PSI - Issue 13

Milan Micunovic et al. / Procedia Structural Integrity 13 (2018) 2158–2163 Micunovic, Kudrjavceva / Structural Integrity Procedia 00 (2018) 000 – 000

2162

5

{ tr D } { tr  T D D T D (11) for “rigid” plastic materials (i.e. materials where elastic strain is so many times smaller than the elastic strain that it may be neglected). In above equality t d D T is the time rate of true stress deviator whereas D P is the plastic stretching tensor. Let the anisotropy direction be along the x 1 axis. Then, we will analyze three stress histories: (a) longitudinal uniaxial stress T 11 ≡  1 , (b) transverse uniaxial stress T 22 ≡  2 and (c) shear stress T 12 = T 21 ≡  . QRI approach Let b 1 + b 2 ≡ b¯ . Then, inserting (4) into the local condition (6) we arrive at the equations }, t d P P d P

   

   

   

  

36exp( ) M

for longitudinal uniaxial stress,

D

1   

0



1

t

1

2

Y

(6 2 3 a b a

3 1 4 ) b

  

1 

 



0

1

2 1

2 Y           0 1



1        

,

for transverse uniaxial stress with

t D

0

2

 



2

18 exp( ) (6 2 3 M a b a

b  

3 2 5 ) ,

 

  

where 1

1

2 2

2

2 (36 36 

2    2 2 9 a

a a a

a b a a b

24 24 60

 

2 



  

2

1

1 2 2

1

1

1 3 2

2

2 8 8

2 2 3 2 22 ) , b 

a b

a b

b bb

12 

30







 

3 2 



2 2

2 3 2

   

   



   

  

3exp( ) M

3



for shear.

D

3

1   

0





t

3

Y

2 3 a b 

(3

)



0

It is noteworthy that in the above three stress history cases we have to solve algebraic equations. This is important and compatible with QRI modeling which covers strain rate in the quite wide range of six orders of magnitude 3 [10 ,  (cf. [11]). Thus we do not need to solve any differential equation. Classic J 2 approach As already mentioned, by comparing (9) and (10) we see that 1 1 2 1/ , 2 / a h b R h   . Now, inserting (10) into the local condition (11) we arrive at the following differential equations: 2 2 1 1 1 1 1 1 1 2 1 9 (6 4 ) 18 24 8 0 t D a b a a b b        for longitudinal uniaxial stress, 2 2 2 2 1 1 1 1 1 1 2 2 9 (6 ) 18 6 0 4 t D a b a a b b b ac         for transverse uniaxial stress and 1 2 (18 ) 0 t D a    for shear. If the material functions , h R are known, then t he above three differential equations could be solved. It should be underlined here that the J 2 approach allows diffuse instability only for small strain rates.