# Ch:20 Flat belt drive¶

## exa 20-1 - Page 565¶

In [1]:
from __future__ import division
from math import sqrt, pi, asin, exp, degrees
b=0.2#
P=50*10**3#
v=20#
m=1.95#
d=0.3#
D=0.9#
C=5.8#
u=0.4#
#Let density be rho
rho=1000#
E=40#
#Let T1-T2 = T
T=P/v#
#Let the centrifugal tension be Tc
Tc=m*v**2#
alpha=degrees(asin((D+d)/(2*C)))#
theta=180+(2*alpha)#
theta=theta*pi/180#
x = exp(u*theta)#
T2=(((1-x)*Tc)-T)/(1-x)#
#T1=T+T2#
T1=T+T2#
t=m/(b*rho)*10**3#
#Let maximum stress be sigmax
b=200#
d=300#
sigmax=(T1/(b*t)+((E*t)/d))#
sigmin=(T2/(b*t))#
print "T1 is %0.1f N "%(T1)#
print "\nT2 is %0.1f N "%(T2)#
print "\nt is %0.2f mm "%(t)
print "\ntheta is %0.2f rad "%(theta)
print "\nsigmax is %0.2f N/mm**2 "%(sigmax)#
print "\nsigmin is %0.3f N/mm**2 "%(sigmin)#
#The answer for T1 is miscalculated in the book.

T1 is 4167.4 N

T2 is 1667.4 N

t is 9.75 mm

sigmax is 3.44 N/mm**2

sigmin is 0.855 N/mm**2


## exa 20-2 - Page 566¶

In [2]:
from math import degrees,asin,pi,exp,sqrt
P=12*10**3#
d=0.2#
D=0.5#
C=2#
sigmax=2*10**6#
t=8*10**-3#
#Let density be rho
rho=950#
u=0.38#
N=1500#
#Let angle of contact = thetad
v=(2*pi*N*d)/(60*2)#
#Let T1-T2=T
T=P/v#
b=(T*x)/((1-x)*t*((rho*v**2)-(sigmax)))#
b=b*10**3#
#Let breadth of the pulley be b1
b1=b*10**3+13# #Table 20-3
# Let pulley crown for d=h1, D=h2
h1=0.6#  #Table 20-4
h2=1#
print "b is %0.2f mm "%(b)
print "\nL is %0.2f m "%(L)
print "\nb1 is %0.2f mm "%(b1)#
print "\nh1 is %0.1f mm "%(h1)#
print "\nh2 is %0.1f mm "%(h2)#


b is 79.64 mm

L is 5.10 m

b1 is 79650.98 mm

h1 is 0.6 mm

h2 is 1.0 mm


## exa 20-3 - Page 567¶

In [3]:
from math import exp, degrees,asin,pi,sqrt
P=11#
N=1440#
n=480#
C=2.4#
#Let power transmitte dfrom high speed belt =P1
P1=0.0118#
V=5#
Ks=1.2#
v=15#
d=v*10**3*60/(2*pi*N)#
d=0.2#
D=N/n*d#
#Let angle of contact =thetaA
thetaA=180-(2*degrees(asin((D-d)/(2*C))))#
thetaA=thetaA*pi/180#
v=(2*pi*N*d)/(60*2)#
#Let the arc of contact correction factor be Ka
Ka=1.05#
Pd=P*Ka*Ks#
Pc=P1*v/V#
b=Pd/(Pc*4)#
thetaB=(2*pi)-thetaA#
L=sqrt((4*C**2)-((D-d)**2))+((d*thetaA/2)+(D*thetaB)/2)#

print "v is %0.2f m/s "%(v)
print "\nb is %0.3f mm "%(b)
print "\nL is %0.4f m "%(L)#

v is 15.08 m/s

b is 97.364 mm

L is 6.0733 m


## exa 20-4 - Page 568¶

In [4]:
from math import exp, degrees,asin,pi,sqrt
N=1440#
i=2.5#
C=3600#
LF=1.3#
P=12*10**3#
n=N/i#
V=16#
d=V*10**3*60/(2*pi*N)#
d=220#
D=d*i#
V=2*pi*N*d/(2*60*1000)#
v=5#
#Let power transmitte dfrom high speed belt =P1
P1=0.0118#
#Let LR be the load rating of belt
LR=P1/v*V#
theta=180+(2*degrees(asin((D-d)/(2*C))))#
theta=theta*pi/180#
#Let Arc of contact connection factor be CF
CF=1-(0.03/2)#
Pd=P*LF*CF#
b=Pd/(LR*5)#
b=80#
L=sqrt((4*C**2)-(D+d)**2)+(theta*(D+d)/2)#
L=L*10**-3#
print "V is %0.1f m/s "%(V)
print "\nb is %0.0f mm "%(b)
print "\nL is %0.3f m "%(L)#


