Chapter 3 The Constitutive Relations The major difference between the theories of plasticity and the elasticity is the difference in their constitutive relations, so the constitutive relation is the primary subject of the study of the plasticity.

3.1 Generalized Hooke’s Law In three-dimensional problems, the state of the stress and strain of a point should be denoted by nine components of them. In linear elastic stage, the relation between

stress and strain is linear. Because jiij σσ = and jiij εε = （ ji ≠ ）, there will be six

independent components of stresses and strains respectively.

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

+++++====

+++++=+++++=

zxyzxyzyxzx

yz

xy

z

zxyzxyzyxy

zxyzxyzyxx

dddddd

dddddddddddd

γγγεεετττσ

γγγεεεσγγγεεεσ

666564636261

262524232221

161514131211

（3-1）

where （ ）are the elastic constants and independent of the

coordinates

nmd , 6,,2,1, =nm

x , and . y z

Using the notation of tensor

klijklij D εσ = （ 3,2,1,,, =lkji ） （3-2）

where is the symmetric and positive definite fourth-order tensor which is of the

correspondence with the elastic constants as

ijklD

， ，111111 Dd = 112212 Dd = 113313 Dd = ， 111214 Dd = ， 112315 Dd = ， 113116 Dd =

， ，……，221121 Dd = 222222 Dd = 223126 Dd = ，……

The symmetric case of the tensor is

klijijlkjiklijkl DDDD ===

There are only two elastic constants for the isotropic materials：

)( jkiljlikklijijklD δδδδμδλδ ++= （3-3）

where λ and μ are the Lamé constants and can be denoted by the elastic modulus

and the Poisson’s ratio E ν :

)21)(1( νννλ

−+=

E，

)1(2 νμ

+=

E（Shear elastic modulus ） （3-4） G

The commonly used form of the generalized Hooke’s law is

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

=+−=

=+−=

=+−=

zxzxyxzz

yzyzxzyy

xyxyzyxx

E

E

E

τμ

γσσνσε

τμ

γσσνσε

τμ

γσσνσε

1)],([1

1)],([1

1)],([1

（3-5）

or the tensor denotation

ijkkijijmijeij EE

δσνσνδσννσ

με −

+=⎟

⎠⎞

⎜⎝⎛

+−=

113

21 （3-6）

The law also can be written as：

⎪⎩

⎪⎨

⎧

=+==+==+=

zxzxzz

yzyzyy

xyxyxx

eee

μγτμελσμγτμελσμγτμελσ

,2,2,2

（3-7）

where

mzyxe εεεεεεε 3321 =++=++= （3-8）

or the tensor denotation

ijijij e μελδσ 2+= （3-9）

Another form using the deviatoric tensors of the stresses and strains is

ijij es μ2= （ 3,2,1, =ji ） （3-10）

and because , five of the six equations are independent. In addition 0=kks

kkkk Kεσ 3= （3-11）

where

μλν 3

2)21(3

+=−

=EK （3-12）

is the elastic volume modulus. Note that kkm σσ31

= and kke ε= , then

eKm =σ （3-13）

where the mean stress mσ is also called the volume stress and is the volume

strain.

e

3.2 Some Assumptions on Material Properties

3.2.1 The stable material hypothesis For the hardening materials, if the strain increment 0>εd when 0>σd , as

shown in Fig. 3-1(a), and thus the work done of σd on εd , 0>εσ dd , the

material is then stable. If 0>εd when 0<σd （Fig. 3-1（b））, or 0<εd when 0>σd （Fig. 3-1

（c））, both yield 0<εσ dd , the material is then unstable.

In general, for the stable materials

0>ijijdd εσ （3-14）

3.2.2 The Drucker’s postulate Considering a stress cycle with loading and unloading shown in Fig. 3-2:

ACBA →→→ or 0 * *ijij ij ij ijd 0σ σ σ σ σ→ → + →

where 0ijσ is certain elastic state, *

ijσ is a

yielding state, ijdσ is the stress increment

applied on the basis of *ijσ . Let ijσ be the

instant stress in the stress cycle and 0ij ijσ σ− is

called the appending stress. Then we have the Drucker’s postulate:

During a full cycle of loading and then unloading, the net work done by the appending stress can never be negative. The postulate can be expressed as

O 0>εd

0>σd

ε

σ

O0>εd

0<σd

ε

σ σ

O0<εd

0>σd

ε

(a) (b) (c)

Fig. 3-1

Loading surface

Initial yield surface

ijdσC

B *ij

σ0ijσA

Fig. 3-2

0( )ij ij ijdσ σ ε 0− ≥∫ （3-15）

The circle sign denotes the integration is along the path of cyclic stress.

