Title HPGE 2 Solutions.pdf ✅

B = (5736 m3 ) ( P 760 m3 ) = P 4,359,360 C = (6175 m3 ) ( P 720 m3 ) = P 4,446,000 The most economical is the borrow pit with the least cost, which is Borrow Pit B. ▣ 4. How much is needed to acquire soil from that borrow pit? [SOLUTION] B = P 4,359,360 SITUATION 3. A sand cone equipment is used to perform a density test on a compacted earth fill. The soil sample dug from the test hole was 20.6 N. The sample has a dry weight of 17.92 N. Ottawa sand weighing 16.05 N and having a density of 15.74 kN/m3 was used to fill the hole. Determine the following: ▣ 5. Water content of the tested soil. [SOLUTION] ω = wm − wd wd ω = 20.6 − 17.92 17.92 ω = 14.9% ▣ 6. In-place dry density of the soil. [SOLUTION] From the unit weight formula V = W γ V = wsand γsand = wsoil γsoil 16.05 N 15.74 kN m3 = 17.92 N γsoil γsoil = 17.57 kN/m3 ▣ 7. Percentage of compaction if the dry unit weight at optimum water content is 18.1 kN/m3 .
[SOLUTION] R = γd γd(optimum) R = 17.57 kN m3 18.1 kN m3 R = 97% SITUATION 4. A sample of dry, coarse-grained material was taken through a layer of sieves and the following results were obtained: Sieve No. Opening (mm) Mass of soil retained (g) 4 4.76 46 10 2 165.6 20 0.84 92 40 0.42 55.2 60 0.25 23 100 0.149 36.8 200 0.074 23 Pan - 18.4 ▣ 8. Determine the effective particle size. [SOLUTION] Sieve No. Opening (mm) Mass of soil retained (g) Mass of soil finer (g) % Finer 4 4.76 46 414.00 90 10 2 165.6 248.40 54 20 0.84 92 156.40 34 40 0.42 55.2 101.20 22 60 0.25 23 78.20 17 100 0.149 36.8 41.40 9 200 0.074 23 18.40 4 Pan - 18.4 By interpolation, logD10 − log 0.149 log 0.25 − log 0.149 = 10 − 9 17 − 9 D10 = 0.159 mm ▣ 9. Determine the uniformity coefficient. [SOLUTION]