# Ch:30 Chain drive¶

## exa 30-1 - Page 778¶

In :
from __future__ import division
from math import sqrt, pi
n1=17#
n2=51#
C=300#
p=9.52#
Ln=(2*C/p)+((n1+n2)/2)+((((n2-n1)/(2*pi))**2)*(p/C))#
x=(Ln-((n2+n1)/(2)))**2#
y=8*(((n2-n1)/(2*pi))**2)#
z=Ln-((n1+n2)/2)#
C=(p/4)*(z+(sqrt(x-y)))

# printing data in scilab o/p window
print "C is %0.2f mm "%(C)#


C is 300.00 mm


## exa 30-2 - Page 778¶

In :
from math import tan
G=4#
n1=17#
n2=n1*G#
N1=2300#
Kc=1.2# #from table 30-2
p=12.7# #fom table 30-1
D1=p*n1#
D2=p*n2#
phi=2*10.6#
x=tan(phi/2)# #phi/2 = 10.6deg, from table 30-3
Da1=(p/x)+(0.6*p)#
Da2=(p/x*4)+(0.6*p)#
Cmin=Kc*((Da1+Da2)/2)#
Ln1=(2*Cmin/p)+((n1+n2)/2)+((((n2-n1)/(2*pi))**2)*(p/Cmin))#
Ln1=80#
print "Ln is %0.0f  "%(Ln1)#

Ln is 80


## exa 30-3 - Page 779¶

In :
from math import sqrt, pi
N1=1000#
N2=500#
P=2.03*10**3# #from table 30-8
K1=1.26#
Ks=1#
#let Pc be the power transmitting capacity of the chain
Pc=P*K1/Ks#
p=9.52#
n1=21#
n2=42#
V=n1*p*N1/(60*10**3)#
#Let the chain tension be T
T=Pc/V#
#Let the breaking load be BL
BL=10700#
FOS=BL/T#
C=50*p#
Ln=(2*C/p)+((n1+n2)/2)+((((n2-n1)/(2*pi))**2)*(p/C))#
L=Ln*p#
Pc=Pc*10**-3#
print " Pc is %0.2f KW  "%(Pc)#
print "\n V is %0.3f m/s  "%(V)#
print "\n T is %0.1f N  "%(T)#
print "\n FOS is %0.2f   "%(FOS)#
print "\n L is %0.2f mm  "%(L)#

#The difference in the value of L and T is due to rounding-off the values.

 Pc is 2.56 KW

V is 3.332 m/s

T is 767.6 N

FOS is 13.94

L is 1254.01 mm


## exa 30-5 - Page 780¶

In :
from math import sqrt, pi
G=2#
P=5000#
Ks=1.7#
Pd=P*Ks#
K2=1.7#
p=15.88#
n1=17#
n2=n1*G#
D1=n1*p#
D2=n2*p#
C=40*p#
Ln=(2*C/p)+((n1+n2)/2)+((((n2-n1)/(2*pi))**2)*(p/C))#
L=Ln*p#
print "L is %0.2f mm  "%(L)#
#The difference in the value of L is due to rounding-off the values.

L is 1678.25 mm