Manufacturing Technologies- ME303

1. 3.1
Chapter 3 – Material Properties (II)
Questions & Answers
Slide 3.7
Question: Would you give an example on the application of the additive property of true
strains?
Answer: Consider the problem (Example 1) in this slide. Let us create a table showing
the true- and engineering strains at each stage:
Notice that the total true strain can be directly computed as
𝜀𝑡𝑜𝑡𝑎𝑙 = ln (
200
100
) = 0.6931
As expected, the sum of true strains given in the table is in accordance with this result.
Let us do the same calculation for engineering strain:
𝜀𝑒𝑛𝑔,𝑡𝑜𝑡𝑎𝑙 =
200 − 100
100
= 1
This result does not match with the one given in the table. Consequently, the additive
property is only valid for true strains.
Slide 3.10
Question: Is it possible to find the yield strength of an annealed material (i.e. initial flow
stress, f0) using Ludwik-Hollomon curve? That is, since the plastic (engineering) strain,
when the stress on the specimen reaches to f0, is 0.2% by definition, the corresponding
true strain becomes
𝜀 = ln(1 + 𝜀𝑒𝑛𝑔) = ln(1 + 0.002) ≅ 0.002
Stage True Strain Engineering Strain
1 𝜀1 = ln (
120
100
) = 0.1823 𝜀𝑒𝑛𝑔,1 =
120 − 100
100
= 0.2
2 𝜀2 = ln (
150
120
) = 0.2231 𝜀𝑒𝑛𝑔,2 =
150 − 120
120
= 0.25
3 𝜀3 = ln (
200
150
) = 0.2877 𝜀𝑒𝑛𝑔,3 =
200 − 150
150
= 0.3333
 ∑ 𝜀𝑖
3
𝑖=1
= 0.6931 ∑ 𝜀𝑒𝑛𝑔,𝑖
3
𝑖=1
= 0.7833

2. 3.2
Therefore, is it true that
𝜎𝑓0 ≅ 𝑌𝑆 = 𝐾𝜀𝑛
= 𝐾(0.002)𝑛
?
Answer: Unfortunately, NO. According to one of the axioms for this course, the Ludwik-
Hollomon curve does not accurately capture the material’s behavior when the plastic
strains (𝜀𝑝𝑙) are small. Therefore, the initial flow stress (f0) must accompany the
strength coefficient (K) and the strain-hardening exponent (n) to characterize the
plastic behavior of the material entirely.
Note that some textbooks such as Ref. [1] take 𝜎𝑓0 ≅ 𝐾(0.002)𝑛
owing to the fact that K
and n coefficients are adjusted such that the curve passes through the points f0 and UTS
(i.e. the actual stress when the load hits its maximum)1. Such treatments do naturally
violate one of our axioms on the slope of the load curve. Please refer to the Slide 3.11 that
elaborates the issue.
Question: How do we obtain K and n coefficients using the uniaxial tensile test data?
Answer: We start out by arranging the test data in homogeneous deformation region (i.e.
up to the onset of necking). The engineering quantities are transformed into the true ones
by employing the expressions in the Slides 3.8 and 3.9. That is, a table containing the
transformed data is formed:
Point 1 2 … N
i 1 2 … N
i 1 2 … N
To fit a curve to the given data, we write the equation of power law:
𝜎 = 𝐾𝜀𝑛
(3.1)
This is not a linear equation: The unknowns (i.e. K and n) are not expressed as the linear
combinations of  and . The natural logarithm of Eqn. (3.1) could be more useful:
ln(𝜎)
⏟
𝑦
= ln[𝐾(𝜀)𝑛] = ln(𝐾)
⏟
𝑐
+ 𝑛 l n(𝜀)
⏟
𝑥 (3.2)
Hence, it is more convenient for us to fit a curve to the logarithm of the data. Note that in
Eqn. (3.2), a new parameter is defined for sake of convenience:
𝑐 ≜ ln(𝐾) (3.3)
Accordingly, the new table becomes
Point 1 2 … N
yi = ln(i) ln(1) ln(2) … ln(N)
xi = ln(i) ln(1) ln(2) … ln(N)
Ideally, we want all the data in the table above to satisfy Eqn. (3.2):
Point 1 2 … N
Equation 𝑦1 = 𝑛𝑥1 + 𝑐 𝑦2 = 𝑛𝑥2 + 𝑐 … 𝑦𝑁 = 𝑛𝑥𝑁 + 𝑐
1 For instance, see Example 3.3 of Ref. [1].

3. 3.3
Here, one can arrange all these equations into a matrix form:
[
𝑥1 1
𝑥2 1
⋮ ⋮
𝑥𝑁 1
]
⏟
𝐀(𝑁×2)
[
𝑛
𝑐
]
⏟
𝐗(2×1)
= [
𝑦1
𝑦2
⋮
𝑦𝑁
]
⏟
𝐁(𝑁×1)
(3.4)
Unfortunately, Eqn. (3.4) cannot be solved directly since A is not a square matrix. In
other words, there are more equations than the number of unknowns (N > 2). Hence, a
neat manipulation called the Pseudo Inverse Method (PIM) can be utilized to this end.
First, both sides of Eqn. (3.4) are multiplied by AT:
𝐀T
𝐀𝐗 = 𝐀T
𝐁 (3.5)
Note that since ATA is now a square matrix (22), X in Eqn. (3.5) could be solved:
𝐗 = (𝐀T
𝐀)−1
𝐀T
𝐁 (3.6)
where [(𝐀T
𝐀)−1
𝐀T
] is called the pseudo inverse of the matrix A. In fact, the solution
provided by Eqn. (3.6) does not satisfy the equations in the table above [or Eqn. (3.4) for
that matter!]. It yields a least squares solution where an exponential curve is fit to the
given data to minimize the following error (i.e. sum of squares) function:
𝐸(𝑛, 𝑐) = ∑[𝑦𝑖 − (𝑛𝑥𝑖 + 𝑐)]2
𝑁
𝑖=1
(3.7)
Consequently, the unknown coefficients are obtained as
𝑛 = 𝐗[1] (3.8a)
𝐾 = exp(𝐗[2]) (3.8b)
Slide 3.11
Question: In this slide, it looks as if the derivative of the load with respect to the true
strain is computed in order to find true strain at the necking point. Doesn’t it make more
sense to compute the derivative of the load with respect to the engineering strain? After
all, we get to chart the engineering strain (or elongation) versus the load in uniaxial
tensile test.
Answer: Each semester, we get this question a lot. For some reason, the differential
calculus in this slide does not satisfy most of our students! We shall take a more direct
approach now. Let us write the load as a function of engineering strain:
𝑃 = 𝜎𝑓𝐴 (3.9)
The true stress 𝜎𝑓 in Eqn. (3.9) can be expressed as
𝜎𝑓 = 𝐾𝜀𝑛
(3.10)
Employing volume constancy, the area A at a particular instant in time can be written as
𝐴 ⋅ 𝑙 = 𝐴0 ⋅ 𝑙0 (3.11a)
∴ 𝐴 = 𝐴0 ⋅
𝑙0
𝑙
=
𝐴0
𝑙 𝑙0
⁄
⏟
𝑒𝜀
= 𝐴0𝑒−𝜀
(3.11b)

