PSI - Issue 14

Filin V.Yu. et al. / Procedia Structural Integrity 14 (2019) 758–773 Filin V.Yu, Ilyin A.V. / Structural Integrity Procedia 00 (2018) 000–000

765

8

The task of assessment is material selection for a welded joint in respect of fracture toughness.

Figure 3 – Welded joint of a pressure vessel. Assessment points are shown by arrows.

(i) We determine the yield strength at the temperature T d ,

5

491 1.8 10 

189 . (ii) We decide the shape and size of a reference flaw (normally the nondestructive inspection type and resolution are considered, hints are included into the general procedure), here the flaw depth is taken equal to a = 0.65 S 0.5 [mm, mm], and the length l = 3 a . These dimensions are recommended after the statistical analysis of nick-break test data for production welds and a position that a chosen flaw shape gives the maximum SIF at its deepest point among all flaws of the same area. When the scanning sensitivity of nondestructive inspection is proportional to the thickness S , it also appears proportional to a flaw area and this gives such a change of measurement units. After that a nondestructive inspection error is added, e.g. 0.5 mm. Totally we have a = 9.7 mm, l = 29.1 mm. (iii) We find the geometry factor and relative load from formula (15):  = 0.0033, L r = 0.73. (iv) We find SIF related to service stress, Y Y a K t t b b π ) + ( d 1    , (21) where Y-functions are taken from known nomographs, in the outcome we have Y t = 0.80, Y b = 0.74, 59 MPa m d 1  K . (v) We find SIF related to residual stress (20 C) Y Y       d T , MPa 522 Y  

r a Y σ π aY Y

,

(22)

K



1res

where Y-functions for the examined flaw size and position are



1.65 1/ 2

   

   

l a

  

,

(23)

(1  

/ ) 1 4.6

Y

a S

   

a

that is an approximation of a complete elliptic integral of the second kind given in PNAE G-7-007-86 “Equipment and pipelines strength analysis norms for nuclear power plants”, 1.6exp( 8.1 / ) r a S Y   , (24) that is an approximation of FEM simulation results for a weld toe flaw in account of residual stress field change as a crack grows, so we have Y r =1.08, Y a =0.75, 74 MPa m 1res  K . (vi) We find the required J-integral as     2 2 1res 1 d 2 1 2 ) (1 f f K K E J          , (25) where r L f 1 0.67 1   , (26) that accounts for residual stress decrease under a high load, f 2 is taken from (18), so we have f 1 = 0.51, f 2 = 1.25,