Chapter - 03, Forming Processes

Problem 1 on page no. 112

In [2]:
from __future__ import division
from math import tan, sqrt, atan, log
# Given that
A = 150*6 # Cross-section of strips in mm**2
ti = 6 # Thickness in mm
pA = 0.20 # Reduction in area
d = 400 # Diameter of steel rolls in mm
Ys = 0.35# Shear Yield stress of the material before rolling in KN/mm**2
Ys_ = 0.4# Shear Yield stress of the material after rolling in KN/mm**2
mu = 0.1 # Cofficient of friction

tf =0.8*ti
Ys_a = (Ys + Ys_)/2
r=d/2
thetaI = sqrt((ti-tf)/r)
lambdaI=2*sqrt(r/tf)*atan(thetaI *sqrt(r/tf))
lambdaN = (1/2)*((1/mu)*(log(tf/ti)) + lambdaI)
thetaN  =(sqrt(tf/r))*(tan((lambdaN/2)*(sqrt(tf/r))))
print """The final srip thickness is %0.2f mm,\nThe avg shear yield stress during the process is %0.3f KN/mm**2,
The angle subtended by the deformation zone at the roll centre is %0.4f rad, 
The location of neutral point is %0.3f rad."""%(tf,Ys_a,thetaI,thetaN)

The final srip thickness is 4.80 mm,
The avg shear yield stress during the process is 0.375 KN/mm**2,
The angle subtended by the deformation zone at the roll centre is 0.0775 rad, 
The location of neutral point is 0.023 rad.

Problem 2 on page no. 113

In [3]:
from math import exp, ceil
# Given that
A = 150*6 # Cross-section of strips in mm**2
w = 150 # Width of the strip in mm
ti = 6 # Thickness in mm
pA = 0.20 # Reduction in area
d = 400 # Diameter of steel rolls in mm
Ys = 0.35# Shear Yield stress of the material before rolling in KN/mm**2
Ys_ = 0.4# Shear Yield stress of the material after rolling in KN/mm**2
mu = 0.1 # Cofficient of friction
v = 30 # Speed of rolling in m/min
tf =0.8*ti
Ys_a = (Ys + Ys_)/2
r=d/2
thetaI = sqrt((ti-tf)/r)
lambdaI=2*sqrt(r/tf)*atan(thetaI *sqrt(r/tf))
lambdaN = (1/2)*((1/mu)*(log(tf/ti)) + lambdaI)
thetaN  =(sqrt(tf/r))*(tan((lambdaN/2)*(sqrt(tf/r))))
Dtheta_a = thetaN/4
Dtheta_b = (thetaI- thetaN)/8
print "The values of P_after are:\n"
for i in range(0,5,1):
    theta = i*Dtheta_a
    y = (1/2)* (tf+r*theta**2)
    lamda = 2*sqrt(r/tf)*atan(theta*sqrt(r/tf))
    p_a = 2*Ys_a*(2*y/tf)*(exp(mu*lamda))
    print "%0.3f \t\t"%p_a,

I1 = (Dtheta_a/3) *(0.75+.925+4*(.788+.876)+2*.830)# By Simpson's rule
print "\n\nThe values of P_before are:\n"
for i in range(0,8,1):
    theta1 = i*Dtheta_b + thetaN
    y = (1/2)* (tf+r*theta1**2)
    lamda = 2*sqrt(r/tf)*atan(theta1*sqrt(r/tf))
    p_b = 2*Ys_a*(2*y/ti)*(exp(mu*(lambdaI-lamda)))
    print " %0.3f \t\t"%p_b,

I2 = (Dtheta_b/3)*(0.925+.75+4*(.887+.828+.786+.759) + 2*(.855+.804+.772))#By Simpson's rule
F = r*(I1 + I2)
F_ = F*w
T = (r**2)*mu*(I2-I1)
T_ =T*w
W = v*(1000/60)/r
P = 2*T_*W
print "\n\nThe roll separating force = %d kN, The power required in the rolling process = %0.2f kW"%(ceil(F_),P/1000)
# Answer in the book for the power required in the rolling process is given as 75.6 kW
 

The values of P_after are:

0.750 		0.787 		0.829 		0.874 		0.924 		

The values of P_before are:

 0.924 		 0.887 		 0.855 		 0.828 		 0.805 		 0.786 		 0.771 		 0.759 		

The roll separating force = 1908 kN, The power required in the rolling process = 77.38 kW

