Chapter 7: Dislocation and Strengthening Mechanisms

Example 7.1 Page no 183

In [19]:
u1=1
v1=1
w1=0
u2=0
v2=1
w2=0
u3=-1
v3=1
w3=1

import math
phi=math.acos((u1*u2+v1*v2+w1*w2)/(math.sqrt((u1**2+v1**2+w1**2)*(u2**2+v2**2+w2**2))))
lam=math.acos((u3*u2+v3*v2+w3*w2)/(math.sqrt((u3**2+v3**2+w3**2)*(u2**2+v2**2+w2**2))))
sigma=52       #in MPa, Tensile stress
tr=sigma*math.cos(phi)*math.cos(lam)
trc=30         #in MPa Critical resolved shear stress
sy=trc/(math.cos(phi)*math.cos(lam))

print"(a)The resolved shear stress is ",round(tr,2),"MPa"
print"(b)Yield strength is",round(sy,1),"MPa"

(a)The resolved shear stress is  21.23 MPa
(b)Yield strength is 73.5 MPa

Example 7.2 Page no 194

In [6]:
df=12.2  #Final dia in mm
di=15.2  #Initial dia in mm

CW = ((di**2-df**2)/di**2)*100
ts=340  #in Mpa  tensile strength, from fig 7.19 (b)
duc=7   #in %  Ductility from fig 7.19 (c)

print"Percent Cold Work is ",round(CW,1),"%"
print"Tensile strength is",ts,"MPa"
print"Ductility is ",duc,"%"

Percent Cold Work is  35.6 %
Tensile strength is 340 MPa
Ductility is  7 %

Design Example 7.3 Page no 199

In [11]:
di=6.4  #Initial dia in mm
df=5.1  #Final dia in mm

CW = ((di**2-df**2)/di**2)*100
dmid = math.sqrt(df**2/(1-0.215))

print"Cold work is ",round(CW,1),"%"
print"But required ductility and yield strength is not matched at this cold work"
print"Hence required Cold work is 21.5 %"
print"Hence original diameter for second drawing is ",round(dmid,1),"mm"

Cold work is  36.5 %
But required ductility and yield strength is not matched at this cold work
Hence required Cold work is 21.5 %
Hence original diameter for second drawing is  5.8 mm
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