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
r1=30 #cm #radius
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

#Radii of pulley 2
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
theta=165*pi*180**-1 #radians
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
r1=0.6 #m #Radius
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
theta=165*pi*180**-1 #radians
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
theta=2.79 #radians
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
r1=30 #cm #Radius
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
r1=0.75 #m #Radius
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
theta=170*pi*180**-1 #radians
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
theta=160*pi*180**-1 #radians

#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