The void ratio of the compacted fill is specified as 065 Four borrow pits are

The void ratio of the compacted fill is specified as

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of compacted soil. The void ratio of the compacted fill is specified as 0.65. Four borrow pits are available as described in the following table, which lists the respective void ratios of the soil and the cost per m 3 for moving the soil to the proposed construction site. Assume the specific gravity of the soil in each borrow pit are equals. A. Which of the following borrow pit will give the lowest amount of volume to fill the required embankment (indicate the volume)? B. Which of the following borrow pit will give the lowest cost for the embankment (indicate the amount)? BORROW PIT VOID RATIO COST(PESOS/m 3 ) A 0.85 855 B 1.20 645 C 0.95 765 D 0.75 945 SOLUTION : Given : V = 5000m 3 e = 0.65 (A.) For A : For C : ? 5000 = 1+0.85 1+0.65 ? 5000 = 1+0.95 1+0.65 V = 5606.0606m 3 V = 5909.0909m 3 For B : For D: ? 5000 = 1+1.20 1+0.65 ? 5000 = 1+0.75 1+0.65 V = 6666.6667m 3 V = 5303.0303m 3 (B.) For A: (855)(5606.0606) = 𝑃ℎ?. 4,793,181.813 For B: (645)(6666.6667) = 𝑃ℎ?. 4,3000,000.022 For C: (765)(5909.0909) = 𝑃ℎ?. 4,520,454.539 For D: (945)(5303.0303) = 𝑃ℎ?. 5,001,363.643 Therefore, most economical is BORROW PIT B.
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Page 56 of 72 Problem No. 2: Sand cone equipment is used to determine an in-place unit weight (field density test) on a compacted earth fill. Ottawa sand is used in the cone and is known to have a bulk density of 1604 kg/m 3 . The laboratory moisture unit weight curve indicates a dry unit weight of 19.20 kN/m 3 and an optimum water content of 13%. Determine: a. The moisture content at the field b. The field dry unit weight c. The relative compaction SOLUTION: γ = 1604 kg/m 3 γ dMAX = 19.20 kN/m 3 w opt = 13% w @field = ? w = ?? ?? = 2200−1930 1930 → w = 13.9896% ANSWER for V hole 1604 kg/m 3 = 1.636 kg ? V = 1.01995 x 10 -3 m 3 γ = 2.2 (9.81) 1.01995 γ = 21.1599 kN/m 3 γ dFIELD = γ 1+? = 21.1599 1+0.139896 → γ dFIELD = 18.5630 kN/m 3 ANSWER R = γ dFIELD x 100 = 18.5630 19.2 ? 100 → R = 96.6822% ANSWER γ dMAX Soil Sample from the Hole Weight of Wet Soil Sample 2200g Weight of Dried Soil Sample 1930g Weight of Sand to fill the test hole 1636g
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Page 57 of 72 Problem No.4 . Determine the dry unit weight of the soil sample in a field density test conducted by core-cutter method. The following data was obtained: Weight of empty core-cutter = 22.80 N Weight of soil and core-cutter = 50.05 N Inside diameter of core-cutter = 90.0 mm Height of core-cutter = 180.0 mm Weight of wet sample for moisture determination = 0.5450 N Weight of over-dry sample = 0.5112 N Specific gravity of soil grains = 2.72 Given: ? ? = 27.25 𝑁 ? ? = 2.27 ? ??? = 0.5512𝑁 ? ? = 0.0293𝑁 ? = 180.0?? ? ? = 90.0?? Solution: 𝜔 = ? ? ? ? 𝜔 = 0.0293𝑁 0.5112𝑁 𝜔 = 0.0573 ? = (?? 2 )ℎ ? = (?(4.5?10 −3 ?) 2 )(0.18) ? = 1.1451?10 −3 ? 3 ? ? = ? 1 + 𝜔 ? ? = 27.25?10 −3 𝑘𝑁 1.1415?10 −3 ? 3 1 + 0.0573 ? ? = 22 . 5072 𝑘𝑁 ? 3
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Page 58 of 72 Problem No. 5 An embankment for a highway 30m wide and 1.5m in a compacted thickness is to be constructed from a sandy soil trucked from a borrow pit. The water content of the sandy soil in the borrow pit is 11% and its void ratio is 0.69. The specification requires dry unit weight of 18 kN/m 3 , the specific gravity of soil is 2.70. For a 1 km length of embankment, compute a. The weight of soil required to provide embankment. b. The number of truck required to fill the embankment if the capacity of truck is 10m 3 c. The total volume of water in liters for moisture conditioning at degree of saturation of 80% for the required embankment.
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