HW05-CE357-Compaction_Solution_SI

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University of Washington *

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Civil Engineering

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Apr 3, 2024

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The University of Texas at Austin Department of Civil, Architectural and Environmental Engineering Homework #5 Solution CE 357 Geotechnical Engineering Problem 1. Given; Gs = 2.68, W mold = 2.38kg, V mold = 1000 cm 3 Find; γ t , γ d , w c , S, γ d ,max , w opt , Compaction curve Solution; (a) Step 1. compute weights of compacted soils: (Weight of compacted soil and mold, kg) – (Weight of empty mold, 2.38kg) = Weight of compacted soil Sample 1: 4.09kg – 2.38kg = 1.71kg Sample 2: 4.11kg – 2.38kg = 1.73kg Sample 3: 4.06kg – 2.38kg = 1.68kg Sample 4: 4.14kg – 2.38kg = 1.76kg Sample 5: 4.11kg – 2.38kg = 1.73kg Sample 6: 4.14kg – 2.38kg = 1.76kg Step 2. compute water contents: (Mass of can and moist soil, g) - (Mass of can and dry soil, g) = Mass of water, W w (Mass of can and dry soil, g) – (Mass of can, g) = Mass of solids (soil particle), W s w c = (W w ÷ W s ) × 100 (%) weight of water (g) weight of solids (g) w c = W w ÷ W s × 100 (%) Sample 1 252.1 - 211.9 = 40.2 211.9 - 21.52 = 190.38 (40.2÷190.38)×100 = 21 Sample 2 239.8 - 202.8 = 37.0 202.8 - 20.28 = 182.52 (37.0÷182.52)×100 = 20 Sample 3 245.5 - 221.6 = 23.9 221.6 - 22.13 = 199.47 (23.9÷199.47)×100 = 12 Sample 4 297.7 - 255.1 = 42.6 255.1 - 23.36 = 231.74 (42.6÷231.74)×100 = 18 Sample 5 223.8 - 199.4 =24.4 199.4 - 25.26 = 174.14 (24.4÷174.14)×100 = 14 Sample 6 277.8 - 242.3 = 35.5 242.3 - 19.74 = 222.56 (35.5÷222.56)×100 = 16 Step 3. compute total unit weight: (Weight of compacted soil) ÷ (Volume of empty mold, 1000 cm 3 ) Sample 1: 1.71kg ÷ 1000 cm 3 = 0.00171 kg/cm 3 Sample 2: 1.73kg ÷ 1000 cm 3 = 0.00173 kg/cm 3 Sample 3: 1.68kg ÷ 1000 cm 3 = 0.00168 kg/cm 3 Sample 4: 1.76kg ÷ 1000 cm 3 = 0.00176 kg/cm 3 Sample 5: 1.73kg ÷ 1000 cm 3 = 0.00173 kg/cm 3 Sample 6: 1.76kg ÷ 1000 cm 3 = 0.00176 kg/cm 3 Step 4. compute dry unit weight: 1/4

The University of Texas at Austin Department of Civil, Architectural and Environmental Engineering Homework #5 Solution CE 357 Geotechnical Engineering γ d = γ t ( 1 + w c ) Sample 1: 0.00171 kg / cm 3 ( 1 + 0.21 ) = 0.00141 kg/cm 3 Sample 2: 0.00173 kg / cm 3 ( 1 + 0.20 ) = 0.00144 kg/cm 3 Sample 3: 0.00168 kg / cm 3 ( 1 + 0.12 ) = 0.00150 kg/cm 3 Sample 4: 0.00176 kg / cm 3 ( 1 + 0.18 ) = 0.00149 kg/cm 3 Sample 5: 0.00173 kg / cm 3 ( 1 + 0.14 ) = 0.00152 kg/cm 3 Sample 6: 0.00176 kg / cm 3 ( 1 + 0.16 ) = 0.00152 kg/cm 3 Table 1. Results of question (a) Sample Total unit weight (kN/m 3 ), γ t Water content (%), w c Dry unit weight (kN/m 3 ), γ d Saturation (%), S 1 17.1 21 14.1 63 2 17.3 20 14.4 63 3 16.8 12 15.0 41 4 17.6 18 14.9 61 5 17.3 14 15.2 49 6 17.6 16 15.2 56 (b) Plot the compaction curve From the relationship, Se = G s w, e = (G s w) ÷ S γ d = G s γ w ( 1 + e ) = G s γ w ( 1 + G s w S ) , γ w = 10kN/m 3 S=100% (Zero air void curve) S=80% S=60% 2/4