4. 3.4
Substituting Eqn. (3.11b) and (3.10) into (3.9) yields the load as a function of true strain:
𝑃 = 𝐾𝐴0𝜀𝑛
𝑒−𝜀
(3.12)
Since
𝜀 = ln(1 + 𝜀𝑒𝑛𝑔), (3.13)
Eqn. (3.12) becomes
𝑃 = 𝐾𝐴0
[ln(1 + 𝜀𝑒𝑛𝑔)]
𝑛
1 + 𝜀𝑒𝑛𝑔
(3.14)
Now that we have expressed the load as function of engineering strain, we are ready to
compute its derivative:
𝑑𝑃
𝑑𝜀𝑒𝑛𝑔
= 𝐾𝐴0
[𝑛 − l n(1 + 𝜀𝑒𝑛𝑔)][ln(1 + 𝜀𝑒𝑛𝑔)]
𝑛−1
(1 + 𝜀𝑒𝑛𝑔)
2 (3.15)
By setting Eqn. (3.15) to zero, one can find the engineering strain that maximizes the
load:
[𝑛 − ln(1 + 𝜀𝑒𝑛𝑔,𝑈𝑇𝑆)]
⏟
=0
[ln(1 + 𝜀𝑒𝑛𝑔,𝑈𝑇𝑆)]
𝑛−1
(1 + 𝜀𝑒𝑛𝑔,𝑈𝑇𝑆)
2
⏟
≠0
= 0
(3.16)
Assuming that in Eqn. (3.16)
[ln(1 + 𝜀𝑒𝑛𝑔,𝑈𝑇𝑆)]
𝑛−1
(1 + 𝜀𝑒𝑛𝑔,𝑈𝑇𝑆)
2 ≠ 0, (3.17)
the leftmost term (evaluated at the UTS point) must be equal to zero in order to have a
solution for Eqn. (3.16):
𝑛 − ln(1 + 𝜀𝑒𝑛𝑔,𝑈𝑇𝑆 ) = 0 (3.18)
Consequently,
𝜀𝑒𝑛𝑔,𝑈𝑇𝑆 = 𝑒𝑛
− 1 (3.19)
Note that using Eqn. (3.13), the corresponding true strain takes the following form:
𝜀𝑈𝑇𝑆 = ln(1 + 𝜀𝑒𝑛𝑔,𝑈𝑇𝑆) = 𝑛 (3.20)
Not surprisingly, this is the same result offered by the slide.
Question: If we experimentally determine K and n coefficients (as explained in the Slide
3.10), do we have a control over the condition that 𝜀𝑈𝑇𝑆 = 𝑛 (i.e. the slope of the load curve
at the UTS point is zero)?
Answer: No, not really. You can select more data points near the UTS region. That could
help to some extent. However, the slope of the estimated load curve is likely not to be flat
at the UTS point.

5. 3.5
Actually, the condition (𝜀𝑈𝑇𝑆 = 𝑛) helps us to determine K and n coefficients through
analytical means very easily. From the test data, we have Pmax and the corresponding
elongation at that point (lUTS = lUTS – l0). Therefore,
𝜀𝑈𝑇𝑆 = ln(1 + 𝜀𝑒𝑛𝑔,𝑈𝑇𝑆) = ln (1 +
Δ𝑙𝑈𝑇𝑆
𝑙0
) = 𝑛 (3.21)
Since the true stress at the UTS point is
𝜎𝑈𝑇𝑆 = 𝜎𝑒𝑛𝑔,𝑈𝑇𝑆(1 + 𝜀𝑒𝑛𝑔,𝑈𝑇𝑆) =
𝑃𝑚𝑎𝑥
𝐴0
(1 +
Δ𝑙𝑈𝑇𝑆
𝑙0
)
⏟
𝑒𝑛
,
(3.22)
the strength coefficient becomes
𝜎𝑈𝑇𝑆 = 𝐾 (𝜀𝑈𝑇𝑆
⏟)
𝑛
𝑛
⇒ 𝐾 =
𝜎𝑈𝑇𝑆
𝑛𝑛
=
𝑃𝑚𝑎𝑥
𝐴0
(
𝑒
𝑛
)
𝑛
(3.23)
One could argue that K can be determined by considering the initial flow stress:
𝜎𝑓0 = 𝐾(0.002)𝑛
⇒ 𝐾 =
𝜎𝑓0
0.002𝑛
=
𝑃0.2%
𝐴0
⋅
1.002
0.002𝑛 (3.24)
where P0.2% refers to the load leading to a permanent deformation of 0.2%. Note that with
this K selection, Eqn. (3.23) will not be satisfied:
𝜎𝑈𝑇𝑆 ≠ 𝐾(𝑛)𝑛
(3.25)
In this course, we definitely employ Eqn. (3.21) and (3.23) to compute K and n owing to
the fact that the underlying procedure is compatible with our axioms. It is all a matter of
consistency.
Slide 3.13
Question: What is the difference between the definitions associated with the ductility
and the area reduction?
Answer: They both define the shrinkage in the area2. The area reduction is described by
𝑞 =
𝐴𝑟𝑒𝑓 − 𝐴
𝐴𝑟𝑒𝑓
(3.26)
where Aref refers to the area at the reference state (usually initial). On the other hand, the
ductility, which indicates the ability of a particular material to sustain deformation
without fracturing, utilizes the same definition except that the area of interest (A) is now
replaced by the area at fracture (Afrac):
𝑞∗
=
𝐴𝑟𝑒𝑓 − 𝐴𝑓𝑟𝑎𝑐
𝐴𝑟𝑒𝑓
(3.27)
Note that as a notational convenience, the superscript (*) is added to the ductility.
2 These two definitions are valid for uniaxial tensile test.

6. 3.6
Question: How do we define the amount of cold-work? How is it related to the
engineering (or true) strain?
Answer: The amount of cold-work, which is usually expressed as percentage, uses the
definition in Eqn. (3.26). Let us first show its association with the true strain:
𝑞 =
𝐴0 − 𝐴
𝐴0
= 1 −
𝐴
𝐴0
(3.28a)
∴
𝐴
𝐴0
= 1 − 𝑞 (3.28b)
Recall that the true strain is
𝜀 = ln (
𝑙
𝑙0
) = ln (
𝐴0
𝐴
) (3.29)
Substituting Eqn. (3.28b) into (3.29) gives
𝜀 = ln (
1
1 − 𝑞
) = − ln (1 − 𝑞) (3.30a)
or
𝑞 = 1 − 𝑒−𝜀
(3.30b)
Next is the engineering strain. From Eqn. (3.13), we have
𝜀𝑒𝑛𝑔 = 𝑒𝜀
− 1 (3.31)
Plugging Eqn. (3.30a) into (3.31) yields
𝜀𝑒𝑛𝑔 =
𝑞
1 − 𝑞 (3.32a)
Alternatively,
𝑞 =
𝜀𝑒𝑛𝑔
1 + 𝜀𝑒𝑛𝑔
(3.32b)
Slide 3.14
Question: Yield strength is an engineering definition. This slide claims that the yield
strength of a cold-worked material equals to the true stress associated with a particular
strain inflicted at cold-work. How is that possible?
Answer: To explain this concept, let us take a look at the load-strain diagram in Fig. 3.1.
Here, the specimen, which is initially annealed, is strained via the Path (o-a-b). We shall
presume that the test is interrupted at Point (b) and that the load is released. It is
obvious that the strength of the cold-worked test specimen at Point (c) will be much
higher than its annealed state due to strain-hardening. If the test is resumed, the
specimen is expected to follow the remainder of the curve. Notice that the cold-worked
specimen at Point (c) is to yield again when the load reaches to Pb (not Pa).

