Chapter 4:Strain Energy and Impact Loading

Problem 4.1,page no.145

In [1]:
import math

#Given
#Variable declaration
P=60*10**3    #Load in N
d=4*10        #diameter in mm
L=5*10**3     #Length of rod in mm
E=2e5         #Young's Modulus in N/sq.mm


#Calculation
A=(math.pi/4)*d**2           #Area in sq.mm
V=int(A*L)                   #Volume of rod in cubic.mm
#case (ii):stress in the rod
sigma=round(P/A,3)           #stress in N/sq.mm

#case (i):stretch in the rod
x=round((sigma/E)*L,2)       #stretch or extension in mm

#case (iii):strain energy absorbed by the rod
U=round((sigma**2/(2*E)*V),-1)*1e-3     #strain energy absorbed by the rod in Nm


#Result
print "stress in the rod =",sigma,"N/mm^2"
print "stretch in the rod =",x,"mm"
print "strain energy absorbed by the rod =",U,"N-m"
stress in the rod = 47.746 N/mm^2
stretch in the rod = 1.19 mm
strain energy absorbed by the rod = 35.81 N-m

Problem 4.3,page no.146

In [2]:
#Given
#Variable declaration
A=10*10**2         #Area of bar in sq.mm
L=3*10**3          #Length of bar in mm
x=1.5              #Extension due to suddenly applied load in mm
E=2e5              #Young's Modulus in N/sq.mm

#Calculation
sigma=int(x*E/L)            #Instantaneous stress due to sudden load in N/sq.mm     
P=int((sigma*A)/2*1e-3)     #Suddenly applied load in kN

#Result
print "Instantaneous stress produced by a sudden load =",sigma,"N/mm^2"
print "Suddenly applied load =",P,"kN" 
Instantaneous stress produced by a sudden load = 100 N/mm^2
Suddenly applied load = 50 kN

Problem 4.4,page no.147

In [5]:
import math

#Given
#Variable declaration
L=2*10**3            #Length in mm
d=50                 #Diameter in mm
P=100*10**3          #Suddenly applied load in N
E=200e3              #Young's Modulus in N/sq.mm

#Calculation
A=(math.pi/4)*d**2        #Area in sq.mm
sigma=round(2*P/A,2)      #Instantaneous stress induced in N/sq.mm
dL=(sigma*L)/E            #Elongation in mm

#Result
print "Instantaneous stress induced =",sigma,"N/mm^2"
print "Instantaneous elongation =",dL,"mm"
Instantaneous stress induced = 101.86 N/mm^2
Instantaneous elongation = 1.0186 mm

Problem 4.5,page no.147

In [4]:
#Given
#Variable declaration
A=700              #Area in sq.mm
L=1.5*10**3        #Length of a metal bar in mm
sigma=160          #Stress at elastic limit in N/sq.mm
E=2e5              #Young's Modulus in N/sq.mm


#Calculation
V=A*L                           #Volume of bar in sq.mm
Pr=(sigma**2/(2*E)*V)*1e-3      #Proof resilience in N-m
P=int(sigma*A/2*1e-3)           #Suddenly applied load in kN
P1=int(sigma*A*1e-3)            #gradually applied load in kN

#Result
print "Proof resilience =",Pr,"N-m"
print "Suddenly applied load =",P,"kN"
print "Gradually applied load =",P1,"kN"
Proof resilience = 67.2 N-m
Suddenly applied load = 56 kN
Gradually applied load = 112 kN

Problem 4.9,page no.154

In [7]:
import math

#Given
#Variable declaration
P=10*10**3             #Falling weight in N
h=30                   #Falling height in mm
L=4*10**3              #Length of bar in mm
A=1000                 #Area of bar in sq.m
E=2.1e5                #Young's modulus in N/sq.mm

#Calculation
sigma=float(str((P/A)*(1+(math.sqrt(1+((2*E*A*h)/(P*L))))))[:5])
delL=round(sigma*L/E,3)

#Result                           
print "Instantaneous elongation due to falling weight =",delL,"mm"
Instantaneous elongation due to falling weight = 3.575 mm