Fig 3-3

CB

A

σ

σ

pdε

σσ d+

*σ

ε

Because the elastic deformation is recoverable, the work done by the appending stress on the elastic

deformation is zero, namely eijdε

0( ) eij ij ijdσ σ ε 0− =∫ （3-16）

thus 0( ) p

ij ij ijdσ σ ε 0− ≥∫ （3-17）

Because the plastic deformation occurred in the loading phase B C→ of the appending stress cycle, the Eq. (3-17) can also be written as

0( ) pij ij ijdσ σ ε 0− ≥ （3-18）

If the initial stress 0ijσ corresponds to point on the loading surface, then we

have for the loading cycle and：

B

BCB →→

0≥pijijdd εσ （3-19）

Knowing from Eq. (3-18) that 0 cos 0p

ij ij ijdσ σ ε θ− ≥ （3-20）

which infers that 22πθπ

≤≤− , i.e., the angle between

vectors 0ij ijσ σ− and d is acute. p

ij yield surface

B

A

θ

pijdε

ijσ

0ijσ

Fig. 3-4

ε

Two important conclusions can be deduced from the Drucker’s postulate: Conclusion 1：The yield surface should be convex.

A0ijσ

Bijσ

pijdε

Fig. 3-5

If the yield surface is not convex toward outside, then

however oblique the vector is, one can always find a

point which leads to a obtuse angle between

pijdε

A 0ij ijσ σ− and

, which is conflict with Eq. (3-20). pijdε

Conclusion 2：The direction of is identical to the

outer normal of the yield surface.

pijdε

For the convex yield surface, the condition 0( ) pij ij ijdσ σ ε 0− ≥ can be satisfied

only for the case that is vertical to the yield surface,

otherwise it always be possible to find an obtuse angle between both vectors.

pijdε

3.2.3 The single curve hypothesis (1) Simple loading The loading case that the components of the stress tensor increasing proportionally with the increase of parameter t

0ijij tσσ = （3-21）

is called the simple loading. is a monotonically increasing parameter. Because the influence of hydrostatic pressure on the plastic deformation process is negligible, the simple loading condition can also be expressed as

t

0ijij sts ⋅= （3-22）

in this way the principal stress axes do not change. (2) The theorem of simple loading Iliushin proved that the following conditions will result in the simple loading： ① External loadings increase with the same parameter, namely the proportional

loading； ② The material is incompressible； ③ The relation between equivalent stress (stress intensity) and equivalent strain

(strain intensity) is described as power function: nAεσ = （3-23）

where and are the material constants； A n④ small deformation.

(3) The “single curve” hypothesis Many test results have proved that in simple loading cases, the relation between σ and ε （or T and Γ ）is approximately unique and independent of the stress state. So we have the single curve hypothesis： In the case of simple loading, the σ ～ε (or T～ Γ ) curve is unique, no matter how the stress state is, which means

)(εσ Φ= （3-24）

the above relation can be determined by tension test.

yield surface

α

pε ijd

ijσ 0ijσ

Fig. 3-6

n

In practice the single curve hypothesis is used occasionally in the cases deviating from the simple loading in which the principal stress axes is rotating and the rule of the stress state similarity is slightly violated. The reason for doing so is that the single curve hypothesis has been verified experimentally for a few of the more general loading cases.

3.3 The Increment Theory of Plasticity（The theory of Plastic Flow） The plastic strain is dependent on the loading history, therefore the equations describing the plastic deformation (the constitutive relation) is by its nature the increment relations. The stresses and strains can not be associated by their final states. The theory describing the relations between the increments of stresses and strains is called the increment theory. Materials usually have a larger range of the plastic deformation after yielding, behaving like “flow”. The “plastic flow” is analogy to the flow equations in hydrodynamics. So in the increment theory the relations between the strain increments (strain rate) and the stresses is also called the “flow theory”. The flow theory establishes the relations between the increment strain and increment stress as well as certain parameters denoting the plastic state.

3.3.1 The plastic potential

Assuming there exists the plastic potential function )( ijσΦ in the plastic

deformation field and the equations of the plastic flow can be expressed as

ij

pij dd

σλε∂Φ∂

= — Plastic potential theory （3-25）

where 0≥λd is a scalar factor to be determined; the plastic potential is the scalar function of the stress components.

