# Chapter 2 : Stress and Strain—Axial Loading¶

## Example 2.01, Page number 69¶

In [1]:
import math

#Variable declaration
E=29*(pow(10,6))                                                   # Modulus of elasticity(psi)
L1=12                                                              # Length(in)
L2=12                                                              # Length(in)
L3=16                                                              # Length(in)
A1=0.9                                                             # Area(in**2)
A2=0.9                                                             # Area(in**2)
A3=0.3                                                             # Area(in**2)
P1=60*(pow(10,3))                                                  # Internal force(lb)
P2=15*(pow(10,3))                                                  # Internal force(lb)
P3=30*(pow(10,3))                                                  # Internal force(lb)

#Calculation
Delta=round((1/E)*(((P1*L1)/A1)+(-(P2*L2)/A2)+((P3*L3)/A3)),4)     # deformation of the steel rod(in)

#Result
print ('Deformation of the steel rod = %lf in' %Delta)

Deformation of the steel rod = 0.000000 in


## SAMPLE PROBLEM 2.1, Page number 70¶

In [1]:
import math
from sympy import symbols,solve

#Variable declaration
Eab=70*(pow(10,9))                                                 # Modulus of elasticity of AB(GPa)
Aab=500*(pow(10,-6))                                               # Cross sectional area of AB(mm**2)
Ecd=200*(pow(10,9))                                                # Modulus of elasticity of CD(GPa)
Acd=600*(pow(10,-6))                                               # Cross sectional area of CD(mm**2)
Fe=30                                                              # Force(kN)
Pb=-60*(pow(10,3))                                                 # Internal force in link AB(kN)
n=-1
Pd=90*(pow(10,3))                                                  # Internal force in link CD(kN)
Lab=0.3                                                            # Length of AB(m)
Lcd=0.4                                                            # Length of CD(m)
x=symbols('x')                                                     # Variable declaration
Se=symbols('Se')                                                   # Variable declaration

#Calculation
#Free Body: Bar BDE
Fcd=(30*0.6)/(0.2)                                                 # Force of tension(kN)
Fab=n*(30*0.4)/(0.2)                                               # Force of compression(kN)

#Case(a)
#Deflection of B.
Sb=((Pb*Lab)/(Aab*Eab))*(1000)                                              # Deflection of B(mm)

#Case(b)
#Deflection of D.
Sd=((Pd*Lcd)/(Acd*Ecd))*(1000)                                              # Deflection of D(mm)

#Case(c)
#Deflection of E.
x=solve(((200-x)/x)-((0.514)/(0.300)),x)                           # Distance HD(mm)
Se=solve(((400+73.7)/73.7)-(Se/0.300),Se)                          # Deflection of E(mm)

#Result
print('Case(a): Deflection of B = %1f psi' %Sb)
print('Case(b): Deflection of D = %lf psi' %Sd)
print('Case(c): Deflection of E = %lf psi' %Se[0])

Case(a): Deflection of B = -0.514286 psi
Case(b): Deflection of D = 0.300000 psi
Case(c): Deflection of E = 1.928223 psi


## SAMPLE PROBLEM 2.2, Page number 71¶

In [3]:
import math
from sympy import symbols,solve

#Variable declaration
Lb=18                                                        # Length of Bolt CD(in)
Pb=symbols('Pb')                                             # Force(kips)
Ab=((1/4.0)*(math.pi)*pow(0.75,2))                             # Area of cross section of CD and GH(in**2)
Eb=(29*(pow(10,6)))                                          # Modulus of elasticity of steel(psi)
Lr=12                                                        # Length of Bolt EF(in)
Pr=symbols('Pr')                                             # Force(kips)
Ar=((1/4.0)*(math.pi)*pow(1.5,2))                              # Area of cross section of EF(in**2)
Er=((10.6)*(pow(10,6)))                                      # Modulus of elasticity of aluminium(psi)
n=-1

#Calculation
# Bolts CD and GH
Sb=((Pb*Lb)/(Ab*Eb))                                         # Deformation

