Chapter 8:Transformation of stress and strain and Yield and Fracture criteria

Example 8.1 page number 405

In [2]:
#Given 
import math 
from math import radians
o = 22.5     #degrees , The angle of infetisimal wedge 
A = 1        #mm2 The area of the element 
A_ab = 1*(math.cos(radians(o))) #mm2 - The area corresponds to AB
A_bc = 1*(math.sin(radians(o))) #mm2 - The area corresponds to BC
S_1 = 3 #MN The stresses applying on the element 
S_2 = 2 #MN
S_3 = 2 #MN
S_4 = 1 #MN 
F_1 = S_1*A_ab # The Forces obtained by multiplying stress by their areas 
F_2 = S_2*A_ab
F_3 = S_3*A_bc
F_4 = S_4*A_bc
#sum of F_N = 0 equilibrim in normal direction 
N = (F_1-F_3)*(math.cos(radians(o))) + (F_4 - F_2)*(math.sin(radians(o)))

#sum of F_s = 0 equilibrim in tangential direction 

S = (F_2-F_4)*(math.cos(radians(o))) + (F_1 - F_3)*(math.sin(radians(o)))

Stress_Normal = N/A #Mpa - The stress action in normal direction on AB
Stress_tan = S/A    #Mpa - The stress action in tangential direction on AB
print "The stress action in normal direction on AB",round(Stress_Normal,2),"Mpa"
print "The stress action in tangential direction on AB",round(Stress_tan,2),"Mpa"

The stress action in normal direction on AB 1.29 Mpa
The stress action in tangential direction on AB 2.12 Mpa

Example 8.2 page number 413

In [3]:
#Given
o = -22.5 #degrees , The angle of infetisimal wedge 
A = 1     #mm2 The area of the element 
import math 
from math import radians
from numpy import array
A_ab = 1*(math.cos(radians(o))) #mm2 - The area corresponds to AB
A_bc = 1*(math.sin(radians(o))) #mm2 - The area corresponds to BC
S_1 = 3.0 #MN The stresses applying on the element 
S_2 = 2.0 #MN
S_3 = 2.0 #MN
S_4 = 1.0 #MN
#Caliculations 

F_1 = S_1*A_ab # The Forces obtained by multiplying stress by their areas 
F_2 = S_2*A_ab
F_3 = S_3*A_bc
F_4 = S_4*A_bc
#sum of F_N = 0 equilibrim in normal direction 
N = (F_1-F_3)*(math.cos(radians(o))) + (F_4 - F_2)*(math.sin(radians(o)))

#sum of F_s = 0 equilibrim in tangential direction 

S = (F_2-F_4)*(math.cos(radians(o))) + (F_1 - F_3)*(math.sin(radians(o)))

Stress_Normal = N/A #Mpa - The stress action in normal direction on AB
Stress_tan = S/A    #Mpa - The stress action in tangential direction on AB
print "a) The stress action in normal direction on AB",round(Stress_Normal,2),"Mpa"
print "a) The stress action in tangential direction on AB",round(Stress_tan,2),"Mpa"

#Part- b

S_max = (S_4+S_1)/2 + (((((S_4-S_1)/2)**2) + S_3**2)**0.5)   #Mpa - The maximum stress
S_min = (S_4+S_1)/2.0 - (((((S_4-S_1/2))**2) + S_3**2)**0.5) #Mpa - The minumum stress
k = 0.5*math.atan(S_3/((S_1-S_4)/2))                         #radians The angle of principle axis
k_1 = math.degrees(k)
k_2 = k_1+90 #The principle plane angles
print "b) The principle stress ",round(S_max,1),"Mpa tension"
print "b) The principle stress ",round(S_min,2),"Mpa compression"
print "b) The principle plane angles are",round(k_1,0),",",round(k_2,0),"degrees"

#part-c
#The maximum shear stress case
t_xy = (((((S_4-S_1)/2)**2) + S_3**2)**0.5) #Mpa - The maximum shear stress case
K = 0.5*math.atan((-(S_1-S_4)/(2*S_3)))     #radians The angle of principle axis
K_0 = math.degrees(K)
if K_0<0:
    K_1 = K_0+90
else:
    K_1 = K_0
K_2 = K_1+90 #PRinciple plain angles
T_xy = -((S_1-S_4)/2)*(math.sin(radians(2*K_1))) + ((S_4+S_1)/2)*(math.cos(radians(2*K_1))) # Shear stress
print "c) The maximum shear is ",round(T_xy,2),"Mpa" 
S_mat_a = array([round(S_max,1),round(S_min,1),0])                       #MPa maximum stress matrix
S_mat_b = array([(S_4+S_1)/2,round(T_xy,2),round(T_xy,2),(S_4+S_1)/2])   #MPa maximum stress matrix at maximum shear
print "a)",S_mat_a,"Mpa"
print "b)",S_mat_b,"Mpa"

a) The stress action in normal direction on AB 4.12 Mpa
a) The stress action in tangential direction on AB 0.71 Mpa
b) The principle stress  4.2 Mpa tension
b) The principle stress  -0.06 Mpa compression
b) The principle plane angles are 32.0 , 122.0 degrees
c) The maximum shear is  -2.24 Mpa
a) [ 4.2 -0.1  0. ] Mpa
b) [ 2.   -2.24 -2.24  2.  ] Mpa

