SOLVED EXAMPLE IN TOOL DESIGN

SOLVED EXAMPLE:-1

A washer with a 12.7mm internal hole and an
outside diameter of 25.4mm is to be made from 1.5mm
thickness of strip of 0.2% carbon steel. Considering the
elastic recovery of the material.

Find (a) the clearance
(b) blank die opening size
(c) blank punch size
(d) piercing punch size
(e) piercing dieopening size

Solution:
(a)clearance for soft steel is taken as
C = 5% of t
= 5/100 X 1.5
clearance = 0.075mm/side.

(b)Blank dieopening size is equal to the blank size.
but to allow for the expansion of the blank,the die
opening should be made smaller, Thus,

Blank dieopening = Blank size – Elastic recovery
= 25.40 – 0.05
= 25.35mm.

(c)Blank punch size = Blank dieopening size – 2C
= 25.35 – 2 X 0.075
= 25.20mm.

(d)Piercing punch size is equal to the hole size.
But to allow for the contraction of the hole due
to elastic recovery, the punch is made larger,
Thus,

Piercing punch size = hole size + elastic recovery
= 12.70 + 0.05
= 12.75mm.

(e)Piercing die opening = Piercing punch size + 2C
= 12.75 + 2 X 0.075
= 12.90mm.



SOLVED EXAMPLE:-2

The strip thickness is 2.0mm and the length of
blank is 10mm and height is 20mm. Strip length is 1.0m.
Find (a) the value for front scrap
(b) the value for scrap bridge
(c) width of strip
(d) length of one part of stock needed to produce
one part.
(e) Number of parts which canbe produced in strip.
(f) Scrap maintaining at the end of strip.

Solution:















A = front scrap ; b = back scrap(bridge thickness)
l = length of part ; h = height of part
T = thickness of part; w = width of scrap
C = distance from part to part
Y = end of scrap; L = total length of scrap

Solution:

Length of part l = 10mm
Height of part h = 20mm
Thickness of part = 2.0mm
Total length of scrap L = 1.0m (= 1000mm)
(a)Front scrap and back scrap
a = t + 0.015h
= 2.0 + 0.015 X 20
= 2.30mm.
(b)Scrap bridge thickness = It is taken as one times
of sheet thickness
= 1 X 2.0
= 2.0mm.

(C) Width of strip W = Height of part + 2bridgethick
= 20 + 2 X 2
= 24mm.

(d) Length of one piece of stock needed to produce one
blank is,
C = l + b
= 10 + 2
C = 12.0mm.

(e) Number of parts in the strip
N = Total length of strip – bridge thickness
C
= L – b
C
= 1000 –2
12
N = 83 blanks produced.

(f) Scrap remaining at the end
Y = L – (Nc+b)
= 1000 – (83 X 12 + 2)
= 2.0mm.

SOLVED EXAMPLE:-3

A washer with a12.7mm hole and an outside
diameter of 25.4mm is to be made from 1.50mm
hickness Of strip of 0.2% carbon steel. The ultimate
shearing Strength of the material is 2800Kg/C
(1) Find the total cutting force if both punches act at
the same time and no shear is applied to either
punch or die.

(2) What will be the cutting force if the punches are
staggered. So that only one punch acts at a time.

(3) Taking 60% penetration and shear on punch of
1.0mm. What will be the cutting force if both
punches act together.

Solution:

(1)Cutting force F = П(D + d)st
D=25.4mm; d=12.7mm; t=1.5mm
Shear strength S=2800Kg/Cm2 = 28Kg/mm2.
F = П(25.4+12.7)X1.5X28.0
= 5.027 tonnes.


(2)When the punches are staggered, the punch taking
the largest cut will require the greatest force.

F=ПDst
=П X 25.4 X 28 X 1.5
=3.35 tonnes.

(3) F = t X K X Fmax
K X t X I
K = percentage of penetration = 0.6
I = shear on punch = 1.0mm
Fmax = 5.027 tonnes.

F = 1.5 X 0.6 X 5.027
(0.6 X 1.5)+ 1.0
= 2.38tonnes.

SOLVED EXAMPLE:-4

A hole of 60mm diameter is to be produced in
steel plate 2.5mm thick. The ultimate shear strength
of the material is 45Kg/mm2. If the punching force
is reduced to half of the force using a punch without
shear. Estimate the amount of shear on the punch.
Take % of penetration as 40%.

Solution:

The punching force with non-sheared punch.
Fmax = ПDst
= П X 60 X 45 X 2.5
= 21.20 tonnes.

Work done = Fmax X penetration(punch travel)
= 21.20 X 0.4t
= 21.20 X 0.4 X 2.5
= 21.20tonnes.

Now workdone remains the same with a sheared and
A non-sheared punch,
Punch travel = penetration + shear
= K.t + I

If F is the blanking force, then comparing the workdone,
Fmax X K X t = F(K.t + I)
I= K X t (Fmax – F)
F
Now, F = ½ Fmax
= 10.6tonnes.

I = 0.4 X 2.5 X 10.6
10.6
I = 1.0mm
So the amount of shear on punch is 1.0mm.