# Chapter 6: Friction¶

## Example 1, Page 185¶

In [23]:
import math

#Variable declaration
theta=60.#degrees
u1=0.15#between surfaces A annd B
u2=0.10#for the guides

#Calculations
phi=math.degrees(math.atan(u1))
phi1=math.degrees(math.atan(u2))
alpha=(theta+phi+phi1)/2#from 6.22, maximum efficiency is obtained at alpha
#from 6.23, maximum efficiency is given by nmax=(cos(theta+phi+phi1)+1)/(cos(theta-phi-phi1)+1)
nmax=(math.cos((theta+phi+phi1)*math.pi/180)+1)/(math.cos((theta-phi-phi1)*math.pi/180)+1)

#Results
print "Maximum efficiency = %.2f %% and it is obtained when alpha = %.2f degrees"%((nmax*100),alpha)

Maximum efficiency = 74.90 % and it is obtained when alpha = 37.12 degrees


## Example 3, Page 193¶

In [14]:
import math

#Variable declaration
#from equation 6.36 we know, M=(2/3)*u*W*(ri^3-r2^3)/(r1^2-r2^2)
u=0.04
W=16#tons
w=W*2240#lbs
r1=8#in
r2=6#in
N=120
P=50#lb/in^2

#Calculations
M=(2./3)*u*w*(r1**3-r2**3)/(r1**2-r2**2)
hp=M*2*math.pi*N/(12*33000)#horse power absorbed
#from fig 137,effective bearing surface per pad is calsulate from the dimensions to be 58.5 in^2
A=58.5#in^2
n=w/(A*P)
x=math.floor(n)

#Results
print "Horsepower absorbed = %.2f\nNumber of collars required = %.f"%(hp,x)

Horsepower absorbed = 19.24
Number of collars required = 12


## Example 4, Page 195¶

In [15]:
import math

#Variable declaration
ratio=1.25
u=.675
P=12#hp

#Calculations
#W=P*%pi*(r1^2-r2^2); Total axal thrust.
#M=u*W*(r1+r2); Total friction moemnt
#reducing the two equations and using ratio=1.25(r1=1.25*r2) we get, M=u*21.2*r2^3
ReqM=65#lb ft
RM=ReqM*12#lb in
r2=(RM/(u*P*math.pi*(1.25**2-1)))**(1./3)
r1=1.25*r2
d1=r1*2
d2=r2*2

#Results
print "The dimensions of the friction surfaces are:\nOuter Diameter= %.1f in\nInner Diameter= %.1f in"%(d1,d2)

The dimensions of the friction surfaces are:
Outer Diameter= 9.5 in
Inner Diameter= 7.6 in


## Example 5, Page 196¶

In [16]:
import math

#Variable declaration
P=20#lb/in^2
u=0.07#friction coefficient
N=3600#rpm
H=100#hp
r1=5#in

#Calculations
r2=0.8*r1#given
A=math.pi*(r1**2-r2**2)#the area of each friction surface
W=A*P#total axial thrust on plates
M=(1./2)*u*W*(r1+r2)#friction moment for each pair of contacts
T=H*33000*12/(2*math.pi*N)#total torque to be transmitted
x=(T/M)#effective friction surfaces required

#Results
print "\nNumber of effective friction surfaces required= %.f"%x

Number of effective friction surfaces required= 10


## Example 7, Page 198¶

In [21]:
import math

#Variable declaration
P=6 #tons
u=0.05
theta=60#degrees
CP=80
Stroke=16#in
OC=Stroke/2
r1=7#in
r2=15#in
r3=4.4#in

#Calculations
ro=u*r1
rc=u*r2
rp=u*r3
phi=math.degrees(math.asin(OC*math.sin(theta*math.pi/180)/CP))
alpha=math.degrees(math.asin((rc+rp)/CP))
#a) without friction
Qa=P/math.cos((phi)*math.pi/180)
Xa=OC*math.cos((30-phi)*math.pi/180)#tensile force transmitted along the eccentric rod when friction is NOT taken into account
Ma=Qa*Xa/12
#b) with friction
Qb=P/math.cos((phi-alpha)*math.pi/180)#tensile force transmitted along the eccentric rod when friction is taken into account
Xb=OC*math.cos((30-(phi-alpha))*math.pi/180)-(rc+ro)
Mb=Qb*Xb/12
n=Mb/Ma

#Results
print "Turning moment applied to OC:\na)Without friction= %.2f ton.ft\nb)With friction(u=0.05)= %.2f ton.ft"%(Ma,Mb)
print "\nThe efficiency of the mechanism is %.2f %%"%(n*100)

Turning moment applied to OC:
a)Without friction= 3.64 ton.ft
b)With friction(u=0.05)= 3.06 ton.ft

The efficiency of the mechanism is 84.17 %


## Example 8, Page 201¶

In [22]:
import math
import numpy as np

#Variable declaration
stroke=4#in
d=11.5#in
ds=4#in
dp=14#in
theta=math.pi
u1=.25
T1=350#lb
u2=0.1

#Calculations
k=math.e**(u1*theta)
T2=T1/k
Tor=(T1-T2)*(dp/2)#total resisting torque
#total resisting torque is also given by P*(r+2*(cos%pi/6))+u2*R*(ds/2)
#equating and putting values we get the following quadratic equation
p = [1,-1163,334200]
a=np.roots(p)

#Results
print "P=",a
print "The larger of two values is inadmissible. \n It corresponds to a negative sign in front of the second" \
"term on the \n right hand side of equation (1)"

P= [ 644.28733949  518.71266051]
The larger of two values is inadmissible.
It corresponds to a negative sign in front of the secondterm on the
right hand side of equation (1)