Ch:22 Friction clutches

exa 22-1 - Page 588

In [1]:
from __future__ import division
from math import sqrt, pi
u=0.28 #(coefficient of friction)
N=300 #(Engine rpm)
I=7.2 
Pmax= 0.1# 
R1=70#
R2=110#
n=2# #(Both sides of the plate are effective)
#Using Uniform Wear Theory
#Axial Force W
W=n*pi*Pmax*R1*(R2-R1)#
#Frictional Torque Tf
Tf=u*W*(R1+R2)/2*(10**-3)#
w=2*pi*N/60#
#Power P
P=Tf*w#
#Torque = Mass moment of inertia*angular acceleration
a=Tf/I#
t=w/a# 
#Angle turned by driving shaft theta1 through which slipping takes place
theta1=w*t#
#angle turned by driven shaft theta2
theta2=a*(t**2)/2#
E=Tf*(theta1-theta2)#
print "\nThe force is %0.1f N"%(W)#
print "\nThe Torque is %0.2f Nm"%(Tf)#
print "\nThe Power is %0.0f W"%(P)#
print "\nThe angular acceleration is %0.2f rad/sec**2"%(a)#
print "\nThe time taken is %0.1f sec"%(t)#
print "\nThe energy is %0.2f Nm"%(E)#

#The difference in the answer of energy 'E' is due to rounding-off of values.

The force is 1759.3 N

The Torque is 44.33 Nm

The Power is 1393 W

The angular acceleration is 6.16 rad/sec**2

The time taken is 5.1 sec

The energy is 3553.06 Nm

exa 22-2 - Page 589

In [2]:
from math import sqrt, pi
#Power P
P=80*10**3# #(Watt)
N=3000# #(Engine rpm)
w=2*pi*3*10**3/60
Tf=8*10**4/w#
Rm=100##(mm)
p=0.2 #N/mm**2
u=0.22 
# let width b= (R1-R2). 
#Axial force W=2*pi*Rm*b*p
#Torque T=u*W*Rm
b=Tf/(u*2*pi*(Rm**2)*p)#
b=50# 
R2=Rm+b#
R1=Rm-b#
Di=2*R1# #inner diameter
W=2*pi*Rm*b*p#
n=8# #n is number of springs
#Axial force per spring W1
W1=W/n#
W1=W1+15#
#axial deflection del
Del=10# 
#stiffness k
k=W1/Del#
# Spring index C
C=6#
#number of coils n1
n1=6# #Assumption
d=k*n*n1*(C**3)/(80*10**3)#
d=11# # Rounding off to nearest standard value
D=C*d#
clearance=2#
FL=((n1+2)*d)+(2*Del)+clearance# # two end coils, therefore (2*del)

print "\nThe Torque is %0.2f Nm"%(Tf)#
print "\nThe width is %0.0f mm"%(b)#
print "\nThe force is %0.0f N"%(W)#
print "\nThe Axial force per spring is %0.0f N"%(W1)#
print "\nThe Spring stiffness is %0.0f N/mm"%(k)#
print "\nThe Spring wire diameter is %0.0f mm"%(d)#
print "\nThe Mean coil diameter is %0.0f mm"%(D)#
print "\nThe Free length is %0.0f mm"%(FL)#

The Torque is 254.65 Nm

The width is 50 mm

The force is 6283 N

The Axial force per spring is 800 N

The Spring stiffness is 80 N/mm

The Spring wire diameter is 11 mm

The Mean coil diameter is 66 mm

The Free length is 110 mm

exa 22-3 - Page 589

In [3]:
#Power P
P=40*10**3 #Watt
n1=100# #rpm
n2=400# #rpm
#Speed factor Ks
Ks=0.9+0.001*n2#
#Clutch power Pc
Pc=P*n2/(n1*Ks)*10**-3#
print "\nThe Speed factor is %0.1f "%(Ks)#
print "\nThe clutch poweris %0.0f KW"%(Pc)#

The Speed factor is 1.3 

The clutch poweris 123 KW

exa 22-4 - Page 590

In [4]:
%matplotlib inline
from matplotlib.pyplot import plot, xlabel, ylabel, show, grid

# plot Torque vs Ro/Ri
#x=Ro/Ri
#According to Uniform Wear theory
x=[0, 0.2, 0.4, 0.577, 0.6, 0.8, 1.0]#
n=len(x)#
Tf = range(0,n)
for i in range(0,n):
    Tf[i]=(x[i]-(x[i]**3))#

plot (x,Tf)#
xlabel(' Ro/Ri ')#
ylabel('Tf')#
grid()#
show()

exa 22-5 - Page 591

In [5]:
from math import pi
n1=4#
n2=3#
n=(n1+n2-1)#
R2=80#
R1=50#
#According to Uniform Pressure Theory
#W=p*pi*((R2**2)-(R1**2)) T=n*2*u*W*((R2**3)-(R1**3))/(((R2**2)-(R1**2))*3)
P=15*10**3#
N=1400#
u=0.25#
w=2*pi*N/60#
T=P/w#
W=T*3*((R2**2)-(R1**2))/(n*2*u*((R2**3)-(R1**3)))*10**3#
p=W/(pi*((R2**2)-(R1**2)))#
print "\nThe angular speed is %0.2f rad/sec"%(w)#
print "\nThe Torque is %0.3f Nm"%(T)#
print "\nThe uniform pressure is %0.3f N/mm**2"%(p)#
print "\nThe Force is %0.1f N"%(W)#

