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Exam Problem on Dry Density and Borrow Pit

Exam Problem on Dry Density and Borrow Pit

I'm studying for my PE exam in Geotechnical Materials and Analysis and have come across a problem that I seem to be able to do the first part but a bit confused on the second.  The abridged question is this

Need to fill a volume of 2100 m3 using material to be taken from one of two borrow pits.  When placed the material to be compacted to e=0.6 and w=14%.  Unit rates are provided for transportation, cost of material and cost to add water.  Which soil represents the more cost effective solution?

the two materials are defined as

Soil 1 = unit weight gamma @ 17.5 kn/m3 and w=12%   (Gamma Dry = 15.6 kN/m3)
Soil 2 = unit weight gamma @ 17.65 kn/m3 and w = 10% (gamma dry = 16 kN/m3)

The first thing I did was bring the materials to a common denominator by calculating the dry unit weights of each materials as shown in the brackets above.

Using the above dry unit weights I then calculated the respect buck unit weights of each material if placed as fill to e=0.6 and 14% (calculated Gs first) with the result of soil 1 = 15.35 kN/m3 and soil 2 = 15.9 kN/m3.  Not much difference.

I then calculated the volume of material to transport based on 2100 m3 * 15.35/15.6 = 2066 m3
and soil 2 at 2100 m3 x 15.9/16 = 2086.  THere fore soil one is slightly cheaper on supply and haul.

To calculate water cost  I use the respective volumes and

Water Vol 1 = {M-M/(1+w) }/9.81 KN/m3 = 253m3  and for Vol 2 = {2100 - 2100/(1+14%)} = 257 m3

Comments would be appreciated.

RE: Exam Problem on Dry Density and Borrow Pit

Are all the costs involved based on in-place volumes?  Borrow pit material will bulk when excavated and loaded into truck and then "shrink" when compacted.  So the volume of soil in the truck will be greater than the volume of soil compacted - I am assuming volumes and not trucking costs in mass.   

RE: Exam Problem on Dry Density and Borrow Pit

I'm assuming in-place volumes with no account for bulking.

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