# Rod EF
Sr=n*((Pr*Lr)/(Ar*Er))                                       # Deformation

# Displacement of D Relative to B
Eq=(1/4.0)*(0.1)-Sb+Sr                                         # Because displacements are equal to the deformations of the bolts and of the rod

# Free Body: Casting B
#Pr-2*Pb=0                                                   # Submission of forces is zero

# Forces in Bolts and Rod
Pb=(solve((Eq).subs(Pr,(2*Pb)))[0])/(1000.0)                   # Force in the bolt(kips)
Pr=2*Pb                                                      # Force in th rod(kips)

# Stress in Rod
er=Pr/Ar                                                     # Stress in rod(ksi)

#Result
print('Stress in rod = %1f ksi' %er)

Stress in rod = 10.533169 ksi


## Example 2.02, Page number 78¶

In [4]:
import math
from sympy import symbols, solve

#Variable declaration
P1=symbols('P1')                                            # Axial force
L=symbols('L')                                              # Length of rod
A1=symbols('A1')                                            # Cross section of rod
E1=symbols('E1')                                            # Modulus of elasticity
P2=symbols('P2')                                            # Axial force
A2=symbols('A2')                                            # Cross section of rod
E2=symbols('E2')                                            # Modulus of elasticity
P=symbols('P')                                              # Total axial force

#Calculation
S1=(P1*L)/(A1*E1)                                           # Deforamation in rod
S2=(P2*L)/(A2*E2)                                           # Deformation in rod

P1=((P2)/(A2*E2))*(A1*E1)                                   # Equating deformations S1 and S2

P2=solve(P1-P+P2,P2)                                        # Solving the above equation for P2
P1=((P2[0]/(A2*E2))*(A1*E1))                                # Solving for P1
S1=(P1*L)/(A1*E1)
S2=(P2[0]*L)/(A2*E2)

#Result
print("Deformation in the rod :-")
print(S1)
print("Deformation in the tube :-")
print(S2)

Deformation in the rod :-
L*P/(A1*E1 + A2*E2)
Deformation in the tube :-
L*P/(A1*E1 + A2*E2)


## Example 2.03, Page number 79¶

In [5]:
import math
from sympy import symbols,solve

#Variable declaration
Ra=symbols('Ra')                                                # Reaction at A
Rb=symbols('Rb')                                                # Reaction at B
P=symbols('P')                                                  # Net reaction
S=symbols('S')                                                  # Total elongation
S1=symbols('S1')                                                # Elongation of AC
S2=symbols('S2')                                                # Elongation of BC
L1=symbols('L1')                                                # Length of AC
L2=symbols('L2')                                                # Length of BC
P1=symbols('P1')                                                # Internal force
P2=symbols('P2')                                                # Internal force
A=symbols('A')                                                  # Area of cross section
E=symbols('E')                                                  # Modulus of elasticity

#Calculation
S1=(P1*L1)/(A*E)                                                # Elongation of AC
S2=(P2*L2)/(A*E)                                                # Elongation of BC
S=S1+S2                                                         # Total elongation of the bar
S=0                                                             # Total elongation of the bar is zero

x=solve((Ra*L1-Rb*L2, Ra+Rb-P), Ra, Rb)                         # Solving the equations to find reactions at A and B.

e1=(x[Ra])/A                                                    # Stress in AC
e2=(x[Rb])/A                                                    # Stress in BC

#Result
print("Stress in AC :-")
print(S1)
print("Stress in BC :-")
print(S2)

Stress in AC :-
L1*P1/(A*E)
Stress in BC :-
L2*P2/(A*E)


## Example 2.04, Page number 80¶

In [6]:
import math
from sympy import symbols, solve

#Variable declaration
P1=0                                                                # Force in the first portion(N)
P2=600*(pow(10,3))                                                  # Force in the second portion(N)
P3=600*(pow(10,3))                                                  # Force in the third portion(N)
P4=900*(pow(10,3))                                                  # Force in the forth portion(N)
E=symbols('E')                                                      # Modulus of elasticity
Rb=symbols('Rb')                                                    # Reaction at B(N)
n=-1

