# Chapter 10:Dislocation and Strengthing Mechanism of Metals¶

## Example 10.1 Page 356¶

In [1]:
#initiation of variable
from math import acos, asin, sqrt, cos, sin
u1 = 0.0 # Bravais index
v1 = 1.0 # Bravais index
w1 = 0.0 # Bravais index
u2 = 1.0 # Bravais index
v2 = 1.0 # Bravais index
w2 = 0.0 # Bravais index
u3 = -1.0 # Bravais index
v3 = 1.0 # Bravais index
w3 = 1.0 # Bravais index
tau_r1 = 30.0 # Critical resolved shear stress
sigma = 52.0 # Tensile strength in MPa

#Part A:
#calculation
phi = acos((u1*u2+v1*v2+w1*w2)/sqrt((u1**2+v1**2+w1**2)*(u2**2+v2**2+w2**2)))
Lambda = acos((u3*u1+v3*v1+w3*w1)/sqrt((u1**2+v1**2+w1**2)*(u3**2+v3**2+w3**2)))
tau_r = sigma*cos(phi)*cos(Lambda)

#results
print" Resolved shear stress is %.1f MPa" %tau_r
print "Answer in book is 21.3 MPa which is due to approximation"

#Part B
sigma1 = tau_r1/(cos(phi)*cos(Lambda))

#result
print" Applied tensile force to initiate yielding is %.1f MPa" %sigma1
print "Answer in book is 73.4 MPa which is due to approximation"

 Resolved shear stress is 21.2 MPa
Answer in book is 21.3 MPa which is due to approximation
Applied tensile force to initiate yielding is 73.5 MPa
Answer in book is 73.4 MPa which is due to approximation


## Example 10.2 Page 366¶

In [2]:
#initiation of variable
d1 = 15.2 # Initial diameter in mm
d2 = 12.2 # Final diameter in mm

#calculation
per_CW = (d1**2 - d2**2)*100/d1**2
# Some values are deduced from figures

#result
print" The tensile strength is read directly from the curve for copper(figure 10.9b) as 340 MPa From figure 10.19c, the Ductility at %0.1f CW is about 7%% EL." %per_CW

 The tensile strength is read directly from the curve for copper(figure 10.9b) as 340 MPa From figure 10.19c, the Ductility at 35.6 CW is about 7% EL.


## Example 10.3 (Design Problem 10.1) Page 371¶

In [3]:
#initiation of variable
from math import acos, asin, sqrt, cos, sin
d1 = 6.4 # Initial diameter in mm for first drawing
sigma = 345.0 # tensile strength in MPa
el = 20.0 # ductility in percent
d2 = 5.1 # Final diameter in mm
per_cw = 21.5 # deformation

#calculation
per_CW = (d1**2 - d2**2)*100/d1**2
d0 = sqrt((d2**2*100)/(100.0-el))

#result
print" Theoretical %% Cold Work is %.1f" %per_CW
print" Original Diameter for second drawing is %.1f mm" %d0
print "Answer in book is 5.8 mm which is due to approximation at intermediate steps"

 Theoretical % Cold Work is 36.5
Original Diameter for second drawing is 5.7 mm
Answer in book is 5.8 mm which is due to approximation at intermediate steps