Chapter10-Brakes and Dynamometers

Ex1-pg268

In [1]:
##CHAPTER 10 ILLUSRTATION 1 PAGE NO 268
##TITLE:Brakes and Dynamometers
import math
#calculate torque transmitted by the block brake
##===========================================================================================
##INPUT DATA
d=0.32;##Diameter of the drum in m
qq=90.;##Angle of contact in degree
P=820.;##Force applied in N
U=0.35;##Coefficient of friction


U1=((4.*U*math.sin(45/57.3))/((qq*(3.14/180.))+math.sin(90./57.3)));##Equivalent coefficient of friction
F=((P*0.66)/((0.3/U1)-0.06));##Force value in N taking moments
TB=(F*(d/2.));##Torque transmitted in N.m

print'%s %.4f %s'%('Torque transmitted by the block brake is ',TB,' N.m')

Torque transmitted by the block brake is  120.4553  N.m

Ex2-pg269

In [2]:
##CHAPTER 10 ILLUSRTATION 2 PAGE NO 269
##TITLE:Brakes and Dynamometers
import math
#calculate The bicycle travels a distance and makes  turns before it comes to rest
##===========================================================================================
##INPUT DATA
m=120.;##Mass of rider in kg
v=16.2;##Speed of rider in km/hr
d=0.9;##Diameter of the wheel in m
P=120.;##Pressure applied on the brake in N
U=0.06;##Coefficient of friction

F=(U*P);##Frictional force in N
KE=((m*(v*(5./18.))**2.)/2.);##Kinematic Energy in N.m
S=(KE/F);##Distance travelled by the bicycle before it comes to rest in m
N=(S/(d*3.14));##Required number of revolutions

print'%s %.1f %s %.1f %s'%('The bicycle travels a distance of ',S,' m'and'',N,'turns before it comes to rest')

The bicycle travels a distance of  168.8  59.7 turns before it comes to rest

Ex3-pg270

In [3]:
##CHAPTER 10 ILLUSRTATION 3 PAGE NO 270
##TITLE:Brakes and Dynamometers
import math
#evaluvate maximum torque absorbed
##===========================================================================================
##INPUT DATA
S=3500.;##Force on each arm in N
d=0.36;##Diamter of the wheel in m
U=0.4;##Coefficient of friction 
qq=100.;##Contact angle in degree

qqr=(qq*(3.14/180));##Contact angle in radians
UU=((4*U*math.sin(50/57.3))/(qqr+(math.sin(100./57.3))));##Equivalent coefficient of friction
F1=(S*0.45)/((0.2/UU)+((d/2.)-0.04));##Force on fulcrum in N
F2=(S*0.45)/((0.2/UU)-((d/2.)-0.04));##Force on fulcrum in N
TB=(F1+F2)*(d/2.);##Maximum torque absorbed in N.m

print'%s %.2f %s'%('Maximum torque absorbed is ',TB,' N.m')

Maximum torque absorbed is  1412.67  N.m

Ex4-pg271

In [4]:
##CHAPTER 10 ILLUSRTATION 4 PAGE NO 271
##TITLE:Brakes and Dynamometers
import math
#calculate The maximum braking torque on the drum
##===========================================================================================
##INPUT DATA
a=0.5;##Length of lever in m
d=0.5;##Diameter of brake drum in m
q=(5/8.)*(2*3.14);##Angle made in radians
b=0.1;##Distance between pin and fulcrum in m
P=2000.;##Effort applied in N
U=0.25;##Coefficient of friction

T=math.exp(U*q);##Ratios of tension
T2=((P*a)/b);##Tension in N
T1=(T*T2);##Tension in N
TB=((T1-T2)*(d/2.))/1000.;##Maximum braking torque in kNm

print'%s %.2f %s'%('The maximum braking torque on the drum is',TB,' kNm')

The maximum braking torque on the drum is 4.17  kNm

Ex5-pg271

In [5]:
##CHAPTER 10 ILLUSRTATION 5 PAGE NO 271
##TITLE:Brakes and Dynamometers
import math
#caculate the brake is self -locking and tension in the side 
##===========================================================================================
##INPUT DATA
q=220.;##Angle of contact in degree
T=340.;##Torque in Nm
d=0.32;##Diameter of drum in m
U=0.3;##Coefficient of friction

