GEOM-210 · Lesson 5 of 12 · 25 min

Profiles, cross-sections and earthwork volumes

Roads, pipelines, channels and ponds are designed in three dimensions, but engineers usually think about them in slices. A profile is a vertical section along the centerline of an alignment, showing the existing ground and the proposed grade. Cross-sections are vertical slices at right angles to the centerline, showing the ground across the corridor and the proposed shape of the road, ditch or embankment. Comparing the existing ground with the design in these slices gives the cuts and fills that contractors must build and the volumes of earth to be moved, which often control the cost of a project. This lesson covers profile and cross-section surveys, stationing, plotting, cut and fill, and the standard methods for computing earthwork volumes.

On the Coulee Ridge Drive extension, Leah helps run a profile along the proposed centerline and cross-sections every 50 ft. Theo's drone survey, covered in lesson 11, gives a surface model of the whole site, but the field profile is a useful check and is still the standard method for many city and rural projects. The same methods are used for the stormwater pond, where a grid survey gives the volume of excavation.

Stationing

Positions along an alignment are given by stations, measured from the beginning of the alignment along the centerline. In US customary practice, stations are written as hundreds of feet plus feet, so station 1+50 is 150 ft from the start and station 12+34.56 is 1,234.56 ft from the start. Full stations are every 100 ft, with intermediate stations often every 50 ft on roads, and additional stations, called plus stations, at breaks in the ground, intersections, culverts and the beginning and end of curves. Metric projects write stations as kilometers plus meters, such as 1+234.560. An offset is the perpendicular distance left or right of the centerline, so a point described as station 1+00, 28 ft right, is 28 ft to the right of the centerline at 100 ft from the start, looking in the direction of increasing stations.

Profile leveling

Profile leveling is differential leveling along the centerline with many intermediate sights. From each setup, the crew takes rod readings at every station and at every significant break in the ground within reach, then moves forward on turning points. Each intermediate sight elevation equals the height of instrument minus the rod reading. The run begins and ends on benchmarks so that it can be checked for closure, just like any level run; the intermediate sights do not affect the closure, which is why a gross mistake in an intermediate reading can pass unnoticed unless the profile is plotted and checked for odd spikes.

Profiles are plotted with a large vertical exaggeration, often ten times, so that small changes in elevation are visible. The horizontal axis is stationing and the vertical axis is elevation. The designer draws a proposed grade line made of straight tangents joined by vertical curves (lesson 12), and the difference between ground and grade at each station is the centerline cut or fill.

\[\text{Grade elevation} = \text{Elev}_{0} + g\,(s - s_0) \qquad \text{cut} = \text{ground} - \text{grade}\]

A grade \(g\) is a slope expressed as a decimal or percentage, positive uphill in the direction of increasing stations and negative downhill. A grade of −1.5 percent falls 1.5 m per 100 m.

Worked problem

Centerline cut and fill

The proposed grade of Coulee Ridge Drive starts at elevation 2999.80 ft at station 0+00 and falls at −1.5 percent. Ground elevations from the profile survey are 3001.2 at 0+00, 3002.3 at 0+50, 3003.4 at 1+00, 3002.1 at 1+50, 3000.2 at 2+00 and 2998.3 at 2+50. Find the cut or fill at each station.

Show solution
  1. Grade elevations fall 0.75 ft every 50 ft: 2999.80, 2999.05, 2998.30, 2997.55, 2996.80, 2996.05.
  2. Cuts are ground minus grade: 1.40, 3.25, 5.10, 4.55, 3.40 and 2.25 ft.

Answer: Every station is in cut, with the deepest cut of 5.10 ft at 1+00 where the road passes through a low rise. Positive values are cuts and negative values would be fills.

Cross-sections

At each station, the crew measures ground elevations at right angles to the centerline, out to beyond the limits of construction, at every break in slope. The result is a ground line for each cross-section. The designer overlays a template: the proposed road surface, shoulders, ditches and side slopes. Side slopes are written as a ratio of horizontal to vertical, such as 3:1 for a gentle slope that can be mowed, or 2:1 for a steeper embankment. The points where the side slopes meet the existing ground are the catch points, or slope stakes, which define the limits of the earthwork and are marked in the field during construction (lesson 12).

A road cross-section showing the existing ground, the road template with ditches and side slopes, the cut area shaded and the catch points.
A cross-section in cut. The area between the ground line and the template is the cut area at this station; the catch points mark where the side slopes meet the ground.

The area of each cross-section, in cut and in fill separately, is computed from the coordinates of its corners, with offset as the horizontal coordinate and elevation as the vertical coordinate. The coordinate method, explained for parcels in lesson 8, works for any polygon. For simple shapes, the section can also be divided into trapezoids and triangles. Software does this automatically from digital terrain models, but the logic is the same.