Φ

when 0>∂Φ∂

ijijdσ

σ， 0>λd ； when 0<

∂Φ∂

ijijdσ

σ， 0=λd

The vector consisting of ijσ∂Φ∂ is the gradient vector of the potential function：

1 2

grad i j kσ σ σ 3

∂Φ ∂Φ ∂ΦΦ = + +

∂ ∂ ∂

so the vector being composed of the components of the plastic strain increments is parallel to the gradient vector of the potential function. The geometric curved surfaces depicted by are called the “potential iso-surfaces” and the gradient vectors take the direction of the outer normals of the iso-surfaces.

consts=Φ

Some studies reveal that, if the yield function 0=f is continue and has

derivativeness, it must be the same function of stresses as the plastic potential . In fact, considering the Drucker’s postulate

Φ

0( ) pij ij ijdσ σ ε 0− ⋅ ≥

and noting that is normal to the yield surface, then pijdε

ij

pij

fddσ

λε∂∂

= （3-26）

is the plastic flow principle associated with the yield condition.

3.3.2 The increment theory for the perfect plastic materials For the perfect plastic materials, the initial yield surface is their limit surface as well. The flow principle for the perfect plastic materials with limit function

0)( =ijf σ is

0 0 0

0, 0 0 0

pij

ij

ijij

ijij

fd d

fwhen f and d d

fwhen f or f and d d

ε λσ

σ λσ

σ λσ

⎫∂= ⎪

∂ ⎪⎪∂ ⎪= = ≥ ⎬∂ ⎪⎪∂ ⎪< = <

∂=

⎪⎭

，

，

（3-27）

3.3.2.1 The flow principle associated with von Mises yield criterion (1) Levy-Mises relation Assuming that the material is obedient to the von Mises yield criterion, namely

031 2

2 =−= sJf σ

Since )3(21

21 2

2 mijijijij ssJ σσσ −== , and ijijij

sJf=

∂∂

=∂∂

σσ2 , then we have

ijpij sdd ⋅= λε （3-28）

is called the Levy-Mises equation. If elastic deformation can be ignored (Levy and

Mises considered as the total strain initially) pijdε

ijij sdd ⋅= λε （3-29）

Since the volume change is elastic, , thus 0321 =++= ppppii dddd εεεε

ijpij sdde ⋅= λ （3-30）

There are two ways to denote λd :

1）Using the increment of equivalent plastic strain

The Levy-Mises equation is multiplied by itself at both sides, i.e.

ijijpij

pij ssddd ⋅=⋅ 2)( λεε

Noticing that 222 3

2322 sijij Jss σσ === , 2)(

23 pp

ijpij ddd εεε =⋅

and

21222222 )](23)()()[(

32 p

zxpyz

pxy

px

pz

pz

py

py

px

p dddddddddd γγγεεεεεεε +++−+−+−=

（3-31） is the increment of the equivalent plastic strain, we get

22232)()(

23

sp dd σλε ⋅=

thus

s

pp dddσε

σελ

23

23

== （3-32）

2）Using the increment of plastic work

The increment of the specific plastic work is

22 JdssddsddW ijijpijij

pijijp ⋅=⋅=== λλεεσ

Upon yielding, 222 3

1ssJ τσ == , so

22 s

pdWd

τλ = （3-33）

(2) Prandtl-Reuss relation

For linear elastic deformation, ijij es μ2= and kkkk Kεσ 3= . Adding the elastic

portion into Levy-Mises equation

ijpij sdde ⋅= λ

we obtain

ijijij sddsde ⋅+= λμ21 — Prandtl-Reuss equation （3-34）

Substituting 22 s

pdWd

τλ = into the above equation:

ijs

pijij sdW

dsde ⋅+= 2221

τμ （3-35）

3.3.2.2 The plastic flow law associated with Tresca yield criterion The above plastic flow principle and the associated flow law require that the yield surface is smooth, that is to say the yield surface must have the continuum rotating tangent plane. But for the yield surface with edges and corners, such as the Tresca yield surface, the normals are uncertain along the edges. Prager and Koiter assume that the plastic flow along the edges shall be the linear combination of those of the left side and the right side:

ijij

pij

fdfddσ

λσ

λε∂∂

+∂∂

= 22

11 （3-36）

here and are the

equations of the yield surfaces at both sides of

edge and

constf =1 constf =2

021 ≥λλ dd ， . Therefore the direction of

the plastic flow is between the normals of the two neighboring yield surfaces.