Problem 3 on page no. 115

In [4]:
from math import pi
# Given that
A = 150*6 # Cross-section of strips in mm**2
w = 150 # Width of the strip in mm
ti = 6 # Thickness in mm
pA = 0.20 # Reduction in area
d = 400 # Diameter of steel rolls in mm
Ys = 0.35# Shear Yield stress of the material before rolling in KN/mm**2
Ys_ = 0.4# Shear Yield stress of the material after rolling in KN/mm**2
mu = 0.1 # Cofficient of friction
mu_ = 0.005 # Cofficient of friction in bearing 
D = 150 # The diameter of bearing in mm
v = 30 # Speed of rolling in m/min
tf =0.8*ti
Ys_a = (Ys + Ys_)/2
r=d/2
thetaI = sqrt((ti-tf)/r)
lambdaI=2*sqrt(r/tf)*atan(thetaI *sqrt(r/tf))
lambdaN = (1/2)*((1/mu)*(log(tf/ti)) + lambdaI)
thetaN  =(sqrt(tf/r))*(tan((lambdaN/2)*(sqrt(tf/r))))
Dtheta_a = thetaN/4
Dtheta_b = (thetaI- thetaN)/8
for i in range(0,5,1):
    theta = i*Dtheta_a
    y = (1/2)* (tf+r*theta**2)
    lamda = 2*sqrt(r/tf)*atan(theta*(pi/180) *sqrt(r/tf))*180/pi
    p_a = 2*Ys_a*(2*y/tf)*(exp(mu*lamda))

I1 = (Dtheta_a/3) *(0.75+.925+4*(.788+.876)+2*.830)
for i in range(0,8,1):
    theta1 = i*Dtheta_b + thetaN
    y = (1/2)* (tf+r*theta1**2)
    lamda = 2*sqrt(r/tf)*atan(theta1*(pi/180) *sqrt(r/tf))*180/pi
    p_b = 2*Ys_a*(2*y/ti)*(exp(mu*(lambdaI-lamda)))
I2 = (Dtheta_b/3)*(0.925+.75+4*(.887+.828+.786+.759) + 2*(.855+.804+.772))
F = r*(I1 + I2)
F_ = F*w
T = (r**2)*mu*(I2-I1)
T_ =T*w
W = v*(1000/60)/r
P_ = 2*T_*W
Pl = mu_*F_*D*W
P = Pl+P_
print "The mill power = %0.2f kW"%(P/1000)
# Answer in the book is given as 79.18 kW  whcih is wrong.
 

The mill power = 80.95 kW

Problem 4 on page no. 118

In [5]:
# Given that
mu = 0.25 # Cofficient of friction between the job and the dies 
Y = 7 # Avg yield stress of the lead in N/mm**2
h = 6 # Height of die in mm
L = 150 # Length of the strip in mm
V1 = 24*24*150 # Volume of the strip in mm**3
V2 = 6*96*150 # Volume of the die in mm**3
w= 96 # Weidth of the die in mm
from sympy.mpmath import quad
K = Y/sqrt(3)
x_ = (h/(2*mu))*(log(1/(2*mu)))
l = w/2
p1= lambda x:(2*K)*exp((2*mu/h)*x)
I1 = quad(p1,[0,x_])
p2 = lambda y:(2*K)*((1/2*mu)*(log(1/(2*mu))) + (y/h))
I2 = quad(p2,[x_,l])
F = 2*(I1+I2)
F_ = F*L
print "The maximum forging force = %0.2e N" %F_ 
# Answer in the book is given as 0.54*10**6 N which is wrong.
 

The maximum forging force = 4.89e+05 N

Problem 5 on page no. 119

In [6]:
# Given that
mu = 0.08# Cofficient of friction between the job and the dies 
Y = 7 # Avg yield stress of the lead in N/mm**2
h = 6 # Height of die in mm
L = 150 # Length of the strip in mm
V1 = 24*24*150 # Volume of the strip in mm**3
V2 = 6*96*150 # Volume of the die in mm**3
w= 96 # Weidth of the die in mm
K = Y/sqrt(3)
x_ = (h/(2*mu))*(log(1/(2*mu)))
l = w/2
p1 =lambda x:(2*K)*exp((2*mu/h)*x)
I = quad(p1,[0,l])
F = 2*(I)
F_ = F*L
print "The maximum forging force = %0.2e N"%F_ 

 

The maximum forging force = 2.36e+05 N

Problem 6 on page no. 122

In [7]:
# Given that
r = 150 # Radius of the circular disc of lead in mm
Ti = 50 # Initial thickness of the disc in mm
Tf = 25 # Reduced thickness of the disc in mm
mu = 0.25# Cofficient of friction between the job and the dies 
K = 4 # Avg shear yield stress of the lead in N/mm**2