7. 3.7
l0 A0
lcw Acw
Annealed Strain (εeng)
o
a
b
Pb
Pa
c
Load (P)
Cold-worked
fail
Figure 3.1: The effect of cold-work on yield strength.
Now, let us find the yield strength of the cold-worked material. By definition,
𝑌𝑆𝑐𝑤 =
𝑃𝑏
𝐴𝑐𝑤
(3.33)
The load Pb can be expressed as the product of true stress and the instantaneous area at
Point (b):
𝑌𝑆𝑐𝑤 =
𝜎𝑓𝐴𝑏
𝐴𝑐𝑤
(3.34)
By assuming that the elastic effects are negligible (i.e. 𝐴𝑏 ≅ 𝐴𝑐𝑤; 𝜀𝑏 ≅ 𝜀𝑐𝑤) and that the
true stress is depicted by the power law, one can write
𝑌𝑆𝑐𝑤(𝜀𝑐𝑤) ≅ 𝜎𝑓 ≅ 𝐾(𝜀𝑐𝑤)𝑛
(3.35)
where 𝜀𝑐𝑤 = 𝑙𝑛 (
𝑙𝑐𝑤
𝑙0
) = 𝑙𝑛 (
𝐴0
𝐴𝑐𝑤
). This is the result presented by the Slide 3.14. In fact, the
flow stress (represented by the power law) can be interpreted as the geometric loci of the
yield strength for cold-worked material.
By employing (3.13), Eqn. (3.35) can rearranged as a function of engineering strain:
𝑌𝑆𝑐𝑤(𝜀𝑒𝑛𝑔,𝑐𝑤) ≅ 𝐾[ln(1 + 𝜀𝑒𝑛𝑔,𝑐𝑤)]
𝑛
(3.36)
where 𝜀𝑒𝑛𝑔,𝑐𝑤 =
𝑙𝑐𝑤−𝑙0
𝑙0
=
𝐴0−𝐴𝑐𝑤
𝐴𝑐𝑤
. Similarly, with the utilization of Eqn. (3.30a) and (3.35),
the yield strength of a cold-worked material could be expressed in terms of qcw:
𝑌𝑆𝑐𝑤(𝑞𝑐𝑤) ≅ 𝐾 [ln (
1
1 − 𝑞𝑐𝑤
)]
𝑛
(3.37)
where 𝑞𝑐𝑤 =
𝐴0−𝐴𝑐𝑤
𝐴0
.

8. 3.8
Slide 3.15
Question: Is it possible to find the UTS of the annealed material when the parameters of
the power law (e.g. K and n) are specified.
Answer: Yes, it is. We shall employ the basic definition of UTS:
𝑈𝑇𝑆𝑎𝑛𝑛 =
𝑃𝑚𝑎𝑥
𝐴0
=
𝜎𝑈𝑇𝑆𝐴𝑈𝑇𝑆
𝐴0
(3.38)
The true stress at the UTS (necking) point is
𝜎𝑈𝑇𝑆 = 𝐾 (𝜀𝑈𝑇𝑆
⏟
𝑛
)
𝑛
= 𝐾𝑛𝑛
(3.39)
Recall that
𝜀𝑈𝑇𝑆 = ln (
𝑙𝑈𝑇𝑆
𝑙0
) = ln (
𝐴0
𝐴𝑈𝑇𝑆
) = 𝑛 (3.40)
From Eqn. (3.40), we get
𝐴0
𝐴𝑈𝑇𝑆
= 𝑒𝑛
(3.41)
Substituting Eqns. (3.39) and (3.41) into (3.38) yields
𝑈𝑇𝑆𝑎𝑛𝑛 =
𝐾𝑛𝑛
𝑒𝑛
= 𝐾 (
𝑛
𝑒
)
𝑛
(3.42)
Likewise, the UTS of a cold-worked material is described by
𝑈𝑇𝑆𝑐𝑤 = 𝑈𝑇𝑆𝑎𝑛𝑛
𝐴0
𝐴𝑐𝑤
(3.43)
(see the slide). Since
𝜀𝑐𝑤 = ln (
𝑙𝑐𝑤
𝑙0
) = ln (
𝐴0
𝐴𝑐𝑤
) ⇒
𝐴0
𝐴𝑐𝑤
= 𝑒𝜀𝑐𝑤, (3.44)
Eqn. (3.43) becomes
𝑈𝑇𝑆𝑐𝑤 = 𝑈𝑇𝑆𝑎𝑛𝑛𝑒𝜀𝑐𝑤 = 𝐾 (
𝑛
𝑒
)
𝑛
𝑒𝜀𝑐𝑤
(3.45)
Question: What is the underlying mechanism behind the neck formation? Why doesn’t
the test specimen exhibit uniform plastic deformation until fracture?
Answer: The underlying reason for the neck formation (i.e. the source for plastic
instability) is due to the deviation in the ideal geometry of the test specimen as well as
the inhomogeneity associated with the material. Roughly speaking, the weakest cross-
section of the specimen is expected to deform more than the rest (and fail eventually) in
tensile straining.
To understand (and to quantify) this concept, let us consider the (multi-segmented) test
specimen shown in Fig. 3.2. In this model, each segment (say, the ith segment) of the

9. 3.9
specimen can be perceived as a perfect disc whose initial diameter (D0i) is essentially
constant (i.e. no diametric deviation) throughout its axis. Furthermore, each disc is
presumed to be made out of an isotropic material which is free from any significant
defects3. Notice that if each disc were subjected to uniaxial tensile test individually, it
would have deformed uniformly until its fracture.
The flow curve for each disc can be expressed as
𝜎𝑓𝑖 = 𝐾𝑖(𝜀𝑖)𝑛
(3.46)
where n is the strain-hardening exponent (which is presumably common to all elements);
𝐾𝑖 (𝑖 ∈ {1, 2, … , 𝑁}) refers to the strength coefficient associated with a particular cross-
section while 𝜀𝑖 denotes the true plastic strain inflicted on that segment at a specific
instant during the tensile testing. The weakest section of the specimen can be determined
as follows:
𝐾𝑤𝐷𝑤
2
= min {𝐾1𝐷01
2
, 𝐾2𝐷02
2
, … , 𝐾𝑁𝐷0𝑁
2 } (3.47)
where w denotes the index of the weakest section. It is self-evident that the weakest link
in the chain will dictate the force transmitted through the whole test specimen. Thus,
using Eqn. (3.12), the load acting on each disc can be estimated by considering the plastic
deformation taking place at the weakest portion:
𝑃 = 𝐾𝑤 (
𝜋𝐷0𝑤
2
4
)
⏟
≜ 𝐴0𝑤
(𝜀𝑤)𝑛
𝑒−𝜀𝑤
(3.48)
Here, 𝜀𝑤 is the true plastic deformation experienced by the weakest section throughout
the straining regime and ranges between 0 and 𝜀𝑓𝑟𝑎𝑐. Fig. 3.3 shows the resulting load as
a function of 𝜀𝑤. As can be seen, the force attains its maximum value when 𝜀𝑤 = 𝑛.
L0
P
P
D
01
D
02
D
03
D
0N
D
0N-1
D
0N-2
L0/N L0/N L0/N L0/N L0/N L0/N εw
P
Pmax
n εfrac
0
Figure 3.2: The geometric model for a non-ideal
test specimen.
Figure 3.3: Load profile dictated by
the weakest section.
Note that there are two products in Eqn. (3.48): The first product 𝐾𝑤(𝜀𝑤)𝑛
depicts the
strength increase due to strain-hardening while the second one (𝐴0𝑤𝑒−𝜀𝑤) describes the
shrinkage in the area as the plastic deformation proceeds. Since the corresponding
3 Fracture occurs as a consequence of crack initiation/propagation. If the tested material is said to be ideal,
it will be theoretically free from all defects such as voids and surface scratches that are known to instigate
the crack formation. In that case, the perfect specimen is expected to elongate indefinitely (without
breaking) in the form of a filament!

