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20.0 1 Milton 2 Berryman-Milton 15.0 1 3 Gibiansky-Torquato 10.0 2 5.0

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Presentation on theme: "20.0 1 Milton 2 Berryman-Milton 15.0 1 3 Gibiansky-Torquato 10.0 2 5.0"— Presentation transcript:

1 20.0 1 Milton 2 Berryman-Milton 15.0 1 3 Gibiansky-Torquato 10.0 2 5.0 3 0.0 0.0 5.0 10.0 15.0 20.0 Figure 2-1

2 1.00 1 Bristow 0.75 2 Milton 3 Gibiansky-Torquato 0.50 3 2 0.25 1 0.00 0.00 0.25 0.50 0.75 1.00 Figure 2-2

3 Equivalent for conductivity
Not equivalent for elasticity Figure 3-1

4 Reff /a =a/c Figure 4-2 0 = 0.10 0 = 0.25  0 = 0.50 0.00 0.25
0.75 1.00 0 = 0.10 0 = 0.25 0 = 0.50 Reff /a =a/c r = a r = a - c a r c s Figure 4-2

5 15 2 10 2 -3 4 5 -6 1 3 4 -9 1 3 -5 -12 0.01 0.1 1 0.01 0.1 1 B1, D1 B3, D3 1 3 B2, D2 B4, D4 2 4 Figure 4-5

6 Figure 4-6 Rigid inclusions 100 crack-like 0.01 1 100 needle-like 0.01
Soft inclusions Rigid inclusions Rigid inclusions 100 100 crack-like 0.01 1 100 needle-like crack-like 0.01 1 100 needle-like 0.01 0.01 Soft inclusions Soft inclusions Figure 4-6

7 Figure 4-7 Rigid inclusions 100 crack-like 0.01 1 100 needle-like 0.01
Soft inclusions Rigid inclusions Rigid inclusions 100 100 crack-like 0.01 1 100 needle-like crack-like 0.01 1 100 needle-like 0.01 0.01 Soft inclusions Soft inclusions Figure 4-7

8 10 1 4 5 6 5 2 3 -5 -10 A1 g -15 0.01 0.1 1.0 10 100 6 6 4 5 1 2 3 2 4 -2 A2 g -4 0.01 0.1 1.0 10 100 Figure 4-9

9  Reff /a Elasticity problem Conductivity problem Figure 4-10 1.00
0.75 0.50 Conductivity problem 0.25 0.00 0.00 0.25 0.50 0.75 1.00 Figure 4-10

10 2 0.5 2 4 1 1 3 -0.5 1 -1 2 4 3 g g -1 -1.5 0.01 0.1 1 10 100 0.01 0.1 1 10 100 2 0.5 2 4 1 1 1 3 -0.5 2 4 g g 3 -1 -1 0.01 0.1 1 10 100 0.01 0.1 1 10 100 Figure 5-1

11 g g g g 3 4 2 1 1 2 4 3 Copper reinforced with diamond particles 4 2 3
0.6 0.1 4 3 0.4 0.0 2 0.2 1 -0.1 1 0.0 -0.2 2 -0.2 -0.3 4 3 g g -0.4 -0.4 0.01 0.1 1 10 100 0.01 0.1 1 10 100 Copper reinforced with diamond particles 0.6 0.1 4 0.4 0.0 2 0.2 3 1 -0.1 0.0 1 2 -0.2 -0.2 3 g 4 g -0.4 -0.3 0.01 0.1 1 10 100 0.01 0.1 1 10 100 Poly(phenylene sulfide) reinforced with glass particles Figure 5-2

12 g g 1 1 2 2 Poly(phenylene sulfide) reinforced with glass particles
0.8 1 0.6 1.5 0.4 1 1 2 0.2 2 g g 0.5 0.01 0.1 1 10 100 0.01 0.1 1 10 100 Poly(phenylene sulfide) reinforced with glass particles Porous aluminum Exact connection 1 Coefficient at Young’s modulus Coefficient at shear modulus Approximate connection 2 Figure 5-3

