Tiêu đề 03 Atterberg Limits Solutions.pdf ✅

03 ATTERBERG LIMITS SOLUTIONS SITUATION 1. ▣ 1. PI = LL − PL = 41 − 21.1 = 19. 9 ▣ 2. LI = ω − PL LL − PL = 30 − 21.1 41 − 21.1 = 0. 447 ▣ 3. LI < 0 Brittle Solid 0 < LI < 1 Plastic LI > 1 Liquid 0 < 0.447 < 1 ∴ Plastic SITUATION 2. MT = 11.3 g MW = 38.51 g MD = 32.81 g V1 = 15.26 cm3 V2 = 12.83 cm3 ▣ 4. M1 = MW − MT = 38.51 − 11.3 = 27.21 g M2 = MD − MT = 32.81 − 11.3 = 21.51 g ω = M1 − M2 M2 (100%) = 27.21 − 21.51 21.51 (100%) = 26. 50% ▣ 5. SL = ω − [ (V1 − V2 )ρw M2 ] (100%) = 26.5% − [ (15.26 − 12.83)(1) 21.51 ] (100%) = 15. 2% ▣ 6. SR = M2 V2ρw = 21.51 12.83(1) = 1. 68

Cc = (D30) 2 D60D10 = (0.70) 2 2.3(0.17) = 1.25 < 3 The soil is SW, 10% < 15% gravel ∴ SW well − graded sand ▣ 10. passing 200 = 33% % retained #200 = 100 − 33 = 67% > 50% ∴ coarse − grained coarse fraction = 100 − 33 = 67% 50% coarse fraction = 67%(0.50) = 33.5% < 100% ∴ sand % passing 200 = 33% > 12% ∴ SM or SC PI = LL − PL = 46 − 29 = 17 > 7 ∴ SC clayey sands