Earthwork volumes by average end area and the pyramid rule: 16-Civ-B7 Transportation Planning and Engineering, May 2015, Q1

Question 1: Earthwork volumes by average end area and the pyramid rule

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A run of six surveyed cross-sections between station 250+00 and station 251+50, each reported as a cut end area and a fill end area. A blank cell in the printed table means that branch has no area at that section, i.e. zero.

End areas read from the examination paper (m²)
Station	Chainage (m)	Cut area	Fill area
250+00	25 000	—	60
250+50	25 050	—	50
250+75	25 075	0	25
251+00	25 100	10	5
251+15	25 115	15	0
251+50	25 150	35	—

Find. (a) the volume of cut and the volume of fill over the 150 m of alignment, each computed by whichever of the average-end-area or pyramid formula suits the interval; (b) with a shrinkage of 10 percent, the resulting excess or shortfall of material.

End areas plotted against chainage. The alignment starts entirely in fill, passes through a transition where the fill dies out and the cut begins, and finishes entirely in cut. The dashed line marks the section at which the cut area is still zero — the interval on either side of a zero end area is a pyramid, not a prismoid.

Approach. Take each interval between consecutive sections in turn, apply the average-end-area rule where both end areas are non-zero and the pyramid rule where one of them is zero, then sum the cut and fill branches separately before comparing them on a common (compacted) basis.

Fix the interval lengths from the stationing. The paper states that the distance between (full) stations is 100 m, so a station equals 100 m and the plus-value is the metres beyond it. The five intervals are therefore $$L_1 = 50\ \text{m},\quad L_2 = 25\ \text{m},\quad L_3 = 25\ \text{m},\quad L_4 = 15\ \text{m},\quad L_5 = 35\ \text{m}$$ which sum to 150 m, the distance from 250+00 to 251+50.
Select the formula interval by interval. The average-end-area (trapezoidal) rule and the pyramid rule are $$V_{\text{AEA}} = \frac{A_1 + A_2}{2}\,L , \qquad V_{\text{pyr}} = \frac{A\,L}{3}$$ where $A_1$ and $A_2$ are the two end areas and $A$ is the single non-zero end area. The pyramid form applies wherever one end area vanishes, because the solid between a finite section and a point is a pyramid, and the trapezoidal rule would overstate it by half.
Work the fill branch. Fill is present from 250+00 through to 251+15, where it dies out, so four of the five intervals carry fill and only the last of those is a pyramid. $$\begin{aligned}V_{f,1} &= \tfrac{1}{2}(60 + 50)(50) = 2\,750\ \text{m}^3 \\V_{f,2} &= \tfrac{1}{2}(50 + 25)(25) = 937.5\ \text{m}^3 \\V_{f,3} &= \tfrac{1}{2}(25 + 5)(25) = 375\ \text{m}^3 \\V_{f,4} &= \tfrac{1}{3}(5)(15) = 25\ \text{m}^3\end{aligned}$$ The final interval, 251+15 to 251+50, has zero fill at both ends and contributes nothing.
Total the fill. Adding the four contributions back from their own components, $$V_{\text{fill}} = 2\,750 + 937.5 + 375 + 25 = \boxed{4\,087.5\ \text{m}^3}$$
Work the cut branch. Cut first appears at 250+75, where its end area is still zero, so the interval from 250+75 to 251+00 is a pyramid; the two intervals beyond it are ordinary prismoids. $$\begin{aligned}V_{c,3} &= \tfrac{1}{3}(10)(25) = 83.33\ \text{m}^3 \\V_{c,4} &= \tfrac{1}{2}(10 + 15)(15) = 187.5\ \text{m}^3 \\V_{c,5} &= \tfrac{1}{2}(15 + 35)(35) = 875\ \text{m}^3\end{aligned}$$ The first two intervals are wholly in fill and contribute no cut.
Total the cut. $$V_{\text{cut}} = 83.33 + 187.5 + 875 = \boxed{1\,145.83\ \text{m}^3}$$ Both totals are bank (in-place) measure, which is how surveyed end areas always report them.
Part (b) — bring the cut onto a compacted basis. A shrinkage of 10 percent means that material excavated from the cut occupies only 90 percent of its bank volume once it has been placed and compacted in the fill: $$V_{\text{cut, placed}} = (1 - s)\,V_{\text{cut}} = 0.90 \times 1\,145.83 = 1\,031.25\ \text{m}^3$$
Compare supply against demand. The embankment needs 4 087.5 m³ of compacted fill and the cut can supply only 1 031.25 m³ of it, so there is no excess cut at all — there is a shortfall: $$\Delta V = V_{\text{fill}} - V_{\text{cut, placed}} = 4\,087.5 - 1\,031.25 = \boxed{3\,056.25\ \text{m}^3\ \text{of compacted fill short}}$$
Express the shortfall as material to be won. Borrow is measured in the borrow pit in bank measure and shrinks by the same 10 percent when placed, so the quantity actually to be excavated and hauled in is $$V_{\text{borrow}} = \frac{\Delta V}{1 - s} = \frac{3\,056.25}{0.90} = \boxed{3\,395.83\ \text{m}^3\ \text{(bank measure)}}$$