A1=400*(pow(10,-6))                                                 # Area of cross section of first portion(m**2)
A2=400*(pow(10,-6))                                                 # Area of cross section of second portion(m**2)
A3=250*(pow(10,-6))                                                 # Area of cross section of third portion(m**2)
A4=250*(pow(10,-6))                                                 # Area of cross section of forth portion(m**2)

L1=0.150                                                            # Length of first portion(m)
L2=0.150                                                            # Length of second portion(m)
L3=0.150                                                            # Length of third portion(m)
L4=0.150                                                            # Length of forth portion(m)

#Calculation
Sl=((600*(pow(10,3)))/(400*(pow(10,-6))) + (600*(pow(10,3)))/(250*(pow(10,-6))) + (900*(pow(10,3)))/(250*(pow(10,-6))))*(0.150/E)        # Deformation(m)
P1=n*Rb                                                             # Force on first portion(N)
P2=n*Rb                                                             # Force on second portion(N)
A1=400*(pow(10,-6))                                                 # Area of cross section of first portion(m**2)
A2=250*(pow(10,-6))                                                 # Area of cross section of second portion(m**2)
L1=0.300                                                            # Length of first portion(m)
L2=0.300                                                            # Length of second portion(m)

Sr=(P1*L1)/(A1*E) + (P2*L2)/(A2*E)                                  # Deformation(m)

Rb=solve(Sl+Sr,Rb)                                                  # Reaction at B(N)

Ra=900+(Rb[0])                                                      # Reaction at A(N)

#Result
print('Reaction at A = %lf' %Ra)
print('Reaction at B = %lf' %Rb[0])

Reaction at A = 577823.076923
Reaction at B = 576923.076923


## Example 2.05, Page number 81¶

In [7]:
import math
from sympy import symbols, solve

#Variable declaration
E=200*(pow(10,9))                                                   # Modulus of elasticity(GPa)
Rb=symbols('Rb')                                                    # Reaction at B(N)
Ra=symbols('Ra')                                                    # Reaction at A(N)
#Calculation
Rb=(solve(((1.125*(pow(10,9)))/(200*(pow(10,9))))-(((1.95*(pow(10,3)))*Rb)/(200*(pow(10,9))))-(4.5*(pow(10,-3))),Rb))[0]/(10**3) # Solving for Rb(N)
Ra=solve(Ra-900+Rb,Ra)                                              # Solving for Ra(N)

#Result
print('Reaction at A = %lf' %Ra[0])
print('Reaction at B = %lf' %Rb)

Reaction at A = 784.615385
Reaction at B = 115.384615


## Example 2.06, Page number 84¶

In [6]:
import math
from sympy import symbols, solve

#Variable declaration
E=29*(pow(10,6))*(1.0)                                                           # Modulus of elasticity(psi)
Alpha=6.5*(pow(10,-6))                                                     # Constant(/F)
n=-1
L1=12                                                                      # Length of AC(m)
L2=12                                                                      # Length of CB(m)
A1=0.6                                                                     # Area of cross section of AC(m**2)
A2=1.2                                                                     # Area of cross section of BC(m**2)
Rb=symbols('Rb')                                                           # Reaction at B(N)
P1=Rb                                                                      # Internal force(N)
P2=Rb                                                                      # Internal force(N)
DELTAt=n*(50+75)                                                           # Diff in temperature(F)
AC=12                                                                      # Length of AC(m)
CB=12                                                                      # Length of CB(m)

#Calculation
St=Alpha*DELTAt*24                                                         # Deformation(in)
Sr=((P1*L1)/(A1*E)) +  ((P2*L2)/(A2*E))                                    # Deformation(in)

Rb=(solve(St+Sr,Rb))[0]/(1000.0)                                             # Reaction at B(kips)
S1=(Rb)/A1                                                                 # Stress in portion AC(ksi)
S2=(Rb)/A2                                                                 # Stress in portion CB(ksi)

Et=Alpha*DELTAt                                                            # Thermal strain(in./in.)
Ratio=S1/E                                                                 # Other component of eAC(in./in.)
Eac=Et + S1/E                                                              # Component of strain in AC(in)
Ecb=Et + S2/E                                                              # Component of strain in CB(in)

