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![11
difference between Ai and A0 is and difference between Ai and A0 is
significant. The true stress−strains negligibly small. But this is not true in
give accurate picture and hence it is the plastic region and especially when
more useful to metal working designs. maximum load is reached.
4. It is not easy to obtain values of σ 4. It is easy to obtain these values
from test since the force Fi and cross through test and convenient−less
sectional area (Ai) must be measured costly. These values are widely
simultaneously. True stress ( σ ) is available and documented.
important in metal working
calculation because of its
fundamental significance.
5. It is more consistent with the 5. It is less consistent with physical
phenomenon of metal deformation. phenomenon of metal deformation.
Idealisation of stress – strain curves
The solutions to the plasticity problems are quite complex. To obtain solution to these problems,
stress – strain curves are idealized by [i] neglecting elastic strains and/or [ii] ignoring the effect work
hardening. Idealization and simplification restrict its field of application.
1. Elastic perfectly plastic
It considers elastic strains and neglects effects of work hardening; it yields more difficult constitutive
relations. As a consequence, it also leads to greater mathematical difficulties in practical applications.
It must be used for those processes in which elastic and plastic strains are of the same order. This is the
case in structural engineering or for bending.
2. Rigid, perfectly plastic
In most metal forming operations, the permanent strains are much longer than the elastic. One therefore
in air no great error by assuming the metal to behave as a rigid body prior to yielding. It is for this
reason that one mainly employs perfectly plastic material idealisation.](https://image.slidesharecdn.com/ch-2-stress-strainsandyieldcriterion-110923212911-phpapp02/85/Ch-2-stress-strains-and-yield-criterion-11-320.jpg)
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![17
Fig.
P = load at any instance
A = cross section of specimen.
P = A σ = A . k (∈) n σ = k (∈) n (1)
Ao A
∈ = log ∴ e∈ = 0
A A
⇒ A = A 0 e −∈ (2)
Substituting value in equation (1)
P = A 0 e −∈ K (∈) n
P = K A 0 [e −∈ (∈) n ]
At maximum load point on engg.
stress – strain curve dP = 0
∈ = ∈u
When true strain
dP = K A 0 [e −∈N (∈u ) n −1 + (−1) e −∈ ∈u n ] = 0
n ∈u n −1 =∈u n
∴
∈u = n
Problem 1:
Hollomon equation for a material is given as σ = 1400 (∈) 0.33 . Find the ultimate tensile strength of the
material.
Solution:](https://image.slidesharecdn.com/ch-2-stress-strainsandyieldcriterion-110923212911-phpapp02/85/Ch-2-stress-strains-and-yield-criterion-17-320.jpg)
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![18
Ultimate tensile strength of a material is measured at maximum load point and where necking begins.
Upto the necking point, deformation is uniform throughout its gauge length. It is a engineering stress
(S u ).
True strain for uniform elongation is equal to strain hardening exponent. Therefore ∈u = n.
σu
Ultimate tensile strength = = S u = [σ = (1 + e) × S]
1 + σu
∈u = log e (1 + e u )
∴ ∈u = (1 + e u )
e = 2.71 (logarithmic base)
1 + eu = en
n n
σu K . ∈u n ∈ n
∴ Su = = = K u = K n
e e
en en
=
0.33
0.33
1400
2.71
= 698.1 N / mm 2
∴ UTS = 698. 1 N / mm 2
This shows that ultimate strength of a material can be calculated from the value of K and n.
Problem 2:
A metal obeys Hollomon relationship and has a UTS of 300 MPa. To reach the maximum load requires
an elongation of 35%. Find strain hardening exponent (n) and strength coefficient (K).
Solution:
UTS = S u = 300 MPa = 300 N / mm 2
Engineering elongation strain = e u = 35% = 0.35](https://image.slidesharecdn.com/ch-2-stress-strainsandyieldcriterion-110923212911-phpapp02/85/Ch-2-stress-strains-and-yield-criterion-18-320.jpg)
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![21
A0 d
∈ = ln = 2 ln 0
A d
12.7
= 2 ln = 0.199
11.5
Calculate the total volume of bar
πd2 π (0.0127 m 2
V= ×l= × 0.1 m = 1.26 × 10 − 5 m 3
4 4
For AA–1100, K = 140 MPa and n = 0.25. Note that, as ∈< n, the deformation is homogeneous
∈1 +1
n
Wi = K × ×V
n +1
1.25
= 140 × 10 6 N × 0.199 × 1.26 × 10 − 5 m 3 = 187.5 N (J )
m 6 1.25
(b) Mean stress during deformation
n
∈1
σm = K ×
n +1
0.199 0.25
= 140 MPa × = 74.8 MPa
1.25
(c) Peak (maximum) stress applied, from Hollomon equation
σ1 = K ∈1 = 140 MPa × 0.199 0.25 = 93.5 MPa
n
Yield Criterion
Yield point under simplified condition of uniaxial tension is widely known and documented. But such
simplified conditions [1 – Pure uniaxial tension 2 – Pure shear] are rare in reality. In many situations
complex and multiaxial stresses are present and in this situation it is necessary to know when a material
will yield. Mathematically and empirically, the relationships between the yield point under uniaxial
tensile test and yield strength under complex situations have been found out. These relationships are
known as yield criteria. Thus yield criterion is defined as mathematical and empirically derived](https://image.slidesharecdn.com/ch-2-stress-strainsandyieldcriterion-110923212911-phpapp02/85/Ch-2-stress-strains-and-yield-criterion-21-320.jpg)
![22
relationship between yield strength under uniaxial tensile load and yielding under multiaxial complex
stress situation.
