Chapter 4:Torsion

Example 4.2 page number 183

In [1]:
#Given
dia = 10   #diameter of shaft(A-C)
c = dia/2  #mm - Radius
T = 30     #N/mm -Torque in the shaft 
#Caliculations

J = 3.14*(dia**4)/32      #mm4
shear_T = T*c*pow(10,3)/J # The torsion shear in the shaft AC
import numpy as np 
print "The maximum shear due to torsion is ",round(shear_T,2),"Mpa"
arr_T = np.zeros((3,3))
arr_T[0][1]=round(shear_T,1) #arranging the elements in array
arr_T[1][0]=round(shear_T,1)
print "stress tensor matrix",(arr_T),

The maximum shear due to torsion is  152.87 Mpa
stress tensor matrix [[   0.   152.9    0. ]
 [ 152.9    0.     0. ]
 [   0.     0.     0. ]]

Example 4.3 page number 184

In [2]:
#Given 
dia_out = 20      #mm- outer diameter of shaft
dia_in  = 16      #mm- inner diameter of shaft 
c_out = dia_out/2 #mm - outer Radius of shaft  
c_in  = dia_in/2  #mm - inner radius of shaft 
T = 40            #N/mm -Torque in the shaft 
#caliculations

J = 3.14*((dia_out**4)- (dia_in**4))/32 #mm4
shear_T_max = T*c_out*pow(10,3)/J       # The maximum torsion shear in the shaft
shear_T_min = T*c_in*pow(10,3)/J        # The maximum torsion shear in the shaft
print "The maximum shear due to torsion is ",round(shear_T_max,2),"Mpa"
print "The minimum shear due to torsion is ",round(shear_T_min,2),"Mpa"

The maximum shear due to torsion is  43.15 Mpa
The minimum shear due to torsion is  34.52 Mpa

Example 4.4 page number 187

In [3]:
#Given
hp = 10         # horse power of motor 
f = 30          # given 
shear_T = 55    #Mpa - The maximum shearing in the shaft 
#caliculations

T = 119*hp/f            # N.m The torsion in the shaft 
#j/c=T/shear_T=K
k = T*pow(10,3)/shear_T #mm3
#c3=2K/3.14
c = pow((2*k/3),0.33)   #mm - The radius of the shaft 
diamter = 2*c           #mm - The diameter of the shaft
print "The Diameter of the shaft used is",round(diamter,2),"mm"

The Diameter of the shaft used is 15.26 mm

Example 4.5 page number 188

In [4]:
#Given 
hp = 200          #Horse power
stress_sh = 10000 #psi- shear stress
rpm_1 = 20.0      # The rpm at which this shaft1 operates 
rpm_2 = 20000.0   # The rpm at which this shaft2 operates
T_1= hp*63000.0/rpm_1 #in-lb Torsion due to rpm1
T_2= hp*63000/rpm_2   #in-lb Torsion due to rpm1
#caliculations 

#j/c=T/shear_T=K
k_1= T_1/stress_sh       #mm3
#c3=2K/3.14
c_1= pow((2*k_1/3),0.33) #mm - The radius of the shaft 
diamter_1 = 2*c_1        #mm - The diameter of the shaft
print "The Diameter of the shaft1 is",round(diamter_1,2),"mm"

#j/c=T/shear_T=K
k_2= T_2/stress_sh       #mm3
#c3=2K/3.14
c_2= pow((2*k_2/3),0.33) #mm - The radius of the shaft 
diamter_2 = 2*c_2        #mm - The diameter of the shaft
print "The Diameter of the shaft2 is",diamter_2,"mm"

The Diameter of the shaft1 is 6.87 mm
The Diameter of the shaft2 is 0.702590481015 mm

Example 4.7 page number 193

In [5]:
#Given 
T_ab = 0    #N.m - torsion in AB 
T_bc = 150  #N.m - torsion in BC
T_cd = 150  #N.m - torsion in CD
T_de = 1150 #N.m - torsion in DE
l_ab = 250 #mm - length of AB
l_bc = 200 #mm - length of BC
l_cd = 300 #mm - length of cd 
l_de = 500.0 #mm - length of de
d_1 = 25 #mm - outer diameter 
d_2 = 50 #mm - inner diameter
G = 80 #Gpa -shear modulus
#Caliculations 

J_ab = 3.14*(d_1**4)/32           #mm4
J_bc = 3.14*(d_1**4)/32           #mm4
J_cd = 3.14*(d_2**4 - d_1**4)/32  #mm4
J_de = 3.14*(d_2**4 - d_1**4)/32  #mm4
rad =  T_ab*l_ab/(J_ab*G)+ T_bc*l_bc/(J_bc*G)+ T_cd*l_cd/(J_cd*G)+ T_de*l_de/(J_de*G) # adding the maximum radians roteted in each module
print "The maximum angle rotated is ",rad,"radians " 