Td=(T/(d/2.));##Difference in tensions in N
Tr=math.exp(U*(q*(3.14/180.)));##Ratio of tensions
T2=(Td/(Tr-1.));##Tension in N
T1=(Tr*T2);##Tension in N
P=((T2*(d/2.))-(T1*0.04))/0.5;##Force applied in N
b=(T1/T2)*4.;##Value of b in cm when the brake is self-locking

print'%s %.2f %s  %.2f %s  %.2f %s '%('The value of b is ',b,' cm' 'when the brake is self-locking ' 'Tensions in the sides are ',T1,' N and',T2,' N')

The value of b is  12.65  cmwhen the brake is self-locking Tensions in the sides are   3107.70  N and  982.70  N

Ex6-pg272

In [6]:
##CHAPTER 10 ILLUSRTATION 6 PAGE NO 272
##TITLE:Brakes and Dynamometers
import math
#calculate torque required and thickness necessary to limit the tensile stress to 70 and secton of the lever taking stress to 60 mpa
##===========================================================================================
##INPUT DATA
d=0.5;##Drum diamter in m
U=0.3;##Coefficient of friction
q=250;##Angle of contact in degree
P=750;##Force in N
a=0.1;##Band width in m
b=0.8;##Distance in m
ft=(70*10**6);##Tensile stress in Pa
f=(60*10**6);##Stress in Pa
b1=0.1;##Distance in m

T=math.exp(U*(q*(3.14/180.)));##Tensions ratio
T2=(P*b*10.)/(T+1.);##Tension in N
T1=(T*T2);##Tension in N
TB=(T1-T2)*(d/2.);##Torque in N.m
t=(max(T1,T2)/(ft*a))*1000.;##Thickness in mm
M=(P*b);##bending moment at fulcrum in Nm
X=(M/((1/6.)*f));##Value of th**2
##t varies from 10mm to 15 mm. Taking t=15mm,
h=math.sqrt(X/(0.015))*1000.;##Section of the lever in m

print'%s %.1f %s %.1f %s %.1f %s'%('Torque required is ',TB,' N.m' 'Thickness necessary to limit the tensile stress to 70 MPa is ',t,' mm ''Section of the lever taking stress to 60 MPa is ',h,' mm')

Torque required is  861.7  N.mThickness necessary to limit the tensile stress to 70 MPa is  0.7  mm Section of the lever taking stress to 60 MPa is  63.2  mm

Ex7-pg273

In [7]:
##CHAPTER 10 ILLUSRTATION 7 PAGE NO 273
##TITLE:Brakes and Dynamometers
#calculate value of x and value of power/bd ratio 
import math
##===========================================================================================
##INPUT DATA
P1=30.;##Power in kW
N=1250.;##Speed in r.p.m
P=60.;##Applied force in N
d=0.8;##Drum diameter in m
q=310.;##Contact angle in degree
a=0.03;##Length of a in m
b=0.12;##Length of b in m
U=0.2;##Coefficient of friction
B=10.;##Band width in cm
D=80.;##Diameter in cm

T=(P1*60000.)/(2.*3.14*N);##Torque in N.m
Td=(T/(d/2.));##Tension difference in N
Tr=math.exp(U*(q*(3.14/180.)));##Tensions ratio
T2=(Td/(Tr-1.));##Tension in N
T1=(Tr*T2);##Tension in N
x=((T2*b)-(T1*a))/P;##Distance in m;
X=(P1/(B*D));##Ratio

print'%s %.3f %s %.3f %s'%('Value of x is ',x,' m '' Value of (Power/bD) ratio is ',X,'')

Value of x is  0.155  m  Value of (Power/bD) ratio is  0.037

Ex8-pg274

In [8]:
##CHAPTER 10 ILLUSRTATION 8 PAGE NO 274
##TITLE:Brakes and Dynamometers
import math
#calculate time required to bring the shaft to the rest from its running condition
##===========================================================================================
##INPUT DATA
m=80.;##Mass of flywheel in kg
k=0.5;##Radius of gyration in m
N=250;##Speed in r.p.m
d=0.32;##Diamter of the drum in m
b=0.05;##Distance of pin in m
q=260.;##Angle of contact in degree
U=0.23;##Coefficient of friction
P=20;##Force in N
a=0.35;##Distance at which force is applied in m

