# Chapter 11:Belts,Ropes And Chain Drives¶

## Example 11.1,Page No.386¶

In [1]:
import math

#Initilization of Variables

N1=200 #r.p.m
d1=51 #cm #Dia. of engine
d2=30 #cm #Dia. of driven shaft

#Calculations

#Speed of driven shaft
N2=d1*d2**-1*N1 #r.p.m

#Result
print"Speed of driven shaft is",round(N2,2),"r.p.m"

Speed of driven shaft is 340.0 r.p.m


## Example 11.2,Page No.386¶

In [2]:
import math

#Initilization of Variables

t=1 #cm #thickness
N1=200 #r.p.m
d1=51 #cm
d2=30 #cm

#Calculations

#Speed of shaft
N2=(d1+t)*(d2+t)**-1*N1 #r.p.m

#Result
print"speed of shaft is",round(N2,2),"r.p.m"

speed of shaft is 335.48 r.p.m


## Example 11.3,Page No.388¶

In [3]:
import math

#Initilization of Variables

N1=200 #r.p.m
N2=300
d1=60 #cm
t=0.5 #cm
s=4 #%

#Calculations

#Diameter of pulley
d2=N1*N2**-1*d1 #cm

#Taking belt thickness
d2_2=(d1+t)*N1*N2**-1-t

#Also considering slip
d2_3=(d1+t)*N1*N2**-1*(1-s*100**-1)-t #cm

#Result
print"Diameter of belt is:Neglecting belt thickness",round(d2,2),"cm"
print"                   :Belt thickness only",round(d2_2,2),"cm"
print"                   :Considering belt thickness and slip",round(d2_3,2),"cm"

Diameter of belt is:Neglecting belt thickness 40.0 cm
:Belt thickness only 39.83 cm
:Considering belt thickness and slip 38.22 cm


## Example 11.4,Page No.389¶

In [4]:
import math

#Initilization of Variables

d1=1 #m #dia. of driver pulley
N1=200 #r.p.m #Speed of driver pulley
d2=2.5 #m #Dia. of driven pulley
f1=1.44 #N/mm**2 #strress
f2=0.49 #N/mm**2
E=100 #N/mm**2 #Young's Modulus

#Calculations

#Speed of driven pulley
N2=d1*d2**-1*(E+round((f2)**0.5,2))*(E+round((f1)**0.5,2))**-1*N1

#Speed if creep is neglected
N2_2=d1*d2**-1*N1 #r.p.m

#Speed lost by driven pulley due to creep
N=N2_2-N2

#Result
print"speed Lost by driven pulley due to creep is",round(N,3),"r.p.m"

speed Lost by driven pulley due to creep is 0.395 r.p.m


## Example 11.5,Page No.390¶

In [5]:
import math

#Initilization of Variables

d1=1 #m #Diameter
N1=200 #r.p.m
d2=2.5 #m
b=500 #mm #width
t=10 #mm #thickness
E=100 #N/mm**2

#Calculations

#Area
A=b*t #mm**2

#Tension on tight side
T1=10*b

#Tension on slack side
T2=4*b #N

#Stress on tight side
f1=T1*A**-1 #N/mm**2

#Stress on slack side
f2=T2*A**-1 #N/mm**2

#Speed of driven pulley
N2=d1*d2**-1*(E+(f2)**0.5)*(E+(f1)**0.5)**-1*N1

#Speed if creep is neglected
N2_2=d1*d2**-1*N1 #r.p.m

#Speed lost by driven pulley due to creep
N=N2_2-N2

#Result
print"speed Lost by driven pulley due to creep is",round(N,2),"r.p.m"

speed Lost by driven pulley due to creep is 0.29 r.p.m


## Example 11.7,Page No.397¶

In [2]:
import math
from math import sin, cos, tan, radians, pi

#Initilization of Variables

x=600 #cm #Distance between shafts
r2=20 #cm

#Calculations

#If belt is open
L1=pi*(r1+r2)+(r1-r2)**2*x**-3+2*x #cm

#If belt is crossed
L2=pi*(r1+r2)+(r1+r2)**2*x**-1+2*x #cm

#Result
print"If belt is open Length is",round(L1,2),"cm"
print"If belt is crossed length is",round(L2,2),"cm"

