Issue 62

V. Shlyannikov et alii, Frattura ed Integrità Strutturale, 62 (2022) 1-13; DOI: 10.3221/IGF-ESIS.62.01

E LASTIC GENERALIZED PARAMETER GPK1

T

he values of the elastic SIF K 1 were obtained in accordance with standard ASTM E399 [12] by the following equations. For the compact tension C(T) configuration:

q P

a      

K

1 Y B W W

1

2

3

4

   

   

a W

2 

a       W

a      

a      

a      

a      

0.886 4.64 

(4)

Y

13.31

14.72

5.6

1

1.5

W W

W W

a W

1 /

and for the single-edge-notched bend (SENB) configuration:

1 3 2 1 q P L a Y BW W

     

K

1

2

3

4

a       W

a      

a      

a      

a      

(5)

 

Y

1.93 3.07

14.53

25.11

25.8

1

W W

W W

where a is the crack length, B is the specimen thickness, W is the specimen width, L 1 is the span of the bending specimen, and Y 1 is the geometry-dependent SIF correction factor. The values of the P Q loads were obtained using the typical load versus load-line crack opening displacement curves for the C(T) and SENB configurations.

HRR- PLASTIC GENERALIZED PARAMETER GP K P

T

he classical HRR singular solution [13,14] for an infinite size cracked body of a strain-hardening material was completed by Shlyannikov and Tumanov as numerical method [15] for plastic stress intensity factor determination applied to mixed mode plane strain/plane stress problems and general three-dimensional (3D) structural element configurations. According to this method, the plastic SIF K p can be expressed directly in terms of the corresponding elastic SIF K 1 :     1 1 2 2 1 n K a      

yn    I

W      

0        

 

K

(6)

P

n

where α and n are the strain hardening parameters,  yn is the nominal stress,  0 is the yield stress, and I n is the governing parameter of the elastic–plastic stress–strain fields in the form of dimensionless factor:

 

  n a w d

    

FEM

FEM

( , , 

n I

n a w

, ,

  

   

FEM

FEM

  

  

  

  

du 

du 

n

  1 n e   

FEM

 

FEM

FEM FEM

FEM FEM

r

rr     

r  

n a w

u 

u 

, ,

cos

sin

r

n

d

d

1

(7)

1 

FEM FEM FEM FEM rr r r u u         

cos

n

1

4

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