Sac=Eac*(AC)                                                               # Deformation of rod AC(in)
Scb=Ecb*(CB)                                                               # Deformation of rod CB(in)

#Result
print('Stress in portion AC = %lf' %S1)
print('Stress in portion CB = %lf' %S2)

Stress in portion AC = 31.416667
Stress in portion CB = 15.708333


## SAMPLE PROBLEM 2.3, Page number 86¶

In [7]:
import math
from sympy import symbols

#Variable declaration
Ldf=30                                                                 # Length of DF(in)
Adf=(1/4.0)*(math.pi)*(3/4.0)*(3/4.0)                                        # Cross sectional area of DF(in**2)
Lce=24                                                                 # Length of CE(in)
Ace=(1/4.0)*(math.pi)*(1/2.0)*(1/2.0)                                        # Cross sectional area of DF(in**2)
E=(10.6*pow(10,6))                                                     # Molulus of elasticity(psi)
Fce=symbols('Fce')                                                     # Force on CE(N)
Fdf=symbols('Fdf')                                                     # Force on DF(N)
Sd=symbols('Sd')                                                       # Force on CE(N)

#Calculation
# Case(a)
# Statics
Eq=Fce*12 +  Fdf*20  - 180                                             # Because submission of Mb is 0

# Geometry
Sc=0.6*Sd                                                              # By using similar triangles
Sa=0.9*Sd                                                              # By using similar triangles

# Deformations
Sc=(Fce*Lce)/(Ace*E)                                                   # Deformation at C(in)
Sd=(Fdf*Ldf)/(Adf*E)                                                   # Deformation at D(in)
Fce=0.333*Fdf                                                          # By solving Sc= 0.6*Sd we get this result

# Force in Each Rod
Fdf=(180/(((12)*(0.333))+20))                                          # By using Eq and the value of Fce
Fce=0.333*Fdf                                                          # Force in CE(N)

# Case(b)
# Deflections
Sd=(Fdf*(pow(10,3))*Ldf)/(Adf*E)                                       # Deflection of point D(in)
Sa=0.9*Sd                                                              # Deflection of point A(in)

#Result
print('Case(a): Force in rod DF = %.1f psi' %Fdf)
print('Case(a): Force in rod CE = %lf psi' %Fce)
print('Case(b): Deflection of point A = %lf psi' %Sa)

Case(a): Force in rod DF = 7.5 psi
Case(a): Force in rod CE = 2.497916 psi
Case(b): Deflection of point A = 0.043249 psi


## SAMPLE PROBLEM 2.4, Page number 87¶

In [8]:
import math
from sympy import symbols

#Variable declaration
Ldf=30                                                                 # Length of DF(in)
Adf=(1/4.0)*(math.pi)*(3/4.0)*(3/4.0)                                        # Cross sectional area of DF(in**2)
Lce=24                                                                 # Length of CE(in)
Ace=(1/4.0)*(math.pi)*(1/2.0)*(1/2.0)                                        # Cross sectional area of DF(in**2)
E=(10.6*pow(10,6))                                                     # Molulus of elasticity(psi)
Fce=symbols('Fce')                                                     # Force on CE(N)
Fdf=symbols('Fdf')                                                     # Force on DF(N)
Sd=symbols('Sd')                                                       # Force on CE(N)
Rb=symbols('Rb')                                                       # Reaction at B(m)

#Calculation
Ra=0.4*Rb                                                              # By considering the free body motion
L=0.3                                                                  # Length of rod DE(m)
DeltaT=30                                                              # Change in temperature(C)
Alpha=(20.9*pow(10,-6))                                                # Constant(/C)
A=(1/4)*(math.pi)*(pow(0.22,2))                                        # Area of cross-section of rod AC(m**2)
E=200                                                                  # Molulus of elasticity(psi)

St=L*(DeltaT)*Alpha                                                    # Deflection(m)

temp=(L/(A*E))
Sc=str(temp)+'*Ra'                                                     # Deflection(m)

