Chapter 9 : Mechanisms with Lower Pairs

Example 9.1 Page No : 245

In [1]:
import math 

# Variables:
c = 1.2
b = 2.7 			#m

#Solution:
#Calculating the inclination of the track arm to the longitudinal axis
alpha = math.tan(c/(2*b))*180/math.pi 			#degrees

#Results:
print " Inclination of the track arm to the longitudinal axis, alpha  =  %.1f degrees."%(alpha)

 Inclination of the track arm to the longitudinal axis, alpha  =  12.9 degrees.

Example 9.2 Page No : 251

In [12]:
import math

# variables
a = 180. - 160.    # degrees
N = 1500.          # r.p.m.; 
m = 12.            # kg ; 
k =  0.1           # m

# calculations
w = round(2*math.pi*N/60)
I = m*k**2
cos2theta = 2*math.sin(math.radians(a))**2/(2 - math.sin(math.radians(a))**2)
theta = math.degrees(math.acos(cos2theta))/2
dw1bydt = w**2*math.cos(math.radians(a)) * math.sin(math.radians(2*theta)) * math.sin(math.radians(a))**2 / ( 1 - math.cos(math.radians(theta))**2 * math.sin(math.radians(a))**2)**2
max_t = I * dw1bydt

# results
print "Maximum angular acceleration of the driven shaft : %.f rad/s**2"%dw1bydt
print "maximum torque required : %.f N-m"%max_t


# answers are different because of rounding error. please check using calculator.

Maximum angular acceleration of the driven shaft : 3080 rad/s**2
maximum torque required : 370 N-m

Example 9.3 Page No : 252

In [2]:
import math 

# Variables:
alpha = 18*math.pi/180 			#radians

#Solution:
#Maximum velocity is possible when
theta1 = 0.
theta2 = 180. 			#degrees

#Calculating the angle turned by the driving shaft when the velocity ratio is unity
theta3 = math.cos(math.sqrt((1-math.cos(alpha))/(math.sin(alpha)**2)))*180/math.pi 			#degrees
theta4 = 180-theta3 			#degrees

#Results:
print " Angle turned by the driving shaft when the velocity ratio is maximum, theta  =  %d degrees\
 or %d degrees."%(theta1,theta2)
print " Angle turned by the driving shaft when the velocity ratio is unity, theta  =  %.1f degrees or\
 %.1f degrees."%(theta3,theta4)

 Angle turned by the driving shaft when the velocity ratio is maximum, theta  =  0 degrees or 180 degrees.
 Angle turned by the driving shaft when the velocity ratio is unity, theta  =  43.2 degrees or 136.8 degrees.

Example 9.4 Page No : 252

In [15]:
import math 

# Variables:
N = 500. 			#rpm

#Solution:
#Calculating the angular velocity of the driving shaft
omega = 2*math.pi*N/60.0 			#rad/s
#Calculating the total fluctuation of speed of the driven shaft
q = 12./100*omega 			#rad/s
#Calculating the greatest permissible angle between the centre lines of the shafts
#alpha = math.cos((-(q/omega)+math.sqrt(0.12**2+4))/2.0)*180/math.pi			#degrees
cosalpha =((-(q/omega)+math.sqrt(0.12**2+4))/2.0)			#degrees
alpha = math.degrees(math.acos(cosalpha))

#Results:
print " Greatest permissible angle between the centre lines of the shafts, alpha  =  %.2f degrees."%(alpha)

 Greatest permissible angle between the centre lines of the shafts, alpha  =  19.64 degrees.

Example 9.5 Page No : 252

In [24]:
import math 

# Variables:
N = 1200.
q = 100. 			#rpm
#Solution:
#Calculating the greatest permissible angle between the centre lines of the shafts
cosalpha = ((-(100./1200)+math.sqrt(0.083**2+4))/2)
alpha = math.degrees(math.acos(cosalpha))                    #degrees
#Calculating the maximum speed of the driven shaft
N1max = N/cosalpha 			#rpm
#Calculating the minimum speed of the driven shaft
N1min = N*cosalpha			#rpm

#Results:
print " Greatest permissible angle between the centre lines of the shafts, alpha  =  %.1f degrees."%(alpha)
print " Maximum speed of the driven shaft, N1max  =  %d rpm."%(N1max)
print " Minimum speed of the driven shaft, N1min  =  %d rpm."%(N1min)

 Greatest permissible angle between the centre lines of the shafts, alpha  =  16.4 degrees.
 Maximum speed of the driven shaft, N1max  =  1251 rpm.
 Minimum speed of the driven shaft, N1min  =  1151 rpm.

Example 9.6 page no : 253

In [3]:
import math

# variables
N = 240.       # r.p.m 
w = 2 * math.pi * 240./60       #rad/s 
alpha = 20    
m = 55.          # kg ;
k = .150         
mm = 0.15        #m ; 
T1 = 200.         #N-m ; 
theta = 45.       #  ° ; 
q = 24.          # r.p.m.

# calculations
I = round(m * k**2,2) 
dw1bydt = round(-(w**2)*math.cos(math.radians(alpha))*math.sin(math.radians(2*theta))*math.sin(math.radians(alpha))**2 / (1- math.cos(math.radians(theta))**2 * math.sin(math.radians(alpha))**2)**2,2)
T2 = I * dw1bydt
T = T1 + T2
Tdash = T*math.cos(math.radians(alpha))/(1-math.cos(math.radians(theta))**2 * math.sin(math.radians(alpha))**2)
cosapha = (-0.1+math.sqrt((0.1**2)+4))/2
alpha = math.degrees(math.acos(cosapha))

# result
print "T' = %.1f N-m"%Tdash
print "Alpha a = %.1f degrees"%alpha

# rounding off error

T' = 102.7 N-m
Alpha a = 18.0 degrees

Example 9.7 Page No : 254

In [27]:
import math 

# Variables:
alpha = 20. 			#degrees
NA = 500.    			#rpm

#Solution:
#Calculating the maximum speed of the intermediate shaft
NBmax = NA/math.cos(math.radians(alpha)) 			#rpm
#Calculating the minimum speed of the intermediate shaft
NBmin = NA*math.cos(math.radians(alpha)) 			#rpm
#Calculating the maximum speed of the driven shaft
NCmax = NBmax/math.cos(math.radians(alpha)) 			#rpm
#Calculating the minimum speed of the driven shaft
NCmin = NBmin*math.cos(math.radians(alpha)) 			#rpm

#Results:
print " Maximum speed of the intermediate shaft( NBmax)  =  %.1f rad/s."%(NBmax)
print " Minimum speed of the intermediate shaft( NBmin)  =  %.2f rad/s."%(NBmin)
print " Maximum speed of the driven shaft( NCmax)  =  %.2f rad/s."%(NCmax)
print " Minimum speed of the driven shaft( NCmin)  =  %.1f rad/s."%(NCmin)

 Maximum speed of the intermediate shaft( NBmax)  =  532.1 rad/s.
 Minimum speed of the intermediate shaft( NBmin)  =  469.85 rad/s.
 Maximum speed of the driven shaft( NCmax)  =  566.24 rad/s.
 Minimum speed of the driven shaft( NCmin)  =  441.5 rad/s.