S0997-7538(00)00186-8/FLA
The contraction ratios of perfect elastoplasticity under biaxial controls
Chein-Shan Liua,Hong-Ki Hongb
aDepartment of Mechanical and Marine Engineering, National Taiwan Ocean University, Keelung, 202-24, Taiwan bDepartment of Civil Engineering, National Taiwan University, Taipei 106-17, Taiwan
(Received 16 July 1999; revised and accepted 22 February 2000)
Abstract – We studied contraction ratios, one rate form and one total form, of the Prandtl–Reuss model under combined axial and torsional controls. In the transition point of elasticity and plasticity, the rate form contraction ratio may undergo a discontinuous jump, which, depending on the control paths and initial stresses, may be positive, zero, or negative. For the total form contraction ratio, no similar jump phenomenon is observed in the elasticity– plasticity transition point. Depending on initial stresses both ratios may be greater than 1/2. In the simulations of the axial–torsional strain control tests, the hoop and radial strains are not known a priori and hence can not be viewed as inputs. This greatly complicates the constitutive model analyses because the resulting differential equations become highly non-linear. To tackle this problem, we devise a new parametrization of the axial and shear stresses, deriving a first order differential equation to solve for the parameter variable, with which the consistency condition and initial conditions are fulfilled automatically. For mixed controls, the responses can be expressed directly in terms of the parameter without solving the first order differential equation. In particular, when control paths are rectilinear exact solutions can be obtained.2000 Éditions scientifiques et médicales Elsevier SAS perfect elastoplasticity / contraction ratio / plasticity / compressibility / biaxial control
1. Introduction
The combined axial and torsional testing of thin-walled tubes is ideal for the study of constitutive equations of metals; see, for example, Nadai (1950) and Hill (1950). The thin-walled tubular specimen is usually subjected to a combination of axial load P (t) and torque T (t), and with an appropriate feedback arrangement the length
Z(t) and the relative twist angle 2(t) as well as P (t) and T (t) can serve as control variables. Thus the
axial-torsional testing may have (P , T ), (Z, 2), (P , 2), and (Z, T ) as control input pairs, as illustrated in figure 1. Under the assumption of uniform deformation and stress distribution in the main parallel segment of the thin wall of the specimen, the four control pairs can be correspondingly related to (σzz, σzθ), (εzz, εzθ), (σzz, εzθ),
and (εzz, σzθ). They are, respectively, pure stress control, strain control, and mixed controls; see, for example,
Klisinski et al. (1992). The stress and strain paths for the considered tests are such that their rates, or the tangents to the paths, are of the forms:
˙σ = ˙σ˙σ1112 ˙σ012 00 0 0 0 , ˙ε = ˙ε˙ε1112 ˙ε˙ε1222 00 0 0 ˙ε22 , (1)
where the superimposed dot denotes differentiation with respect to time t , d/ dt . Notice that ˙σ21 = ˙σ12,
˙ε21= ˙ε12, and ˙ε33= ˙ε22. In cylindrical coordinates (z, θ, r) as shown in figure 1, ε11= εzz is the axial strain,
ε22= εθ θ is the hoop (or circumferential) strain, ε33= εrris the radial strain, and ε12= εzθ is the shear strain in
the thin wall of the tube, whereas σ11= σzz is the axial stress and σ12= σzθ is the shear stress in the thin wall
Figure 1. The main parallel segment of a thin-walled tubular specimen under the combined axial and torsional controls (P , T ), (Z, 2), (P , 2), or (Z, T ).
In any control case, ˙ε33= ˙ε22is the quantity induced by the axial control on a compressible material and is
not necessarily equal to zero1. We usually use the contraction ratios
νr:= −˙ε22 ˙ε11 , νt:= −ε22 ε11 (2) to measure the compressibility of the material, because
tr˙ε = (1 − 2νr)˙ε11, tr ε= (1 − 2νt)ε11+ ε33(ti)− ε22(ti), (3)
where ti is an initial time. In definitions (2) the first is a rate form and the second is a total form. These two
forms are different in the plastic phase but have the following relation:
νr= νt+ ε11
dνt
dε11
. (4)
Note that all the three quantities νr, νtand ε22are functions of time t and depend upon ε11. There are two ways
to derive the above formula: taking the time derivative of equation (3)2, one obtains
tr˙ε = (1 − 2νt)˙ε11− 2ε11
dνt
dε11
˙ε11,
where the chain rule was used in the derivative ˙νt. Equating the above equation with equation (3)1, one obtains
(1− 2νr)˙ε11= (1 − 2νt)˙ε11− 2ε11
dνt
dε11
˙ε11,
which, through some arrangement, yields equation (4) if˙ε116= 0. Instead of the above derivation, we may take
the differential of equation (2)2with respect to ε11directly. It follows that
dνt dε11 =−ε11dε22/ dε11 ε2 11 +ε22 ε2 11 .
Noting that dε22/ dε11= −νr and ε22/ε11 = −νt from equation (2) and rearranging the above equation, we
obtain equation (4) again.
For isotropic linear elasticity, νr = νt = ν is a material constant known as Poisson’s ratio, and the bulk
modulus is given by
K=2G(1+ ν)
3(1− 2ν),
where G is the shear modulus. The value ν= 1/2 implies 1/K = 0, i.e., elastic incompressibility.
