Calculating excavation and backfill is not as easy as getting the volume of soil. Angle of repose and compaction need to be considered.
A. Angle of repose
Angle of repose is the angle loose materials form from the horizontal when it is placed freely on a flat surface. So when excavations are made, it is not only the volume of the soil right under the specified area that needs to be calculated but also the surrounding area around the perimeter that forms ramps because of angle of repose of the soil that became loose after excavation.
Angle of repose depends on the materials’ coefficient of internal friction. Here are the approximate average values of angle of repose for specific types of materials.
Average Angle of Repose of Different Materials | Calculating excavation and backfill
| Type of Material | Angle of repose (degrees) |
| Dry Sand | 27.5 |
| Moist Sand | 37.5 |
| Wet Sand | 30 |
| Ordinary Dry Earth | 32.5 |
| Ordinary Moist Earth | 35 |
| Ordinary Wet Earth | 27.5 |
| Gravel | 39 |
| Gravel, Sand and Clay | 25 |
You can actually measure the angle of repose of a material. Get a sample of the loose material. Pour it on a flat surface and measure the angle it makes from the horizontal.
By applying the law of tangents, we can calculate the horizontal distance X that will be added to the volume calculation.
tanθ= DEPTH/x
Example:
Calculate the total volume of soil to be excavated for 10m x 20m pit with the depth of 2m. Assume the soil to be ordinary moist.
Given:
Depth = 2m
θ = 35 degrees for ordinary moist earth
Area = 10m x 20m
Solution:
x = DEPTH / tan θ
= 2m / tan35
x = 2.86 m
Base dimension: 10m by 20m
Top dimension: (10m+2x) by (20m+2x)
Base area = A1 = 10 x 20 = 200 m2
Top area = A2 = (10+2*2.86)(20+2*2.86) = 404.3 m2
V = 2/3(200+404.3+sqrt(200+404.3)
V = 592.44 m3
B. Compaction
Backfills are usually compacted up to 25% of the fill volume. On the other hand, excavated materials are expected to expand up to 25% in volume when loaded for transport. By considering compaction and loosening up, we could get the idea.
Volumes of fills need to be multiplied by 1.25 to allow compaction.
Example:
Find the required volume of fill for 10m x 10m x 3m pit. Assume the pit to have sheet piles on the edges to prevent soil from slipping down.
Given:
V actual = 10 x 10 x 3 = 300 m3
Solution
V fill = V actual x 1.25 = 300 x 1.25 = 375m3