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Chapter 5

Detail Quantity Estimate for Road, Bridge and Irrigation Works

8 teaching hours 10 marks Estimating and Costing (ENCE 351)

5.1 Pipe culverts, slab culverts and box culverts

A culvert is a cross-drainage structure of span up to about 6 m carrying the road over a small stream. Three kinds are examined:

TypeDescriptionUsed when
Pipe culvertOne or more Hume (RCC) pipes bedded in concrete between headwallsSmall discharge; the cheapest option
Slab culvertAn RCC slab spanning between two abutments, on a concrete or masonry foundationSpan 1–6 m, firm foundation
Box culvertA monolithic RCC box — base slab, two walls and top slab acting togetherPoor foundation, or where the base must resist uplift and scour

Items for a slab culvert

S.N.ItemUnit
1Earthwork in excavation in foundation, including dewateringm³
2PCC 1:3:6 levelling course under the footingsm³
3Stone masonry or RCC in abutments and wing wallsm³
4RCC in the deck slabm³
5Reinforcement in slab and abutmentkg
6Formwork to the slab soffit and sidesm²
7Wearing coat over the slabm² or m³
8Parapet / railingm
9Backfilling behind the abutments, with a granular filterm³
10Apron, floor protection and pitching upstream and downstreamm²
11Weep holesnumber
volume of deck slab = (clear span + 2 × bearing) × width of deck × thickness abutment volume = length × mean thickness × height ( use the mean for a battered face )

Worked estimate — an RCC slab culvert

Clear span 4.00 m, clear width 1.50 m of waterway, deck 2.10 m wide overall.
Abutments 500 mm thick brick masonry in 1:4, 1.57 m high overall (0.45 footing + 0.90 clear + 0.22 slab), with wing walls 1.40 m long.
Foundation trench 0.80 m wide, 0.75 m deep, on a 300 mm PCC 1:3:6 bed.
Deck slab 220 mm RCC 1:2:4, bearing 500 mm on each abutment.
Reinforcement 16 mm main bars at 100 mm c/c, 10 mm distribution bars at 220 mm c/c, clear cover 25 mm.
Parapets 500 mm and 400 mm wide with a string course; 80 mm PCC wearing coat.

Item 1 — Earthwork in excavation

Description of itemNo.L (m)B (m)H / D (m)QuantityRemarks
Abutment walls25.300.800.756.3605.30 = 5.00 + 2 × 0.15
Wing walls41.400.800.753.3600.75 = 0.45 + 0.30
Total — Earthwork in excavation in foundation9.720m³

Item 2 — PCC 1:3:6 in foundation

Description of itemNo.L (m)B (m)H / D (m)QuantityRemarks
Under the abutment walls25.300.800.302.544
Under the wing walls41.400.800.301.344
Total — PCC 1:3:6 in foundation3.888m³

Item 3 — Brickwork in abutments, wing walls and parapets

Description of itemNo.L (m)B (m)H / D (m)QuantityRemarks
Abutment walls up to slab top25.000.501.577.8501.57 = 0.45 + 0.90 + 0.22 slab
Wing walls41.400.501.574.396
Parapet wall, 50 cm25.300.500.301.5900.30 = 0.08 + 0.22
Parapet wall, 40 cm25.300.400.401.696
String course25.300.120.080.10176
Slab bearing inside the abutments25.000.300.22−0.660
Total — First class brickwork in 1:4 cement mortar14.97376m³

The slab bearing must be deducted from the abutment masonry — that volume is occupied by the RCC slab and is measured as concrete, not as masonry. Omitting the deduction counts the same volume twice.

