Section 8.6
Strain Hardening In the applications discussed in the preceding sections, the material was assumed to be perfectly plastic. The issue of strain hardening materials is addressed in this section.
8.6.1
Strain Hardening
In the one-dimensional (uniaxial test) case, a specimen will deform up to yield and then generally harden, Fig. 8.6.1. 8.6.1. Also shown in the figure is the perfectly-plastic idealisation (horizontal line). In the perfectly plastic case, once the stress reaches the yield point (A), plastic deformation ensues, so long as the stress is maintained at Y . If the stress is reduced, elastic unloading occurs. In the strain-hardening strain-hardenin g case, once yield occurs, the stress needs to be continually increased in order to drive the plastic deformation. If the stress is held constant, for example at B, no further plastic deformation will occur; at the same time, no elastic unloading will occur. Note that this condition cannot occur in the perfectly-plastic case, where there is one of plastic deformation or elastic unloading. stress
Yield point Y
B
strain-hardening
• perfectly-plastic
A
•
elastic unload
0
strain
Figure 8.6.1: uniaxial stress-strain curve (for a typical metal)
These ideas can be extended to the multiaxial case, where one now has a yield surface rather than a yield point. In the perfectly plastic case, the yield surface remains the same size and shape. For plastic deformation, the stress state must be on the yield surface and remain on the yield surface. For elastic unloading, the stress state must move back inside the yield surface. For the strain-hardening material, the yield surface must change in some way so that an increase in stress is necessary to induce further plastic deformation. This can be done in a number of ways. Before looking at how the yield surface might change, consider first the related topic of the loading function.
The Loading Function The yield surface is in general described by a function of the form
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Section 8.6
f (σ ij ) = 0
(8.6.1)
Suppose that the stress state, represented by the vector σ in stress space, is such that one is “on” the yield surface, Fig. 8.6.2. The normal vector to the surface is n. An increment in stress d σ now takes place. The notions of (plastic) loading, neutral loading and (elastic) unloading are then defined through:
⇒ ⇒ ⇒
n ⋅ d σ < 0 n ⋅ d σ = 0 n ⋅ d σ > 0
unloading neutral loading loading
(8.6.2)
As in 8.3.19-20, a normal to the surface is ∂ f / ∂σ , so this scalar product can be expressed as n ⋅ d σ =
∂ f ∂ f ∂ f d σ 1 + d σ 2 + d σ , ∂σ 1 ∂σ 2 ∂σ 3 3
(8.6.3)
or, for a general 6-dimensional stress space,
∂ f d σ ij ∂σ ij
(8.6.4)
As mentioned above, neutral loading does not occur for the perfectly plastic material. In this case, the criteria for loading and unloading can be expressed as n ⋅ d σ < 0 n ⋅ d σ = 0
⇒ ⇒
unloading loading
(8.6.5)
Because of its importance (definition) in describing the loading pattern, the yield surface function f is also called the loading function. neutral loading
d σ d σ loading
n σ
d σ unloading
f Figure 8.6.2: Loading and unloading in stress space
Initially, the loading function is the initial yield surface. Increments of stress inducing loading in the strain hardening case then produce a new loading function, as illustrated in 8.6.3. Further increments in stress must again satisfy n ⋅ d σ > 0 , where f and n now refer to the new loading function.
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Section 8.6
d σ d σ initial yield surface
n new loading surface
σ
f
Figure 8.6.3: A new loading surface, due to stressing to “outside” the initial yield surface
Strain Softening Materials can also strain soften, for example soils. In this case, the stress-strain curve “turns down”, as in Fig. 8.6.4. The loading function for such a material will in general decrease in size with further straining. stress
strain
0
Figure 8.6.4: uniaxial stress-strain stress-strain curve for a strain-softening material
8.6.2
Changes in the Loading Function
Isotropic Hardening The simplest means by which the loading function (yield surface) can change is through isotropic hardening. This is where the loading function remains the same shape but expands with increasing stress, Fig. 8.6.5.
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Section 8.6
elastic unloading
σ 2
•
plastic deformation (hardening) stress at initial yield
•
subsequent yield surface initial yield surface
elastic loading
σ 1
Figure 8.6.5: isotropic hardening
Kinematic Hardening The isotropic model implies that, if the yield strength in tension and compression are initially the same, i.e. the yield surface is symmetric about the stress axes, they remain equal as the yield surface develops with plastic strain. In order to model the Bauschinger effect, and similar responses, where a strain hardening in tension will lead to a softening in a subsequent compression, one can use the kinematic hardening rule. This is where the yield surface remains the same shape and size but merely translates in stress space, Fig. 8.6.6.
elastic loading
plastic deformation (hardening)
elastic unloading
σ 2
•
•
stress at initial yield
σ 1
initial yield surface
subsequent yield surface
Figure 8.6.6: kinematic hardening
Other Hardening Rules More complex models can be used, for example the mixed hardening rule, which combines features of both the isotropic and kinematic hardening models.
