The document elaborates on the derivations of the energy components in an extrusion model. It discusses four components: 1) Ideal work to plastically deform the material, 2) Friction work at the die-shoulder and inside the container, 3) Shearing work to plastically shear the material, and 4) Bending work which does not apply to direct extrusion. It then provides detailed derivations of the ideal work and friction work components using concepts of stress, strain, and energy methods. The derivations make assumptions about mean flow stresses and friction to obtain equations for the specific work and total work of each component.

1. 6.1
Chapter 6 – Bulk Forming Processes (II)
Questions & Answers
Slides 6.9-6.18
Question: Would you elaborate the derivations in these slides?
Answer: The extrusion model in these slides was adapted from a well-known German
handbook in three volumes [1]1. However, the model presented by this book (see Vol. 2)
is, by and large, based on the works of Siebel [3-4]. Anyway, the given extrusion model
essentially makes good use of energy methods. The energy/work required to perform
extrusion process can be divided into four components:
1) Ideal Work: Energy required to convert the billet (area: A0; height: h0) into the
final shape of the product (area: A1; height: h1  h0A0/A1). In other words, it refers
to the energy necessary to plastically deform the billet into the final product.
2) Friction Work: Energy to overcome:
a. Friction in the die-shoulder
b. Friction inside the container
3) Shearing Work: Total energy required to shear the material plastically as it
enters and exits the die-shoulder.
4) Bending Work: It does not exist in direct/forward extrusion.
Before we proceed, a quick review on work/energy might helpful at this point. As you
might remember from your Physics courses, the work done by a force is a scalar quantity
and is measured in [J]. Work/energy is defined as the inner product of a force vector (F)
and a displacement vector (s). That is,
𝑊 = 𝑭 ⋅ 𝒔 = |𝑭||𝒔| cos(𝜃) = 𝐹𝑠 cos(𝜃) (6.1)
where  is the angle between the vectors. If the force is a function of displacement, the
work done by the force takes the following form:
𝑊 = ∫ 𝑭(𝒔) ⋅ 𝑑𝒔 (6.2)
Recall that
 If W > 0, force transfers some energy to the system under study;
 If W = 0, no energy is exchanged;
 If W < 0, some energy is absorbed or taken away from the system.
Armed with this information, let us elaborate the above-mentioned energy components:
1 In 1994, the Society of Manufacturing Engineers (SME) published the English translation of this handbook:
See Ref. [2].

2. 6.2
1. Ideal Work (Slides 6.9-6.11):
To compute the ideal work component, we shall consider an infinitesimal disc element
inside the die shoulder (where plastic deformation is taking place). Under the action of
stresses acting on this element (i.e. radial- and axial stresses: r and z), it will be
displaced while changing its overall shape as illustrated in Fig. 6.1.
A0
A1
Deformation
Region
Initial Position
Final Position
Die
r
z
Δz
Δz+d(Δz)
r
r-dr
Area: A
Figure 6.1: Slab model for the die shoulder [1].
The stresses on the disc is shown in Fig. 6.2. In our derivations, we shall assume that
 z > 0 is the maximum principal stress component (1)2;
 r < 0 is the minimum one (3).
σz
σz
σr σr
dr
d(Δz)/2
d(Δz)/2
Final
Initial
Figure 6.2: Stresses and deformations on the disc elements.
Now, let us write the energy required to deform this disc element to its final shape. For
this purpose, we shall compute the work done by every force component acting on the disc:
𝑑𝑊ideal = (𝜎𝑟2𝜋𝑟Δ𝑧)
⏟
radial force
𝑑𝑟
⏟
rad.
disp.
+ (𝜎𝑧𝐴)
⏟
axial
force (up)
𝑑(Δ𝑧)
2
⏟
axial
disp.
+ (𝜎𝑧𝐴)
⏟
axial
force (down)
𝑑(Δ𝑧)
2
⏟
axial
disp.
(6.3)
Thus, it becomes
2 In extrusion, z is known to be compressive. However, the assumption that z > 0 does not alter the nature
of the derivations as long as one carries out the computations consistently in accordance with the assumptions
adopted.

3. 6.3
𝑑𝑊ideal = 𝜎𝑟2𝜋𝑟Δ𝑧 𝑑𝑟 + 𝜎𝑧𝐴 𝑑(Δ𝑧) (6.4)
Recall that the area of disc element initially is
𝐴 = 𝜋𝑟2
(6.5)
As the disc deforms to its final shape, the change in the area3 (dA) can be calculated by
differentiating Eqn. (6.5):
𝑑𝐴 = 2𝜋𝑟 𝑑𝑟 (6.6)
Substituting Eqn. (6.6) into (6.4) actually eliminates r:
𝑑𝑊ideal = 𝜎𝑟Δ𝑧𝑑𝐴 + 𝜎𝑧𝐴 𝑑(Δ𝑧) (6.7)
Now, we can calculate the change in thickness [d(z)] by considering the volume
constancy:
Δ𝑉 = 𝐴Δ𝑧 = constant (6.8)
Taking the differential of Eqn. (6.8) yields
d(Δ𝑉) = 𝐴 𝑑(Δ𝑧) + 𝑑𝐴 Δ𝑧 = 0 (6.9)
By rearranging Eqn. (6.9), we obtain
d(Δ𝑧) = −
𝑑𝐴
𝐴
Δ𝑧 (6.10)
Plugging Eqn. (6.10) into (6.7) leads to
𝑑𝑊ideal = (𝜎𝑟 − 𝜎𝑧)Δ𝑧 𝑑𝐴 (6.11)
It would be convenient to express Eqn. (6.11) in terms of V (rather than z). By
employing Eqn. (6.8), we get
𝑑𝑊ideal = (𝜎𝑟 − 𝜎𝑧)
𝑑𝐴
𝐴
ΔV (6.12)
Now, let us define infinitesimal specific ideal work4:
𝑑𝑤ideal ≜
𝑑𝑊ideal
ΔV
= (𝜎𝑟 − 𝜎𝑧)
𝑑𝐴
𝐴
(6.13)
Since the disc element is plastically deforming, Tresca yield criterion must be satisfied:
𝜎1 − 𝜎3 = 𝜎𝑓 ≅ 𝑌𝑆 (6.14)
That is,
𝜎𝑧 − 𝜎𝑟 = 𝜎𝑓 (6.15)
3 In our case, dA (or dr) must be a negative quantity indicating a decrease in the final area. We accommodated
that fact by changing the sign of dr in Fig. 6.1. That is, we have modified the direction of radial displacement
vector in Fig. 6.2 such that r and dr point the same direction (i.e. towards the center of the die).
4 This definition is very similar to those in Thermodynamics. For instance, specific heat capacity (c) is defined
by dividing the heat capacity of the substance (C) by the sample’s mass (m).