Sd=str(0.4)+'*Sc'                                                      # Deflection(m)
Sd=4.74*pow(10,-9)                                                     # After computing the above equation

Sbd=str(4.04*pow(10,-9))+'*Ra'                                         # Deflection(m)

S1=str(5.94*pow(10,-9))+'*Rb'                                          # Deflection(m)

Rb=((188.1*pow(10,-6))/(5.94*pow(10,-9)))/(pow(10,3))                  # Reaction at B(m)

db=((round(Rb,1))/((1/4.0)*(math.pi)*(pow(0.03,2))))/(pow(10,3))         # Stress in cylinder(MPa)

#Result
print('Stress in cylinder = %.1f MPa' %db)

Stress in cylinder = 44.8 MPa


## Example 2.07, Page number 94¶

In [10]:
import math

#Variable declaration
P=12*(pow(10,3))                                                                  # Axial load(kN)
r=8*(pow(10,-3))                                                                  # Radius of the rod(m)
n=-1

#Calculation
A=(math.pi)*(r**2)                                                                # Cross sectional area of rod(m**2)
Sx=(P/A)                                                                          # Stress in cylinder(MPa)
Ex=(300/500.0)                                                                      # Strain()
Ey=(n*(2.4))/16.0                                                                   # Strain()

E=Sx/Ex                                                                           # Modulus of elasticity(GPa)
v=n*(Ey/Ex)                                                                       # Poissons ratio()

#Result
print('Modulus of elasticity = %.1f GPa' %E)
print('Poissons ratio = %.1f ' %v)

Modulus of elasticity = 99471839.4 GPa
Poissons ratio = 0.2


## Example 2.08, Page number 96¶

In [11]:
import math
from sympy import symbols, solve

#Variable declaration
E=29*(pow(10,6))                                                     # Modulus of elasticity(GPa)
v=0.29                                                               # Poissons ratio()
Sx=symbols('Sx')                                                     # Strain variable
Sy=symbols('Sy')                                                     # Strain variable
Sz=symbols('Sz')                                                     # Strain variable
Ex=symbols('Ex')                                                     # Stress variable
Ey=symbols('Ey')                                                     # Stress variable
Ez=symbols('Ez')                                                     # Stress variable
p=symbols('p')                                                       # pressure
n=-1
Sx=n*p                                                               # Strain at x
Sy=n*p                                                               # Strain at y
Sz=n*p                                                               # Strain at z
Ey=Ez=Ex                                                             # Equating all the stresses

#Calculation
# Case(a)
Ex=((n*p)*(1-(2*v)))/E                                               # Stress
Ex=(n*1.2*(pow(10,3)))/4                                             # Stress(in./in)
DELTAy=n*(300*(pow(10,-6)))*(2)                                       # Change in length(in)
DELTAz=n*(300*(pow(10,-6)))*(3)                                       # Change in length(in)
# Case(b)
p=(solve(p+((29*(pow(10,6)))*(n*300*(pow(10,-6))))/(1-0.58),p)[0])/(pow(10,3))          # pressure(ksi)

#Result
print('Change in length in y direction = %1f in' %DELTAy)
print('Change in length in z direction = %1f in' %DELTAz)
print('Pressure = %.1f ksi' %p)

Change in length in y direction = -0.000600 in
Change in length in z direction = -0.000900 in
Pressure = 20.7 ksi


## Example 2.09, Page number 98¶

In [17]:
import math

#Variable declaration
p=180                                                               # Hydrostatic pressure(MPa)
E=200                                                               # Modulus of elasticity(GPa)
v=0.29                                                              # Poissons ratio()

#Calculation
k=E/(3*(1-(2*v)))                                                   # Bulk modulus of steel(GPa)
e=-p/k                                                              # Dialation
V=80*40*60                                                          # Volume of block in unstressed state(mm**3)
DELTAv=(e*V)/(pow(10,3))                                            # change in volume per unit volume

