PSI - Issue 13

B.A. Stratula et al. / Procedia Structural Integrity 13 (2018) 1402–1407 Author name / Structural Integrity Procedia 00 (2018) 000 – 000

1404

3

12 1 2         , 2 3         . We define the orientation of the critical plane with the components of the normal: 2 1 1 0 x n   , 2 2 2 0 x n   , 2 3 3 0 x n   , 1 2 3 1 x x x    . It can be shown that the problem of determining the critical plane for a triaxial stress state reduces to determining the maximum of the function     2 2 2 2 3 12 2 13 3 12 2 13 3 1 12 2 13 3 ( , ) 2 2 F F F x x x x x x x x                     under constraints 2 3 0 1 x x    , 2 0 x  , 3 0 x  . Let us present the results of solving this problem for all possible values of the maxima and variations of the principal stresses. I. If 12 12 13 13 / /        , 12 0    , 13 0    , then I-a. For the case 12 13      conditions of extremum for 2 3 ( , ) F x x : 13 1 3         , 23

x

2       , 3 3 x 12 23 / ( )

13 23       / (

)

2

2 3 ( , ) F x x :

0       conditions of extremum for

I-b. For the case

12

13

2 3       12 / ( ) S x x

Here is denoted

  13 12 2 0      

12 F   4

12     , 13 12 /

4 F  

/     ,

13

13





13 T       12 2

12 13 S          , 13 12

  2 2 2 / S T            , 2 13 / S





2 2 S T S           2 3 12 / /

12 13 0       , 12 13    .

12        then 13 13 / /

II. If 12

2 3 ( , ) F x x :

Conditions of extremum



 2

x x

12 1 / 4 

/ 2

  





2 3

12

Found values ( 2 x , 3 x ) should satisfy inequalities

2 2 3 1 x x   and conditions of maximum 0 x  , 3 0 x  ,

2 12 13 23 3 0 x         4

2 13 12 23 2 0 x         . 4

2 3 ( , ) F x x :

,

These inequalities follow from the condition that the quadratic form is negative 2 2 2 2 2 2 2 3 3 2 2 2 3 2 3 ( ) 2 ( ) 0 F F F dx dx dx dx x x x x           If such values do not exist, then it needs to find maximum

2 3 ( , ) F x x on boundaries: 2 0 x  , or 3 0 x  , or 2

3 1 x x   .

2 3 ( , ) F x x is reached for values:

III. In this case maximum





2

0 x  ,

x

13 1 / 4 

/ 2

 





3

13

2





2

3 0 x  ,

x

12 1 / 4 

/ 2

 





2

12