Yield Criterion is a law defining the limit of elastic behaviour under any possible combination of
stresses is called yield criterion. Yield criterion is a mathematical expression which unites experimental
observations with mathematical expressions n a phenomenological manner. Yield criteria is primarily
used to predict if or when yielding
will occur under combined stress states in terms of particular properties of the metal being stressed [
σ 0 , K] .
Any yield criterion is a postulated mathematical expression of the stress that will induce yielding or the
onset of plastic deformation. The most general form is
f (σ x , σ y , σ z, Txy , Tyz , Tzx ) = a constant.
or in terms of principal stresses
f( f (σ1 , σ 2 , σ 3 ) = C
For most ductile metals that are isotropic, the following assumptions are invoked:
1. There is no Bauschinger effect, thus the yield strengths in tension and compression are equivalent.
Bauschinger effect
The lowering of yield stress for a material when deformation in one direction is followed by deformation
in the opposite direction, is called Bauschinger effect.](https://image.slidesharecdn.com/ch-2-stress-strainsandyieldcriterion-110923212911-phpapp02/85/Ch-2-stress-strains-and-yield-criterion-22-320.jpg)
![23
Fig. 10
2. The constancy of volume prevails so that plastic equivalent of poison’s ratio 0.5.
σ1 + σ 2 + σ 3
3. The magnitude of the mean normal stress σ m = does not cause yielding. The
3
assumption that yielding is independent of σm (also called hydrostatic component of the total state
of stress) is reasonable if plastic flow depends upon shear mechanism such as slip or twinning. In
this context, yield criterion is written as
F[(σ1 − σ2), (σ2 − σ3), (σ3 − σ1)] = C
which implies that yielding depends upon the size of the Mohr’s circle and not their position. It is
shown that if a stress state (σ1, σ2, σ3) will cause yielding, an equivalent stress state
(σ1′ , σ 2 ′ , σ 3′ ) will cause yielding, if,
Two widely used yield criterion:
1. Tresca criterion or maximum shear stress criterion.
2. Von Mises criterion or distortion energy criterion.
1. Tresca criterion
Tresca found that plastic flow in a metal begins when tangential stress attains a value.
Assume that a body is subjected to triaxial stresses. σ1 , σ 2 , σ 3 are principal stresses and σ1 > σ 2 > σ 3
(algebraically).
Then maximum shear stress
σ − σ3
Tmax = 1
2
when Tmax exceeds a certain value ‘c’, specific to that material, yielding will occur. To find the value
of ‘c’, the material is subjected to uniaxial tensile test and find out yield point strength (σ 0 ).](https://image.slidesharecdn.com/ch-2-stress-strainsandyieldcriterion-110923212911-phpapp02/85/Ch-2-stress-strains-and-yield-criterion-23-320.jpg)
![26
Von Mises Criterion
According to this criterion, yielding will occur when shear strain energy per unit volume reaches a
critical value. The shear strain energy per unit volume is expressed terms of three principal stresses:
e=
1
σG
[
(σ1 − σ 2 ) 2 + (σ 2 − σ 3 ) 2 + (σ 3 − σ1 ) 2 ]
G = modulus of shear which is a constant.
(σ1 − σ 2 ) 2 + (σ 2 − σ 3 ) 2 + (σ 3 − σ1 ) 2 = Constant.
