# Example 10.1 Cable subjected to concentrated loads¶

In [7]:
import math,numpy
# Initilization of variables
W1=400 # N # vertical load at pt C
W2=600 # N # vertical load at pt D
W3=400 # N # vertical load at pt E
l=2 # m # l= Lac=Lcd=Lde=Leb
h=2.25 # m # distance of the cable from top
L=2 # m # dist of A from top
# Calculations
# Solving eqn's 1&2 using MATRIX for Xb & Yb
A=numpy.matrix([[-L,4*l],[-h,2*l]])
B=numpy.matrix([[((W1*l)+(W2*2*l)+(W1*3*l))],[(W1*l)]])
C=numpy.linalg.inv(A)*B
# Now consider the F.B.D of BE, Take moment at E
y_e=(C[1]*l)/C[0] # m / here y_e is the distance between E and the top
theta_1=math.degrees(math.atan(y_e/l)) # degree # where theta_1 is the angle between BE and the horizontal
T_BE=C[0]/math.cos(theta_1*math.pi/180) # N (T_BE=T_max)
# Now consider the F.B.D of portion BEDC
# Take moment at C
y_c=((C[1]*6)-(W3*4)-(W2*2))/(C[0]) # m
theta_4=math.degrees(math.atan(((y_c)-(l))/(l))) # degree
T_CA=C[0]/math.cos(theta_4*math.pi/180) # N # Tension in CA
# Results
print('(i) The horizontal reaction at B (Xb) is %f N'%C[0])
print('(i) The vertical reaction at B (Yb) is %f N'%C[1])
print('(ii) The sag at point E (y_e) is %f m'%y_e)
print('(iii) The tension in portion CA (T_CA) is %f N'%T_CA)
print('(iv) The max tension in the cable (T_max) is %f N'%T_BE)
print('(iv) The max slope (theta_1) in the cable is %f degree'%theta_1)

(i) The horizontal reaction at B (Xb) is 1600.000000 N
(i) The vertical reaction at B (Yb) is 1100.000000 N
(ii) The sag at point E (y_e) is 1.375000 m
(iii) The tension in portion CA (T_CA) is 1627.882060 N
(iv) The max tension in the cable (T_max) is 1941.648784 N
(iv) The max slope (theta_1) in the cable is 34.508523 degree


# Example 10.2 Cables subjected to concentrated loads¶

In [9]:
from __future__ import division
import math
# Initiization of variables
W1=100 # N # Pt load at C
W2=150 # N # Pt load at D
W3=200 # N # Pt load at E
l=1 # m # l=Lac=Lcd=Lde=Leb
h=2 # m # dist between Rb & top
Xa=200 # N
Xb=200 # N
# Calculations
# consider the F.B.D of entire cable
# Take moment at A
Yb=((W1*l)+(W2*2*l)+(W3*3*l)-(Xb*h))/(4*l) # N
Ya=W1+W2+W3-Yb # N # sum Fy=0
# Now consider the F.B.D of AC
# Take moment at C,
y_c=(Ya*l)/Xa # m
theta_1=math.degrees(math.atan(y_c/l)) # degree
T_AC=Xa/math.cos(theta_1*math.pi/180) # N # T_AC*cosd(theta_1)=horizontal component of tension in the cable
# here, T_AC=T_max
T_max=T_AC # N
# Now consider the F.B.D of portion ACD
y_d=((Ya*2*l)-(W1*l))/(Xa) # m # taking moment at D
theta_2=math.degrees(math.atan(((y_d)-(y_c))/(l))) # degree
T_CD=Xa/(math.cos(theta_2*math.pi/180)) # N
# Results
print('(i) The component of support reaction at A (Ya) is %f N'%Ya)
print('(i) The component of support reaction at B (Yb) is %f N'%Yb)
print('(ii) The tension in portion AC (T_AC) of the cable is %f N'%T_AC)
print('(ii) The tension in portion CD (T_CD) of the cable is %f N'%T_CD)
print('(iii) The max tension in the cable is %f N'%T_max)

(i) The component of support reaction at A (Ya) is 300.000000 N
(i) The component of support reaction at B (Yb) is 150.000000 N
(ii) The tension in portion AC (T_AC) of the cable is 360.555128 N
(ii) The tension in portion CD (T_CD) of the cable is 282.842712 N
(iii) The max tension in the cable is 360.555128 N