Item 4 — RCC 1:2:4 in the deck slab

Description of itemNo.L (m)B (m)H / D (m)QuantityRemarks
Deck slab, 220 mm15.002.100.222.3105.00 = 4.00 clear + 2 × 0.50 · 2.10 = 1.50 + 2 × 0.30
Total — RCC 1:2:4 in the deck slab2.310m³

Item 5 — Bar bending schedule

Bar bending schedule — deck slab (one-way, B < L/2)

Barø (mm)No.Cutting length (m)Total length (m)kg/mWeight (kg)Remarks
Main straight, 16 mm16262.3460.841.58096.14n = 5.00/0.10 + 1 = 51, of which 26 straight · L = 2.10 − 2 × 0.025 cover + 2 × 9d
Main bent-up, 16 mm16252.4962.251.58098.37the other 25 bars · adds 2 × 0.42 D with D = 0.22 − 2 × 0.015 − 0.016 = 0.174 m
Distribution bottom, 10 mm10115.1356.430.61734.83n = 2.10/0.22 + 1 = 11 · L = 5.00 − 2 × 0.025 + 2 × 9d
Top holder bars, 10 mm1045.1320.520.61712.672 on each side
Total weight of steel242.01kg
number of main bars = clear span / spacing + 1 = 5.00 / 0.10 + 1 = 51 bars alternate bars are bent up, so 26 straight and 25 bent-up cutting length, straight bar = 2.10 − 2 × 0.025 cover + 2 × 9 × 0.016 hook = 2.050 + 0.288 = 2.338 m say 2.34 m cutting length, bent-up bar = 2.34 + 2 × 0.42 D D = 0.22 − 2 × 0.015 − 0.016 = 0.174 m = 2.34 + 2 × 0.42 × 0.174 = 2.486 m say 2.49 m total weight of steel = 242.01 kg = 2.42 quintal

Weight per metre is taken as w = d²/162. Some textbooks use d²/160, which gives figures about 1.2% higher — 1.600 instead of 1.580 kg/m for a 16 mm bar. Either is acceptable provided you state which you have used.

Item 6 — Formwork

Description of itemNo.L (m)B (m)H / D (m)QuantityRemarks
Slab soffit15.001.50—7.500
Slab edges2—2.100.220.924
Total — Formwork to the deck slab8.424m²

Item 7 — Wearing coat

Description of itemNo.L (m)B (m)H / D (m)QuantityRemarks
Over the deck14.002.500.080.8002.50 = 1.50 + 2 × 0.50
Total — PCC 1:2:4 wearing coat, 80 mm0.800m³

For a box culvert take the four elements as one unit: base slab, two vertical walls, top slab, plus the haunches. Its great advantage in the estimate is that no separate foundation is needed — the base slab is the foundation.

Geometry after Dawadi, RCC slab culvert; every quantity recomputed here from the dimensions.

5.2 T-beam bridge (superstructure, abutment and pier)

A T-beam bridge has a deck slab cast monolithically with longitudinal girders, so that each girder and the slab above it act as a T. It is the standard medium-span (10–25 m) RCC road bridge.

Superstructure

ElementHow measured
Deck slaboverall length × width × thickness, measured over the girders
Longitudinal girdersnumber × length × web width × depth below the slab (the slab is already counted)
Cross girders / diaphragmsnumber × clear length between girders × width × depth
Kerb and footpathlength × width × thickness
Wearing coatcarriageway area × thickness
Railing / parapetrunning metre, both sides
Bearings and expansion jointscounted
Reinforcementkg, from the bar schedule of each element
Formworkm² — soffit and sides of slab, girders and cross girders

The single commonest error is double counting the flange. The deck slab is measured across the full width including the part over the girders, so each girder must be measured only for its web below the slab — not its full T-depth.