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Section 8.6
8.6.3
The Flow Curve
In order to predict and describe the possible changes to the loading function outlined in the previous section, one can introduce the concept of the flow curve. Strain hardening in the uniaxial tension test can be described using a relationship of the form
( )
σ = h ε p
(8.6.6)
A typical plot, the flow curve, of this function for a strain-hardening material is shown in Fig. 8.6.7. The slope of this flow curve is the plastic modulus, Eqn. 8.1.9, H =
d σ
(8.6.7)
d ε p
σ
( )
σ = h ε p Y
perfectly-plastic H ≡
d σ p d ε
ε p
0
Figure 8.6.7: uniaxial stress – plastic strain curve (for a typical metal)
In the multi-axial case, one needs again a flow curve, of the form 8.6.6, but one which relates a complex three-dimensional stress state to a corresponding three dimensional state of plastic strain. This formidable task is usually tackled by defining an effective stress and an effective strain , which describe in a simple way the “amount” of stress and plastic strain, and then by relating these effective parameters using an expression equivalent to 8.6.6.
Effective Stress ˆ , some function of the stresses, which reduces to the Introduce an effective stress σ stress σ 1 in the uniaxial case. It is to be a measure of the “amount” of stress in the general 3D stress state. Since the loading function determines whether additional plastic flow takes place, the effective stress can be defined through f .
The yield function can usually be expressed in the form f (σ ij , k ) = F (σ ij ) − k = 0
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(8.6.8)
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Section 8.6
where k is a material parameter. Consider first the case of isotropic hardening (kinematic hardening will be considered in a later section). As plastic strain accumulates, the shape of the yield surface, as described by F (σ ij ) , remains the same. If one writes ˆn F (σ ij ) = C σ
(8.6.9)
then the effective stress is guaranteed to reduce to σ 1 in the uniaxial case. For example, for the Von Mises material, 1/ 2
⎧1 2 2 2 ⎫ ⎨ (σ 1 − σ 2 ) + (σ 2 − σ 3 ) + (σ 3 − σ 1 ) ⎬ ⎩6 ⎭
[
]
σˆ n = C σ
(8.6.10)
ˆ , σ 2 = σ 3 = 0 , one has n = 1, c = 1 / 3 and With σ 1 = σ ˆ = σ
1
(σ 1 − σ 2 )2 + (σ 2 − σ 3 )2 + (σ 3 − σ 1 )2
2
= 3 J 2 =
3 2
(8.6.11)
sij sij
This is the Von Mises stress 8.3.11, and equals the yield stress in uniaxial tension at first yield, but it must increase in some way with strain hardening in order to continue to drive plastic deformation. Similarly, the effective stress for the Drucker-Prager yield criterion is {▲Problem 1} ˆ= σ
α I 1 + J 2 α + 1 / 3
(8.6.12)
which reduces to 8.6.11 when α = 0 .
Effective Plastic Strain The idea now is to introduce an effective plastic strain so a plot of the effective stress against the effective plastic strain can be used to determine the multi-axial hardening behaviour. The two most commonly used means of doing this are to define an effective plastic strain increment: (i)
which is a similar function of the plastic strains as the effective stress is of the deviatoric stresses
(ii) by equating the plastic work (per unit volume), also known as the plastic dissipation,
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Section 8.6
dW p = σ ij d ε ij p
(8.6.13)
to the plastic work done by the effective stress and effective plastic strain: ˆd ε ˆ p dW p = σ
(8.6.14)
Consider first method (i), which is rather intuitive and non-rigorous. The deviatoric stress and plastic strain tensors are of a similar character. In particular, their traces are zero, albeit for different physical reasons; J 1 = 0 because of independence of hydrostatic pressure, the first invariant of the plastic strain tensor is zero because of material incompressibility in the plastic range: d ε ε ii p = 0. For this reason, one chooses chooses ˆ is of the effective plastic strain (increment) d ε εˆ p to be a similar function of d ε ε ij p as σ the s ij . ˆ = For example, for the Von Mises material one has 8.6.11, σ
3 2
s ij s ij . Thus one
chooses d ε ˆ p = C d ε ij p d ε ij p , where the constant C is to ensures that the expression reduces to d ε ˆ p = d ε 1 p in the uniaxial case. Considering this uniaxial case, p p p d ε 11 = d ε 1 p , d ε 22 = d ε 33 = − 12 d ε 1 p , one finds that
d ε ˆ p =
=
2 3
d ε ij p d ε ij p
2
(d ε
p 1
3
) + (d ε
p 2 2
− d ε
p 2
) + (d ε
p 2 3
− d ε
p 3
)
p 2 1
− d ε
(8.6.15)
Consider now method (ii). Consider also the Prandtl-Reuss flow rule, Eqn. 8.4.1, 8.4.1, d ε i p = s i d λ (other flow rules will be examined more generally in §8.7). In that case, working with principal stresses, the plastic work increment is (see Eqns. 8.2.7-10) dW p = σ i d ε i p
= σ i si d λ =
(8.6.16)
1
[(σ − σ ) 3 1
2
2
]
+ (σ 2 − σ 3 )2 + (σ 3 − σ 1 )2 d λ
Using the effective stress 8.6.11 and 8.6.14 then gives, again with d ε i p = s i d λ , d ε ˆ p =
=
2 3 2 3
(σ 1 − σ 2 )2 + (σ 2 − σ 3 )2 + (σ 3 − σ 1 )2 d λ (8.6.17)
(d ε
p 1
− d ε 2 p ) + (d ε 2 p − d ε 3 p ) + (d ε 3 p − d ε 1 p ) 2
2
2
This is the same expression as derived using method (i), Eqn. 8.6.15, but this is so only for the Von Mises yield condition; it will not be so in general.
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Section 8.6
Note also that, in this derivation, the Von Mises term J 2 conveniently appeared in the Prandtl-Reuss work expression 8.6.16. It will be shown in the next section that this is no coincidence, and that the Prandtl-Reuss flow-rule is indeed naturally associated with the Von-Mises criterion.
Prandtl-Reuss Relations in terms of Effective Parameters With the definitions 8.6.11, 8.6.15 for effective stress and effective plastic strain, one can now write {▲Problem 2} 3 d ε ˆ p
d λ =
(8.6.18)
ˆ 2 σ
and the Prandtl-Reuss (Levy-Mises) plastic strain increments can be expressed as p d ε xx = d ε ˆ p / σ ˆ σ xx −
1 2
yy
+ σ zz )]
p d ε yy
1 2
zz
+ σ xx )]
xx
+ σ yy )]
d ε zz p p xy
d ε
p d ε yz p d ε zx
(σ ( )[ = (d ε ˆ / σ ˆ )[σ − (σ = (d ε ˆ / σ ˆ )[σ − (σ = (d ε ˆ / σ ˆ )σ = (d ε ˆ / σ ˆ )σ = (d ε ˆ / σ ˆ )σ p
yy
p
1 2
zz
3 2
p
3 2
p
3 2
p
(8.6.19)
xy yz zx
or 3 d ε ˆ p d ε = sij . ˆ 2 σ p ij
(8.6.20)
A relation between the effective stress and the effective plastic strain will now make equations 8.6.19 complete.
The Flow Curve The flow curve can now be plotted for any test and any conditions, by plotting the effective stress against the effective plastic strain. The idea (hope) is that such a curve will coincide with the uniaxial flow curve. If so, the strain hardening behaviour for new conditions can be predicted by using the uniaxial flow curve, that is, it is taken that the effective stress and effective plastic strain for any conditions are related through 8.6.6, ˆ = h ε ˆ p σ
(8.6.21)
and the effective plastic modulus is given by
ˆ) = H (σ
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ˆ d σ d ε ˆ p
(8.6.22)
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Section 8.6
The total accumulated effective plastic strain is ˆ d σ
∫
∫ H (σ ˆ )
ε ˆ p = d ε ˆ p =
(8.6.23)
which is a function of effective stress only. The inverse of this expression will be
( )
∫
ˆ = σ ˆ ε ˆ p = σ ˆ d ε σ εˆ p
(8.6.24)
