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Introduction Characteristic properties of granular matters They have a liquid and solid region. "https://www.youtube.com/watch?v=50_-zqsgda4" In the liquid region we use the viscosity η. σxy η := γ σxy: shear stress shear rate γ: y Shear Velocity In the solid region we use the shear modulus G. G := σxy γ γ: shear strain Ly!2 γ = yvx x δx Shear γ = δx/ly
G = G + ig. σ G = lim γ γ G G = ω lim γ 0 (σ 0) σ γ, η = G ω G (, μ) G (ϕ ϕ J ) 1/2 ( 0) ΔZ = Z ΔZ 3 = Z 3 C. S. O Hern, et al., Phys. Rev. Lett. 88, 7 (2002). E. Somfai, et al., Phys. Rev. E 75, 020301 (2007). M. Otsuki & H. Hayakawa, Phys. Rev. E 95, 062902 (2017). G ΔZ ( 0) G/G (lin) = ( μ b 1(ϕ ϕ c ) b 2 )!3
f ij = ( f ij,n + f ij,n = k n ξ ij f ij,t = { k t ζ ij n ij η n μ c f ij,n f ij,t ) Θ(r ij x ij ) t ij η t t ij v ij,n r i v ij,t ( f ij,t < μ c f ij,n ) (otherwise) ω i t ij ξ ij ω j n ij r j S. Luding, Granular Matter 10, 235 (2008). η n / mk n = 1 (restitution coefficient 0.043) k t /k n = 0.25 η t /η n = 0.5 μ c = 0 1!4
Oscillatory shear system We explain how to apply oscillatory shear. Lx P A sin(ωt) In 2 dimensional N = 4000 (tetradisperse) 2 10 5 P/kn 6 10 2 1 10 6 γ0 1 A γ0 := (A : amplitude) Lx /2 A sin(ωt) y x P We press the walls with a pressure P and they move according to ± A sin(ωt)!5
= 6 10 1, P/k n = 2 10 3. = 6 10 1, P/k n = 2 10 3.!6
1 10 1 G = lim γ σ γ. ( σ = σ σ ) μ = 1 μ = 1 G = ω lim γ 0( σ 0) σ γ. ( σ = σ σ ) G /G res 1 10-5 1 10-6 P=2 10-5 G res = lim 0 /P β 1 1 1 10 2 1 10 4 G > 0, β 1 = 1. M. Otsuki & H. Hayakawa, Phys. Rev. E 95, 062902 (2017). G P (ϕ ϕ J )? P M. Otsuki & H. Hayakawa, Phys. Rev. E 80, 011308 (2009).!7 G /P α 1 1 10-5 1 10-6 1 10-7 P=2 10-5 /P β 1 1 1 10 2 1 10 4 α 1 = 0.2, β 1 = 1.05. G
G G = lim γ σ γ. ( σ = σ σ ) G = ω lim γ 0( σ 0) P = 2 10 3 1 µ=0 µ=0.1 µ=0.25 µ=0.5 µ=0.75 µ=1 1 10-6 G G G μ=0 μ=0 μ G μ α μ 0 1 10-1 σ γ. ( σ = σ σ ) P = 2 10 3 /μ β μ γ 1 0 µ=0 µ=0.1 1 10-5 1 10-5 µ=0.25 µ=0.5 µ=0.75 1 10-6 µ=1 1 10-6 1 10-4 1 10-2 α μ = 0.3, β μ = 0.75. μ G G μ 0.1 μ 0.1!8
G γ 1 0 σ σ / lim 0 1 10 5 1 10 4 1 10 3 1 10 2 1 10 1 P=2 10-5 1 10-6 σ(t) σ(t) P = 2 10 2 γ(t)/ P = 2 10 4 γ(t)/ max σ(t) const ( >,c ). G = lim γ σ γ 1!9
G < 0 P = 2 10 2 σ(t) P = 2 10 4 γ(t)/ω G eff /Pα 1 1 10-5 1 10-6 1 10-7 P=2 10-5 1 1 10 2 1 10 4 σ(t) γ /P β 0 1 γ(t)/ω G < 0 G G eff = ω α 1 = 0.2, β 1 = 1.05. lim γ 0( σ 0) eff σ γ. ( σ = σ σ )!10
G G G (Z) G /k n G eff /k n Z 1 10-5 1 10-6 P=2 10-5 1 10-6 5.5 5 4.5 4 1 10-5 1 10-6 1 10-7 1 10-8 P=2 10-5 1 10-6 3.5 3 2.5 2 P=2 10-5 1.5 1 10-6!11
G ϕ<ϕj > 0.!12 1 10 1 UJ DI 1 10-5 1 10-6 P=2 10-5 0.74 0.76 0.78 0.8 0.82 0.84 0.86 0.88 0.9 ϕ J
G G /G res 1 10 1 1 10-5 1 10-6 /P β 1 1 1 10 2 1 10 4 G eff /Pα 1 1 10-5 1 10-6 P=2 10-5 1 10-7 P=2 10-5!13 /P β 1 1 1 10 2 1 10 4
!14
G /P α 0 1 10 1 1 10-5 G = lim γ σ γ. /P β 0 α 0 = 0.5, β 0 = 1. G ( σ = σ σ ) μ = 0 μ = 0 P=2 10-5 1/2 1 1 10 2 1 10 4 G /P α 0 M. Otsuki, H. Hayakawa, Phys. Rev. E 90, 042202 (2014). M. Otsuki, H. Hayakawa, Phys. Rev. E 95, 062902 (2017). P!15 1 10-5 1 10-6 P=2 10-5 1 10 2 1 10 4 γ /P β 0 0 α 0 = 0.35, β 0 = 1.2. P (ϕ ϕ J )? M. Otsuki, H. Hayakawa, Phys. Rev. E 80, 011308 (2009). 1
1 10 1 SJ J 1 10 1 SJ J UJ 1 10-5 1 10-6 P=2 10-5 0.74 0.76 0.78 0.8 0.82 0.84 0.86 0.88 0.9 ϕ UJ 1 10-5 1 10-6 P=2 10-5 0.8 0.82 0.84 0.86 0.88 0.9 ϕ G ϕ<ϕj > 0.!16
G res G res := lim 0 G. 0.55 0.5 0.45 0.4 G res a log P + b, G res 0.35 0.3 0.25 a = 0.04 ± 0.01, b = 0.62 ± 0.03. 0.2 0.15 0.1 P G res!17