Chapter 1 :Simple Stresses and Strains

Problem 1.1,page no.9

In [1]:
import math
#Given
#Variable declaration
L=150           #Length of the rod in cm
D=20            #Diameter of the rod in mm
P=20*10**3      #Axial pull in N
E=2.0e5         #Modulus of elasticity in N/sq.mm

#Calculation
A=(math.pi/4)*(D**2)    #Area in sq.mm
 #case (i):stress
sigma=P/A               #Stress in N/sq.mm
 #case (ii):strain
e=sigma/E               #Strain
 #case (iii):elongation of the rod
dL=e*L                  #Elongation of the rod in cm

#Result
print "Stress =",round(sigma,3),"N/mm^2"
print "Strain =",round(e,6)
print "Elongation =",round(dL,4),"cm"
Stress = 63.662 N/mm^2
Strain = 0.000318
Elongation = 0.0477 cm

Problem 1.2,page no.10

In [2]:
import math

#Given
#variable declaration
P=4000         #Load in N
sigma=95       #Stress in N/sq.mm

#Calculation
D=round(math.sqrt(P/((math.pi/4)*(sigma))),2)  #Diameter of steel wire in mm

#Result
print "Diameter of a steel wire =",D,"mm"
Diameter of a steel wire = 7.32 mm

Problem 1.3,page no.10

In [4]:
import math

#Given
#Variable declaration
D=25        #Diameter of brass rod in mm
P=50*10**3  #Tensile load in N
L=250       #Length of rod in mm
dL=0.3      #Extension of rod in mm

#Calculation
A=(math.pi/4)*(D**2)    #Area of rod in sq.mm
sigma=round(P/A,2)      #Stress in N/sq.mm
e=dL/L                  #Strain
E=(sigma/e)             #Young's Modulus in N/sq.m

#Result
print "Young's Modulus of a rod,E =",round(E*(10**-3),3),"GN/m^2"    #Young's Modulus in GN/sq.m
Young's Modulus of a rod,E = 84.883 GN/m^2

Problem 1.4,page no.11

In [2]:
import math

#Given
#Variable Declaration
D=3          #Diameter of the steel bar in cm    
L=20         #Gauge length of the bar in cm
P=250        #Load at elastic limit in kN 
dL=0.21      #Extension at a load of 150kN in mm
Tot_ext=60   #Total extension in mm
Df=2.25      #Diameter of the rod at the failure in cm

#Calculation
A=round((math.pi/4)*(D**2),5)    #Area of the rod in sq.m

#case (i):Young's modulus
e=round((150*1000)/(7.0685),1)   #stress in N/sq.m
sigma=dL/(L*10)                  #strain 
E=round((e/sigma)*(10**-5),3)    #Young's modulus in GN/sq.m

#case (ii):stress at elastic limit
stress=int(round((P*1000)/A,0))*1e-2   #stress at elastic limit in MN/sq.m

#case (iii):percentage elongation
Pe=(Tot_ext*1e2)/(L*10)

#case (iv):percentage decrease in area
Pd=(D**2-Df**2)/D**2*1e2


#Result
print "NOTE:The Young's Modulus found in the book is incorrect.The correct answer is,"
print "Young's modulus,E =",E,"GN/m^2"
print "Stress at the elastic limit,Stress =",stress,"MN/m^2"
print "Percentage elongation = %d%%"%Pe
print "Percentage decrease in area = %.2f%%"%Pd
NOTE:The Young's Modulus found in the book is incorrect.The correct answer is,
Young's modulus,E = 202.104 GN/m^2
Stress at the elastic limit,Stress = 353.68 MN/m^2
Percentage elongation = 30%
Percentage decrease in area = 43.75%

Problem 1.5,page no.12

In [6]:
import math

#Given
#Variable declaration
sigma=125*10**6    #Safe stress in N/sq.m
P=2.1*10**6        #Axial load in N
D=0.30             #External diameter in m

