# Chapter 3:Stresses And Strains¶

## Problem 3.1,Page no.54¶

In [65]:
import math

#Initilization of Variables

P=40 #mm #Force applied to stretch a tape
L=30 #m #Length of steel tape
A=6*1 #mm #Cross section area
E=200*10**9*10**-6 #KN/m**2 #Modulus of Elasticity

#Calculations

sigma_L=(P*L*10**3)*(A*E)**-1 #mm

#Result
print"The Elongation of steel tape is",round(sigma_L,4),"mm"

The Elongation of steel tape is 1.0 mm


## Problem 3.2,Page no.54¶

In [4]:
import math

#Initilization of VAriables

#D=(D_0-2) #cm #Inside Diameter of cyclinder
#A=(pi*(D_0-1)) #cm**2 #Area of cross-section
#L=(pi*(D_0-1)*5400) #N #Crushing load for column
F=6 #Factor of safety
T=1 #cm #wall thickness of cyclinder

#S=L*F**-1
#After Simplifying,we get
S=600*10**3

#Calculations

D_0=(S*F)*(pi*54000)**-1+1 #cm #Outside diameter of cyclinder

#Result
print"The outside Diameter of cyclinder is",round(D_0,2),"cm"

The outside Diameter of cyclinder is 22.22 cm


## Problem 3.3,Page no.56¶

In [25]:
import math

#Initilization of variables

P=800 #N #force applied to steel wire
L=150 #m #Length of steel wire
E=200 #GN/m**2 #Modulus of Elasticity
d=10 #mm #Diameter of steel wire
W=7.8*10**4 #N/m**3 #Weight Density of steel
#A=(pi*4**-1)*(d)**2 #m**2

#After simplifying Area,we get
A=7.85*10**-5 #m**2

#calculation (Part-1)

dell_L_1=(P*L*10**-3)*(A*E*10**9*10**-6)**-1 #mm

#calculation (Part-2)

#Elongation due to Weight of wire
dell_L_2=((pi*4**-1)*150*W*L*10**-3)*(2*A*E*10**7)**-1 #mm

#calculation (Part-3)

#Total Elongation of wire
dell_L_3=dell_L_1+dell_L_2

#Result
print"The Elongation due to 800N Load is",round(dell_L_1,2),"mm"
print"The Elongation due to Weight of wire is",round(dell_L_2,2),"mm"
print"Total Elongation of wire is",round(dell_L_3,2),"mm"

The Elongation due to 800N Load is 7.64 mm
The Elongation due to Weight of wire is 4.39 mm
Total Elongation of wire is 12.03 mm


## Problem 3.4,Page no.55¶

In [29]:
import math

#Intilization of variables

d=10 #mm #Diameter of Punching Hole
t=4 #mm #Thickness of Mild Steel Plate
tou=320 #N/mm**2 #Shear Strength of mild Steel

#Calculations

#Force Required for punching the hole
P=tou*pi*d*t #N

#Area of punch in contact with the plate surface
A=(pi*4**-1*d**2) #mm*2

#Compressive stress
sigma_c=P*A**-1 #N/mm*2

#Result
print"Force Required for punching the hole is",round(P,2),"N"
print"Compressive stress is",sigma_c,"N/mm*2"

Force Required for punching the hole is 40212.39 N
Compressive stress is 512.0 N/mm*2


## Problem 3.6,Page no.57¶

In [67]:
import math

#Initilization of variables

P=200*10**3 #N
L_1=0.10 #mm #Length of of portin AB
L_2=0.16 #mm #Length of of portin BC
L_3=0.12 #mm #Length of of portin CD
E=200*10**9 #N
d_1=0.1 #cm
d_2=0.08 #cm
d_3=0.06 #cm
A_1=(pi*4**-1)*(0.1)**2 #mm**2
A_2=(pi*4**-1)*(0.08)**2 #mm**2
A_3=(pi*4**-1)*(0.06)**2 #mm**2

#Calculations

dell_L_1=(P*L_1*10**3)*(A_1*E)**-1 #mm
dell_L_2=(P*L_2*10**3)*(A_2*E)**-1 #mm
dell_L_3=(P*L_3*10**3)*(A_3*E)**-1 #mm
dell_L=dell_L_1+dell_L_2+dell_L_3 #mm

#Result
print"Total Elongation of steel bar is",round(dell_L,3),"mm"

Total Elongation of steel bar is 0.087 mm


## Problem 3.7,Page no.57¶

In [44]:
import math

#Initilization of variables

#from F.B.D,we get
P_1=50 #KN
P_2=20 #KN
P_3=40 #KN

d=0.02 #mm #Diameter of steel bar
L_1=0.4 #mm
L_2=0.3 #mm
L_3=0.2 #mm
E=210*10**9 #N