If belt is open Length is 1357.08 cm
If belt is crossed length is 1361.25 cm


## Example 11.8,Page No.397¶

In [8]:
import math
from math import sin, cos, tan, radians, pi

#Initilization of Variables

#speed of shafts
N1=N3=N5=160 #r.p.m
N2=60 #r.p.m
N4=80 #r.p.m
N6=100 #r.p.m
x=180 #cm #Distance between shafts
r1=15 #cm #Radius of smallest pulley

#Calculations

r2=r1*N1*N2**-1 #cm

#using same above equation for radii of pulley 3 we get and further simplifying we get
#r4=2*r3   ..................1

#but for crossed belt equation is
#r1+r2=r3+r4=r5+r6   .................2
#After further simplifying we get
r3=(r1+r2)*3**-1 #cm
r4=2*r3 #cm

#Using same above equation and further simplifying we get
#r6=1.6*r5    ...............3

#sub value of r6 in equation we get
r5=(r1+r2)*2.6**-1 #cm
r6=1.6*r5 #cm

#Length of open belt
L=pi*(r1+r2)+(r1-r2)**2*x**-1+2*x #cm

#For pulley 3 and 4 equation is
#L=pi*(r3+r4)+(r3-r4)**2*x**-1+2*x
#sub value in above equation we get an equation as
#r3**2+1696.5*r3-31710.6=0
a=1
b=1696.5
c=-31710.6

X=b**2-4*a*c

r3_2=(-b+X**0.5)*2**-1 #cm
r4_2=2*r3_2 #cm

#Sim for r5 & r6

#L=pi*(r6+r5)+(r6-r5)**2*x**-1+2*x
#sub value in above equation we get an equation as
#r5**2+4084*r5-88085=0
a=1
b=4084
c=-88085

X=b**2-4*a*c

r5_2=(-b+X**0.5)*2**-1 #cm
r6_2=1.6*r5_2 #cm

#Result
print"Radii of two stepped pulleys is:For crossed belt:r3",round(r3,2),"cm"
print"                                                :r4",round(r4,2),"cm"
print"                                                :r5",round(r5,2),"cm"
print"                                                :r6",round(r6,2),"cm"
print"Radii of two stepped pulleys is:For open belt:r3_2",round(r3_2,2),"cm"
print"                                             :r4_2",round(r4_2,2),"cm"
print"                                             :r5_2",round(r5_2,2),"cm"
print"                                             :r6_2",round(r6_2,2),"cm"

Radii of two stepped pulleys is:For crossed belt:r3 18.33 cm
:r4 36.67 cm
:r5 21.15 cm
:r6 33.85 cm
Radii of two stepped pulleys is:For open belt:r3_2 18.49 cm
:r4_2 36.98 cm
:r5_2 21.46 cm
:r6_2 34.33 cm


## Example 11.9,Page No.403¶

In [3]:
import math
from math import sin, cos, tan, radians, pi

#Initilization of Variables

d=1.2 #m #Diameter
N=200 #r.p.m #Speed
mu=0.3 #Coefficient of friction
T1=3000 #N #MAx Tension

#Calculations

#Velocity
v=pi*d*N*60**-1 #m/s

#From ration of tensions we get
#T1*T2=e**mu*theta
#After simplifying we get
#e**mu*theta =2.3714
T2=T1*2.3714**-1 #N

#Power transmitted
P=(T1-T2)*v*1000**-1 #KW

#Result
print"Power transmitted is",round(P,2),"KW"

Power transmitted is 21.8 KW


## Example 11.10,Page No.403¶

In [5]:
import math
from math import sin, cos, tan, radians, pi
import numpy as np

#Initilization of Variables

d1=1.20 #m #Diameter
r2=0.25 #m
x=4 #m #Distance between shafts
T1=1855.3 #N #Max TRension
mu=0.3 #Coefficient of friction
N1=200 #r.p.m

#Calculations

#Velocity
v=pi*d1*N1*60**-1 #m/s

##Let sin(alpha)=X
X=(r1-r2)*x**-1
alpha=np.arcsin(X)*(pi**-1*180)

#Angle of contact
theta=180-2*alpha

#From equation of max tension and further simplifying we get
T2=1855.3*2.435**-1 #N

#Power transmitted
P=(T1-T2)*v*1000**-1 #KW

#Torque
t1=(T1-T2)*r1 #N*m
t2=(T1-T2)*r2 #Nm

#Result
print"Power transmitted is",round(P,2),"KN"
print"Torque Exerted on driving shaft is:t1",round(t1,2),"N*m"
print"                                  :t2",round(t2,2),"N*m"

Power transmitted is 13.74 KN
Torque Exerted on driving shaft is:t1 656.02 N*m
:t2 273.34 N*m


