# Chapter 2 - Press Tool Design¶

## Exa 2.1 : page 132¶

In [4]:
from __future__ import division
from math import pi, sqrt
D = 50 # Diameter of washer in mm
t = 4 # thickness of material in mm
d = 24 # diameter of hole in mm
p = 360 # shear strength of material in N/mm**2
F1 = pi*D*t*p # blanking pressure in N
F2 = pi*d*t*p # piercing pressure in N
F = F1 + F2 # total pressure in N
d1 = d + 0.4 # piercing die diameter in mm
d2 = D - 0.4 # blank punch diameter in mm
c = 0.8*F # press capacity in N
print " Blanking pressure = %d kN\n Piercing pressure = %0.3f KN\n Total pressure required = %0.1f KN"%(F1/1000,F2/1000,F/1000)
print " piercing punch diameter = %0.2f cm\n blanking punch diametre = %0.2f cm \n press capacity = %0.2f KN\n"%(d1/10 , d2/10 , c/1000)
# Answers vary due to round off error

 Blanking pressure = 226 kN
Piercing pressure = 108.573 KN
Total pressure required = 334.8 KN
piercing punch diameter = 2.44 cm
blanking punch diametre = 4.96 cm
press capacity = 267.81 KN



## Exa 2.2 : page 133¶

In [5]:
from math import sqrt, pi
h = 10 # height of cup in cm
d = 5 # diameter of cup in cm
D = sqrt(d**2 + 4*d*h) # blank diameter in cm
# height daimeter ratio is 2 . Therefore from table 2.9 total reductions are 3
r1 = 0.45*D # first reduction is 45%
d1 = D - r1 # diameter at first draw in cm
r2 = d1*0.25 # second reduction is 25%
d2 = d1 - r2 # diameter at second draw in cm
r3 = d2*0.2 # third reduction is 20%
d3 = d2 - r3 # diameter at third draw in cm
print " Diameter at first draw  = %0.2f cm\n Diameter at second draw  = %0.2f cm\n Diameter at third draw  =%0.3f cm"%( d1 , d2 , d3)


 Diameter at first draw  = 8.25 cm
Diameter at second draw  = 6.19 cm
Diameter at third draw  =4.950 cm


## Exa 2.3 : page 133¶

In [6]:
K = 1.20 # die-opening factor
L = 37.5 # Length of strip in cm
T = 2.5 # thickness of strip in mm
sigma_ut = 630 # tensile strength in N/mm**2
W = 16*T # width of die opening in mm
F = (K*L*10*sigma_ut*T**2)/W # bending force in N
print "bending force = %0.1f KN"%( F/1000)

bending force = 44.3 KN


## Exa 2.4 : page 134¶

In [7]:
b = 25 # width of blank in mm
l = 30 # length of blank in mm
tau = 450 # ultimate shear stress of material in N/mm**2
t = 1.5 # thickness of metal strip in mm
p = 2*(l + b) # perimeter of blank in mm
f = p*t*tau # blanking force in N
Pt = 0.25*t # punch travel in mm
w = f*Pt # work done in Nmm
print " blanking force  = %0.2f KN\n work done  = %0.2f Nm"%(f/1000,w/1000)

 blanking force  = 74.25 KN
work done  = 27.84 Nm


## Exa 2.5 : page 134¶

In [8]:
t = 1.5 # thickness in mm
c = 0.05*t # clearance in mm
D = 25.4 # outside diameter in mm
D_o = D - 0.05 # blank die opening in mm
B_s = D_o - 2*c # blanking punch size in mm
d = 12.7 # internal diameter in mm
P_s = d + 0.05 # piercing punch size in mm
D_s = P_s + 2*c # piercing die size in mm
print " clearance = %0.3f mm\n blank die opening size = %0.2f mm "%(c ,D_o)
print " blanking punch size = %0.2f mm\n piercing punch size = %0.2f mm\n piercing die size = %0.2f mm"%(B_s,P_s,D_s )

 clearance = 0.075 mm
blank die opening size = 25.35 mm
blanking punch size = 25.20 mm
piercing punch size = 12.75 mm
piercing die size = 12.90 mm


