# Chapter15-Fracture Mechanics¶

## Ex2-pg559¶

In [1]:
#calculate the magnitude of the fracture of load
## initialization of variables
import math
d=250. ##mm
c=30. ##mm
t=25. ##mm
## part (a)
a=5 ##mm
lambd=a/(2.*c)
f1l=1.22 ##from the tble
f2l=1.02
##We don't know P yet so say P=1
P=1
Sfl=P/(t*2.*c)*f1l+3.*280.*P*f2l/(2.*t*c**2)
K_IC=59.*math.sqrt(1000.)
P=K_IC/(Sfl*math.sqrt(a*math.pi))
print('part (a)')
print'%s %.2f %s'%('\n P = ',P/10**3,' kN')
## part (b)
a=10. ##mm
lambd=a/(2.*c)
f1l=1.33 ##from the tble
f2l=1.05
## We don't know P yet so say P=1
P=1.
Sfl=P/(t*2.*c)*f1l+3.*280.*P*f2l/(2.*t*c**2)
K_IC=59.*math.sqrt(1000.)
P=K_IC/(Sfl*math.sqrt(a*math.pi))
print('\n part (b)')
print'%s %.2f %s'%('\n P = ',P/10**3,' kN')

part (a)

P =  23.71  kN

part (b)

P =  16.25  kN


## Ex3-pg560¶

In [2]:
#calculate the  nominal stress
## initialization of variables
import math
a=100./2. ##mm
Y=1500. ##MPa
t=6. ##mm
w=800. ##mmm
c=200. ##mm
a_c=a/c
fl=1.045
w=w*10**-3
t=t*10**-3
a=a*10**-3
A=w*t
Sigma=1./A
K_I=Sigma*math.sqrt(math.pi*a)*fl
print('part (a)')
print'%s %.2f %s'%('\n K_I = ',K_I,' MPa sqrt(m)')

part (a)

K_I =  86.28  MPa sqrt(m)


## Ex4-pg560¶

In [3]:
#calculate the  maximum allowable crack length of central crack in the sheet of steel
## initialization of variables
import math
S_u=1300. ##MPa
K_C=69. ## MPa sqrt(m)
SF=2.2
##calculations
S_c=S_u/2.2
a=1./math.pi*(K_C/S_c)**2
print'%s %.2f %s'%('a = ',a*10**3,' mm')

a =  4.34  mm


## Ex5-pg561¶

In [4]:
#calculate the aveage fracture stress
## initialization of variables
import math
## For 30 mm crack
a=30./2. ## mm  crack
S_30 =600. ##MPa
a=a*10**-3
C=S_30*math.sqrt(a)
## For 120 mm crack
a=120/2.
a=a*10**-3
S_120=C/math.sqrt(a)
print'%s %.2f %s'%('Sigma_120 = ',S_120,' MPa')

Sigma_120 =  300.00  MPa