# Chapter 13 : Analysis of Metal Forming Process¶

## Exa 13.1 : page 515¶

In [10]:
from __future__ import division
from math import sqrt, pi, tan, exp
sigma_0 = 240 # N/mm**2
d1 = 5 # initial wire diameter in mm
d0 = 5.5 # final wire diameter in mm
x = d1/d0 # mm
alpha = 8 # angle of contact
alpha = alpha*pi/180
mu = 0.1 # coefficient of friction
B = mu/tan(alpha)
sigma_d  = (sigma_0*(1+B)*(1-(x)**(2*B)))/B # N/mm**2
l = 3 # die land in mm
mu = 0.1 # coefficient of friction
r1 = d1/2 # mm
sigma_t = sigma_0 - (sigma_0 - sigma_d)/exp((2*mu*l)/r1) # N/mm**2
dl = sigma_t*pi*(r1)**2 # drawing load in N
print "Total drawing load = %0.1f N"%(dl)
# Answers vary due to round off error

Total drawing load = 2136.5 N


## Exa 13.2 : page 517¶

In [11]:
from math import log10, tan
alpha = 15 # angle of contact
alpha = alpha*pi/180
bita = 0 # degree
mu = 0.1 # coefficient of friction
mu1 = mu
mu2 = mu
h1 = 1.75 # mm
h0 = 2.5 # mm
B = (mu1+mu2)/(tan(alpha)-tan(bita))
y1 = (1+B)*(1-(h1/h0)**B)/B  #sigma_d/sigma_0 for plug mandrels in N/mm**2
z = 1/((h0/h1)-1)
y2 = log10(z)# sigma_d/sigma_0 for movable mandrels in N/mm**2
print " The pipe drawing force force on plug mandrels = %0.3f \n The pipe drawing forcw on mandrels = %0.3f"%(y1,y2)
print " Use of movable mandrel substantially reduces drawing force"

 The pipe drawing force force on plug mandrels = 0.547
The pipe drawing forcw on mandrels = 0.368
Use of movable mandrel substantially reduces drawing force


## Exa 13.3 : page 517¶

In [12]:
from math import acos, atan, log, tan, sqrt, exp
h0 = 25 # thickness of plate in mm
h1 = 20 # mm
delta_h = h0-h1 # mm
sigma = 100 # maximum pressure in N/mm**2
D = 500 # rolled diameter in mm
r = D/2 # rolled radius in mm
alpha = acos(1-(delta_h/D)) # angle of contact in radians
mu = tan(alpha) # coefficient of friction
Ho = 2*sqrt(r/h1)*atan(sqrt(r/h1)*mu)
Hn = (Ho - (log(h0/h1))/mu)/2
hn = h1 + r*theta**2 # neutral section in mm
x = hn/h0
bs = (1-x)*100 # backward slip
y = hn/h1
fs = (y-1)*100 # forward slip
sigma0 = 2*sigma/sqrt(3)
pn = sigma0*hn*exp(mu*Hn)/h1 #N/mm**2
print "Neutral section = %0.1f mm"%(hn)
print "Backward slip = %0.1f percent\nForward slip = %0.1f percent"%(bs,fs)
print "Maximum pressure = %0.1f N/mm**2"%(pn)
# Answers vary due to round off error

Neutral section = 20.3 mm
Backward slip = 18.8 percent
Forward slip = 1.5 percent
Maximum pressure = 132.6 N/mm**2


## Exa 13.4 : page 519¶

In [13]:
from math import sqrt, pi ,exp
Do = 250 # diameter in mm
ho = 250 # hieght in mm
delta_h = 100 # mm
h = 150 # mm
sigma0 = 55 # N/mm**2
d = Do*sqrt(ho/(ho-delta_h)) # diameter in mm
mu = 0.42 # coefficient of friction
R = 162.5 # mm
pa = sigma0/2*(h/(mu*R))**2*(exp(2*mu*R/h)-2*mu*R/h-1) # N/mm**2
p = pa*pi*(R)**2 # force in kN
print "Force = %d kN"%(p/1000)

