Compaction test – dry unit weight from mould measurement: If W is the wet (as-compacted) weight of soil at water content w (decimal) placed in a Proctor mould of volume V, what is the dry unit weight gamma_d of the soil?
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Agamma_d = (W / V) * (1 / (1 + w))
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Bgamma_d = (W / V) * (1 + w)
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Cgamma_d = (W / V) * (1 − w)
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Dgamma_d = (W / V) / (1 − w)
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Egamma_d = (W / V) * (1 / w)
Answer
Correct Answer: gamma_d = (W / V) * (1 / (1 + w))
Explanation
Introduction / Context:Compaction control relies on converting measured wet unit weight in a mould to dry unit weight. This requires correcting for the water present at the time of compaction using the measured water content w (decimal), yielding gamma_d for comparison with Proctor curves.
Given Data / Assumptions:
- W is total wet weight of compacted specimen.
- V is volume of Proctor mould (e.g., 944 cc for standard mould).
- w is water content as a decimal (w = Mw / Ms).
Concept / Approach:Wet unit weight gamma = W / V. Since W = Ws + Ww and Ws = W / (1 + w), the dry unit weight gamma_d equals the unit weight associated with solids only: gamma_d = (W / (1 + w)) / V = (W / V) * 1 / (1 + w).
Step-by-Step Solution:
Compute gamma = W / V (wet).Find dry mass: Ws = W / (1 + w).Compute gamma_d = Ws / V = (W / V) / (1 + w).Report gamma_d in kN/m^3 if W is a force; in Mg/m^3 if masses are used consistently.Verification / Alternative check:Example: W/V = 20 kN/m^3, w = 0.10 → gamma_d = 20 / 1.10 ≈ 18.18 kN/m^3, matching standard lab conversions.
Why Other Options Are Wrong:
- Multiplying by (1 + w) would increase gamma_d above wet unit weight, impossible.
- (1 − w) forms are dimensionally and conceptually incorrect.
- Dividing by w alone would blow up at small w.
Common Pitfalls:Using w in percent instead of decimal; mixing mass-density with unit weight without g conversion; incorrect mould volume.
Final Answer:gamma_d = (W / V) * (1 / (1 + w))