Let\'s pretend you are an earthwork construction control inspector and you are c

Question

Let's pretend you are an earthwork construction control inspector and you are checking the field compaction of a layer of soil. The laboratory compaction curve for the soil is shown in Fig. P5-9. Specifications call for the compacted density to be at least 95% of the maximum laboratory value and within plusminus 2% of the optimum water content. When you did the sand cone test, the volume of soil excavated was 1153 cm3. It weighed 2209 g wet and 1879 g dry. What is the compacted dry density? What is the field water content? What is the relative compaction? Docs the test meet specifications? What is the degree of saturation of the field sample? If the sample were saturated at constant density, what would be the water content? (rhox = 2.64 Mg/m3)

Explanation / Answer

Volume of the soil excavated = 1153 cc weight of dry soil excavated = 1879 gms (a)Dry density of the soil = 1.63 g/cc = 1.63 Mg/m^3 (b)Field water content = wt. of water in soil / total weight of the wet soil =(2209-1879)/2209 = 14.94% (c) Maximum dry density from graph = 1.73 g/cc Actual dry density = 1.63 g/cc So relative compaction = actual dry density / maximum dry density =1.63/1.73 = 94.2% (d) It did not meet the specification (e) Actual water content = 14.94% Volume of voids in the field sample = volume of soil - volume of solids = 1153 - (1879/2.64) = 441.26 cc Weight of water = 2209 - 1879 =330 gms Volume of water = 330cc Degree of saturation = 330/441.26 = 74.8% (f) Given sample is saturated at constant density So volume of Water = 441.26 cc So Water content = 441.26/(1879+441.26) = 19%

Related Questions

Navigate

• Browse (All)
• Browse L
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.