Force = 6328 kN


## Exa 13.5 : page 519¶

In [14]:
from sympy import symbols, diff, solve, exp
from math import log
d = 150 # diameter in mm
h = 10 # thickness in mm
R = d/2 # radius in mm
mu = 0.2 # coefficient of friction
sigma_0 = 200 # N/mm**2
Rs = R - (h/(2*mu))*log(1/(sqrt(3)*mu)) # sticking radius in mm
Ps = sigma_0*exp(2*mu*(R-Rs)/h) # pressure at sticking radius in N/mm**2
y=lambda r:2*pi*r*sigma_0*exp(2*mu/h*(R-r))
L_sld = L_sld/1000 # load on sliding portion in kN
Pc = Ps + (2*sigma_0*Rs)/(h*sqrt(3)) # pressure at centre in N/mm**2
L_sp = (Pc+Ps)*pi*(Rs)**2/(2*1000) # load on sticking portion in kN
F_l = L_sld + L_sp # total forging load in kN
print " Sticking radius = %0.1f mm \n Total forging load = %0.3f MN"%(Rs ,F_l/1000)
# Answers vary due to round off error

 Sticking radius = 48.5 mm
Total forging load = 11.927 MN


## Exa 13.7 : page 522¶

In [15]:
#from math import sqrt, pi
RA = 0.30
d = 12 # diameter in mm
alpha = 6  # angle of contact in degree
alpha = 6*pi/180 # angle of contact in radian
mu = 0.10 # coefficient of friction
sigma_0 = 240  # N/mm**2
B = mu/tan(alpha)
x = 1 - RA
sigma_d  = (sigma_0*(1+B)*(1-(x)**B))/B # N/mm**2
r1  = sqrt(x)*(d/2) # mm
l = sigma_d*pi*(r1)**2 # load in kN
ita = 98 # efficiency
ita = ita/100
s = 2.3 # drawing speed in m/s
P = (l*s)/ita # kW
print "Drawing load = %0.2f kN\nPower of motor = %0.2f kW"%(l/1000 ,P/1000 )
# Answers vary due to round off error

Drawing load = 11.21 kN
Power of motor = 26.32 kW


## Exa 13.8 : page 522¶

In [16]:
from math import sqrt, pi, tan
mu1 = 0.15 # coefficient of friction
mu2 = 0.18 # coefficient of fricton
alpha = 14 # angle of contact in degree
alpha = alpha*pi/180
bita = 10 # semi-cone angle in degree
bita = bita*pi/180
sigma_0 = 1.40 # kN/mm**2
h0 = 1.5 #mm
h1 = 1 # mm
B = (mu1+mu2)/(tan(alpha)+tan(bita))
sigmad = (sigma_0*(1+B)*(1-(h1/h0)**B))/B # drawing stress in kN/mm**2
d1 = 11 # outside diameter in mm
t = 1 # thickness in mm
d2 = d1-2*t # mm
a = (pi*((d1)**2-(d2)**2))/4 # area in mm**2
s = 0.65 # drawing speed in m/s
w = l*s # work in kJ/s
p = w # power in kW
print " Drawing load = %0.3f kN\n Power rating of motor = %0.2f kW"%(l , p)
# Answers vary due to round off error

 Drawing load = 27.166 kN
Power rating of motor = 17.66 kW


## Exa 13.9 : page 523¶

In [17]:
sigma_0 = 50 # pressure at start in MPa
B = 0.9 # width in m
h1 = 0.2 # thickness in m
b = 0.3 # tool bite in m
# At commencement of forging
FL = sigma_0*B*b*(1+(b/(4*h1))) # forging load in MN
# At completion of forging
h2 = 0.1 # thickness in m
sigma_0c = 150 # pressure at completion in MPa
FL2 = sigma_0c*B*b*(1+(b/(4*h2))) # forging load in MN
print " Forging load at start of forging = %0.4f MN\n Forging load at completion of forging = %0.3f MN"%(FL , FL2)

 Forging load at start of forging = 18.5625 MN
Forging load at completion of forging = 70.875 MN