4. 6.4
Combining Eqn. (6.13) and (6.15) results in
𝑑𝑤ideal = −𝜎𝑓
𝑑𝐴
𝐴
(6.16)
If Eqn. (6.16) is integrated, the specific ideal work is calculated:
𝑤ideal = ∫ −𝜎𝑓
𝐴1
𝐴0
𝑑𝐴
𝐴
(6.17)
Due to strain-hardening, the flow stress of the material (f) is to change as a function of
the area: 𝜎𝑓 = 𝐾 [ln (
𝐴0
𝐴
)]
𝑛
. Thus, evaluating the definite integral in Eqn. (6.17) would be
a challenging task (at least, analytically!). For all practical purposes, Siebel [3-4]
presumes that a constant mean flow stress (fm) can be utilized instead of a flow stress
that changes constantly. Hence, Eqn. (6.17) becomes
𝑤ideal = −𝜎𝑓𝑚 ∫
𝑑𝐴
𝐴
𝐴1
𝐴0
= 𝜎𝑓𝑚𝑙𝑛 (
𝐴0
𝐴1
) = 𝜎𝑓𝑚𝜀𝑒𝑞 (6.18)
Here, the mean flow stress5 is computed as follows:
𝜎𝑓𝑚 ≅
𝜎𝑓0 + 𝜎𝑓1
2
(6.19)
where f0 is the initial flow stress of the (annealed) material while f1 = K(eq)n refers to
the strength of work-hardened material at the exit. Finally, the overall ideal work turns
out to be
𝑊ideal = 𝑉𝑤ideal = 𝑉𝜎𝑓𝑚𝜀𝑒𝑞 (6.20)
Notice that V here denotes the volume of the material that has passed through the die-
shoulder. That is, it intrinsically refers to the volume of the extruded material at a
particular position of the ram and can be expressed as a function of the ram’s stroke (s):
𝑉(𝑠) = 𝐴0𝑠 (6.21)
where 0  s  h0.
2.a Friction Work at Die-shoulder (Slides 6.12-6.14):
As highlighted in Slide 6.12, Siebel [3-4] makes the following assumptions to compute
work done against friction:
 Pressure acting on the walls of the container is in the order of initial flow stress:
f0
 Pressure on the walls of the die shoulder is about the mean flow stress: fm
5 This definition is pivotal in his derivation as it will be also employed in the later stages (i.e. shear energy
computation). Therefore, it is not recommended to utilize any other mean flow stress formulations when using
this mathematical model.

5. 6.5
With these assumptions, the work done by the friction forces inside the die shoulder can
be conveniently estimated. To this end, we shall utilize the friction model shown in Fig.
6.3. The infinitesimal work done against the friction on the die-shoulder can be written
as
𝑑𝑊friction
die shoulder
= [𝜇𝜎𝑓𝑚
2πrΔ𝑠
cos(𝛼)
]
⏟
friction force
𝑑𝑧
cos(𝛼)
⏟
displacement
(6.22)
Note that the work done by the friction force itself is less than zero because the angle
between the friction force vector and the corresponding displacement vector is 180 in Fig.
6.3 [also refer to Eqn. (6.1)]. That is, the friction process is absorbing energy as expected.
On the other hand, since we are interested in the work done against the friction, the sign
of Eqn. (6.22) is modified accordingly.
A0
A1
Deformation
Region
Initial Position
Final Position
Die
Δs
dz
α
Δs/cos(α)
dz/cos(α)
r
Figure 6.3: Friction model for the die-shoulder [1].
To eliminate dz in Eqn. (6.22), the differential change in the area (dA) is considered:
𝑑𝐴 = 𝜋[𝑟 − 𝑑𝑧 ⋅ 𝑡𝑎𝑛(𝛼)]2
⏟
final
− 𝜋𝑟2
⏟
initial
≅ −2𝜋𝑟 ⋅ 𝑑𝑧 ⋅ 𝑡𝑎𝑛(𝛼) (6.23)
Hence,
𝑑𝑧 = −
𝑑𝐴
2𝜋𝑟 ⋅ 𝑡𝑎𝑛(𝛼)
(6.24)
Note that the thickness of the disc can be expressed in terms of its volume (V):
Δ𝑠 =
Δ𝑉
𝐴
(6.25)
Plugging Eqns. (6.24) and (6.25) into (6.22) yields
𝑑𝑤friction
die shoulder
≜
𝑑𝑊friction
die shoulder
Δ𝑉
= −
𝜇𝜎𝑓𝑚
sin(𝛼) cos(𝛼)
𝑑𝐴
𝐴
(6.26)
Similarly, integration of Eqn. (6.26) leads to

6. 6.6
𝑤friction
die shoulder
= −
𝜇 𝜎𝑓𝑚
sin(𝛼) cos(𝛼)
⏟
sin(2𝛼)/2
∫
𝑑𝐴
𝐴
𝐴1
𝐴0
=
2𝜇 𝜎𝑓𝑚
sin(2𝛼)
𝑙𝑛 (
𝐴0
𝐴1
)
⏟
𝜀𝑒𝑞
= 𝜎𝑓𝑚𝜀𝑒𝑞
2𝜇
sin(2𝛼) (6.27)
Consequently, total friction work at the die-shoulder takes the following form:
𝑊friction
die shoulder
= 𝑉 𝑤friction
die shoulder
= 𝑉𝜎𝑓𝑚𝜀𝑒𝑞
2𝜇
sin(2𝛼)
(6.28)
2.b Friction Work at Container (Slide 6.15):
Next, we shall take a close look at the container friction. The derivation is somewhat
similar to the one discussed in the previous section. Consider the friction model in Fig.
6.4.
Initial Position
Final Position
ds
σf0
μσf0 μσf0
σf0
d0
RAM
h
0
h
s
dz
z
Figure 6.4: Friction model for the container [1].
The infinitesimal work done against the friction inside the container can be written as
𝑑𝑊friction
container
= ∫ 𝜇𝜎𝑓0πd0dz
ℎ
𝑧=0
⏟
friction force
ds
⏟
incr. disp.
of the ram
(6.29)
Again, the discussion in Section 2.a (about the sign of the work done) is applicable here
as well. Hence, Eqn. (6.29) becomes
𝑑𝑊friction
container
= 𝜇𝜎𝑓0πd0h ds (6.30)
Since the displacement of the ram is, by definition, s = h0 – h; differentiating this term
yields ds = –dh. By substituting ds = –dh into Eqn. (6.30) and integrating the resulting
expression, we get
𝑊friction
container
(ℎ) = −𝜇𝜎𝑓0πd0 ∫ ℎ𝑑ℎ
ℎ
ℎ0
=
1
2
πd0(𝜇𝜎𝑓0) (ℎ0
2
− ℎ2
)
⏟
(ℎ0−ℎ)(ℎ0+ℎ)
(6.31)

7. 6.7
Eqn. (6.31) takes the following form in terms of s:
𝑊friction
container
(𝑠) =
1
2
πd0(𝜇𝜎𝑓0)(2ℎ0 − 𝑠)𝑠 (6.32)
3. Shearing Work (Slides 6.16-6.17):
When the material enters the die-shoulder area, the cylindrical shape quickly morphs into
a cone. Similarly, the cone inside the die-shoulder must be transformed back to a cylinder
at the die exit. Hence, a shearing work is necessary to carry out these shape
transformations. For this purpose, let us consider a (differential) disc element that has
entered to the die shoulder as demonstrated in Fig. 6.5.
Δs
α
γ
r
s
dA
τmax0
r0 Δs tan(γ)
γ
τmax0
Die
dr
Deformation
Region
cone center
Figure 6.5: Shear model for the inlet [1].
On this disc, we shall take into consideration a differential ring element as illustrated in
the figure. As the disc enters the inlet of the die-shoulder, the ring will distort under the
action of shear stresses. The work done by the shearing forces can be simply expressed as
Δ𝑊shear0 = ∫ Δ𝑠 tan(𝛾)
⏟
radial
displacement
τmax02πr dr
⏟
shearing force
𝑟0
0
(6.33)
Since the material is plastically shorn at the inlet, (according to Tresca) the maximum
shear stress becomes
τmax0 = τyield ≅ 𝜎𝑓0/2 (6.34)
From the given geometry, we have
tan(𝛾) =
𝑟
𝑠
(6.35)
Substituting Eqns. (6.34) and (6.35) into (6.33) and integrating the resulting expression
lead to
Δ𝑊shear0 =
1
3
π𝜎𝑓0Δ𝑠
𝑟0
3
𝑠
=
1
3
𝜎𝑓0Δ𝑠 π𝑟0
2
⏟
𝐴
𝑟0
𝑠
⏟
tan(𝛼)
=
1
3
𝜎𝑓0 Δ𝑠 𝐴 tan(𝛼) (6.36)

