# CHAPTER 2.5: MECHANICAL DESIGN OF OVERHEAD LINES¶

## Example 2.5.1, Page number 198¶

In :
#Variable declaration
u = 5758.0       #Ultimate strength(kg)
S = 2.0          #Sag(m)
s = 2.0          #Factor of safety
L = 250.0        #Span length(m)

#Calculation
T = u/s                                #Allowable max tension(kg)
w = S*8.0*T/L**2                       #weight(kg/m)
l = L/2                                #Half span length(m)
half_span = l+(w**2*l**3/(6*T**2))     #Half span length(m)
total_length = 2*half_span             #Total length(m)
weight = w*total_length                #Weight of conductor(kg)

#Result
print('Weight of conductor = %.2f kg' %weight)

Weight of conductor = 184.29 kg


## Example 2.5.2, Page number 198¶

In :
import math

#Variable declaration
L = 250.0             #Span length(m)
h = 10.0              #Difference in height(m)
r = 1.0               #Radius of conductor(cm)
w = 2.5               #Weight of conductor(kg/m)
s = 3.0               #Factor of safety
tensile = 4300.0      #Maximum tensile strength(kg/sq.cm)

#Calculation
W = (w**2+wind**2)**0.5       #Total pressure on conductor(kg/m)
f = tensile/s                 #Permissible stress in conductor(kg/sq.cm)
a = math.pi*r**2              #Area of the conductor(sq.cm)
T = f*a                       #Allowable max tension(kg)
x = (L/2)-(T*h/(L*W))         #Point of maximum sag at the lower support(m)

#Result
print('Point of maximum sag at the lower support, x = %.2f metres' %x)

Point of maximum sag at the lower support, x = 60.05 metres


## Example 2.5.3, Page number 198-199¶

In :
import math

#Variable declaration
a = 2.5             #Cross-sectional area(sq.cm)
L = 250.0           #Span(m)
w_c = 1.8           #Weight of conductor(kg/m)
u = 8000.0          #Ultimate strength(kg/cm^2)
s = 3.0             #Factor of safety

#Calculation
d = (4.0*a/math.pi)**0.5       #Diameter(cm)
T = u*a/s                      #Allowable max tension(kg)
w_w = wind*d/100.0             #Horizontal wind force(kg)
w_r = (w_c**2+w_w**2)**0.5     #Resultant force(kg/m)
S = w_r*L**2/(8*T)             #Slant sag(m)
vertical_sag = S*(w_c/w_r)     #Vertical sag(m)

#Result
print('Vertical sag = %.3f metres' %vertical_sag)

Vertical sag = 2.109 metres


## Example 2.5.4, Page number 199¶

In :
#Variable declaration
a = 110.0               #Cross-sectional area(sq.mm)
w_c = 844.0/1000        #Weight of conductor(kg/m)
U = 7950.0              #Ultimate strength(kg)
L = 300.0               #Span(m)
s = 2.0                 #Factor of safety
wind = 75.0             #Wind pressure(kg/m^2)
h = 7.0                 #Ground clearance(m)
d = 2.79                #Diameter of copper(mm)
n = 7.0                 #Number of strands

#Calculation
dia = n*d                        #Diameter of conductor(mm)
w_w = wind*dia/1000.0            #Horizontal wind force(kg)
w = (w_c**2+w_w**2)**0.5         #Resultant force(kg)
T = U/2.0                        #Allowable tension(m)
l = L/2.0                        #Half-span(m)
D = w*l**2/(2*T)                 #Distance(m)
height = h+D                     #Height above ground at which the conductors should be supported(m)

#Result
print('Height above ground at which the conductors should be supported = %.2f metres' %height)

Height above ground at which the conductors should be supported = 11.78 metres


## Example 2.5.5, Page number 199¶

In :
#Variable declaration
w_w = 1.781          #Wind pressure on conductor(kg/m)
w_i = 1.08           #Weight of ice on conductor(kg/m)
D = 6.0              #Maximum permissible sag(m)
s = 2.0              #Factor of safety
w_c = 0.844          #Weight of conductor(kg/m)
u = 7950.0           #Ultimate strength(kg)

#Calculation
w = ((w_c+w_i)**2+w_w**2)**0.5       #Total force on conductor(kg/m)
T = u/s                              #Allowable maximum tension(kg)
l = ((D*2*T)/w)**0.5                 #Half span(m)
L = 2.0*l                            #Permissible span between two supports(m)

#Result
print('Permissible span between two supports = %.f metres' %L)
print('\nNOTE: ERROR: Horizontal wind load, w_w = 1.781 kg/m, not 1.78 kg/m as mentioned in problem statement')

Permissible span between two supports = 270 metres

NOTE: ERROR: Horizontal wind load, w_w = 1.781 kg/m, not 1.78 kg/m as mentioned in problem statement