10. 3.10
decrease in cross-sectional area of the segment is faster than the apparent increase in the
strength such that the decreasing trend in the load does not change when 𝜀𝑤 ≥ 𝑛. Since
every section is subjected to the same load, one can find the true plastic strains inflicted
upon each segment utilizing Eqn. (3.48):
𝑃 = 𝐾𝑖 (
𝜋𝐷0𝑖
2
4
) (𝜀𝑖)𝑛
𝑒−𝜀𝑖 = 𝐾𝑤 (
𝜋𝐷0𝑤
2
4
) (𝜀𝑤)𝑛
𝑒−𝜀𝑤
(3.49a)
(𝜀𝑖)𝑛
𝑒−𝜀𝑖 = (
𝐾𝑤𝐷0𝑤
2
𝐾𝑖𝐷0𝑖
2 )
⏟
≜ 𝑐𝑖
(𝜀𝑤)𝑛
𝑒−𝜀𝑤
(3.49b)
Solving Eqn. (3.49b) for 𝜀𝑖 leads to
𝜀𝑖 = −𝑛 ⋅ 𝑊 (−
𝜀𝑤 √𝑐𝑖𝑒−𝜀𝑤
𝑛
𝑛
) (𝑖 ≠ 𝑤) (3.50)
where W refers to the Lambert W function4. The similarity factor in Eqn. (3.50) is
defined as
𝑐𝑖 ≜
𝐾𝑤𝐷0𝑤
2
𝐾𝑖𝐷0𝑖
2 < 1(𝑖 ≠ 𝑤) (3.51)
Fig. 3.4 plots 𝜀𝑖 given in Eqn. (3.50) as
𝜀𝑤 varies from 0 to 1 for n = 0.2. For
𝑐𝑖 = 1, the curve should boil down to a
line: 𝜀𝑖 = 𝜀𝑤. Notice that when 𝜀𝑤 > 𝑛,
the true plastic strains for each
segment (𝜀𝑖) tend to drop below the
maximum ones (𝜀𝑖
∗
) that were
attained when 𝜀𝑤 = 𝑛. This means
that when 𝜀𝑤 ≥ 𝑛, the plastic
deformation of that particular
segment will terminate (if and only if
𝑐𝑖 < 1) since the stresses induced at
each segment are not capable of
generating further plastic
deformation. From this point on, the
plastic deformation will cease for all
the segments except the weakest one.
Figure 3.4: The true plastic strains for a
particular segment as a function of 𝜀𝑤 (n = 0.2).
In an actual tensile test, the necking region exhibits the plastic deformation pattern for
the weakest portion of the specimen. Hence, Eqn. (3.50) can be rearranged to
accommodate this plastic deformation limit for each segment (𝑖 ≠ 𝑤):
𝜀𝑖 = {
−𝑛 ⋅ 𝑊 (−
𝜀𝑤 √𝑐𝑖𝑒−𝜀𝑤
𝑛
𝑛
) , 𝜀𝑤 < 𝑛
−𝑛 ⋅ 𝑊(− √𝑐𝑖𝑒−𝑛
𝑛
), 𝜀𝑤 ≥ 𝑛
(3.52)
Employing Eqn. (3.52), the engineering strain at a particular instant in the tensile test
can be written. That is, the new length of each section is
4 Consider the real-valued function 𝑓(𝑥) = 𝑥𝑒𝑥
. The Lambert W function is simply 𝑊(𝑥) = 𝑓−1
(𝑥).
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0.16
w

i
ci
= 0.9
ci
= 0.99

11. 3.11
𝜀𝑖 = ln (
𝐿𝑖
𝐿0𝑖
) = ln (
𝐿𝑖
𝐿0/𝑁
) (3.53a)
∴ 𝐿𝑖 =
𝐿0
𝑁
𝑒𝜀𝑖
(3.53b)
Total length of the deformed specimen then becomes
𝐿𝑡𝑜𝑡𝑎𝑙 = ∑ 𝐿𝑖
𝑁
𝑖=1
=
𝐿0
𝑁
∑ 𝑒𝜀𝑖
𝑁
𝑖=1
(3.54)
Since 𝜀𝑒𝑛𝑔 =
𝐿𝑡𝑜𝑡𝑎𝑙
𝐿0
− 1, we have
𝜀𝑒𝑛𝑔(𝜀𝑤) =
1
𝑁
∑ 𝑒𝜀𝑖(𝜀𝑤)
𝑁
𝑖=1
− 1 (3.55)
Notice that (3.48) (i.e. P) and (3.55) (i.e. 𝜀𝑒𝑛𝑔) are the parametric equations of 𝜀𝑤.
Evidently, plotting (3.48) against (3.55) yields the load as a function of engineering strain.
Similarly, the new diameter for each segment can be calculated as
𝜀𝑖 = ln (
𝐴0𝑖
𝐴𝑖
) = ln (
𝐷0𝑖
2
𝐷𝑖
2
) = ln (
𝐷0𝑖
𝐷𝑖
)
2
= 2 ln (
𝐷0𝑖
𝐷𝑖
) (3.56a)
∴ 𝐷𝑖(𝜀𝑤) = 𝐷0𝑖𝑒−
𝜀𝑖(𝜀𝑤)
2 (3.56b)
Based on the presented mathematical model, a simulation of the uniaxial tensile test is
conducted using MATLAB. Table 3.1 tabulates the MATLAB script developed for the
simulation while its outputs are presented in Figs. 3.5 and 3.6. The load curve given in
Fig. 3.5 resembles the expected one except that the load quickly drops in the overstraining
section. Furthermore, the maximum load does not occur when 𝜀𝑒𝑛𝑔,𝑈𝑇𝑆 = 𝑒𝑛
− 1 = 0.2214
due to the underlying simplifications incorporated to the model. Note that the specimen
progressively thins out at its weakest site as can be seen from Fig. 3.6 since the smallest
diameter is artificially set at the 17th segment.
Figure 3.5: Normalized eng versus eng in
the simulated test.
Figure 3.6: Diametric changes in the
simulated test.
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18
0
0.1
0.2
0.3
0.4
0.5
eng

eng
/K
0 5 10 15 20 25 30 35 40 45 50
0.55
0.6
0.65
0.7
0.75
0.8
0.85
0.9
0.95
1
Segment No.
D/D
0

12. 3.12
Table 3.1 MATLAB script to simulate the uniaxial tensile test.
N = 50; n = .2; q = 0.65; eps_f = -log(1-q);
eps_w = linspace(0,eps_f,N)';
c = .99 + .01*cosd(360*eps_w/eps_f+60);
[cmin,w] = min(c); f = cmin*exp(-eps_w).*(eps_w.^n);
eps_eng = zeros(N,1); eps = zeros(N,1);
h = waitbar(0,'Please wait...'); close all
for time = 1:N
epsw = eps_w(time);
for seg = 1:N
if (seg==w)
eps(seg) = epsw;
else
if (epsw<n)
eps(seg) = -n*lambertw(-(c(seg)*exp(-epsw)*epsw^n)^(1/n)/n);
else
eps(seg) = -n*lambertw(-(c(seg)*exp(-n))^(1/n));
end
end
end
eps_eng(time) = mean(exp(eps))-1;
D = sqrt(c.*exp(-eps));
figure(1); plot(D); hold on
waitbar(time/N,h)
end
xlabel('Segment No.'); ylabel('D/D_0'); grid on; close(h)
figure(2)
plot(eps_eng,f); grid on; axis('tight')
xlabel('epsilon_{eng}'); ylabel('sigma_{eng}/K')
Question: Why does the test specimen appear to fracture right in the middle in an actual
tensile test?
Answer: This critical question is raised once in a blue moon. As explained before, the
weakest portion of the test specimen plastically deforms continuously until its demise!
The location of the weakest site is highly affected by the following factors: i) raw material;
ii) material processing techniques; iii) manufacturing methods involved in the production
of the test specimen. Long-, round-, and slender specimens are frequently manufactured
via turning5 where the specimen is held securely between the tailstock and the chuck. In
such a scheme, the workpiece typically attains a barrel-like shape (see Ref. [4]) due to the
elastic deformations caused by machining forces. Hence, the thickest portion usually lies
at the center and the chances of breaking the specimen in the middle become quite slim
in that case. In fact, the smallest diameter (i.e. weakest segment) is frequently
encountered right after the shoulders. Fig. 3.7 shows fractured test specimens made out
of various engineering materials. As can be seen from these examples, they did not
necessarily fail near the center.
5 The geometry- and tolerances of the specimens are commonly specified by international standards like ISO
6892-1: "Metallic materials. Tensile testing. Method of test at ambient temperature."

13. 3.13
Figure 3.7: Various fractured test specimens (Copyrighted images are compiled from the
Internet).
Question: The test specimen may contain a large number of weak spots. What is the
reason for not observing multiple-necks in tensile tests?
Answer: From the standpoint of statistics, there will be only one weakest spot among
the ones already present inside the specimen [see Eqn. (3.47)]. The chances of finding two
weakest spot with absolutely the same attributes (i.e. the same geometry, surface
texture, grain structure/distribution, dislocation densities, point defects, material
impurities, and more) are next to impossible. That is why, (to the best of our knowledge)
two or more necks have never been registered in tensile tests.
Question: In tensile test, the neck forms when 𝜀𝑒𝑛𝑔,𝑈𝑇𝑆 = 𝑒𝑛
− 1. From this point on, the
specimen deforms locally around the weakest site until its fracture. Does this fact mean
that the plastic strains to be inflicted in a particular bulk-deformation process should not
exceed 𝜀𝑈𝑇𝑆 = 𝑛?
Answer: This is a common misconception. The plastic instability in tensile test is a
unique/special case manifested by both uniaxial tension and the geometric / strength
deviation of the specimen along its major axis. Actually, the uniaxial test accurately
captures the strain-hardening ability of the tested material up to the instant where the
local neck commences to develop. Beyond this point, we get to observe the deformation
pattern in a small localized region where the material continues to strain-harden further.
To characterize the material’s behavior at high strains (i.e. 𝜀 ≫ 𝑛), some researchers
measure (and model) the geometry changes (e.g. radius of curvature around the neck, min.
radius of the neck, etc.) in the necking region during tests (refer to Ref. [5]). However,
such approaches are not too practical. Thus, the 𝜎 − 𝜀 curve in uniform deformation region
is commonly extrapolated for 𝜀 > 𝑛.
It is critical to note that in many bulk-deformation processes (such as forward extrusion,
closed-die forging, rolling), the strains can well exceed this critical strain (𝜀𝑈𝑇𝑆 = 𝑛)
registered in uniaxial tensile test. Despite a significant reduction in ductility, the
workpiece does not exhibit any defect in such circumstances. Note that when the plastic
strains go beyond 𝜀𝑈𝑇𝑆, the material is said to be over-strained (see also Slide 3.23).

14. 3.14
Slide 3.17
Question: Would you give an example on the application of the topics covered so far?
Answer: Let us do the example (Example 2) in this slide. The given data are as follows:
 d0 = 12 [mm]
 l0 = 50 [mm]
 l1 = 60 [mm]
 P1 = 32.2 [kN]
a) All stresses and strains:
𝜀1 = ln (
𝑙1
𝑙0
) = ln (
60
50
) = 0.1823
𝜀𝑒𝑛𝑔,1 =
𝑙1 − 𝑙0
𝑙0
=
60 − 50
50
= 0.2
𝜎𝑒𝑛𝑔,1 =
𝑃1
𝐴0
=
32200
𝜋(0.012)2
4
= 284.711 [MPa]
𝜎1 = 𝜎𝑒𝑛𝑔,1(1 + 𝜀𝑒𝑛𝑔,1) = 284.711(1 + 0.2)
𝜎1 = 341.653 [MPa]
b) Final diameter (d1):
𝐴0𝑙0 ≅ 𝐴1𝑙1 ⇒ 𝑑1 = √
𝑑0
2
𝑙0
𝑙1
= √
12250
60
𝑑1 = 10.95 [mm]
Let us cross-check our result in (a):
𝜎1 =
𝑃1
𝐴1
=
32200
𝜋(0.01095)2
4
= 341.653 [MPa]
They match!
c) UTS of cold-worked material (UTS1):
New data on the material is presented in this part: K = 800 [MPa]; n = 0.5. Let us find
the UTS of the annealed material first using Eqn. (3.42):
𝑈𝑇𝑆𝑎𝑛𝑛 = 𝐾 (
𝑛
𝑒
)
𝑛
= 800 (
0.5
2.7183
)
0.5
= 343.1056 [MPa]
Employing Eqn. (3.45) yields the UTS of cold-worked material:
𝑈𝑇𝑆𝑐𝑤 = 𝑈𝑇𝑆1 = 𝑈𝑇𝑆𝑎𝑛𝑛𝑒𝜀1 = 343.1056 ⋅ 𝑒0.1823
𝑈𝑇𝑆𝑐𝑤 = 411.7178 [MPa]

15. 3.15
d) Yield strength and ultimate tensile strength as a function of qcw: The solution of this
part is given in the Slides 3.18-3.19. Also refer to the Q&A comments for the Slides
3.14-3.15.
Slide 3.23
Question: Would you give an example on the application of the topics covered so far?
Answer: Let us do the example
(Example 3) in this slide. The given
data are as follows:
 K = 700 [MPa]
 n = 0.3
 𝑞𝑐𝑤1 =
𝐴0−𝐴1
𝐴0
= 0.15
 𝑞𝑐𝑤2 =
𝐴1−𝐴2
𝐴1
= 0.1
In this problem, the workpiece is
strained in two stages as illustrated
in the figure.
l0 A0
l2 A2
Strain (εeng)
Load (P)
A1
l1
Af
a) Yield strength at each stage: The amount of cold-work is specified (rather than the
true strains inflicted). Therefore, we start with converting these quantities into the
true strains with the utilization of Eqn. (3.30a):
𝜀0→1 = ln (
1
1 − 𝑞𝑐𝑤1
) = ln (
1
1 − 0.15
) = 0.1625
𝜀1→2 = ln (
1
1 − 𝑞𝑐𝑤2
) = ln (
1
1 − 0.1
) = 0.1054
Employing the additive property of true strains, we can calculate the total strain at
each stage:
𝜀𝑐𝑤1 ≜ 𝜀0→1 = 0.1625
𝜀𝑐𝑤2 ≜ 𝜀0→2 = 𝜀0→1 + 𝜀1→2 = 0.1625 + 0.1054 = 0.2679
Note that these strains gives us the strains inflicted on the material with respect to the
annealed state. Consequently, using Eqn. (3.35) yields
𝑌𝑆𝑐𝑤1 ≅ 𝐾 (𝜀𝑐𝑤1)𝑛
= 700(0.1625)0.3
= 405.8393 [MPa]
𝑌𝑆𝑐𝑤2 ≅ 𝐾 (𝜀𝑐𝑤2)𝑛
= 700(0.2679)0.3
= 471.5089 [MPa]
A word of caution: The direct application of Eqn. (3.37) [or Eqn. (3.36) for that
matter!] will get you in trouble owing to the fact that the amount of cold-work in this
equation is defined with respect to the annealed state. Consequently, applying the
formula for the second/last stage will give you the wrong answer because the additional
cold-work (e.g. %qcw2 = 10%) is defined with respect to the intermediate state (not the
annealed state!). That is the reason why we get to employ true strains in the first place
because their additive property enables us to compute the total strains with respect to

16. 3.16
the annealed state. Thanks to strain hardening hypothesis (see the Slides 3.20-
3.22), the yield strength at any cold-worked state can be calculated by employing the
uniaxial tensile test data (i.e. K and n) available for a specific annealed material.
b) UTS of each state: Let us find the UTS of the annealed material first using Eqn. (3.42):
𝑈𝑇𝑆𝑎𝑛𝑛 = 𝐾 (
𝑛
𝑒
)
𝑛
= 700 (
0.3
2.7183
)
0.3
= 361.3650 [MPa]
Employing Eqn. (3.45) yields the UTS of cold-worked material at stage 1:
𝑈𝑇𝑆𝑐𝑤1 = 𝑈𝑇𝑆𝑎𝑛𝑛𝑒𝜀𝑐𝑤1 = 361.3650 ⋅ 𝑒0.1625
𝑈𝑇𝑆𝑐𝑤1 = 425.1272 [MPa]
Likewise, the UTS of cold-worked material at stage 2 becomes
𝑈𝑇𝑆𝑐𝑤2 = 𝑈𝑇𝑆𝑎𝑛𝑛𝑒𝜀𝑐𝑤2 = 361.3650 ⋅ 𝑒0.2679
𝑈𝑇𝑆𝑐𝑤2 = 472.3823 [MPa]
c) Ductility of each state: In this part, the ductility of the annealed part is specified:
𝑞𝑎𝑛𝑛
∗
=
𝐴0 − 𝐴𝑓
𝐴0
= 0.5 ⇒
𝐴𝑓
𝐴0
= 0.5
Similarly, the ductility of cold-worked material at stage 1 is defined as
𝑞𝑐𝑤1
∗
=
𝐴1 − 𝐴𝑓
𝐴1
= 1 −
𝐴𝑓
𝐴0
⏟
0.5
⋅
𝐴0
𝐴1
Recall that
𝜀𝑐𝑤1 = ln (
𝐴0
𝐴1
) ⇒
𝐴0
𝐴1
= 𝑒𝜀𝑐𝑤1
Therefore,
𝑞𝑐𝑤1
∗
= 1 − 0.5
𝐴0
𝐴1
⏟
𝑒𝜀𝑐𝑤1
= 1 − 0.5 ⋅ 𝑒0.1625
= 0.4118
Ductility of cold-worked material at stage 2 can be defined as
𝑞𝑐𝑤2
∗
=
𝐴2 − 𝐴𝑓
𝐴2
= 1 −
𝐴𝑓
𝐴0
⏟
0.5
⋅
𝐴0
𝐴2
𝜀𝑐𝑤2 = ln (
𝐴0
𝐴2
) ⇒
𝐴0
𝐴2
= 𝑒𝜀𝑐𝑤2
Finally,
𝑞𝑐𝑤2
∗
= 1 − 0.5
𝐴0
𝐴2
⏟
𝑒𝜀𝑐𝑤2
= 1 − 0.5 ⋅ 𝑒0.2679
= 0.3464

17. 3.17
The figure below plots the quantities computed in this exercise. As can be seen, the
yield strength is approaching to UTS as the plastic strain approaches to n. Note that
when (𝜀 > 𝑛), 𝑌𝑆 ≅ 𝑈𝑇𝑆. This case is commonly referred to as overstraining.
Slide 3.25
Question: Would you explain the graph in this slide?
Answer: This graphs shows the effect of temperature on mechanical properties of
materials. On the abscissa, the homologous temperature is displayed. In this (unitless)
temperature scale:
 0 corresponds to absolute zero: 0 [K];
 0.5 refers to the recrystallization temperature;
 1 denotes the melting temperature of a particular material;
 The range between 0 and 0.3 is the cold-working region;
 The range between 0.3 and 0.5 is the warm-working region;
 The range between 0.5 and 0.7 is the hot-working region.
As can be seen, when the temperature of the workpiece (i.e. working temperature) is
elevated, the yield strength of the material tends to decrease while the ductility
commences to improve. In hot working regime (TH > 0.5), the strength of the material
actually depends on the strain-rate (i.e. 𝜀̇ = 𝑑𝜀/𝑑𝑡) rather the strain itself. Higher the
straining rate (i.e. faster the inflicted deformation), higher the flow stress. The underlying
reasons behind this behavior are further explained in the Slides 3.27 and 3.30. Note that
in hot working regime, the flow stress can be characterized by
𝜎𝑓 = 𝐶(𝜀̇)𝑚
(3.57)
where C is a material coefficient that is a function of temperature; m is the strain-rate
sensitivity exponent. The effect of strain in Eqn. (3.57) is negligible for all practical
purposes.
0
10
20
30
40
50
60
0
50
100
150
200
250
300
350
400
450
500
0 0.05 0.1 0.15 0.2 0.25 0.3
Ductility
[%]
Stress
[MPa]
True Strain
YS UTS q*