#Calculation
    
d=round(math.sqrt((D**2)-P*4/(math.pi*sigma)),4)*1e2  #internal diameter in cm

#Result
print "internal diameter =",d,"cm"      
internal diameter = 26.19 cm

Problem 1.6,page no.13

In [8]:
import math

#Given
#Variable declaration
stress=480      #ultimate stress in N/sq.mm
P=1.9*10**6     #Axial load in N
D=200           #External diameter in mm
f=4             #Factor of safety

#Calculation
sigma=stress/f                                        #Working stress or Permissable stress in N/sq.mm
d=str(math.sqrt((D**2)-((P*4)/(math.pi*sigma))))[:6]  #internal diameter in mm

#Result
print "internal diameter =",d,"mm"
internal diameter = 140.85 mm

Problem 1.15,page no.26

In [5]:
import math

#Given
#Variable declaration
D1=40      #Larger diameter in mm
D2=20      #Smaller diameter in mm
L=400      #Length of rod in mm
P=5000     #Axial load in N
E=2.1e5    #Young's modulus in N/sq.mm

#Calculation
dL=float(str((4*P*L)/(math.pi*E*D1*D2))[:7])   #extension of rod in mm

#Result
print "Extension of the rod =",dL,"mm"
Extension of the rod = 0.01515 mm

Problem 1.16,page no.27

In [1]:
import math

#Given
#Variable declaration
D1=30          #Larger diameter in mm
D2=15          #Smaller diameter in mm
L=350          #Length of rod in mm
P=5.5*10**3    #Axial load in N
dL=0.025       #Extension in mm

#Calculation
E=int((4*P*L)/(math.pi*D1*D2*dL))   #Modulus of elasticity in N/sq.mm

#Result
print "Modulus of elasticity,E = %.5e"%E,"N/mm^2" 
Modulus of elasticity,E = 2.17865e+05 N/mm^2

Problem 1.17,page no.29

In [4]:
import math

#Given
#Variable declaration
L=2.8*10**3      #Length in mm
t=15             #Thickness in mm
P=40*10**3       #Axial load in N
a=75             #Width at bigger end in mm
b=30             #Width at smaller end in mm
E=2e5            #Young's Modulus in N/sq.mm

#Calculation
dL=round((round((P*L)/(E*t*(a-b)),4)*(round(math.log(a)-math.log(b),4))),2)    #extension of rod in mm

#Result
print "Extension of the rod,dL =",dL,"mm"
Extension of the rod,dL = 0.76 mm

Problem 1.18,page no.29

In [9]:
import math

#Given
#Variable declaration
dL=0.21          #Extension in mm
L=400            #Length in mm
t=10             #Thickness in mm
a=100            #Width at bigger end in mm
b=50             #Width at smaller end in mm
E=2e5            #Young's Modulus in N/sq.mm

#Calculation
P=int(dL/(round((L)/(E*t*(a-b)),6)*(round(math.log(a)-math.log(b),4))))*1e-3    #Axial load in kN

#Result
print "Axial load =",P,"kN"
Axial load = 75.746 kN

Problem 1.20,page no.32

In [1]:
import math

#Given
#Variable declaration
Di_s=140     #Internal diameter of steel tube in mm 
De_s=160     #External diameter of steel tube in mm
Di_b=160     #Internal diameter of brass tube in mm    
De_b=180     #External diameter of brass tube in mm
P=900e3      #Axial load in N
L=140        #Length of each tube in mm
Es=2e5       #Young's modulus for steel in N/sq.mm
Eb=1e5       #Young's modulus for brass in N/sq.mm

#Calculation
As=round(math.pi/4*(De_s**2-Di_s**2),1)     #Area of steel tube in sq.mm
Ab=round(math.pi/4*(De_b**2-Di_b**2),1)     #Area of brass tube in sq.mm
sigmab=round(P/(2*As+Ab),2)                 #Stress in steel in N/sq.mm
sigmas=2*sigmab                             #Stress in brass in N/sq.mm
Pb=int(sigmab*Ab)*1e-3                      #Load carried by brass tube in kN
Ps=(P*1e-3)-(Pb)                            #Load carried by steel tube in kN
dL=round(sigmab/Eb*(L),4)                   #Decrease in length in mm

