Degree of saturation from compaction data: A compacted soil has moisture content w = 10% and bulk density rho = 2.0 g/cm^3. Given Gs = 2.7 and rho_w = 1.0 g/cm^3, determine the degree of saturation S.
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A55.6%
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B11.1%
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C69.6%
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D33.3%
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ENone of these
Answer
Correct Answer: 55.6%
Explanation
Introduction / Context:
The degree of saturation S indicates how much of a soil’s void space is filled with water. It is a critical parameter for compaction control, permeability assessment, and predicting volume change behavior. From routine compaction data (bulk density and water content), along with specific gravity, S can be calculated using standard index relationships.
Given Data / Assumptions:
- w = 10% = 0.10.
- rho = 2.0 g/cm^3 (bulk density).
- Gs = 2.7, rho_w = 1.0 g/cm^3.
- Representative sample with no unusual chemical effects.
Concept / Approach:
Use two key relations. First, rho_d = rho / (1 + w). Second, rho_d = (Gs / (1 + e)) * rho_w, which gives 1 + e = (Gs * rho_w) / rho_d. With e known, use w = (S * e) / Gs, so S = (w * Gs) / e.
Step-by-Step Solution:
Compute dry density: rho_d = rho / (1 + w) = 2.0 / 1.10 = 1.818 g/cm^3.Find void ratio: 1 + e = (Gs * rho_w) / rho_d = (2.7 * 1.0) / 1.818 = 1.485 -> e = 0.485.Compute saturation: S = (w * Gs) / e = (0.10 * 2.7) / 0.485 = 0.556 -> 55.6%.Verification / Alternative check:
Back-calculate w using w = (S * e) / Gs = (0.556 * 0.485) / 2.7 = 0.10 = 10%, confirming consistency.
Why Other Options Are Wrong:
- 11.1% and 33.3% are too low given the computed e and w.
- 69.6% is higher than the value indicated by the mass-volume relationships.
- None of these is incorrect because 55.6% matches the calculation.
Common Pitfalls:
- Confusing porosity n with void ratio e; here formulas use e.
- Forgetting to convert percentage to decimal form before substitution.
Final Answer:
55.6%