## Exa 2.6 : page 135¶

In [9]:
from math import pi
D = 25.4 # outside diameter in mm
d = 12.7 # internal diameter in mm
t = 1.5 # thickness in mm
tau = 280 # ultimate shearing strength in N/mm**2
F = pi*(D + d)*t*tau # total cutting force in N
F_s = pi*D*t*tau # cutting force when punches are staggered in N
k = 0.6 # penetration
i = 1 # shear of punch in mm
F_p = (t*k*F)/(k*t +i)# cutting force when both punches act together in N
print "shear force when both punch act at same time and no shear is applied = %0.2f kN"%( F/1000)
print "cutting force when punches are staggered = %0.1f kN"%(F_s/1000)
print "cutting force when there is penetration and shear on punch = %0.1f kN"%(F_p/1000)

shear force when both punch act at same time and no shear is applied = 50.27 kN
cutting force when punches are staggered = 33.5 kN
cutting force when there is penetration and shear on punch = 23.8 kN


## Exa 2.7 : page 135¶

In [10]:
from math import pi
D = 60 # hole diameter in mm
tau = 450 # ultimate shear strength in N/mm**2
t = 2.5 # thickness in mm
F = pi*D*t*tau # maximum punch force in N
w = F*0.4*t # work done in Nm
f = F/2 # punching force in N
k = 0.4 # penetration percentage
i = k*t*(F-f)/f # shear on punch in mm
print "Amount of shear on punch = %d mm"%( i)

Amount of shear on punch = 1 mm


## Exa 2.8 : page 136¶

In [11]:
from math import ceil
# from fig 2.27
w = 2.5 # cm
t = 3.2 # strip thickness in mm
h = 10 # cm
a = t + 0.015*h*10 # back scrap and front scrap in mm
b = t # scrap bridge in mm
W = h + (2*a)/10 # width of strip in cm
W = ceil(W) # cm
s = w + b/10 # length of one piece of stock in cm
L = 240 # length of strip in cm
N = (L-b)/s # number of parts
y = L - (N*s + b/10)  # scrap remaining at the end in mm
print " Value of back scrap = %0.1f mm\n Value of scrap bridge = %0.1f mm "%( a , b  )
print " Width of strip = %0.2f cm\n Length of one piece of stock needed to produce one part = %0.2f cm"%( W , s)
print " Number of parts = %0.1f blanks\n Scrap remaining at the end = %0.2f mm"%( N , y)
# Answers vary due to round off error

 Value of back scrap = 4.7 mm
Value of scrap bridge = 3.2 mm
Width of strip = 11.00 cm
Length of one piece of stock needed to produce one part = 2.82 cm
Number of parts = 84.0 blanks
Scrap remaining at the end = 2.88 mm


## Exa 2.9 : page 137¶

In [12]:
from math import pi, sqrt
# from figure 2.73
t = 0.8 # thickness in mm
d = 50 # shell diameter in mm
r = 1.6 # radius of bottom corner in mm
h = 50 # height in mm
D = sqrt(d**2 + 4*d*h) # shell blank size in mm
el = 6.4 # extra length required to add in shell blank size
D = D + el # mm
pr = 100*(1-(d/D)) # percentage reduction
ratio = h/d
n = 2 # number of draws
R1 = 45 # first reduction
D1 = D - R1*D/100 # diameter at first reduction in mm
R2 = 100*(1-(d/D1)) # second reduction
PR = 4*t # punch radius in mm
PR = ceil(PR)
DR = 6 # die radius in mm
DC1 = 0.87 # die clearance for first draw in mm
DC2 = 0.88 # die clearance for second draw in mm
PD2 = d - 2*t # punch diameter for second draw in mm
DD2 = PD2 + 2*DC2 # Die opening diameter for second draw in mm
PD1 = D1 - 2*t # punch diameter for first draw in mm
DD1 = D1 + 2*DC1 # Die opening diameter for first draw in mm
# Drawing pressure
c = 0.65 # constant
sigma = 427 # N/mm**2
F = pi*d*t*sigma*(D/d-c) # Drawing pressure in mm
Bhp = 30.8 # blanking holding pressure in kN
pc = 150 # press capacity in kN
print "(i) size of blank = %0.2f mm \n(ii) Percentage reduction = %0.1f percent \n(iii) Number of draws = %d \n(iv) Radius on punch = %d mm \n Die Radius = %d mm \n(v) Die clearance for first draw = %0.2f mm \n die clearance for second draw = %0.2f mm"%( D , pr,n,PR,DR,DC1,DC2)
print " Punch diameter for second draw = %0.1f mm \n Die opening diameter for second draw = %0.2f mm \n Punch diameter for first draw  = %0.3f mm \n Die opening diameter for first draw = %0.3f mm\n(vi) Drawing pressure = %0.2f mm \n(vii) Blank holding pressure = %d kN \n(viii) Press capacity = %d kN"%( PD2 , DD2 ,PD1 ,DD1 , F/1000 ,Bhp , pc)
# Answers vary due to round off error