# Results
print('Change in volume = %1f mm**3' %DELTAv)

Change in volume = -217.728000 mm**3


## Example 2.10, Page number 101¶

In [19]:
import math

#Variable declaration
G=90                                                                      # Modulus of rigidity(ksi)
disp=0.04                                                                 # Displacement of upper rod(in)
Lda=2                                                                     # Height of bar(in)
A=8*2.5                                                                   # Area of cross section(in**2)

#Calculation
Yxy=(disp/Lda)                                                            # Shearing strain(rad)
Txy=(90*(pow(10,3)))*(0.020)                                              # Shearing stress(psi)
P=(Txy*A)/(pow(10,3))                                                     # Force exerted on the upper plate(kips)

# Results
print('Shearing strain in rod=%1f rad' %Yxy)
print('Force exerted on the upper plate=%1f kips' %P)

Shearing strain in rod=0.020000 rad
Force exerted on the upper plate=36.000000 kips


## Example 2.11, Page number 105¶

In [12]:
import math

#Variable declaration
Ex=155.0                                                                      # Modulus of elasticity in x direction(GPa)
Ey=12.10                                                                      # Modulus of elasticity in y direction(GPa)
Ez=12.10                                                                      # Modulus of elasticity in z direction(GPa)
Vxy=0.248                                                                     # Poissons ratio in xy direction
Vxz=0.248                                                                     # Poissons ratio in xz direction
Vyz=0.458                                                                     # Poissons ratio in yz direction
n=-1
F=140*(pow(10,3))                                                             # Compressive load(kN)
L=0.060                                                                       # Length of cube(m)

#Calculation
#(a) Free in y and z Directions
Sx=(n*F)/(0.060*0.060)                                                        # Stress in x direction(MPa)
Sy=0                                                                          # Stress in y direction(MPa)
Sz=0                                                                          # Stress in z direction(MPa)
ex=Sx/Ex                                                                      # Lateral strains
ey=n*((Vxy*Sx)/Ex)                                                            # Lateral strains
ez=n*((Vxy*Sx)/Ex)                                                            # Lateral strains
DELTAx=ex*L                                                                   # Change in cube dimension in x direction(um)
DELTAy=ey*L                                                                   # Change in cube dimension in y direction(um)
DELTAz=ez*L                                                                   # Change in cube dimension in z direction(um)
#(b) Free in z Direction, Restrained in y Direction
Sx=n*38.89                                                                    # Stress in x direction(MPa)
Sy=(Ey/Ex)*(Vxy)*(Sx)                                                         # Stress in y direction(MPa)
Vyx=(Ey/Ex)*(Vxy)                                                             # Poissons ratio
ex=(Sx/Ex)-(((Vyx)*(Sy))/Ey)                                                  # Lateral strains in x direction
ey=0                                                                          # Lateral strains in y direction
ez=n*((Vxz*Sx)/Ex)-(((Vyz)*(Sy))/Ey)                                          # Lateral strains in z direction
DELTAx=ex*L*1000                                                              # Change in cube dimension in x direction(um)
DELTAy=ey*L                                                                   # Change in cube dimension in y direction(um)
DELTAz=ez*L*1000                                                              # Change in cube dimension in z direction(um)

# Results
print('Change in cube dimension in x direction=%1f um' %DELTAx)
print('Change in cube dimension in y direction=%1f um' %DELTAy)
print('Change in cube dimension in z direction=%1f um' %DELTAz)

Change in cube dimension in x direction=-14.981914 um
Change in cube dimension in y direction=0.000000 um
Change in cube dimension in z direction=5.443356 um


## SAMPLE PROBLEM 2.5, Page number 107¶

In [17]:
import math

#Variable declaration
d=9                                                                              # Diameter of the rod(in)
t=3/4.0                                                                            # Thickness of the rod(in)
ex=12                                                                            # Normal stresses(ksi)
ez=20                                                                            # Normal stresses(ksi)
E=(10*pow(10,6))                                                                 # Moduluus of elasticity(psi)
v=(1/3)                                                                          # Poissons ratio
V=15*15*(3/4.0)                                                                    # Volume(in**3)
n=-1