## Example 2.5.6, Page number 199-200¶

In :
import math

#Variable declaration
a = 0.484           #Area of conductor(sq.cm)
d = 0.889           #Overall diameter(cm)
w_c = 428/1000.0    #Weight(kg/m)
u = 1973.0          #Breaking strength(kg)
s = 2.0             #Factor of safety
L = 200.0           #Span(m)
t = 1.0             #Ice thickness(cm)
wind = 39.0         #Wind pressure(kg/m^2)

#Calculation
#Case(i)
l = L/2.0                                         #Half span(m)
T = u/s                                           #Allowable maximum tension(kg)
D_1 = w_c*l**2/(2*T)                              #Maximum sag due to weight of conductor(m)
#Case(ii)
w_i = 913.5*math.pi*t*(d+t)*10**-4                #Weight of ice on conductor(kg/m)
w = w_c+w_i                                       #Total weight of conductor & ice(kg/m)
D_2 = w*l**2/(2*T)                                #Maximum sag due to additional weight of ice(m)
#Case(iii)
D = d+2.0*t                                       #Diameter due to ice(cm)
w_w = wind*D*10**-2                               #Wind pressure on conductor(kg/m)
w_3 = ((w_c+w_i)**2+w_w**2)**0.5                  #Total force on conductor(kg/m)
D_3 = w_3*l**2/(2*T)                              #Maximum sag due to (i), (ii) & wind(m)
theta = math.atan(w_w/(w_c+w_i))*180.0/math.pi    #θ(°)
vertical_sag = D_3*math.cos(theta*math.pi/180)    #Vertical sag(m)

#Result
print('Case(i)  : Maximum sag of line due to weight of conductor, D = %.2f metres' %D_1)
print('Case(ii) : Maximum sag of line due to additional weight of ice, D = %.2f metres' %D_2)
print('Case(iii): Maximum sag of line due to (i),(ii) plus wind, D = %.2f metres' %D_3)
print('           Vertical sag = %.2f metres' %vertical_sag)

Case(i)  : Maximum sag of line due to weight of conductor, D = 2.17 metres
Case(ii) : Maximum sag of line due to additional weight of ice, D = 4.92 metres
Case(iii): Maximum sag of line due to (i),(ii) plus wind, D = 7.54 metres
Vertical sag = 4.92 metres


## Example 2.5.7, Page number 200¶

In :
#Variable declaration
W = 428/1000.0      #Weight(kg/m)
u = 1973.0          #Breaking strength(kg)
s = 2.0             #Factor of safety
l = 200.0           #Span(m)
h = 3.0             #Difference in tower height(m)

#Calculation
T = u/s                               #Allowable maximum tension(kg)
x_2 = (l/2.0)+(T*h/(W*l))             #Point of minimum sag from tower at higher level(m)
x_1 = l-x_2                           #Point of minimum sag from tower at lower level(m)

#Result
print('Point of minimum sag, x_1 = %.1f metres' %x_1)
print('Point of minimum sag, x_2 = %.1f metres' %x_2)

Point of minimum sag, x_1 = 65.4 metres
Point of minimum sag, x_2 = 134.6 metres


## Example 2.5.8, Page number 200-201¶

In :
#Variable declaration
h_1 = 50.0          #Height of tower P1(m)
h_2 = 80.0          #Height of tower P2(m)
L = 300.0           #Horizontal distance b/w towers(m)
T = 2000.0          #Tension in conductor(kg)
w = 0.844           #Weight of conductor(kg/m)

#Calculation
h = h_2-h_1                           #Difference in height of tower(m)
x_2 = (L/2.0)+(T*h/(w*L))             #Point of minimum sag from tower P2(m)
x_1 = (L/2.0)-(T*h/(w*L))             #Point of minimum sag from tower at lower level(m)
P = (L/2.0)-x_1                       #Distance of point P(m)
D = w*P**2/(2*T)                      #Height of P above O(m)
D_2 = w*x_2**2/(2*T)                  #Height of P2 above O(m)
mid_point_P2 = D_2-D                  #Mid-point below P2(m)
clearance = h_2-mid_point_P2          #Clearance b/w conductor & water(m)
D_1 = w*x_1**2/(2*T)                  #Height of P1 above O(m)
mid_point_P1 = D-D_1                  #Mid-point above P1(m)
clearance_alt = h_1+mid_point_P1      #Clearance b/w conductor & water(m)

#Result
print('Clearance b/w conductor & water at a point midway b/w towers = %.2f m above ground' %clearance)
print('\nALTERNATIVE METHOD:')
print('Clearance b/w conductor & water at a point midway b/w towers = %.2f m above ground' %clearance_alt)

Clearance b/w conductor & water at a point midway b/w towers = 60.25 m above ground