8. 6.8
Recall that the volume of the disc element is
ΔV = Δ𝑠 𝐴 (6.37)
Hence, Eqn. (6.37) boils down to
Δ𝑊shear0 =
1
3
𝜎𝑓0 Δ𝑉 tan(𝛼) (6.38)
Similarly, the specific work at the inlet turns out to be
Δ𝑤shear0 ≜
Δ𝑊shear0
Δ𝑉
=
1
3
𝜎𝑓0 tan(𝛼) (6.39)
It is self-evident that the analysis conducted above can be easily extended to the die exit.
The result would be quite identical to Eqn. (6.39) except that the flow stress at the exit is
f1 (rather than f0). Consequently, the specific work at the exit becomes
Δ𝑤shear1 =
1
3
𝜎𝑓1 tan(𝛼) (6.40)
Adding Eqns. (6.39) and (6.40) gives the combined specific shearing energy:
Δ𝑤shear = Δ𝑤shear0 + Δ𝑤shear1 =
2
3
(
𝜎𝑓0 + 𝜎𝑓1
2
)
⏟
𝜎𝑓𝑚
tan(𝛼) =
2
3
𝜎𝑓𝑚 tan(𝛼)
(6.41)
Hence, total shearing energy boils down to
𝑊shear = Δ𝑤shear𝑉 =
2
3
𝑉 𝜎𝑓𝑚 tan(𝛼) (6.42)
Total Work and Extrusion Force (Slide 6.18)
Total work in forward extrusion as function of s can be obtained by combining Eqns. (6.20),
(6.28), (6.32), and (6.42):
𝑊total(s) = A0𝑠
⏟
𝑉
𝜎𝑓𝑚 [
2
3
tan(𝛼) + (1 +
2𝜇
sin(2𝛼)
) 𝜀𝑒𝑞] +
1
2
𝜋𝑑0(𝜇𝜎𝑓0)(2ℎ0 − 𝑠)𝑠 (6.43)
Taking the derivative of Wtotal with respect to s yields the extrusion force:
𝑑𝑊total
𝑑𝑠
= Ftotal(s) = A0𝜎𝑓𝑚 [
2
3
tan(𝛼) + (1 +
2𝜇
sin(2𝛼)
) 𝜀𝑒𝑞]
⏟
𝐹
̅
+ 𝜋𝑑0(𝜇𝜎𝑓0)(ℎ0 − 𝑠)
⏟
𝐹friction
container
(𝑠)
(6.44)
Note that since s = h0 – h,
𝐹friction
container
= 𝜋𝑑0ℎ𝜇𝜎𝑓0 (6.45)
In fact, Eqn. (6.45) could be directly obtained with the utilization of Eqn. (6.30).
Consequently, we have
Ftotal(h) = A0𝜎𝑓𝑚 [
2
3
tan(𝛼) + (1 +
2𝜇
sin(2𝛼)
) 𝜀𝑒𝑞]
⏟
𝐹
̅ (constant)
+ 𝜋𝑑0ℎ𝜇𝜎𝑓0
⏟
𝐹friction
container
(ℎ)
(6.46)

9. 6.9
Slide 6.19
Question: Would you give an example on the application of the presented extrusion
model?
Answer: Let us do the example (Example 1) in this slide. The given data are as follows:
 f0 = 240 [MPa]
 K = 700 [MPa]
 n = 0.24
  = 0.06
  = 45
 d0 = 75 [mm]
 d1 = 45 [mm]
 h0 = 110 [mm]
Let us start our calculations by finding the necessary quantities for Eqn. (6.46):
𝜀𝑒𝑞 = 𝑙𝑛 (
𝐴0
𝐴1
) = 𝑙𝑛 (
𝑑0
2
𝑑1
2) = 𝑙𝑛 (
752
452) = 1.0217
𝐴0 =
𝜋𝑑0
2
4
= 𝜋
752
4
= 4418 [mm2
]
𝜎𝑓1 = 𝐾(𝜀𝑒𝑞)
𝑛
= 700(1.0217)0.24
≅ 704 [MPa]
𝜎𝑓𝑚 =
𝜎𝑓0 + 𝜎𝑓1
2
=
240 + 704
2
≅ 472 [MPa]
For h = h0 (the worst case!), we can compute the extrusion force using Eqn. (6.46):
Ftotal(h0) = A0𝜎𝑓𝑚 [
2
3
tan(𝛼) + (1 +
2𝜇
sin(2𝛼)
) 𝜀𝑒𝑞] + 𝜋𝑑0ℎ0𝜇𝜎𝑓0
Ftotal = 4418 ⋅ 472 [
2
3
tan(45𝑜) + (1 +
2 ⋅ 0.06
sin(90𝑜)
) 1.0217] + 𝜋 ⋅ 75 ⋅ 110 ⋅ 0.06 ⋅ 240
Ftotal = 4418 ⋅ 472 [
2
3
tan(45𝑜) + (1 +
2 ⋅ 0.06
sin(90𝑜)
) 1.0217]
⏟
F
̅ = 3.775 [MN]
+ 𝜋 ⋅ 75 ⋅ 110 ⋅ 0.06 ⋅ 240
⏟
Ffriction
container
= 0.373 [MN]
Consequently, we have
Ftotal = 4.148 [MN]
Here are some interesting facts about the extrusion process [5]:
1. There is no limit imposed on the extrusion ratio (A0/A1) because all principal stress
components are compressive (see Chapter 5 – the discussion about bulk
formability). The strength of the die material often times becomes the limiting
factor.
 For hard materials, the extrusion ratio in practice is 20:1
 For aluminum, it can be as high as 100:1
2. For good efficiency at large reductions, the die angle () must be increased.
3. Good lubrication is needed to improve the efficiency of the extrusion process:
 For cold-extrusion of steel, the lubricant is generally graphite.
 For hot-extrusion of steel, glass based lubricants are utilized.

