##CHAPTER 3 ILLUSRTATION 1 PAGE NO 102
##TITLE:FRICTION
##FIRURE 3.16(a),3.16(b)
import math
##===========================================================================================
##INPUT DATA
P1=180.## PULL APPLIED TO THE BODY IN NEWTONS
theta=30.## ANGLE AT WHICH P IS ACTING IN DEGREES
P2=220.## PUSH APPLIED TO THE BODY IN NEWTONS
##Rn= NORMAL REACTION
##F= FORCE OF FRICTION IN NEWTONS
##U= COEFFICIENT OF FRICTION
##W= WEIGHT OF THE BODY IN NEWTON
##==========================================================================================
##CALCULATION
F1=P1*math.cos(theta/57.3)## RESOLVING FORCES HORIZONTALLY FROM 3.16(a)
F2=P2*math.cos(theta/57.3)## RESOLVING FORCES HORIZONTALLY FROM 3.16(b)
## RESOLVING FORCES VERTICALLY Rn1=W-P1*sind(theta) from 3.16(a)
## RESOLVING FORCES VERTICALLY Rn2=W+P1*sind(theta) from 3.16(b)
## USING THE RELATION F1=U*Rn1 & F2=U*Rn2 AND SOLVING FOR W BY DIVIDING THESE TWO EQUATIONS
X=F1/F2## THIS IS THE VALUE OF Rn1/Rn2
Y1=P1*math.sin(theta/57.3)
Y2=P2*math.sin(theta/57.3)
W=(Y2*X+Y1)/(1-X)## BY SOLVING ABOVE 3 EQUATIONS
U=F1/(W-P1*math.sin(theta/57.3))## COEFFICIENT OF FRICTION
##=============================================================================================
##OUTPUT
print'%s %.1f %s %.1f %s '%('WEIGHT OF THE BODY =',W,'N''THE COEFFICIENT OF FRICTION =',U,'')
##CHAPTER 3 ILLUSRTATION 2 PAGE NO 103
##TITLE:FRICTION
##FIRURE 3.17
import math
##===========================================================================================
##INPUT DATA
THETA=45## ANGLE OF INCLINATION IN DEGREES
g=9.81## ACCELERATION DUE TO GRAVITY IN N/mm**2
U=.1## COEFFICIENT FRICTION
##Rn=NORMAL REACTION
##M=MASS IN NEWTONS
##f=ACCELERATION OF THE BODY
u=0.## INITIAL VELOCITY
V=10.## FINAL VELOCITY IN m/s**2
##===========================================================================================
##CALCULATION
##CONSIDER THE EQUILIBRIUM OF FORCES PERPENDICULAR TO THE PLANE
##Rn=Mgcos(THETA)
##CONSIDER THE EQUILIBRIUM OF FORCES ALONG THE PLANE
##Mgsin(THETA)-U*Rn=M*f.............BY SOLVING THESE 2 EQUATIONS
f=g*math.sin(THETA/57.3)-U*g*math.cos(THETA/57.3)
s=(V**2-u**2)/(2*f)## DISTANCE ALONG THE PLANE IN metres
##==============================================================================================
##OUTPUT
print'%s %.1f %s'%('DISTANCE ALONG THE INCLINED PLANE=',s,' m')
##CHAPTER 3 ILLUSRTATION 3 PAGE NO 104
##TITLE:FRICTION
##FIRURE 3.18
import math
##===========================================================================================
##INPUT DATA
W=500.## WEGHT IN NEWTONS