R = r*sqrt(2)
rs = (R - ((Tf/(2*mu)) * log(1/(mu*sqrt(3)))))
p1 = lambda x:(((sqrt(3))*K)*exp((2*mu/Tf)*(R-x)))*x
I = quad(p1,[rs,R])
p2 = lambda y:((2*K/Tf)*(R-y) + ((K/mu)*(1+log(mu*sqrt(3)))))*y
I_ = quad(p2,[0,rs])
F = 2*pi*(I+I_)
print "The maximum forging force = %0.3e N"%F 

 

The maximum forging force = 3.649e+06 N

Problem 7 on page no. 126

In [8]:
# Given that
Di = 12.7 # Intial diameter in mm
Df = 10.2 # Final diameter in mm
v = 90 # Drawn speed in m/min
alpha=6 # Half angle of dia in degree
mu = 0.1# Cofficient of friction between the job and the dies 
Y = 207 # Tensile  yield stress of the steel specimen in N/mm**2
Y_ = 414 # Tensile yield stress of the similar specimen at strain 0.5 in N/mm**2
e = 0.5 # Strain

e_ =2* log(Di/Df)
Y_e = Y + (Y_ - Y)*e_/e
Y__ = (Y+Y_e)/2
phi = 1 + (mu/tan(alpha*pi/180))
Y_f = Y__ * ((phi/(phi-1)) *(1-((Df/Di)**(2*(phi-1)))))
p = Y_f * (pi/4)*(Df**2)*v/60
Dmax = 1- (1/(phi**(1/(phi-1))))
print "Drawing power = %0.3f kW, \nThe maximum passible reduction with same die = %0.2f mm"%(p/1000,Dmax)

Drawing power = 25.530 kW, 
The maximum passible reduction with same die = 0.50 mm

Problem 8 on page no. 130

In [9]:
# Given that
Ri = 30 # Inside radius of cup in mm
t = 3 # Thickness in mm
Rb = 40 # Radius of the blank in mm
K = 210 # Shear yield stress of the material in N/mm**2
Y = 600 # Maximum allowable stress in N/mm**2
Beta = 0.05
mu = 0.1# Cofficient of friction between the job and the dies 

Fh = Beta*pi*(Rb**2)*K
Y_r = (mu*Fh/(pi*Rb*t))+(2*K*log(Rb/Ri))
Y_z = Y_r*exp(mu*pi/2)
F = 2*pi*Ri*t*Y_z
Y_r_ = Y/exp(mu*pi/2)
Rp = (Rb/exp((Y_r_/(2*K)) - ((mu*Fh)/(2*pi*K*Rb*t))))-t
print "Drawing force = %d N,\nMinimum passible radius of the cup which can drawn from the given blank without causing a fracture = %0.2f mm"%(F,Rp)
# Answer in the book given as 62680 N

Drawing force = 89210 N,
Minimum passible radius of the cup which can drawn from the given blank without causing a fracture = 9.20 mm

Problem 9 on page no. 135

In [13]:
from math import cos, sin
# Given that
L_ = 20 # Length of the mild steel product in mm
h = 50 # Height of the mild steel product in mm
L = 50 # Horizontal length of the mild steel product in mm
t = 5 # Thickness in mm
l=25 # Length of the bend in mm
E = 207 # Modulus of elasticity in kN/mm**2
n = 517 # Strain hardening rate in N/mm**2
Y = 345 # Yield stress in N/mm**2
mu = 0.1# Cofficient of friction
e = 0.2 # Fracture strain
theta = 20 # Bend angle in degree

Rp = ((1 /((exp(e) - 1)))-0.82)*t/1.82
Y_1 = Y+n*e
Y_2 = Y + n*(log(1+(1/(2.22*(Rp/t)+1))))
M = ((0.55*t)**2)*((Y/6)+(Y_1/3)) + ((0.45*t)**2)*((Y/6)+(Y_2/3))
Fmax = (M/l)*(1+(cos((atan(mu))+mu*sin(atan(mu)))))
Fmax_ = L_*Fmax
alpha = 90 /((12*(Rp+0.45*t)*M/(E*(10**3)*(t**3)))+1)
Ls = 2*(((Rp+0.45*t)*pi/4) + 50-(Rp+t))
print "Maximum bending force = %d N,\nThe required puch angle = %0.2f°,\nThe stock length = %0.2f mm"%(Fmax_,alpha,Ls)
# Answer in the book for maximum bending force is given as 4144 N