Work Hardening The hardening rule 8.6.21 describes how the yield surface evolves. It is a function of the effective plastic strain, hence the term strain hardening. An alternative procedure to describe the hardening process is to plot stress, not against plastic strain, but against the plastic work. Directly from Fig. 8.6.1, by evaluating the area beneath the stress – plastic strain curve, one can obtain the plot shown in Fig. 8.6.8. Here, the stress is expressed in the form σ = w(W
p
) = w(∫ σ d ε ) p
(8.6.25)
ˆ = w(W p ) = w σ d ε p . Eqn. The flow curve for arbitrary loading conditions is then σ
∫
8.6.25 is called a work hardening rule. σ
( )
σ = w W p Y d σ σ dW
p
∫
W = σ d ε p
0
p
Figure 8.6.8: uniaxial stress – plastic work curve (for a typical metal)
. When the effective stress and effective plastic strain are defined using Eqns. 8.6.13 p ˆd ε ˆ p and the strain hardening and work hardening rules are 8.6.14, then dW = σ equivalent. In that case the plastic modulus is ˆ)= H (σ
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ˆ d σ p
d ε ˆ
=
p
ˆ dW d σ p
dW
p
d ε ˆ
227
= σ ˆ
ˆ d σ p
dW
(8.6.26)
Kelly
Section 8.6
8.6.4
Application: Combined Tension/Torsion of a thin walled tube with Isotropic Hardening
Consider again the thin-walled tube, now brought to the point of yield through tension and then subjected to a twist whilst maintaining the axial stress constant, at the initial tensile yield stress. The Prandtl-Reuss equations in terms terms of effective stress and effective plastic strain, 8.6.19, reduce to d ε xx =
1 E
d σ xx +
d ε ˆ p
ˆ σ
σ xx
1 d ε ˆ p σ xx d ε yy = d ε zz = − d σ xx − ˆ E 2 σ 1 + ν 3 d ε ˆ p d ε xy = d σ xy + σ xy ˆ E 2 σ ν
(8.6.27)
Maintaining σ xx at a value Y 0 and introducing the plastic modulus 8.6.22, d ε xx =
ˆ 1 d σ Y 0 ˆ H σ
ˆ 1 1 d σ Y 0 ˆ 2 H σ ˆ 1 + ν 3 1 d σ d ε xy = d σ xy + σ xy ˆ E 2 H σ
d ε yy = d ε zz = −
(8.6.28)
Using the terminology of Eqn. 8.6.8, the Von Mises condition is f = F (σ ,τ ) − k = 0,
F =
1 3
2 2 Y 0 + 3τ ,
k =
Y
(8.6.29)
3
ˆ = 3 F = Y 02 + 3τ 2 . The expansion of the yield surface and the effective stress is σ
is shown in Fig. 8.6.9 (see Fig. 8.3.2).
τ plastic loading
Y 0 / 3
Y 0
σ
Figure 8.6.9: expansion of the yield locus for a thin-walled tube under isotropic hardening
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Section 8.6
Thus d ε xx =
Y 0
τ d τ
H τ 2 + Y 02 / 3
d ε yy = d ε zz = − d ε xy =
1 + ν E
1 Y 0
τ d τ
(8.6.30)
2 H τ 2 + Y 02 / 3
d τ +
τ 2 d τ
3 1
2 H τ 2 + Y 02 / 3
These equations can now be integrated. If the material is linear hardening, so H is constant, then they can be integrated exactly using
∫ x
x 2
+a
dx =
2
1 2
(
)
ln x + a , 2
2
∫ x
x 2 2
+a
2
⎛ x ⎞ ⎟ ⎝ a ⎠
dx = x − a arctan⎜
(8.6.31)
leading to { ▲Problem 3}
⎛ τ 2 ⎞ ⎜ ε xx = 1 + ln⎜1 + 3 2 ⎟⎟ Y 0 2 H ⎝ Y 0 ⎠ E E E ⎛ τ 2 ⎞ ε yy = ε zz = − ln⎜⎜1 + 3 2 ⎟⎟ Y 0 Y 0 4 H ⎝ Y 0 ⎠ ⎛ τ ⎞ 3 E ⎡ τ ⎛ τ ⎞⎤ E 1 ε xy = (1 + ν )⎜⎜ ⎟⎟ + arctan⎜⎜ 3 ⎟⎟⎥ ⎢ − Y 0 3 ⎝ Y 0 ⎠ 2 H ⎣ Y 0 ⎝ Y 0 ⎠⎦ 1 E
E
(8.6.32)
Results are presented in Fig. 8.6.10 for the case of ν = 0.3, E / H = 10 . The axial strain grows logarithmically and is eventually dominated by the faster-growing shear strain.
E Y 0
8
ε 6
ε xx
4
ε xy
2
0
0. 2
0 .4
0 .6
0.8
1
τ Y 0
Figure 8.6.10: Stress-strain curves for thin-walled tube with isotropic linear li near strain hardening
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Section 8.6
8.6.5
Kinematic and Mixed Hardening
In the above, hardening rules have been discussed and used for the case of isotropic hardening. In kinematic hardening, the yield surface translates in stress-space, in which case Eqn. 8.6.8 take the general form f (σ ij , k , α ij ) = F (σ ij − α ij ) − k = 0
(8.6.33)
The stress α ij is known as the back-stress ; the yield surface is shifted relative to the stress-space axes by α ij , Fig. 8.6.11.