The angular speed is 146.61 rad/sec

The Torque is 102.314 Nm

The uniform pressure is 0.084 N/mm**2

The Force is 1031.1 N

exa 22-6 - Page 592

In [6]:
from math import pi
P=5*10**3#
N=1000#
w=2*pi*N/60#
Rm=50#
pm=0.3#
Tf=P/w#
u=0.1#
R2=50*2/(0.6+1)#
R1=0.6*R2#
#According to uniform Wear theory
W=pm*Rm*(R2-R1)*2*pi#
n=Tf*(10**3)/(u*W*Rm)#
pmax=pm*Rm/R1#
print "\nThe angular speed is %0.2f rad/sec"%(w)#
print "\nThe Torque is %0.3f Nm"%(Tf)#
print "\nThe Inner radius is %0.1f mm"%(R1)#
print "\nThe Outer radius is %0.1f mm"%(R2)#
print "\nThe number of contacting surfaces is %0.0f "%(n)#
print "\nThe max. pressure is %0.1f N/mm**2"%(pmax)#

The angular speed is 104.72 rad/sec

The Torque is 47.746 Nm

The Inner radius is 37.5 mm

The Outer radius is 62.5 mm

The number of contacting surfaces is 4 

The max. pressure is 0.4 N/mm**2

exa 22-7 - Page 593

In [7]:
from math import pi
P=12*10**3#
N=750 #Speed=N
w=2*pi*N/60#
Tf=P/w#
p1=0.12#
a=12.5##Semi-cone angle
u=0.3#
k=u*0.18246*1.121/0.21644#
R1=(Tf*(10**3)/k)**(1/3)#
R2=R1*1.242#
Rm=1.121*R1#
W=2*pi*p1*R1*(R2-R1)#
print "\nThe angular speed is %0.2f rad/sec"%(w)#
print "\nThe Torque is %0.1f Nm"%(Tf)#
print "\nThe Inner radius is %0.1f mm"%(R1)#
print "\nThe Outer radius is %0.1f mm"%(R2)#
print "\nThe mean radius is %0.2f mm"%(Rm)#
print "\nThe axial force is %0.0f N"%(W)#

#The difference in the answer is due to rounding-off of values.

The angular speed is 78.54 rad/sec

The Torque is 152.8 Nm

The Inner radius is 81.4 mm

The Outer radius is 101.1 mm

The mean radius is 91.23 mm

The axial force is 1208 N

exa 22-8 - Page 594

In [8]:
from math import sin,pi
#semi-cone angle is given as 15 degree
k=sin(15*pi/180)#
u=0.3#
W=300#
Rm=90/2#
Tf=u*W*Rm/k#
Tf=Tf*(10**-3)#
I=0.4#
a=Tf/I#
N=1440#
w=2*pi*N/60#
t=w/a#
#During Slipping
theta1=w*t#
theta2=theta1/2#
U=Tf*(theta1-theta2)#
print "\nThe Torque is %0.3f Nm"%(Tf)#
print "\nThe angular acceleration is %0.3f rad/sec**2"%(a)#
print "\nThe angular speed is %0.1f rad/sec"%(w)#
print "\nThe time taken is %0.2f sec"%(t)#
print "\nThe Energy lost in friction is %0.0f Nm"%(U)#

The Torque is 15.648 Nm

The angular acceleration is 39.120 rad/sec**2

The angular speed is 150.8 rad/sec

The time taken is 3.85 sec

The Energy lost in friction is 4548 Nm

exa 22-9 - Page 595

In [9]:
from math import pi, sin
P=15*10**3#
Ka=1.25#
N=1500#
w=2*pi*N/60#
Tf=P/w#
d=(Tf*16/(50*pi))**(1/3)#
d=25#
Rm=5*d#
Pav=0.12#
u=0.22#
b=Tf/(pi*u*Pav*(Rm**2))#
b=40#
R1=Rm-(b*sin(15*pi/180)/2)#
R2=Rm+(b*sin(15*pi/180)/2)#
print "\nThe Torque is %0.2f Nm"%(Tf)#
print "\nThe shaft diameter is %0.0f mm"%(d)#
print "\nThe width is %0.0f mm"%(b)#
print "\nThe Inner radius is %0.1f mm"%(R1)#
print "\nThe Outer radius is %0.1f mm"%(R2)#

The Torque is 95.49 Nm

The shaft diameter is 25 mm

The width is 40 mm

The Inner radius is 119.8 mm

The Outer radius is 130.2 mm

exa 22-10 - Page 596

In [10]:
from math import pi
w2=2*pi*1400/60#
w1=0.8*w2#
P=40*10**3#
T=P/w2#
n=4#
T1=T/4#
R=0.16##Inner radius of drum
r=0.13##radial distance of each shoe from axis of rotation
u=0.22##coefficient of friction
x=u*r*R*((w2**2)-(w1**2))
m =T1/x#
l=R*pi/3#
N=T1/(R*u)#
p=1*10**5#
b=N/(p*l)*10**3#
print "\nThe full speed is %0.1f rad/sec"%(w2)#
print "\nThe engagement speed is %0.2f rad/sec"%(w1)#
print "\nThe number of shoes is %0.0f "%(n)#
print "\nThe Torque is %0.1f Nm"%(T)#
print "\nThe Torque per shoe is %0.1f Nm"%(T1)#
print "\nThe mass per shoe is %0.2f kg"%(m)#
print "\nThe length of friction lining is %0.5f m"%(l)#
print "\nThe width is %0.1f mm"%(b)#

The full speed is 146.6 rad/sec

The engagement speed is 117.29 rad/sec

The number of shoes is 4 

The Torque is 272.8 Nm

The Torque per shoe is 68.2 Nm

The mass per shoe is 1.93 kg

The length of friction lining is 0.16755 m

The width is 115.7 mm