## Example 11.11,Page No.407¶

In [11]:
import math

#Initilization of Variables

theta=2.88
v=28.33 #m/s #velocity
b=20 #cm #Width
t=0.8 #cm #thickness
rho=10**-3 #Kg/cm**3 #density
f=250 #N/cm**2 #Stress
mu=0.25 #coefficient of friction

#Calculations

#Max Tension
Tm=f*b*t #N

#mass
m=rho*b*t*100 #Kg

#Centrifugal Tension
Tc=m*v**2 #N

#Tension on tight side
T1=Tm-Tc #N

#From ratio of tension equation we get
T2=T1*2.056**-1 #N

#MAx Power
P=(T1-T2)*v*1000**-1 #KW

#Result
print"Max Power transmitted is",round(P,2),"KW"

Max Power transmitted is 39.52 KW


## Example 11.12,Page No.408¶

In [12]:
import math

#Initilization of Variables

rho=10**-3 #kg/cm**3 #density
f=250 #N/cm**2 #stress
b=20 #cm #width
t=1.2 #cm #thickness

#Calculations

#MAss
m=rho*b*t*100

#MAx tension
Tm=f*b*t #N

#Velocity
v=(Tm*(3*m)**-1)**0.5 #m/s

#From equation of max tension and further simplifying we get
T1=2*3**-1*Tm #N
T2=T1*2**-1 #N

#Power
P=(T1-T2)*v*(1000)**-1 #KW

#Result
print"Max Power transmitted is",round(P,2),"KW"

Max Power transmitted is 57.74 KW


## Example 11.13,Page No.409¶

In [6]:
import math
from math import sin, cos, tan, radians, pi

#Initilization of Variables

P=9 #KW #Power
d1=1.2 #m #Diameter
N1=200 #r.p.m
mu=0.3 #Coefficient of friction
f=140 #N/cm**2 #Stress
rho=10**-3 #kg/cm**3
t=1 #cm #thickness

#Calculations

#From Ratio of tension equation we get
#T1*T2=e**mu*theta
#After simplifying we get
#e**mu*theta =2.3714
#T1=2.3714*T2   ................1

#Max tension in belt
#Tm=f*b*t    ..............2

#Centrifugal tension
#Tc=m*v**2     .....................3

#Velocity
v=pi*d1*N1*60**-1 #m/s

#mass
#m=rho*b*t*100
#After simplifying we get
#m=b*10**-1

#Sub value of m in equation 2 and further simplfying we get
#T1-T2=716.5

#After further simplifying equations 1,2,3 we get
T2=716.5*1.3714**-1 #N
T1=2.3714*T2 #N

#sub value in MAx tension and further simplifying we get
b=1238.96*124**-1 #cm

#Result
print"Width of belt is",round(b,2),"cm"

Width of belt is 9.99 cm


## Example 11.14,Page No.411¶

In [14]:
import math

#Initilization of Variables
b=100 #mm #Width
t=10 #mm #thickness
rho=10**-6 #kg/mm**3
mu=0.25 #coefficient of friction
f=1.5 #N/mm**2
g=9.81

#Calculations

#MAx tension
Tm=f*b*t #N

#From Ratio of tension equation we get
#T1*T2=e**mu*theta
#After simplifying we get
#e**mu*theta =2

#For Max power
Tc=Tm*3**-1 #N

#From max transmissiom equation
T1=Tm-Tc
T2=T1*2**-1 #N

#MAss
m=rho*b*t*1000 #Kg

#Weight
W=m*g #N

#Velocity
v=(Tm*(3*m)**-1)**0.5 #m/s

#Power transmitted
P=(T1-T2)*v*10**-3 #KW

#Result
print"Max Power that can be transmitted is",round(P,2),"KW"

Max Power that can be transmitted is 11.18 KW


## Example 11.15,Page No.412¶

In [7]:
import math
from math import sin, cos, tan, radians, pi
import numpy as np

#Initilization of Variables

d1=60 #cm #diameter
d2=24 #cm
r2=12 #cm
x=300 #cm #dist between two shafs
N2=300 #r.p.m #speed of small pulley
mu=0.3 #coefficient of friction
m=0.6703 #kg
t=100 #N per cm width #Safe working tension

#Calculations

#LEt sin(Alpha)=X
alpha=np.arcsin((r1-r2)*x**-1)*(180*pi**-1)