THETA=30.## ANGLE OF INCLINATION IN DEGRESS
U=0.2## COEFFICIENT FRICTION
S=15.## DISTANCE IN metres
##============================================================================================
Rn=W*math.cos(THETA/57.3)## NORMAL REACTION IN NEWTONS
P=W*math.sin(THETA/57.3)+U*Rn## PUSHING FORCE ALONG THE DIRECTION OF MOTION
w=P*S
##============================================================================================
##OUTPUT
print'%s %.1f %s'%('WORK DONE BY THE FORCE=',w,' N-m')
##CHAPTER 3 ILLUSRTATION 4 PAGE NO 104
##TITLE:FRICTION
##FIRURE 3.19(a) & 3.19(b)
import math
##===========================================================================================
##INPUT DATA
P1=2000.## FORCE ACTING UPWARDS WHEN ANGLE=15 degrees IN NEWTONS
P2=2300.## FORCE ACTING UPWARDS WHEN ANGLE=20 degrees IN NEWTONS
THETA1=15.## ANGLE OF INCLINATION IN 3.19(a)
THETA2=20.## ANGLE OF INCLINATION IN 3.19(b)
##F1= FORCE OF FRICTION IN 3.19(a)
##Rn1= NORMAL REACTION IN 3.19(a)
##F2= FORCE OF FRICTION IN 3.19(b)
##Rn2= NORMAL REACTION IN 3.19(b)
##U= COEFFICIENT OF FRICTION
##===========================================================================================
##CALCULATION
##P1=F1+Rn1 RESOLVING THE FORCES ALONG THE PLANE
##Rn1=W*cosd(THETA1)....NORMAL REACTION IN 3.19(a)
##F1=U*Rn1
##BY SOLVING ABOVE EQUATIONS P1=W(U*cosd(THETA1)+sind(THETA1))---------------------1
##P2=F2+Rn2 RESOLVING THE FORCES PERPENDICULAR TO THE PLANE
##Rn2=W*cosd(THETA2)....NORMAL REACTION IN 3.19(b)
##F2=U*Rn2
##BY SOLVING ABOVE EQUATIONS P2=W(U*cosd(THETA2)+sind(THETA2))----------------------2
##BY SOLVING EQUATIONS 1 AND 2
X=P2/P1
U=(math.sin(THETA2/57.3)-(X*math.sin(THETA1/57.3)))/((X*math.cos(THETA1/57.3)-math.cos(THETA2/57.3)))## COEFFICIENT OF FRICTION
W=P1/(U*math.cos(THETA1/57.3)+math.sin(THETA1/57.3))
##=============================================================================================
##OUTPUT
##print'%s %.1f %s'%('%f',X)
print'%s %.1f %s %.1f %s '%('COEFFICIENT OF FRICTION=',U,'' 'WEIGHT OF THE BODY=',W,' N')
##CHAPTER 3 ILLUSRTATION 5 PAGE NO 105
##TITLE:FRICTION
import math
##===========================================================================================
##INPUT DATA
d=5.## DIAMETER OF SCREW JACK IN cm
p=1.25## PITCH IN cm
l=50.## LENGTH IN cm
U=.1## COEFFICIENT OF FRICTION
W=20000.## LOAD IN NEWTONS
PI=3.147
##=============================================================================================
##CALCULATION
ALPHA=math.atan((p/(PI*d)/57.3))
PY=math.atan(U/57.3)
P=W*math.tan((ALPHA+PY)*57.)
