Chapter 2:Elastic Constants

Problem 2.1,page no.60

In [2]:
from __future__ import division
import math

#Given
#Variable declaration
L=4*(10**3)       #Length of the bar in mm
b=30              #Breadth of the bar in mm
t=20              #Thickness of the bar in mm
P=30*(10**3)      #Axial pull in N
E=2e5             #Young's modulus in N/sq.mm
mu=0.3            #Poisson's ratio

#Calculation
A=b*t                        #Area of cross-section in sq.mm
long_strain=P/(A*E)          #Longitudinal strain 
delL=long_strain*L           #Change in length in mm
lat_strain=mu*long_strain    #Lateral strain
delb=b*lat_strain            #Change in breadth in mm
delt=t*lat_strain            #Change in thickness in mm

#Result
print "change in length =",delL,"mm"
print "change in breadth =",delb,"mm"
print "change in thickness =",delt,"mm"
change in length = 1.0 mm
change in breadth = 0.00225 mm
change in thickness = 0.0015 mm

Problem 2.2,page no.61

In [2]:
#Given
#Variable declaration
L=30                #Length in cm
b=4                 #Breadth in cm
d=4                 #Depth in cm
P=400*(10**3)       #Axial compressive load in N
delL=0.075          #Decrease in length in cm
delb=0.003          #Increase in breadth in cm

#Calculation
A=(b*d)*1e2                      #Area of cross-section in sq.mm
long_strain=delL/L               #Longitudinal strain
lat_strain=delb/b                #Lateral strain
mu=lat_strain/long_strain        #Poisson's ratio
E=int((P)/(A*long_strain))       #Young's modulus

#Result
print "Poisson's ratio =",mu
print "Young's modulus = %.e N/mm^2"%E
Poisson's ratio = 0.3
Young's modulus = 1e+05 N/mm^2

Problem 2.3,page no.63

In [3]:
#Given
#Variable declaration
L=4000        #Length of the bar in mm
b=30          #Breadth of the bar in mm
t=20          #Thickness of the bar in mm
mu=0.3        #Poisson's ratio
delL=1.0      #delL from problem 2.1

#Calculation
ev=(delL/L)*(1-2*mu)         #Volumetric strain 
V=L*b*t                      #Original volume in mm^3
delV=ev*V                    #Change in volume in mm^3
F=int(V+delV)                #Final volume in mm^3

#Result
print "Volumetric strain =",ev
print "Final volume =",F,"mm^3"
Volumetric strain = 0.0001
Final volume = 2400240 mm^3

Problem 2.4,page no.63

In [5]:
from __future__ import division
#Given
#Variable declaration
L=300              #Length in mm
b=50               #Width in mm
t=40               #Thickness in mm
P=300*10**3        #Pull in N
E=2*10**5          #Young's modulus in N/sq.mm
mu=0.25            #Poisson's ratio

#Calculation
V=L*b*t                           #Original volume in mm^3
Area=b*t                          #Area in sq.mm   
stress=P/Area                     #Stress in N/sq.mm 
ev=(stress/E)*(1-2*mu)            #Volumetric strain 
delV=int(ev*V)                    #Change in volume in mm^3                     

#Result
print "Change in volume =",delV,"mm^3"
 Change in volume = 225 mm^3

Problem 2.7,page no.69

In [7]:
import math

#Given
#Variable declaration
L=5*10**3                #Length in mm
d=30                     #Diameter in mm
P=50*10**3               #Tensile load in N
E=2e5                    #Young's modulus in N/sq.mm
mu=0.25                  #Poisson's ratio

#Calculation
V=int(round((math.pi*d**2*L)/4,-2))        #Volume in mm^3 
e=P*4/(math.pi*(d**2)*E)                   #Strain of length
delL=round(e*L,3)                          #Change in length in mm
lat_strain=round(mu*round(e,7),7)          #Lateral strain 
deld=lat_strain*d                          #Change in diameter in mm
delV=round(V*(0.0003536-(2*lat_strain)),2) #Change in volume in mm^3

#Result
print "Change in length =",delL,"mm"
print "Change in diameter =",deld,"mm"
print "Change in volume =",delV,"mm^3"
Change in length = 1.768 mm
Change in diameter = 0.002652 mm
Change in volume = 624.86 mm^3

Problem 2.10,page no.79

In [8]:
#Given
#Variable declaration
E=1.2e5                  #Young's modulus in N/sq.mm
C=4.8e4                  #Modulus of rigidity in N/sq.mm

#Calculation
mu=(E/(2*C))-1           #Poisson's ratio 
K=int(E/(3*(1-2*mu)))    #Bulk modulus in N/sq.mm

#Result
print "Poisson's ratio =",mu
print "Bulk modulus = %.0e N/mm^2"%K
Poisson's ratio = 0.25
Bulk modulus = 8e+04 N/mm^2

Problem 2.11,page no.79

In [9]:
#Given
#Variable declaration
A=8*8         #Area of section in sq.mm
P=7000        #Axial pull in N
Ldo=8         #Original Lateral dimension in mm
Ldc=7.9985    #Changed Lateral dimension in mm
C=0.8e5       #modulus of rigidity in N/sq.mm

#Calculation
lat_strain=(Ldo-Ldc)/Ldo                                        #Lateral strain
sigma=P/A                                                       #Axial stress in N/sq.mm
mu=round(1/((sigma/lat_strain)/(2*C)-1),3)                      #Poisson's ratio
E=round((sigma/lat_strain)/((sigma/lat_strain)/(2*C)-1),-1)     #Modulus of elasticity in N/sq.mm

#Result
print "Poisson's ratio =",mu
print "Modulus of elasticity = %.4e N/mm^2"%E
Poisson's ratio = 0.378
Modulus of elasticity = 2.2047e+05 N/mm^2
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