10. 6.10
Slide 6.20
Question: Would you explain the extrusion force calculations using empirical techniques?
Answer: The empirical model given in this slide is presented in Section 4.5.5 of Ref. [6].
However, there are some minor notational changes. We start our computation by
calculating the extrusion ratio:
𝑅𝑒 =
𝐴0
𝐴1
(6.47)
Next is the equivalent strain:
𝜀𝑒𝑞 = ln(𝑅𝑒) (6.48)
Then, the mean flow stress needs to be determined. For cold-extrusion,
𝜎𝑓𝑚 =
𝐾(𝜀𝑒𝑞)
𝑛
𝑛 + 1
(6.49)
For hot extrusion, it is
𝜎𝑓𝑚 = 𝐶(𝜀̇𝑚)𝑚
(6.50)
where 𝜀̇𝑚 is the mean strain rate. To calculate this quantity, let us find the material’s
velocity in the axial direction (z) inside the deformation zone. By considering volume
constancy [e.g. 𝑣(𝑧) ⋅ 𝐴(𝑧) = 𝑣𝑝𝑢𝑛𝑐ℎ𝐴0], we have
𝑣(𝑧) =
𝑣𝑝𝑢𝑛𝑐ℎ𝑑0
2
[𝑑0 − 2𝑧 𝑡𝑎𝑛(𝛼)]2
(6.51)
where  is the cone angle and 0 ≤ 𝑧 ≤
𝑑0−𝑑1
2tan(𝛼)
. The travel time of the material between the
inlet and the exit can be expressed as follows:
∫ 𝑑𝑡
Δ𝑡
0
= ∫
𝑑𝑧
𝑣(𝑧)
𝑑0−𝑑1
2tan(𝛼)
0
=
1
𝑣𝑝𝑢𝑛𝑐ℎ𝑑0
2 ∫ [𝑑0 − 2𝑧 ⋅ 𝑡𝑎𝑛(𝛼)]2
𝑑𝑧
𝑑0−𝑑1
2tan(𝛼)
0
(6.52)
Hence, evaluating the integral in Eqn. (6.52) yields
Δ𝑡 =
𝑑0
3
− 𝑑1
3
6 𝑣𝑝𝑢𝑛𝑐ℎ𝑑0
2
𝑡𝑎𝑛(𝛼)
(6.53)
As an alternative approach, one can consider the time to fill up the volume of the
deformation zone (i.e. the volume of a truncated cone) when the material is presumed to
be flowing at constant rate of 𝑄 = 𝑣𝑝𝑢𝑛𝑐ℎ𝐴0 [m3/s]:
Δ𝑡 =
𝑉
𝑄
=
1
3 ∙
𝜋𝑑0
2
4
⏞
𝐵𝑎𝑠𝑒 𝑎𝑟𝑒𝑎
∙
𝑑0
2 tan(𝛼)
⏞
𝐻𝑒𝑖𝑔ℎ𝑡
−
1
3 ∙
𝜋𝑑1
2
4 ∙
𝑑1
2 tan(𝛼)
⏞
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑐𝑜𝑛𝑒
𝜋𝑑0
2
4 ∙ 𝑣𝑝𝑢𝑛𝑐ℎ
=
𝑑0
3
− 𝑑1
3
6 𝑣𝑝𝑢𝑛𝑐ℎ𝑑0
2
𝑡𝑎𝑛(𝛼)
(6.54)

11. 6.11
Consequently, Eqn. (4-33) of Ref. [6] is obtained:
𝜀̇𝑚 ≅
Δ𝜀𝑒𝑞
Δ𝑡
=
𝜀𝑒𝑞 − 0
Δ𝑡
=
6 𝑣𝑝𝑢𝑛𝑐ℎ𝑑0
2
𝑡𝑎𝑛(𝛼)
𝑑0
3
− 𝑑1
3 𝜀𝑒𝑞 (6.55)
Once the mean flow stress is computed, the average extrusion found is found:
𝐹
̅ = 𝜎𝑓𝑚 Qe A0 (6.56)
where the strain factor Qe is given as
Qe = 0.8 + 1.2𝜀𝑒𝑞 (6.57)
Note that Eqn. (6.57) is generally referred to as Johnson’s extrusion strain formula
(see Eqn. (18.22) of Ref. [7]). If desired, the container friction force [i.e. Eqn. (6.45)] can be
added to Eqn. (6.56) for direct/forward extrusion.
Please note that the backward extrusion process resembles the constrained piercing
operation (Slide 6.26). Consequently, this condition must be checked by computing the
corresponding piercing force. If the piercing force is found to be smaller than its
counterpart in Eqn. (6.56), the piercing will essentially take place.

12. 6.12
Slide 6.21
Question: Is there an analytical model to calculate the force in wire/bar drawing?
Answer: Yes, indeed. The model elaborated in the previous slides (e.g. Siebel’s forward
extrusion model) can be directly adopted for this purpose. The result can be given as
follows:
Fdraw = A1𝜎𝑓𝑚 [
2
3
tan(𝛼) + (1 +
2𝜇
sin(2𝛼)
) 𝜀𝑒𝑞] (6.58)
Note that apart from wire/rod drawing, we can easily extend this basic model to the other
extrusion processes as well. Please refer to Slide 6.28 for more information.
Question: Would you explain the draw force calculations using empirical techniques?
Answer: The empirical model given in this slide is presented in Section 4.7.2 of Ref. [6].
However, there are some minor notational changes. Anyway, consider the wire/rod
drawing process in Fig. 6.6. The empirical expression for the axial stress at the exit (z1)
is
𝜎𝑧1 = 𝜎𝑓𝑚𝜀𝑒𝑞𝜑[1 + 𝜇 ⋅ cot(𝛼)] (6.59)
where 𝜀𝑒𝑞 = ln (
𝐴0
𝐴1
) = 2 ln (
𝑑0
𝑑1
). The deformation factor  in Eqn. (6.59) is defined as
𝜑 = {
0.88 + 0.12
ℎ
𝐿
, for axisymmetric / round parts
0.8 + 0.2
ℎ
𝐿
, for plane strain / rectangular parts
(6.60)
where is ℎ =
𝑑0+𝑑1
2
is the mean diameter/thickness while 𝐿 =
𝑑0−𝑑1
2 sin(𝛼)
denotes the length of
the draw die. Consequently, the draw force becomes
𝐹𝑑𝑟𝑎𝑤 = 𝐴1𝜎𝑧1 (6.61)
Notice that h/L plays a key role on the performance of the process. If (h/L > 2), the center-
burst (Chevroning or Arrowhead) defect as shown in Fig. 6.7 is commonly observed
(also see Chapter 5 Q&A – the discussion for Slide 6.10).
Die
Deformation
Region
d0
α
d1
σz1
Figure 6.6: Wire/rod drawing process. Figure 6.7: Center-burst defect.

13. 6.13
Slide 6.22
Question: Would you give an example on the application of the presented empirical model
on wire drawing process?
Answer: Consider the problem (Example 2) in this slide. The given data are as follows:
 K = 1300 [MPa]
 n = 0.3
  = 0.04
  = 6
 d0 = 3 [mm]
 A1 = 5 [mm2]
 v1 = 2 [m/s]
First of all, we shall take a look at the analytical model given by Eqn. (6.47). The unknown
quantities are calculated as follows:
𝑑1 = √
4𝐴1
𝜋
= 2.52 [mm]
𝜀𝑒𝑞 = 2 𝑙𝑛 (
𝑑0
𝑑1
) = 2 𝑙𝑛 (
3
2.52
) = 0.3462
Since f0 is not specified in the problem statement, we shall (reluctantly!) compute the
mean flow stress using the expression in Slide 4.17:
𝜎𝑓𝑚 =
𝐾(𝜀𝑒𝑞)
𝑛
𝑛 + 1
=
1300(0.3462)0.3
1.3
= 727 [MPa]
Hence, the drawing force becomes
Fdraw = 5 ⋅ 727 [
2
3
tan(6°) + (1 +
2 ⋅ 0.04
sin(12°)
) 0.3462]
Fdraw ≅ 2000 [N]
Similarly, the drawing power can be calculated as
Pdraw = Fdrawv1 = 2000 ⋅ 2 ≅ 4 [kW]
Now, let us employ the empirical model. First, calculate all the unknown quantities:
ℎ =
𝑑0 + 𝑑1
2
=
3 + 2.52
2
= 2.76 [mm]
𝐿 =
𝑑0 − 𝑑1
2 sin(𝛼)
=
3 − 2.52
2 sin(6°)
= 2.3 [mm]
For round part, we have
𝜑 = 0.88 + 0.12
ℎ
𝐿
= 0.88 + 0.12
2.76
2.3
= 1.024
Consequently,
𝜎𝑧1 = 𝜎𝑓𝑚𝜀𝑒𝑞𝜑[1 + 𝜇 ⋅ cot(𝛼)]
𝜎𝑧1 = 727 ⋅ 0.3462 ⋅ 1.024[1 + 0.04 ⋅ cot(6°)] = 356 [MPa]