P1=P*d/(2.*l)
##=============================================================================================
##OUTPUT
print'%s %.1f %s '%('THE AMOUNT OF EFFORT NEED TO APPLY =',P1,' N')
##CHAPTER 3 ILLUSRTATION 6 PAGE NO 106
##TITLE:FRICTION
import math
##===========================================================================================
##INPUT DATA
d=50.## DIAMETER OF SCREW IN mm
p=12.5## PITCH IN mm
U=0.13## COEFFICIENT OF FRICTION
W=25000.## LOAD IN mm
PI=3.147
##===========================================================================================
##CALCULATION
ALPHA=math.atan((p/(PI*d))/57.3)
PY=math.atan(U/57.3)
P=W*math.tan((ALPHA+PY)/57.3)## FORCE REQUIRED TO RAISE THE LOAD IN N
T1=P*d/2.## TORQUE REQUIRED IN Nm
P1=W*math.tan((PY-ALPHA)/57.3)## FORCE REQUIRED TO LOWER THE SCREW IN N
T2=P1*d/2.## TORQUE IN N
X=T1/T2## RATIOS REQUIRED
n=math.tan((ALPHA/(ALPHA+PY))/57.3)## EFFICIENCY
##============================================================================================
print'%s %.1f %s'%('RATIO OF THE TORQUE REQUIRED TO RAISE THE LOAD,TO THE TORQUE REQUIRED TO LOWER THE LOAD =',X,'')
##CHAPTER 3 ILLUSRTATION 7 PAGE NO 107
##TITLE:FRICTION
import math
##===========================================================================================
##INPUT DATA
d=39.## DIAMETER OF THREAD IN mm
p=13.## PITCH IN mm
U=0.1## COEFFICIENT OF FRICTION
W=2500.## LOAD IN mm
PI=3.147
##===========================================================================================
##CALCULATION
ALPHA=math.atan((p/(PI*d))/57.3)
PY=math.atan(U/57.3)
P=W*math.tan((ALPHA+PY)*57.3)## FORCE IN N
T1=P*d/2.## TORQUE REQUIRED IN Nm
T=2.*T1## TORQUE REQUIRED ON THE COUPLING ROD IN Nm
K=2.*p## DISTANCE TRAVELLED FOR ONE REVOLUTION
N=20.8/K## NO OF REVOLUTIONS REQUIRED
w=2.*PI*N*T/100.## WORKDONE BY TORQUE
w1=w*(7500.-2500.)/2500.## WORKDONE TO INCREASE THE LOAD FROM 2500N TO 7500N
n=math.tan(ALPHA/57.3)/math.tan((ALPHA+PY)/57.3)## EFFICIENCY
##============================================================================================
##OUTPUT
print'%s %.1f %s %.1f %s %.1f %s '%('workdone against a steady load of 2500N=',w,' N' 'workdone if the load is increased from 2500N to 7500N=',w1,' N' 'efficiency=',n,'')
##CHAPTER 3 ILLUSRTATION 8 PAGE NO 107
##TITLE:FRICTION
import math
##===========================================================================================
##INPUT DATA
W=50000.## WEIGHT OF THE SLUICE GATE IN NEWTON
P=40000.## POWER IN WATTS
N=580.## MAX MOTOR RUNNING SPEEED IN rpm
d=12.5## DIAMETER OF THE SCREW IN cm
p=2.5## PITCH IN cm
PI=3.147
U1=.08## COEFFICIENT OF FRICTION for SCREW
U2=.1## C.O.F BETWEEN GATES AND SCREW
Np=2000000.## NORMAL PRESSURE IN NEWTON
Fl=.15## FRICTION LOSS
n=1.-Fl## EFFICIENCY
ng=80.## NO OF TEETH ON GEAR
##===========================================================================================
##CALCULATION
TV=W+U2*Np## TOTAL VERTICAL HEAD IN NEWTON
ALPHA=math.atan((p/(PI*d))/57.3)##
PY=math.atan(U1/57.3)##
P1=TV*math.tan((ALPHA+PY)*57.3)## FORCE IN N
T=P1*d/2./100.## TORQUE IN N-m
Ng=60000.*n*P*10**-3./(2.*PI*T)## SPEED OF GEAR IN rpm
np=Ng*ng/N## NO OF TEETH ON PINION
##=========================================================================================
##OUTPUT
print'%s %.1f %s %.1f %s '%('NO OF TEETH ON PINION =',np,' say ',np+1,'')
##CHAPTER 3 ILLUSRTATION 9 PAGE NO 108
##TITLE:FRICTION