#After simplifying Area,we get
A=pi*10**-4 #m**2 #Area of cross section

#Calculations

sigma_AB=P_1*1000*A #N/m**2
sigma_BA=P_2*1000*A #N/m**2
sigma_CD=P_3*1000*A #N/m**2
dell_L=((P_1*L_1+P_2*L_2+P_3*L_3)*(A*E)**-1)*10**6 #mm

#Result
print"Total Elongation of Steel bar is",round(dell_L,3),"mm"

Total Elongation of Steel bar is 0.515 mm


## Problem 3.8,Page no.58¶

In [26]:
import math
import numpy as np

#Initilization of variables

#R_a+R_c=25 #KN #R_a,R_b are reactions at supports A and C respectively
L_ab=2 #m
L_bc=3 #m

#Calculation

#From F.B.D,we get
#dell_L_AB=(R_a*L_AB)*(A*E)**-1 #Elongation of portion AB
#dell_L_BC=(R_c*L_BC)*(A*E)**-1 #Compression of portion BC

#After simplifying above equations we get,
#R_a=(1.5)*R_c #KN
#R_a+R_c=25 #KN
#Solving the above simultaneous equations using matrix method
A=np.array([[1,-1.5],[1,1]]) #Here the coefficients of the first equations of unknowns are setup
B=np.array([0,25]) #Here the RHS of both equations are setup
C=np.linalg.solve(A,B)

#print C[0] #Prints the first element in the vector C
#print C[1] #Prints the second element in the vector C

#Result
print"The reaction at support A is",round(C[0],2),"KN"
print"The reaction at support C is",round(C[1],2),"KN"

The reaction at support A is 15.0 KN
The reaction at support C is 10.0 KN


## Problem 3.9,Page no.59¶

In [17]:
import math

#Initilization of variables

#P is the force acting on the bar BC compressive in nature and force on AB is (100-P) Tensile in nature
E=200*10*9 #N
A_1=3*10**-4 #cm**2 #Area of AB
A_2=4*10**-4 #cm**2 #Area of BC
L=1.5 #cm #Length of bar

#Calculations

#The total elongation of bar
#(((100-P)*10**3*1.5)*(3*10**-4*E)**-1)-((P*10**3*1.5)*(4*10**-4*E)**-1)=0

#The total elongation of bar is limited to 1
#(25-0.4375*P)*10**-4=1*10**-3

#After simplifying above equation we get,
P=-(10-25)*0.4375**-1 #KN #Total elongation of bar
F_AB=100-P #KN #force in AB
F_BC=P #KN #Force in BC
sigma_AB=(((F_AB)*(3*10**-4)**-1)*10**-3) #KN #Stress in AB
sigma_BC=((F_BC)*(4*10**-4)**-1*10**-3) #KN #Stress in Bc

#Result
print"F_AB",round(F_AB,2),"KN"
print"F_BC",round(F_BC,2),"KN"
print"sigma_AB",round(sigma_AB,2),"KN"
print"sigma_BC",round(sigma_BC,2),"KN"

F_AB 65.71 KN
F_BC 34.29 KN
sigma_AB 219.05 KN
sigma_BC 85.71 KN


## Problem 3.12,Page no.61¶

In [11]:
import math

#Initilization of variables

d=20 #mm #steel rod diameter
n=4 #number of steel rod
sigma_c=4 #N/mm**2 #stress in concrete
#E_S*E_c**-1=15

#Calculations

A_s=4*pi*4**-1*d**2 #mm**2 #Area os steel rod
sigma_s=15*sigma_c #N/mm**2 #stress in steel

#P=sigma_s*A_s+sigma_c*A_c

#After substituting and simplifying above equation we get,

A_c=(P*10**3-sigma_s*1256)*(sigma_c)**-1 #mm**2 #Area of the concrete
X=(A_s+A_c)**0.5 #mm #Total cross sectional area
P_s=A_s*sigma_s*10**-3 #KN #Load carried by steel

#Result
print"stress induced in steel is",round(sigma_s,3),"KN"
print"cross sectional area of column is",round(X,2),"mm"

Load carried by steel is 75.4 KN
stress induced in steel is 60.0 KN
cross sectional area of column is 327.74 mm


## Problem 3.13,Page no.62¶

In [1]:
import math

#Initilization of variables

A_s=500 #mm**2
E_s=200000
E_al=80000
A_al=1000

#Calculations

#(P_al*L_al)*(A_al*E_al)**-1+(P_s*L_s)*(A_s*E_s)**-1=1*2**-1

P=1*1000**-1*((A_s*E_s*A_al*E_al)*(A_s*E_s+A_al*E_al)**-1) #N
P_s=P_al=P #N
sigma_t=P_s*A_s**-1 #N/mm**2 #Tensile stress in bolt
sigma_c=P_al*A_al**-1 #N/mm**2 #Compressive stress in Aluminium tube