ALTERNATIVE METHOD:
Clearance b/w conductor & water at a point midway b/w towers = 60.25 m above ground


## Example 2.5.9, Page number 201¶

In :
import math

#Variable declaration
L = 300.0          #Span(m)
a = 226.0          #Area(mm^2)
d = 19.53/10       #Overall diameter(cm)
w_2 = 0.844        #Weight of conductor(kg/m)
u = 7950.0         #Ultimate strength(kg)
t = 0.95           #Ice thickness(cm)
wind = 39.0        #Wind pressure(kg/m^2)

#Calculation
w_i = 915.0*math.pi*t*(d+t)*10**-4      #Weight of ice on conductor(kg/m)
w_w = wind*(d+2*t)*10**-2               #Wind load of conductor(kg/m)
w_1  = ((w_2+w_i)**2+w_w**2)**0.5       #Total force on conductor(kg/m)
T_2 = 1710.0                            #Tension in conductor(kg). Obtianed directly from textbook
l = L/2.0                               #Half span(m)
sag = w_2*l**2/(2*T_2)                  #Sag at erection(m)

#Result
print('Sag at erection = %.2f metres' %sag)
print('Tension of the line, T_2 = %.f kg' %T_2)

Sag at erection = 5.55 metres
Tension of the line, T_2 = 1710 kg


## Example 2.5.10, Page number 201-202¶

In :
import math

#Variable declaration
L = 250.0          #Span(m)
d = 1.42           #Diameter(cm)
wind = 37.8        #Wind pressure(kg/m^2)
r = 1.25           #Ice thickness(cm)
f_m = 1050.0       #Maximum working stress(kg/sq.cm)

#Calculation
w_i = 913.5*math.pi*r*(d+r)*10**-4      #Weight of ice on conductor(kg/m)
w_w = wind*(d+2*r)*10**-2               #Wind load of conductor(kg/m)
w_r  = ((w+w_i)**2+w_w**2)**0.5         #Resultant pressure(kg/m)
a = math.pi*d**2/4.0                    #Area(cm^2)
T_0 = f_m*a                             #Tension(kg)
S = w_r*L**2/(8*T_0)                    #Total sag(m)
vertical_sag = S*(w+w_i)/w_r            #Vertical component of sag(m)

#Result
print('Case(i) : Sag in inclined direction = %.f m' %S)
print('Case(ii): Sag in vertical direction = %.2f m' %vertical_sag)

Case(i) : Sag in inclined direction = 12 m
Case(ii): Sag in vertical direction = 9.62 m


## Example 2.5.11, Page number 202-203¶

In :
import math

#Variable declaration
a = 120.0                 #Area(mm^2)
ds = 2.11                 #Diameter of each strand(mm)
W = 1118.0/1000           #Weight of conductor(kg/m)
L = 200.0                 #Span(m)
stress = 42.2             #Ultimate tensile stress(kg/mm^2)
wind = 60.0               #Wind pressure(kg/m^2)
t = 10.0                  #Ice thickness(mm)

#Calculation
n = 3.0                                          #Number of layers
d = (2*n+1)*ds                                   #Overall diameter of conductor(mm)
u = stress*a                                     #Ultimate strength(kg)
T = u/4.0                                        #Working stregth(kg)
#Case(a)
S_a = W*L**2/(8*T)                               #Sag in still air(m)
#Case(b)
area = d*100*10.0*10**-6                         #Projected area to wind pressure(m^2)
w_r = (W**2+w_w**2)**0.5                         #Resultant weight/m(kg)
S_b = w_r*L**2/(8*T)                             #Total sag with wind pressure(m)
w_i = 0.915*math.pi/4*((d+2*t)**2-(d**2))/1000.0 #Weight of ice on conductor(kg/m)
area_i = (d+2*t)*1000.0*10**-6                   #Projected area to wind pressure(m^2)
w_r_c = ((W+w_i)**2+w_n**2)**0.5                 #Resultant weight/m(kg)
S_c = w_r_c*L**2/(8*T)                           #Total sag with wind pressure and ice coating(m)
S_v = S_c*(W+w_i)/w_r_c                          #Vertical component of sag(m)

#Result
print('Case(a) : Sag in still air, S = %.2f m' %S_a)
print('Case(b) : Sag with wind pressure, S = %.2f m' %S_b)
print('          Sag with wind pressure and ice coating, S = %.2f m' %S_c)
print('          Vertical sag, S_v = %.2f m' %S_v)
print('\nNOTE: ERROR: calculation mistake in the textbook')

Case(a) : Sag in still air, S = 4.42 m
Case(b) : Sag with wind pressure, S = 5.63 m
Sag with wind pressure and ice coating, S = 10.96 m
Vertical sag, S_v = 7.23 m

NOTE: ERROR: calculation mistake in the textbook