The maximum angle rotated is  0.0232628450106 radians

Example 4.9 Pagenumber 196

In [6]:
#given 
#its a statistally indeterminant 
#we will take of one of the support 
#Given 
T_ab = 0    #N.m - torsion in AB 
T_bc = 150  #N.m - torsion in BC
T_cd = 150  #N.m - torsion in CD
T_de = 1150 #N.m - torsion in DE
l_ab = 250  #mm - length of AB
l_bc = 200  #mm - length of BC
l_cd = 300  #mm - length of cd 
l_de = 500.0#mm - length of de
d_1 = 25 #mm - outer diameter 
d_2 = 50 #mm - inner diameter
#Caliculations 

J_ab = 3.14*(d_1**4)/32          #mm4
J_bc = 3.14*(d_1**4)/32          #mm4
J_cd = 3.14*(d_2**4 - d_1**4)/32 #mm4
J_de = 3.14*(d_2**4 - d_1**4)/32 #mm4
G = 80 #Gpa -shear modulus
rad =  T_ab*l_ab/(J_ab*G)+ T_bc*l_bc/(J_bc*G)+ T_cd*l_cd/(J_cd*G)+ T_de*l_de/(J_de*G) 
#now lets consider T_A then the torsion is only T_A
# T_A*(l_ab/(J_ab*G)+ l_bc/(J_bc*G)+ l_cd/(J_cd*G)+ l_de/(J_de*G)) +rad = 0
# since there will be no displacement 
T_A =-rad/(l_ab/(J_ab*G)+ l_bc/(J_bc*G)+ l_cd/(J_cd*G)+ l_de/(J_de*G)) #Torsion at A
T_B = 1150 - T_A                                                        #n-m F_X = 0 torsion at B
print "The Torsion at rigid end A is",round(T_A,2),"N-m"
print "The Torsion at rigid end B is",round(T_B,2),"N-m"

The Torsion at rigid end A is -141.72 N-m
The Torsion at rigid end B is 1291.72 N-m

Example 4.12 Pagenumber 202

In [7]:
#Given
dai_bc =  240   #mm- daimeter of '8'bolt circle 
dia =  dai_bc/8 #Diameter of each bolt
A =  0.25*(dia**2)*3.14 # Area of a bolt
S_allow  = 40           #Mpa - The maximum allowable allowable shear stress 
P_max =  (S_allow)*A    #N - The maximum allowable force 
D = 120.0               #mm - the distance from central axis 
T_allow =P_max*D*8      #N-m The allowable torsion on the 8 bolt combination 
print "The allowable torsion on the 8 bolt combination",T_allow ,"N-m"

The allowable torsion on the 8 bolt combination 27129600.0 N-m

Example 4.15 page number 211

In [8]:
#Given 
#AISC MANUALS
#approximated by three narrow tubes 
#J = Bbt^3
B = 0.33 # constant mentiones in AISC
#three rods 

#rod_1
t_1 = 0.605 #inch - Thickness 
b =  12.0 #inches - width 
J_1 = B*b*(t_1**3) #in4 - Torsion constant 

#rod_2
t_2 = 0.605 #inch - Thickness 
b =  12 #inches - width 
J_2 = B*b*(t_2**3) #in4 - Torsion constant 

#rod_3
t_3 = 0.390 #inch - Thickness 
b =  10.91 #inches - width 
J_3 = B*b*(t_3**3) #in4 - Torsion constant 

#Equivalent
J_eq = J_1+J_2+J_3  #in4 - Torsion constant 
print "the Equivalent Torsion constant is ",round(J_eq,2), "in4"

the Equivalent Torsion constant is  1.97 in4

Example 4.16 page number 214

In [9]:
#Given 
dia_out = 10       #mm- outer diameter of shaft
dia_in  = 8     #mm- inner diameter of shaft 
c_out = dia_out/2 #mm - outer Radius of shaft  
c_in  = dia_in/2  #mm - inner radius of shaft 
T = 40            #N/mm -Torque in the shaft 
#caliculations

J = 3.14*((dia_out**4)- (dia_in**4))/32 #mm4
shear_T_max = T*c_out*pow(10,3)/J       # The maximum torsion shear in the shaft
shear_T_min = T*c_in*pow(10,3)/J        # The maximum torsion shear in the shaft
print "The maximum shear due to torsion is ",round(shear_T_max,2),"Mpa"
print "The minimum shear due to torsion is ",round(shear_T_min,2),"Mpa"

The maximum shear due to torsion is  345.23 Mpa
The minimum shear due to torsion is  276.18 Mpa