elastic region

constf =2

constf =1

Fig. 3-7 If the plastic flow is considered in the space of the principal stresses, then

ii

pi

fdfddσ

λσ

λε∂∂

+∂∂

= 22

11

For Tresca yield criterion 3σ ′

2

⎪⎩

⎪⎨

⎧

±=−±=−±=−

s

s

s

σσσσσσσσσ

13

32

21

A

Taking point B in Fig. 3-8 as example. If the stress point is located on the intersection edge of the surfaces AB and BC：

AB： 0211 =−−= sf σσσ

BC： 0312 =−−= sf σσσ

then for ： 01 =f 11

1 =∂∂σf

， 12

1 −=∂∂σf

，

03

1 =∂∂σf

，we get

1σ ′ σ ′f

01 =

02 =

f

C

B

Fig. 3-8

13ε= 21

011λεε dddd ppp

==−

for ： 02 =f 11

2 =∂∂σf

， 02

2 =∂∂σf

， 13

2 −=∂∂σf , we get

2321

101λd=

−==

For the points located on edge B:

， εd p =

or

εεε ddd ppp

211 λλε ddd p += 1λd− ，2 23 λε dd p −=

)1(::1:: 321 λλεεε −−−=ppp ddd ， 10 1 <=<21 + λλ

λλ d dd

3.3.3 The generalized increment theory of the hardening medium According to Drucker’s postulate, the plastic strain is denoted by

⎪⎪≤

∂=< 0,0 σdfandforfhen ij

⎪

⎭

⎪⎪⎪

⎬

⎫

=∂

≥>∂∂

=

∂∂

=

00

000

λσ

λσσ

σλε

dw

ddfandfwhen

fdd

ij

ijij

ij

pij

，

， （3-37）

We then know that λd is in direct proportion to the term ijij

dσσ∂f∂ , i.e.

⎪⎭

⎪⎬

≥∂

=

0λ

σσ

λ

d

dhd klkl （3-38）

where

⎫∂f

),( βξσ ijhh = is a positive scalar function called the hardening modulus.

Therefore the increment theory of the hardening materials can generally be written as：

⎪⎪⎭

≤∂

= 0,0 ijij

ij dwhend σσ

ε

⎪⎬∂p

ijkl

f （3-3⎪⎫

∂∂

∂∂

= klpij

fdfhdσ

σσ

ε9）

rmined only with the hardening law used.

3.3.3.1 The increment theory for isotropic h

The hardening modulus h can be dete

ardening materials

Fo isotropic hardening medium r

（3-40）

where

0)()(* =−= ξψσ ijff

ξ is the history recording parameter (internal variable).

In practical, ξ is often taken as the cumulative equivalent plastic strain：

∫= ppp dddd εεεξ == （3-41） pdεξ ， ijij32

the material is then the strain hardened. ξ c ed as the cumulative an also be consider

plastic work：

∫∫ == pp ddW εσξ ， pp dWd == εσξ （3-42） ijij ijijd

so the material is work hardened. To determine the hardening modulus , make the sel ultiplication of the equation (3-39)：