18. 3.18
Slides 3.30
Question: Would you explain the annealing process?
Answer: Let’s start with the cold-work. Here, the plastic deformation essentially takes
place at low temperatures (i.e. TH < 0.3). As the amount of plastic deformation increases,
the grains tend to be more distorted while the dislocation density multiplies. Due to
entanglement of dislocations, their motions become progressively more inhibited.
Evidently, the strength of the cold-worked material increases while the ductility drops as
shown in graph on the left.
If the temperature of this cold-worked part is elevated to, say, warm-working range (i.e.
0.3 < TH < 0.5), recovery annealing can be performed. As can be seen from the figure in
the middle, the strength drops slightly while the ductility improves marginally as time
progresses. This is due to the fact that the dislocations commence to disentangle and
migrate slowly towards the grain boundaries. Note that there are no changes in the grain
structure in recovery annealing. As a consequence, internal residual stresses produced by
prior cold-work could be reduced dramatically. That is why, this process is referred to as
stress-relief annealing (or stress-relieving heat treatment).
Similarly, if this cold-worked part is heated up above the recrystallization temperature
(i.e. TH > 0.5) and is held at that temperature for about an hour or so, the distorted grains
will begin to reform. That is, new grains will develop at nucleation sites. These sites are
located at highly distorted points on the original lattice as a consequence of prior cold-
work. Eventually, the new grains will completely take over and replace the old ones. Their
average size depends on the amount of prior cold-work, annealing temperature, and time.
As shown in the figure on the right, this is the primary phase of the recrystallization.
Note that since the new grains have very low dislocation density, the corresponding
strength is reduced while its ductility is improved. However, if ample time is given, some
of the grains will start to grow by coalescence with their neighbors. This grain-coarsening
process, which is accelerated by high annealing temperatures, forms the secondary
phase. Due to coarse grains, the yield strength is expected to drop. Additionally,
hydrogen embrittlement, where the absorbed hydrogen creates brittle compounds (like
hydrides) at grain boundaries, is observed. This further reduces the ductility in the
secondary phase. For more information on this issue, please refer to Refs. [2] and [3].
Slides 3.31-3.32
Question: Would you explain the annealing temperature?
Answer: Annealing temperature is not a single value. It simply refers to a range of
temperatures. For instance, at a high temperature, the material can be annealed within
a short period. Likewise, the annealing process takes a bit longer at a moderate
temperature as can be seen from the Slide 3.31. Finally, at a relatively low temperature6,
more time is required to anneal the material. For short, a certain amount of energy must
be expended to complete the metallurgical state transformation in annealing at a
particular temperature.
6 Note that in all these three cases, the annealing temperature is greater than 0.5 in homologous temperature
scale.

19. 3.19
It is critical to notice that the annealing temperature also depends on the amount of prior
cold-work and the average size of the resulting grains. As mentioned in the Slide 3.30,
heavier cold-work leads to the formation of smaller- and distorted grains. Evidently, more
nucleation sites are produced and the recrystallization process commences at lower
temperatures. Roughly speaking, higher the number of nucleation sites, lower the
annealing temperature as depicted in the Slide 3.32. If there is no prior cold-work, no
recrystallization occurs.
Slide 3.36
Question: What is the relationship between Tresca- and von Mises yield criteria?
Answer: von Mises yield criterion, which is based on maximum (distortion) strain energy
principle, is presumed to be the most general form of all yield criteria. In fact, Tresca yield
criterion appears to be a piecewise approximation of its counterpart. To understand this
issue, let us apply both yield criteria to a planar stress state case as illustrated in Fig. 3.8.
In this particular example, the third principal stress component (3) is set to zero while
the remaining ones (1, 2) are presumed to vary within the four quadrants of the stress-
plane. It is obvious that the name of each principle stress components needs to change in
this exercise owing to the fact that 1, 2, 3 (by definition) refer to the maximum-, the
medium-, and the minimum principle stress components respectively. However, assigning
new names to the variables creates a mess in the representation. Therefore, while
applying Tresca’s yield criterion, we select the maximum (max) and the minimum (min)
components from the set {1, 2, 3} depending on the yield condition under study (a total
of 6 conditions). Then, we get to sketch the boundary created by |𝜎𝑚𝑎𝑥 − 𝜎𝑚𝑖𝑛| = 𝑌𝑆 . The
result is the line segments forming a distorted hexagon as illustrated in Fig. 3.8.
σ
1
–0
=
YS
σ2–0 = YS
0-σ2 = YS
0-σ
1
=
YS
σ1
σ2
YS
YS
-YS
-YS
Von Mises
Ellipse
σ3 = 0
σ2>σ1>σ3
σ1>σ2>σ3
σ1>σ3>σ2
σ3>σ1>σ2
σ3>σ2>σ1
σ2>σ3>σ1
Figure 3.8: Yield boundaries of different yield criteria.

20. 3.20
With respect to von Mises yield criterion, we don’t need to reassign the roles of varying
principle stress components due to the fact that every possible combination of differences
between the principle stress components is evaluated in the expression. That is,
𝜎𝑒𝑞 ≜ 𝜎
̅ = √
(𝜎1 − 𝜎2)2 + (𝜎2 − 𝜎3)2 + (𝜎3 − 𝜎1)2
2
= 𝑌𝑆 (3.58)
Since 𝜎3 = 0, simplifying Eqn. (3.58) leads to
𝜎1
2
− 𝜎1𝜎2 + 𝜎2
2
𝑌𝑆2
=
(𝜎1 + 𝜎2)2
2(√2 𝑌𝑆)
2 +
(𝜎1 − 𝜎2)2
2 (√2/3 𝑌𝑆)2
= 1 (3.59)
Notice that Eqn. (3.59) in fact represents a rotated ellipse. In general, such an ellipse can
be expressed as
[𝜎1 cos(𝜃) +𝜎2 sin(𝜃)]2
(√2 𝑌𝑆)
2 +
[−𝜎1 sin(𝜃) +𝜎2 cos(𝜃)]2
(√2/3 𝑌𝑆)2
= 1 (3.60)
where  refers to the rotation angle in between the major axis of the ellipse and the 1
axis. By setting  = /4, Eqn. (3.60) could be turned into Eqn. (3.59). As can be seen from
Fig. 3.8, the transformed ellipse in Eqn. (3.59) encircles the boundaries of the Tresca’s
criterion. In other words, the von Mises yield criterion, which conforms to the
experimental studies better than its counterpart, estimates higher stresses for the onset
of plastic deformation. One can claim that the Tresca yield criterion is much more
conservative than its counterpart.
Slide 3.40-3.41
Question: Would you explain this slide?
Answer: Levy-Mises flow rule defines how materials plastically deform under the action
of stresses. Hence, it gives us the ability to relate plastic strains to the stresses or vice
versa. In fact, the rule performs a job similar to that of the Hooke’s law except that it
depicts the plastic deformations of materials. Let us elaborate this issue.
For this purpose, consider the Levy-Mises flow rule7 for the x-axis:
𝑑𝜀𝑥𝑥
𝜎𝑥′
=
𝑑𝜀𝑥𝑥
𝜎𝑥𝑥 − 𝜎ℎ
= 𝑑𝜆 (3.61)
Here, the hydrostatic stress is defined as
𝜎ℎ ≜
𝜎𝑥𝑥 + 𝜎𝑦𝑦 + 𝜎𝑧𝑧
3 (3.62)
Plugging Eqn. (3.62) into (3.61) gives
𝑑𝜀𝑥𝑥 = 𝑑𝜆 (𝜎𝑥𝑥 −
𝜎𝑥𝑥 + 𝜎𝑦𝑦 + 𝜎𝑧𝑧
3
) (3.63a)
7 In Ref. [6] (see Section 2.6), the derivation of the flow rule is elaborated. It is interesting to note that the
flow rule in Slide 3.40 is in fact related to the yield boundary dictated by von Mises yield criterion [i.e. Eqn.
(3.58)]:
𝑑𝜀𝑖𝑗
𝑑𝜆
= 𝜕𝜎
̅
𝜕𝜎𝑖𝑗
.

