# Chapter16-Clutches and Brakes¶

## Ex1-pg358¶

In [1]:
import math

print('MACHINE DESIGN \n Timothy H. Wentzell, P.E. \n EXAMPLE-16.1 Page No.358\n');

##Torque capacity
f=0.3;
N=120.;
ro=12.;
ri=9.;
Tf=f*N*(ro+ri)/2.;

print'%s %.2f %s '%('\n Torque capacity = ',Tf,' in-lb.');
n=2000.;
##Power

Pf=Tf*n/63000.;

print'%s %.2f %s '%('\n Power = ',Pf,' hp.');

MACHINE DESIGN
Timothy H. Wentzell, P.E.
EXAMPLE-16.1 Page No.358

Torque capacity =  378.00  in-lb.

Power =  12.00  hp.


## Ex2-pg359¶

In [2]:
import math

print('MACHINE DESIGN \n Timothy H. Wentzell, P.E. \n EXAMPLE-16.2 Page No.359\n');

##Normal force
W=100.;
L=20.;
a=4.;
N=(W*L)/a;

print'%s %.2f %s '%('\n Normal force = ',N,' lb.');

##Torque friction
f=0.4;
D=12.;
Tf=f*N*D/2.;

print'%s %.2f %s '%('\n Torque friction = ',Tf,' in-lb.');

MACHINE DESIGN
Timothy H. Wentzell, P.E.
EXAMPLE-16.2 Page No.359

Normal force =  500.00  lb.

Torque friction =  1200.00  in-lb.


## Ex3-pg360¶

In [3]:
import math

print('MACHINE DESIGN \n Timothy H. Wentzell, P.E. \n EXAMPLE-16.3 Page No.360\n');

##For alpha=20 deg.
alpha=20.*(math.pi/180.);
f=0.35;
rm=12./2.;
Fa=75.;
Tf=(f*rm*Fa)/(math.sin(alpha)+f*math.cos(alpha));

print'%s %.2f %s '%('\n Torque capacity (alpha=20 deg.) = ',Tf,' in-lb.');

##For alpha=10 deg.
alpha=10.*(math.pi/180.);
Tf=(f*rm*Fa)/(math.sin(alpha)+f*math.cos(alpha));

print'%s %.2f %s '%('\n Torque capacity (alpha=10 deg.) = ',Tf,' in-lb.');

MACHINE DESIGN
Timothy H. Wentzell, P.E.
EXAMPLE-16.3 Page No.360

Torque capacity (alpha=20 deg.) =  234.75  in-lb.

Torque capacity (alpha=10 deg.) =  303.86  in-lb.


## Ex4-pg361¶

In [4]:
import math

print('MACHINE DESIGN \n Timothy H. Wentzell, P.E. \n EXAMPLE-16.4 Page No.361\n');

##Stopping rate
V=60.*5280./3600.;
Va=0.5*V;
D=400.;
t=D/Va;
a=V/t;

print'%s %.2f %s '%('\n Stopping rate = ',a,' ft/sec^2.');

##Stopping force
W=40000.;
g=32.2;
F=W*a/g;

##Torque
r=36./2.;
T=F*r;

print'%s %.2f %s '%('\n Torque = ',T,' in-lb.');

##For each wheel
T=T/10.;

##Braking normal force
rm=10.;
f=0.4;
N=T/(f*rm);

print'%s %.2f %s '%('\n Braking normal force = ',N,' lb.');

##Note-There is an error in the answer given in textbook

MACHINE DESIGN
Timothy H. Wentzell, P.E.
EXAMPLE-16.4 Page No.361

Stopping rate =  9.68  ft/sec^2.

Torque =  216447.20  in-lb.

Braking normal force =  5411.18  lb.


## Ex5-pg365¶

In [5]:
import math

print('MACHINE DESIGN \n Timothy H. Wentzell, P.E. \n EXAMPLE-16.5 Page No.365\n');
W=3500.;
V=73.;
g=32.2;
V=50.*5280./3600.;
V=round(V);

##Kinetic energy to be absorbed
KE=W*V**2/(2.*g);

print'%s %.2f %s '%('\n Kinetic energy to be absorbed = ',KE,' ft-lb.');

##Temperature rise
Uf=KE;
Wb=40.;
c=93.;
deltaT=Uf/(Wb*c);

print'%s %.2f %s '%('\n Temperature rise = ',deltaT,' deg.');

##Stopping time
a=20.;
t=V/a;

print'%s %.2f %s '%('\n Stopping time = ',t,' sec.');

##Frictional power
t=round(t*10.)/10.;
fhp=Uf/(550.*t);

print'%s %.2f %s '%('\n Frictional power = ',fhp,' hp.')

MACHINE DESIGN
Timothy H. Wentzell, P.E.
EXAMPLE-16.5 Page No.365

Kinetic energy to be absorbed =  289619.57  ft-lb.

Temperature rise =  77.85  deg.

Stopping time =  3.65  sec.

Frictional power =  142.32  hp.