(i) For uniaxial tensile test, yielding will occur when σ1 = σ 0 ; σ 2 = σ 3 = 0
(σ y ) 2 + (−σ y ) 2 = cons tan t = 2 σ 0 2
Therefore Von Mises criterion can be stated as
(σ1 − σ 2 ) 2 + (σ 2 − σ 3 ) 2 + (σ 3 − σ1 ) 2 = 2 σ 0 2
(σ x − σ y ) 2 + (σ y − σ z ) 2 + (σ z − σ x ) 2 + σ (T 2 + T 2 + T 2 ) = 2σ 0 2
x y y z z x
i) For plane stress: σ 2 = 0
σ + σ3
ii) For plane strain: σ 2 = 1
2
iii) For pure shear stress condition:
σ1 = k σ2 = 0 σ 3 = −k
(σ1 − σ 2 ) 2 + (σ 2 − σ 3 ) 2 + σ 3 − σ1 ) 2 = 2 σ 0 2
(k 0 − 0) 2 + (0 + k 0 ) 2 (− k 0 − k 0 ) 2 = 2 σ 0 2
σ k 02 = 2 σy2
σy
∴k = = 0.557 σ 0
3](https://image.slidesharecdn.com/ch-2-stress-strainsandyieldcriterion-110923212911-phpapp02/85/Ch-2-stress-strains-and-yield-criterion-26-320.jpg)
![27
This is the relationship between shear yield strength and tensile yield strength of the material as per Von
Mises criterion.
k = 0.5 σ 0 2 Tresca criterion
k 0 = 0.577 σ 0 Von Mises criterion.
σy
Von Mises criterion satisfy the experimental data better than Tresca and therefore k = value is
3
normally used.
Advantages of Von Mises criterion
1. It overcomes major deficiency of Tresca criterion. Von Mises criterion implies that yielding is not
dependent on any particular normal stress but instead, depends on all three principal shearing
stresses.
2. Von Mises criterion conforms the experimental data better than Tresca and therefore more
realistic.
3. Since it involves squared terms, the result is independent of sign of individual stresses. This is an
important since it is not necessary to know which is the largest and the smallest principal stress in
order to use this criterion.
Von Mises yield criteria:
[(σ x − σy )
2
+ (σ y − σ z ) 2 + (σ z − σ x ) 2 + 6 (T 2 xy + T 2 yz + T 2 zx ) = 2 σ 0 2 ]
Effective stress
With the yield criterion, it is useful to define an effective stress denoted as σ which is function of the
applies stresses. If the magnitude of σ reaches a critical value, then the applied stress will cause
yielding.
For Von Mises criterion
σ=
1
2
[(σ1 − σ2 )
2
+ (σ 2 − σ 3 ) 2 + (σ 3 − σ1 ) 2 ] 1/ 2](https://image.slidesharecdn.com/ch-2-stress-strainsandyieldcriterion-110923212911-phpapp02/85/Ch-2-stress-strains-and-yield-criterion-27-320.jpg)
![28
For Tresca criterion
σ = σ1 − σ 3
σ ≥ σ 0 ……………………. For both the criteria.
σ ≥ 3 k …………………. Von Mises
σ ≥ 2 k …………………… Tresca
Plane stress condition Plane strain condition
1. In plane stress condition, there is no stress 1. In plane strain condition, the strain
in third direction is absent.
in third direction.
1
∈2 = [ σ 2 − υ(σ 3 + σ1 )]
E
∴ σ 2 = υ ( σ1 + σ 3 )
Near yield point and in plastic zone
1
υ = (For plastic defo)
2
σ + σ3
σ2 = 1
But there is strain in third direction. Two 2
principal stresses
2
σx + σy σx + σy
σ1 =
2
+
2
+ Txy 2
( )
2
σx + σy σx + σy )
σ2 =
2
−
2 ( )
+ Txy 2
1
∈1 = [ σ1 − υ (σ 2 )]
E
1
∈2 = [ σ 2 − υ (σ1 )]
E
1
∈3 = [ 0 − υ (σ1 + σ 2 )]
E
Plane strain condition](https://image.slidesharecdn.com/ch-2-stress-strainsandyieldcriterion-110923212911-phpapp02/85/Ch-2-stress-strains-and-yield-criterion-28-320.jpg)
![29
In majority of metal forming operations the problem can be simplified by assuming a condition of plane
strain is one. One of the principal strains is zero.
1
∈1 = [ σ1 − υ (σ 2 + σ 3 )]
E
1
∈2 = [ σ 2 − υ (σ1 + σ 3 )]
E
1
∈3 = [ σ 3 − υ (σ1 + σ 2 )]
E
let ∈2 = 0 ⇒ σ 2 = υ (σ1 + σ 3 ) for plastic region, Nadai has shown that υ = 0.5
σ + σ3
σ2 = 1
2
σ + σ3
Thus, for Tresca criterion: σ1 , 1 , σ3
2
σ1 − σ 3 = σ 0
Von Mises criterion in plane strain:
2 2
σ + σ3 σ + σ3
σ1 − 1 + 1 − σ 3 + (σ 3 − σ1 ) 2 = 2 σ 0 2
2 2
3
( σ1 . σ 3 ) 2 = 2 σ 0 2
2
2
σ1 − σ 3 ) 2 =
σ 0 = σ '0
3
σ '0 = 1.155 σ 0 = constrained yield strength of the material.](https://image.slidesharecdn.com/ch-2-stress-strainsandyieldcriterion-110923212911-phpapp02/85/Ch-2-stress-strains-and-yield-criterion-29-320.jpg)
![31
σ1 − σ 3 = σ 0
σ0 =
1
[(σ 1 − σ2 )
2
+ (σ 2 − σ 3 ) 2
]
2
where σ1, σ 2 , σ 3 are principal
+ (σ 3 − σ1 ) 2 1 / 2
stresses, and σ1 > σ 2 > σ 3 .