#Result
print "Stress in brass =",sigmab,"N/mm^2"
print "Stress in steel =",sigmas,"N/mm^2"
print "Load carried by brass tube =",Pb,"kN"
print "Load carried by stress tube =",Ps,"kN"
print "Decrease in the length of the compound tube=",dL,"mm"
Stress in brass = 60.95 N/mm^2
Stress in steel = 121.9 N/mm^2
Load carried by brass tube = 325.515 kN
Load carried by stress tube = 574.485 kN
Decrease in the length of the compound tube= 0.0853 mm

Problem 1.28,page no.43

In [2]:
#Given
#Variable declaration
L=2*10**2        #Length of rod in cm
T1=10            #Initial temperature in degree celsius
T2=80            #Final temperature in degree celsius
E=1e5*10**6      #Young's Modulus in N/sq.m
alpha=0.000012   #Co-efficient of linear expansion 

#Calculation
T=T2-T1                          #Rise in temperature in degree celsius
dL=alpha*T*L                     #Expansion of the rod in cm
sigma=int((alpha*T*E)*1e-6)      #Thermal stress in N/sq.mm

#Result
print "Expansion of the rod =",dL,"cm"
print "Thermal stress =",sigma,"N/mm^2"
Expansion of the rod = 0.168 cm
Thermal stress = 84 N/mm^2

Problem 1.29,page no.43

In [2]:
import math

#Given
#Variable declaration
d=3*10           #Diameter of the rod in mm
L=5*10**3        #Area of the rod in sq.mm
T1=95            #Initial temperature in degree celsius
T2=30            #Final temperature in degree celsius
E=2e5*10**6      #Young's Modulus in N/sq.m
alpha=12e-6      #Co-efficient of linear expansion in per degree celsius

#Calculation
A=math.pi/4*(d**2)        #Area of the rod
T=T1-T2                   #Fall in temperature in degree celsius

#case(i) When the ends do not yield 
stress1=int(alpha*T*E*1e-6)     #Stress in N/sq.mm
Pull1=round(stress1*A,1)        #Pull in the rod in N

#case(ii) When the ends yield by 0.12cm
delL=0.12*10
stress2=int((alpha*T*L-delL)*E/L*1e-6)      #Stress in N/sq.mm
Pull2=round(stress2*A,1)                    #Pull in the rod in N

#Result
print "Stress when the ends do not yield =",stress1,"N/mm^2"
print "Pull in the rod when the ends do not yield =",Pull1,"N"
print "Stress when the ends yield =",stress2,"N/mm^2"
print "Pull in the rod when the ends yield =",Pull2,"N"
Stress when the ends do not yield = 156 N/mm^2
Pull in the rod when the ends do not yield = 110269.9 N
Stress when the ends yield = 108 N/mm^2
Pull in the rod when the ends yield = 76340.7 N

Problem 1.30,page no.45

In [7]:
from __future__ import division
import math
#Given
#Variable declaration
Ds=20               #Diameter of steel rod in mm
Di_c=40             #Internal diameter of copper tube in mm
De_c=50             #External diameter of copper tube in mm
Es=200*10**3        #Young's modulus of steel in N/sq.mm
Ec=100*10**3        #Young's modulus of copper in N/sq.mm
alpha_s=12e-6       #Co-efficient of linear expansion of steel in per degree celsius
alpha_c=18e-6       #Co-efficient of linear expansion of copper in per degree celsius
T=50                #Rise of temperature in degree celsius

