Tsch?tsch H., Koth A. Metal Forming Practise: Processes - Machines - Tools
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5.9 Achievable precision 41
Example:
Calculate the junction diameter and the required die clearance for the following data.
Where:
p
i
= 1000 N/mm
2
(assumed value)
D = 120 mm (die holder bore fitted in the machine)
d = 40 mm
R
e
1
= 2100 N/mm
2
(for the die material S 6-5-2)
R
e
2
= 1400 N/mm
2
(for the reinforcement material X 40 CrMoV 51).
Solution:
d
J
ӽ 0.9 · 0.9 120 mm 40 mm 62.3 mmDd
d
J
= 65 mm is selected
p
c
=
1
2
i
2
e
100 N/ mm
2100 N/ mm
p
R
= 0.476
K
1
=
1
2
2
e
2
e
2100 N/ mm
1.5
1400 N/ mm
R
R
Q
1
= c
1
11 11
1 1 0.476 0.59
221.5
p
K
§·
§·
¨¸
¨¸
©¹
©¹
z
1
=
1
2
Je
2
1
2
1
1 65 mm 2100 N/ mm 1
0.592
1.5
210000 N/ mm
dR
Q
EK
§·
§·
¨¸
¨¸
©¹
©¹
z
1
= 0.20 mm
If there is total freedom as to the external diameter of the die, then d
J
can be determined from
equation 8 and
D from equation 9; with these values, z
1
can then be calculated.
In this way, however, considerably larger dimensions for the die usually result.
42 5 Extrusion
5.9 Achievable precision
The accuracies which can be achieved with cold extrusion are shown in Tables 5.4.1 and 5.4.2.
Table 5.4.1 Diameter tolerances for cold extrusions made of steel
Range in mm
Applies to
up to 10
over
10 to 16
over
16 to 25
over
25 to 40
over
40 to 63
over
63 to 100
Mass in kg
0.05 0.06 0.06 0.07 0.08 up to 0.1
0.08 0.09 0.10 0.11 0.12 0.14 – 0.5
0.10 0.11 0.12 0.14 0.16 0.18 – 4.0
Inside diameter
0.12 0.14 0.16 0.18 0.20 0.22 – 25
0.09 0.11 0.14 0.18 0.22 0.28 up to 0.1
0.14 0.18 0.22 0.28 0.35 0.45 – 0.5
0.18 0.22 0.28 0.35 0.45 0.56 – 4.0
Outside diameter
0.22 0.28 0.35 0.45 0.56 0.71 – 25
Table 5.4.2 Wall thickness tolerances in backward cup extrusion for l/d < 2.5
l in mm bore length
d in mm bore diameter
Wall thickness in mm Tolerance in mm
0.3 to 0.6
± 0.05
0.6 to 1.0
± 0.075
1.0 to 2.5
± 0.1
>2.5 ± 0.2
Surface quality
Cold extrusions have high surface quality and therefore good wear characteristics.
Presuming that
a) the surface quality of the tools is good,
b) the lubricant and the method of applying it have been correctly chosen,
c) the deformation remains within permissible limits,
5.11 Sequence of operations diagram 43
it is possible to achieve roughness in the region of
R
t
= 5 to 10 Pm
on the surfaces where material flow takes place.
5.10 Defects during extrusion
The main defects which can occur during extrusion are summed up in Table 5.5.
Table 5.5 Defects and their causes during extrusion
Defect Cause Steps to be taken
Internal surface cracks
Deformability exceeded Split deformation into two
operations and anneal be-
tween them
Shear cracks below 45 ° Deformability exceeded during setting
process (upsetting)
– Work operation before extrusion to
produce a precise billet
Choose a larger initial
diameter
External surface cracks Wrong lubrication.
Too much liquid lubricant which can not
escape from the die during extrusion re-
sults in the lubricant exploding. This pro-
duces cracks.
Use less lubricant
5.11 Sequence of operations diagram
In most cases several operations are required to produce an extrusion. The results of each step
in the process and the dimensions of the intermediate forms are put together in a diagram
known as a
“sequence of operations diagram”.
As well as the dimensions for each step, the sequence of operations diagram also includes data
on
heat and surface treatment,
the extent of the deformation,
the force and mechanical work required for forming.
44 5 Extrusion
The sequence of operations diagram is the most important manufacturing document in establish-
ing the metal forming machines required, the construction of the tools and the time needed for
production.
Figure 5.12 shows a sequence of operations diagram for an M12 x 60 bolt.
Figure 5.12 Steps in the production of a bolt with a pressed hexagonal head
5.12 Example calculations
Example 1
The aim is to produce bolts as in Figure 5.13 out of 42
CrMo 4. The following must be established:
operating method, stock, force and mechanical work,
where:
K
= 0.7.
Solution:
1. Operating method: forward extrusion
2. Stock
Figure 5.13 Collar bolt
2.1. Volume of the finished item
22
22
F12 3
ʌʌ ʌ
412 4
Dd
VhhDDddh
=
2
22 22
13
ʌ
()
43
h
Dh D D d d dh
ªº
«»
¬¼
5.12 Example calculations 45
22 2
3mm
0.785 (30 mm) 16 mm (900 mm 30 mm 20 mm 400 mm )
3
ª
«
¬
+ (20 mm)
2
· 37 mm
º
»
¼
V
F
= 24413 mm
3
.
Extra for melt and pickling losses: 2 %
V
Me
=
2
100
· 24 413 mm
3
= 488 mm
3
.
Volume of the stock V
B
= V
F
+ V
Me
= 24 413 mm
3
+ 488 mm
3
= 24 901 mm
3
.
2.2. Stock dimensions
D
0
= 30.0 mm selected, A
0
=
2
(30 mm) ʌ
4
= 706.5 mm
2
h
0
=
B
0
V
A
=
2
2
706.5 mm
314 mm
= 35.24 mm, h
0
= 35.0 mm selected.
3. Force
F
3.1.
M
p
= ln
0
1
A
A
= ln
2
2
706.5 mm
314 mm
= ln 2.25 = 0.81,
M
p
= 81 %.
3.2.
k
str
0
= 420 N/mm
2
, k
str
l
= 1080 N/mm
2
, k
str
m
=
01
str str
2
kk
750 N/mm
2
.
3.3.
F =
m
0str p
F
Ak
M
K
=
22
706.5 mm 750 N/ mm 0.81
0.7
= 613141 N = 613 kN.
4. Work
W
s
w
=h
0
– h
k
= 35 mm – 16 mm = 19 mm = 0.019 m
W = F · s
w
· x = 613 kN · 0.019 m · 1 = 11.6 kNm.
Example 2
The aim is to produce casings as in Figure 5.14 out of Al 99.5.
where:
K
F
= 0.7
Figure 5.14 Casing
46 5 Extrusion
The following must be established:
operating method, stock, force and mechanical work.
Solution 1:
1. Operating method: backward extrusion
2. Stock
2.1. Volume of the finished item
V
F
=
22
2
b
1
ii
ʌʌ
()
44
dsDdh
= (28 mm)
2
ʌ
4
1.5 mm + (30
2
– 28
2
) mm
2
ʌ
4
60 mm = 6386.7 mm
3
Extra to make up for melt and pickling losses: 1 %
V
Me
=
1
100
6336 mm
3
# 64 mm
3
.
Volume of the stock
V
B
= V
F
+ K
Me
= 6386.7 mm
3
+ 64 mm
3
= 6451 mm
3
.
2.2. Stock dimensions
D
0
= 30
– 0.1
mm, A
0
= 706.5 mm
2
h
0
=
3
F
2
0
6451 mm
9.13 mm,
706.5 mm
V
A
h
0
= 9.0 mm selected.
3. Force
F
3.1.
M
p
=
0
0
30 mm
ln 0.16 ln
30 mm 28 mm
D
Dd
– 0.16 = ln 15 – 0.16
= 2.70 – 0.16 = 2.54
M
p
= 254 %.
3.2. Flow stress from the flow stress curve or Table 1 (Part III)
k
str
0
= 60 N/mm
2
, k
str
1
= 184 N/mm
2
,
k
str
m
=
01
22
str str
60 N/ mm 184 N/mm
22
kk
= 122 N/mm
2
3.3.
F =
m
22
str
0
p
F
(28 mm) ʌ 122 N/mm 9
2 0.25 2 0.25
40.7 1
k
h
A
c
K
§· §·
¨¸ ¨¸
©¹ ©¹
= 455836 N = 456 kN.
4.1. D
eformation displacement
s
w
=h
0
– h
1
= 9 mm – 1.5 mm = 7.5 mm = 0.0075 m
5.12 Example calculations 47
4.2.
W = F · s
w
· x = 456 kN · 0.0075 m · 1 = 3.4 kNm.
Example 3
The aim is to produce workpieces as in Figure 5.15. The stock diameter must be selected so
that initial calculations result in an upsetting ratio of
s = 2. The calculated diameter, d
0
, must
be rounded up to the nearest mm. The calculated stock length,
h
0
, must also be rounded up to
the nearest mm.
where: Material Ck 15
,
K
= 0.7,
P
= 0.15.
Create the sequence of operations diagram.
Solution:
1.
Volume
1.1. Head volume
V
HD
=
32
4
ʌ
33
h
rhr
S
§·
¨¸
©¹
V
HD
=
32
45
ʌ 25 5 25
33
ªº
§·
¨¸
«»
©¹
¬¼
V
HD
=
3
63617 mm
1.2. Volume of both shanks
V
1
+ V
2
=
4
S
(30
2
· 35 + 20
2
· 25)
= 24740 + 7854.
Figure 5.15 Round-headed bolt
1.3. Total volume
V
tot
= V
HD
+ V
1
+ V
2
=
3
96211 mm
2. Establish the required
d
0
from the upsetting ratio
d
0
=
HD
3
4
ʌ
V
s
d
0
=
3
3
4 63617 mm
ʌ 2
= 34.34 mm
d
0
= 35 mm selected .
3. Stock length
h
0
h
0
=
3
tot
22
0
96211 mm
35 ʌ/4 mm
V
A
= 99.99 mm , h
0
= 100 mm selected .
48 5 Extrusion
3.1. Required stock length for the round head
h
0
HD
HD
3
HD
0
2
0
63617 mm
962.11 mm
V
h
A
= 66.1 mm
4. Forward extrusion
4.1.
M
p
= ln
0
1
A
A
= ln
2
0
2
1
d
d
= ln
2
2
35
20
= 1.12 o 112% ,
M
p
perm
= 120 %.
4.2.
k
str
0
= 280, k
str
1
= 784, k
str
m
= 532 N/mm
2
4.3.
F =
m
p
0str
F
Ak
M
K
=
22
962.11 mm 532 N/mm 1.12
0.7
= 818948 N = 819 kN
4.4.
s
W
= h
0
– h
0
HD
= 100 – 66.1 = 33.9 mm = 0.0339 m
4.5.
W = F · s
W
· x = 819 kN · 0.0339 m · 1 = 27.8 kNm.
Figure 5.16 Sequence of operations for the production of a round-headed bolt
Se
q
uence of o
p
erations dia
g
ram
5.13 Shape classification 49
5. Upsetting (heading)
5.1
M
p
= ln
HD0
HD
66.1
ln 0.384 38.4%
45
h
h
o
M
p
= 150 %.
5.2
k
str
0
= 280, k
str
1
= 670, k
str
m
= 475 N/mm
2
5.3
F = A
1
· k
str
1
1
1
1
1
3
d
h
P
§·
¨¸
©¹
=
2
22
50 ʌ 150mm
mm 670 N/mm 1 0.15
4345mm
§·
¨¸
©¹
F = 1 387 896.7 N = 1388 kN
5.4 W =
m
32
HD str p
F
63617 mm 475 N / mm 0.384
0.7
Vk
M
K
W = 16 576 771 Nmm = 16.6 kN m
5.13 Shape classification
For operations scheduling, during which the sequence of operations plans for the extrusions
are produced, it is important to collect and record empirical values.
This is why a shape classification table is used in which the various extrusion shapes are or-
dered according to operating method, sequence of operations, degree of difficulty etc. Table
5.6 shows a shape classification table. The advantages of this kind of classification are:
– Production of sequence of operations plans for new parts.
If sequence of operations plans already exist in the shape classification table for similar
parts, this makes the production of a new sequence of operations plan easier as it can be ba-
sed on similar shapes.
This means the induction of new inexperienced workers is made far simpler and the need
for specialists is reduced.
– Furthermore, groups of the same shape have the same or a similar amount of difficulty
during pressing.
Depending on the group of shapes, the extent of the deformation efficiency to be expected
can also be deduced (see Table 5.7).
– There are also advantages for tool manufacture. Similar shapes also produce similar ele-
ments as concerns tools; these elements can be standardised within a company, simplified
and thus produced at less expense.
50 5 Extrusion
Table 5.6
Shape classification according to length