Chapter 8 Strain Hardening and Annealing

Example 8_1 pgno:300

In [1]:
# Initialisation of Variables
t0=1;#Thickness of Copper plate in cm
tf=0.50;#Cold reducetion of coopper in cm in step1
tf2=0.16;# Further Cold reduction of cooper in cm in step2
#CALCULATIONS
CW1=((t0-tf)/t0)*100;#Amount of Cold work accomplished in step1
CW2=((tf-tf2)/tf)*100;#Amount of Cold work accomplished in step2
CW=((t0-tf2)/t0)*100;#Actual Total Cold work in percent
print "Amount of Cold work accomplished in step1:",CW1 
print "Amount of Cold work accomplished in step2:",CW2
print "Actual Total Cold work in percent:",CW

Amount of Cold work accomplished in step1: 50.0
Amount of Cold work accomplished in step2: 68.0
Actual Total Cold work in percent: 84.0

Example 8_2 pgno:301

In [2]:
# Initialisation of Variables
tf=0.1;#Thickness of cooper to produce in cm
CW1=40.;#cold work to produce a tensile strengthof 65,000 psi
CW2=45.;#cold work to produce a tensile strengthof 60,000 psi
#CALCULATIONS
Tmax=(tf/(1-(CW1/100)));#Maximum thicknessproduced in step1 in cm
Tmin=(tf/(1-(CW2/100)));#Minimum thicknessproduced in step2 in cm
print "Maximum thicknessproduced  in cm:",round(Tmax,3)
print "Minimum thicknessproduced  in cm:",round(Tmin,3)

Maximum thicknessproduced  in cm: 0.167
Minimum thicknessproduced  in cm: 0.182

Example 8_4 pgno:307

In [3]:
from math import pi
# Initialisation of Variables
D0=0.40;#Let's assume that the starting diameter of the copper wire in in.
Df=0.20;# Diameter of the copper wire to be produced in in.
sigma1=22000;#Yeidl strength at 0% cold work
#CALCULATIONS
CW=((D0**2-Df**2)/D0**2)*100;#The fianal Cold Work in percent
F=sigma1*(pi/4)*D0**2;#The draw force required to deform the initial wire in lb
sigma2=F/((pi/4)*Df**2);# The stress acting on the wire after passing through the die in psi
print CW,"The fianal Cold Work in percent:"
print "The draw force required to deform the initial wire in lb:",F
print "The stress acting on the wire after passing through the die in psi:",sigma2

75.0 The fianal Cold Work in percent:
The draw force required to deform the initial wire in lb: 2764.60153516
The stress acting on the wire after passing through the die in psi: 88000.0

Example 8_5 pgno:313

In [8]:
# Initialisation of Variables
t0=5;#Assming we are able to purchase only 5-cm thick stock
t02=1;#Thickness of strip in cm
tf=0.182;#Final thickness of strip in cm
CW2=80;#cold work  of a strip in percent
M=1085;# The melting point of copper in degree celsius
#CALCULATIONS
CW=((t0-tf)/t0)*100;#Cold work between from 5 to 0.182 cm in percent
tf2=(1-(CW2/100))*t0;# Final Thickness of strip in cm
Tr=0.4*(M+273);# Recrystallization temperature By using 0.4Tm relationship in degree celsius
CW3=((t02-tf)/t02)*100;#Cold work of the strip of 1 cm thickness 
print "Cold work between from 5 to 0.182 cm in percent:",CW
print "1. Final Thickness of strip in cm",tf2
print "2. Recrystallization temperature By using 0.4Tm relationship in degree celsius:",Tr-273
print "3. Cold work of the strip of 1 cm thickness :",CW3

Cold work between from 5 to 0.182 cm in percent: 96.36
1. Final Thickness of strip in cm 5
2. Recrystallization temperature By using 0.4Tm relationship in degree celsius: 270.2
3. Cold work of the strip of 1 cm thickness : 81.8

Example 8_6 pgno:316

In [6]:
# Initialisation of Variables
t0=5;#We are able to purchase strip of 5cm thickness in cm
tf=0.182;#Thickness to be produced in cm
tf2=0.167;#Thickness to procedure in cm
#CALCULATIONS
HW=((t0-tf)/t0)*100;#Hot work for a strip from 5cm to 0.182 cm in percent
HW2=((t0-tf2)/t0)*100;#Hot work for a strip from 5cm to 0.167 cm in percent
print "Hot work for a strip from 5cm to 0.182 cm in percent:",round(HW,1)
print "Hot work for a strip from 5cm to 0.167 cm in percent",round(HW2,1)

Hot work for a strip from 5cm to 0.182 cm in percent: 96.4
Hot work for a strip from 5cm to 0.167 cm in percent 96.7