Earthwork and Compaction – Borrow volume for specified embankment dry density A borrow pit soil has dry density 17 kN/m^3. How many cubic metres of this soil are required to build an embankment of 100 m^3 at dry density 16 kN/m^3 (ignore wastage and shrinkage)?

Difficulty: Easy

Correct Answer: 94 m^3

Explanation:


Introduction / Context:
In earthwork, material quantity is governed by conservation of dry mass. When compacting soil from a borrow pit into an embankment at a different dry density, the required borrow volume changes accordingly. This is a frequent calculation in planning haul quantities and estimating cost and time for construction.


Given Data / Assumptions:

  • Borrow soil dry density ρ_d,borrow = 17 kN/m^3.
  • Embankment target dry density ρ_d,emb = 16 kN/m^3.
  • Required embankment compacted volume V_emb = 100 m^3.
  • No losses (no bulking, no shrinkage, no wastage) and same dry mass basis.


Concept / Approach:

Conserve dry mass M_d. Dry mass placed in embankment equals dry mass removed from borrow: M_d = ρ_d,borrow * V_borrow = ρ_d,emb * V_emb. Solve for V_borrow. This avoids the need for water-content data when only dry densities are involved.


Step-by-Step Solution:

Set equality of dry mass: ρ_d,borrow * V_borrow = ρ_d,emb * V_emb.Plug values: 17 * V_borrow = 16 * 100.Compute: V_borrow = (16 * 100) / 17 = 1600 / 17 ≈ 94.12 m^3.Rounded practical value ≈ 94 m^3.


Verification / Alternative check:

Because the embankment has a lower dry density than the borrow, a slightly smaller borrow volume than 100 m^3 is needed; the 94 m^3 result matches that expectation.


Why Other Options Are Wrong:

106 m^3 assumes the opposite density relation; 100 m^3 ignores density differences; 90 m^3 underestimates mass requirements.


Common Pitfalls:

Using bulk (wet) densities or moisture contents unnecessarily, or confusing which density is higher. The conservation should always be done on dry mass when densities are given as dry.


Final Answer:

94 m^3

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