import math
##===========================================================================================
##INPUT DATA
d=5.## MEAN DIAMETER OF SCREW IN cm
p=1.25## PITCH IN cm
W=10000.## LOAD AVAILABLE IN NEWTONS
dc=6.## MEAN DIAMETER OF COLLAR IN cm
U=.15## COEFFICIENT OF FRICTION OF SCREW
Uc=.18## COEFFICIENT OF FRICTION OF COLLAR
P1=100.## TANGENTIAL FORCE APPLIED IN NEWTON
PI=3.147
##============================================================================================
##CALCULATION
ALPHA=math.atan((p/(PI*d))/57.3)##
PY=math.atan(U/57.3)##
T1=W*d/2*math.tan((ALPHA+PY)/100)*57.3## TORQUE ON SCREW IN NEWTON
Tc=Uc*W*dc/2./100.## TORQUE ON COLLAR IN NEWTON
T=T1+Tc## TOTAL TORQUE
D=2.*T/P1/2.*100.## DIAMETER OF HAND WHEEL IN cm
##============================================================================================
##OUTPUT
print'%s %.1f %s'%('SUITABLE DIAMETER OF HAND WHEEL =',D,' cm')
##CHAPTER 3 ILLUSRTATION 10 PAGE NO 108
##TITLE:FRICTION
import math
##===========================================================================================
##INPUT DATA
PI=3.147
d=2.5## MEAN DIA OF BOLT IN cm
p=.6## PITCH IN cm
beeta=55/2.## VEE ANGLE
dc=4.## DIA OF COLLAR IN cm
U=.1## COEFFICIENT OF FRICTION OF BOLT
Uc=.18## COEFFICIENT OF FRICTION OF COLLAR
W=6500.## LOAD ON BOLT IN NEWTONS
L=38.## LENGTH OF SPANNER
##=============================================================================================
##CALCULATION
##LET X=tan(py)/tan(beeta)
##y=tan(ALPHA)*X
PY=math.atan(U)*57.3
ALPHA=math.atan((p/(PI*d)))*57.3
X=math.tan(PY/57.3)/math.cos(beeta/57.3)
Y=math.tan(ALPHA/57.3)
T1=W*d/2.*10**-2*(X+Y)/(1.-(X*Y))## TORQUE IN SCREW IN N-m
Tc=Uc*W*dc/2.*10**-2## TORQUE ON BEARING SERVICES IN N-m
T=T1+Tc## TOTAL TORQUE
P1=T/L*100.## FORCE REQUIRED BY @ THE END OF SPANNER
##=============================================================================================
##OUTPUT
print'%s %.1f %s'%('FORCE REQUIRED @ THE END OF SPANNER=',P1,' N')
##CHAPTER 3 ILLUSRTATION 11 PAGE NO 109
##TITLE:FRICTION
import math
##===========================================================================================
##INPUT DATA
d1=15.## DIAMETER OF VERTICAL SHAFT IN cm
N=100.## SPEED OF THE MOTOR rpm
W=20000.## LOAD AVILABLE IN N
U=.05## COEFFICIENT OF FRICTION
PI=3.147
##==================================================================================
T=2./3.*U*W*d1/2.## FRICTIONAL TORQUE IN N-m
PL=2.*PI*N*T/100./60.## POWER LOST IN FRICTION IN WATTS
##==================================================================================
##OUTPUT
print'%s %.1f %s'%('POWER LOST IN FRICTION=',PL,' watts')
##CHAPTER 3 ILLUSRTATION 12 PAGE NO 109
##TITLE:FRICTION
import math
##===================================================================================
##INPUT DATA
PI=3.147
d2=.30## DIAMETER OF SHAFT IN m
W=200000.## LOAD AVAILABLE IN NEWTONS
N=75.## SPEED IN rpm
U=.05## COEFFICIENT OF FRICTION
p=300000.## PRESSURE AVAILABLE IN N/m**2
P=16200.## POWER LOST DUE TO FRICTION IN WATTS
##====================================================================================
##CaLCULATION
T=P*60./2./PI/N## TORQUE INDUCED IN THE SHFT IN N-m
##LET X=(r1**3-r2**3)/(r1**2-r2**2)
X=(3./2.*T/U/W)
r2=.15## SINCE d2=.30 m
c=r2**2.-(X*r2)
b= r2-X
a= 1.
r1=( -b+ math.sqrt (b**2. -4.*a*c ))/(2.* a);## VALUE OF r1 IN m
d1=2*r1*100.## d1 IN cm
n=W/(PI*p*(r1**2.-r2**2.))