Example 8.3 page number 421

In [4]:
#Given 
import math 
from math import radians 
S_x = -2 #Mpa _ the noraml stress in x direction
S_y = 4 #Mpa _ the noraml stress in Y direction
c = (S_x + S_y)/2 #Mpa - The centre of the mohr circle 
point_x = -2 #The x coordinate of a point on mohr circle
point_y = 4  #The y coordinate of a point on mohr circle
Radius = pow((point_x-c)**2 + point_y**2,0.5) # The radius of the mohr circle
S_1  = Radius +1#MPa The principle stress
S_2 = -Radius +1 #Mpa The principle stress
S_xy_max = Radius #Mpa The maximum shear stress
print "The principle stresses are",S_1 ,"Mpa",S_2,"Mpa"
print "The maximum shear stress",S_xy_max,"Mpa"

The principle stresses are 6.0 Mpa -4.0 Mpa
The maximum shear stress 5.0 Mpa

Example 8.4 page number 423

In [5]:
#Given
import math 
S_x = 3.0 #Mpa _ the noraml stress in x direction
S_y = 1.0 #Mpa _ the noraml stress in Y direction
c = (S_x + S_y)/2 #Mpa - The centre of the mohr circle 
point_x = 1 #The x coordinate of a point on mohr circle
point_y = 3  #The y coordinate of a point on mohr circle
#Caliculations 

Radius = pow((point_x-c)**2 + point_y**2,0.5) # The radius of the mohr circle
#22.5 degrees line is drawn 
o = 22.5 #degrees 
a = 71.5 - 2*o #Degrees, from diagram 
stress_n = c + Radius*math.sin(math.degrees(o)) #Mpa The normal stress on the plane 
stress_t =  Radius*math.cos(math.degrees(o)) #Mpa The tangential stress on the plane
print "The normal stress on the 221/2 plane ",round(stress_n,2),"Mpa"
print "The tangential stress on the 221/2 plane ",round(stress_t,2),"Mpa"

The normal stress on the 221/2 plane  4.82 Mpa
The tangential stress on the 221/2 plane  1.43 Mpa

Example 8.7 page number 437

In [1]:
import math
e_x = -500   #10-6 m/m The contraction in X direction
e_y = 300   #10-6 m/m The contraction in Y direction
e_xy = -600 #10-6 m/m discorted angle
centre = (e_x + e_y)/2  #10-6 m/m 
point_x = -500 #The x coordinate of a point on mohr circle
point_y = 300  #The y coordinate of a point on mohr circle
Radius = 500   #10-6 m/m - from mohr circle
e_1  = Radius +centre    #MPa The principle strain
e_2 = -Radius +centre    #Mpa The principle strain
k = math.atan(300.0/900) # from geometry
k_1 = math.degrees(k)
print "The principle strains are",e_1,"um/m",e_2,"um/m"
print "The angle of principle plane",round(k_1,2) ,"degrees"

The principle strains are 400 um/m -600 um/m
The angle of principle plane 18.43 degrees

Example 8.8 page number 441

In [10]:
#Given
e_0 = -500 #10-6 m/m 
e_45 = 200 #10-6 m/m 
e_90 = 300 #10-6 m/m
E = 200    #Gpa - youngs modulus of steel 
v = 0.3    # poissions ratio 
#Caliculations 

e_xy = 2*e_45 - (e_0 +e_90 ) #10-6 m/m from equation 8-40 in text
# from example 8.7
e_x = -500        #10-6 m/m The contraction in X direction
e_y = 300         #10-6 m/m The contraction in Y direction
e_xy = -600       #10-6 m/m discorted angle
centre = (e_x + e_y)/2  #10-6 m/m 
point_x = -500          #The x coordinate of a point on mohr circle
point_y = 300           #The y coordinate of a point on mohr circle
Radius = 500            #10-6 m/m - from mohr circle
e_1  = Radius +centre #MPa The principle strain
e_2 = -Radius +centre #Mpa The principle strain

stress_1 = E*(10**-3)*(e_1+v*e_2)/(1-v**2) #Mpa the stress in this direction 
stress_2 = E*(10**-3)*(e_2+v*e_1)/(1-v**2) #Mpa the stress in this direction 
print"The principle stresses are ",round(stress_1,2),"Mpa",round(stress_2,2),"MPa" 

The principle stresses are  48.35 Mpa -105.49 MPa