14. 6.14
The drawing force becomes
Fdraw = σz1A1 = 356 ⋅ 5 = 1780 [N]
Likewise, the power is
Pdraw = Fdrawv1 = 1780 ⋅ 2 ≅ 3.6 [kW]
As can be seen, the analytical results and the empirical ones are in good agreement.
Now, a feasibility check is in order. After pointing (see page 265 of Ref. [6] and page 473
of Ref. [7]), the material may fail to deform plastically inside the die due to high friction.
Under the action of the capstan pulling the wire, the cold-worked wire might yield and
plastically elongate outside of the die (just like a specimen under uniaxial tension).
To check this condition, let us compute the yield strength of the cold-worked wire at the
exit:
𝑌𝑆𝑐𝑤 ≅ 𝜎𝑓1 = 𝐾(𝜀𝑒𝑞)
𝑛
= 1300(0.3462)0.3
≅ 946 [MPa]
The minimum drawing force, which starts to deform the drawn wire, becomes
Fmin = σf1A1 = 946 ⋅ 5 = 4.74 [kN]
Since Fdraw (1.78 kN) < Fmin (4.74 kN), the wire deforms inside the die before its gets to
elongate at the exit. Therefore, the operation is feasible.
Here are some limitations of the wire drawing process [6]:
1. To avoid center-burst defect, h/L must be less than 2. (In this example, it is only
1.2.)
2. To reduce the frequent breaking of the wire, the area reduction (per die) must be
kept below 30%.
3. Good lubrication is needed to decrease the drawing force.

15. 6.15
Slides 6.23-6.25
Question: Would you elaborate the derivations in these slides?
Answer: The backward can impact extrusion model in these slides was again adapted
from Ref. [1]. This time, Lange [1] presents a model that comes from the Ph.D.
dissertation of Dipper [8]. The analytical model, which makes good use of slab method, is
quite easy to understand. Despite its sheer simplicity, the force estimates of the model
are in very good agreement with experimental ones especially for thin-walled extrusions.
Before we continue, let us briefly discuss the limitation of this model: In backward can
extrusion, the relative area reduction is defined as
𝜀𝐴 ≜
𝐴0 − 𝐴1
𝐴0
= 1 −
1
𝑅𝑒
(6.62)
where A0 is the area of the billet; A1 refers to the cross-section of the extruded product; Re
denotes the reduction ratio in Eqn. (6.47). Hence, Eqn. (6.62) boils down to
𝜀𝐴 ≜
𝑑𝑒
2
− (𝑑𝑒
2
− 𝑑𝑖
2
)
𝑑𝑒
2
=
𝑑𝑖
2
𝑑𝑒
2
(6.63)
Dipper’s backward can extrusion model yields accurate results when 0.5 ≤ 𝜀𝐴 ≤ 0.6.
Notice that as 𝜀𝐴 increases, the wall thickness of the extruded can (~
𝑑𝑒−𝑑𝑖
2
) decreases.
Therefore, as confirmed by experimental studies, the analytical model presented here is
mostly applicable to thin-walled cans.
As illustrated in Fig. 6.8, the deformation zone is divided into three portions:
1. Zone 1: Material is upset between ram/punch and the die.
2. Zone 2: Material, which is compressed sideways, is plastically flowing in the
upwards direction.
3. Zone 3: Material in this zone essentially forms the wall of the can. Since the
plastic deformation is presumed to terminate at the interface between Zone 2 and
Zone 3, this zone is free from any plastic deformation and does not bear any
load/stress.
Ram
Zone 1
di
Die
d0 = de
Zone
3
Zone
3
Zone 2
Zone 2
b
s
s
z
dz
r dr
Figure 6.8: Nomenclature for the backward can extrusion process.

16. 6.16
The analysis starts with Zone 2 and works its way up to Zone 1. In Zone 2, a differential
ring element is considered as illustrated in Figs. 6.8 and 6.9.
dz
σr2
σr2
σr2
σz2+dσz2
σz2
σr2
σf2/2
μ1σf2 σf2/2 μ1σf2
σz2
σz2+dσz2
di/2
de/2
Figure 6.9: Stresses on the differential ring element (Zone 2).
The axial force equilibrium for this element can be written as
[(𝜎𝑧2 + d𝜎𝑧2) − 𝜎𝑧2]
⏟
Net axial stress
[
𝜋(𝑑𝑒
2
− 𝑑𝑖
2
)
4
]
⏟
Area
= μ1𝜎𝑓2(𝜋𝑑𝑒𝑑𝑧)
⏟
Friction force
(Die side)
+
𝜎𝑓2
2
⏞
𝜏𝑦𝑖𝑒𝑙𝑑
(𝜋𝑑𝑖𝑑𝑧)
⏟
Friction force
(Zone 1 boundary)
(6.64)
where μ1 refers to the kinematic friction coefficient between the die and the workpiece
material. The inner surface of the ring (i.e. the boundary between Zones 1 and 2) is
subjected to plastic shearing. Consequently, the shear stress is taken as 𝜏𝑦𝑖𝑒𝑙𝑑 =
𝜎𝑓2
2
. It is
critical to note that while estimating Coulomb friction forces, the normal stresses (like
𝜎𝑟2) are not employed for the sake of simplicity. Just like the assumptions in Slide 6.12,
the pressures on the walls of the die/punch are presumed to be on the order of the flow
stress of the material at a specific zone.
Rearranging Eqn. (6.64) yields
d𝜎𝑧2
𝜋 (𝑑𝑒 − 𝑑𝑖)
⏞
2𝑠
(𝑑𝑒 + 𝑑𝑖)
⏞
2(𝑑𝑖+𝑠)
4
= π𝜎𝑓2dz [ μ1 (𝑑𝑖 + 2𝑠)
⏞
𝑑𝑒
+
1
2
𝑑𝑖]
(6.65)
That is,
d𝜎𝑧2𝜋 2𝑠
𝑑𝑖 + 𝑠
2
= π𝜎𝑓2dz (μ1𝑑𝑖 + μ1𝑠 + μ1𝑠
⏞
~
1
2
𝑠
+
1
2
𝑑𝑖) (6.66)
For the sake of simplifying Eqn. (6.66), the term μ1𝑠 is replaced by 1
2
𝑠. Consequently, we
get
d𝜎𝑧2𝜋 2𝑠
𝑑𝑖 + 𝑠
2
= 2π𝜎𝑓2dz
μ1 + 1
2
2
⏟
≜ 𝜇
∙
𝑑𝑖 + 𝑠
2
⏟
average
radius
∙ 2
(6.67)
Here, 𝜇 ≜
1
2
(μ1 + 1
2
) is called the effective coefficient of friction. Finally, Eqn. (6.67)
boils down to the following differential equation:
d𝜎𝑧2𝑠 = 2𝜇𝜎𝑓2dz (6.68)
Integrating Eqn. (6.68) yields

