Chapter 3: Mechanics Properties of Materials¶

Example 3.1 Page no 94¶

In [7]:
#Given
offset = 0.2 #
a_x = 0.0016 #in/in
a_y = 50     #ksi

#Calculations
#Modulus of Elasticity
E = a_y/(a_x*10**3) #E is the slope in GPa.
#Yield Strength
sigma_ys = 68       #ksi, Graphically, for a strain of 0.0016 mm/mm
#Ultimate Stress
sigma_u = 108       #ksi B is the peak of stress strain graph.
#Fracture Stress
ep_f = 0.23        #in/in
sigma_f = 90       #ksi from the graph.

#Display
print"The Modulus of Elasticity is   ",E,"ksi"
print"The Yield Strength from the graph ",sigma_ys,"ksi"
print"The Ultimate Stress from the graph is ",sigma_u,"ksi"
print"The Fracture Stress from the graph is ",sigma_f,"ksi"

The Modulus of Elasticity is    31.25 ksi
The Yield Strength from the graph  68 ksi
The Ultimate Stress from the graph is  108 ksi
The Fracture Stress from the graph is  90 ksi


Example 3.2 Page No 95¶

In [11]:
#Given
stress_b = 600          #MPa, stress
strain_b = 0.023       #mm/mm, strain
stress_a = 450         #Mpa, stress
strain_a = 0.006      #mm/mm, strain

#Calculations
#Permanent Strain
E = stress_a/strain_a
strain_cd = stress_b/E              #The recovered elastic strain
perm_strain = strain_b - strain_cd  #mm/mm

#Modulus of Resilience
ur_initial = (0.5*stress_a*strain_a)
ur_final = (0.5*stress_b*strain_cd)

#Display
print"The Permanent Strain is              =",perm_strain,"mm/mm"
print"The Initial Modulus of Resilience is = ",ur_initial,"MJ/mm**3"
print"The Final Modulus of Resilience is   = ",ur_final,"MJ/mm**3"

The Permanent Strain is              = 0.015 mm/mm
The Initial Modulus of Resilience is =  1.35 MJ/mm**3
The Final Modulus of Resilience is   =  2.4 MJ/mm**3


Example 3.3 Page No 96¶

In [13]:
#Given
E_al = 70*(10**3) #MPa,pressure
#The given dimension are
l_ab = 600.0     #mm
d_ab = 20.0     #mm
l_bc = 400.0    #mm
d_bc = 15.0     #mm

#Calculations
import math
a_ab = (math.pi/4.0)*(d_ab**2)     # Area of AB
a_bc = (math.pi/4.0)*(d_bc**2)
stress_ab = p/a_ab
stress_bc = p/a_bc
e_ab = stress_ab/E_al
e_bc = 0.045                        #mm/mm . From the graph for stress_bc
elongation = (l_ab*e_ab)+ (l_bc*e_bc)
strain_rec = stress_bc/E_al
e_og = e_bc-strain_rec
rod_elong = e_og*l_bc

#Display
print"The elongation of the rod when load is applied is",round(elongation,1),"mm"

The elongation of the rod when load is applied is 18.3 mm


Example 3.4 Page No 103¶

In [19]:
#Given
#Dimensions are
l_z = 1.5	 #m
l_y = 0.05	#m
l_x = 0.1 	#m

#Calculations
Aa= l_x*l_y
normal_stress_z = (P*(10**3))/Aa
Est = 200 #GPa - from the tables.
strain_z = (normal_stress_z)/(Est*(10**9))
axial_elong = strain_z*l_z          #elongation in the y direction
nu_st = 0.32                        #Poisson's Ratio - from the tables.
strain_x = -(nu_st)*(strain_z)     #strain in the x direction.
strain_y = strain_x
#Elongations
delta_x = strain_x*l_x
delta_y = strain_y*l_y

#Display
print"The change in the length (z direction)       =",axial_elong*10**6,"micrometer"
print"The change in the cross section (x direction)=",delta_x*10**6,"micrometer"
print"The change in the cross section (y direction)=",delta_y*10**6,"micrometer"

The change in the length (z direction)       = 120.0 micrometer
The change in the cross section (x direction)= -2.56 micrometer
The change in the cross section (y direction)= -1.28 micrometer


Example 3.5 Page No 105¶

In [23]:
#Given
#Refer to the graph of shear stress-strain of titanium alloy.
x_A = 0.008     #rad - x co-ordinate of A
y_A = 52	   #MPa - y co-ordinate of A
height = 2   	#mm
l = 3.0 		    #mm
b = 4.0 	        #mm

#Calculations
#Shear Modulus
G = y_A/x_A
#Proportional Limit
tou_pl = 52            #ksi Point A
#Ultimate Stresss
tou_u = 73             #ksi - Max shear stress at B
#Maximum Elastic Displacement
tanA= x_A               # tan theta is approximated as theta.
d = tanA*height
#Shear Force
A = l*b
V = tou_pl*A

#Display
print"The Shear Modulus                = ",G,"ksi"
print"The Proportional Limit           = ",tou_pl,"ksi"
print"The Ultimate Shear Stress        =  ",tou_u,"ksi"
print"The Maximum Elastic Displacement = ",d,"inch"
print"The Shear Force                  =  ",V,"kip"

The Shear Modulus                =  6500.0 ksi
The Proportional Limit           =  52 ksi
The Ultimate Shear Stress        =   73 ksi
The Maximum Elastic Displacement =  0.016 inch
The Shear Force                  =   624.0 kip


Example 3.6 Page No 106¶

In [2]:
#Given
d_o = 0.025 		#m, outside diameter
l_o =0.25 		    #m
delta = 1.2 		#mm
G_al = 26 		    #GPa
sigma_y = 440 		#MPa

#Calculations
import math
#Modulus of Elasticity
A = (math.pi/4)*(d_o**2)
avg_normal_stress = (F*10**3)/A
avg_normal_strain = delta/l_o
E_al = avg_normal_stress/ avg_normal_strain
E_al = E_al/10**6
#Contraction of Diameter
nu = (E_al/(2*G_al))-1
strain_lat = nu*(avg_normal_strain)
d_contraction = strain_lat* d_o

#Display
print"The Modulus of Elasticity is = ",round(E_al,0),"GPa"
print"The contraction in diameter due to the force = ",round(d_contraction,4),"mm"

70.0281749604
The Modulus of Elasticity is =  70.0 GPa
The contraction in diameter due to the force =  0.0416 mm