Chapter 7 Yielding criteria

Chapter 7 Yielding criteria

1. Criteria for yielding

(1) What is the meaning about yield criterion?

In this case the stress is un-axial and this point can readily be determined. But what if there are several stress acting at a point in different direction ？

⇒The criteria for deciding which combination of multi-axial stress will cause yielding are called criteria.

(2). Theory of yield criterion- (A) Tresa criterion

Yielding will occur when the maximum shear stress reaches the values of the maximum shear stress occurring under simple tension.

The maximum shear stress in multi-axial stress = the maximum shear stress in simple tension

1 3 2 3 0 1 2 max , , 2 2 2 2 σ σ σ σ σ σ σ− − − ⎧ _{⎫ =} ⎨ ⎬ ⎩ ⎭ 換言之,最大剪應力為1 ₀ 2σ 材料就降伏

For pure shear (k) 1 2 1 2 1 2 1 2 0 2 k σ σ σ σ σ σ _{σ σ} + = ⇒ = − − = = = 又 1 2 0 2 2 k =σ σ− =σ =k 0 2 k σ ∴ =

⇒For pure shear( )k state, the yielding is happened if 0 2 k =σ (2)The von-Mises yield criterion

Yielding begin when the octahedral shear stress reaches the octahedral shear stress at yield in simple tension.

τoct =τoct o_,

2 2 2 2 2 2 1 ( ) ( ) ( ) 6( ) 3 oct x y y z x z xy yz zx

τ

σ

σ

σ

σ

σ

σ

τ

τ

τ

For the principal stress

2 2 2 1 3 1 2 2 3 0 1 2 ( ) ( ) ( ) 3 σ σ− + σ σ− + σ −σ = 3 σ 又 2 2 2 2 2 2 2 1 [( ) ( ) ( ) 6( )] 6 x y y z x z xy xz yz J = σ −σ + σ −σ + σ −σ + τ +τ +τ 得 2 2 3 o J =σ

1 2 0 0 , 0 6 2 3 3 1 3 z k k k σ σ σ σ σ = − = = = ∴ = 註: k 為純剪應力之大小 故 von-Mises yield 可簡化為: 2 2 J =k

Discussion: For Tresca criterion For Von-Mises Yield criteion 0 1 2 T k = σ =0.5σ₀ 1 ₀ 3 V k = σ =0.577σ₀ ∴kV >kT

(3)Yield surface and Haigh-westergaard stress space

From the yielding criterion, the shear condition in multi-axial stress = the shear condition in simple tension

( ) ( ) 1 : The stress state

: obtain from simple tension ij ij F K k k σ σ = − − − − − − < >

(3) Yield surface And Haigh-Westergaard stress space From the form of yielding criterion. That is

The shear condition in multiaxial stress = The shear condition in simple tension

F( )σ_ij =K k( ) ---(1)

↑ ↖obtain from simple tension The stress state

(A). Represents a hyper surface in the six-dimensional stress space, any point on this surface represents a points a point at which yielding can begin and function (1) is called the yielding function.

The surface in the stress space is called the yield surface. Since the rotating the axes does not affect the yielding state, we can choose the principal axes for the coordinates.

F( ,σ σ σ_{1 2}, ₃)=K

1 2 3 1 2 3 1 1 2 3 2 2 2 2 1 2 3 3 3 3 3 1 2 3

and , , can be writen in terms of the invariants , , 0 1 Where ( ) 2 1 ( ) 3 s s s J J J J s s s J s s s J s s s ⎧ ⎪ = + + = ⎪ ⎪ = + + ⎨ ⎪ ⎪ _{= + +} ⎪⎩ 2 3 ( , ) ( ) f J J K k ∴ =

For von-mises criterial

2 2 2 1 2 3 m 1 2 2 2 2 2 1 2 ( 3 3 2 2 3 3 1

, and is the yield in pure shear. 3 oct o o o s s s J J k k τ σ σ σ σ = + + = = = ∴ = = 和 無關)

For Tresca criterion:

1 2 1 2 1 2 m ( ) ( ) ( 2 2 2 2 m m s s o σ σ σ σ σ σ σ− − − − − _σ = = = 和 無關) 3 2 2 2 4 6 2 3 2 2 4J −27J −36k J +96k J −64k =0

B. Haigh-Wester-gaard stress space. => yielding criterion can be expressed as function of ( ,σ σ σ_{1 2}, ₃)

故可以 σ σ σ₁, ₂, ₃為座標軸,可得函數圖.

The principal ( ,σ σ σ_{1 2}, ₃) coordinate system represents a stress space called the Haigh-wWester-gaard stress space.

Consider a line ON uuur

which passing through the origin, and having equal angles

With the coordinate’s axes, then every point on this line is

1 2 3 1 2 3 3 m σ σ σ σ =σ =σ =σ = + + (即該線之點皆為靜水壓應力)

and σ σ₁+ ₂+σ₃ =0,is the plane called π-plane. And this is the pure shear stress condition.

And 1 ₁ 1 ₂ 1 ₃ 3 3 3 3 m ON A A OP ON σ σ σ σ = = ⋅ = + + = uuur ur uuur uuur Bur = − =Pur urA (σ σ₁− _m)ir+(σ₂−σ_m)rj+(σ σ₃− _m)kr 2 2 2 2 1 2 3 ( m) ( m) ( m) B σ σ σ σ σ σ ∴ = − + − + − 2 2 2 1 2 3 2 2 2 1 2 3 1 1 ( 2 ( ) ) 2 2 ij ij s s s J s s s s s = + + = + + = Q 2 2 2 # 2 0 1 2 2 0 1 2 3 1 2 and ( ) 3

2 => the components of are therefore the stress 2 deriators , , 3 B J J von Mises B J B s s s σ σ ∴ = = − ∴ = = ur => 1 2 2

3 m 2 (at Haigh - Wester - gaard stress space)

ON B P A B J ON B σ = + = + uuur ur ur ur ur uuur ur

Since it is assumed that yielding is determined by the deriatoric state of stress only, it follows that if one of the points on the line through p parallel to ON

uuur

lies on the yield surface => they must all lie on the yield surface, since they all have the same deriatoric stress components. Hence the yield surface must be composed of lines parallel to ONuuur; i.e, it must be a cylinder with generators parallel to

ON uuur

Note:

(a) The intersection of this yield cylinder with any plane perpendicular to it will produce a curve called the yield locus. Since this curve will be the same for all planes perpendicular to the cylinder. => For this purpose we choose the π-plane whichσ_m=0.

(b) If, as usual, isotropy is assumed so that rotating the axes does not affect the yielding. That means a line perpendicular to

1, 1; 2, 2; ,3 3;

σ σ σ− −σ σ −σ are therefore lines of symmetry and we now

have six symmetric sectors. (1,-1, 2,-2, 3,-3)

(c) The yield surface must be symmetric in the principal stress since it certainly does not matter.

=> Hence, we have divided the yield locus into 12 symmetric sectors, each of30o. and we need only consider the stress states lying in one of these sectors.

C. The stress in π-plane

0 0 2 1 2 1 0 0 3 2 1 3 2 1 2 2 cos30 cos30 ( ) / 2 3 3 2 2 2 1 sin 30 sin 30 (2 ) 3 3 3 6 a b σ σ σ σ σ σ σ σ σ σ = − = − = − − = − − 2 2 2 2 2 2 1 2 2 3 3 1 2 2 2 1 2 3 2 1 ( ) ( ) ( ) 3 ( ) ( ) ( ) 2 m m m r a b J σ σ σ σ σ σ σ σ σ σ σ σ ⎡ ⎤ = + = _⎣ − + − + − _⎦ ⎡ ⎤ =_⎣ − + − + − _⎦ = 3 2 1 1 1 1 3 2 2 1 2 1 1 (2 ) 2 1 6

tan tan tan

1 _{3( )} ( ) 2 b a σ σ σ _{σ σ} θ σ σ σ σ − − − − − − − = = = − − 3 2 1 3 2 1 2 1 2 3 tanθ σ σ σ (ifσ σ σ ) σ σ − − ∴ = > > −

2 2 2 2 2 1 1 3 2 2 3 o o J r r σ σ = = ∴ =

The yield locus is therefore a circle of radius r = 2 3σo NoTe: (a).for 0

θ

at -plane 0 0, That means =0 is pure shear state m m σ σ σ θ σ σ σ σ σ σ σ σ σ σ σ σ π σ σ θ − − = = − ∴ = + = + + = + = = ∴ = (b) for 0 0 3 2 1 2 1 2 30 , from 3 tan 30 σ σ σ 1 θ σ σ − − = = = − 3 2 => uniaxial stress σ σ ∴ =

(c) If the yield locus is assumed to be convex, the bounds on yield loci will be between C' A B' and C A B.

A yielding curve below CAB Which pass the C, A, B point will not be convex is call lower bound.

a yielding curve outside C C' A B' B which pass the C,A,B point will not be convex also is call upper bound.

(4) Subsequent yield surfaces, Loading and unloading

前面所討論的是 ＂initial yield＂之問題, 但若 yielding 後,其 yielding surface 會如何呢？即＂strain hardens＂之問題 ｡

For a yield Function

F( )σ_ij =k (加工硬化方程式)

And

ij

k is a value wich defind a yield surface. and strain-harden function F( ) is loading function.σ

⎧ ⎨ ⎩

After yielding has occurred, k take on a new value, depending on the strain-hardening properties of the material.

After yielding has occurred, k takes on a new value, depending on the strain-hardening properties of the material.

A. Loading and unloading for a strain-hardening material Three cases for a strain-hardening material:

(A) Loading⇒plastic flow is occurring.

, ij 0 ij F F k dF dσ σ ∂ = = > ∂

(B) Neutral loading⇒stress state moving on yield surface.

, _ij 0 ij F F k dF dσ σ ∂ = = = ∂ (C) unloading , ij 0 ij F F k dF dσ σ ∂ = = < ∂ Note:若取( ,σ σ σ_{1 2}, ₃)為座標軸 , _j ( _j) ( _{i j}) j j F F dF dσ e dσ e σ σ ∂ ∂ = = ⋅ ∂ ∂ uur uur = ∇ ⋅F dσiej ur uur F ∇ur :表示垂直yield surface之向量 dσi ej uur :任意增量向量 ⇒故 _j 0 j F dF dσ σ ∂ = < ∂ 表示曾增加向量為unloading

B. Subsequent yield loci (A) Isotropic hardening

If ' 0 0

σ >σ , then the new yield locus is a circle of radius 2 ' 3σ for von-Misses criterion, which is larger than, but concentric with , the origin yield circle. ⇒ The material is called strain harden

(B) Bauschinger effect

註: 可發現當剪應力走不同方向時,橢圓不在! Prager Kinematic model

Assume :

(a) The rigid frame having the shape of the yield surface

(b)The frame is assumed to be constrained against rotation and to be perfectly smooth, so that only forces normal to the frame can be transmitted to it.

(c)The state of stress and the state of strain are represented in the model in different ways, For example, for a rigid strain-hardening

material, the displacement of center of the frame relative to the origin is proportional to the total strain, and the state of stress is represented by the position of the pin relative to the origin.

Note: the isotropic hardening assumption is still generally used. For

small plastic strain it probably gives answers that are sufficiently accurate.

6,Plastic stress-strain relations-

在彈性區之應力和應變之關係已經討論過現在,則更進一步討

論塑性區之 stress-strain .

(1) General derivation of plastic stress-strain relations.

(note: 所有的運算皆為向量法則)

For obtaining general stress-strain relation, two definition and two assumption are needed.

Definition:

ij

(a)Positive work is done by external agency during the application of the set of stress. 0

(b)The net work performed by it over the cycle of application and removal i ij

dσ εd >

ij

s zero or position. p 0(delcte elastic strain engergy)

ij dσ εd ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ _≥ ⎩

Assume:(a)A loading function exists. At each stage of the plastic deformation there

exists a function ( )σij so that further plastic deformation takes place only for ( )σij >k

Both and k may depend on the existing state of stress and on the strain history.

(b)The relation between infinitesimals of stress and plastic strain is linear: Gij =dF p ij ijk k dε =c _ldσ _l

Note: (a)cijkl may be functions of stress, strain, and history of loading that implies they are independent of the dσij

(b)From assumption 1, it follows that for plastic may be applied to the stress and strain increments.

If ' ij

dσ and dσ''_ij are two increments producing plastic strain increments, p' ij dε and dεp''_ij.And '' ' ij : incremental stress perpendicular to f : incremental stress tangent to f( ) B B d d σ σ σ ⎧⎪ ⎨ ⎪⎩ l l ⇒ ' B

dσ _l produce no plastic flow(延著硬化方程式,繞著走依然是

彈性範圍) From assumption 2. p ₍ ' '' ₎ '' ij ijk k k ijk k dε c dσ dσ c dσ ⇒∴ = l l + l = l l From assumption 1. _{⇒∴ =} ∂F _{σ =} ∂F _{( σ}' _{+ σ}'' _{) 0>}

0 k k dF a F F σ σ ∴ = > ∂ ∂ ∂ l ∂ l From p '' _{( )}

ij ijk k ijk ijk

k k k k F F dF d c d c a c F F ε σ σ σ σ σ ∂ ∂ = = = ∂ ∂ ∂ ∂ ∂ ∂ l l l l l l l l = _ijk ( ) k k dF c dF F F σ σ ∂ ∂ ∂ ∂ l l l =Gij =dF ' '' ij ( ) and ( )( ) is called potential function

From definition 2. 0 ( ) 0 Beca p ke ij ij ij ijke ke ke p p ij ij ij ij F d G dF G C F F d d d d d σ ε σ σ σ ε σ σ ε ∂ ∂ ⇒ = = _{∂ ∂} ∂ ∂ ↑ ≥ ⇒ + ≥ '

use produce no p We can chose any constc.

ij ij dσ dε ∴ ' '' ' ij p ij ' positive or negative,

will produce the same ( ) 0 d 0 plastic increment d d p p ij ij ij ij ij i d d d d G σ σ ε σ ε ε σ ⎫ ⎪ ⇒ + < ⇒ = ⎬ ⎪ ⎭ ⇒ jdF =0 ' ' 0 0 G 0 p ij ij ij ij ij ij dF d d F d G dF σ ε σ σ ⎫ > ⇒ = _{⎪ ⇒} ∂ ⎬ ⇒ = _⎪⎭ ∂ Q

( is a constant which may

( )

be function of stress, strain

( )( ) _{and history of loading)} ke ij ijke ij ke ke F G F G G C F F σ σ σ σ ∂ ∂ ∂ ⇒ = = _{∂ ∂} ⇒ ∂ ∂ ∂

Hardening value yield locus

(d : is a nonnegative constant which may vary through out

the loading history p ij ij ij ij f d G dF F GdF F d ε σ σ λ λ σ ∂ ∴ = → ∂ ∂ = ∂ ∂ = ∂ ^ 垂直 之方向梯度 )

=> The plastic strain increment vector must be normal to the yield surface. ( Note : From p ₍ '' ₎ ij kl ij kl F F dε G dF dF dσ σ σ ∂ ∂ = = ∂ ∂ '' kl ij kl F F G dσ σ σ ∂ ∂ = ∂ ∂ dF之增加由 '' kl dσ 控制,而dF則為yield locus半徑增加量)

一、 The flow rules associated with von-Mises and Tresca (一) For the von-Mises yield function

2 2 2 2 2 1 2 2 3 1 3 1 1 1 ( ) ( ) ( ) ( ) 6 2 3 ij ij ij o F σ =J = _⎣⎡ σ σ− + σ −σ + σ σ− _⎦⎤= s s = σ From εp ₌ ∂F ₌ ∂F ∂sij ( ₌σ ₋σ _⇒ ∂sij ₌1)

(二) For Tresca yield function –

Assuming it is known which is the maximum principal stress 1

σ and minimum principal stress σ σ₃, ₁ >σ₂ >σ₃

1( _{1 3}) 1 ₀ 2 2 F= σ σ− = σ 1 3 1 1 , 0 , 2 o 2 F F F σ σ σ ∂ ∂ ∂ ∴ = = = − ∂ ∂ ∂ 1 2 3 1 2 0 1 2 p p p d d d d d ε λ ε ε λ ⎧ ₌ ⎪ ⎪ ⇒_⎨ = ⎪ ⎪ = − ⎩

(1)(2) are known as the flow rules associated with the von-Mises and Tresca criteria.

二.Perfectly Plastic Material

For ideal plasticity it is also assumed that F( )σ_ij exists and is function of stress only , and that plastic flow take place without limit when F( )σij =k, and the material behaves elastically when F( )σij <k.

For plastic flow.

// p p ij ij ij ij F F dε dε dλ σ σ ∂ ∂ ∴ ⇒ = ⇒ ∂ ∂ 0 0 ij ij p ij ij F dF d d d σ σ σ ε ∂ ⎧ _{= =} ⎪ _∂ ⎨ ⎪ ₌ ⎩

// p p ij ij ij ij F F dε dε dλ σ σ ∂ ∂ ∴ ⇒ = ∂ ∂ where dλ is a scalar.

三.Determination of the function G.. Effective stress and strain (一)Effective stress σe

From yield criterion ⇒F( )σij =K( )σ₀ or f( )σij =σ0

↑From uni-axial tensile test.

即左式為多軸壓力狀態,可相對某一值σ_e,若該值等於σ₀則降 伏。σ_e稱之 effective stress. Definition 2 2 1 ( ) ( ) 3 ij e e F σ =K σ =J = σ For Von-Mises 0 ' 3 3 2 2 , ij ij p p p e ij e ij e e k k k k k k ij F F d s d d d F F F F F F s σ σ σ ε σ ε ε σ σ σ σ σ σ σ ∂ ∂ ∂ ∂ ∴ == ∴ = = ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∴ = ∂ ∂ l l l l l l 3 3 2 2 e s sij ij σ ∴ = = Note: 當降伏時σe=σ0,即應力之降伏相當於單軸應力之降伏.

(二)Effective plastic strain p e dε

= 2 3 3 2 p p p p e e ij ij p p ij ij d d d d d d ε ε _{ε ε} ε ε = 3 2 p p p e ij ij dε dε εd

∴ = (For Von-mises criterion )

= 3 _{( )}2 _{( )}2 _{( )}2 _{2( )}2 _{( )}2 _{( )}2 2

p p p p p p

x y z xy yz zx

dε + dε + dε + dε + dε + dε

For uniaxial tens it test 1 , 1

2 2 p p p p y x z x dε dε dε dε ⇒ = − = − p p e x dε dε ∴ = 即降伏之塑性應變當於單軸應力之降伏｡ 即 p p ₀ p ij e e e x s dε =σ εd =σ εd ⇒塑性功能不變法則⇒達到塑性狀態所所需之功和過程無關｡

(三)Determination of the Function G.

From p ij ij F dε GdF σ ∂ = ∂ and 2 3 p p p e ij ij dε = dε εd 2 3 p e ij ij F F dε GdF σ σ ∂ ∂ ∴ = ∂ ∂ 3 2 p e ij ij d dFG F F ε σ σ ∴ = ∂ ∂ ∂ ∂ ' 3 3 2 _ij 2 _ij p p e ij e e k k k k F F d d d F F F F σ σ σ ε ε σ σ σ σ σ ∂ ∂ ∂ ∂ ∴ = = ∂ ∂ ∂ ∂ ∂ l ∂ l ∂ l ∂ l ' ( e) e p e d d σ σ ε

= ←The slope of the uniaxial stress-plastic strain curve at the current value of σe

For Von-Mises criterion ⇒ _e ' , 3 3 2 _{( )} 2 3 3 2 2 k ij k ij ij p p ij e ij ij k k ij p ij e e e e e F F s s s d d s s s s s s d d σ σ ε ε σ σ ε σ σ σ ∂ ₌ ∂ ₌ ∂ ∂ ∴ = = = = l l l l Q k , ij k ij F F s s σ σ ∂ ∂ = = ∂ l ∂ l e 3 3 2 _{( )} 2 ij p p ij e ij ij k k s d d s s s s ε ε σ ∴ = = l l Q ' 3 3 2 2 ij p ij e e e e e s s d dε σ σ σ σ

= = ←(The flow rule associated with von mises

yield criterion) or 3 2 2 3 1 , ( ) 2 ij e p ij p e ij ij p e j i ij i e j j s s v v v x x

σ

ε

σ ε

ε

σ

ε

⇒s_ij will be know, If in plastic state. (?) Incremental and deformation theories

For 3 2 ij p p ij e e s dε dε σ = 3 2 ij p p ij e e s dε dε σ

= are called incremental

stress-strain relations because they relate the increments of plastic strain to the stress.

For the case of proportion or radial loading i.e if all the stress are increasing in ratio (stressed disk or cylinders), the incremental theory reduces of the deformation theory.

IF 0 0 0 0 3 3 2 2 ij e ij ij p p p p ij e ij e e e s K K s s d d σ ε ε ε ε σ σ = ⎧ ⎨ ₌ ⎩ ∴⇒ =

∫

monotonically increasing function of Then ij ij0₀ e e s Ks K σ σ ⎧ = ⎨ ₌ ⎩ 0 0 0 0 3 3 2 2 ij ij p p p p ij e ij e e e s s dε dε dε dε σ σ ∴ = ⇒

_∫

⇒The plastic strain is a function only of the current of stress and is independent of the loading path.

(?)Convexity of yield surface. Singular points.

一. Convexity of yield surface

Let some external agency add stresses along some arbitrary path inside the surface until a state of stress dεijis reached which is on the

yield surface. Now suppose the external agency to add a very small outward pointed stress increment dσ_ij which produces small plastic strain increments dε_ij, as coell as elastic increments.

The work done by the external agency over the cycle is

(1) Elastoplastic problems of spheres and cylinders

一、 Spherical coordinates – The problems of spheres ⎩ ⎨ ⎧ = = φ θ φ θ ε ε σ σ

(a) The equilibrium equation 0 sin 2 sin sin ) ( ) )( (σr+dσr r+dr dφ r+dr φdθ −σrrdφr φdθ − σθr φdθdrdφ = ⇒ σr2rdr+dσrr2dr−2σθrdr=0 Fr r dr d r r − = − +2(σ σθ) σ _and

⎪ ⎪ ⎪ ⎩ ⎪⎪ ⎪ ⎨ ⎧ − = ⇒ = + + = − + ⇒ = = > < > < = = = p p r p p p r r r r r dr d dr r d dr du From r u dr du θ θ φ θ θ θ φ θ ε ε ε ε ε ε ε ε ε ε ε ε ε 2 0 0 ) ( , 1 1 ~~~~~ ,

(c) The stress-strain relation

⎪ ⎩ ⎪ ⎨ ⎧ + − − = + + − = + − = p r p r p r r r E E E θ θ θ φ θ θ θ ε µσ σ µ ε σ σ µ σ ε ε µσ σ ε ] ) 1 [( 1 )] ( [ 1 ) 2 ( 1

Ⅰ. Von-Miese yield function

[( ) ( ) ( ) ] 6 1 2 1 , 3 1 2 1 3 2 3 2 2 2 1 2 2 0 2 = σ J = SijSij = σ −σ + σ −σ + σ −σ J and σ₁ =σr,σ₂ =σ₃ =σ_θ 2 0 2 ) (σ −σ_θ =σ ∴ r ⇒σ0 = σr −σθ Ⅱ.Prandfl-reuss Equation From ij e p e p ij S r d dε ε 2 3 = p r p ij p ij p e d d d dε = ε ε = ε 3 2 ∴ ε = ε sgn(σr −σ_θ) p e p r d d

二、 Polar coordinates-For cylinders problems (1) The equation of equilibrium of stress

Fr r dr d _{r r} − = − +σ σθ σ

(2) The strain-displacement relations or compatibility equation r u dr du r = εθ = ε , ∴ + − =0 r dr dε_θ ε_θ ε_r + + p =0 z p p r ε ε ε _θ

(3) The stress-strain relation ⎪ ⎪ ⎪ ⎩ ⎪⎪ ⎪ ⎨ ⎧ + − + − = + + − = + + − = ) ( )] ( [ 1 )] ( [ 1 )] ( [ 1 p r p r z z p r z p r z r r E E E ε ε σ σ µ σ ε ε σ σ µ σ ε ε σ σ µ σ ε θ θ θ θ θ θ

NOTE: for the case of plane stress，σz =0，and for the case of plane strain εz =0 or εz =const. for generalized strain. In both cases the stear stresses and strains are zero.

三、 Thick Hollow sphere with internal pressure

Consider a sphere with inner radius a and outer radius b , subjected to an internal pressure P. It is obvious that complete symmetry about the center will exist so that the radial and any two tangential direction will be principal direction.

(1) Elastic solution ⎪ ⎩ ⎪ ⎨ ⎧ − − = − = ] ) 1 [( 1 ) 2 ( 1 r r r E E From µσ σ µ ε µσ σ ε θ θ θ and + − =0 r dr dε_θ ε_θ ε_r

dr d dr d σ_r µ µ σ µ µ _θ + − = + − ⇒ 1 1 1 ) 1 ( 2 r c r dr d c r _r r 1 1 2 3 2 1 _{+ ⇒ + =} − = ∴σ_θ σ σ σ 1 1 _{3 2} 2 3 c dr d r r c r dr d From r _r r r + σ = ⇒ σ + σ = σ let rD dr d r e dr d dz dr dr d dz d e r dz d = z ∴ ≡ = = z = = − = ϑ ϑ , 原式⇒(ϑ +3)σr =2c1 2 3 2 ₃ 1 r c e c y_h = − z = 2 ₁ 1 1 1 ₃ 2 ) ) 3 ( 3 1 ( 3 2 ) 3 1 ( 3 2 2 ) 3 ( 1 c c c c y_p = − + − = + = + = _ϑ ϑ ϑ Λ ϑ 3 1 2 1 3 2 3 2 3 2 c r c c r c y y y= _h + _p = + ∴ _r = + ∴ σ