Worked estimate — superstructure of a T-beam bridge

Clear span 12.00 m, bearing 300 mm each end · overall deck width 8.40 m with a 7.20 m carriageway.
Four longitudinal girders at 2.40 m centres, 300 mm web, 1.30 m total depth · deck slab 220 mm.
Five lines of cross girders, 250 mm × 1.10 m deep · kerbs 600 × 300 mm · wearing coat 75 mm.
Description of itemNo.L (m)B (m)H / D (m)QuantityRemarks
Deck slab, 220 mm112.608.400.2223.2848measured right across, over the girders
Longitudinal girders — web below the slab412.600.301.0816.3296web 300 mm × (1.30 − 0.22) deep
Cross girders / diaphragms152.100.250.886.9305 lines × 3 bays · clear 2.40 − 0.30
Kerb and footpath212.600.600.304.536
Wearing coat, 75 mm112.607.200.0756.804carriageway width only
Total — RCC in the superstructure of a T-beam bridge57.8844m³
girder depth measured = 1.30 total − 0.22 slab = 1.08 m cross girder depth = 1.10 total − 0.22 slab = 0.88 m had the girders been measured to their full 1.30 m depth : error = 4 × 12.60 × 0.30 × 0.22 = 3.326 m³ counted twice

Abutment and pier

Both are measured as a stack of prisms, bottom upward, each with its own dimensions:

battered element : volume = length × ½(ttop + tbottom) × height circular pier : volume = π/4 × d² × height

One pier

Description of itemNo.L (m)B (m)H / D (m)QuantityRemarks
Footing16.003.000.9016.200
Circular shaft, 1.20 m dia., 5.00 m highπ/4 × 1.20² × 5.005.65487
Pier cap18.401.500.607.560
Bed blocks under the bearings40.600.600.200.288
Total — RCC in one pier29.70287m³

One abutment

Description of itemNo.L (m)B (m)H / D (m)QuantityRemarks
Footing19.004.001.0036.000
Stem — battered face, mean thickness18.401.055.0044.100mean of 1.50 at the bottom and 0.60 at the top
Abutment cap18.401.500.607.560
Dirt wall18.400.301.303.276
Return / wing walls, mean thickness24.000.603.5016.800
Total — RCC in one abutment107.736m³

The abutment stem is battered from 1.50 m at the bottom to 0.60 m at the top, so its mean thickness is ½(1.50 + 0.60) = 1.05 m. Using either the top or the bottom thickness alone would be wrong by about 43%.

Dawadi, RCC T-beam, well, septic, caisson, pier; quantities computed here from the stated dimensions.

5.3 Well foundation

A well foundation is a deep foundation sunk through a river bed to a firm stratum below the scour level. It is used where an open foundation cannot be taken down — the standard foundation for bridge piers in Nepal's braided rivers.

Parts, and how each is measured

PartFunctionMeasurement
Cutting edgeThe sharpened steel-shod bottom that helps the well sinkweight of steel in kg; concrete in m³
Well curbThe wedge-shaped RCC ring above the cutting edgem³ — volume of the tapering ring
SteiningThe main cylindrical wall of the wellm³ — the annular volume × height
Bottom plugConcrete seal at the base, cast under waterm³
Sand fillingFills the dredge hole above the bottom plugm³
Top plugConcrete seal over the sand fillingm³
Well capThe RCC slab that transfers the pier load to the wellm³
steining volume = π/4 ( Douter² − Dinner² ) × height dredge hole = π/4 × Dinner² × depth ( sand filling, plugs ) well cap = π/4 × Dcap² × thickness

Worked estimate — one well foundation

External diameter 6.00 m, internal (dredge hole) 4.00 m — steining 1.00 m thick, 18.00 m high.
Well curb 2.00 m high · bottom plug 2.00 m · sand filling 13.00 m · top plug 1.00 m · well cap 7.00 m dia. × 1.50 m thick.
Description of itemNo.L (m)B (m)H / D (m)QuantityRemarks
Well curb — tapering ring, 2.00 m high, mean thickness 1.00 mπ × 5.00 mean dia. × 1.00 × 2.0031.41593cutting edge shod with steel
Steining — 6.00 m outer, 4.00 m inner, 18.00 m highπ/4 ( 6.00² − 4.00² ) × 18.00282.74334
Bottom plug, 2.00 m thick, cast under waterπ/4 × 4.00² × 2.0025.13274
Sand filling in the dredge hole, 13.00 mπ/4 × 4.00² × 13.00163.36282
Top plug, 1.00 m thickπ/4 × 4.00² × 1.0012.56637
Well cap, 7.00 m dia., 1.50 m thickπ/4 × 7.00² × 1.5057.72677
Total volume of concrete and filling572.94796m³
steining = π/4 ( 6.00² − 4.00² ) × 18.00 = π/4 ( 36.00 − 16.00 ) × 18.00 = π/4 × 20.00 × 18.00 = 282.74 m³ dredge hole area = π/4 × 4.00² = 12.566 m² bottom plug = 12.566 × 2.00 = 25.13 m³ sand filling = 12.566 × 13.00 = 163.36 m³ top plug = 12.566 × 1.00 = 12.57 m³ check : 2.00 + 13.00 + 1.00 = 16.00 m of the 18.00 m depth is filled; the remaining 2.00 m is the curb at the bottom. ✓

