## Exam Problem on Dry Density and Borrow Pit

## Exam Problem on Dry Density and Borrow Pit

(OP)

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.

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

## RE: Exam Problem on Dry Density and Borrow Pit