Problem 4.10,page no.155

In [12]:
import math
#Given
#Variable declaration
P=100           #Impact load in N
h=2*10          #Height in mm
L=1.5*1000      #Length of bar in mm
A=1.5*100       #Area of bar in sq.mm
E=2e5           #Modulus of elasticity in N/sq.mm

#Calculation
V=A*L           #Volume in mm^3
#case(i):Maximum instantaneous stress induced in the vertical bar
sigma=round((P/A)*(1+(math.sqrt(1+((2*E*A*h)/(P*L))))),2)
#case(ii):Maximum instantaneous elongation
delL=round(sigma*L/E,3)
#case(iii):Strain energy stored in the vertical rod
U=round(sigma**2/(2*E)*V*1e-3,3)

#Result
print "NOTE:The answer in the book for instantaneous stress is incorrect.The correct answer is,"
print "Maximum instantaneous stress =",sigma,"N/mm^2"
print "Maximum instantaneous elongation =",delL,"mm"
print "Strain energy =",U,"N-m"
NOTE:The answer in the book for instantaneous stress is incorrect.The correct answer is,
Maximum instantaneous stress = 60.3 N/mm^2
Maximum instantaneous elongation = 0.452 mm
Strain energy = 2.045 N-m

Problem 4.11,page no.155

In [1]:
import math
#Given
#Variable declaration
delL=2.1          #Instantaneous extension in mm
L=3*10**3         #Length of bar in mm
A=5*100           #Area of bar in mm
h=4*10            #Height in mm
E=2e5             #Modulus of elasticity in N/sq.mm

#Calculation
V=A*L                           #Volume of bar in mm^3

#case(i):Instantaneous stress induced in the vertical bar
sigma=int(E*delL/L)         

#case(ii):Unknown weight
P=round(((sigma**2)/(2*E)*V)/(h+delL),1)    

#Result
print"Instantaneous stress =",sigma,"N/mm^2"
print"Unknown weight =",P,"N"
Instantaneous stress = 140 N/mm^2
Unknown weight = 1745.8 N

Problem 4.13,page no.157

In [9]:
import math
#Given
#Variable declaration
d=12        #Diameter of bar in mm              
delL=3      #Increase in length in mm
W=8000      #Steady load in N
P=800       #Falling weight in N
h=8*10      #Vertical distance in mm
E=2e5       #Young's modulus in N/sq.mm

#Calculation
A=round((math.pi/4)*d**2,1)           #Area of bar in sq.mm
L=round(E*A*delL/W,1)                 #Length of the bar in mm
sigma=round(round(P/A,4)*float(str(1+(math.sqrt(1+round((2*E*A*h)/(P*L),2))))[:7]),4)  
sigma=float(str(sigma)[:7])           #Stress produced by the falling weight in N/sq.mm

#Result
print "Stress produced by the falling weight =",sigma,"N/mm^2"
Stress produced by the falling weight = 170.578 N/mm^2

Problem 4.14,page no.158

In [10]:
import math
#Given
#Variable declaration
d=12.5           #Diameter of the rod in mm
delL=3.2         #Increase in length in mm
W=10*1000        #Steady load in N
P=700            #Falling load in N
h=75             #Falling height in mm
E=2.1e5          #Young's modulus in N/sq.mm

#Calculation
A=round((math.pi/4)*d**2,2)                                 #Area of rod in sq.mm   
L=round(E*A*delL/W,1)                                       #Length of the rod in mm
sigma=round((P/A)*(1+(math.sqrt(1+((2*E*A*h)/(P*L))))),2)   #Stress produced by the falling weight in N/mm^2

#Result
print "NOTE:The given answer for stress is wrong.The correct answer is,"
print "Stress = %.2f N/mm^2"%sigma
NOTE:The given answer for stress is wrong.The correct answer is,
Stress = 153.42 N/mm^2

Problem 4.15,page no.159

In [1]:
import math

#Given
#Variable declaration
L=1.82*1000      #Length of rod in mm
h1=30            #Height through which load falls in mm
h2=47.5          #Fallen height in mm
sigma=157        #Maximum stress induced in N/sq.mm
E=2.1e5          #Young's modulus in N/sq.mm

#Calculation
U=sigma**2/(2*E)      #Strain energy stored in the rod in N-m
delL=sigma*L/E        #Extension of the rod in mm
Tot_dist=h1+delL      #Total distance in mm