σ 2
α ij
initial yield surface
• •
σ 1
subsequent loading surface
Figure 8.6.11: kinematic hardening; a shift by the back-stress
There are many hardening rules which define how the back stress depends on development of plastic strain. The simplest is the linear kinematic (or Prager’s ) hardening rule, α ij = cε ij p
or d α ij = cd ε ij p
(8.6.34)
where c is a material constant. Thus the yield surface is translated in the same direction as the plastic strain increment. This is illustrated in Fig. 8.6.12, where the principal directions of stress and plastic strain are superimposed.
σ 2 , d ε 2 p
d ε
p
• • d α = cd ε
p
σ 1 , d ε 1 p
Figure 8.6.12: Linear kinematic hardening rule
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Section 8.6
Ziegler’s hardening rule is
( )(
p d α ij = da ε ij σ ij − α ij
)
(8.6.35)
where a is some scalar function of the plastic strain, for example da = η d ε ˆ p , where p ε ˆ is the effective plastic strain and η is a material constant. Here, then, the loading
function translates in the direction of σ ij − α ij , Fig. 8.6.13.
σ 2
•σ d α
•
• −α
σ
σ 1
Figure 8.6.13: Ziegler’s kinematic hardening rule
When there is a combination of isotropic and kinematic hardening, then the hardening rule will be of the form
( )
p f = F (σ ij − α ij ) − k ε ˆ = 0
8.6.6
(8.6.36)
The Consistency Condition
It has been seen that the loading function can in general be expressed in the form f (σ ij , κ ) = 0
(8.6.37)
where κ represents one or more hardening parameters, which are zero when there is no plastic loading. For example, in isotropic hardening, 8.6.37 can be written in the form 8.6.8 through f (σ ij , k ) = F (σ ij ) − k
= F (σ ij ) − (Y + κ ) = 0
(8.6.38)
Alternatively, for kinematic hardening, the hardening parameter is related to the α ij in 8.6.33 (see §8.8). There are two hardening parameters in the mixed mixed hardening rule 8.6.36. The hardening parameters themselves depend on other variables, for example the plastic strain. The increment in f can now be described by
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Section 8.6
df =
∂ f ∂ f d σ ij + d κ ∂σ ij ∂κ
(8.6.39)
The second term here is zero when there is no plastic straining or perfect plasticity. When there is plastic deformation, then, for the stress to remain on the yield surface, i.e. for the yield criterion to remain satisfied, one must satisfy the following consistency condition : df =
∂ f ∂ f d σ ij + d κ = 0 ∂σ ij ∂κ
(8.6.40)
Thus the stress state and also the hardening parameters change to ensure the yield criterion remains satisfied.
8.6.7
Problems
ˆ n , to derive an expression for the 1. Use the general formula 8.6.9, F (σ ij ) = C σ
Drucker-Prager material’s effective stress, Eqn. 8.6.12. 3 d ε ˆ p 2. Derive Eqns. 8.6.20, d λ = ˆ 2 σ 3. Integrate Eqns. 8.6.30 and use the initial (first yield) conditions to get Eqns. 8.6.32. 4. Consider the combined tension-torsion of a thin-walled cylindrical tube. The tube is made of an isotropic hardening Von Mises metal with uniaxial yield stress Y 0 . The strain-hardening is linear with plastic modulus H . The tube tube is loaded, keeping the ratio σ / τ = 3 at all times throughout the elasto-plastic deformation (i) Show that the stresses and strains at first yield are given by 1 1 1 Y 0 1 + v Y 0 Y Y Y σ Y = Y 0 , τ = Y 0 , ε xx = , ε xy = 2 6 2 E 6 E (ii) Use the hardening rule 8.6.18 to express the Prandtl-Reuss equations ˆ and σ only. Eliminate τ using 8.6.18 in terms of effective stress σ (iii)
σ / τ = 3 . Eliminate the effective stress to obtain 1 1 d ε xx = d σ + d σ E H d ε xy =
1 1 + ν
d σ +
3 1
(iv)
d σ 2 H 3 E Solve the differential equations and evaluate any constants of integration
(v)
Hence, show that the strains at the final stress values σ = Y 0 , τ = Y 0 / 3 are given by
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Section 8.6
E Y 0 E Y 0
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ε xx = 1 + ε xy =
E ⎛ 1 ⎞ ⎜1 − ⎟ H ⎝ 2 ⎠
1 + ν
+
3
233
3 E ⎛ 1 ⎞ ⎜1 − ⎟ 2 H ⎝ 2 ⎠
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