# Chapter20-Ball and Roller Bearings¶

## Ex1-pg431¶

In [1]:
import math

print('MACHINE DESIGN \n Timothy H. Wentzell, P.E. \n EXAMPLE-20.1 Page No.431\n');

##L10 design life
Cd=5050.;
Pd=2400.;
k=3.;
Ld1=(Cd/Pd)**k*10**6;
Ld=Ld1/(1750.*60.);

print'%s %.2f %s '%('\n L10 design life = ',Ld,' hr.');

MACHINE DESIGN
Timothy H. Wentzell, P.E.
EXAMPLE-20.1 Page No.431

L10 design life =  88.73  hr.


## Ex2-pg432¶

In [2]:
import math

print('MACHINE DESIGN \n Timothy H. Wentzell, P.E. \n EXAMPLE-20.2 Page No.432\n');

T=200.;
n=1750.;
L=T*n*60./10**6;
Pd=2400.;
Ld=21.;
Lc=1.;
k=1/3.;

Cd=Pd*(Ld/Lc)**k

print'%s %.2f %s '%('\n Dynamic load capacity required = ',Cd,' lb.');

print('\n Bearing 6211 meets this criterion.');

MACHINE DESIGN
Timothy H. Wentzell, P.E.
EXAMPLE-20.2 Page No.432

Dynamic load capacity required =  6621.42  lb.

Bearing 6211 meets this criterion.


## Ex3-pg434¶

In [3]:
import math

print('MACHINE DESIGN \n Timothy H. Wentzell, P.E. \n EXAMPLE-20.3 Page No.434\n');

R=1200.;
Ft=500.;
n=1500.;
L10=5000.;

##Assume thrust factor=1.6

Y=1.6;

Pd=0.56*R+Y*Ft;

Ld=n*L10*60./10**6;
Lc=1.;
k=3.;
Cd=Pd*(Ld/Lc)**(1./k);

##For bearing number 6215

Cd1=11400.;
Cs1=9700.;

##Verify the assumption for Y
Ft_Cs1=Ft/Cs1;

Y=(0.056-Ft_Cs1)*(1.99-1.71)/(0.056-0.028)+1.71;

Pd=0.56*R+Y*Ft;

Cd=Pd*(Ld/Lc)**(1./k);

if Cd>Cd1:
print('\n Since Cd of bearing < Cd required, So bearing number 6215 is not acceptable.');

##For bearing number 6216
Cd2=12600.;
Cs2=10500.;

Ft_Cs2=Ft/Cs2;
Y=(0.056-Ft_Cs2)*(1.99-1.71)/(0.056-0.028)+1.71;

Pd=0.56*R+Y*Ft;

Cd=Pd*(Ld/Lc)**(1/k);

if Cd<Cd2:
print('\n Since Cd of bearing > Cd required, So bearing number 6215 meets the design criteria.');

MACHINE DESIGN
Timothy H. Wentzell, P.E.
EXAMPLE-20.3 Page No.434

Since Cd of bearing < Cd required, So bearing number 6215 is not acceptable.

Since Cd of bearing > Cd required, So bearing number 6215 meets the design criteria.


## Ex4-pg436¶

In [4]:
import math

print('MACHINE DESIGN \n Timothy H. Wentzell, P.E. \n EXAMPLE-20.4 Page No.436\n');

##Thrust factor
Ft=300.;
Cs=2320.;
Ft_Cs=Ft/Cs;

Y=(0.17-Ft_Cs)*(1.45-1.31)/(0.17-0.11)+1.31;

print'%s %.2f %s '%('\n Thrust factor =  ',Y,'');

V=1.2;
X=0.56;
R=1000.;

P=V*X*R+Y*Ft;

Cd=3350.;
Pd=1095.;
k=3.;

Ld=(Cd/Pd)**k*10**6;

print'%s %.2f %s '%('\n Life = ',Ld,' revolutions.');

MACHINE DESIGN
Timothy H. Wentzell, P.E.
EXAMPLE-20.4 Page No.436

Thrust factor =   1.40

Life =  28634662.16  revolutions.