##================================================================================
##OUTPUT
print'%s %.1f %s %.1f %s %.1f %s'%('EXTERNAL DIAMETER OF SHAFT =',d1,' cm''NO OF COLLARS REQUIRED =',n,'' '0 or ',n+1,'')
##CHAPTER 3 ILLUSRTATION 13 PAGE NO 111
##TITLE:FRICTION
import math
##===================================================================================
##INPUT DATA
PI=3.147
W=20000.## LOAD IN NEWTONS
ALPHA=120./2.## CONE ANGLE IN DEGREES
p=350000.## INTENSITY OF PRESSURE
U=.06
N=120.## SPEED OF THE SHAFT IN rpm
##d1=3d2
##r1=3r2
##===================================================================================
##CALCULATION
##LET K=d1/d2
k=3.
Z=W/((k**2.-1.)*PI*p)
r2=Z**.5## INTERNAL RADIUS IN m
r1=3.*r2
T=2.*U*W*(r1**3.-r2**3.)/(3.*math.sin(60/57.3)*(r1**2.-r2**2.))## total frictional torque in N
P=2.*PI*N*T/60000.## power absorbed in friction in kW
##================================================================================
print'%s %.1f %s %.1f %s %.1f %s'%('THE INTERNAL DIAMETER OF SHAFT =',r2*100,' cm' 'THE EXTERNAL DIAMETER OF SHAFT =',r1*100,' cm' 'POWER ABSORBED IN FRICTION =',P,' kW')
##CHAPTER 3 ILLUSRTATION 14 PAGE NO 111
##TITLE:FRICTION
import math
##===========================================================================================
##INPUT DATA
PI=3.147
P=10000.## POWER TRRANSMITTED BY CLUTCH IN WATTS
N=3000.## SPEED IN rpm
p=.09## AXIAL PRESSURE IN N/mm**2
##d1=1.4d2 RELATION BETWEEN DIAMETERS
K=1.4## D1/D2
n=2.
U=.3## COEFFICIENT OF FRICTION
##==========================================================================================
T=P*60000./1000./(2.*PI*N)## ASSUMING UNIFORM WEAR TORQUE IN N-m
r2=(T*2./(n*U*2.*PI*p*10**6.*(K-1.)*(K+1.)))**(1./3.)## INTERNAL RADIUS
##===========================================================================================
print'%s %.1f %s %.1f %s '%('THE INTERNAL RADIUS =',r2*100,' cm' 'THE EXTERNAL RADIUS =',K*r2*100,' cm')
##CHAPTER 3 ILLUSRTATION 14 PAGE NO 111
##TITLE:FRICTION
##===========================================================================================
##INPUT DATA
PI=3.147
n1=3.## NO OF DICS ON DRIVING SHAFTS
n2=2.## NO OF DICS ON DRIVEN SHAFTS
d1=30.## DIAMETER OF DRIVING SHAFT IN cm
d2=15.## DIAMETER OF DRIVEN SHAFT IN cm
r1=d1/2.
r2=d2/2.
U=.3## COEFFICIENT FRICTION
P=30000.## TANSMITTING POWER IN WATTS
N=1800.## SPEED IN rpm
##===========================================================================================
##CALCULATION
n=n1+n2-1.## NO OF PAIRS OF CONTACT SURFACES
T=P*60000./(2.*PI*N)## TORQUE IN N-m
W=2.*T/(n*U*(r1+r2)*10.)## LOAD IN N
k=W/(2.*PI*(r1-r2))
p=k/r2/100.## MAX AXIAL INTENSITY OF PRESSURE IN N/mm**2
##===========================================================================================
## OUTPUT
print'%s %.3f %s'%('MAX AXIAL INTENSITY OF PRESSURE =',p,' N/mm^2')