W_T = %.4f lb/ft "%(W_T) ;
print " c) Horizontal wind force in pounds per feet , P = %.4f lb/ft "%(P) ;
print " d) Effective load acting in pounds per feet , w_e = %.4f lb/ft "%(w_e) ;
print " e) Sag in feet , d = %.2f ft "%(d) ;
print " f) Vertical Sag in feet  = %.2f ft "%(d_v) ;

EXAMPLE :13.2 : SOLUTION :-
a) Weight of ice in pounds per feet , w_i = 0.9956 lb/ft
b) Total vertical load on conductor in pounds per feet , W_T = 2.0182 lb/ft
c) Horizontal wind force in pounds per feet , P = 0.6977 lb/ft
d) Effective load acting in pounds per feet , w_e = 2.1354 lb/ft
e) Sag in feet , d = 4.68 ft
f) Vertical Sag in feet  = 4.43 ft