Ch-9, Mechanical Properties

example-9.1 page no-259

In [6]:
from __future__ import division
#given
#bonding characteristics of a material are
n=1
m=9
A=7.56*10**-29   #J m
#initial bond length
r_0=2.3*10**-10   #m
#this bond length exceed by 15 %
#so extension is 
e=15/100*r_0   #m
#new bond length will be
r=r_0+e   #m
#the axial tensile starin is given by
epsilon_t=e/r_0   #(change in dimension /original dimension)
#according to P-r function we have
B=A*r_0**8/9   #J m**9
#we have double deravative of P as
double_derivative_P=-2*A/r_0**3+90*B/r_0**11   #/J/m**2
#so youngs modulus will be given by
E=double_derivative_P/r_0*10**-9   #GPa  (some approximation is done in book)
print  "the axial stain is %.2f \nand youngs modulus of elasticity is %0.3e GPa "%(epsilon_t,E)
the axial stain is 0.15 
and youngs modulus of elasticity is 2.161e+02 GPa 

example-9.4 page no-274

In [7]:
#given
#according to the data given in the question we can have the graph shown in the fig 9.11 in book
#part(a):- tangent modulus of elasticity at 200 MPa
E_tangent=(222-168)*10**9/(1.90-1.42)/10**9   #G Pa ( values from graph)
#part(b):- secant modulus of elasticity at 180 MPa
E_secant= (180-0)*10**9/(1.46-0)/10**9   #G Pa  (values from graph)
#part(c):-youngs modulus of elasticity at 85 MPa
E=(85-0)*10**6/((0.68-0)*10**-3)/10**9   #G Pa  (values from graph)
print """tangent modulus of elasticity at 200 MPa is %0.2f Pa
secant modulus of elasticity at 180 MPa is %d Pa
youngs modulus of elasticity at 85 MPa is %d Pa"""%(E_tangent,E_secant,E)
tangent modulus of elasticity at 200 MPa is 112.50 Pa
secant modulus of elasticity at 180 MPa is 123 Pa
youngs modulus of elasticity at 85 MPa is 124 Pa

example-9.5 page no-280

In [4]:
#given
#stress
sigmamean=70  #MPa
#stress range
sigmarange=210  #MPa
#as we know that
#sigmamean=(sigmamax+sigmamin)/2
#from this we get
#sigmamax+sigmamin=140    -----------(1)
#also
#sigmarange=sigmamax-sigmamin
#so we get
#sigmamax-sigmamin=210  ----------(2)
#solving 1 and 2
#we get
sigmamax=(140+210)/2   #MPa
#and
sigmamin=140-sigmamax  #MPa
#stress ratio
R=sigmamin/sigmamax
#stress ranges from sigmamin to sigmamax
#so total sum will be
totalsum=-sigmamin+sigmamax  #MPa
print """the maximum and minimum stresses are %d MPa and %d MPa resp,
stress ratio is %f
and total sum is %d MPa"""%(sigmamax,sigmamin,R,totalsum)
the maximum and minimum stresses are 175 MPa and -35 MPa resp,
stress ratio is -0.200000
and total sum is 210 MPa

example-9.7 page no-289

In [8]:
#given
#temp 
T=600  #degree C
#tightenening stress
sigmai=750    #kgf/cm**2
#minimum creep rate
vcr=2.8*10**-8  #cm/cm/hour
#stress
sigma=300  #kgf/cm**2
#young's modulus
E=2*10**6   #kgf/cm**2
#constant
n=3
#we knnow that minimum creep rate
#vcr=A*sigma**n
#so
A=vcr/sigma**n
#total time involved in creep of bolt is 1 year
t=365*24  #hours
#the stress relaxation in bolt due to creep is expresed by:-
#1/(sigmaif)**(n-1)=1/(sigmai)**(n-1)+A*E*(n-1)*t
#we have to find sigmaif
#so
sigmaif=((1/(sigmai)**(n-1)+A*E*(n-1)*t)**(1/(n-1)))**-1  
print  "the stress relaxation in bolt to creep is %0.3f kgf/cm**2"%(sigmaif)
the stress relaxation in bolt to creep is 161.975 kgf/cm**2