In [1]:

```
from math import pi
# Initialisation of Variables
f=1.12;#Geometry factor for the specimen and flaw
sigma=45000.;#Applied stress on Steel in psi
K=80000.;#The stress intensity factor
#CALCULATIONS
a=(K/(f*sigma))**2/pi;#Depth of crank in in
print "Depth of crank that will propagate in the steel in:",round(a,1)
```

In [2]:

```
# Initialisation of Variables
T=60000.;#Tensile strength Of Sialon (acronym for silicon aluminum oxynitride) in psi
sigma=500.;#The stress at which the part unexpectedly fails in psi
a=0.01;#Depth of thin crack in in
#CALCULATIONS
r=a/(T/(2*sigma))**2;#The radius of the crack tip in in
print "The radius of the crack tip in Angstroms",round(r*2.54*10**8)
#difference in answer is due to erronous caluctions
```

In [3]:

```
from math import sqrt,pi
# Initialisation of Variables
F=40000;# Maximum Tensile load in lb
K=9000;#Fracture toughness of Ceramic
w=3;# plate made of Sialon width
#CALCULATIONS
A=F*sqrt(pi)/K;#Area of ceramic
T=A/w;# Thickness of Ceramic
print "THickness of ceramic :",round(T,2)
```

In [4]:

```
from math import log,exp
# Initialisation of Variables
m=9;#Weibull modulus of an ceramic
sigma1=250;#The flexural strength in MPa
F1=0.4;#probability of failure
F2=0.1;#Expected the probability of failure
#CALCULATIONS
sigma2=exp(log(sigma1)-(log(log(1/(1-F1)))/m ));# The characteristic strength of the ceramic
sigma3=exp((log(log(1/(1-F2)))/m)+log(sigma2));#Expected level of stress that can be supported in MPa
print "The characteristic strength of the ceramic in MPa:",round(sigma2,1)
print "Expected level of stress that can be supported in MPa:",round(sigma3,1)
```

In [5]:

```
from math import log
# Initialisation of Variables
Ln1=0.5
Ln2=-2.0
sigma1=52;#the maximum allowed stress level on ceramic at one point in MP.
sigma2=23.5;#the maximum allowed stress level on ceramic at another point in MP.
#CALCULATIONS
m=(Ln1-Ln2)/(log(sigma1)-log(sigma2));#Weibull modulus of ceramic
print "Weibull modulus of ceramic:",round(m,2)
```

In [6]:

```
from math import pi
# Initialisation of Variables
N=5.256*10**5;#No. of cycles that the shaft will experience in one year
F=12500.;#applied load on shaft in lb
L=96.;#Length of Kliin produced from tool steel in in.
sigma1=72000.;#the applied stress on Shaft
f=2.;#Factor of saftey of shaft
sigma2=sigma1/f;#the maximum allowed stress level
#CALCULATIONS
d1=(16*F*L/(sigma1*pi))**(1./3.);#The Diameter of Shaft in in.
d2=(16*F*L/(sigma2*pi))**(1./3.);#The minimum diameter required to prevent failure
print "The Diameter of Shaft in in.:",round(d1,2)
print "The minimum diameter required to prevent failure in in.:",round(d2,2)
```