Chapter 7:Torsion

Problem no 7.1,Page no.183

In [17]:
import math

#Initilization of variables

G=84 #Gpa #Modulus of Rigidity
N=110 #no. of revolution
#d*D**-1=0.6 #Ratio of inner diameter to outer diameter
sigma_s=63 #MPa #shear stress
L=3 #m #Length of shaft
P=590 #KW #Power

#Calculation

#P=2*pi*N*T_mean*60000**-1 #KW #Power
T_mean=P*60000*(2*pi*N)**-1 #N*m #Mean Torque

#I_p=p*32**-1*(D**4-d**4)

#After substituting value of d in above equation we get
#I_p=0.0272*pi*D**4 #m**4 #Polar moment of Inertia

T_max=1.2*T_mean #N*m #Max torque

#Using Relation
#T_max*T_p**-1=sigma_s*R**-1=G*theta*L**-1 

#After substituting values and simplifying we get

D=(5.7085*10**-3)**0.3333 #m #Diameter of shaft

theta=1.4*pi*180**-1 #radians

#theta=((T_max*L)*(G*10**9*I_p)) #radians

#After substituting values and simplifying we get
D_1=(1.0513*10**-3)**0.25

#Result
print"The Minimum external diameter is",round(D_1,2),"m"

The Minimum external diameter is 0.18 m

Problem no 7.2,Page no.184

In [54]:
import math

#Initilization of variables

P=295 #KW #Power
N=100 #R.p.m 
sigma_s=80 #MPa #shear stress


#Calculations

T_mean=((P*60000)*(2*pi*N)**-1) #N*m

#T_max=T_mean=(pi*D**3*sigma_s)*16**-1
D=((T_mean*16)*(pi*sigma_s*10**6)**-1)**0.333 #m #Diameter of solid shaft

#For hollow shaft
#I_p_h=pi*32**-1*(D_1**4-d_1**4) (equation 1)

#Now d_1=0.6*D_1
#substituting above value in equation 1,we get,

#I_p_h=0.0272*pi*D_1**4

#For solid shaft
#I_p_s=pi*32**-1*D**4

#T and sigma_s being the same then I_p*R**-1 will be the same for the two shafts
#Using relation  I_p_h*R_1**-1=I_p_s*R**-1

#Substituting values and simplifying we get

D_1=(D**3*0.8704**-1)**0.3333333 #m #External diameter of hollow shaft
d_1=0.6*D_1 #cm #Internal diameter of hollow shaft

A_s=pi*4**-1*(D*10**2)**2 #cm**2 #Area of solid shaft
A_h=pi*4**-1*(((D_1*10**2)**2)-((d_1*10**2)**2))

W=(A_s-A_h)*A_s**-1*100 #Percentage #Percentage saving in weight


#Result
print"Diameter of solid shaft is ",D,"m"
print"Percentage saving in weight is",round(W,2),"%"

Diameter of solid shaft is  0.121751208998 m
Percentage saving in weight is 29.8 %

Problem no 7.7,Page no.188

In [1]:
import math

#Initilization of variables

P_C=45 #KW Power aplled at C
P_B=15 #KW Power taken off at B
P_BA=30 #KW #Power transmitted across BA
G=85 #GPa

#Calculations (Part-1)

#For BC
P_1=45 #KW #Power across BC
N_1=200  #r.p.m
d_1=0.075 #m #diameter of shaft BC
L_BC=2 #m #Length of shaft BC


T_BC=60000*P_1*(2*pi*N_1)**-1 #N*m #Torque transmitted across BC
sigma_s_BC=16*T_BC*((pi*(d_1)**3)**-1)*10**-6 #N/m**2 #max shear stress in BC
I_p_BC=pi*32**-1*d_1**4 #m**4 #Polar M.I of BC
theta_1=T_BC*L_BC*(G*10**9*I_p_BC)**-1 #Radian #Max angle of twist theta_1 in BC of B relative to C

#For AB
P_2=30 #KW #Power across AB
N_2=200  #r.p.m
d_2=0.05 #m #diameter of shaft AB
L_BC=4 #m #Length of shaft AB


T_AB=60000*P_2*(2*pi*N_2)**-1 #N*m #Torque transmitted across AB
sigma_s_AB=16*T_AB*(pi*(d_2)**3)**-1*10**-6 #N/m**2 #max shear stress in AB
I_p_AB=pi*32**-1*d_2**4 #m**4 #Polar M.I of AB
theta_2=T_AB*L_BC*(G*10**9*I_p_AB)**-1 #Radian #Max angle of twist theta_1 in AB of A relative to B
C=(theta_1+theta_2)*180*pi**-1 #radian #Angle of Twist of gear


