소성재료역학 (metal plasticity) chapter 12: yield function 8... · 2019. 9. 5. · 446.631a...

446.631A

소성재료역학(Metal Plasticity)

Chapter 12: Yield Function

Myoung-Gyu Lee

Office: Eng building 33-309

Tel. 02-880-1711

[email protected]

TA: Chanyang Kim (30-522)

mailto:[email protected]

Three-dimensional plasticity

1-D constitutive law Stresses and strains in 3-D

General state : multiple components existTensionCompressionBending….

ε

σ

σxx σyy


Stress and Strain tensorTensor : mathematical notation of physical quantities

Stress : symmetric 2nd order tensor

• 0 th order Tensor : scalar value – energy, etc• 1st order Tensor : vector - force, displacement, velocity, etc• 2nd order Tensor : linear transformation from vector to the other vector

- stress, strain, etc.

xx yx zx

xy yy zy

xz yz zz

σ


1-D : point 2-D : line 3-D : surface

I

II

( ,0,0)ST Y

2 or 2(2 , ,0)PS PLS K K

( , ,0)BB B B

3 or 3( , ,0)PS PLS K K

ε

σ

For isotropic case,

Basic features of the yield surface

Concept of material coordinates

(or materially embedded coordinates)

𝑥′

𝑦′ 𝑦

𝑥

p

Global coordinates system

(or laboratory coordinates system)

𝑥

𝑦

Material coordinates system

• Since the constitutive law describes material properties, components of

vector or tensor quantities are defined with respect to material coordinates


Convexity of yield surface

(a) (b)

Convex Non-convex

• The yield surface is considered to be convex

• In other words, yield surface should be bulged out

• Or, any straight line connecting two points located inside the surface stays

inside the surface


• For the simplicity, consider a 2-D imaginary yield surface

( ) ( , ) xx yyf f constant C

• Size of yield function f(σ) is defined by constant C

• For the isotropic case, yield function can be assumed as a circle

• For the anisotropic case, ellipse can be an yield function

2 2 1/2

1

2 2 2 2

2

( ) ( ) c

( ) ( ) c

xx yy

xx yy

f

f

1

2 2 21 1

1

2 2 22 2

2 2 2

3

2 2 2

4

( ) ( ) ( ( ) )

( ) ( ) (( ) )

( ) ( ( ) )

( ) (( ) )

yy

xx

xxyy

yy

xx

xxyy

af a

b

bf b

a

af a

b

bf b

a


n-th order homogeneous function

( ) ( )nf f x x Bold : vector or tensor

Two properties of n-th order homogeneous function

i) (n-1)-th order homogeneous function

or

ii)

f

x

( ) ( )i ij

i ij

f f fx or x nf

x x

x x

x

1 1 1

1 1

1 1 1

( )and ( )

( )

n n nn n

n n n

nx n x nxf

ny n y ny

g g gxx


Proof (1)

( ) ( ) nf fx x

( ) ( ) ( ) ( )( )

( ) ( )

( x )( x )

x

f x f f

fwhere

g

g

x x xI x

x x x x

Left hand side

Right hand side

( )( )n n

f

g

xx

x

1( ) ( ) ng gx x


Proof (2)

( ) ( ) nf fx x

( ) ( ) ( ) ( )( )

( ) ( )

f x f f

g

x x xx x x

x x

Left hand side

Right hand side

1

( )( )

n

nf

n f

xx

( )( )

fnf

xx x

x


• Linear transformation of isotropic yield surface into anisotropic yield

surface – preserve the convexity

/ 0

0 /

xx xx

yy yy

c a

c b

• Effective stress or equivalent stress –

the yield function of the first order homogeneous function


Expansion, translation, shape change of yield surface

Expansion Translation Shape change(distortion)

Appendix

Principal stress and invariants

Coordinates transformation Principal

axis

Generalcoordinates

Appendix

Principal values and directions of symmetric tensor

= Eigenvalues and eigenvectors

( ) 0

,

det 0

ij ij ior n

for the non trivial solution

σn n σ I n 0

σ I

Where λ : principal values (or eigenvalues) of tensorn : principal directions(or eigenvectors) of tensor

Appendix

Principal values & vectors and Invariants

= Do not change with any coordinates system

3 2 2

1 2 3

1 2 3

1

2 2

2

3

det 0,charactoristicequation is

0

where , , areinvariants of tensor

( )

1( ) ( )

2

det( )

from

I I I

I I I

I trace

I trace trace

I

σ I

σ

σ σ

σ

Yield function - IsotropyIsotropic generalization

1 2 3

( )

( , , , , , )

( , , )

= ( , , )

ij

I II III I II III

I II III

f

f

f as a symmetric function

f I I I

n n n

General case

Isotropic case

Denote with invariants

1 2 3

where , , :

, , :

, , : the threeinvariants of

I II III

I II III

pricipal stresses

principal directions

I I I

n n n

Yield function - Incompressibility

Hydrostatic & deviatoric stresses

• Decomposition of the stress into two components: the hydrostatic

and deviatoric components

11 22 33

1 where

3

1( )

3

:deviatric stress

ij ij kk ij kkS

trace

S I

S

11 22 33

12 1311 22 33

22 33 11

21 23

11 22 33

33 11 22

31 32

11 22 33

2

0 03

2 1

0 03 3

2

3 0 0

ij


Deviatoric strains

Similarly,1

( )3

: ( )

: ( )

( ) 0 ( )

p p p

p

p

p p

kk

d d trace d

d plastic strain increment

d deviatoric plastic strain increment

trace d de for crystaline materials

e I

e

e

Yield function - IncompressibilityHydrostatic & deviatoric stresses - example

III 33,

I 11,

II 22,

P

O

H

HL(Hydrostatic Line) (1, 1, 1)

H

S

Deviatoric plane(Octahedral plane) (1, 1, 1)

11 12 13 11 12 13

21 22 23 21 22 23

31 32 33 31 32 33

23 23 3211 11

22 22 31 31 13

33 33 12 12 21

1 0 0

0 1 0

0 0 1

1

1 with

1

S S S

S S S A

S S S

S SS

S A S S

S S S

, , S HNow from the figure

T

1 11 1

1 13 3

1 1

H

1 1

1 13

1 1

ii A

1

3A trace [111]H

11 22 33 11

11 22 33 22

11 22 33 33

21

23

2

S

S

S

S H

T

11 22 33

1

0 1 0

1

SS S S

[111]S


Yield surface for incompressible case

• For crystal materials, plastic deformation is incompressible and

the hydrostatic stress does not affect the plastic deformation

since the plastic deformation is incurred by shear stress.

0

( ) is independent of

( ) ( , ) ( )

p

kk kk

p p

ij ij

ij kk

ij ij kk ij

d for arbitrary

dW S de

f f

f f f S f f S

Isotropic & Incompressible yield surface

1 2 3

2 3

( )

( )

( , , , , , )

= ( , , ) as a symmetric function

= ( , , )

= ( , )

ij

ij

S S S

I II III I II III

I II III

f

f S

f S S S n n n

f S S S

f J J J

f J J

General case

Isotropic case

Denote with invariants

incompressible case

Since J1 = 0

1 2 3

1

2 2

2

3

where , , areinvariantsof deviatoric tensor

( ) 0

1( ) ( )

2

det( )

J J J

J trace

J trace trace

J

S

S S

S

Incompressibility

π-diagram

I II

III

A

BB

A

I( )Pd II( )

Pd

III( )Pd

O

Incompressibility

π-diagram

A way to find the deviatoric stress component: S1, S2 and S3 (with the condition

Sii=0):3,III 3,III( )S

2,II 2,II( )S1,I 1,I( )S

(2,1,0)

'(3,2,1)

(1,0, 1)SThey all share the same position on the

deviatoric plane (therefore, in the

diagram), but with different

hydrostatic component

Von Mises isotropic yield function

2 2 2 2

2 3 2

1 1 1( ) ( ) ( , ) ( ) .

2 2 2 ij ij I II IIIf f f J J J S S S S S ConstS S

(or J2 plasticity)

From above, yield function can be defined like in below

( ) ( ) ( ) ij ijf S S c S

Where 𝜎 : equivalent(or effective) stress

Reference stress state

( ) ( ) ( ) ij ijf S S cS

How can we define the size of yield function c? From reference state

I

II

( ,0,0)ST Y

2 or 2(2 , ,0)PS PLS K K

( , ,0)BB B B

3 or 3( , ,0)PS PLS K K


I II

III

2: ( , , )

3 3 3

Y Y YS

Y

: ( ,0,0)Y

I II

III

B

B

2: ( , , )

3 3 3

B B BS

: ( , ,0)B B

I II

III

K

: ( , ,0)S K K

: ( , ,0)K K

I II

III

: ( ,0, )S K K

K

: (2 , ,0)K K

Simple tension Pure shear(PS2)

Pure shear(PS3) Balanced biaxial

I

II

( ,0,0)ST Y

2 or 2(2 , ,0)PS PLS K K

( , ,0)BB B B

3 or 3( , ,0)PS PLS K K


( ) ij ijf S S

Reference state : simple tension case

For the reference state to calibrate the yield criterion is the simple tension,

1

3

0 0 2 3 0 0 3 0 0

0 0 0 0 3 0 0 3 0

0 0 0 0 0 3 0 0 3

( yield stress of the simple tension)

ij ij kk ijS

Y Y Y

Y Y

Y Y

f Y

2 2 24 1 1

9 9 9

3

2

f Y Y Y Y

3

2

ij ijS S Y


( ) ij ijf S S

Reference state : pure shear case

For the reference state to calibrate the yield criterion is the pure shear,

1

3

0 0 0 0 0 0 0

0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0

( yield stress of the pure shear)

ij ij kk ijS

K K

K K

f K

2 2

1

2

f K K K

1 1

2 3ij ijS S K Y


( ) ij ijf S S

Reference state : balanced biaxial case

For the reference state to calibrate the yield criterion is the balanced biaxial,

1

3

0 0 3 0 0 2 3 0 0

0 0 0 3 0 0 2 3 0

0 0 0 0 0 2 3 0 0 2 3

( yield stress of balanced biaxial tension)

ij ij kk ijS

B B B

B B B

B B

f B

2 2 21 1 4

9 9 9

3

2

f B B B B

3

32

ij ijS S B Y K

Von Mises yield criterion

2 22 2 2 2

11 22 22 33 33 11 12 23 31

1 6 6 62

Y

22 2 2 2 2 2

11 22 33 12 23 31 11 22 33

2 2 2 2 2

11 22 22 33 33 11 12 23

( )=

3 =

2

3 3 1 1 ( )( )2 2 3 3

3 1 2 3

3 2 2 2 1 32

1 6 6

2

ij ij

ij ij

ij ij ij kk ij ij pp ij

ij ij kk pp

f f f S

f S S

f S S

2316

In summary,

Tresca isotropic yield function

(or maximum shear criterion)

max min max min( ) ( ) ( , , )2 2

I II III

S Sf f S S S KS

2

2

2

I II I II

II III II III

I III I III

S S K

S S K

S S K

3 2 2 2 4 6

2 3 2 2

1 27 9 3J - J - 0

16 64 16 2 or K J K J K

Tresca yield function

max min max min

2 2

S Sf A A

Different reference state cases

I II

III

Y

2: ( , , )

3 3 3

Y Y YS

: ( ,0,0)Y

I II

III

B

B

2: ( , , )

3 3 3

B B BS

: ( , ,0)B B

I II

III

K

: ( , ,0)S K K

: ( , ,0)K K

I II

III

: ( ,0, )S K K

K

: (2 , ,0)K K

Pure shear(PS3) Pure shear(PS2)

Simple tension Balanced biaxial

Tresca yield functionReference state : pure shear case

For the reference state to calibrate the yield criterion is the pure shear,

max min max min

2 2

S Sf A A

max min .2

f A

0 0 0 0

0 0 0 0

0 0 0 0 0 0

K K

K K

( yield stress of the pure shear)K

( )

2

1

K Kf A K

A

max min 2 2

YK

Tresca yield functionReference state : simple tension case

For the reference state to calibrate the yield criterion is the simple tension,

max min max min

2 2

S Sf A A

0 0

0 0 0

0 0 0

Y

0

2

2

Yf A Y

A

max min ( )Y

Tresca yield functionReference state : balanced biaxial

max min max min

2 2

S Sf A A

max min 2

f A B

0 0

0 0

0 0 0

B

B

0

2

2

Bf A B

A

max min B

Drucker isotropic yield function

26 3 63

2 3 2 3 2 3

2

( ) ( ) ( , ) ( , ) (1 )J

f f f J J J J J KJ

S

• Incompressible, isotropic, and symmetric for tension and compression,

• If ξ=0 von Mises yield criterion

• So, Drucker yield criterion is extension of von Mises yield criterion, which

modifies the shape of yield surface from the von Mises yield surface by

material constant ξ

Drucker isotropic yield function2

6 3 632 3 2 3 2 3

2

( ) ( ) ( , ) ( , ) (1 )J

f f f J J J J J KJ

S

• 𝐽32/𝐽2

2 from 𝛼 = −1(pure shear) to 𝛼 = 0(simple tension)

22

2

3

3

22

3

332

2

13

2 2 1 127

2 2 1 1

27 1

I

I

J

J

J

J

• Let 𝜎1, 𝜎2, 0 → 𝜎1 1, 𝛼, 0where 𝛼 = 𝜎2/𝜎1 (−1 ≤ 𝛼 ≤ 0)


26 3 63

2 3 2 3 2 3

2

( ) ( ) ( , ) ( , ) (1 )J

f f f J J J J J KJ

S

22

2

3

3

22

3

332

2

13

2 2 1 127

2 2 1 1

27 1

I

I

J

J

J

J

1

62

32

32

2 2 1 1( , ) 3 1 1

27 1I K

• Principal stress σI can be derived as follows

( 0, ) ( )I Y B • Since and , ( 1, )I K

6 6 62 2

27 4 27 4

27 27K Y B


( , )I • By normalizing with which is von Mises yield function case, ( , 0)I

11

2 62 63

3 32

2

2 2 1 1( , )1 1

( , 0) 27 1

I

I

J

J

• From convexity condition, 27 9

8 4


• From previously driven equations, Drucker isotropic yield function can be plotted

In plane stress condition In π diagram

Non-quadratic isotropic yield functiongeneralized from von Mises yield function

1

2 2 2 2

I II II III III I S S S S S S

Von Mises yield function

generalization

1

I II II III III I

M M M MS S S S S S

Hosford non-quadratic isotropic yield function

Non-quadratic isotropic yield functiongeneralized from von Mises yield function

1

I II II III III I

M M M MS S S S S S

Hosford non-quadratic isotropic yield function in plane stress condition

Case 1 : 1 ≤ M ≤ 2 Case 2 : 2 ≤ M ≤ ∞

Hill 1948 quadratic anisotropic yield function

2 2 2 2

2 2 2

( ) F( ) G( ) H( )

2L 2M 2N

yy zz zz xx xx yy

yz zx xy

f

• Anisotropic expansion of von Mises yield criterion• This will be further discussed in Chapter 14

Von Mises

Drucker-Prager compressibleisotropic yield function

Deviatoric plane

oII

III

I

Hydrostatic Line (1, 1, 1)

oII

III

I


Deviatoric plane

1 2 3 1 2 3 2 1( , , ) ( , , ) 2 +I I I I J I J I

• Drucker-Prager yield function can be utilized for Soil, Concrete, or other hydrostatic stress dependent materials

Case 1 : β>0 Case 2 : β

Drucker-Prager compressibleisotropic yield function• Variation of Drucker-Prager yield criterion

2 12 +J I

Double cone shape

Deviatoric plane

o

III

I


II

Ellipsoid shape

2

2 12 +J I

Deviatoric plane

o

III

I


II

소성재료역학 (metal plasticity) chapter 12: yield function 8... · 2019. 9. 5. · 446.631a...

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