21. 3.21
∴ 𝑑𝜀𝑥𝑥 =
2
3
𝑑𝜆 [𝜎𝑥𝑥 −
1
2
(𝜎𝑦𝑦
+ 𝜎𝑧𝑧)] (3.63b)
Eqn. (3.63b) resembles the Hooke’s law that describes elastic deformations (see Slide
3.39). For instance,
𝜀𝑥𝑥 =
1
𝐸
[𝜎𝑥𝑥 − 𝜈 (𝜎𝑦𝑦
+ 𝜎𝑧𝑧)] (3.64)
Let us make a brief comparison between Eqn. (3.63b) and (3.64):
 In Levy-Mises rule, the Poisson’s ratio () is replaced by ½. Recall that volume is
conserved in plastic deformation whereas in elastic region, it is not conserved
unless  =½.
 In Hooke’s law, the total strain (𝜀𝑥𝑥) is associated with the stresses. On the other
hand, the incremental strain (𝑑𝜀𝑥𝑥) is employed by the Levy-Mises rule.
 The reciprocal of the Young’s modulus (1/E) in Eqn. (3.64) is substituted by the
term (
2
3
𝑑𝜆) in the Levy-Mises rule. Roughly speaking, 𝑑𝜆 in Eqn. (3.63a) is treated
as a material parameter that changes throughout the straining regime.
In fact, Hooke’s law depicts a linear relationship whereas the Levy-Mises rule describes a
nonlinear one. To understand this concept, consider a linear spring element. Its behavior
can be depicted by
𝐹 = 𝑘𝑥 (3.65)
where F is the force applied to the spring; k is the spring constant; x denotes the
displacement of the spring. Naturally, Eqn. (3.65) is analogous to the one-dimensional
application of the Hooke’s law:
𝜎𝑥 = 𝐸𝜀𝑥 (3.66)
Here, x, E, and x correspond to F, k, and x in Eqn. (3.65) respectively. For the sake of
argument, now let us consider a nonlinear spring element whose force-displacement curve
is as illustrated in Fig. 3.9. It is self-evident that Eqn. (3.65) cannot be employed to depict
this nonlinear element.
In engineering applications, nonlinear functions are often times linearized around a
point of interest using Taylor’s series expansion. Let’s do this for the function F(x):
𝐹(𝑥) = 𝐹(𝑥0) +
1
1!
⋅
𝑑𝐹
𝑑𝑥
|
𝑥=𝑥0
(𝑥 − 𝑥0) +
1
2!
⋅
𝑑2
𝐹
𝑑𝑥2
|
𝑥=𝑥0
(𝑥 − 𝑥0)2
+ ⋯
⏟
Higher Order Terms
(3.67)
Presuming that (𝑥 − 𝑥0) is small (i.e. the region of interest around point x0 is small), we
can neglect the higher order terms in Eqn. (3.67):
𝐹(𝑥) − 𝐹(𝑥0)
⏞
≜𝐹0
⏟
≜∆𝐹
=
𝑑𝑓
𝑑𝑥
|
𝑥=𝑥0
⏟
≜𝑘0
(𝑥 − 𝑥0)
⏟
≜∆𝑥
(3.68)
Evidently, Eqn. (3.68) boils down to
∆𝐹 = 𝑘0∆𝑥 (3.69)

22. 3.22
Eqn. (3.69) now resembles Eqn. (3.65). However, there are some major differences:
 In Eqn. (3.69), k0 is not a constant. Depending on the point of interest (x0), it varies.
 The spring force in Eqn. (3.69) is associated with the incremental displacement
(x) (not the displacement x itself).
Consequently, by using this spring element analogy, one can draw simple (and rather
didactic) conclusions about the Levy-Mises flow rule and the Hooke’s law. They are
summarized in Table 3.2.
Table 3.2: Comparison of models.
Entity Analogous to
Levy-Mises Rule ∆𝐹 = 𝑘0∆𝑥
∆𝐹 𝜎𝑥
′
𝐹(𝑥0) = 𝐹0 𝜎ℎ
𝑘0 𝑑𝜆
∆𝑥 𝑑𝜀𝑥𝑥
Hooke’s Law 𝐹 = 𝑘𝑥
𝐹 𝜎𝑥
𝑘 𝐸
𝑥 𝜀𝑥
x
F
x0
F0
F = F(x)
k0
Δx
ΔF
Figure 3.9: Nonlinear function.
Question: How is d of the flow rule related to the material coefficients in power law (e.g.
K and n)?
Answer: Recall that in uniaxial tension test, the principal stresses take the following
form:
𝜎1 = 𝜎𝑥; 𝜎2 = 𝜎3 = 0 (3.70)
The Levy-Mises flow rule for this special case [see Eqn. (3.63b)] becomes
2
3
𝑑𝜆 =
𝑑𝜀1
𝜎1 − 1
2
(𝜎2 + 𝜎3)
=
𝑑𝜀𝑥
𝜎𝑥
(3.71)
With the utilization of the power law (𝜎𝑥 = 𝐾𝜀𝑥
𝑛
), we get
𝑑𝜆 =
𝑑𝜀𝑥
2
3
𝐾𝜀𝑥
𝑛
= 𝑓(𝜀𝑥) (3.72)
It is evident from Eqn. (3.72) that the non-negative d parameter varies with 𝜀𝑥 (or the
elongation) throughout the plastic deformation regime. Roughly speaking, d in Levy-
Mises flow rule is employed to characterize the strain-hardening behavior of materials.
The discussion above specifically focusses on the uniaxial tension case. For multiaxial-
stress cases, the effective-stress (𝜎
̅) [see Eqn. (3.58)] and (total) effective-strain (𝜀̅) must
be utilized to compute d parameter. Consequently, Eqn. (3.72) can be generalized as
𝑑𝜆 =
𝑑𝜀̅
2
3
𝐾𝜀̅ 𝑛
= 𝑓(𝜀̅) (3.73)

23. 3.23
where
𝜀̅ =
√2[(𝜀1 − 𝜀2)2 + (𝜀2 − 𝜀3)2 + (𝜀3 − 𝜀1)2]
3
(3.74)
It is critical to notice that since 𝜎
̅ = 𝜎𝑥; 𝜀̅ = 𝜀𝑥 in uniaxial tensile test, 𝜎
̅ − 𝜀̅ curve
essentially boils down to the true stress-true strain (𝜎𝑥 − 𝜀𝑥) curve.
References
[1] Groover, M. P., Fundamentals of Modern Manufacturing, 5th Edition, John Wiley, NY,
2013.
[2] Schey, J. A., Introduction to Manufacturing Processes, 2nd Edition, McGraw Hill, NY, 1987.
[3] Tlusty, G., Manufacturing Processes and Equipment, Prentice Hall, NJ, 2000.
[4] Altıntaş, Y., Manufacturing Automation, 2nd Edition, Cambridge University Press,
Cambridge, UK, 2012.
[5] Lange, K., Lehrbuch der Umformtechnik: Grundlagen, Band 1, Springer-Verlag, Berlin, 1972.
[6] Hosford, W. H., Caddell, R. M., Metal Forming: Mechanics and Metallurgy, 4th Edition,
Cambridge University Press, Cambridge, UK, 2011.