#Result
print"Angle of Twist of gear is",round(C,2),"Degree"
print"The maximum shear stress developed in the shaft AB is",round(sigma_s_AB,2)

 

Angle of Twist of gear is 7.23 Degree
The maximum shear stress developed in the shaft AB is 58.36

Problem no 7.8,Page no.189

In [16]:
import math

#Initilization of variables

L_BC=1.8 #m #Length of BC
L_AB=1.2 #m #Length of AB
sigma_s=70 #MPa #shear stress
d_1=0.05 #m #diameter of BC
d_2=0.1 #m #diameter of AB
r_BC=0.025 #cm #Radius of BC

#Calculations

I_p_BC=pi*32**-1*d_1**4 #m**4 #Polar M.I of BC
I_p_AB=pi*32**-1*d_2**4 #m**4 #Polar M.I od AB

#For BC
#theta_1=T*L_BC*(G*10**9*I_p_BC)**-1 #Angle of Twist of C relative to B
#After substituting and simplifying value, we get

#theta_1=3.4923*10**-5*T

#For AB
#theta_2=T*L_AB*(G*10**9*I_p_AB)**-1 #Angle of Twist of B relative to A
#After substituting and simplifying value, we get

#theta_2=1.45513*T

#sigma_s=T*R*(I_P)**-1 #The max shear stress in BC

#After substituting and simplifying value in above equation, we get

T=sigma_s*10**6*I_p_BC*r_BC**-1 
theta_1=3.4923*10**-5*T
theta_2=1.45513*10**-6*T
theta_c=theta_1-theta_2 #radian #total angle of twist

#Result
print"Total angle of Twist is",round(theta_c,3),"radian"

Total angle of Twist is 0.057 radian

Problem no 7.9,Page no.190

In [6]:
import math

#Initilization of variables

D=0.05 #m #Diameter of shaft
sigma_s_a=55 #MPa #shear stress of alloy
sigma_s_s=80 #MPa #shear stress of steel
P=185 #KW #Power 

#Calculations

#For alloy shaft,
#theta*L**-1=T*(G_A*I_p_A)**-1

#For steel shaft,
#theta*L*-1=I*(G_S*I_p_S)**-1

#After substituting and simplifying we get
d=(246.2*10**-8)**0.25 #m #Internal diameter of steel shaft

T_1=pi*16**-1*D**3*sigma_s_s*10**6 #N*m #For alloy shaft max torque
T_2=pi*16**-1*((D**4-d**4)*D**-1)*sigma_s_s*10**6 #N*m #For steel shaft max torque

#Permissible torque,T_2

#P=2*pi*N*T_2*(60000)**-1 

#After substituting we get
N=P*60000*(2*pi*T_2)**-1 #r.p.m #Speed

#Result
print"The speed at which the shafts to be driven is",round(N,0),"r.p.m"

The speed at which the shafts to be driven is 1485.0 r.p.m

Problem no 7.10,Page no.190

In [36]:
import math

#Initilization of variables

sigma_s=90 #MPa #shear stress of steel
sigma_d=60 #MPa #shear stress of duralumin
G_d=28 #GPa #modulus of rigidity of duralumin
G_s=84 #GPa #modulus of rigidity of steel
L=1 #m #Length of shaft

#Calculations

#theta*L**-1=sigma_s*(G_s*R_s)**-1=sigma_d*(G_d*R_d)**-1

#After substituting and simplifying,we get,
#D=2*d

#T_s=pi*16**-1*d**3*sigma_s #N*m #torque of steel
#T_d=pi*16*(((D**4-d**4)*D**4)**-1)*sigma_d #N*m #torque of duralumin

#After substituting and simplifying above two equations,we get,

#T_s=17.6714*10**6*d**3 #N*m
#T_d=88.3572*d**3 #N*m

#T=T_s+T_d #Total torque

#T=106.02875*10**6*d**3 

d=(700*(106.02875*10**6)**-1)**0.333 #m 
D=2*d #m
R_s=d*2**-1 #m

theta=(sigma_s*10**6*L*(G_s*10**9*R_s)**-1)*180*pi**-1 #degree #Angle of twist

#Result
print"The Angle of Twist is",round(theta,2),"Degree"

The Angle of Twist is 6.52 Degree

Problem no 7.11,Page no.191

In [65]:
import math

#Initilization of variables

P=4415 #KW #Power transmitted
N=110 #r.p.m
sigma_s=75 #MPs #shear stress
G=85 #GPa