The University of Texas at Austin Department of Civil, Architectural and Environmental Engineering Homework #5 Solution CE 357 Geotechnical Engineering w (%) e (S=1) γ d (S=1) (kN/m 3 ) e (S=0.8) γ d (S=0.8) (kN/m 3 ) e (S=0.6) γ d (S=0.6) (kN/m 3 ) 10 (2.68 × 0.10) ÷ 1 = 0.27 2.68 × 10 ( 1 + 0.27 ) = 21.1 (2.68 × 0.10) ÷ 0.8 = 0.34 2.68 × 10 ( 1 + 0.34 ) = 20.1 (2.68 × 0.10) ÷ 0.6 = 0.45 2.68 × 10 ( 1 + 0.45 ) = 18.5 12 (2.68 × 0.12) ÷ 1 = 0.32 2.68 × 10 ( 1 + 0.32 ) = 20.3 (2.68 × 0.12) ÷ 0.8 = 0.40 2.68 × 10 ( 1 + 0.40 ) = 19.1 (2.68 × 0.12) ÷ 0.6 = 0.54 2.68 × 10 ( 1 + 0.54 ) = 17.4 14 (2.68 × 0.14) ÷ 1 = 0.38 2.68 × 10 ( 1 + 0.38 ) = 19.5 (2.68 × 0.14) ÷ 0.8 = 0.47 2.68 × 10 ( 1 + 0.47 ) = 18.2 (2.68 × 0.14) ÷ 0.6 = 0.63 2.68 × 10 ( 1 + 0.63 ) = 16.4 16 (2.68 × 0.16) ÷ 1 = 0.43 2.68 × 10 ( 1 + 0.43 ) = 18.8 (2.68 × 0.16) ÷ 0.8 = 0.54 2.68 × 10 ( 1 + 0.54 ) = 17.4 (2.68 × 0.16) ÷ 0.6 = 0.71 2.68 × 10 ( 1 + 0.71 ) = 15.6 18 (2.68 × 0.18) ÷ 1 = 0.48 2.68 × 10 ( 1 + 0.48 ) = 18.1 (2.68 × 0.18) ÷ 0.8 = 0.60 2.68 × 10 ( 1 + 0.60 ) = 16.7 (2.68 × 0.18) ÷ 0.6 = 0.80 2.68 × 10 ( 1 + 0.80 ) = 14.9 20 (2.68 × 0.20) ÷ 1 = 0.54 2.68 × 10 ( 1 + 0.54 ) = 17.4 (2.68 × 0.20) ÷ 0.8 = 0.67 2.68 × 10 ( 1 + 0.67 ) = 16.0 (2.68 × 0.20) ÷ 0.6 = 0.89 2.68 × 10 ( 1 + 0.89 ) = 14.1 22 (2.68 × 0.22) ÷ 1 = 0.59 2.68 × 10 ( 1 + 0.59 ) = 16.9 (2.68 × 0.22) ÷ 0.8 = 0.74 2.68 × 10 ( 1 + 0.74 ) = 15.4 (2.68 × 0.22) ÷ 0.6 = 0.98 2.68 × 10 ( 1 + 0.98 ) = 13.5 24 (2.68 × 0.24) ÷ 1 = 0.64 2.68 × 10 ( 1 + 0.64 ) = 16.3 (2.68 × 0.24) ÷ 0.8 = 0.80 2.68 × 10 ( 1 + 0.80 ) = 14.9 (2.68 × 0.24) ÷ 0.6 = 1.07 2.68 × 10 ( 1 + 1.07 ) = 12.9 3/4

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The University of Texas at Austin Department of Civil, Architectural and Environmental Engineering Homework #5 Solution CE 357 Geotechnical Engineering (c) From the compaction curve, Maximum dry unit weight: 15.25 kN/m 3 , optimum water content: 15% (d) From the compaction curve, it appears that the line of optimums is approximately equal to the 60% saturation line. You would need another compaction curve for the same soil generated from higher compaction energy (e.g., Modified Proctor) to know for sure. 4/4 γ d =15.25 kN/m 3 γ d =15.25 kN/m 3 w c =15 % w c =15 %

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