Volumes by average end area

The most widely used method for road and pipeline earthwork is the average end area method. The volume between two cross-sections is the average of their areas times the distance between them.

\[V = \frac{L}{2}\,(A_1 + A_2)\]

The method is exact if the cross-sections change linearly between stations, which they rarely do, and it usually overestimates the volume slightly. Where greater accuracy is needed, the prismoidal formula uses the area of a middle section \(A_m\), measured or computed halfway between the end sections (not the average of the end areas):

\[V = \frac{L}{6}\,(A_1 + 4A_m + A_2)\]

When a section changes from cut to fill, the earthwork forms a wedge or pyramid rather than a prism, and the volume is computed as a pyramid, one third of the base area times the length, or by adding a section at the transition point where the cut area becomes zero.

Worked problem

Volume between two stations

At station 1+00 the cut area is 200.0 ft², and at station 1+50 it is 262.0 ft². Find the volume in cubic yards by the average end area method.

Show solution
  1. \[V = \frac{50}{2}(200.0 + 262.0) = 11\,550\ \text{ft}^3\]
  2. \[V = \frac{11\,550}{27} = 427.8\ \text{yd}^3\]

Answer: The volume is about 428 yd³. There are 27 ft³ in a cubic yard, and US earthwork quantities are almost always paid in cubic yards.

Volumes by the grid method

For ponds, building pads, borrow pits and stockpiles, a grid method is often more convenient than cross-sections. The area is divided into squares or rectangles of equal size, and the depth of cut or fill is determined at every grid corner. Each grid cell is treated as a prism whose volume is its area times the average of its four corner depths. Interior corners are shared by several cells, so the total volume can be computed by weighting each corner depth by the number of cells that use it:

\[V = \frac{A}{4}\Big(\sum h_1 + 2\sum h_2 + 3\sum h_3 + 4\sum h_4\Big)\]

where \(A\) is the area of one cell and \(h_n\) are depths at corners shared by \(n\) cells. Corners at the outside of the grid belong to one cell, edge points to two, and interior points to four; points at an inside corner of an irregular grid belong to three.

Worked problem

Excavation volume for the stormwater pond

The pond area is covered by a grid of 3 × 2 cells, each 25 ft × 25 ft, giving 4 × 3 corner points. The depths of cut, row by row, are 3.9, 4.8, 5.3, 4.5; 5.1, 6.0, 6.5, 5.2; and 4.3, 5.0, 5.6, 4.7 ft. Find the excavation volume in cubic yards.

Show solution
  1. Corner points (used once): \(3.9 + 4.5 + 4.3 + 4.7 = 17.4\).
  2. Edge points (used twice): \(4.8 + 5.3 + 5.1 + 5.2 + 5.0 + 5.6 = 31.0\).
  3. Interior points (used four times): \(6.0 + 6.5 = 12.5\).
  4. \[V = \frac{625}{4}(17.4 + 2 \times 31.0 + 4 \times 12.5) = 156.25 \times 129.4 = 20\,219\ \text{ft}^3 = 749\ \text{yd}^3\]

Answer: The excavation is about 749 yd³ of in-place (bank) material.

# Cut and fill along the profile, end-area volumes and the grid method (US customary)
stations = [0, 50, 100, 150, 200, 250]
ground = [3001.2, 3002.3, 3003.4, 3002.1, 3000.2, 2998.3]
z0, g = 2999.80, -0.015
for s, e in zip(stations, ground):
    grade = z0 + g * s
    print(f"{s // 100}+{s % 100:02d}  ground {e:7.1f}  grade {grade:8.2f}  cut {e - grade:5.2f}")

A1, A2, L = 200.0, 262.0, 50
v = L / 2 * (A1 + A2)
print(f"average end area: {v:.0f} ft3 = {v / 27:.1f} yd3")

def grid_volume(h, cell_area):
    rows, cols = len(h), len(h[0])
    total = 0
    for i in range(rows):
        for j in range(cols):
            n = sum(1 for di in (0, -1) for dj in (0, -1)
                    if 0 <= i + di < rows - 1 and 0 <= j + dj < cols - 1)
            total += n * h[i][j]
    return cell_area / 4 * total

pond = [[3.9, 4.8, 5.3, 4.5], [5.1, 6.0, 6.5, 5.2], [4.3, 5.0, 5.6, 4.7]]
bank = grid_volume(pond, 25 * 25) / 27
print(f"pond excavation: {bank:.0f} bank yd3, about {bank * 1.25:.0f} loose yd3 to haul at 25% swell")
Output
0+00  ground  3001.2  grade  2999.80  cut  1.40
0+50  ground  3002.3  grade  2999.05  cut  3.25
1+00  ground  3003.4  grade  2998.30  cut  5.10
1+50  ground  3002.1  grade  2997.55  cut  4.55
2+00  ground  3000.2  grade  2996.80  cut  3.40
2+50  ground  2998.3  grade  2996.05  cut  2.25
average end area: 11550 ft3 = 427.8 yd3
pond excavation: 749 bank yd3, about 936 loose yd3 to haul at 25% swell

Bank, loose and compacted volumes

A cubic meter of soil in the ground, called bank or in-place volume, does not stay a cubic meter. When excavated it loosens and occupies more space, typically 10 to 40 percent more depending on the material; this increase is called swell, and it governs how many truckloads are needed. When placed and compacted in an embankment, it usually occupies less space than in the bank, typically by 5 to 20 percent; this is called shrinkage. Rock behaves differently and usually swells permanently. Earthwork balances must therefore convert all volumes to a common basis, and contracts must state whether payment is for bank, loose or compacted volume.