#Calculations

#D=2*d 

T=P*60000*(2*pi*N)**-1 #N*m #Torque Transmitted

#T=pi*16**-1*(D**4-d**4)*D**-1*sigma_s #N*m

#After substituting and simplifying above equations,we get,

D=(T*16*pi**-1*(sigma_s*10**6)**-1)**(1*3**-1)
d=D*2**-1
X=5*(sigma_s*10**6)**2*(16*G*10**9)**-1

#U*V**-1 #Energy stored
#X=U*V**-1 #Energy stored #Notations has been changed

#Result
print"Diameter of shaft is:D",round(D,2),"cm"
print"                    :d",round(d,2),"cm"
print"Energy stored per cubic meter is",round(X,2),"N/m**2"

Diameter of shaft is:D 0.3 cm
                    :d 0.15 cm
Energy stored per cubic meter is 20680.15 N/m**2

Problem no 7.12,Page no.192

In [83]:
import math

#Initilization of variables

P=3680 #KW #Power transmitted
N=110 #r.p.m 
X=20000 #N*m #Energy stored
G=85 #GPa

#Calculations


#U*V**-1=X #Strain Energy per unit volume #Notification has been changed
#X=sigma_s**2*(4*G)**-1*((D**2+d**2)*(D**2)**-1)

T=P*60000*(2*pi*N)**-1 #N*m #Torque transmitted by shaft
sigma_s=(20000*3*G*10**9)**(1*2**-1) #MPa #shear stress of shaft

#T=pi*16**-1*((D**4-d**4)*D**-1)*sigma_s 

#After substituting and simplifying above equations,we get,

d=((T*16*3**0.5)*(pi*8*sigma_s)**-1)**(1*3**-1)
D=3**0.5*d 

#Result
print"Diameter of shaft is D=",round(D,2),"m"
print"                     d=",round(d,2),"m"

Diameter of shaft is D= 0.29 m
                     d= 0.17 m

Problem no 7.13,Page no.193

In [78]:
import math

#Initilization of variables

D=8 #cm #Diameter of bronze
d=5 #cm #diameter of steel shaft
R_b=4 #cm #Radius of bronze
R_s=2.5 #cm #Radius of steel shaft
sigma_b=40 #MPa #shear stress of bronze
sigma_s=65 #MPa #shear stress of steel shaft
N=500 #r.p.m
G_s=85 #GPa #Modulus of rigidity of steel
G_b=45 #GPa #Modulus of rigidity of bronze

#Calculations

I_p_s=pi*32**-1*(5*10**-2)**4 #m**4 #Polar M.I of Steel shaft
I_p_b=pi*32**-1*((8*10**-2)**4-(5*10**-2)**4) #m**4 #Polar M.I of Bronze shaft

#T*(G_b*I_p_b)**-1=T_s*(G_s*I_s)**-1

#After substituting and simplifying above equations,we get

#T_b=2.94*T_s

T_b=I_p_b*sigma_b*10**6*(R_b*10**-2)**-1 #N*m #Torque carried by bronze
T_s=I_p_s*sigma_s*10**6*(R_s*10**-2)**-1 #N*m #Torque carried by steel shaft
T_s_1=T_b*2.94**-1 #N*m 

T=T_b+T_s_1 #N*m #Total Torque
P=(2*pi*N*T*(60000)**-1) #KW #Power transmitted

#Result
print"Power transmitted by compound shaft is",round(P,2),"KW"

Power transmitted by compound shaft is 239.11 KW

Problem no 7.14,Page no.194

In [90]:
import math

#Initilization of variables

d=10 #cm #Diameter of shaft
r=5 #cm #radius of shaft
P=100 #KW #Power 
N=120 #r.p.m 
n=6 
L_k=14 #cm #Length of key
B_k=2.5 #cm #width of key
n=6 
d_b=2 #cm #Diameter of bolt
D_b=30 #cm #Diameter of bolt circle
R_b=15 #cm #radius 

#Calculations

T=(P*60000*(2*pi*N)**-1)*10**2 #N*m #Torque
I_p=pi*32**-1*d**4 #Polar M.I of shaft
sigma_s=T*r*(I_p)**-1 #N/cm**2
sigma_k=T*(L_k*B_k*r)**-1 #N/cm**2
sigma_b=T*4*(n*pi*d_b**2*R_b)**-1 #N/cm**2

#Result
print"shear stress in shaft",round(sigma_s,2),"N/cmm**2"
print"                Key",round(sigma_k,2),"N/cm**2"
print"              bolts",round(sigma_b,2),"N/cm**2" 

shear stress in shaft 4052.85 N/cmm**2
                Key 4547.28 N/cm**2
              bolts 2814.48 N/cm**2