#case(i):Stress induced in the rod if the load is applied gradually
sigma1=round((U/Tot_dist)*L,1)

#case(ii):Maximum stress if the load had fallen from a height of 47.5 mm
sigma2=round((sigma1)*(1+(math.sqrt(1+((2*E*h2)/(sigma1*L))))),2)

#Result
print "Stress induced in the rod = %.1f N/mm^2"%sigma1
print "NOTE:The given answer for stress(2nd case) in the book is wrong.The correct answer is,"
print "Maximum stress if the load has fallen = %.2f N/mm^2"%sigma2
Stress induced in the rod = 3.4 N/mm^2
NOTE:The given answer for stress(2nd case) in the book is wrong.The correct answer is,
Maximum stress if the load has fallen = 196.48 N/mm^2

Problem 4.17,page no.162

In [2]:
import math

#Given
#Variable declaration
L=4*10**3     #Length of bar in mm
A=2000        #Area of bar in sq.mm
P1=3000       #Falling weight in N(for 1st case)
h1=20*10      #Height in mm(for 1st case)
P2=30*1000    #Falling weight in N(for 2nd case)
h2=2*10       #Height in mm(for 2nd case)
E=2e5         #Young's modulus in N/sq.mm

#Calculation
V=A*L             #Volume of bar in mm^3

#case(i):Maximum stress when a 3000N weight falls through a height of 20cm
sigma1=round(((math.sqrt((2*E*P1*h1)/(A*L)))),1)

#case(ii):Maximum stress when a 30kN weight falls through a height of 2cm
sigma2=round((P2/A)*(1+(math.sqrt(1+((2*E*A*h2)/(P2*L))))),2)

#Result
print"Maximum stress induced(when a weight of 3000N falls through a height of 20cm)=",sigma1,"N/mm^2"
print"Maximum stress induced(when a weight of 30kN falls through a height of 2cm)=",sigma2,"N/mm^2"
Maximum stress induced(when a weight of 3000N falls through a height of 20cm)= 173.2 N/mm^2
Maximum stress induced(when a weight of 30kN falls through a height of 2cm)= 188.85 N/mm^2

Problem 4.18,page no.163

In [3]:
from __future__ import division
import math

#Given 
#Variable declaration
A=6.25*100     #Area in sq.mm
W=10*10**3     #Load in N
V=(40/60)      #Velocity in m/s
L=10000        #Length of chain unwound in mm
E=2.1e5        #Young's modulus in N/sq.mm
g=9.81         #acceleration due to gravity

#Calculation
K_E=round(((W/g)*(V**2))/2,1)*1e3          #K.E of the crane in N mm
sigma=round(math.sqrt(K_E*E*2/(A*L)),2)    #Stress induced in the chain in N/sq.mm

#Result
print "Stress induced in the chain due to sudden stoppage =",sigma,"N/mm^2"
Stress induced in the chain due to sudden stoppage = 123.37 N/mm^2

Problem 4.19,page no.164

In [4]:
import math

#Given
#Variable declaration
W=60*10**3      #Weight in N
V=1             #Velocity in m/s
L=15*10**3      #Free length in mm
A=25*100        #Area in sq.mm
E=2e5           #Young's modulus in N/sq.mm
g=9.81          #acceleration due to gravity

#Calculation
K_E=((W/g)*(V**2))/2*1e3                  #Kinetic Energy of the cage in N mm
sigma=round(math.sqrt(K_E*E*2/(A*L)),2)   #Maximum stress in N/sq.mm

#Result
print"Maximum stress produced in the rope =",sigma,"N/mm^2"
Maximum stress produced in the rope = 180.61 N/mm^2

Problem 4.20,page no.166

In [5]:
#Given 
#Variable declaration
tau=50       #Shear stress in N/sq.mm
C=8e4        #Modulus of rigidity in N/sq.mm

#Calculation
ste=(tau**2)/(2*C)    #Strain energy per unit volume in N/sq.mm

#Result
print"Strain energy per unit volume =",ste,"N/mm^2"
Strain energy per unit volume = 0.015625 N/mm^2
In [ ]: