Title HPGE Solutions.pdf ✅

01 Soil Properties Situation 1 ▣ 1. Moist Unit Weight of the Soil (kN/m3 ) [SOLUTION] Formula for Moist Unit Weight γm = Wm V γm = mmg V γm = (30.6 kg) (9.81 m s 2 ) 0.0183 m3 γm = 16. 4 kN m3 ▣ 2. Dry Unit Weight of the Soil [SOLUTION] Formula for Dry Unit Weight γd = Wd V = mdg V γd = (27.2 kg) (9.81 m s 2 ) 0.0183 m3 γd = 14. 6 kN m3 ▣ 3. Moisture Content [SOLUTION] Formula for Moisture Content ω = Wm − Wd Wd = Mm − Md Md
ω = 30.6 kg − 27.2 kg 27.2 kg ω = 0. 125 = 12. 5% ▣ 4. Void Ratio [SOLUTION] Formula for Moist Unit Weight γm = G(1 + ω) 1 + e ∗ γw 16.4 kN m3 = 2.65(1 + 0.125) 1 + e ∗ (9.81 kN m3 ) e = 0. 783 ▣ 5. Porosity [SOLUTION] Formula for Porosity given Void Ratio η = e 1 + e η = 0.783 1 + 0.783 η = 0. 439 ▣ 6. Degree of Saturation [SOLUTION] Relationship of Soil Properties Se = ωG S ∗ (0.783) = (0.125)(2.65) S = 0. 423 = 42. 3%
Situation 2 ▣ 7. Calculate the natural water content of the sample in percent. [SOLUTION] Formula for Water Content ω = Mm − Md Md ω = 1526 g − 1053 g 1053 g ω = 0. 4492 = 44. 92% ▣ 8. Calculate the void ratio in percent. [SOLUTION] Relationship of Soil Properties Se = ωG Fully Saturated, S = 1 1(e) = (0.4492)(2.7) e = 1. 213 = 121. 3% ▣ 9. Calculate the porosity in percent [SOLUTION] Formula for porosity η = e 1 + e η = 1.213 1 + 1.213 η = 0. 5481 = 54. 81%
Situation 3 ▣ 10. What is the void ratio of the soil sample? [SOLUTION] Given: ρ = 1600 kg m3 ; ρs = 2500 kg m3 ; ω = 12% Formula for soil unit weight ρ = ρs (1 + ω) 1 + e 1600 kg m3 = (2500 kg m3 ) (1 + 0.12) 1 + e e = 0. 75 ▣ 11. What is the degree of saturation? [SOLUTION] Specific Gravity G = ρs ρw G = 2500 kg m3 1000 kg m3 G = 2.5 Relationship of Soil Properties Se = ωG S ∗ (0.75) = (0.12)(2.5) S = 0. 40 ▣ 12. What is the dry unit weight in kg/m3 ? [SOLUTION] Dry Unit Weight