z From boundary condition ⎪ ⎩ ⎪ ⎨ ⎧ = + = = + = − = 0 3 2 0 ) ( 3 2 ) ( 1 3 2 1 3 2 c b c b c a c P a r r σ σ ) ( , 2 3 3 3 3 3 2 3 3 3 1 a b b Pa c a b Pa c − − = − = ∴ ⎪ ⎪ ⎩ ⎪⎪ ⎨ ⎧ − + − = − − = − + − − = ∴ 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 2 3 2 1 ) 1 ( 1 ) ( a b Pa a b a b Pa a b Pa r a b b a r r σ σ σ θ

For convenience the following dimensionless quantities are now introduced: , , a r a b = = ρ β 0 0 0 , , σ σ σ σ σ = θ = θ = P S r S r P ⎪ ⎪ ⎩ ⎪⎪ ⎨ ⎧ − + = − − = = ∴ ) 1 ( 2 2 ) 1 ( 3 3 3 3 3 3 3 3 0 β ρ β ρ β ρ β ρ σ σ θ P P S S r r ( ) p p ij ij ij ij ij

the loading surface

0 cos 0 (acute angle)

2 2

( ) 0 cos 0 (acute angle)

2 2 p p ij ij ij ij p ij ij ij ij ij d d d d d π π σ ε σ ε θ θ π π σ σ∗ σ σ∗ ε ψ θ ⎧ _{≥ ∴ ≥ ∴− ≤ ≤} ⎪⎪ ⎨ ⎪ _{− ≥ − ≥ ∴− ≤ ≤} ⎪⎩ Q Q

For convex surface No vector σij σij

∗

− can pass outside the surface intersecting the surface twice. The surface must therefore be convex.

( at

2 2

π _ψ π

For surface is not convex

If the surface is not convex , there exist, some points σ σ_ij, _ij∗ _such

that the vector σ_ij−σ_ij∗ _{form an obtuse angle.}

二. Singular point-

The yield surface has vertices or corners where the gradient is not defined (Tresca hexagon). Such point can be treated by introducing an auxiliary parameter.

7.Application

To solve any plasticity problem, four sets of relations must be satisfied as:

(a) The equation of equilibrium of stress

⇒ ij 0 j j f x σ ∂ + = ∂

(b) The strain-displacement or compatibility relations:

1 ( ) 2 0 for plasticity j i ij j i ii u u x x ε ε ∂ ⎧ ₌ ∂ ₊ ⎪ _{∂ ∂} ⎨ ⎪ = ⎩

(c) The stress-strain relations I、Von-Mises yield function

2 2 0 2 1 1 , 3 2 ij ij J = σ J = s s II、 Prandtl-Reuss Equations 3 2 p p e ij ij e d dε ε s σ = and 3 2 2 3 e ij ij p p p e ij ij s s d d d σ ε ε ε ⎧ ⎫ = ⎪ ⎪ ⎪ ⎪ ⎨ ⎬ ⎪ ₌ ⎪ ⎪ ⎪ ⎩ ⎭ Test by tensidle-tet

(d) The boundary conditions stress-boundary i= j displacement-boundary ij j J u U σ τ ⎧ ⎨ ₌ ⎩ l p26

p31

⇒If we know σ θm, then any point stress could be know.

If curves are now drawn in the xy plane such that at every point of each curve the tangent coincides with one of maximum shear direction, The two families of curves called shear lines, or slip lines.

line line α β − ⎧ ⎨ − ⎩

Note:<a>α β, are merely parameters or curvilinear coordinates used to designate the point under consideration, just as x and y designate the point.