17. 6.17
∫ d𝜎𝑧2 =
2𝜇𝜎𝑓2
𝑠
∫ dz (6.69a)
∴ 𝜎𝑧2 =
2𝜇𝜎𝑓2
𝑠
𝑧 + C (6.69b)
To compute the integration constant C in Eqn. (6.69b), a boundary condition is needed.
Recall that Zone 3 is presumed not to carry any axial stress. Therefore, 𝜎𝑧2 = 0 at the
boundary between these two neighboring zones (i.e. z = b). Thus,
C = −
2𝜇𝜎𝑓2
𝑠
𝑏 (6.70)
Evidently, Eqn. (6.69b) becomes
𝜎𝑧2 = −
2𝜇𝜎𝑓2
𝑠
(𝑏 − 𝑧) (6.71)
Note that since the ring element inside Zone 2 is plastically deforming altogether, we can
determine the radial stress component (𝜎𝑟2) by applying Tresca yield criterion. As shown
in Slide 6.23 (see estimated stress distributions at each zone), Dipper assumes that
𝜎1 = 𝜎𝑧2 and 𝜎3 = 𝜎𝑟2. Hence,
𝜎1 − 𝜎3 = 𝜎𝑧2 − 𝜎𝑟2 = 𝜎𝑓2 (6.72a)
∴ 𝜎𝑟2 = 𝜎𝑧2 − 𝜎𝑓2 (6.72b)
Using Eqn. (6.71), the mean axial stress6 can be computed:
𝜎𝑧2|𝑧=𝑏 2
⁄ = 𝜎𝑧2𝑚 = −𝜇𝜎𝑓2
𝑏
𝑠
(6.73)
Plugging Eqn. (6.73) into (6.72b) yields the mean radial stress in Zone 2:
𝜎𝑟2|𝑧=𝑏 2
⁄
⏟
𝜎𝑟2𝑚
= 𝜎𝑧2|𝑧=𝑏 2
⁄
⏟
𝜎𝑧2𝑚
− 𝜎𝑓2
(6.74a)
∴ 𝜎𝑟2𝑚 = −𝜎𝑓2 (1 + 𝜇
𝑏
𝑠
) (6.74b)
Now, we can proceed to Zone 1. To this end, a differential ring element in Fig. 6.8 is
considered. Fig. 6.10 shows the stresses acting on this element. The radial force
equilibrium for the differential ring takes the following form:
[(𝜎𝑟1 + d𝜎𝑟1) − 𝜎𝑟1]
⏟
Net radial stress
2𝜋𝑟𝑏
⏟
Area
= 2μ1𝜎𝑓1(2𝜋𝑟𝑑𝑟)
⏟
Total friction force
(6.75)
Simplifying this equation leads to
d𝜎𝑟1 =
2μ1𝜎𝑓1
𝑏
𝑑𝑟 (6.76)
By integrating Eqn. (6.76), we have
6 Since z2 changes linearly, the mean value (z2m) resides in the middle of its domain (z = b/2).

18. 6.18
𝜎𝑟1 =
2μ1𝜎𝑓1
𝑏
𝑟 + 𝐶 (6.77)
To determine the integration constant C in Eqn. (6.77), we shall consider the interface
between the adjacent zones.
σ
r1
+dσ
r1
σz1
σr1
μ1σf1
r σz1
μ1σf1
σz1
μ1σf1
σz1
μ1σf1
σ
r1
+dσ
r1
b
dr
σr1
Figure 6.10: Stresses on the differential ring element (Zone 1).
For that purpose, the transposition of the average radial stresses from Zone 2 to Zone 1 is
assumed. That is,
𝜎𝑟1|𝑟=𝑑𝑖 2
⁄ =
2μ1𝜎𝑓1
𝑏
⋅
𝑑𝑖
2
+ 𝐶 = 𝜎𝑟2𝑚 (6.78a)
∴ 𝐶 = 𝜎𝑟2𝑚 − μ1𝜎𝑓1
𝑑𝑖
𝑏
(6.78b)
Evidently,
𝜎𝑟1 =
2μ1𝜎𝑓1
𝑏
(𝑟 −
𝑑𝑖
2
) + 𝜎𝑟2𝑚 (6.79)
The axial stress component (i.e. the pressure on the ram/punch) can be determined by
applying Tresca yield criterion to the ring element. With the assumption 𝜎1 = 𝜎𝑟1 and
𝜎3 = 𝜎𝑧1, we obtain
𝜎1 − 𝜎3 = 𝜎𝑟1 − 𝜎𝑧1 = 𝜎𝑓1 (6.80a)
∴ 𝜎𝑧1 = 𝜎𝑟1 − 𝜎𝑓1 (6.80b)
Now, let us compute the average stresses for Zone 1. Using Eqn. (6.79), the mean radial
stress can be easily computed:
𝜎𝑟1𝑚 =
4
𝜋𝑑𝑖
2
⏟
1/Area
∫ 𝜎𝑟1(𝑟) 2𝜋𝑟 𝑑𝑟
𝑑𝑖/2
0
(6.81a)
𝜎𝑟1𝑚 =
2μ1𝜎𝑓1
𝑏
∫ (𝑟 −
𝑑𝑖
2
) 2𝜋𝑟𝑑𝑟
𝑑𝑖/2
0
−𝜎𝑓2 (1 + 𝜇
𝑏
𝑠
)
⏟
𝜎𝑟2𝑚
(6.81b)
Consequently, Eqn. (6.81b) becomes
𝜎𝑟1𝑚 = −
1
3
μ1𝜎𝑓1
𝑑𝑖
𝑏
− 𝜎𝑓2 (1 + 𝜇
𝑏
𝑠
) (6.82)

19. 6.19
Similarly, plugging Eqn. (6.82) into (6.80b) yields the average axial stress (i.e. the average
pressure on the ram):
𝜎𝑧1𝑚 = 𝜎𝑟1𝑚 − 𝜎𝑓1 (6.83a)
∴ 𝜎𝑧1𝑚 ≜ −𝑝𝑝𝑢𝑛𝑐ℎ = −𝜎𝑓1 (1 +
1
3
μ1
𝑑𝑖
𝑏
) − 𝜎𝑓2 (1 + 𝜇
𝑏
𝑠
) (6.83b)
Finally, the extrusion force becomes
𝐹𝑝𝑢𝑛𝑐ℎ = 𝑝𝑝𝑢𝑛𝑐ℎ𝐴𝑝𝑢𝑛𝑐ℎ = − 𝜎𝑧1𝑚 (
𝜋𝑑𝑖
2
4
) (6.84)
Note that the flow stresses in Eqn. (6.83b) are defined as
𝜎𝑓1 = 𝐾(𝜀𝑒𝑞1)
𝑛
(6.85a)
𝜎𝑓2 = 𝐾(𝜀𝑒𝑞,𝑡𝑜𝑡𝑎𝑙)
𝑛
(6.85b)
where the equivalent strains can be expressed as follows:
𝜀𝑒𝑞1 = 𝑙𝑛 (
ℎ𝑜
𝑏
) (6.86a)
𝜀𝑒𝑞,𝑡𝑜𝑡𝑎𝑙 = 𝜀𝑒𝑞1 + 𝜀𝑒𝑞1
𝑑𝑖
8𝑠
⏟
𝜀𝑒𝑞2
= 𝑙𝑛 (
ℎ𝑜
𝑏
) (1 +
𝑑𝑖
8𝑠
)
(6.86b)
Here, h0 refers to the initial thickness of the billet.
Slides 6.26
Question: Would you elaborate the empirical piercing model in this slide?
Answer: This slide was actually adapted from Ref. [6]. There is a very nice explanation
about piercing process in Section 4.4.3 (Open-die Forging). Please refer to pages 236-
237.

20. 6.20
Slide 6.27
Question: Would you give an example on the application of the presented model on
backward can extrusion process?
Answer: Consider the problem (Example 3) in this slide. The given data are as follows:
 K = 700 [MPa]
 n = 0.24
 1 = 0.06
 h0 = 35 [mm]
 d0 = de = 70 [mm]
 di = 58 [mm]
 b = 5 [mm]
 s = (de –di)/2 = 6 [mm]
First of all, we shall take a look at the analytical model given by Eqn. (6.83b). The
unknown quantities are calculated as follows:
𝜀𝑒𝑞1 = 𝑙𝑛 (
ℎ0
𝑏
) = 𝑙𝑛 (
35
5
) = 1.95
𝜀𝑒𝑞,𝑡𝑜𝑡𝑎𝑙 = 𝑙𝑛 (
ℎ𝑜
𝑏
) (1 +
𝑑𝑖
8𝑠
) = 1.95 (1 +
58
8 ⋅ 6
) = 4.31
With the strains at hand, we can calculate the corresponding flow stresses:
𝜎𝑓1 = 𝐾(𝜀𝑒𝑞1)
𝑛
= 700(1.95)0.24
≅ 821 [MPa]
𝜎𝑓2 = 𝐾(𝜀𝑒𝑞,𝑡𝑜𝑡𝑎𝑙)
𝑛
= 700(4.31)0.24
≅ 994 [MPa]
Similarly, the effective coefficient of friction becomes
𝜇 =
1
2
(μ1 +
1
2
) =
0.06 + 0.5
2
= 0.28
The mean axial stress on the punch [Eqn. (6.83b)] can be expressed as
𝜎𝑧1𝑚 = −821 (1 +
1
3
0.06
58
5
) − 994 (1 + 0.28
5
6
)
∴ ppunch = −σz1m ≅ 2237 [MPa]
The punch force is
𝐹punch = 𝑝punch𝐴punch = 2237 (
𝜋 582
4
) = 5.910 [MN]
Let us check the validity of the result:
𝜀𝐴 =
𝑑𝑖
2
𝑑𝑒
2 =
582
702
= 0.68
Since 0.5 ≤ 𝜀𝐴 ≤ 0.6, the result we have just obtained (0.68) is a little bit out of the range.
To cross-check our result, we can employ empirical techniques in Slide 6.20. Let us
compute all the required quantities:
𝐴0 =
𝜋 702
4
= 3848 [mm2
]

21. 6.21
𝐴1 =
𝜋(702
− 582
)
4
= 1206 [mm2
]
Extrusion ratio:
𝑅𝑒 =
𝐴0
𝐴1
= 3.19
Equivalent strain:
𝜀𝑒𝑞 = ln(𝑅𝑒) = 1.16
Mean flow stress:
𝜎𝑓𝑚 =
𝐾(𝜀𝑒𝑞)
𝑛
𝑛 + 1
=
700(1.16)0.24
1.24
= 585 [MPa]
Strain factor:
Qe = 0.8 + 1.2𝜀𝑒𝑞 = 0.8 + 1.2 ⋅ 1.16 = 2.19
Consequently, the average punch force is estimated as
𝐹
̅ = 𝜎𝑓𝑚 Qe A0 = 585 ⋅ 2.19 ⋅ 3848 = 4.934 [MN]
This punch force, which does not constitute any container friction, is actually in good
agreement with the one we have found earlier.
As the final step, let us also compute the constrained piercing force (see Slide 6.26):
𝐹piercing = (3 … 5)𝜎𝑓𝑚 (A0 − 𝐴1)
⏟
𝐴punch
= 5 ⋅ 585(3848 − 1206) = 7.727 [MN]
Note that in constrained piercing, the pressure factor starts with 3 (see equation above).
As the piercing tool (i.e. indenter) descends, this coefficient quickly reaches up to 5 due to
strain-hardening. Therefore, we have chosen to use the highest factor of 5 in our
computation.
Anyway, since 𝐹
̅ < 𝐹𝑝𝑖𝑒𝑟𝑐𝑖𝑛𝑔, inhomogeneous deformation (i.e. piercing) never takes
place.

22. 6.22
Slide 6.28-6.34
Question: Would you explain Table 1 in Slide 6.28?
Answer: This table, which was adapted from Lange [1] (see Tab. 5.2 on page 250),
tabulates the force components in extrusion/drawing processes. Most results given in this
table are obtained through the Siebel’s extrusion model that we have elaborated in the
previous pages. We shall show the transposed version of this table to enhance the
readability of this document.
Table 6.1: Calculation of force components for extrusion and drawing processes.
Process:
Direct/forward
extrusion of rods
Direct/forward
extrusion of tubes
Wire/rod- & bar
drawing
Ironing & tube
drawing over
fixed mandrel
Fideal 𝜎𝑓𝑚𝜀𝑒𝑞A0 𝜎𝑓𝑚𝜀𝑒𝑞A0 𝜎𝑓𝑚𝜀𝑒𝑞A1 𝜎𝑓𝑚𝜀𝑒𝑞A1
Fshear
2
3
tan(𝛼) 𝜎𝑓𝑚A0
1
2
tan(𝛼) 𝜎𝑓𝑚A0
2
3
tan(𝛼) 𝜎𝑓𝑚A1
1
2
tan(𝛼) 𝜎𝑓𝑚A1
Ffriction
(Container)
𝜋𝑑0ℎ𝜇𝜎𝑓0 𝜋𝑑0ℎ𝜇𝜎𝑓0
Ffriction
(Die-shoulder)
2𝜇𝜎𝑓𝑚𝜀𝑒𝑞A0
sin(2𝛼)
2𝜇𝜎𝑓𝑚𝜀𝑒𝑞A0
sin(2𝛼)
2𝜇𝜎𝑓𝑚𝜀𝑒𝑞A1
sin(2𝛼)
2𝜇𝑆𝜎𝑓𝑚𝜀𝑒𝑞A1
sin(2𝛼)
Ffriction
(Mandrel-die)
𝜇𝜎𝑓𝑚𝜀𝑒𝑞A1
tan(𝛼)
±
𝜇𝑀𝜎𝑓𝑚𝜀𝑒𝑞A1
tan(𝛼)
Ffriction
(Mandrel-exit)
𝜋𝑑2ℎ𝑟𝜇𝑝̅𝑟
Let us explain each process presented on a column of this table and discuss the exceptions
that go with them:
Column 1
This column refers to the direct/forward extrusion model in Eqn. (6.46). There are a few
exceptions to consider:
i) If reducing type extrusion (sometimes referred to as open-die extrusion) is
performed as shown in Slide 6.29, the friction force for the container must be
omitted [i.e. Ffriction (Container) = 0] owing to the fact that there is no container here!
The following slide (Slide 6.30) is not actually related to the comments for this table.
It reminds us that in backward extrusion process (as illustrated in this slide),
there is no container friction due to the fact that the material inside the container
is not moving at all. That is, there is no relative motion between the workpiece and
the die. Since the friction develops as a resistance to motion, the workpiece material
(under huge pressures) does not experience any frictional force whatsoever.
ii) The rod extrusion in practice is generally performed with a die angle (2) of 180. In
that case, Eqn. (6.46) obviously breaks down and yields infinite extrusion force! In
this situation, a dead-zone at an angle () of 45 (i.e. the direction where maximum
shear stress occurs) develops, the material is then plastically shorn along those
boundaries as illustrated in Slide 6.31. Hence, the shear stress becomes

23. 6.23
𝜏 = 𝜏𝑦𝑖𝑒𝑙𝑑 =
1
2
𝜎𝑓𝑚 = 𝜇′
𝜎𝑓𝑚 (6.88)
Consequently, the model is modified for this special case by defining an artificial die
angle () of 45 and the effective friction coefficient (𝜇′
) of
1
2
. Eqn. (6.46) then takes
the following form:
Ftotal′(h) = A0𝜎𝑓𝑚 [
2
3
tan(𝛼′) + (1 +
2𝜇′
sin(2𝛼′)
) 𝜀𝑒𝑞] + 𝜋𝑑0ℎ𝜇𝜎𝑓0 (6.89)
With 𝛼′
= 45° and 𝜇′
=
1
2
, it boils down to
Ftotal′(h) = A0𝜎𝑓𝑚 (
2
3
+ 2 𝜀𝑒𝑞) + 𝜋𝑑0ℎ𝜇𝜎𝑓0 (6.90)
Notice that since a dead-zone is formed inside the container, the height of the
material in the container must be modified accordingly. That is, the range of h in
Eqn. (6.90) becomes 0 ≤ ℎ ≤ ℎ0
′
where ℎ0
′
≜ ℎ0 −
𝑑0−𝑑1
2 tan(𝛼′)
.
It is critical to note that apart from 2𝛼 = 180°, the dead-zone might also emerge
when 0° < 2𝛼 < 180°. If the extrusion force computed by Eqn. (6.46) is less than that
of Eqn. (6.90), the material will favor to slide on the surface of the die-shoulder.
Otherwise, it is expected to be plastically shorn at an angle of 45. Under the
presumption that the friction forces at the container are somewhat negligible, this
artificial angle can be simply calculated as
𝛼′
= {
𝛼, 𝑓(𝛼) < 0
𝜋/4, 𝑓(𝛼) ≥ 0
(6.91)
where the function describing the sliding condition in Eqn. (6.91) is
𝑓(𝛼) = tan(𝛼) − 1 −
3
2
(1 −
2𝜇
sin(2𝛼)
) 𝜀𝑒𝑞 (6.92)
The pictorial description of Eqn. (6.91) is given in Fig. 6.11. Here, the constants 𝛼−
and 𝛼+
are essentially the roots of the nonlinear function 𝑓(𝛼) = 0. The MATLAB
function (dzone.m) listed in Table 6.2 numerically solves the above-mentioned
equation and returns these roots. It is important to note that the shallow dies (except
for the dies with 2𝛼 = 180°) are very rare in practice.
The question that might be raised at this point is whether there exists an optimum
die angle that will minimize the extrusion force. To this end, let us take the
derivative of Eqn. (6.46) with respect to  and set it to zero. Since the container
friction is not a function of , the resulting expression becomes
𝑑
𝑑𝛼
[
2
3
tan(𝛼) +
2𝜇𝜀𝑒𝑞
sin(2𝛼)
] = 0 (6.93a)

24. 6.24
∴
cos(2𝛼)
sin2(𝛼)
=
2
3𝜇𝜀𝑒𝑞
(6.93b)
Solving Eqn. (6.93b) for the optimum die angle * leads to
𝛼∗
= tan−1
(√
3𝜇𝜀𝑒𝑞
3𝜇𝜀𝑒𝑞 + 2
) (6.94)
When a simulation7 with the parameters given in Tables 2 and 3 of Slide 6.28 is
performed, one can discover that the optimum angle ranges in between 11 and 37
while the average value is 28. Note that in all of these cases, 𝛼−
< 𝛼∗
< 𝛼+
. Thus,
the dead-zone formation is effectively avoided.
π/2
π/4
π/4
α+
α-
α-
α+
α
α´
f(α)<0 f(α)>0
Figure 6.11: Artificial die angle.
Column 2
This column refers to the direct/forward extrusion of tubes with a moving mandrel (its
diameter is dm = d2) as shown in Slide 6.33. The process is quite similar to the forward
extrusion of rods except that the friction forces acting on the mandrel should be taken into
consideration:
 Inside the container, the relative speed of the material with respect to the mandrel
is zero. Therefore, there is no friction at this portion of the mandrel.
7 This simulation is frequently referred to as Monte-Carlo simulation. Here, the variables (, eq) are
assigned randomly within their prescribed range. The output (*) is then computed and stored for a large
number of trials (like 106). Consequently, the statistical attributes of the outcome (e.g. maximum & minimum
values, average, variance, probability distribution, etc.) are reported at the end of a typical simulation.

25. 6.25
 In the die shoulder, the friction force on the mandrel is as specified in Table 6.1
(see the 5th row). Interested students are encouraged to refer to Lange [1] that
includes a derivation of this result.
 Slide 6.33 elaborates the computation of the friction force acting on the mandrel at
the die exit (see the 6th row). The frictional shear stress (i.e. 𝜇𝑝̅𝑟) is presumed to be
in the range of 10 to 12 [MPa].
The rest of the components are as expected. However, the shear force decreases (e.g. the
corresponding multiplier changes from 2/3 to ½) owing to the fact that mandrel occupies
some portion of the die center. Thus, the effective area of the shorn material (at the inlet
and exit) is essentially reduced.
Table 6.2: MATLAB program to compute the artificial die angle.
mu = .125*rand;
epsilon = .5 + 6.5*rand;
as = atan(sqrt(3*mu*epsilon/(3*mu*epsilon+2)));
disp(['Optimum Die Angle: ' num2str(rad2deg(as)) ' [deg]'])
[a2,a1] = dzone(mu,epsilon);
disp(['Upper Limit of Die Angle: ' num2str(rad2deg(a2)) ' [deg]'])
disp(['Lower Limit of Die Angle: ' num2str(rad2deg(a1)) ' [deg]'])
alpha = linspace(0,pi/2,100)';
cond = tan(alpha)-1-1.5*(1-2*mu./sin(2*alpha))*epsilon;
alpha_p = alpha.*(cond<0) + .25*pi*(cond>=0);
plot(rad2deg(alpha),rad2deg(alpha_p)); grid on
xlabel('alpha [circ]'); ylabel('alphaprime [circ]')
function [alpha2,alpha1] = dzone(mu,epsilon)
fcn = @(x) tan(x)-1-1.5*(1-2*mu/sin(2*x))*epsilon;
alpha1 = fsolve(fcn,.1,optimoptions('fsolve','Display','off'));
alpha2 = fsolve(fcn,pi/4,optimoptions('fsolve','Display','off'));
end
Column 3
The expressions in this column give the force components for rod/bar drawing. An
application of this model was presented earlier through an example in Slide 6.22.
Column 4
The model presented in this column essentially neglects the changes in the inner diameter
of the drawn tube. As shown in Slide 6.32, when the inner diameter of the tube commences
to change, a bending moment (i.e. the energy required to bend the tube) must be taken
into consideration. The given model obviously lacks that component.

26. 6.26
The last comment is about ironing process. As shown in Slide 6.34, the frictional force
acting directly on the punch reduces the force transmitted to the base. In ironing, the base
force (i.e. Fp – Ff) is, by definition, the component drawing the material out. Therefore,
this friction force needs to be subtracted from the total punch force. The minus (–) sign in
the fifth row does serve for this purpose. On the other hand, in tube drawing with fixed
mandrel, the mandrel friction force adds on to the draw force. Hence, (+) sign is employed
for this case. Here, the friction coefficient 𝜇𝑀 denotes the friction coefficient between the
workpiece and the mandrel/punch while 𝜇𝑆 refers to the one between the die and the
workpiece.
References
[1] Lange, K., Lehrbuch der Umformtechnik, Band 1, 2, 3, Springer-Verlag, Berlin, 1975.
[2] Lange, K., Handbook of Metal Forming, SME Publications, Michigan, 1994.
[3] Siebel, E., Die Formgebung im bildsamen Zustand, Düsseldorf:Verlag Stahleisen, 1932.
[4] Siebel, E., “Untersuchungen über bildsame Formänderung unter besonderer
Berücksichtigung des Schmiedens,” Maschinenbau/Betrieb 9, 307-312, 1923.
[5] Tlusty, G., Manufacturing Processes and Equipment, Prentice Hall, NJ, 2000.
[6] Schey, J. A., Introduction to Manufacturing Processes, 2nd Edition, McGraw Hill, NY, 1987.
[7] Groover, M. P., Fundamentals of Modern Manufacturing, 5th Edition, John Wiley, NY,
2013.
[8] Dipper, M., Das Fließpressen von Hülsen in Rechnung und Versuch, Dr.-Ing.-Diss.,
Hochschule für Technik Stuttgart, 1949, s. auch Archiv f. d. Eisenhüttenwesen 20, 275-286,
1949.