13 1.5 1.5 4 4 1.0 1.0 (a) (b) 0.5 0.5 1 1 0.0 0.0 2 3 2 3 -0.5 -0.5 1 2 3 4 1 2 3 4 1.5 1.0 (d) 4 1.0 (c) 0.5 0.5 1 0.0 2 3 -0.5 0.0 1 2 3 4 0.0 2.0 4.0 Figure 5-4

14 1.5 1.5 1.0 1.0 1 1 2 2 (a) (b) 0.5 0.5 4 4 0.0 0.0 3 3 -0.5 -0.5 1 2 3 4 1 2 3 4 1.5 1.0 1 2 (c) 0.5 4 0.0 3 -0.5 1 2 3 4 Figure 5-5

15 1.00 1.00 E1/E2 E1/E2 0.80 0.90 k3 / k0=0.6 k3 / k0=0.8 0.60 0.80 k1/k2 k1/k2 0.70 0.40 0.80 0.85 0.90 0.95 1.00 0.60 0.70 0.80 0.90 1.00 1.00 1.00 E1/E3 E1/E3 0.80 0.80 0.60 k3 / k0=0.8 k3 / k0=0.6 0.60 0.40 k1/k3 k1/k3 0.20 0.40 0.50 0.60 0.70 0.80 0.90 1.00 0.60 0.70 0.80 0.90 1.00 g = 0 (cracks) g = 3 (prolate pores) g = 1/3 (oblate pores) g (cylinders) Figure 5-6

16 E1 / E3 E1 / E3 3.0 1.5 k3 / k0=0.5 k3 / k0=0.7 2.0 1.0 1.0 0.5 (a) (b) 0.0 0.0 0.5 1.0 1.5 2.0 0.0 0.5 1.0 1.5 k1 / k3 k1 / k3 E1 / E3 1.5 g = 0 (cracks) k3 / k0=0.9 1.0 g = 1/3 (oblate pores) g = 3 (prolate pores) 0.5 g (cylinders) (c) 0.0 0.0 0.5 1.0 1.5 k1 / k3 Figure 5-7

17 1.0 0.0 -1.0 -1.0 0.0 1.0 Figure 6-6

18 Figure 7-1

19 1.0 0.9 0.8 0.7 0.6 0.5 p 0.4 0.5 0.6 0.7 0.8 0.9 1.0 Figure 7-2

20 E/E0 p Figure 7-3 0.2 Direct measurements Prediction via
cross-property connection 0.1 p 0.0 0.6 0.7 0.8 0.9 1.0 Figure 7-3

21 30 60 1.5 1.6 0.5 1.0 2.0 2.0 2.2 0.5 1.0 1.5 20 40 10 20 γ γ 0.01 0.1 1 10 0.01 0.1 1 10 Figure 7-4

22 p Figure 7-5 4 Non-interaction approximation Mori-Tanaka method 3
Levin-Kanaun method Differential scheme 2 Formula (6.4) Experiment 1 p 0.5 0.6 0.7 0.8 0.9 Figure 7-5

23 k, x106S/m E, GPa 25 40 20 30 15 20 10 10 5 p p 0.5 0.6 0.7 0.8 0.9 1.0 0.5 0.6 0.7 0.8 0.9 1.0 Non-interaction approximation Differential scheme Mori-Tanaka method Experiment Levin-Kanaun method Figure 7-6

24 E, GPa 70 65 60 55 10 100 1000 Number of cycles Figure 7-7 1 2 3 15%
20% 25% 65 1 2 60 3 55 10 100 1000 Number of cycles Figure 7-7

25 (a) (b) (c) (d) Figure 7-8

26 1.0 2.5 1 3 (a) (b) 0.9 2.0 2 2 1 4 0.8 1.5 3 0.7 1.0 5 10 15 20 25 5 10 15 20 25 Aging time (days) Aging time (days) 1 1 2 2 3 3 4 Figure 7-9