(i) size of blank = 118.20 mm
(ii) Percentage reduction = 57.7 percent
(iii) Number of draws = 2
(iv) Radius on punch = 4 mm
(v) Die clearance for first draw = 0.87 mm
die clearance for second draw = 0.88 mm
Punch diameter for second draw = 48.4 mm
Die opening diameter for second draw = 50.16 mm
Punch diameter for first draw  = 63.412 mm
Die opening diameter for first draw = 66.752 mm
(vi) Drawing pressure = 91.97 mm
(vii) Blank holding pressure = 30 kN
(viii) Press capacity = 150 kN


## Exa 2.10 : page 138¶

In [13]:
from math import pi
# from figure 2.74
l1 = 76 - ( 2.3 + 0.90) # length1 in mm
l2 = 115 - (2.3 + 0.90) # length2 in mm
t = 2.3 # mm
r = 0.90 # inner radius in mm
k = t/3 # mm
B = 0.5*pi*(r + k) # bending allowance in mm
d = l1 + l2 + B # developed length in mm
print "Developed length = %0.2f mm"%( d)
# Answers vary due to round off error

Developed length = 187.22 mm


## Exa 2.11 : page 139¶

In [14]:
from math import floor
t = 3.2 # thickness of blank in mm
l = 900 # bending length in mm
sigma = 400 # ultimate tensile strength in N/mm**2
k = 0.67 # die opening factor
r1 = 9.5 # punch radius in mm
r2 = 9.5 # die edge radius in mm
c = 3.2 # clearance in mm
w = r1 + r2 + c # width between contact points batween die and punch in mm
F= (k*l*sigma*t**2)/w # bending force in N
F=floor(F/10)*10 # N
print "bending force = %0.2f kN"%(F/1000)

bending force = 111.25 kN


## Exa 2.12 : page 139¶

In [15]:
k = 1.33 # die opening factor
l = 1200 # bend length in mm
sigma = 455 # ultimate tensile strength in N/mm**2
t = 1.6 # blank thickness in mm
w = 8*t # width of die opening in mm
F = k*l*sigma*t**2/w # bending force in N
print "bending force  = %0.2f kN"%(F/1000)

bending force  = 145.24 kN


## Exa 2.13 : page 140¶

In [16]:
c = 1.25 # clearance in mm
r1 = 3 # die radius in mm
r2 = 1.5 # punch radius in mm
sigma = 315 # ultimate tensile strength in MPa
t = 1 # thickness in mm
l = 50 # width at bend in mm
w = r1 + r2 +c # width between contact points on die and punch in mm
F = 0.67*l*sigma*t**2/w # bending force in N
F_p = 0.67*sigma*l*t # pad force in N
sigma_c = 560 # setting pressure in MPa
b1 = 2 # beads on punch
b = b1*r1 # mm
F_b = sigma_c*l*b # bottoming force in N
F_o = F_p + F_b # Force required when bottoming is used in N
F_n = F +F_p # Force required when bottoming is not used in N
print "Force required when bottoming is used  = %0.1f tonnes"%(F_o/(9.81*1000))
print "Force required when bottoming is not used  = %0.3f tonnes"%( F_n/(9.81*1000))

Force required when bottoming is used  = 18.2 tonnes
Force required when bottoming is not used  = 1.263 tonnes