V is 16.6 m/s

b is 80 mm

L is 8.404 m


## exa 20-5 - Page 569¶

In [5]:
from math import exp, degrees,asin,pi,sqrt
i=3.6#
N=1440#
d=220#
Ks=1.2#
Kf=1.1#
C=5000#
u=0.8#
D=i*d#
#From table 20-7, the following data is available
t=5#
b=120#
Fa=30.64#
#let weight density be w
w=0.106*10**5#
Cp=0.71# #From table 20-6
Cv=1#
T1=Fa*b*t*Cp*Cv#
m=w*b*t/10**6#
V=2*pi*N*d/(2*60*1000)#
Tc=m*V**2/9.81#
theta=180+(2*degrees(asin((D-d)/(2*C))))#
theta=theta*pi/180#
x=u*theta#
T2=Tc+((T1-Tc)/exp(x))#
Pd=(T1-T2)*V*10**-3#
P=Pd/(Ks*Kf)#
print "V is %0.2f m/s "%(V)#
print "\nPd is %0.2f KW "%(Pd)#
print "\nP is %0.1f KW "%(P)#
#The value of T2 is calculated incorrectly, therefore there is a difference in the values of Pd and P.

V is 16.59 m/s

Pd is 197.77 KW

P is 149.8 KW


## exa 20-6 - Page 570¶

In [6]:
from math import exp, degrees,asin,pi,sqrt
i=2.5#
C=4500#
N=960#
P=20*10**3#
Ks=1.15#
Kf=1.10#
t=8#
#let weight density be w
w=0.110*10**5#
m=w*t/10**6#
Fa=8.75#
d=200#
D=i*d#
u=0.4#
V=2*pi*N*d/(2*60*1000)#
Pd=P*Ks*Kf#
Cp=1#
Cv=0.6#
#to find b
T1=Fa*t*Cp*Cv#
Tc=m*V**2/9.81#
theta=180-(2*degrees(asin((D-d)/(2*C))))
theta=theta*pi/180#
x=u*theta#
T2=Tc+((T1-Tc)/exp(x))#
T=Pd/V#
b=T/(T1-T2)#
#b=90#
L=sqrt((4*C**2)-(D+d)**2)+(theta*(D+d)/2)#
L=L*10**-3#
print "V is %0.2f m/s "%(V)
print "\nb is %0.3f mm "%(b)
print "\nL is %0.3f m "%(L)#


V is 10.05 m/s

b is 86.537 mm

L is 10.049 m


## exa 20-7 - Page 571¶

In [7]:
from math import exp, degrees,asin,pi,sqrt
b=160#
t=7#
P=3*10**3#
Ks=1.2#
d=160#
N=1440#
D=480#
C=2400#
w=11200#
u=0.4#
Fa=7.2#
m=w*b*t/10**6#
V=2*pi*N*d/(2*60*1000)#
Tc=m*V**2/9.81#
Cp=0.6# #from table 20-6
Cv=0.98# #from table 20-7
Ta=Fa*b*Cp*Cv#
T=P/V#
theta=180-(2*degrees(asin((D-d)/(2*C))))
theta=theta*pi/180#
x=u*theta#
#T2=Tc+((T1-Tc)/exp(x))#
T2=(T+((exp(x)*Tc)-Tc))/(exp(x)-1)#
T1=T+T2#
Kf=Ta/T1#
Pd=P*Ks*Kf#
Pd=Pd*10**-3#
print "Tc is %0.0f N "%(Tc)#
print "\nT1 is %0.2f N "%(T1)#
print "\nT2 is %0.2f N "%(T2)#
print "\nKf is %0.2f  "%(Kf)#
print "\nPd is %0.1f KW "%(Pd)#
#The difference in values of T1 and T2 is due to rounding-off of values.


Tc is 186 N

T1 is 541.46 N

T2 is 292.78 N

Kf is 1.25

Pd is 4.5 KW