Earthwork volumes by average end area, and the shrinkage balance: 16-Civ-B7 Transportation Planning and Engineering, December 2014, Q1

Question 1: Earthwork volumes by average end area, and the shrinkage balance (15 + 5 marks)

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. Six surveyed cross-sections between station 351+00 and station 352+50, with the cut and fill end areas tabulated above. The statement that “the distance between the stations is 100 m” fixes the stationing as metric (full stations 100 m apart), so the plus-values read directly as metres: 351+50 lies 50 m beyond 351+00, 351+75 lies 75 m beyond it, and so on. The material shrinks 12 percent between the borrow condition and the compacted condition.

Find. (a) The volume of cut and the volume of fill over the 150 m reach by the average-end-area method; (b) after allowing for 12 percent shrinkage, whether the reach is in surplus or deficit and by how much.

Figure 1.1 — End areas plotted against station. Fill is shown below the datum and cut above it. The tabulated zeros at 351+75 (cut) and 352+14 (fill) are the grade points, so each branch begins and ends at a genuine zero area and no interval straddles a transition.

Approach. Apply the average-end-area rule interval by interval, keeping the cut and the fill branches in separate running totals, then convert the bank volume of cut to its compacted equivalent through the shrinkage factor and compare it with the compacted fill required.

Part (a) — set out the interval lengths from the metric stationing. The six sections are not equally spaced, so the spacing of each of the five intervals must be taken from the stationing rather than assumed: $$L_1 = 50\ \text{m},\ L_2 = 25\ \text{m},\ L_3 = 25\ \text{m},\ L_4 = 14\ \text{m},\ L_5 = 36\ \text{m}$$ These sum to 150 m, which is the distance from 351+00 to 352+50 and confirms that the stationing has been read correctly.
State the average-end-area rule. For two consecutive sections of end area $A_i$ and $A_{i+1}$ separated by a distance $L$, the enclosed volume is $$V = \frac{A_i + A_{i+1}}{2}\,L$$ The rule is exact when the section varies linearly with chainage and is the standard basis for progress payment on Canadian highway contracts. Cut and fill are never netted section by section: a station carrying both areas is contributing to two different earthwork operations, and combining them would conceal the true haul.
Integrate the fill branch. Fill is present from 351+00 through to 352+14, where the fill area falls to zero: $$V_F = \tfrac{57.93+52.28}{2}(50) + \tfrac{52.28+23.58}{2}(25) + \tfrac{23.58+3.73}{2}(25) + \tfrac{3.73+0}{2}(14)$$ $$V_F = 2755.25 + 948.25 + 341.375 + 26.11 = \boxed{4071.0\ \text{m}^3}$$
Integrate the cut branch. Cut begins at the grade point at 351+75, where the cut area is zero, and runs to the end of the reach: $$V_C = \tfrac{0+8.40}{2}(25) + \tfrac{8.40+13.80}{2}(14) + \tfrac{13.80+33.34}{2}(36)$$ $$V_C = 105.00 + 155.40 + 848.52 = \boxed{1108.9\ \text{m}^3}$$ The two branches overlap only over the 39 m between the grade points, where the section is part cut and part fill — a side-hill section, which is exactly why the paper tabulates the two areas separately.
Part (b) — convert the cut from bank measure to compacted measure. A shrinkage of 12 percent means that a bank cubic metre of excavated material occupies only 0.88 compacted cubic metres once it has been placed and rolled in the embankment: $$V_{C,\text{comp}} = V_C\,(1 - s) = 1108.9 \times 0.88 = \boxed{975.9\ \text{m}^3}$$
Strike the balance over the reach. Comparing what the cut can deliver with what the embankment demands, both now in compacted measure, $$\Delta V = V_F - V_{C,\text{comp}} = 4071.0 - 975.9 = \boxed{3095.1\ \text{m}^3\ \text{deficit}}$$ The reach is therefore not in surplus. It is short of fill by roughly 3100 compacted cubic metres, and that shortfall must be imported.
Express the shortfall as a borrow quantity. Borrow is paid for and hauled in bank measure, and it shrinks by the same 12 percent on placement, so the volume to be excavated from the borrow pit is $$V_{\text{borrow}} = \frac{\Delta V}{1-s} = \frac{3095.1}{0.88} = \boxed{3517.2\ \text{m}^3\ \text{bank measure}}$$ Rounding for a schedule of quantities, roughly 3520 m³ of borrow is required over these 150 m.

Check — two conventions worth stating on the answer paper. First, the balance above is struck for this reach in isolation; on a real project the deficit would be tested against the mass-haul diagram for the whole grading contract before any borrow pit is opened, since an adjacent reach may be in surplus within economic haul. Second, if the examiner intended shrinkage to be applied as a swell on the fill side instead (“fill needed in bank measure” $= 4071.0/0.88 = 4626.1\ \text{m}^3$), the deficit in bank measure is $4626.1 - 1108.9 = 3517.2\ \text{m}^3$ — identically the borrow quantity found in Step 7. The two readings agree, so the answer is robust to the ambiguity.

Quantity	Value	Basis
Volume of fill, station 351+00 to 352+14	4071.0 m³	compacted
Volume of cut, station 351+75 to 352+50	1108.9 m³	bank
Cut available after 12 percent shrinkage	975.9 m³	compacted
Net balance over the reach	3095.1 m³ deficit	compacted
Borrow required	3517.2 m³ (say 3520)	bank

Topic. Earthwork quantities by the average-end-area method, and the shrinkage factor that links bank measure to compacted measure.

Key relations. Average end area, $V = \tfrac{1}{2}(A_1+A_2)L$; the prismoidal formula, $V = \tfrac{L}{6}(A_1 + 4A_m + A_2)$, which uses the mid-section area $A_m$ (not the mean of the ends); and the shrinkage relation $V_{\text{compacted}} = V_{\text{bank}}(1-s)$.

Why this works. The average-end-area rule is the trapezoidal rule applied along the centreline: it treats the cross-sectional area as varying linearly between surveyed sections, so the solid between them is a prismatoid approximated by a prism of mean area. Its error is second-order in the section spacing, which is why sections are surveyed at every full station and additionally at every grade point and every abrupt change of ground. Here the exam has done exactly that — the sections at 351+75 and 352+14 are inserted precisely because the area of one branch vanishes there — so each branch is integrated from zero to zero and the trapezoidal approximation is applied only where the section really is varying smoothly. The shrinkage factor is a separate, material question: undisturbed soil loosens when excavated and is then compacted to a density higher than it had in the bank, so a bank cubic metre yields less than a cubic metre in the finished embankment. Cohesive soils typically shrink 10 to 15 percent and granular soils rather less, so the 12 percent here is a plausible glacial-till value.

Common pitfall. The commonest error is netting cut against fill at a side-hill station, which here would turn a real 3100 m³ deficit into an apparent balance and leave the contractor without a borrow source. The second is applying shrinkage to the wrong side of the equation — multiplying the fill by 0.88 rather than the cut, which reverses the correction and understates the borrow by about 30 percent.

Source. Garber and Hoel, Traffic and Highway Engineering, 5th ed., Ch. 16 (Highway Surveys and Location) and Ch. 20 (Earthwork and Mass Diagrams); Mamlouk and Zaniewski, Materials for Civil and Construction Engineers, 4th ed., Ch. 6 for compaction and volume-change terminology.

Question 1 of 7: Earthwork volumes by average end area, and the shrinkage balance

Question 1: Earthwork volumes by average end area, and the shrinkage balance (15 + 5 marks)