A mass haul diagram plots the cumulative volume of earthwork along an alignment, with cuts positive and fills negative after applying shrinkage factors. Rising parts of the curve show cut, falling parts show fill, and a horizontal line drawn across the curve, called a balance line, shows stretches where cut and fill balance. The diagram helps designers adjust the grade line to reduce the volume of material that must be imported or wasted, and helps contractors plan haul distances.

Surfaces and modern methods

Most volumes today are computed from digital terrain models. A triangulated irregular network (TIN) connects surveyed points with triangles to represent the ground surface. The volume between two surfaces, such as existing ground and design, or the same stockpile surveyed in two different months, is computed by comparing the surfaces over a common area. Drone photogrammetry and lidar, discussed in lesson 11, can measure millions of points in an hour, which is ideal for stockpiles and large earthworks. Field profiles, cross-sections and grid shots remain important for checking those surfaces, for small jobs and wherever vegetation hides the ground.

Whatever the method, the quality of the volume depends on the quality of the surfaces. Missing a ditch bottom, a crest of a berm or a break in slope can change the result more than the choice of formula. Surveyors therefore pay particular attention to breaklines, lines along which the ground changes slope sharply, and include them in surface models so that triangles do not cut across them.

Common mistakes

  • MistakeIncluding intermediate sight readings in the level run closure check.

    InsteadClosure uses only backsights and foresights on turning points and benchmarks; check intermediate sights by plotting the profile.

  • MistakeUsing the average of the end areas as the middle area in the prismoidal formula.

    InsteadThe middle area must be measured or computed at the midpoint; averaging end areas just reproduces the average end area result.

  • MistakeAdding bank and compacted volumes directly in an earthwork balance.

    InsteadConvert all volumes to a common basis using swell and shrinkage factors.

  • MistakeBuilding a terrain model without breaklines.

    InsteadSurvey and include breaklines such as ditch bottoms and crests so the surface follows the real ground.

Summary and key terms

  • A profile is a section along the centerline; cross-sections are at right angles; stations and offsets locate points.
  • Profile leveling uses many intermediate sights; the run still closes on benchmarks.
  • Grade elevation equals the starting elevation plus the grade times the distance; cut equals ground minus grade.
  • Average end area volume is L(A1 + A2)/2; the prismoidal formula uses a measured middle area.
  • The grid method weights corner depths by the number of cells sharing each corner.
  • Earthwork volumes must account for swell and shrinkage, and modern volumes come from surface models built with breaklines.
Profile
A vertical section along an alignment showing existing ground and proposed grade.
Cross-section
A vertical section at right angles to an alignment.
Station
A distance along an alignment from its beginning.
Offset
A perpendicular distance left or right of an alignment.
Catch point
The point where a design side slope meets the existing ground.
Average end area method
A volume method that multiplies the average of two cross-section areas by the distance between them.
Swell
The increase in volume of soil when it is excavated and loosened.
Breakline
A line along which the ground surface changes slope sharply, used to control a terrain model.

Practice problems

Work each problem on paper first, then reveal the answer to check.

  1. A grade of +2.0 percent starts at elevation 3010.00 ft at station 10+00. What is the grade elevation at station 12+50?

    Show answer

    \(3010.00 + 0.02 \times 250 = 3015.00\) ft.

  2. End areas of 35 m² and 41 m² are 20 m apart. Find the volume by average end areas.

    Show answer

    \(10 \times 76 = 760\) m³.

  3. A cut area of 12 m² shrinks to zero over 15 m, with both its width and depth shrinking so that it tapers to a point like a pyramid. Estimate the volume.

    Show answer

    As a pyramid, one third of the base area times the length: \(12 \times 15/3 = 60\) m³. The average end area method would give 90 m³ and overestimate.

  4. How many 10 m³ truckloads are needed for 986 bank m³ with 25 percent swell?

    Show answer

    \(986 \times 1.25/10 = 123.3\), so 124 loads.

  5. Why is a profile plotted with vertical exaggeration?

    Show answer

    Elevation changes are small compared with horizontal distances and would be invisible at true scale.

  6. What does a horizontal balance line on a mass haul diagram show?

    Show answer

    A stretch of the alignment over which cut and fill volumes balance.

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