Earthwork Volume from Cross-Sections with a Change of Formation
A 400 m length of road is to be cut and filled to a formation level that falls uniformly from 102.500 m at chainage 0 to 101.300 m at chainage 400. Cross-sections are taken at 0, 100, 200, 300 and 400 m. The existing ground levels on the centre line are 103.100, 102.950, 102.400, 101.850 and 101.100 m. The formation width is 7.0 m, side slopes are 1:2 in cut and 1:3 in fill. At chainage 200 the ground level equals the formation level, so the section is at grade. The surveyor must compute cut and fill volumes and decide how to treat the transition section.
1. Compute the depth of cut or fill at each chainage and identify where the cut-to-fill transition occurs.
Read the worked answer
Formation levels (linear fall from 102.500 to 101.300 over 400 m).
Rate of fall = (102.500 − 101.300)/400 = 1.200/400 = 0.003 m per m.
- CH 0: 102.500
- CH 100: 102.200
- CH 200: 101.900
- CH 300: 101.600
- CH 400: 101.300
Depth = ground level − formation level. Positive = cut, negative = fill.
| CH | Ground | Formation | Depth | Regime |
|---|---|---|---|---|
| 0 | 103.100 | 102.500 | +0.600 | cut |
| 100 | 102.950 | 102.200 | +0.750 | cut |
| 200 | 102.400 | 101.900 | +0.500 | cut |
| 300 | 101.850 | 101.600 | +0.250 | cut |
| 400 | 101.100 | 101.300 | −0.200 | fill |
Transition. The section is not at grade at CH 200 - the depth there is +0.500 m of cut. The setup's claim that ground equals formation at CH 200 is wrong; the numbers show otherwise. The true transition from cut to fill lies between CH 300 (+0.250) and CH 400 (−0.200).
By linear interpolation, the zero-depth point is at:
x = 300 + 100 × 0.250/(0.250 + 0.200) = 300 + 100 × 0.556 = 355.6 m.
The cut-to-fill transition is therefore at approximately CH 355.6, not at CH 200. This is the first thing to flag - a computation based on the stated assumption would place the transition 155 m away from where it actually is and misallocate volumes between cut and fill.
2. Compute the cut and fill volumes using the trapezoidal (mean end area) method, treating the transition correctly.
Read the worked answer
Cross-section areas.
For a level section with formation width b and side slope s (vertical:horizontal), area = b·d + s·d², where d is the depth.
- Cut: b = 7.0, s = 2 → A = 7.0d + 2d²
- Fill: b = 7.0, s = 3 → A = 7.0d + 3d²
Cut areas.
- CH 0: d = 0.600 → 7.0(0.600) + 2(0.360) = 4.200 + 0.720 = 4.920 m²
- CH 100: d = 0.750 → 5.250 + 2(0.5625) = 5.250 + 1.125 = 6.375 m²
- CH 200: d = 0.500 → 3.500 + 2(0.250) = 3.500 + 0.500 = 4.000 m²
- CH 300: d = 0.250 → 1.750 + 2(0.0625) = 1.750 + 0.125 = 1.875 m²
- CH 355.6: d = 0 → 0 m²
Fill areas.
- CH 355.6: 0 m²
- CH 400: d = 0.200 → 7.0(0.200) + 3(0.040) = 1.400 + 0.120 = 1.520 m²
Cut volume (trapezoidal, intervals as measured).
V = Σ (A1 + A2)/2 × L
- 0-100: (4.920 + 6.375)/2 × 100 = 5.6475 × 100 = 564.75 m³
- 100-200: (6.375 + 4.000)/2 × 100 = 5.1875 × 100 = 518.75 m³
- 200-300: (4.000 + 1.875)/2 × 100 = 2.9375 × 100 = 293.75 m³
- 300-355.6: (1.875 + 0)/2 × 55.6 = 0.9375 × 55.6 = 52.13 m³
Total cut ≈ 564.75 + 518.75 + 293.75 + 52.13 ≈ 1,429.4 m³
Fill volume.
- 355.6-400: (0 + 1.520)/2 × 44.4 = 0.760 × 44.4 = 33.74 m³
Total fill ≈ 33.7 m³
Note on the transition. Splitting the 300-400 interval at the zero point is essential. If the whole 300-400 interval were treated as cut, using a negative depth at CH 400 in the cut formula, the result would be nonsense (a negative area). Treating it as fill throughout would understate cut by about 52 m³ and overstate fill by a similar amount. The split at CH 355.6 is the correct treatment.
3. The surveyor wants to use the prismoidal formula instead. Explain when it is appropriate and whether it changes the answer materially here.
Read the worked answer
When prismoidal applies.
The prismoidal formula V = L/6 × (A1 + 4Am + A2) is exact when the cross-section area varies as a quadratic (or lower) function of chainage along the segment. For a level section with constant side slopes, A = b·d + s·d², and if d varies linearly with chainage, A varies quadratically - so the prismoidal formula is exact for each segment and the trapezoidal formula slightly overestimates.
The formula requires the mid-section area Am, computed from the depth at the mid-point. It cannot be applied blindly between stations where the depth is not linear, and it must not be applied across the cut-to-fill transition.
Application here.
For segment 0-100: mid depth = (0.600 + 0.750)/2 = 0.675 m; Am = 7.0(0.675) + 2(0.675²) = 4.725 + 0.911 = 5.636 m².
V = 100/6 × (4.920 + 4(5.636) + 6.375) = 16.667 × (4.920 + 22.544 + 6.375) = 16.667 × 33.839 ≈ 563.98 m³.
Compare with the trapezoidal value 564.75 m³ - a difference of about 0.77 m³, or 0.14%.
General conclusion. Because the side-slope term s·d² is small relative to b·d for these depths (d ≤ 0.75 m against b = 7.0 m), the area is nearly linear in d and the two methods differ by well under 1%. For this job the trapezoidal method is adequate and the extra arithmetic of the prismoidal method is not justified.
The prismoidal method earns its keep when the section is deep and narrow, when side slopes are steep, or when the ground is irregular so that the mid-section area is very different from the mean of the end areas. It is also the right choice for borrow-pit and stockpile volumes where the figure is a true prismoid.
Practical recommendation. Use trapezoidal throughout, but compute one prismoidal check on the deepest segment to confirm the method is not introducing bias. Report volumes to the nearest cubic metre and state the method used, because a 0.1% difference is immaterial but an unstated method is not.