The shrinkage allowance matters less here than the raw imbalance. Ignoring shrinkage altogether the section would still be 2 941.67 m³ short; allowing for it adds only a further 114.58 m³, about 3.9 percent. This alignment is fill-dominated, and the shortfall is roughly 74.8 percent of the total fill requirement, so the governing construction issue is finding a borrow source, not balancing the haul.

Check — assumptions stated under the paper's Note 1. (i) A station is taken as 100 m, as the question itself states; the plus-values then read directly as metres. (ii) A blank cell in the printed table is read as a zero end area on that branch, which is the only reading consistent with the transition the table describes. (iii) The 10 percent shrinkage is applied to excavated material when it is placed and compacted in the embankment, which is the conventional definition; if instead it were read as a swell on excavation the cut would yield more loose volume but no more compacted fill, and the shortfall would be unchanged.

Had every interval been treated by average end area, including the two that run to a zero area, the cut would have come out at 1 187.5 m³ and the fill at 4 100 m³ — overstating the cut by 41.67 m³ and the fill by 12.5 m³. That is the whole point of the question: the error is small in the fill, where the pyramid interval is short and its end area small, but it is 3.6 percent of the cut, and it is always in the unsafe direction because it credits material that is not there.

Final results
Quantity	Value
Total volume of cut (bank measure)	1 145.83 m³
Total volume of fill (compacted, in place)	4 087.5 m³
Cut placed as compacted fill after 10 % shrinkage	1 031.25 m³
Balance	shortfall — there is no excess cut
Shortfall of compacted fill	3 056.25 m³
Borrow required, measured in the pit (bank)	3 395.83 m³

Topic. Earthwork quantities from cross-sectional end areas, and the shrinkage factor that converts bank measure to placed measure.

Key relations. $V_{\text{AEA}} = \tfrac{1}{2}(A_1 + A_2)L$ for a prismoid between two non-zero end areas; $V_{\text{pyr}} = \tfrac{1}{3}AL$ where one end area is zero; the prismoidal (Simpson) formula $V = \tfrac{L}{6}(A_1 + 4A_m + A_2)$ when a mid-section area is available; and $V_{\text{placed}} = (1-s)V_{\text{bank}}$ for a shrinkage $s$.

Why this works. The average-end-area rule is the trapezoidal rule applied to the area function along the alignment, so it is exact when the area varies linearly and errs whenever the section is curved. Between a real section and a point the solid is a cone or pyramid whose volume is one third of the base times the height, not one half, so the trapezoidal rule overstates such an interval by 50 percent. Shrinkage then accounts for the fact that a compacted embankment is denser than the ground the material came from, so a cubic metre of cut never places a cubic metre of fill.

Common pitfall. Netting cut against fill in bank measure and reporting the difference as the haul balance. Cut is bank measure and fill is compacted measure; they cannot be subtracted until one has been converted to the other. The second pitfall is applying the trapezoidal rule blindly across the transition sections, which is exactly what the phrase “or the pyramidal volume as appropriate” is testing.

Source. Garber & Hoel, Traffic and Highway Engineering, chapter on highway earthwork and mass diagrams; TAC, Geometric Design Guide for Canadian Roads, Part 2 (cross-section elements and earthworks).

Question 1 of 7: Earthwork volumes by average end area and the pyramid rule

Question 1: Earthwork volumes by average end area and the pyramid rule