## Exa 2.14 : page 140¶

In [17]:
from math import sin, pi
w = 2 # width in mm
t = 5 # thickness in mm
theta=6 # shear in degrees
tau = 382.5 # ultimate shear stress in MPa
F = w*t*tau*1000 # cutting force in N
l = t/sin(theta*pi/180) # length to be cut in mm
F_i = l*t*tau # cutting force in N
print "cutting force with parallel cutting edges = %0.3f MN\ncutting force when shear is considered = %0.2f kN "%(F/10**6 ,F_i/1000)

cutting force with parallel cutting edges = 3.825 MN
cutting force when shear is considered = 91.48 kN


## Exa 2.15 : page 141¶

In [18]:
from math import pi, sqrt
d1 = 105 # inside diameter in mm
h = 90 # depth in mm
t = 1 # thickness in mm
D = sqrt(d1**2+4*d1*h) # blank diameter in mm
tr = t*100/D # thickness ratio
# from table safe drawing ratio is 1.82
r = 1.82 # draw ratio
d2 = D/r # diameter for first draw in mm
d = 130 # Let diameter for first draw in mm
ratio1 = D/d # D/d for first draw
ratio2 = d/d1 # D/d for second draw
h1 =((D)**2-(d)**2)/(4*d) # Depth of first draw in mm
sigma = 415 # N/mm**2
c = 0.65 # constant
pc = pi*d*t*sigma*(D/d - c) # press capacity in kN
print " Blank diameter = %d mm \n Thickness ratio = %0.3f \n Diameter for first draw = %d mm \n Depth of first draw = %0.2f mm \n Press capacity = %0.2f kN"%(D,tr,d2,h1,pc/1000)
# Answers vary due to round off error

 Blank diameter = 220 mm
Thickness ratio = 0.453
Diameter for first draw = 121 mm
Depth of first draw = 61.39 mm
Press capacity = 177.92 kN


## Exa 2.16 : page 142¶

In [19]:
from math import pi, sqrt
d = 80 # diameter in mm
h = 250 # height in mm
D = sqrt((d**2+4*d*h))/10 # blank diameter in cm
D1 = 0.5*D # diameter after first draw in cm
# let reduction be 40% in second draw
D2 = D1-0.4*D1 # diameter after scond draw in cm
R = (1 - (d/(10*D2)))*100 # percentage reduction for third draw
l1 = ((D)**2-(D1)**2)/(4*D1) # height of cup after first draw in cm
l2 = ((D)**2-(D2)**2)/(4*D2) # height of cup after first draw in cm
l3 = ((D)**2-(d/10)**2)/(4*d/10) # height of cup after first draw in cm
t = 3 # mm
sigma = 250 # N/mm**2
C = 0.66
F = pi*d/10*t*sigma*((D*10/d)-C) # drawing force in kN
print " Diameter after first draw = %0.1f \n Diameter after second draw = %0.2f \n Percentage reduction after third draw = %d percent"%(D1,D2,R)
print " Height of cup after first draw = %0.2f cm\n Height of cup after second draw = %0.2f cm\n Height of cup after third draw = %0.2f cm"%(l1,l2,l3)
print " Drawing force = %0.3f kN"%(F/1000)
# Answers vary due to round off error

 Diameter after first draw = 14.7
Diameter after second draw = 8.82
Percentage reduction after third draw = 9 percent
Height of cup after first draw = 11.02 cm
Height of cup after second draw = 22.29 cm
Height of cup after third draw = 25.00 cm
Drawing force = 56.817 kN


## Exa 2.17 : page 143¶

In [20]:
from math import pi
# from figure 2.75 (a)
r1 = 30 # radius in mm
t = 10 # thickness in mm
h1 = 300 # height in mm
ir1 = r1-t # inner radius of bends in mm
L1 = h1-(ir1+t) # mm
alpha1 = 90 # degree
r2 = 2*t # mm
k = 0.33*t # mm
L2 = alpha1*2*pi*(r2+k)/360 # mm
w = 200 # mm
L3 = w-2*(t+ir1)# mm
L4 = L2 #mm
h2 = 100 # mm
L5 = h2 -(t+ir1) # mm
r3 = 150 #mm
ir2 = r3 - t # inner radius in mm
alpha2 = 180 # degree
L6 = alpha2*2*pi*(ir2+k)/360 # mm
dl = L1+L2+L3+L4+L5+L6 # Total developed length in mm
print "Total developed length = %0.2f mm"%( dl)
# Answers vary due to round off error

Total developed length = 1003.39 mm