#Calculation
As=(math.pi/4)*(Ds**2)                                             #Area of steel rod in sq.mm
Ac=(math.pi/4)*(De_c**2-Di_c**2)                                   #Area of copper tube in sq.mm
sigmac=float(str(((alpha_c-alpha_s)*T)/(((Ac/As)/Es)+(1/Ec)))[:6]) #Compressive stress in copper  
sigmas=round(sigmac*(Ac/As),2)                                     #Tensile stress in steel 

#Result
print "Stress in copper =",sigmac,"N/mm^2"
print "Stress in steel =",sigmas,"N/mm^2"
Stress in copper = 14.117 N/mm^2
Stress in steel = 31.76 N/mm^2

Problem 1.31,page no.47

In [9]:
import math

#Given
#Variable declaration
Dc=15             #Diameter of copper rod in mm
Di_s=20           #Internal diameter of steel in mm
De_s=30           #External diameter of steel in mm
T1=10             #Initial temperature in degree celsius
T2=200            #Raised temperature in degree celsius
Es=2.1e5          #Young's modulus of steel in N/sq.mm
Ec=1e5            #Young's modulus of copper in N/sq.mm
alpha_s=11e-6     #Co-efficient of linear expansion of steel in per degree celsius
alpha_c=18e-6     #Co-efficient of linear expansion of copper in per degree celsius

#Calculation
Ac=(math.pi/4)*Dc**2                #Area of copper tube in sq.mm
As=(math.pi/4)*(De_s**2-Di_s**2)    #Area of steel rod in sq.mm
T=T2-T1                             #Rise of temperature in degree celsius
sigmas=round(((alpha_c-alpha_s)*T)/((round(As/Ac,2)/Ec)+(1/Es)),3)
sigmac=round(sigmas*round(As/Ac,2),2)

#Result
print "NOTE: The answers in the book for stresses are wrong.The correct answers are,"
print "Stress in steel =",sigmas,"N/mm^2"
print "Stress in copper =",sigmac,"N/mm^2"
NOTE: The answers in the book for stresses are wrong.The correct answers are,
Stress in steel = 49.329 N/mm^2
Stress in copper = 109.51 N/mm^2

Problem 1.32,page no.48

In [10]:
import math
#Given
#Variable declaration
Dg=20           #Diameter of gun metal rod in mm
Di_s=25         #Internal diameter of steel in mm
De_s=30         #External diameter of steel in mm
T1=30           #Temperature in degree celsius
T2=140          #Temperature in degree celsius
Es=2.1e5        #Young's modulus of steel in N/sq.mm
Eg=1e5          #Young's modulus of gun metal in N/sq.mm
alpha_s=12e-6   #Co-efficient of linear expansion of steel in per degree celsius
alpha_g=20e-6   #Co-efficient of linear expansion of gun metal in per degree celsius

#Calculation
Ag=(math.pi/4)*Dg**2              #Area of gun metal in sq.mm
As=(math.pi/4)*(De_s**2-Di_s**2)  #Area of steel in sq.mm
T=T2-T1                           #Fall in temperature in degree celsius
sigmag=round(((alpha_g-alpha_s)*T)/(((Ag/As)/Es)+(1/Eg)),2)
sigmas=round(sigmag*(Ag/As),2)

#Result
print "Stress in gun metal rod =",sigmag,"N/mm^2"
print "Stress in steel =",sigmas,"N/mm^2"
Stress in gun metal rod = 51.99 N/mm^2
Stress in steel = 75.62 N/mm^2

Problem 1.33,page no.52

In [11]:
import math

#Given
#Variable declaration
P=600e3         #Axial load in N
L=20e3          #Length in mm
w=0.00008       #Weight per unit volume in N/sq.mm
A2=400          #Area of bar at lower end in sq.mm

#Calculation
sigma=int(P/A2)           #Uniform stress on the bar in N/sq.mm
A1=round(A2*round(math.exp(round(w*L/sigma,7)),5),3)

#Result
print "Area of the bar at the upper end =",A1,"mm^2"
Area of the bar at the upper end = 400.428 mm^2
In [ ]: