PSI - Issue 77

Victor Rizov et al. / Procedia Structural Integrity 77 (2026) 382–388 Author name / Structural Integrity Procedia 00 (2026) 000–000

384

3

1 Q , is 1 l as shown in Fig. 1.

The distance between the application point of the bending moment and beam section,

The beam ends, 1 Q and 4 Q , are constrained with an axially double rod and rigidly fixed, respectively. Our purpose is to derive the strain energy release rate for the longitudinal crack in the beam in Fig. 1 with considering the effect of time. We use the non-linear viscoelastic model displayed schematically in Fig. 2 for dealing with the time-dependent mechanical behavior of the beam construction under consideration. The spring with modulus of elasticity, E , and the dashpot with coefficient of viscosity, η , have linear mechanical behavior, while the spring with modulus of elasticity, β E , and the dashpot with coefficient of viscosity, γ η , exhibit non-linear behavior under a stress, σ , applied on the model (Fig. 2). We use the following constitutive dependences for the components of the model in Fig. 2: α ε σ E E = , (1) where β n and γ n are material parameters characterizing the mechanical behavior of the non-linear spring and dashpot, respectively. In formulas (1), (2), (3) and (4), E σ , η σ , β σ and γ σ are the stresses in the model components, E ε , η ε , β ε and γ ε are the respective strains. We apply formulas (5) and (6) for treating the variation of the modulus of elasticity, β E , and coefficient of viscosity, γ η , with time, t f t E E e 1 0 β β = , (5) f t e 2 0 γ γ η η = , (6) where 0 β E , 0 γ η , 1 f and 2 f are material parameters. For the strain, ε , in the viscoelastic model in Fig. 2 we get (7) We describe the smooth distribution of the material parameters along the beam thickness with the help of the following formulas: 1 g α η σ ηε  = , (2) β β β β ε σ n E = , (3) γ γ γ γ σ η ε n  = , (4) γ β α ε ε ε ε = + + .

1 h E E − g NS VS NS η η −

2 h

  

  

E E

z

+

= +

VS

,

(8)

VS

g

2 h −

  

  

2

z

VS = +

+

η η

,

(9)

g

h

2

g

3 h E E β 0 g NS

2 h

  

  

3

0

VS

β

z

E E

= +

+

,

(10)

0

0

VS

β

β

g

0 NS η η − γ γ

2 h

  

  

4

0

VS

z

= +

+

η η

,

(11)

0

0

VS

γ

γ

g

h

4

where