Tr=math.exp(U*q*(3.14/180.));##Tensions ratio
T2=(P*a)/b;##Tension in N
T1=(Tr*T2);##Tension in N
TB=(T1-T2)*(d/2.);##Torque in N.m
KE=((1/2.)*(m*k**2)*((2.*3.14*N)/60.)**2);##Kinematic energy of the rotating drum in Nm
N1=(KE/(TB*2.*3.14));##Speed in rpm
aa=((2*3.14*N)/60.)**2/(4.*3.14*N1);##Angular acceleration in rad/s**2
t=((2.*3.14*N)/60.)/aa;##Time in seconds

print'%s %.1f %s'%('Time required to bring the shaft to the rest from its running condition is ',t,' seconds')

Time required to bring the shaft to the rest from its running condition is  12.7  seconds

Ex9-pg275

In [9]:
##CHAPTER 10 ILLUSRTATION 9 PAGE NO 275
##TITLE:Brakes and Dynamometers
import math
#calculate Minimum force required  and Time taken to bring to rest 
##===========================================================================================
##INPUT DATA
n=12.;##Number of blocks
q=15.;##Angle subtended in degree
P=185.;##Power in kW
N=300.;##Speed in r.p.m
U=0.25;##Coefficient of friction
d=1.25;##Diamter in m
b1=0.04;##Distance in m
b2=0.14;##Distance in m
a=1.;##Diatance in m
m=2400.;##Mass of rotor in kg
k=0.5;##Radius of gyration in m

Td=(P*60000.)/(2.*3.14*N*(d/2.));##Tension difference in N
T=Td*(d/2.);##Torque in Nm
Tr=((1+(U*math.tan(7.5/57.3)))/(1.-(U*math.tan(7.5/57.3))))**n;##Tension ratio
To=(Td/(Tr-1.));##Tension in N
Tn=(Tr*To);##Tension in N
P=((To*b2)-(Tn*b1))/a;##Force in N
aa=(T/(m*k**2));##Angular acceleration in rad/s**2
t=((2*3.14*N)/60.)/aa;##Time in seconds

print'%s %.1f %s %.1f %s'%('Minimum force required is ',P,' N' 'Time taken to bring to rest is ',t,' seconds')

Minimum force required is  406.1  NTime taken to bring to rest is  3.2  seconds

Ex10-pg275

In [10]:
##CHAPTER 10 ILLUSRTATION 10 PAGE NO 275
##TITLE:Brakes and Dynamometers
import math
#calculate Maximum braking torque and Angular retardation of the drum and Time taken by the system to come to rest 
##===========================================================================================
##INPUT DATA
n=12.;## Number of blocks
q=16.;##Angle subtended in degrees
d=0.9;##Effective diameter in m
m=2000.;##Mass in kg
k=0.5;##Radius of gyration in m
b1=0.7;##Distance in m
b2=0.03;##Distance in m
a=0.1;##Distance in m
P=180.;##Force in N
N=360.;##Speed in r.p.m
U=0.25;##Coefficient of friction

Tr=((1.+(U*math.tan(8/57.3)))/(1.-(U*math.tan(8/57.3))))**n;##Tensions ratio
T2=(P*b1)/(a-(b2*Tr));##Tension in N
T1=(Tr*T2);##Tension in N
TB=(T1-T2)*(d/2.);##Torque in N.m
aa=(TB/(m*k**2.));##Angular acceleration in rad/s**2
t=((2.*3.14*N)/60.)/aa;##Time in seconds

print'%s %.2f %s %.2f %s %.2f %s '%('(i) Maximum braking torque is ',TB,'Nm ''(ii) Angular retardation of the drum is ',aa,' rad/s**2''(iii) Time taken by the system to come to rest is ',t,' s')

(i) Maximum braking torque is  2481.63 Nm (ii) Angular retardation of the drum is  4.96  rad/s**2(iii) Time taken by the system to come to rest is  7.59  s