Issue 49

M. Bannikov et alii, Frattura ed Integrità Strutturale, 49 (2019) 383-395; DOI: 10.3221/IGF-ESIS.49.38

3 2 1 3 2 l l 

2 2 1 3 2 l l 

1 2 1 3 2 l l 

1

 

 

 

 

where:

,

,

,

; E - the unit tensor, d

σ and s

σ - the deviator and spherical

4

2

3

1

l

l l

l l

l l

4

m F , 1 c - 4 c – potential approximation constants, G – shear modulus; λ – the first parameter

parts of the stress tensor;

Lame; : p  p p . In general, kinetic coefficients depend on state parameters and are representable as follows:

1

 

i

i G 

where i  - the characteristic relaxation times, which in general can be represented as dependencies:

( , , , , , , , , ) p p δ T σ ε ε ε ε p p   

U

0

i

i 



i 



exp(

)

,

kT

where: k is the Boltzmann constant,

( ) i U  is the characteristic activation energy. The hypothesis is accepted that for the

studied processes the relaxation times have the following dependencies:

0 n U ε U kT  n U p U kT    0 i 4 0

i  

,

0

1, 2, 3 i 

i 

i 



exp(

)

(20)

0

0









exp(

)

(21)

4

4

2 : 3 c ε 

ε ε  



1 c ε   s-1 is the dimensionless factor; 0

U kT  , i

U , 1...3 i  are the constants that have the meaning

where: 0 ε 

,

of the characteristic energy of overcoming barriers, after which new relaxation mechanisms are activated; 0 : c p ε  p p     ; 1 2 3 , , , n n n n  - constants, characterizing the rate sensitivity of the material. The meaning of this hypothesis is that the physicomechanical characteristics of the material depend on the strain rate, and the evolution of damage under cyclic loading with small stress amplitudes depends only on the parameter, which in this case can be interpreted as the growth rate of a fatigue crack, related to its characteristic scale. Assuming a weak strain rate sensitivity of the relaxation times, we can expand the expression (20) into a Taylor series with an accuracy of the second term:

0 n

0 n

0 n

0 n

0 i U ε U U ε U U ε U U ε kT kT kT kT kT             0 0 ) 1 1 i i i i i i

i

exp(

then we get:

U

0

i 

i 

kT

1

0



 

i 

i 

0 n

0 n

0 n

i U ε 

i U ε 

ε



i

i

i

kT

Similarly, for the expression (21):

390