The axial–torsional control pairs (σ11, σ12), (ε11, ε12), (σ11, ε12), and (ε11, σ12) are the total components
rather than the deviatoric components; therefore, it is important to deal directly with the stress and strain components rather than with their deviatoric parts, even though the volumetric parts are elastic in most metals and hence do not play a dominant role in the study of metal plasticity. Therefore, the constitutive relation of the volumetric parts
tr˙ε =tr ˙σ
3K =
˙σ11
3K (5)
must always be included in the study. Because tr˙ε is not yet known, it may greatly complicate the calculation of model responses as will be shown later. Therefore, the values νr and νt were usually taken to be 1/2 in the
plastic deformation range for simplicity; see, for example, Khan and Huang (1995). Note that the simplified
assumption νr = νt = 1/2 should not be absurdly viewed as a result of plastic incompressibility, a popular
assumption for metal plasticity, which means that there is no volumetric plastic deformation in metals.
The rate form contraction ratio was rarely reported in the literature. In this paper we consider the Prandtl– Reuss model subjected to the above controls and attempt to achieve a deeper understanding of the ratios νr and
νt and to assess the validity of the simplified assumption νr = νt = 1/2 prevailing in the plasticity literature.
Although our consideration is limited to the Prandtl–Reuss model, the raised issue is considered typical in the models of plasticity, and the method of analysis and solution presented here may apply to some classes of plasticity models more complicated than the Prandtl–Reuss model.
2. The model
The elastoplastic model for solid materials proposed by Prandtl (1924) and Reuss (1930) is re-formulated (see, for example, Hong and Liu (1997)) as follows:
˙ε = ˙εe+ ˙εp , (6) ˙σ = 2G˙εe+3K− 2G 3 (tr˙ε)I, (7) ˙λσ−1 3(trσ )I = 2τy˙εp, (8) s 1 2 σ−1 3(tr σ )I · σ−1 3(tr σ )I 6 τy, (9) ˙λ > 0, (10) ˙λ s 1 2 σ−1 3(tr σ )I · σ−1 3(tr σ )I = ˙λτy, (11)
where I is the identity tensor, the symbol tr denotes the trace of the tensor, and a dot between two tensors of the same order denotes their Euclidean inner product. The model has only three experimentally determined
material constants, namely the bulk modulus K, the shear modulus G, and the shear yield strength τy, which
are postulated to be
1
K > 0, G > 0, τy> 0. (12)
The boldfaced ε, εe, εp and σ are, respectively, the strain, elastic strain, plastic strain, and stress tensors, all
symmetric, whereas λ is a scalar. All the ε, εe, εp, σ and λ are functions of one and the same independent
variable, which in most cases is taken either as the ordinary time or as the arc length of the control path; however, for convenience, the independent variable, no matter what it is, will simply be called ‘time’ and given the symbol t . As said earlier, a superimposed dot denotes differentiation with respect to time, d/ dt . Without loss of generality, it is also postulated that with the above differential model, there is a time instant designated as
t0, called the zero-value (or annealed) time, before and at which the material is in the zero-value (or annealed)
state in the sense that the relevant values ε(t0), εe(t0), εp(t0), σ (t0) and λ(t0) are all zero. In general the
zero-value time t0, the initial time ti, and the current time t obey the order t06 ti 6 t.
In summary, the material of the thin-walled tube studied in this paper is modelled by equations (6)–(12) and subjected to certain initial conditions at the initial time ti.
Taking traces of both sides of equation (8) and noting equation (12)3, we have
tr˙εp= 0, (13)
that is, the model implies plastic incompressibility. Taking traces of both sides of equation (7) and noting equations (13) and (6) yield the first equality of equation (5), that is, the model implies an elastic relation for
the volumetric part. The model also implies2 E > 0, −1 < ν 61 2, (14) where E:= 9KG 3K+ G, ν:= 3K− 2G 6K+ 2G (15)
are Young’s modulus and Poisson’s ratio, respectively. Furthermore, the model validates
˙ε22= ˙ε33 (16)
under the axial-torsional controls; however, neither the model alone nor the axial–torsional controls by themselves can lead to˙ε22= ˙ε33.
In this paper it will be made clear that, for the model under the axial–torsional controls, plastic incompressibility does not lead to νr = νt = 1/2, and the contraction ratios νr and νt and Poisson’s ratio ν
are different from one another.
3. Axial–torsional problem 3.1. Governing equations
For a two-dimensional axial–torsional problem the following governing equations can be obtained from equations (6)–(8) and (1), ˙σ11 √ 3 + E 3 ˙λ τy σ11 √ 3 = E 3 √ 3˙ε11 , (17) ˙σ12+ G ˙λ τy σ12= G(2˙ε12), (18)
and the yield condition reduces to
σ11 √ 3 2 + σ2 12= τ 2 y. (19)
We may use equations (17) and (18) to solve for σ11 and σ12, respectively. However, the rate ˙λ remains to be
determined, which is the theme of the next subsection. 3.2. Switch of plasticity
In this subsection we shall see that the complementary trios (9)–(11) enable the model to possess an on– off switch of the plastic mechanism, the switching criterion of which will be derived right below. Multiplying
2That the ranges (14) are the implications of the model can be seen in the following way. Equation (14)
1comes from substituting equations (12)1
and (12)2in equation (15)1, whereas equation (14)2comes from substituting equations (12)1, (12)2, and (14)1in the relations K= E/(3 − 6ν) and
equations (17) and (18) by σ11/
√
3 and σ12, respectively, and adding the results lead to
σ11˙σ11 3 + σ12˙σ12+ E ˙λ 3τy σ11 √ 3 2 +G˙λ τy σ122 =E 3σ11˙ε11+ 2Gσ12˙ε12. Since τy is constant, the above equation implies
σ112 + 3σ 2 12= 3τ 2 y ⇒ ˙λ Eσ112 + 9Gσ 2 12 = 3τy(Eσ11˙ε11+ 6Gσ12˙ε12).
Recalling the positivity of τy, E and G we have
σ112 + 3σ122 = 3τy2⇒ Eσ11˙ε11+ 6Gσ12˙ε12> 0⇔ ˙λ > 0, (20) from which σ112 + 3σ 2 12= 3τ 2 y and Eσ11˙ε11+ 6Gσ12˙ε12> 0 ⇒ ˙λ > 0. (21)
On the other hand, if ˙λ > 0, equation (11) ensures σ112 + 3σ122 = 3τy2, which together with equation (20) leads to
˙λ > 0 ⇒σ112 + 3σ122 = 3τy2 and Eσ11˙ε11+ 6Gσ12˙ε12> 0
. (22)
From equations (21) and (22) we thus conclude that the yield condition σ112 + 3σ122 = 3τy2 and the straining condition Eσ11˙ε11+ 6Gσ12˙ε12> 0 are sufficient and necessary for plastic irreversibility ˙λ > 0. Considering
this and the inequality (10), we thus reveal the following criterion for plastic irreversibility:
˙λ = 3τy(Eσ11˙ε11+ 6Gσ12˙ε12) Eσ112 + 9Gσ122 > 0 if σ2 11+ 3σ122 = 3τy2and Eσ11˙ε11+ 6Gσ12˙ε12> 0, 0 if σ2 11+ 3σ122 < 3τy2or Eσ11˙ε11+ 6Gσ12˙ε126 0. (23)
According to the complementary trios (9)–(11), there are just two phases: (i) ˙λ > 0 and σ112 + 3σ122 = 3τy2, and (ii) ˙λ= 0 and σ2
11+ 3σ122 6 3τy2. From the criterion (23) it is clear that (i) corresponds to the plastic phase
(also called the on phase, or the elastoplastic phase) while (ii) to the elastic phase (also called the off phase). During the plastic phase, ˙λ > 0, the mechanism of plasticity is on, the mechanical process is irreversible, and the material exhibits elastoplastic behavior, while in the elastic phase, ˙λ= 0, the mechanism of plasticity is off, the mechanical process is reversible, and the material responds elastically. Thus equation (23) is called the on–off switching criterion for the on–off switch of the mechanism of plasticity.
3.3. Difficulties
Substituting equation (23)1for ˙λ in equations (17) and (18), we obtain
˙σ11+ E(Eσ11˙ε11+ 6Gσ12˙ε12) Eσ2 11+ 9Gσ122 σ11= E˙ε11, (24) ˙σ12+ 3G(Eσ11˙ε11+ 6Gσ12˙ε12) Eσ2 11+ 9Gσ122 σ12= 2G˙ε12. (25)
Figure 2. In control path (1) only one component is varied at one time while the other component is held fixed. Path (2) is rectilinear and path (3) is general.
For the strain control and the mixed controls the above two equations are coupled, highly non-linear, and hard to solve3, and so it seems desirable to re-examine the conventional strategies for dealing with the axial–torsional problem and the difficulties encountered in their approaches.
Recall that in the process of deriving equation (17) one may as well obtain ˙σ11 √ 3+ G ˙λ τy σ11 √ 3= G √ 3 ˙ε11− 1 3tr˙ε . (26)
This equation has the same corresponding homogeneous equation as equation (18) does, but contains in the
non-homogeneous term tr˙ε, a quantity not known a priori, which is indeed one part of the responses, depending
on the inputs ˙ε11 and ˙ε12 and the initial stresses. One may measure the hoop strain rate or the radial strain
rate or both to obtain tr˙ε. Such measurement is not an easy work but can be taken someway. However, in the
model simulation the quantity tr˙ε is not an independent variable requiring extra measurement and, therefore, should be calculated through the constitutive laws. Accordingly, using equation (5) in equation (26), one obtains equation (17) again, and so one still encounters the difficulty of solving the highly non-linear equations (24) and (25). Generally speaking, this difficulty was treated in the literature of plasticity by two approaches aside from the measurement. The first approach was to restrict the paths by moving only one of the two components at one time, while holding the other component fixed, as illustrated by path (1) in figure 2. Then the problem
can be seen to be exactly solvable. The second approach was to simply neglect the term tr˙ε with various
excuses for the neglecting. Under the model of equations (6)–(12) the following excuses are equivalent and any
one of them can lead correctly to tr˙ε = 0: (i) elastic incompressibility (or ν = 1/2 or 1/K = 0 or E = 3G),
(ii) incompressibility. Be cautious that some excuses given in the literature were too restricted to be consistent with the whole problem. However, the second approach still results in non-linear differential equations, and so in the next subsection we shall analyse the second approach in a new perspective, and then in the next section devise a new formulation to unlock the difficulty.
3For the pure stress control equations (24) and (25) are linear and easy to solve, but, in reality, in the axial–torsional testing, the pure stress control is
It cannot be overemphasized that we have to face the difficulty arising from elastic compressibility and to develop a new formulation if we wish to investigate the theoretical variations of the contraction ratios νr and
νt; otherwise, should we assume elastic incompressibility, we would obtain the trivial answers ν= νr = 1/2
and, if tr ε(ti)= 0, νt= 1/2.
3.4. The incompressible model
Under the additional assumption of elastic incompressibility (or ν= 1/2 or 1/K = 0 or E = 3G), the on–off
switching criterion (23) of the model (6)–(12) reduces to ˙λ = 1 τy (σ11˙ε11+ 2σ12˙ε12) > 0 if σ112 + 3σ122 = 3τy2and σ11˙ε11+ 2σ12˙ε12> 0, 0 if σ2 11+ 3σ122 < 3τy2or σ11˙ε11+ 2σ12˙ε126 0, (27)
and equations (26) and (18) reduce to 1 τy ˙σ 11 √ 3+ ˙X0 X0 σ11 √ 3 = √ 3G τy ˙ε11, (28) 1 τy ˙σ12+ ˙X0 X0σ12 =2G τy ˙ε12, (29) where X0(t):= expGλ(t) τy
is the integrating factor. Concentrating on the plastic phase, we re-write equation (27)1as
˙X0 X0 = G τ2 y (σ11˙ε11+ 2σ12˙ε12). (30)
As a consequence, for the plastic phase equations (28), (29), and (30) can be arranged in the matrix form:
d dt X0√σ11 3τy X0σ12 τy X0 = G τy 0 0 √ 3˙ε11 0 0 2˙ε12 √ 3˙ε11 2˙ε12 0 X0√σ11 3τy X0σ12 τy X0 . (31)
This is a system of linear equations for the so-called augmented stress vector (Hong and Liu, 2000)
X:= X0√σ11 3τy X0σ12 τy X0 ,
and the control tensor
A:= G τy 0 0 √ 3˙ε11 0 0 2˙ε12 √ 3˙ε11 2˙ε12 0
is really an element of the Lie algebra of the proper orthochronous Lorentz group SO0(2, 1), namely A∈
so(2, 1). From the formulation above it is easy to obtain the closed-form solutions of the responses under rectilinear strain paths; see, for example, Hong and Liu (1998). However, this scheme does not directly apply to equations (26) and (18) when tr˙ε does not vanish due to elastic compressibility. Instead of the above exact linearization technique a new formulation will be developed in the next section to tackle this difficulty arising from the non-vanishing of tr˙ε.
4. A new formulation for simulation
In this section we concentrate our investigation on the plastic phase for general axial–torsional control paths, as illustrated by path (3) in figure 2.
4.1. Change of variables Let Q(t):= σ√11(t) 3τy σ12(t) τy , ˙q(t) := "√ 3[˙ε11(t)− 1 3tr˙ε(t)] 2˙ε12(t) # , Q⊥(t):= −σ12(t) τy σ11(t) √ 3τy . (32)
Then equations (26) and (18) can be recast to
˙Q + ˙λ
qy
Q= ˙q
qy
, (33)
where qy:= τy/G is the shear yield strain, and the yield condition (19) becomes
kQk = 1, (34)
wherekQk :=√Q· Q. It is easy to see that
kQk = 1, kQ⊥k = 1, Q⊥· Q = 0 (35)
for all t> ti in a plastic-phase time interval. Thus the two constant orthonormal vectors Q(ti) and Q⊥(ti) span
a two-dimensional space, which lie in all stress paths for the axial–torsional problem.
Since Q(t), ˙q(t), and Q⊥(ti) are two-dimensional vectors, two scalars ˙x(t) and ˙u(t) suffice to ensure
˙q(t)
qy
= ˙u(t)Q(t) + ˙x(t)Q⊥(t
i) (36)
to hold for t > ti until Q rotates 90 degrees so as to coincide with Q⊥(ti). Substituting equation (36) into
equation (33) we obtain ˙Q + ˙y yQ= ˙xQ ⊥(t i), (37) where ˙y y := ˙λ qy − ˙u. (38)
Integrating equation (37) yields Q(t)=y(ti) y(t)Q(ti)+ Q ⊥(t i) Z t ti y(ξ ) y(t) ˙x(ξ) dξ. (39)
To summarize, we have made in this subsection a change of variables from σ11(or ε11), σ12(or ε12), ˙λ, tr˙ε in
equations (26) and (18) to σ11, σ12, x, y in equation (39). This change will soon prove to be rewarding.
4.2. Stress in terms of x
Due to Q· Q = 1 the inner product of equation (37) with Q is
˙y
y = ˙xQ
⊥(t
i)· Q. (40)
Substituting equation (39) for Q in the above equation, using equation (35), and recalling ˙x 6= 0, we obtain
y0(x(t))= Z x(t )
x(ti)
y x(ξ )dx(ξ ), (41)
where the prime denotes differentiation with respect to x. Equation (41) is a linear Volterra integrodifferential equation for y(x). Differentiating equation (41) with respect to x yields
y00= y. (42)
If we assign x(ti)= 0 and y(ti)= 1, then we readily find that
y x(t)= cosh x(t), (43)
where y0(x(ti))= 0 due to equation (41) has been used, and the assignment is legitimate because it is ˙x and
˙y/y rather than x and y that were introduced earlier. Define F= F1 F2 := Q(ti)+ (sinh x)Q⊥(ti)= σ11(ti) √ 3τy −σ12(ti) τy sinh x σ12(ti) τy +σ11(ti) √ 3τy sinh x , (44) F⊥:= −F2 F1 = (−sinh x)Q(ti)+ Q⊥(ti)= −σ12(ti) τy −σ√11(ti) 3τy sinh x σ√11(ti) 3τy −σ12(ti) τy sinh x . (45)
Note that F= F(x(t), σ11(ti), σ22(ti)), F⊥= F⊥(x(t), σ11(ti), σ22(ti)), and F· F⊥ = 0. With the aid of y =
cosh x, y(ti)= 1, and definition (44), equation (39) reduces to
Q(t)= F
Figure 3. In the normalized stress plane the yield locus becomes a unit circleS1. The normalized stress Q is decomposed into (1/ cosh x)Q(ti) and
(tanh x)Q⊥(ti), whereas the normalized strain rate˙q is decomposed into ˙λQ and (qy˙x/ cosh x)Q⊥.
Equation (46) indicates that the normalized stress vector Q(t) always lies in the space spanned by the orthonormal vectors Q(ti) and Q⊥(ti), which is two-dimensional and may therefore be called the normalized
stress plane of the axial–torsional problem; see figure 3 for such decomposition. It is worth double checking kQ(t)k = kQ(ti)k = 1 using the expressions in equations (46) and (44); hence, the method presented here
preserves the so-called consistency condition automatically. 4.3. Time history of x
For the strain control4, before calculating Q we need one more equation to solve for x. This can be done as follows. Substituting equation (46) into equation (36) yields
˙q qy = ˙u cosh xF+ ˙xQ ⊥(t i). (47)
Taking the inner products of equation (47) with Q(ti) sinh x and with Q⊥(ti), subtracting the resulting two
equations, and recalling equation (35) and definition (45), we obtain ˙x = ˙q
qy
· F⊥. (48)
Taking the time derivative of equation (46) we have ˙Q = − ˙x sinh x
cosh2xF+ ˙xQ
⊥(t
i). (49)
Using the first component of the above equation in tr˙ε = ˙σ11/(3K) and recalling equations (32) and (44), we have tr˙ε = − √ 3τy 3K ˙x cosh2xF2, (50)
by which equation (48) becomes
˙x = 3KG cosh2x(2˙ε12F1−
√ 3˙ε11F2)
τy(3K cosh2x+ GF22)
. (51)
Equation (51) and the two components of equation (46) can be written out as follows:
˙x = 6KG cosh 2x[σ11√(ti) 3 − σ12(ti) sinh x] 3Kτ2 ycosh 2 x+ G[σ12(ti)+σ11√(ti)3 sinh x]2 ˙ε12 − 3 √
3KG cosh2x[σ12(ti)+σ11√(ti)3 sinh x]
3Kτ2 ycosh 2x+ G[σ 12(ti)+σ11√(t3i)sinh x]2 ˙ε11, (52) σ11(t)= σ11(ti)− √ 3σ12(ti) sinh x cosh x , (53) σ12(t)= σ12(ti)+σ11√(t3i)sinh x cosh x . (54)
These equations apply to general axial–torsional strain control inputs like path (3) in figure 2. Thus we have succeeded in transforming the highly non-linear coupled equations (24) and (25) to a single first-order differential equation (52), the integral of which can give the time history of x, which can then be plugged into formulae (53) and (54) to calculate the response histories of the axial stress σ11 and the shear stress σ12,
respectively. Finally, the hoop strain and the radial strain can be evaluated by
ε22(t)= ε22(ti)+ 1 2 ε11(ti)− ε11(t) +σ11(t)− σ11(ti) 6K , (55) ε33(t)= ε33(ti)+ 1 2 ε11(ti)− ε11(t) +σ11(t)− σ11(ti) 6K , (56)
respectively. These two formulae have been derived from equations (5) and (16). Note that although ˙ε22(t)=
˙ε33(t) for the specimens made of the Prandtl–Reuss material during the axial–torsional testing, it is not unusual
that manufacturing and handling before testing may cause ε22(ti)6= ε33(ti) in the specimens.
4.4. Dissipation
From equations (8), (11), (13), and (23)1, we can calculate the specific power of dissipation, namely the
dissipated energy per unit time per unit volume, by
σ· ˙εp= τy˙λ =
3τy2(Eσ11˙ε11+ 6Gσ12˙ε12)
Eσ2
11+ 9Gσ122
in the plastic phase. In the elastic phase it is simply zero. Taking the inner product of equation (33) with Q and
usingkQk = 1 we obtain
˙λ = Q · ˙q. (58)
Obviously, it is the pair (σ ,˙εp), or (τy, ˙λ), or (τyQ,˙q), not the pair (σ , ˙ε), that constitutes a dissipation–power
conjugacy. Figure 3 provides a geometric interpretation of equations (48) and (58). 4.5. Comments on the new formulation
Some comments on the new formulation are in order. Comment 1. From equations (53) and (54) it follows that
σ112(t)+ 3σ122(t)= σ112(ti)+ 3σ122(ti)= 3τy2.
Thus we can view equations (53) and (54) as a parametric representation of the responses, which match the initial conditions and satisfy the consistency condition automatically.
Comment 2. We may start directly from equation (46) by substituting equations (53) and (54) into equation (23)1 to get ˙λ, and then substituting ˙λ and equation (53) into equation (17) or substituting ˙λ and
equation (54) into equation (18) to get the governing equation of x. However, these procedures lead to the same result as given by equation (52).
Comment 3. The variables u and y introduced in equations (36) and (38) do not appear in the final equations (52), (53), and (54); however, they may help us understand the derivation of the new formulation.
Comment 4. Equations (37) and (40) can be rearranged in the matrix form:
d dx y√σ11 3τy yσ12 τy y = 1 τy 0 0 −σ12(ti) 0 0 σ11(ti)/ √ 3 −σ12(ti) σ11(ti)/ √ 3 0 y√σ11 3τy yσ12 τy y .
Observe that the structure of the above equation is similar to that of equation (31). Both possess the structures of the Lie algebras so(2, 1) of the proper orthochronous Lorentz groups SO0(2, 1), but the Lorentz group of the
above equation has x as its group parameter, which in turn evolves with t in a complicated way as indicated in equation (52), whereas the Lorentz group of equation (31) has t directly as the group parameter.
Comment 5. For the pure stress control we can directly solve equations (24) and (25) without any difficulty. For the mixed controls, by using equation (53) for the control (σ11, ε12), or equation (54) for (ε11, σ12), we can
directly obtain the time history of x and then via x(t) readily calculate all other responses, and so for the mixed controls we are free of solving the first order differential equation (52).
Comment 6. The methodology developed in the new formulation allows us to easily extend it to solve the
axial–torsional problems of more complicated constitutive models, which render ˙q more complicated than that
defined in equation (32)2. Substituting such ˙q into equation (48) will still result in a first order governing
5. Limit strength and discontinuity of contraction ratio
In this section we consider rectilinear strain paths, like path (2) in figure 2, specified by non-zero constant two-dimensional vectors (˙ε11,˙ε12), namely
ε11(t)= ε11(ti)+ (t − ti)˙ε11, ε12(t)= ε12(ti)+ (t − ti)˙ε12 (59)
with
˙ε11= constant1, ˙ε12= constant2 (60)
being not both zeros. The paths are simple but more general than, for example, path (1) in figure 2 and are frequently encountered in experimental tests and numerical calculations. The initial strain vector
(ε11(ti), ε12(ti)) has no effect on responses, while, to the contrary, the initial stress vector (σ11(ti), σ12(ti))
does have significant effects on responses. 5.1. Switch-on time
Given the strain path (59) and an initial stress point (σ11(ti), σ12(ti)) which is confined by the inequality
σ2
11(ti)+ 3σ122(ti)6 3τy2, if at the initial time ti the mechanism of plasticity is not yet switched on, we want to
predict the switch-on time tonat which it will be turned on. The switch-on time can be determined according to
the criterion (23) as follows. First solve for t the following algebraic equation
σ11(ti)+ E ε11(t)− ε11(ti) 2 +3σ12(ti)+ 2G ε12(t)− ε12(ti) 2 = 3τ2 y, (61)
which has been obtained by substituting the elastic equations:
σ11(t)= σ11(ti)+ E ε11(t)− ε11(ti) , σ12(t)= σ12(ti)+ 2G ε12(t)− ε12(ti)
into the yield condition σ112(t) + 3σ122(t) = 3τy2. However, the solution t of equation (61) must satisfy
Eσ11(t)˙ε11+ 6Gσ12(t)˙ε12> 0 in order to be a switch-on time ton. If no solution exists to equation (61) or
the solution t to equation (61) does not satisfy Eσ11(t)˙ε11+ 6Gσ12(t)˙ε12 > 0, then the strain path will not
switch on the plastic mechanism.
Substituting equation (59) into equation (61) we obtain a quadratic equation for t− ti,
A(t− ti)2+ B(t − ti)+ C = 0, where A:= E2˙ε211+ 12G2˙ε122 , B:= 2Eσ11(ti)˙ε11+ 12Gσ12(ti)˙ε12, C:= σ112(ti)+ 3σ122(ti)− 3τy2. Thus, we have ton= ti if C= 0 and B > 0, ti− B A if C= 0 and B < 0, ti+−B + √ B2− 4AC 2A if C < 0. (62)
5.2. Exact solution of x
Under the strain path (59), equation (52) can be written as
(a0+ a1sinh x+ a2sinh2x) dx (1+ sinh2x)(b0− b1sinh x) = dt, (63) where a0:= Gσ122(ti)+ 3Kτy2, a1:= 2 √ 3Gσ11(ti)σ12(ti), a2:= G 3σ 2 11(ti)+ 3Kτy2, b0:= √ 3GK2σ11(ti)˙ε12− 3σ12(ti)˙ε11 , b1:= 3GK σ11(ti)˙ε11+ 2σ12(ti)˙ε12 .
The left-hand side of equation (63) can be decomposed into
(a0+ a1sinh x+ a2sinh2x) dx (1+ sinh2x)(b0− b1sinh x) =(c0+ c1sinh x) dx 1+ sinh2x + c2dx b0− b1sinh x , (64) where c0:= b0(a0− a2)− a1b1 b20+ b21 , c1:= b1(a0− a2)+ a1b0 b02+ b12 , c2:= a2+ b1c1.
Substituting equation (64) into equation (63) then integrating the left-hand side from 0 to x and the right-hand side from ti to t , we obtain
c0tanh x+ c1− c1 cosh x + 2c2 q b20+ b21 arth b1+ b0tanh x/2 q b20+ b21 − arth b1 q b20+ b12 = t − ti, (65)
arth denoting the hyperbolic arctangent function. This equation provides the exact solution of x in an implicit function form.