Maximum bending force = 4121 N,
The required puch angle = 88.68°,
The stock length = 89.18 mm

Problem 10 on page no. 136

In [14]:
# Given that
L_ = 20 # Length of the mild steel product in mm
h = 50 # Height of the mild steel product in mm
L = 50 # Horizontal length of the mild steel product in mm
t = 5 # Thickness in mm
l=25 # Length of the bend in mm
E = 207 # Modulus of elasticity in kN/mm**2
n = 517 # Strain hardening rate in N/mm**2
Y = 345 # Yield stress in N/mm**2
mu = 0.1# Cofficient of friction
e = 0.2 # Fracture strain
theta = 20 # Bend angle in degree
F = 3000 # Maximum available force in N

Rp = ((1 /((exp(e) - 1)))-0.82)*t/1.82
Y_1 = Y+n*e
Y_2 = Y + n*(log(1+(1/(2.22*(Rp/t)+1))))
M = ((0.55*t)**2)*((Y/6)+(Y_1/3)) + ((0.45*t)**2)*((Y/6)+(Y_2/3))
Fmax = (M/l)*(1+(cos((atan(mu))+mu*sin(atan(mu)))))
Fmax_ = L_*Fmax
lmin  = Fmax_*l/F
Ls = 2*(((Rp+0.45*t)*pi/4) + 50-(Rp+t))
lmax = Ls / 2
Fmax_min  = Fmax_*l/lmax
print "Minimum value of die length = %0.2f mm,\nMinimum required capacity of the machine = %d N"%(lmin,ceil(Fmax_min))
# Answer in the book is give as 2323 N for Minimum required capacity of the machine

 

Minimum value of die length = 34.35 mm,
Minimum required capacity of the machine = 2312 N

Problem 11 on page no. 141

In [15]:
# Given that
d = 50 # Diameter of the billet in mm
L =75 # Length of the billet in mm
D = 10 # Final diameter of billet in mm
Y = 170 # Avg tensile yield stress for aluminium in N/mm**2
mu = 0.15 # Cofficient of the friction

l = L - ((d-D)/2)*tan(45*pi/180)
phi = 1+mu
Y_x = Y*(phi/(phi-1))*(((d/D)**(2*(phi-1)))-1)
F = (pi/4)*(d**2)*Y_x + (pi/sqrt(3))*(d*l*Y)
Pf = pi*Y*(d**2)*((phi/(2*mu))*(((d/D)**(2*mu))-1)-log(d/D)) + (pi/sqrt(3))*Y*d*l
Loss_f = (Pf/F)*100
Y_X = Y*4.31*log(d/D)
F_ = (pi/4)*(d**2)*Y_X + (pi/sqrt(3))*(d*l*Y)
Pf_1 = (pi/sqrt(3))*Y*(d**2)*(log(d/D))
Pf_2 = (pi/sqrt(3))*(d*l*Y)
Pf_ = Pf_1+Pf_2
Loss_f_ = (Pf_/F_)*100
print "Maximum force required for extruding the cylindrical aluminium billet = %d N"%F
print "Percent of the total power input will be lost in friction at the start of the operation = %0.2f %%."%(Loss_f_)
# Answer in the book given as 2436444 N for max force required for extruding the cylindrical aluminium billet

 

Maximum force required for extruding the cylindrical aluminium billet = 2436266 N
Percent of the total power input will be lost in friction at the start of the operation = 66.02 %.

Problem 12 on page no. 149

In [16]:
# Given that
d = 50 # Diameter of the steel sheet in mm
t = 3 # Thickness of the steel sheet in mm
e = 1.75 # True fracture strain
Y = 2.1e3 # True fracture stress for the material in N/mm**2

C_0 = (t/(1.36*exp(e)))*((2*exp(e))-1)/((2.3*exp(e))-1)
p = t*(1/2.45)*((1.9*exp(e))-1)/((2.56*exp(e))-1)
F = Y*C_0*pi*d
W = (1/2)*(F)*(p)*(10**-3)
print "The proper clearance between die and punch = %0.2f mm"%C_0
print "Maximum punching force = %0.2f N, \nEnergy required to punch the hole = %0.2f J"%(F/1000,W)
# Answer in the book given as 45.74 J for energy required to punch the hole
 

The proper clearance between die and punch = 0.33 mm
Maximum punching force = 108.61 N, 
Energy required to punch the hole = 48.10 J