27 0.0 0.0 1 1 -0.1 -0.2 3 -0.2 -0.4 3 2 2 -0.3 -0.6 -0.4 (a) (b) -0.8 -0.5 5 10 15 20 25 5 10 15 20 25 Aging time (days) Aging time (days) 0.0 1 Measured 2 Alternative cross-property connections (6.6b) -0.05 2 1 -0.10 3 Cross-property connection (5.9b) 3 (c) -0.15 5 10 15 20 25 Figure 7-10 Aging time (days)

28 r0 a) b) rint rext Figure 7-12

29 1.0 0.8 0.6 0.4 0.2 (a) 0.0 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.0 0.8 0.8 0.6 0.4 0.4 0.2 (c) (b) 0.0 0.0 0.0 0.2 0.4 0.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 Figure 8-1

30 H R x3 x2 x1 Figure 8-2

31 (b) (a) Figure 8-3

32 C b a Figure 8-4

33 A B Figure 8-5

34 Porous aluminum (numerical simulation)
, MPa , MPa 50 1500 ( a ) ( b ) 4.8 % 11.3 % 15.0 % 40 1000 30 dense 7.6 % 20 500 10 x 10-3 0.0 0.4 0.8 1.2 1.6 0.00 0.02 0.04 0.06 0.08 , MPa 800 ( c ) Porous aluminum (numerical simulation) (b), (c) Dense and porous Ti-6Al-4V (experiment) 600 400 dense 7.6 % 200 0.00 0.05 0.10 0.15 Figure 9-1

35 Slight shift to the right
Increasing pore volume fraction Slight shift to the right Figure 9-2

36 (a) (b) k1/k0 k3/k0 Isotropy p k1/k0 Figure 9-3 1.0 1.0 0.9 0.8 0.8
0.6 0.7 Isotropy 0.4 0.6 (a) (b) p k1/k0 0.2 0.5 0.6 0.7 0.8 0.9 1.0 0.00 0.15 0.30 Figure 9-3

37 (a) (b) k3/k0=0.7 k3/k0=0.8 k1/k0 k1/k0 (c) A1 A4 A2 A5 k3/k0=0.9 A3
-1 -2 2 1 -1 -2 (a) (b) k3/k0=0.7 k3/k0=0.8 1.0 0.9 0.8 0.7 0.6 k1/k0 1.0 0.9 0.8 0.7 0.6 k1/k0 2 1 -1 -2 (c) A1 A4 A2 A5 k3/k0=0.9 A3 1.0 0.9 0.8 0.7 0.6 k1/k0 Figure 9-5

38 (a) (b) k3/k0=0.7 k3/k0=0.8 k1/k0 k1/k0 (c) A1 A4 k3/k0=0.9 A2 A5 A3
-2 -1 1 2 -2 -1 1 2 (a) (b) k3/k0=0.7 k3/k0=0.8 1.0 0.9 0.8 0.7 0.6 k1/k0 1.0 0.9 0.8 0.7 0.6 k1/k0 -2 -1 1 2 (c) A1 A4 k3/k0=0.9 A2 A5 A3 1.0 0.9 0.8 0.7 0.6 k1/k0 Figure 9-6

39 4.0 40  = 0.3  = 1.0  = 4.0  = 0.3  = 1.0  = 4.0 3.0 30 A1 A2 20 2.0 1.0 10 0.0 0.7 0.8 0.9 1.0 0.6 0.5 0.7 0.8 0.9 1.0 0.6 0.5 Figure 9-7

40 1.5 20  = 0.3  = 1.0  = 4.0  = 0.3  = 1.0  = 4.0 15 1.0 A1 A2 10 0.5 5 0.0 0.7 0.8 0.9 1.0 0.6 0.5 0.7 0.8 0.9 1.0 0.6 0.5 Figure 9-8

41 (a) (b) A1 A1 Figure 9-9 Normal distribution of shapes with
1.5 6 (a) (b) 5 1.0 A1 A1 4 3 0.5 2 0.0 1 0.6 0.8 1.0 0.5 0.7 0.9 0.6 0.8 1.0 0.5 0.7 0.9 Normal distribution of shapes with Spherical pores Figure 9-9


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