#Using equation of ratio of tension
#T1*T2**-1=e**mu*theta  ...........1
#Simplifying furter we get value of
#e**mu*theta=2.474
#T1=2.474*T2   ...................1

#Velocity
v=pi*d2*N2*60**-1*10**-2 #m/s

#Sub value of v and P in equation of power transmited and further simplifying we get
#(T1-T2)=994.7    .....................2
#Sub value of T1 from equation 1 we get
T2=994.7*1.474**-1 #N
T1=2.474*T2 #N

#Min width
b=T1*t**-1 #cm

#Initial belt tension
To=(T1+T2)*2**-1 #N

#Length of belt required
L=(pi*(r1+r2)+(r1+r2)**2*x**-1+2*x)*100**-1 #m

#Result
print"Minimum width of belt is",round(b,2),"cm"
print"Initial belt tension is",round(To,2),"N"
print"Length of belt required is",round(L,2),"m"

Minimum width of belt is 16.7 cm
Initial belt tension is 1172.18 N
Length of belt required is 7.38 m


## Example 11.16,Page No.414¶

In [8]:
import math
from math import sin, cos, tan, radians, pi
import numpy as np

#Initilization of Variables

d1=1.5 #m #diameter
d2=1 #m
r2=0.5 #m
x=4.80 #dist between two shafs
To=3000 #N #Initial tension
N2=600 #r.p.m #speed of small pulley
mu=0.3 #coefficient of friction
m=0.6703 #kg

#Calculation

#Velocity
v=pi*d2*N2*60**-1 #m/s

#Centrifugal tension
Tc=m*v**2

#from Initial Tensiom
#T1+T2=4677  ..........1

#Let sin(alpha)=X
X=(r1-r2)*x**-1
alpha=np.arcsin(X)*(pi**-1*180)

#Angle of contact
theta=(180-2*alpha)*pi*180**-1

#From equation of ratio of tension we get
#t1=2.5*T2    ...................2

#sub value in equation 1 we get
T2=4677*3.5**-1 #N
T1=2.5*T2

#Power transmitted
P2=(T1-T2)*v*10**-3

#Result
print"Power transmitted is",round(P2,2),"KW"

Power transmitted is 62.97 KW


## Example 11.18,Page No.418¶

In [9]:
import math
from math import sin, cos, tan, radians, pi

#Initilization of Variables

alpha=25 #degrees  #Angle of groove
Tmax=T1=1500 #N #Max tension
mu=0.27 #coefficient of friction
v=2 #m/s #belt speed

#Calculations

#From ratio of tension
#T1*T2**-1=e**mu*cosec(alpha)
#AFter further simplifying we get
#e**mu*cosec(alpha)=8.109
T2=T1*8.109**-1

#Net driving tension
T3=(T1-T2) #N

#Power transmitted
P=T3*v*10**-3 #W

#Result
print"Net driving tension is",round(T3,2),"N"
print"Power transmitted by the pulley is",round(P,2),"W"

Net driving tension is 1315.02 N
Power transmitted by the pulley is 2.63 W


## Example 11.19,Page No.418¶

In [18]:
import math
from math import sin, cos, tan, radians, pi

#Initilization of Variables

alpha=15 #Degrees
t=2 #cm #Depth pf belt
m=3.5*100*1000 #gm/l #mass
f=140 #N/cm**2 #Allowable stress
theta=140*pi*180**-1
mu=0.15 #coefficient of friction

#Calculations

CF=2*tan(15*pi*180**-1)
GC=1-2*tan(15*pi*180**-1)
BC=2*GC
ED=2
DF=2

#Area of v-belt
A=(ED+BC)*2**-1*DF

#MAx permissible tension
Tmax=f*A #N

#Centrifugal tension
Tc=Tmax*3**-1 #N

#Velocity
v=(Tc*m**-1)**0.5*1000 #m/s

#tension on tight side
T1=Tmax-Tc #N

#From ratio of tensions
#T1*T2**-1=e*mu*thta*cosec(alpha)
#After substituting values and furter simplifying we get value of
#e*mu*thta*cosec(alpha)=4.12
T2=T1*4.12**-1 #N

#Power
P2=(T1-T2)*v*1000**-1

#Result
print"Max Power transmitted is",round(P2,2),"KW"

Max Power transmitted is 4.09 KW


## Example 11.20,Page No.420¶

In [19]:
import math
from math import sin, cos, tan, radians, pi

#Initilization of Variables

P=75 #KW #Power
d1=1.50 #m #Dia. of driver pulley
N1=200 #r.p.m
alpha=22.5 #Angle of groove
mu=0.3 #coefficient of friction
theta=160*pi*180**-1
m=0.6 #kg #Mass
Tmax=800 #N #Max safe