## Exa 13.10 : page 524¶

In [18]:
from math import sqrt, tan, pi
sigma_0 = 250 # N/mm**2
d1 = 5 # initial wire diameter in mm
d0 = 15 # final wire diameter in mm
r0 = d0/2
r1 = d1/2
x = (r0/r1)**2 # mm
alpha = 45 # angle of contact
alpha = alpha*pi/180
mu = 0.1 # coefficient of friction
B = mu/tan(alpha)
sigma_x0  = (sigma_0*(1+B)*(1-(x)**B))/B # N/mm**2
sigma_x0 = -sigma_x0
l = 37.5 # length 0f billet in mm
tau1 = sigma_0/2 # Mpa
Pe = sigma_x0 + (4*tau1*l)/d0 # extrusion pressure in Mpa
el = Pe*pi*(r0)**2 # extrusion load in MN
print "Extrusion load = %d MN"%(el/10000)

Extrusion load = 34 MN


## Exa 13.11 : page 524¶

In [19]:
from math import cos, acos, sqrt, tan, exp, atan
h0 = 4.05 # thickness of plate in mm
h1 = 3.55 # mm
D = 500 # rolled diameter in mm
r = D/2 # rolled radius in mm
mu = 0.04 # coefficient of friction
sigma = 210 # N/mm**2
delta_h = h0-h1 # mm
p = 2*sigma/sqrt(3) # N/mm**2
alpha = acos(1-(delta_h/D)) # angle of contact
Ho = 2*sqrt(r/h1)*atan(sqrt(r/h1)*alpha)
Hn1 = (Ho - (log(h0/h1))/mu)/2
hn = h1 + 2*r*(1-cos(theta)) # mm
pn1 = p*hn*exp(mu*Hn1)/h1 # roll pressure in N/mm**2
# b) roll pressure when coefficient of friction is 0.4
mu2 = 0.4 # coefficient of friction
Hn2 = (Ho - (log(h0/h1))/mu2)/2
hn2 = h1 + r*theta**2 # mm
pn2 = (p*hn2*exp(mu2*Hn2))/h1 # roll pressure in N/mm**2
# c) if tension is applied of 35 N/mm**2
sigma_f = 35 # front tension in N/mm**2
pn3 = (p-sigma_f)*hn*exp(mu*Hn1)/h1 # roll ressure in N/mm**2
print "(a) Roll pressure at enter and exit = %0.1f N/mm**2\n    Roll pressure at neutral plane = %0.2f N/mm**2"%(p , pn1)
print "(b) Roll pressure at neutral point when co-efficient of friction is 0.40 = %0.2f N/mm**2"%(pn2)
print "(c) Roll pressure when 35 N/mm**2 tension is applied at neutral point = %0.2f N/mm**2"%(pn3)
# Answers vary due to round off error

(a) Roll pressure at enter and exit = 242.5 N/mm**2
Roll pressure at neutral plane = 257.81 N/mm**2
(b) Roll pressure at neutral point when co-efficient of friction is 0.40 = 779.97 N/mm**2
(c) Roll pressure when 35 N/mm**2 tension is applied at neutral point = 220.60 N/mm**2


## Exa 13.12 : page 526¶

In [20]:
from math import acos, atan, sqrt, log, pi
h1 = 6.35 # thickness in  mm
mu = 0.2 # coefficient of friction
r = 50 # rolled radius in cm
r = r*10 # mm
R = 30 # reduction in percent
h0 = h1*100/(100-R) # mm
delta_h = h0-h1 # mm
alpha = acos(1-(delta_h/(2*r))) # angle of contact
Ho = 2*sqrt(r/h1)*atan(sqrt(r/h1)*alpha)
Hn = (Ho - (log(h0/h1))/mu)/2
theta = sqrt(h1/r)*tan(sqrt(h1/r)*(Hn/2)) # neutral plane in radians
theta = theta*180/pi # neutral plane in degrees
print "Neutral plane = %0.2f degree"%(theta)
# 'Answers vary due to round off error'

Neutral plane = 1.58 degree