Or
1
σ0 = (σ − σ y ) 2 ) 2 + (σ y − σ z ) 2
x
2
+ (σ z − σ x ) 2 +
σ (τ 2 xy + τ 2 yz + τ 2 zx ) ] 1/ 2
2. Phenomenon of slip is limited to 2. The application of this yield criterion
ductile materials and hence holds good for both ductile and brittle
application of this criterion is limited materials.
to ductile materials. This criterion do
not yield good results for brittle
materials.
3. Tresca criterion ignores the effect of 3. Von Mises criterion take into considera−
intermediate principal stress and this tion the intermediate principal stress and
is a major draw back of this. hence move realistic. The predications
offered by Von Mises criterion conforms
empirical data.
4. 4. The yield stress predicted by Von Mises
criterion is 15. 5% greater than the yield
stress predicted by Tresca criterion.
5. Locus shown in Figure. 5. . Locus shown in Figure.
It is Hexagonal. It is Elliptical.](https://image.slidesharecdn.com/ch-2-stress-strainsandyieldcriterion-110923212911-phpapp02/85/Ch-2-stress-strains-and-yield-criterion-31-320.jpg)
![34
Fig. 12
Comments
1. Yielding will occur if the point representing the given stress is plotted and is on the boundary or
outside the boundary.
2. The yield locus of maximum shear stress criterion [Tresca criterion] fall inside the maximum
distortion energy criterion [Von Mises] yield locus.
3. Two yield criteria predict the same yield stress for conditions of uniaxial stress and balanced
biaxial stress (σ1 = σ 3 ). The greatest divergence between the two criteria occurs for pure shear
(σ1 = −σ 3 ).
4. The yield stress predicted by the Von Mises criterion is 15.5% greater than the yield stress,
predicted by Tresca criterion.
Derive a mathematical expression for Von Mises yield criterion applicable to plane strain stress
condition:
Solution:
Von Mises yield criterion is stated as
(σ1 − σ 2 ) 2 + (σ 2 − σ 3 ) 2 + (σ 3 − σ1 ) 2 = 2 σ 0 2
where σ1 , σ 2 , σ 3 are three principal stresses and σ 0 is the yield strength of material. In plane strain
stress condition, the intermediate principal stress is arithmetic mean of other two. Assuming
σ1 > σ 2 > σ 3 , we can write
σ + σ3
σ2 = 1
2
substituting the value of σ 2 in the above expression
2 2
σ + σ3 σ + σ3
σ1 − 1 + 1 − σ 3 + ( σ 3 − σ1 ) 2 = 2 σ 0 2
2 2 ](https://image.slidesharecdn.com/ch-2-stress-strainsandyieldcriterion-110923212911-phpapp02/85/Ch-2-stress-strains-and-yield-criterion-34-320.jpg)
![37
∴ I2 = 41
I 3 = Tij
10 0 3 0 3 10
= 20 −3 +0
0 −5 0 −5 10 0
= − 1000 + 45
∴ I3 = − 955
f (σ) = σ 3 − I1σ 2 + I 2 σ − I 3 = 0
σ 3 − 25 σ 2 + 41 σ + 955 = 0
Applying standard method to get cubic roots,
f (σ) = σ 3 − 25 σ 2 + 41 σ + 955 = 0
f(y) = y3 + py2 + qy + r = 0
∴ p = − 25
q = 41
r = 955
a=
1
3
( 1
)
3q − p 2 = ( 3 × 41 − 625)
3
a = − 167.3
b=
1
[
27
2p 3 − 9 pq + 27r ]
=
1
27
[2(−25) 3
− 9(−25) ( 41) + 27(955) ]
∴ b = 139.25](https://image.slidesharecdn.com/ch-2-stress-strainsandyieldcriterion-110923212911-phpapp02/85/Ch-2-stress-strains-and-yield-criterion-37-320.jpg)
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Ch 2-stress-strains and yield criterion
Here are the steps to solve this problem: 1) Calculate true stress (σ) and true strain (ε) at each load using the given diameter values: σ = Load/Current cross-sectional area ε = ln(Initial diameter/Current diameter) 2) Plot true stress vs true strain curve 3) Determine values from the curve: 1) True stress at maximum load = Highest stress value on curve 2) True fracture stress = Stress at fracture point 3) True fracture strain = Strain at fracture point 4) True uniform strain = Strain value where necking begins 5) True necking strain = Strain at start of necking region 6) Ultimate
