Ch-11, Theories of inelastic, enelastic and viscoelastic deformation and fracture

example-11.1 page no- 343

In [13]:
from __future__ import division
from math import sqrt
#given
#applied stress
sigmax=3.5  #MPa
#aluminium crystal slips from (111) plane in the direction (110) when the stess is applied to (1-11)
#so
h1=1
k1=1
l1=1
h2=1
k2=-1
l2=1
#magnitude of plane (111)
M1=sqrt(h1**2+k1**2+l1**2)
#magnitude of (1-11)
M2=sqrt(h2**2+k2**2+l2**2)
#direction (110)
h3=1
k3=1
l3=0
#magnitude of direction(110)
M3=sqrt(h3**2+k3**2+l3**2)
#the angle between the planes (111) and (1-11) is 
cosphie=((h1*h2+k1*k2+l1*l2)/(M1*M2))
sinphie=sqrt(1-(cosphie)**2)
#similarly angle between the plane (111) and the direction (110) is given by
costheta=((h1*h3+k1*k3+l1*l3)/(M1*M3))  
#critical resolved shear stress
taucr=sigmax*2*sinphie*cosphie*costheta/2  #MPa
print "the critical resolved shear stress is %0.3f MPa"%(taucr)

the critical resolved shear stress is 0.898 MPa

exaplme-11.2 page no-351

In [14]:
#given
#shera stress
tau=715*10**6  #Pa
#shear modulus
G=25*10**9  #Pa
#atomic radius
b=4.05*10**-10  #m
#as we know that 
#tau=G*b/l
#so
l=G*b/tau  #m
print "the length of frank- read source in aluminium crystal is %.10f m"%(l)

the length of frank- read source in aluminium crystal is 0.0000000142 m

example-11.2 page no-351

In [15]:
#given
#shear modulous
G=25*10**9  #Pa
#shear stress
tau=50*10**6  #Pa
#lattice constant of aluminium is
a=4.05*10**-10  #m
#burger's vector for aluminium is 1/2(110)
h=1
k=1
l=0
#atomic radius of aluminium 
b=a/sqrt(h**2+k**2+l**2) # m
#as we know that
#tau=G*b/L
#so
L=G*b/tau  #m
#disloaction density is rhoD (let)
rhoD =1/L**2  # per m**2   (calculation mistake, there in book formula written is correct but calculation is wrong)
print "the dislocation density is %0.3e per m**2"%(rhoD)

the dislocation density is 4.877e+13 per m**2

example-11.4 page no-352

In [16]:
#given
#burger vector
b=4*10**-10  #m
#density
N=10**13  #lines/m**2
#velocity
v=10**-5  #m/s
#rate of straining
epsilonP=N*b*v  #lines/sec
print "the rate of starining is %0.3f lines/sec"%(epsilonP)

the rate of starining is 0.040 lines/sec

example-11.5 page no-357

In [17]:
from math import log, exp
#given
#conventional stress
sigmac=98.9  #MPa
#conventional strain
epsilonc=0.35  #mm/mm
#as we know that
epsilont=log(1+epsilonc)   #mm/mm
#also
#sigmac=sigmat*exp(-epsilont)  
#so
sigmat=sigmac/exp(-epsilont)  #MPa
print "the true strees is %0.3f MPa\n and true strain is %0.3f mm/mm"%(sigmat,epsilont)

the true strees is 133.515 MPa
 and true strain is 0.300 mm/mm

example-11.6 page no-357

In [19]:
#given
#material constant
K=500  #MPa
#strain hardenening coefficient
n=0.20
#according to true stress-strain relation
#sigmat=d sigmat/d epsilont=d(Kepsilont**n)/d epsilont
#so we get
#sigmat=n*K*epsilont**(n-1)  -----(1)
#also we know that
#sigmat=K*epsilont**n  ------------(2)
#equating (1) and (2)
epsilont=n
#also we know that
#epsilont=log (1+epsilonc)
#so
epsilonc=exp(epsilont)-1  
#now
sigmat=K*epsilont**n  #MPa
print """the maximum tensile strength %0.3f
the linear strain are %0.3f MPa"""%(epsilonc,sigmat)

the maximum tensile strength 0.221
the linear strain are 362.390 MPa

exapmle-11.7 page no-363

In [20]:
#given
#applied strain
epsilon=0.4
#immediate stress
sigmai=10*10**9  #Pa
#after 42 days stress is
sigma=5*10**9  #Pa
t=42  #days
#as we know that
#sigma=sigmai*exp(-t/tr)
#so
tr=t/log(sigmai/sigma)   #days
#stress after 90 days
t90=90 #days
sigma90=sigmai*exp(-t90/tr)*10**-9   #MPa
print """relaxation time for the tensile stress to decrease from 10 to 5 MPa is %0.3f days
the stress after 90 days is %0.3f MPa"""%(tr,sigma90)

relaxation time for the tensile stress to decrease from 10 to 5 MPa is 60.593 days
the stress after 90 days is 2.264 MPa

example-11.8 page no-370

In [22]:
from math import pi
#given
#crack length of glass piece
l=3*10**-6/2  #m divided by 2 because min general we consider 2*l be the length of crack
#young modulus
E=70*10**9  #Pa
#specific surface energy
gammae=1.05  #/J/m**2
#fracture strength
sigmaf=sqrt(2*E*gammae/(pi)/l) #N/m**2
#ratio of strength and young's modulus
R=sigmaf/E   
print """fracture strength is %0.3e N/m**2
the ratio of strength and youngs modulus is %0.3f"""%(sigmaf,R)

fracture strength is 1.766e+08 N/m**2
the ratio of strength and youngs modulus is 0.003

exaple-11.9 page no-370

In [23]:
#given
#young's modulus
E=71*10**9   #Pa
#fracture strength
sigmaf=115*10**6  #Pa
#lenght of crack
#2l=6 micro m
#so
l=6/2*10**-6  #m
#as we know that
#sigmaf=sqrt(2*E*sigmac/(pi)/l)
#so
sigmac=(sigmaf)**2*(pi)*l/2/E  #J/m**2 
print "surface energy of the etched glass is %0.3f J/m**2"%(sigmac)

surface energy of the etched glass is 0.878 J/m**2

example-11.10 page no-370

In [24]:
#given
E=70*10**9  #Pa
gammae=0.85  #J/m**2
l=2*10**-6  #m
#necessary stress to satisfy the griffith's energy
sigmaf=sqrt(2*E*gammae/(pi)/l)/10**6 #Pa
print "the necessary stress is %0.3f MPa"%(sigmaf)

the necessary stress is 137.621 MPa

example-11.11 page 371

In [26]:
#given
E=72*10**9  #Pa
gammae=0.9  #J/m**2
sigmaf=17.5*10**6  #Pa
#as we know that
#sigmaf=sqrt(2*E*gammae/(pi)/l) #N/m**2
#so
l=2*E*gammae/(pi)/(sigmaf)**2*1000  #mm
#length of intenal crack
L=2*l  #m
print """length of crack on the outer surface is %0.3f mm
length of crack internally %0.3f mm"""%(l,L)

length of crack on the outer surface is 0.135 mm
length of crack internally 0.269 mm