Issue 41

F. Berto et alii, Frattura ed Integrità Strutturale, 41 (2017) 260-268; DOI: 10.3221/IGF-ESIS.41.35

1000

   rr  r  FE results

t=1mm d/t=1  =  90°

Theoretical solution considering the parameters a 1 ,…, a 7 Theoretical solution considering the parameters K I , K II and T-stress

100

10

Stress components [MPa]

1

0.001

0.010

0.100

1.000

r [mm]

Figure 4 : Stress field along the direction  =-90° for the case d/t=1.

p W x 10 3 (MJ/m 3 )

e W x 10 3 (MJ/m 3 )

 (MPa)

10 20 40 80

1.120 4.481 17.93 71.72 135.5

1.120 4.485 18.07 79.68

110 171.8 Table 3: Mean values of SED as determined from linear elastic and elasto-plastic analyses (control volume R 0

=0.28 mm).

L IMITATIONS TO THE LINEAR ELASTIC APPROACH

L

et us assume for the material the Ramberg-Osgood law, according to which the uniaxial tensile strain  is related to the uniaxial stress  according to the expression:

n

'

          ' E K 

(10)

where 1  n’   . The analyses were carried out under plane strain conditions by introducing into the stress-strain curve E, K’ and n’=6.66. Tab. 3 summarizes the values of the SED parameter as a function of the applied  t is interesting to observe that for values of  < 40 MPa the elastic and elastic-plastic values are close each other. When  MPa,  that is a typical value for this kind of joints at 10 5 cycles to failure, the SED under elastic conditions is significantly different from the elastic-plastic case. The different role played by plasticity at different number of cycles can also be used as justification of the different slope shown by thin lap joints under shear loading and welded joints under tensile loading.

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