h f m

ijijkl

kl

pij

pij

ffdf σ ∂⎟⎞∂

2

hddσσσ

εε∂∂

⋅∂⎟

⎠⎜⎜⎝

⎛∂

=⋅ 2

and noticing that 2)(23 pp

ijpij ddd εεε =⋅ , we then get

ijijkl

kl

p

ffdfh

σ ∂⋅

∂⎟⎞

⎜⎛ ∂

=

If the amount of total plastic deform

d

σσσ

ε

∂∂⎠⎝ ∂

23

（3-43）

ation is chosen as the parameter, ∫= pdεξ

（Note that ppd εε ≠∫ , both sides are equal only for si

the corresponding consistency condition：

mple loading）, then we have

0*

=′−∂∂

= pijij

ddfdf εψσσ

（3-44）

where ξψψdd

=′ , therefore

pijij

ddf εψσσ

′=∂∂ *

and

ijijijijp

p

ijijkl

kl

p

ijijkl

kl

p

ffffd

d

ffdf

d

ffdf

dh

σσψ

σσεψ

ε

σσσ

σ

ε

σσσ

σ

ε

∂∂

⋅∂∂′

=

∂∂

⋅∂∂′

=

∂∂

⋅∂∂

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

=

∂∂

⋅∂∂

⎟⎠⎞

⎜⎝⎛∂∂

=

****

***

1232

3

23

23

（3-45）

Two forms of von Mises yield conditions have been used：

（1） )(21

2 ξψ== ijij ssJ

Here, , 2* Jf = ij

ijijsJf

=∂∂

=∂∂

σσ2

*, then we get

ψψψψσσ

ψ′

=′

=⋅′

=

∂∂

⋅∂∂′

=1

23

21

231

231

23

2** Jssff

hijij

ijij

（3-46）

（2） 23 (J )σ ψ ξ= =

Here, *f σ= , because 22 3J=σ , ij

ijijsJ 332 2 =

∂∂

=∂∂

σσσσ , thus

σσσ

σij

ijij

sf23*

=∂∂

=∂∂

and

ψψ

σσψ

σσψ

′=

′=

⋅⋅′=

∂∂

⋅∂∂′

=1

32

491

23

23

23

1231

23

2

2**

JJssff

hijij

ijij

（3-47）

In the case of uniaxial tension, σσ = , , and the hardening function

, thus the value of

pεξ =

)( pεψσ = pdd εσψ =′ can be determined by means of the

experimental curve obtained in the uniaxial tension test.

3.3.3.2 The Increment Theory for the Kinematic Hardening

Materials The equation of the loading surface for the kinematic hardening materials can be

written as

*0ˆ( )ij ijf f σ σ ψ= − − 0= （3-48）

where， ˆijσ is the translational tensor of the initial yield surface, or the back stresses.

Except for the flow principle of Eq. (3-39) , we also need models to define the back stresses.

(1) Prager’s kinematic hardening law According to the consistency condition

ˆ 0ˆij ij

ij ij

f fdf d dσ σσ σ∂ ∂

= +∂ ∂

= (a)

Knowing from Eq. (3-48) that

ˆij ij

f fσ σ∂ ∂

= −∂ ∂

(b)

Assuming that the direction of vector ˆijdσ is parallel to the gradient ij

fσ∂∂

,

which means

ˆijij

fdσ λσ∂

=∂

(c)

Substitution of Eqs. (b) and (c) into Eq. (a) yields

ijij ij ij

f fdσ λ fσ σ σ∂ ∂

= ⋅∂ ∂

∂∂

(d)

we obtain 2

ijij ij

f fdλ σσ σ∂ ∂

=∂ ∂

(e)

Substituting Eq. (e) into Eq. (c) , we get

2ˆij

ijij

ij

ij

f dfd

f

σσ

σσ

σ

∂∂ ∂

=∂∂

∂

（3-49）

this is the relation between the back stress vector ˆijdσ and stress increment ijdσ .

In Eq. (3-49), ij ij

f fσ σ∂ ∂

=∂ ∂

n denotes the unit vector along the outer normal of

the loading surface. The equation can be denoted using the vector signs as

ˆ ( )d d=σ n σ n （3-50）

the equation means that the vector of back-stress increment equals the component of the vector of stress increment on the outer normal of the loading surface.

The constitutive equation of the hardening materials is

pij kl

ij kl

f fd h dε σσ σ∂ ∂

=∂ ∂

（3-51）

comparison of Eq. (3-51) and (3-49) yields

21ˆ p

ij ij

ij

dfh

dσ ε

σ

=∂∂

（3-52）

and this the relation between the back-stress increments and the plastic strain increments.

The hardening modulus can be determined with the adoption of von Mises yield criterion. Firstly, the criterion in the uniaxial tension case is denoted as

h

ˆ( ) sf 0σ σ σ= − − =

thus, 1fσ∂

=∂

. From Eq. (3-51) we know

pd hdε σ= , 1p

p

dhd Eεσ

= =

If the Mises criterion is denoted as 2 21 1ˆ( )

3 3 sf σ σ σ 0= − − =

then we have: 2 ˆ( )3

f σ σσ∂

= −∂

, 2 24 4ˆ( )9 9 s

f f σ σ σσ σ∂ ∂

⋅ = − =∂ ∂

from Eq. (3-51) we obtain

249

psd h dε σ σ= ⋅ , 2

94 s p

hEσ

=

For the simplest case, let ( )21 ijh f cσ∂ ∂ = in Eq. (3-52), we then have the

Prager’s linear kinematic hardening model:

ˆ pij ijd c dσ ε= ⋅ （3-53）

therefore *

0( )pij ijf cσ ε ψ− = （3-54）

where is a positive constant. cThe linear kinematic hardening model is generally used for linear hardening

materials, as shown in Fig. 3-9. To be understandable, Fig. 3-10 illustrates the model in one dimensional case.

sε

sσ

σ

ε

E

E’

for( ) for

s

s s s

EE

σ ε ε

ε

σ σ ε ε ε ε= ≤

′= + − >

Fig. 3-9

sσ

σ

ε

sσ−

σ̂ sσ2O

sσ

σ

pε

sσ−

σ̂ sσ2O pε

Fig. 3-10 The linear kinematic hardening model for uniaxial stress state For von Mises yield criterion

spijij

pijij cscs σεε =−− ))((

23 （3-55）

In uniaxial tension case, with the assumption that the plastic deformation does not induce the volume change, we have

σ32

11 =s ， σ31

3322 −== ss ， ，pp εε =11ppp εεε

21

3322 −==

and the other components of and are all zeros, therefore ijspijε

ps cεσσ

23

+=

Note that during uniaxial tension p

s pEσ σ ε= +

where

p pd EEE

E Edσε

′= =

′− （3-56）

is the plastic modulus, then we get

hc32

= （3-57）

(2) Ziegler’s kinematic hardening law In Ziegler’s kinematic model

ijσ

ijσˆijσ O′

O0f =

ˆijdσpijdε

Fig. 3-11

ˆ (ij ij ijd d ˆ )σ η σ σ= − （3-58）

Since , we have the

consistency condition

*0ˆ( )ij ijf f σ σ ψ= − − 0=

* *

*

*

ˆˆ

ˆ( )

ˆ[ ( )]

ij ijij ij

ij ijij

ij ij ijij

f fdf d d

f d d

f d d

σ σσ σ

σ σσ

σ η σ σσ

∂ ∂= +∂ ∂∂

= −∂∂

= − −∂

0=

therefore *

*ˆ( )

ijij

ij ijij

f dd

f

σσ

ησ σ

σ

∂∂

=∂

−∂

（3-59）

3.4 Deformation Theory of Plasticity Used for Simple

Loading

In the theory of plasticity, the magnitude of stress depends on the history of plastic deformation, and the constitutive relation is by its nature incremental, as we have discussed above the incremental theory or flow theory where the relation between the total values of stresses and strains depends on loading path. Whereas the deformation theory of plasticity attempts to construct equations for plastic deformation in the form of finite relations between stress and strain.

3.4.1 The Theory of Small Elastic-Plastic Deformation (Iliushin’s

Theory) The Iliushin’s theory bases its self on the following propositions: (1) The body is isotropic. (2) The volumetric change is elastic. (3) The deviator stress and deviator strain are proportional.

then we have the constitutive relations：

⎪⎩

⎪⎨

⎧

===

)(εσσσ

ψKees

m

ijij

（3-60）

where, ψ is a scalar. Putting μψ 2== const , we arrive at Hooke’s law. Thus the

equation ijij es ψ= represents a natural and simple generalization of this law.

To determine the scalar ψ , we get from the first equation of the above：

ijijijij eess 2ψ=

It’s evident that ijij ss23

=σ and ijijee32

=ε , hence we get 222

23

32 εψσ =

and thus εσψ

32

= . Therefore the Iliushin’s deformation theory is expressed as

ijij esεσ

32

= ; Kem =σ ; )(εσσ = （3-61）

3.4.2 The Hencky’s theory In fact, the deformation theory of plasticity is equivalent to the problem of nonlinear elasticity. With the prerequisite of simple loading, the finite relation of stress and strain can be derived by integration of the incremental form of them. During simple loading：

0ijij tσσ = ， ，0

ijij tss = 0σσ t= （ ）, and thus 0,0 >> dtt dtsds ijij0=

where, , and 0ijσ 0

ijs0σ are the constant stress state and the monotonically

increasing parameter.