5.3. Limit strength
In this subsection we study the limit states, especially the limit (or residual) strengths and the limit dissipation powers, under rectilinear strain paths. In the stress plane (σ11, σ12), the limit strength vector (¯σ11, ¯σ12) is indeed
the fixed point of the dynamical system of equations (24) and (25), and, therefore, can be obtained by letting ˙σ11= ˙σ22= 0 in equations (24) and (25) and solving the resultant equations. Accordingly,
¯σ11= 3τy˙ε11 q 3˙ε2 11+ 4˙ε212 , ¯σ12= 2τy˙ε12 q 3˙ε2 11+ 4˙ε212 . (66)
The corresponding limit value ¯x of x can be obtained by letting ˙x = 0 in equation (52) and solving the resultant equation; thus
¯x = arsh 2σ11(ti)˙ε12− 3σ12(ti)˙ε11
√
3[σ11(ti)˙ε11+ 2σ12(ti)˙ε12]
, (67)
arsh denoting the hyperbolic arcsine function. Note that ¯x is an invariant of equation (52), since substituting it into equations (53) and (54) leads to equation (66) again. Substituting equation (66) into (23)1 we obtain the
limit dissipation power
τy¯˙λ = τy q
3˙ε112 + 4˙ε122. (68)
Figure 4 shows four examples of the responses (σ11, σ12,˙ε22, ˙λ) to the same input strain paths but with
different initial stress points; the same input strain paths were rectilinear with constant strain rate vector
(˙ε11,˙ε12)= (0.002, −0.001) s−1and the four different initial stress points (σ11(ti), σ12(ti)) were
Figure 4. Comparison of the responses for four different initial stresses but with the same rectilinear strain path input: (a) time histories of the axial stress σ11, (b) time histories of the shear stress σ12, (c) paths of (σ11, σ12), (d) time histories of ˙λ, (e) time histories of˙ε22.
0, (2ν− 1)τy˙ε12 (1+ ν) q 3˙ε211+ 4˙ε212 , (0, 300), (−300, 0), √ 3τy q (1+ ν)2˙ε2 11+ 3˙ε122 ,√ −(1 + ν)τy˙ε11 3˙ε12 q (1+ ν)2˙ε2 11+ 3˙ε212 ,
respectively. The first initial stress point was chosen such that the yield point was just the fixed point, whereas the last initial stress point was chosen to be located on the yield surface and to render ˙λ(ti)= 0. In the four
examples the material constants were all taken to be G= 50 000 MPa, τy= 500 MPa, ν = 0.3. The resulting
response paths are marked with path 1, path 2, path 3, and path 4, respectively, in figure 4. It is interesting to observe that the four examples have the same limit strength points (¯σ11,¯σ12)= τy(3/2,−1/2). Figure 4c shows
that the limit strength point is asymptotically stable and figure 4d shows that the corresponding ˙λ tends to a fixed value. For the first example, path 1, the heavy lines in figures 4a, 4b and 4c show that, once switched on, the stress components σ11and σ12do not vary again; that is, the point (¯σ11,¯σ12)= τy(3/2,−1/2) is a fixed
point. However, for the other three examples, the stress paths tend to the fixed point very fast and get closer and closer to it as time elapses, but in fact it will take an infinite time to arrive at that point, so that in this sense the limit strength point is a limit point.
5.4. Discontinuity of rate form contraction ratio
In the limiting state ˙x = 0, it follows from equations (3)1 and (50) that tr˙ε = (1 − 2νr)˙ε11 = 0; that is, the
asymptotic value of the rate form contraction ratio is ¯νr =
1
2. (69)
However, νr is not always equal to 1/2 in the plastic phase. This phenomenon is also reflected in the time
histories of the hoop strain rates as shown in figure 4e for the above four examples. The value νrmay experience
a jump from the elastic phase to the plastic phase as shown in figure 5a. Substituting equation (5) into equation (17) we obtain
3Ktr˙ε +E ˙λ 3τy
σ11= E˙ε11. (70)
With the aid of this equation and definition (2)1the rate form contraction ratio is given by
νr = 1 6K˙ε11 (3K− E)˙ε11+ E 3τy σ11(t)˙λ(t) = ν +(1− 2ν)σ11(t)˙λ(t) 6τy˙ε11 . (71)
Due to the jump from ˙λ= 0 in the elastic phase to the value ˙λ > 0 in the plastic phase given by equation (23)1,
tr˙ε may undergo a jump via equation (70), which thus leads to the jump of the rate form contraction ratio νr,
[νr] := νr+− νr−= νr+− ν, (72)
where νr+ and νr−= ν denote the rate form contraction ratios just after and before, respectively, the transition point from elasticity to plasticity. Using equations (71) and (23)1we obtain
Figure 5. Comparison of the contraction ratios for four different initial stresses but with the same rectilinear strain path input: (a) time histories of the rate form contraction ratio νr, (b) time histories of the total form contraction ratio νt.
νr+= 1 6K˙ε11 (3K− E)˙ε11+ E 3τy σ11(ti)˙λ(ti) = 1 6K˙ε11 (3K− E)˙ε11+ Eσ11(ti)(Eσ11(ti)˙ε11+ 6Gσ12(ti)˙ε12) Eσ2 11(ti)+ 9Gσ122(ti) , (73)
where ti= tonfor the initiation of the plastic deformation. Obviously, when ˙λ= 0 (the elastic phase) it leads to
νr+= ν. The value [νr] may be positive, zero, or negative as shown in figure 5a. For most cases it is positive
or negative, and the rate form contraction ratio undergoes a discontinuity jump from ν to νr+. It needs to be pointed out that the value of νr may be greater than 1/2 as shown by paths 2 and 4 in figure 4. For these cases
we have ˙σ11< 0 in some time intervals even when ˙ε11> 0. Wu (1996) recently studied torsional tests on cast
and extruded aluminum, which results supported the current investigation that νr may be greater than 1/2.