#result
print"Tensile stress in bolt is",round(sigma_t,2),"N/mm**2"
print"Compressive stress in Aluminium tube is",round(sigma_c,2),"N/mm**2"

Tensile stress in bolt is 88.89 N/mm**2
Compressive stress in Aluminium tube is 44.44 N/mm**2


## Problem 3.14,Page no.63¶

In [1]:
import math

#Initilization of variables

A=1600 #mm**2 #Area of the Bar
dell_L=0.4 #mm #Contraction of metal bar
L=200 #mm #Length of metal bar
sigma_t=0.04 #mm #Guage Length
t=40

#Calculations

sigma_L=dell_L*L**-1
E=((P*L)*(A*dell_L)**-1*10**-3) #N/mm**2 #Young's Modulus
m=t*sigma_t**-1*sigma_L

#Result
print"The value of Young's Modulus  is",round(E,2),"N/mm*2"
print"The value of Poissoin's ratio is",round(m,2)

The value of Young's Modulus  is 150.0 N/mm*2
The value of Poissoin's ratio is 2.0


## Problem 3.15,Page no.63¶

In [21]:
import math

#Initilization of variables

A_s=0.003848 #m**2 #Area of steel bar
A_al=0.003436 #m**2 #Area of Aluminium tube
E=220*10*9 #N #Young's modulus of steel
E=70*10*9 #N #Young's modulus of aluminium
P=600*10**3 #N #Load applied to the bar
#dell_L_al-dell_L_s=0.00015 #mm #difference between strain in aluminium bar and steel bar

#Calculations

#Let the aluminium tube be compressed by dell_L_al and steel bar by by dellL_s
#dell_L_al=sigma_al*E_al**-1*L_al
#dell_L_s=sigma_s*E_s**-1*L_s

#After substituting and simplifying above equation we get,
#((sigma_al*70**-1)-(sigma_s*220**-1))=300000               #(equation 1)

#After simplifying above equation we get,
#sigma_al=17462.165*10**4-1.1199*sigma_s                    #(equation 2)

#Now substituting sigma_al in equation(1)
#((17462.165*10**4-1.1199*sigma_s)*(70)**-1)-(sigma_s*220**-1)=300000

#After simplifying above equation we get,

sigma_s=-((300000-249.4594*10**4)*0.0205444**-1)*10**-6 #MN/m**2 #stress developed in steel bar
#sigma_al=17462.165*10**4-1.1199*sigma_s
sigma_al=(17462.165*10**4-1.1199*106822005.02)*10**-6

#Result
print"stress developed in steel bar is",round(sigma_s,2),"MN/m**2"
print"stress developed in aluminium bar is",round(sigma_al,2),"MN/m**2"

stress developed in steel bar is 106.82 MN/m**2
stress developed in aluminium bar is 54.99 MN/m**2


## Problem 3.16,Page no.64¶

In [43]:
import math

#Initilization of variables

E=200 #GN/m**2 #Modulus of elasticity
alpha=11*10**-6 #per degree celsius #coeffecient o flinear expansion of steel bar
L=6 #m #Length of rod

#Calculations

#(Part-1) #IF the walls do not yield

t=58 #degree celsius #Fall in temperature #(t=80-22)
dell=alpha*t #strain
sigma=E*10**9*dell*10**-6 #MN/m**2 #Stress
A=pi*4**-1*6.25*10**-4 #mm**2 #Area of wall and rod
P=sigma*10**6*A*10**-3 #KN #Pull Exerted

#(Part-2) #IF the walls yield together at the two ends is 1.15 mm

L_2=L*(1-alpha*t) #m #Length of rod at 22 degree celsius
L_3=L-L_2 #m #Decrease in Length

#As the walls yield by 1.5 mm, actual decrease in length is
L_4=L_3-0.0015 #m
dell_2=L_4*L**-1 #strain
P_2=E*10**9*dell_2*A*10**-3 #KN

#Result
print"Pull Exerted by the rod:when walls do not yield",round(P,2),"KN"
print"                       :when total yield together at two ends is 1.5 mm",round(P_2,2),"KN"

Pull Exerted by the rod:when walls do not yield 62.64 KN
:when total yield together at two ends is 1.5 mm 38.09 KN


## Problem 3.17,Page no.65¶

In [54]:
import math

#Initilization of variables

D=4.5 #cm #External Diameter of tube
d=3 #cm #Internal diameter of tube
t=3 #mm #thickness of tube
t_1=30 #degree celsius
t_2=180 #degree celsius #when metal heated
L=30 #cm #Original LEngth
alpha_s=1.08*10**-5 #Per degree celsius #coefficient of Linear expansion of steel tube
alpha_c=1.7*10**-5 #Per degree celsius #coefficient of Linear expansion of copper tube
E_s=210 #GPa #Modulus of Elasticity of steel
E_c=110 #GPA #Modulus of Elasticity of copper