t

Hence Prandtl-Reuss equation becomes

⎟⎟⎠

⎞⎜⎜⎝

⎛+=+=+= λ

μλ

μλ

μtddtstdsdt

ssd

dsde ijij

ijij

ijij 2

122

000

integration of the above equation yields

⎟⎟⎠

⎞⎜⎜⎝

⎛+= ∫ λ

μtdtse ijij 2

0

Noticing Eq. (3-32), σελ

23 pdd = , then

ijpij

pij

pij

pijij

ss

dt

ts

tdt

ttsdttse

σε

μ

εσμ

σε

μσε

μ

23

2

23

21

231

21

23

2

00

000

+=

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

⎟⎟⎠

⎞⎜⎜⎝

⎛+=⎟⎟

⎠

⎞⎜⎜⎝

⎛+=

∫

∫∫

（3-62）

Let σεμϕp3

= , then

ijpij

eijij seee

μϕ

21+

=+= （3-63）

3.4.3 Example: Combined tension and torsion of a thin-walled tube As an example to illustrate the behavior of plastic deformations expressed by the

plastic flow theory and deformation theory, we now consider a circular thin-walled tube under a torsional moment and an axial tension, the case called P-M test, Fig. 3-12. The material is perfect plasticity and subject to von Mises yield criterion. We also suppose the material to be incompressible, therefore the volume strain

03 =++== zrme εεεε θ

M

M

P

P

Fig. 3-12

and

21

=ν , 3)1(2EEG =

+==

νμ .

Using the cylindrical coordinate system, stresses in the tube wall under P-M loading are

RtP

z πσ

2= ，

tRM

z 22πτθ =

where: —wall thickness， —radius of the cylinder. The corresponding strains are t R

zε and zθγ respectively. Other stress and strain components are all zero.

Introducing the dimensionless stresses and strains

s

z

s

z

s

z

s

z

γγ

γεεε

ττ

τσσσ

θ

θ

==

==

,

, （a）

where, 3s

sστ = , ss Eεσ = , s

sss EG

εστγ 33

13

=⋅==

In elastic range：

Ez

zσε = ，

Gz

zθ

θτ

γ =

Substituting them into equation（a）, we get the stress-strain relations in linear elastic stage

εσ = ， γτ =

von Mises yield criterion of the example is： 222

2 31

31

szzJ στσ θ =+=

i.e. 222

31)()(

31

sss σττσσ =⋅+⋅

Noting that 3s

sστ = , we then get

122 =+τσ （b）

（1） Solution from flow theory

（using Prandtl-Reuss equation, ijijij sddsde ⋅+= λμ21

）

Because zm σσ31

= ， zmzzs σσσ32

=−= ， 0=mε ， zmzze εεε =−= ，

zze θθ γ21

= and 3EG ==μ , then

⎪⎩

⎪⎨

⎧

+=

+=

λττγ

λσσε

θθθ ddG

d

ddE

d

zzz

zzz

21

21

321

（c）

it is normalized as：

⎪⎪⎩

⎪⎪⎨

⎧

⎟⎠⎞

⎜⎝⎛+=+=

⎟⎠⎞

⎜⎝⎛+=+=

λττ

ττ

λγτ

γτ

γγ

λσσ

σσλ

εσ

εσ

εε

θθθθθ Edd

dd

Gd

EddddE

d

s

z

s

z

s

z

s

z

s

z

s

z

s

z

s

z

s

z

s

z

3221

32

321

or

⎩⎨⎧

′+=

′+=λττγλσσεdddddd

， λλ Edd32

=′ （d）

by eliminating λ′d we get

τσ

τγσε=

−−dddd （e）

from equation（b）we know

0=+ ττσσ dd and 21 στ −= ， 21 τσ −=

hence

21 σ

σστσστ

−−=−=

ddd ， 21 τ

ττσττσ

−−=−=

ddd

By eliminating τd and τ in equation (e), we get

))(1(1 22 εγσσσεσ ′−−−=dd

，εγεγdd

=′ )( （f）

By eliminating σd and σ in equation (e), we get

))(1(1 22 γετττγτ ′−−−=dd

，γεγεdd

=′ )( （g）

If we have known that 0σσ = ， 0ττ = ， 0εε = ， 0γγ = at a moment, then:

For a horizontal path（ in Fig. 3-13）： ba→

0)( =′ const=γ ， εγ ，integration of Eq.（f）gives γ

ε

ba

Fig. 3-13

a′

b′

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

⋅++

=−σσ

σσεε

11

11ln

21 0

00 （h）

For a vertical path（ in Fig. 3-13）： a b′ → ′

const=ε ， 0)( =′ γε ，integration of Eq.（g）gives

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

⋅++

=−ττ

ττγγ

11

11ln

21 0

00 （i）

（2）Solution from deformation theory（Using Iliushin’s Theory ）

Because volume is incompressible, 0=kkε ， 0=mε ， ijij e=ε , zr εεε θ 21

−== .

