PSI - Issue 6
N.S. Selyutina et al. / Procedia Structural Integrity 6 (2017) 77–82 Author name / Structural Integrity Procedia 00 (2017) 000 – 000
81
5
ds
t t y t 0
s ( ) ( )
1
1.
(7)
Let us define the actual stress ( ) t in a deformed specimen by the following form
( ) 2 ( ) ( ), t g t t
(8)
where is an “effective” shear stress that takes into account the plastic relaxation; is a scalar 0 corresponds to the plastic deformation without hardening. Considering the stages of elastic and plastic deformations separately, we can obtain from (8) the following stress-strain relation in the case of a linear growth of strain ( ) ( ) t H t t (the constant strain rate) parameter ( 1 0 ) characterizing the level of hardening; the case ( ) t ( ) g t G 1
2 ( ), G t
* t t t t t
,
2
( ) ( ), t
( ( )) t
(9)
1
G
,
*
t
( ) / t
where we take into account that
.
3. Discussion
The deformation curves of aluminum alloy 2519 A Liu et.al. (2014), plotted on the Johnson-Cook model (Johnson and Cook (1985)) and of the relaxation model of plasticity (Petrov and Borodin (2015), Selyutina et. al. (2016)), are presented in Fig. 3. The relaxation model of plasticity describes dynamic effect plasticity (yield drop phenomenon) at 5542 s -1 . The experimental data of aluminum alloy on a hardening stage is a good correspondence with prediction of the relaxation model of plasticity comparison with classical Johnson-Cook model. Thus, the relaxation model of plasticity is an efficient and convenient tool for calculations in a much wider range of strain rates.
Fig. 3. Prediction of stress-strain curves of aluminum alloy 2519 A Liu et al. (2014) at room temperature for strain rates 0.001 s -1 (3,4 curves) and 5542 s-1 (1,2 curves): the experimental data are shown by lines with symbols (rhombus – 0.001 s -1 , triangles – 5542 s -1 ) and the calculation results on the basis of modified Johnson-Cook model (2) and of the relaxation model of plasticity (5) – (8).
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