ICF13B

13th International Conference on Fracture June 16–21, 2013, Beijing, China -4- In this case the equation of motion (2) has the exact solution [8]: ( ) ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = 0 0 sin t t h t h π , (13) where k mV h 2 0 = and k m t = 0 . Also, there is the similar expression for maximum tensile stress: ( ) ( ) 2 2 1 2 R P t t π ν σ − = , 0 → + r R (14) After substituting the expression of tensile stress (14) into criterion (6), the integral attains its maximum value at time ( ) 2 0 τ = + t t .The dimensionless quantities are introduced by analogy with the case of the spherical particle. And the following expressions of dimensionless velocity can be obtained by using criterion (8) and formulas (12) - (14): ( ) ( ) [ ] ∫ + − ⎟ − − = ⎠ ⎞ ⎜ ⎝ ⎛ 2 1 2 1 1 sin λ λ λ λ π α ds H s H s s Vd d , (15) where ( ) ( ) ( ) ( ) 1 2 1 2 2 2 3 1 1 2 8 − − = − cr p d E c σ ν π ρ ν α is dimensionless parameter, ρ- is material density of the particle. From the expression of the impact duration it is possible to calculate the value of radius in dimensionless form: d dR β λ = , (16) where ( ) ( )1 2 2 2 3 1 2 E cp d ν ρ π π β − = is dimensionless parameter. The threshold energy can be calculated by formula (10) and in dimensionless form by formula (11), where cylinder velocity and radius are determined by (15) and (16) respectively. Figure 2 presents the graphs of dependence of the energy on impact duration (Fig. 2a) and radius (Fig. 2b), where the half-space material is zinc and parameters the value 3 / 3200kg m =ρ . Figure 2. Dependence of the threshold energy of the cylindrical particle on a – the impact duration, b – particle radius Fig. 2 shows that in case of cylindrical particle the threshold energy dependence is a monotone increasing function in contrast with the case of the spherical particle. a dW τ λ / 0t = 0 2 4 6 8 10 5 10 3− × 1 10 2− × 1.510 2− × 0 1 2 2 10 3− × 4 10 3− × 6 10 3− × 8 10 3− × 1 10 2− × b dW dR

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