According to Iliushin’s deformation theory

ijij esεσ

32

=

thus 23z zs eσε

= zz εεσσ

32

32

= （j）

23z zs eθ θσε

= zz θθ γεστ

31

= （k）

For perfect plastic materials, sσσ = after yielding. Noticing the assumption of

incompressible medium, we have 12r θ zε ε= = − ε , and the equivalent strain is

22

21

222222

31

)](23)()()[(

32

32

zz

zrzrrzzrijijee

θ

θθθθ

γε

γγγεεεεεεε

+=

+++−+−+−==

Eq.（j）can be rewritten as:

2222

2222

31

31

31

γε

εσεγε

εσ

εγε

σσεε

γε

σσσε

εσσ

θ

θθ

+=⇒

+=⇒

+=⇒⋅

+=⇒=

zz

s

zz

s

s

z

zz

s

s

zzz

EE

（l）

Eq.（k）can be rewritten as:

2222

2222

313/

31

331

313

131

γε

γτγγε

γτ

γγε

ττγ

γ

γε

σττ

γεστ

θ

θ

θ

θ

θθθ

+=⇒

+=⇒

+=⇒⋅

+=⇒=

zz

s

zz

s

s

z

zz

s

s

zzz

GG

（m）

As long as the final values of ε and γ are given, σ and τ can be determined

from Eq.（l）and Eq.（m）immediately.

Summary to the example：Combined extension and torsion of a thin-walled tube

s

z

σσσ = ；

s

z

ττ

τ θ= ；s

z

εεε = ；

s

z

γγ

γ θ=

ss Eεσ = ，3s

sστ = ， ss εγ 3=

Perfect plasticity，yield criterion 122 =+τσ

（1）Plastic flow theory（Prandtl-Reuss equation）

⎩⎨⎧

′+=

′+=λττγλσσεdddddd

； τσ

τγσε=

−−dddd （a）

))(1(1 22 εγσσσεσ ′−−−=dd

，εγεγdd

=′ )( （b）

))(1(1 22 γετττγτ ′−−−=dd

， γεγεdd

=′ )( （c）

For horizontal path： const=γ ， 0)( =′ εγ

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

⋅++

=−σσ

σσεε

11

11ln

21 0

00 （d）

γ

ε

a′ba

b′

Fig. 3-14

For vertical path： const=ε ， 0)( =′ γε

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

⋅++

=−ττ

ττγγ

11

11ln

21 0

00 （e）

（2）Deformation theory

22 γε

εσ+

= ； 22 γε

γτ+

= （f）

Case study： γ

ε

)1,1(C

B

A

D

③

②

①

O

Fig. 3-15

Find the stresses at point C along three paths starting from point O.

Requirements: Using plastic flow theory for the paths：①②③

Using deformation theory for the paths：③（simple loading） Solution：

In elastic stage

εσ = ， γτ =

and the yield condition is

122 =+τσ 。

therefore the elastic region in strain space can be denoted as

122 ≤+ γε

which illustrates the shaded area in Fig. 3-11. （A）Solution by the flow theory Along path ① O—B—C

At point B： 00 =τ ， 00 =γ

From B to C：From Eq.（e）we know ττγ

−+

=11ln

21 , then

γτ γ

γth

11

2

2=

+−

=ee

At point C： 1=γ ，thus 76.011

2

2≈

+−

=eeτ ， 65.01 2 ≈−= τσ

Along path ② O—A—C

At point A： 00 =ε ， 00 =σ

From A to C：From Eq.（d）we know σσε

−+

=11ln

21 , then

εσ ε

εth

11

2

2=

+−

=ee

At point C： 1=ε ，thus 76.011

2

2≈

+−

=eeσ ， 65.01 2 ≈−= στ

Along path ③ O—D—C At point D：located on the border of elastic region to plastic region where

707.02

1≈==τσ

From D to C： γε dd = . From Eq.（a）we know

λττλσσ ′+=′+ dddd ，or λτστσ ′−=

−− dd )(

Integration of the equation gives

CDCD λλτσ ′−′=− )ln( ，or )exp()()( CDDC λλτστσ ′−′⋅−=−

Since DD τσ = , then CC τσ = . Notice , we finally get 122 =+τσ

707.0== CC τσ

（B）Solution by the deformation theory

At point C， 1== γε . recalling that

22 γε

εσ+

= ； 22 γε

γτ+

= （f）

so we get

707.02

1≈==τσ

Conclusion： Only with the simple loading, the deformation theory and flow theory result in

the same results.