<b>Take a curve element, From monr’s cycles

From 1 1 2 2 1 , , ( ) 2 cos sin , sin cos 0 2 along 2 along y y x x x x y xy y m m v v d du v dt x dt y y x v v v v K c line K c line β ε _{ε ε ε} θ α β _{θ θ} σ θ α σ θ β • _∂ • ∂ • _∂ ∂ ∴ = = = = = + ∂ ∂ ∂ ∂ = − ⎧ ⎨ = + = ⎩ − = − ⎧ ⇒ ⎨ _{+ = −} ⎩

If we choose the α β, curve linear coordinate system

⇒ 0, , x y θ α β ∂ ∂ ∂ ∂ = = = ∂ ∂ ∂ ∂ 1 1 2 2 2 0 2 along -curve 2 along -curve 2 0 m m m m K K c K c K σ θ σ θ α α α σ θ σ θ β β β ∂ ∂ ⎧ _{− =} ⎪ _{∂ ∂ ⎧ − =} ⎪ _⇒ ⎨_{∂ ∂} ⎨ _{+ =} ⎩ ⎪ _{+ =} ⎪ ∂ ∂ ⎩

∗ ⇒Hencky equation, From boundary condition, we obtain c c₁, ₂ If we know θ θ₁, ₂ ⇒σ_m⇒σ σ τ_x, _y, _xy

三. Geiringer Velocities equation

From Prandtl-Reuss equation, and incompressibility condition 3 2 p x y x y x y x y p e ij ij xy xy xy e xy d d d d s d σ σ ε ε σ σ ε ε ε ε _{ε τ τ} σ _ε • • • ⎧ ₋ − − − ⎧ ⎧ _{= ⎪ =} = _⎪ ⎪ _{⇒ ⇒}⎪ ⎨ ⎨ ⎨

( ) ( ) 2 ( ) ( ) 0 y x x y y x xy y x v v x y v v y x v v x y σ σ τ ∂ ⎧ ∂ ₋ ⎪ _{∂ ∂ −} ⎪ = ∂ ∂ ⎪ ₊ ⇒ ⎨ ∂_⎪ ∂ ⎪ ∂ ₊∂ ₌ ⎪ _{∂ ∂} ⎩

Since the principal axes of stress and stress and of plastic strain increment coincide, it follows that the maximum shear stress lines and maximum shear velocity lines coincides, or the stress slip lines are the same as the velocity slip line.

⇒the strain rates normal to the α and β direction are equal to the mean strain rate, that mean.

1 ( ) 0 2 x y d d dt dt β α ε ε _{ε ε}• •

∗ = = + = ⇒ There are no extension, only

shearing flows

in the slip direction.

(在靜水壓,方向無應變率)

Now consider the velocities in the slip direction

And cos sin sin cos x y v v v v v v α β α β θ θ θ θ = − ⎧ ⎨ = + ⎩

Take α β, curve-line coordinate θ =0

and 0 0 ( ) 0 ( ) 0 x y v v v x v v v y α α _{θ β} β β _{θ α} θ ε α α θ ε β β • = • = ∂ ∂ ⎧ _{= = − =} ⎪ _{∂ ∂ ∂} ⎪ ⎨ _{∂ ∂} ⎪ _{= = + =} ⎪ ∂ ∂ ∂ ⎩ 0 along a 0 along a dv v d line dv v d line α β β α θ α θ β − = − ⎧ ⇒ ⎨ _{+ = −}

⎩ →Geiringer velocities equation

Hencky’s first law –the angle between two slip lines of one family at the points where they are cut by a slip line of the other family is constant along their lengths.

From Hencky’s equation 1 1 2 2 2 along 2 along m m K c line K c line σ θ α σ θ β − = − ⎧ ⎨ _{+ = −} ⎩ mD D mC D B D B C A D B C A A B D C

along AD- -line = 2K along DC- -line 2=+ 4K 2 2 2 2 4K 4K 4K 4K 4K mD mA A C C A K K K K α σ θ β σ σ σ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ − ⎧ ⎨ ₋ ⎩ ∴ − = + − ⇒ − − = + − ⇒ + = + ⇒ + = + ⇒ − = − mD D mC C along AD- -line 2 = 2K along DC- -line 2 = +2K mA A mD D K K α σ θ σ θ β σ θ σ θ − − ⎧ ⎨ ₋ ⎩ mC mA 4K D 2K A 2K C σ σ θ θ θ ∴ − = − −

(3)Maximum And Octahedral shear stress 一. Maximum shear stress

Let us take the coordinate axes in the principal direction. And any section direction is v= +i n j+mk =vj r r r r l From σij iv =Tvj 1 1v Tv1, 2 2v Tv2, 3 3v Tv3 σ σ σ ⇒ = = = 2 2 2 2 2 2 2 1 1 2 2 3 3 Tv Tv Tv σ v σ v σ v ∴ =uur uurg = + +