# Ch:18 Rolling bearings¶

## exa 18-1 - Page 507¶

In :
from __future__ import division
from math import sqrt, pi
Pr=16*10**3#
u=0.0011#
F=u*Pr#
r=20*10**-3#
#Let frictional moment be M
M=F*r#
N=1440#
w=2*pi*N/60#
Pf=M*w#
print "Pf is %0.2f W    "%(Pf)#

Pf is 53.08 W


## exa 18-2 - Page 508¶

In :
C=5590#
Ca=2500#
Pa=625#
Pr=1250#
V=1#
X=0.56#
Y=1.2#
P1=(X*V*Pr)+(Y*Pa)#
L1=(C/P1)**3#
V=1.2#
P2=(X*V*Pr)+(Y*Pa)#
L2=(C/P2)**3#
print "L1 is %0.1f million revolutions    "%(L1)#
print "\nL2 is %0.2f million revoltions    "%(L2)#

L1 is 57.3 million revolutions

L2 is 43.46 million revoltions


## exa 18-4 - Page 509¶

In :
P=20*10**3#
Co=22400#
C=41000#
Ln=(C/P)**3#
Lh=Ln*10**6/(720*60)#
print "Lh is %0.3f hrs    "%(Lh)#

Lh is 199.424 hrs


## exa 18-5 - Page 510¶

In :
from math import sqrt, pi
R1x=120#
R1y=250#
R2x=300#
R2y=400#
Lh=8000#
N=720#
Ln=Lh*60*N*10**-6#
R1=sqrt(R1x**2+R1y**2)#
R2=sqrt(R2x**2+R2y**2)#
Ks=1.5#
P1=R1*Ks#
P2=R2*Ks#
C1=P1*(Ln**(1/3))#
C2=P2*(Ln**(1/3))#
#let designation,d,D,B,C at bearing B1 be De1,d1,D1,B1,C1
d1=25#
D1=37#
B1=7#
C1=3120#
De1=61805#
#let designation,d,D,B,C at bearing B2 be De2,d2,D2,B2,C2
d2=25#
D2=47#
B2=8#
C2=7620#
De2=16005#
print "Designation of Bearing B1 is %0.0f     "%(De1)#
print "\nd1 is %0.0f mm    "%(d1)#
print "\nD1 is %0.0f mm    "%(D1)#
print "\nB1 is %0.0f mm    "%(B1)#
print "\nC1 is %0.0f N    "%(C1)#
print "\nDesignation of Bearing B2 is %0.0f     "%(De2)#
print "\nd2 is %0.0f mm    "%(d2)#
print "\nD2 is %0.0f mm    "%(D2)#
print "\nB2 is %0.0f mm    "%(B2)#
print "\nC2 is %0.0f N    "%(C2)#
print 'Bearing 61805 at B1 and 16005 at B2 can be installed.'

Designation of Bearing B1 is 61805

d1 is 25 mm

D1 is 37 mm

B1 is 7 mm

C1 is 3120 N

Designation of Bearing B2 is 16005

d2 is 25 mm

D2 is 47 mm

B2 is 8 mm

C2 is 7620 N
Bearing 61805 at B1 and 16005 at B2 can be installed.


## exa 18-6 - Page 511¶

In :
from math import tan, pi,sqrt
P=7500#
N=1440#
w=2*pi*N/60#
T=P/w#
r=0.2#
#Let T1-T2=t
t=T/r#
T2=t/2.5#
T1=3.5*T2#
R=0.125#
Ft=T/R#
Fr=Ft*tan(20*pi/180)#
# RD & RA are reaction forces calculated in vertical and horizontal directions from FBD by force equilibrium
RDv=186.5#
RAv=236.2#
RDh=36.2#
RAh=108.56#
RA=sqrt(RAv**2+RAh**2)#
RD=sqrt(RDv**2+RDh**2)#
Ks=1.4#
P1=RA*Ks#
P2=RD*Ks#
#let designation,d,D,B,C at bearing B1 be De1,d1,C1
d1=25#
C1=3120#
De1=61805#
#let designation,d,D,B,C at bearing B2 be De2,d2,C2
d2=25#

C2=2700#
De2=61804#
L1=(C1/P1)**3#
Lh1=L1*10**6/(720*60)#
L2=(C2/P2)**3#
Lh2=L2*10**6/(720*60)#
print "Lh1 is %0.0f hrs    "%(Lh1)#
print "\nLh2 is %0.0f hrs    "%(Lh2)#
#Incorrect value of P2 is taken in the book while calculating L2.

Lh1 is 14585 hrs

Lh2 is 24216 hrs


## exa 18-7 - Page 511¶

In :
from math import log
P=3500#
Lh=6000#
N=1400#
R98=0.98#
R90=0.9#
L98=Lh*60*N/10**6#
x=(log(1/R98)/log(1/R90))**(1/1.17)#
L90=L98/x#
C=P*L90**(1/3)#
print "C is %0.0f N    "%(C)#
#The difference in the value of C is due to rounding-off of value of L.

C is 44589 N


## exa 18-8 - Page 512¶

In :
from math import log
n=3#
P=3#
#Let Reliability of system be R
R=0.83#
L94=6#
R94=(R)**(1/n)#
x=(log(1/R94)/log(1/0.90))**(1/1.17)#
L90=L94/x#
C=P*L90**(1/3)#
print "C is %0.3f kN    "%(C)#
#The difference in the value of C is due to rounding-off of value of L.

C is 6.337 kN


## exa 18-9 - Page 512¶

In :
P1=3000#
P2=4000#
P3=5000#
N1=1440#
N2=1080#
N3=720#
t1=1/4#
t2=1/2#
t3=1/4#
n1=N1*t1#
n2=N2*t2#
n3=N3*t3#
N=(n1+n2+n3)#
Pe=(((n1*P1**3)+(n2*P2**3)+(n3*P3**3))/N)**(1/3)#
Lh=10*10**3#
L=Lh*60*N/10**6#
C=Pe*L**(1/3)#
print "C is %0.0f N    "%(C)#
#The difference in the value of C is due to rounding-off of value of Pe

C is 34219 N


## exa 18-10 - Page 513¶

In :
Co=695#
C=1430#
Pa1=200#
Pr1=600#
x=Pa1/Co#
y=Pa1/Pr1#
e=0.37+((0.44-0.37)*0.038/0.28)#
X=1#
Y=0#
P1=600#
Pa2=120#
Pr2=300#
X=0.56#
Y=1.2-(0.2*0.042/0.12)#
P2=(X*Pr2)+(Y*Pa2)#
N1=1440#
N2=720#
t1=2/3#
t2=1/3#
n1=N1*t1#
n2=N2*t2#
N=(n1+n2)#
Pe=(((n1*P1**3)+(n2*P2**3))/N)**(1/3)#
L=(C/Pe)**3#
Lh=L*10**6/(N*60)#
print "Lh is %0.2f hrs    "%(Lh)#
#The difference in the value of Lh is due to rounding-off of value of Pe

Lh is 227.66 hrs