# Example 10.3 Cables uniformly loaded per unit horizontal distance¶

In [10]:
from __future__ import division
import math
# Initilization of variables
w=75 # kg/m # mass per unit length of thw pipe
l=20 # m # dist between A &  B
g=9.81 # m/s**2 # acc due to gravity
y=2 # m # position of C below B
# Calculations
# Let x_b be the distance of point C from B
# In eq'n x_b**2+32*x_b-320=0
a=1
b=32
c=-320
x_b=(-b+math.sqrt(b**2-(4*a*c)))/(2*a) # m # we get x_b by equating eqn's 1&2
# Now tension T_0
T_0=((w*g*x_b**2)/(2*y))*(10**-3) #kN # from eq'n 1
# Now the max tension occurs at point A,hence x is given as,
x=20-x_b # m
w_x=w*g*x*10**(-3) # kN
T_max=math.sqrt((T_0)**2+(w_x)**2) # kN # Maximum Tension
# Results
print('The lowest point C which is situated at a distance (x_b) from support B is %f m'%x_b)
print('The maximum tension (T_max) in the cable is %f kN'%T_max)
print('The minimum tension (T_0) in the cable is %f kN'%T_0)

The lowest point C which is situated at a distance (x_b) from support B is 8.000000 m
The maximum tension (T_max) in the cable is 14.715000 kN
The minimum tension (T_0) in the cable is 11.772000 kN


# Example 10.4 Cables uniformly loaded per unit horizontal distance¶

In [14]:
from __future__ import division
import math
# Initilization of variables
m=0.5 # kg/m # mass of the cable per unit length
g=9.81 # m/s**2
x=30 # m # length AB
y=0.5 # m # dist between C & the horizontal
x_b=15 # m # dist of horizontal from C to B
# Calculations
w=m*g # N/m # weight of the cable per unit length
T_0=(w*x_b**2)/(2*y) # N # From eq'n 1
T_B=math.sqrt((T_0)**2+(w*x/2)**2) # N # Tension in the cable at point B
W=T_B # N # As pulley is frictionless the tension in the pulley on each side is same,so W=T_B
# Slope of the cable at B,
theta=math.degrees(math.acos(T_0/T_B)) # degree
# Now length of the cable between C & B is,
S_cb=x_b*(1+((2/3)*(y/x_b)**2)) # m
# Now total length of the cable AB is,
S_ab=2*S_cb # m
# Results
print('(i) The magnitude of load W is %f N'%W)
print('(ii) The angle of the cable with the horizontal at B is %f degree'%theta)
print('(iii) The total length of the cable AB is %f m'%S_ab)

(i) The magnitude of load W is 1106.074781 N
(ii) The angle of the cable with the horizontal at B is 3.814075 degree
(iii) The total length of the cable AB is 30.022222 m


# Example 10.5 Cables uniformly loaded per unit horizontal distance¶

In [15]:
from __future__ import division
import math
# Initilization of variables
x=30 # m # distance between two electric poles
Tmax=400 # N # Max Pull or tension
w=3 # N/m # weight per unit length of the cable
# Calculations
# The cable is assumed to be parabolic in shape, its eq'n is y=w*x^2/2*T_0.....(eq'n 1).
#Substuting the co-ordinates of point B (l/2,h), where h is the sag in the cable.
#This gives, T_0=(w*(l/2)^2)/(2*h)=wl^2/8*h
# Now the maximum pull or tension occurs at B,
T_B=Tmax # N
# Hence T_B=Tmax=sqrt(T_0^2+(w*l/2)^2). On simplyfyingthis eq'n we get,
h=math.sqrt(x**2/(16*(((Tmax*2)/(w*x))**2-(1)))) # m
# Results
print('The smallest value of the sag in the cable is %f m'%h)

The smallest value of the sag in the cable is 0.849141 m


# Example 10.6 Catenary Cables¶

In [16]:
from __future__ import division
import math
# Initilization of variables
l=200 # m # length of the cable
m=1000 # kg # mass of the cable
S=50 # m # sag in the cable
s=l/2 # m
g=9.81 # m/s^2
# Calculations
w=(m*g)/l # N/m # mass per unit length of the cable
# Substuting the values s=l/2 & y=c+S in eq'n 1 to get the value of c,
c=7500/100 # m
Tmax=math.sqrt((w*c)**2+(w*s)**2) # N # Maximum Tension
# To determine the span (2*x) let us use the eq'n of catenary, y=c*cosh(x/c), where y=c+50. On simplyfying we get y/c=cosh(x/c), here let y/c=A
y=c+50
A=y/c
x=c*(math.acosh(A)) # m
L=2*x # m # where L= span
# Results
print('The horizontal distance between the supports and the max Tension (L) is %f m'%L)

The horizontal distance between the supports and the max Tension (L) is 164.791843 m