#Calculations

#Velocity of rope
v=pi*d1*N1*60**-1 #m/s

#centrifugal Tension
Tc=m*v**2 #N

#Tension in tight side of rope
T1=Tmax-Tc #N

#Ratio of tension in rope
#T1*T2=e**mu*theta*cosec(alpha)
#After further simplifying we get value of e**mu*theta*cosec(alpha
#e**mu*theta*cosec(alpha=8.95
T2=T1*8.95**-1

#Power Transmitted by onr rope
P2=(T1-T2)*v*1000**-1 #KW

#No. of ropes required
n=P*P2**-1

#Initial rope tensuion
To=(T1+T2+2*Tc)*2**-1

#Result
print"No. of ropes required for drive is",round(n,2)
print"Initial Rope tension is",round(To,2),"N"

No. of ropes required for drive is 8.24
Initial Rope tension is 510.44 N


## Example 11.21,Page No.421¶

In [20]:
import math
from math import sin, cos, tan, radians, pi

#Initilization of Variables

d1=0.40 #Dia. of pulley
N1=110 #speed #r.p.m
alpha=22.5 #Angle of groove
mu=0.28 #coefficient of friction
N=10 #No.of ropes
P=23.628 #KW #Power

#Calculations

#velocity
v=pi*d1*N1*60**-1 #m/s

#Power transmited by one rope
P2=P*N**-1 #KW

#Centrifugal Tension
#Tc=0.0281*C**2   ................1

#Ratio of tension in rope
#T1=7.71*T2    ...........................2

#From other formula of power transmited by one rope
#P2=(T1-T2)*v*1000**-1
#After further substituting and simplifying we get
T2=1026*6.71**-1 #N
T1=7.71*T2 #N

#Tmax=T1+T2
#After sub values and further simplifying we get
C=(96.86)**0.5 #cm #girth of rope

Tc=0.0281*C**2 #N

#Initial Tension
To=(T1+T2+2*Tc)*2**-1 #N

#Dia. of each rope
d=C*pi**-1 #cm

#Result
print"Initial Tension is",round(To,2),"N"
print"Dia. of each rope is",round(d,2),"cm"

Initial Tension is 668.63 N
Dia. of each rope is 3.13 cm


## Example 11.22,Page No.422¶

In [21]:
import math
from math import sin, cos, tan, radians, pi

#Initilization of Variables

D=3.6 #m #Dia. of pulley
n=15 #No. of ropes
alpha=22.5 #Degrees
theta=170*pi*180**-1 #Angle of contact
mu=0.28 #angle of friction
Tmax=960 #N #MAx tension
m=1.5 #kg/l #mass of rope

#Calculations

#Centrifugal tension
Tc=Tmax*3**-1 #N

#Velocity
v=(Tmax*(3*m)**-1)**0.5 #m
N=60*v*(pi*D)**-1 #r.p.m

#equation
#T1*T2**-1=e**mu*theta*cosec(alpha)
#After simlifying further we get
#e**mu*theta*cosec(alpha)=8.756

#Tension in tight side of rope
T1=Tmax-Tc #N

#Tension in slack side
T2=T1*8.756**-1

#Max Power
P=(T1-T2)*v*1000**-1

#Total max power
P2=P*n

#Result
print"Speed of the pulley in r.p.m is",round(N,2),"r.p.m"
print"Total max power is",round(P2,2),"KW"

Speed of the pulley in r.p.m is 77.49 r.p.m
Total max power is 124.2 KW


## Example 11.23,Page No.423¶

In [6]:
import math
from math import sin, cos, tan, radians, pi

#Initilization of Variables

W=9000 #N #Weight of casting
n=2.5 #turns
theta=5*pi #Total angle covered
D=0.3 #m #diameter
N=20 #Speed
mu=0.25 #Coefficient of friction

#Calculations

#equation
#W*P**-1=e**mu*theta
#After simlifying further we get
P=W*50.65**-1 #Tension in slack side of rope #N

#Velocity
v=pi*D*N*60**-1 #m/s

#Power to raise casting
P2=(W-P)*v*1000**-1

#Result
print"Force Required by the man is",round(P,2),"N"
print"Power to raise the casting is",round(P2,2),"N"

Force Required by the man is 177.69 N
Power to raise the casting is 2.77 N