#Calculation
STRAINx=(1/(pow(10,7)*(1.0)))*(12-(20/3.0))*(1000)                                         # Strain in x direction(in./in)
STRAINy=n*(1/(pow(10,7)*1.0))*((12/3.0)+(20/3.0))*(1000)                                   # Strain in y direction(in./in)
STRAINz=(1/(pow(10,7)*(1.0)))*(20-(12/3.0))*(1000)                                         # Strain in z direction(in./in)

#Case(a)
DELTAba=(STRAINx)*(d)                                                            # Change in diameter(in)
#Case(b)
DELTAcd=(STRAINz)*(d)                                                            # Change in diameter(in)
#Case(c)
DELTAt=(STRAINy)*(t)                                                             # Change in thickness(in)
#Case(d)
e=(STRAINx+STRAINy+STRAINz)                                                      # Volume of the plate(in**3)
DeltaV=(e*V)

# Results
print('Change in diamter of rod AB =%1f in' %DELTAba)
print('Change in diamter of rod CD =%1f in' %DELTAcd)
print('Change in thickness =%1f in' %DELTAt)
print('Volume of the plate =%1f in**3' %DeltaV)

Change in diamter of rod AB =0.004800 in
Change in diamter of rod CD =0.014400 in
Change in thickness =-0.000800 in
Volume of the plate =0.180000 in**3


## Example 2.12, Page number 116¶

In [22]:
import math

#Variable declaration
D=60                                                                       # Width(mm)
d=40                                                                       # Width(mm)
K=1.82                                                                     # Stress-concentration factor
Smax=165                                                                   # Allowable normal stress(MPa)

#Calculation
eave=(165/1.82)                                                            # Average stress in the narrower portion(MPa)
P=round((40*10*eave)/(1000),1)                                             # Largest Axial Load(kN)

# Results
print('Largest Axial Load =%1f in' %P)

Largest Axial Load =36.300000 in


## Example 2.13, Page number 117¶

In [1]:
import math

#Variable declaration
L=500.0                                                                   # Length of rod(mm)
A=60                                                                      # Cross Sectional area(mm**2)
E=200                                                                     # Modulus of elasticity(GPa)
ey=300                                                                    # Yield Point(MPa)
DELTAc=7                                                                  # Stretch(mm)

#Calculation
ec=DELTAc/L                                                               # Maximum strain permitted on point C
ey=(ey*(pow(10,6)))/(E*(pow(10,9))*(1.0))                                       # Maximum strain permitted on point Y

# Results
print('Permanent set deformation =%1f mm' %DELTAd)

Permanent set deformation =6.250000 mm


## SAMPLE PROBLEM 2.6, Page number 123¶

In [20]:
from sympy import symbols, solve

#Variable declaration
Lce=5                                                          # Length of CE(m)
E=200.0                                                        # Modulus of elasticity(GPa)
ey=300                                                         # Stress(MPa)
Aad=400                                                        # Area of cross section of AD(mm**2)
Ace=500.0                                                      # Area of cross section of CE(mm**2)
STRAINad=ey/E                                                  # Stress(MPa)                                                         #

#Calculation
Pce=symbols('Pce')                                             # Variable declaration
Q=symbols('Q')                                                 # Variable declaration
d=ey                                                           # Maximum elastic deflection of point A
Q=(2*PadM)/(1000.0)                                                        # Maximum force(kN)
Sa1=STRAINad*Lad                                               # Maximum deflection of point A(mm)
Pce=120                                                        # This implies Pad is also 120
ece=Pce/Ace                                                    # Stress in rod CE(MPa)
STRAINce=ece/E                                                 # Strain CE
Sc1=STRAINce*Lce                                               # Deflection of point C(mm)
Sb=(1/2)*(Sa1+Sc1)                                             # Deflection of point B(mm)

# Since we must have Sb 10 mm, we conclude that plastic deformation will occur.

# Plastic Deformation
eAD=ey                              # Equating stresses
Sc=6                                # Deflection in C(mm)
Sa2=symbols('Sa2')                  # Variable declaration
Sa2=solve(((1/2)*(Sa2+6)-10),Sa2)   # The deflection Sa for which Sb = 10


Required maximum value of Q =240.0 mm