Chapter 2 Properties Of Material

Example 1 Page No:19

In [16]:
#Input data
L=5                       #Length of steel bar in m
d=25*10**-3               #Diametr of steel bar in mm
deltaLt=25*10**-3         #Steel 
pt=800
pi=3.142                  #Power load of steel bar in N

#Calculation
A=(((pi/4)*((deltaLt)**2))) #Cross-section area
sigmat=(pt)/(A)           #Stress in  steel bar
et=(deltaLt)/L            #Strain in steel bar
E=((sigmat)/(et))         #Young's modulus

#Output
print("value of Cross-section area=",round(A,8),"m**2")
print("value of tress in  steel bar=",round(sigmat,),"MN/m**2")
print("value of strain in steel bar=",et)
print("value of Young's modulus=",round(E,),"N/m**2")

value of Cross-section area= 0.00049094 m**2
value of tress in  steel bar= 1629535 MN/m**2
value of strain in steel bar= 0.005
value of Young's modulus= 325907066 N/m**2

Example 2 Page No:20

In [21]:
#Input data
L=300*10**-3        #Length of hexagonal prismatic steel bar in mm
A=500*10**-6        #Area of cross section of steel bar mm**2
Pt=500*10**3        #Load of steel bar in KN
E=210*10**9         #Modulus of elasticity GN/m**2

#Calculation
sigmat=((Pt)/(A))   #Stress in steel bar
et=((sigmat)/(E))   #Strain steel bar is
deltaLt=((et)*(L))  #Therefore,elongation of the steel bar is given by

#Output
print('stress in steel bar=',sigmat,"N/m**2")
print('therefore,strain steel bar is given by=',round(et,6),)
print('therefore,elongation of the steel bar is given by=',round(deltaLt,7),"m")

stress in steel bar= 1000000000.0 N/m**2
therefore,strain steel bar is given by= 0.004762
therefore,elongation of the steel bar is given by= 0.0014286 m

Example 3 Page No:21

In [1]:
#Input Data
Pt=600                  #Tensils force in N
d=2*10**-3              #Diameter of steel wire in mm
L=15                    #Length of wire in m
E=210*10**9             #Modulus of elasticity of the material in GN/M**2
pi=3.1482


#Calculation
A=((pi/4)*((d)**2))     #(1)cross section area
sigmat=(Pt)/(A)         #stress in the steel wire 
et=((sigmat)/(E))       #(2)Therefore, strain in steel wire is given by
deltaLt=et*L            #(3)Enlongation of the steel wire is given by 
pe=((deltaLt/L)*100)    #(4)Percentage elongation


#Output
print("cross section area= ",A,"m**2")
print("stress in the steel wire=",round(sigmat,),"GN/m**2")
print("modulus of elasticity=",round(et,5),)
print("strain in steel wire=",round(deltaLt,4),"mm")
print("percentage elongation= ",round(pe,3),"%")

cross section area=  3.1481999999999998e-06 m**2
stress in the steel wire= 190585096 GN/m**2
modulus of elasticity= 0.00091
strain in steel wire= 0.0136 mm
percentage elongation=  0.091 %

Example 4 Page No:22

In [2]:
#Input data
A=30*30*10**-6      #Area of square rod in mm**2
L=5                 #Length of square rod in m
Pc=150*10**3        #Axial comperessive load of a rod  in kN
E=215*10**9         #Modulus of elasticity in GN/m**2


#Calculation
sigmac=((Pc)/(A))   #Stress in square rod
ec=((sigmac)/(E))   #Modulusof elasticity is E=sigmac/ec ,therefore strain in square rod is
deltaLc=ec*5        #Therefore shortening of length of the rod 


#Output
print ("stress in square rod",sigmac,"N/m**2")
print("strain in square rod ec=",round(ec,6),)
print("shortening of length of the rod=",round(deltaLc,6),"m")

stress in square rod 166666666.66666666 N/m**2
strain in square rod ec= 0.000775
shortening of length of the rod= 0.003876 m

Example 5 Page No:23

In [3]:
#input data
d=50*10**-6                #Diameter of metalic rod in mm**2
L=220*10**-3               #Length of metalic rod in mm
Pt=40*10**3                #Load of metalic rod in KN
deltaLt=0.03*10**-3        #Elastic enlongation in mm
ypl=160*10**3              #Yield point load in KN
ml=250*10**3               #Maximum load in KN
lsf=270*10**-3             #Length of specimen at fracture in mm
pi=3.1482

#calculation
A=(((pi)/(4)*((d)**2)))   #(1)Cross section area
sigmat=(Pt/A)              #Stress in metallic rod
et=(deltaLt/L)             #Strain n metallic rod
E=(sigmat/et)              #Young's modulus
ys=(ypl/A)                 #(2)Yeild strength
uts=(ml/A)                 #(3)Ultimate tensile strength
Pebf=((lsf-L)/L)*100       #Percentage elongation before fracture 



#output
print("cross section area",A,"m**2")
print("stress in metallic rod",round(sigmat,),"N/m**2")
print("strain n metallic rod",round(et,6),)
print("young's modulus",round(E,8),"GN/m**2")
print("yeild strength",ys,"MN/m**2")
print("ultimate tensile strength",uts,"MN/m**2")
print("percentage elongation before fracture",round(Pebf,3),"%")

cross section area 1.967625e-09 m**2
stress in metallic rod 20329076932851 N/m**2
strain n metallic rod 0.000136
young's modulus 1.4907989750757046e+17 GN/m**2
yeild strength 81316307731402.08 MN/m**2
ultimate tensile strength 127056730830315.75 MN/m**2
percentage elongation before fracture 22.727 %

Example 6 Page No:24

In [4]:
#Input data
A=50*50*10**-6                 #Area ofsquare metal bar in mm**2
Pc=600*10**3                   #Axial compress laod in KN
L=200*10**-3                   #Gauge length of metal bar in mm
deltaLc=0.4*10**-3             #Contraction length of metal bar in mm
deltaLlateral=0.05*10**-3      #Lateral length of metal bar in mm

#Calculation
sigmac=((Pc)/(A))              #Stress in square metal bar 
ec=((deltaLc)/(L))             #Longitudinal or linear strain in square metal bar
E =((sigmac)/(ec))             #Smodule of elasticity
elateral=((deltaLlateral)/(L)) #Lateral strain in square metal bar
poissonsratio=(elateral)/(ec)


#Output
print("stress in bar=",sigmac,"n/m**2")
print("longitudinal or linear strain in square metal bar=",ec,)
print("module of elasticity=",E,"N/m**2")
print("lateral strain in square metal bar=",elateral,)
print("poissons ratio=",poissonsratio,)

stress in bar= 240000000.0 n/m**2
longitudinal or linear strain in square metal bar= 0.002
module of elasticity= 120000000000.0 N/m**2
lateral strain in square metal bar= 0.00025
poissons ratio= 0.125