#Calculation

#For Equilibrium of the system, Total tension in steel=Total tension in copper

#sigma_s*A_s=sigma_c*A_c (equation 1)

A_c=pi*4**-1*d**2 #cm**2 #Area of copper
A_s=pi*4**-1*(D**2-d**2) #cm**2 #Area of steel

#simplifying equation 1
#sigma_s=1.785*sigma_c

T=t_2-t_1 #change in temperature

#Actual expansion of steel=Actual expansion of copper
#alpha_s*T*L+sigma_s*E_s**-1*L=alpha_c*T*L-sigma_c*E_c**-1*L

#After substituting values in above equation and simplifying we get

sigma_c=(930*10**5*1.7591**-1)*10**-6 #MN/m**2 #Stress in copper
sigma_s=1.785*sigma_c #MN/m**2 #Stress in steel

#Increase in Length of either component
L_2=(alpha_s*T+sigma_s*10**6*(E_s*10**9)**-1)*L

#Result
print"stress in copper bar is",round(sigma_c,2),"MN/m**2"
print"stress in steel bar is",round(sigma_s,2),"MN/m**2"
print"Increase in Length is",round(L_2,3),"cm"

stress in copper bar is 52.87 MN/m**2
stress in steel bar is 94.37 MN/m**2
Increase in Length is 0.062 cm


## Problem 3.18,Page no.66¶

In [60]:
import math

#Initilization of variables

t_1=15 #degree celsius #temperature of steel bar
t_2=315 #degree celsius #raised temperature
E_s=210 #GPa #Modulus of Elasticity of steel bar
E_c=100 #GPa #Modulus of Elasticity of copper bar
dell_L=0.15 #cm #Increase in Length of bar

#Calculation

#For Equilibrium of the system, Tension in steel bar = Tension in copper bar
#sigma_s*A_s = sigma_c*2*A_c
#sigma_S=2*sigma_c

#Actual expansion of steel = Actual expansion of copper
#L*alpha_s*T+sigma_s*E_s**-1*L = L*alpha_c*T-sigma_c*E_c**-1*L (Equation 1)

T=t_2-t_1 #per degree celsius #change in temperature

#After substituting values in above equation and simplifying we get
sigma_c=(1650*10**5*1.9524**-1)*10**-6 #MN/m**2 #Stress in copper
sigma_s=2*sigma_c #MN/m**2 #Stress in steel

#Actual Expansion of steel bar
#L*alpha_s*T+sigma_s*E_s**-1*L = L*alpha_c*T-sigma_c*E_c**-1*L
#After substituting values in above equation and simplifying we get
L=0.15*10**-2*0.0044048**-1 #m

#Result
print"Stress in copper bar is",round(sigma_c,2),"MN/m**2"
print"Stress in steel bar is",round(sigma_s,2),"MN/m**2"
print"Original Length of bar is",round(L,2),"m"

Stress in copper bar is 84.51 MN/m**2
Stress in steel bar is 169.02 MN/m**2
Original Length of bar is 0.34 m


## Problem 3.19,Page no.67¶

In [64]:
import math

#Initilization of variables

L=100 #m #Length of rod
A=2 #cm**2 #cross sectional area
rho=80 #KN/m**3

#Calculatiom
W=A*10**-4*L*rho #KN

sigma_s=10+1.6 #KN #Rod experiencing max tensile stress when it is at top performing upstroke
sigma_s_2=sigma_s*10**3*200**-1 #N/mm**2 #corresponding stress at this moment

sigma_c=1 #KN ##Rod experiencing max compressive stress at its lower end,free from its own weight
sigma_c_2=sigma_c*10**3*200**-1 #corresponding stress at this moment

#Result
print"Tensile stress in bar",round(sigma_s_2,2),"N/mm**2"
print"Compressive stress in bar",round(sigma_c_2,2),"N/mm**2"

Tensile stress in bar 58.0 N/mm**2
Compressive stress in bar 5.0 N/mm**2


## Problem 3.20,Page no.68¶

In [6]:
import math

#Initilization of variables

sigma=0.012 #strain
P=150 #KN   #Total Load on the Post
E=1.4*10**4 #N/mm**2 #modulus of elasticity
#b be the width of the post in mm
#2b is the longer dimension of the post in mm

#Calculations

#We know,
#sigma=(P*(A*E)**-1)

#After substituting values and simplifying, we get
b=((150*10**3)*(0.012*1.4*2*10**4)**-1)**0.5
q=2*b #mm #Longer dimension of post

#Result
print"Width of post is",round(b,2),"mm"
print"Longer dimension of post is",round(q,2),"mm"

Width of post is 21.13 mm
Longer dimension of post is 42.26 mm