5.5. Continuity of total form contraction ratio
Integrating equation (5) from ti to t and substituting equation (3)2for tr ε(t), we obtain
νt= 1 2 − 1 2ε11(t) tr ε(ti)+ ε22(ti)− ε33(ti)+ 1 3K σ11(t)− σ11(ti) . (74)
Obviously, it leads to νt+= ν. Thus, [νt] =: νt+− νt−= νt+− ν = 0, and so the total form contraction ratio
(a)
(b)
Figure 6. The total form contraction ratio νtmay be greater than 1/2. (a) The initial stresses which render νt> 1/2 are marked, (b) time histories of νt.
approaches 1/2 asymptotically as ε11 increases, as shown in figure 5b. Recall that the above is for the axial–
torsional strain controls with rectilinear strain paths. For simple tension or compression it is known that
νt= 1 2− 1 2− ν σ11 Eε11 . (75)
See, for example, Lemaitre and Chaboche (1985) and Khan and Huang (1995). The axial stress σ116
√ 3τy;
thus, the total form contraction ratio νt approaches 1/2 asymptotically as in the axial–torsional case. However,
for the rate form contraction ratio, substituting σ12 = 0 and ˙ε12 = 0 into equation (71) and noting ˙λ =
3τy˙ε11/σ11, we obtain νr = 1/2 for the simple tension or compression, which is different from the axial–
torsional case, for which νr = 1/2 holds only in the limiting state.
In view of equation (75) it should be νt< 1/2 for simple tension or compression because ν < 1/2. But, for
the axial–torsional case, some pre-stress conditions may render νt > 1/2. For demonstration, let us consider
the initial stress (σ11(ti), σ12(ti)) which is confined by the inequality σ112(ti)+ 3σ122(ti)6 3τy2. Upon straining
along the path (59) with ε11(ti)= ε12(ti)= 0 and ε22(ti)= ε33(ti), we have
Figure 7. The differences of the responses σ11and σ12between the exact one and the one with the assumption νr= 1/2: (a) path 2, (b) path 3.
Setting ti = ton in equation (74), substituting the above equation for σ11(ton), and noting tr ε(ton)= (1 −
2ν)ε11(ton), we obtain νt= 1 2− 1 6Kε11(t) σ11(t)− σ11(ti) ∀t > ton. (77)
Depending on[σ11(t)− σ11(ti)]/ε11(t), the value νt may be greater than 1/2. Recall that the limit strength as
given in equation (66) is the limit value of σ11and σ12when t approaches infinity. Thus, when
σ11(ti)> 3τy˙ε11 q 3˙ε211+ 4˙ε212 , ˙ε11> 0, or when σ11(ti)6 3τy˙ε11 q 3˙ε211+ 4˙ε212 , ˙ε11< 0,
it should be νt> 1/2; see figure 6a, where the region with νt > 1/2 is marked. Two cases are plotted in figure 6b
with initial stresses σ11(ti)= 700 MPa and σ12(ti)= 0 MPa, and strain rates ˙ε11= 0.002 s−1and ˙ε12= 0.02 s−1
Figure 8. Comparison of the responses σ11and σ12between the exact one and the one with the assumption νr= 1/2 for a rectangular strain path input:
(a) input strain path, (b) output stress paths, (c) time histories of σ11, (d) time histories of σ12.
5.6. Validity of the assumption νr = 1/2
Under the assumption νr = 1/2, ˙q in equation (32) reduces to
˙q = √ 3˙ε11 2˙ε12 , (78)
and equation (48) changes to
˙x = a + b sinh x, (79) where a:=√G 3τ2 y 2σ11(ti)˙ε12− 3σ12(ti)˙ε11 , b:= −G τ2 y σ11(ti)˙ε11+ 2σ12(ti)˙ε12 .
The solution of equation (79) is
a tanhx 2= b + p a2+ b2tanh arth −b √ a2+ b2 + √ a2+ b2 2 (t− ti) . (80)
According to this formula the responses can be calculated and compared with the results calculated from equation (52); figure 7 shows that the differences of the responses are appreciable in the initial stage of the
plastic deformation; however, for large deformation the differences gradually diminish to zero. Figure 8 also
shows the comparison between the exact one and the one with the above assumption νr = 1/2, but the input
strain control is a rectangular path.
6. Concluding remarks
To analyse thin-walled tubes under combined axial and torsional controls, which lead to highly non-linear differential equations, we proposed a new formulation in which the stresses were parametrized and the consistency condition and initial condition were fulfilled automatically. Upon solving the first order differential equation for the parameter x, we can readily calculate the stresses, and also the hoop and radial strains. For mixed controls, we can directly obtain the responses without solving the first-order differential equation.
With the new formulation we investigated the rate form and total form contraction ratios in the thin-walled tube, showing that the rate form contraction ratio may undergo a discontinuity jump in the transition point of elasticity–plasticity and that, depending on the control paths and on the initial stresses, the jump quantity may be positive, zero, or negative. The value of the rate form contraction ratio may be greater than 1/2, but it eventually tends to 1/2 in the limiting state. For the total form contraction ratio, there exists no similar jump in the elasticity–plasticity transition point; however, its value may still be greater than 1/2 and then approach 1/2 in the limiting state. The conventional simplified assumption of νr = 1/2 may lead to appreciable errors in the
initial stage of plastic deformation; it is acceptable only in the limiting state.
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