Brickwork in well steining is one of the classic "state the unit" questions — it is measured and paid in m³ (§1.2). Sinking of the well is paid separately, per metre of depth, because the effort increases sharply with depth; it is not included in the rate for the steining.

Dawadi, RCC T-beam, well, septic, caisson, pier.

5.4 Earthwork in road construction in plain and hilly areas

Road earthwork means bringing the ground to formation level, by cutting where the ground is high and filling (banking) where it is low. The cross-section is a trapezium.

cross-sectional area A = B·D + s·D² B = formation width of the road D = depth of cutting or height of filling at that chainage s = side slope, expressed as s horizontal to 1 vertical area of cutting Ac = B·Dc + sc·Dc² area of filling Af = B·Df + sf·Df²

The s·D² term is the two side triangles taken together: each has base sD and height D, so the two contribute 2 × ½ × sD × D = sD². Cutting and filling normally take different side slopes, which is why sc and sf are kept apart.

Three methods of finding the volume between two sections

Let two cross-sections a distance L apart have depths d₁ and d₂.

MethodFormulaCharacter
1. Mid-sectional area
(average depth)
dm = (d₁ + d₂)/2
Amid = B·dm + s·dm²
Q = Amid × L
Averages the depth first. Under-estimates, because A is quadratic in d.
2. Mean sectional areaA₁ = B·d₁ + s·d₁²
A₂ = B·d₂ + s·d₂²
Q = ½(A₁ + A₂) × L
Averages the areas. Over-estimates, for the same reason.
3. Prismoidal formulaQ = (L/6)(A₁ + 4Amid + A₂)Combines the two in the ratio 1:4:1. The most accurate, and the one to use unless told otherwise.

Worked estimate — earthwork along a road

Formation width B = 10 m · side slope 2 : 1 in both cutting and filling · cross-sections at 30 m intervals.
Depths read from the longitudinal section are tabulated below.

Earthwork in FILLING (embankment) — B = 10 m, side slope 2:1

Chainage (m)Depth (m)Area (m²)Volume by (m³)
d₁d₂A₁AmidA₂mid-sect.mean-sect.prismoidal
0+000 – 0+0301.001.6012.00016.38021.120491.40496.80493.20
0+030 – 0+0601.601.2021.12017.92014.880537.60540.00538.40
0+060 – 0+0901.200.8014.88012.0009.280360.00362.40360.80
Total volume (m³)1389.001399.201392.40

Earthwork in CUTTING — B = 10 m, side slope 2:1

Chainage (m)Depth (m)Area (m²)Volume by (m³)
d₁d₂A₁AmidA₂mid-sect.mean-sect.prismoidal
0+120 – 0+1501.202.0014.88021.12028.000633.60643.20636.80
0+150 – 0+1802.001.5028.00023.62519.500708.75712.50710.00
0+180 – 0+2101.500.9019.50014.88010.620446.40451.80448.20
Total volume (m³)1788.751807.501795.00
SUMMARY, by the prismoidal formula total filling (embankment) = 1392.40 m³ total cutting = 1795.00 m³ ---------- surplus of cut over fill = 402.60 m³ that surplus must be carried to spoil, or used elsewhere on the project
COMPARISON of the three methods, on the filling mid-sectional = 1389.00 m³ (the smallest) prismoidal = 1392.40 m³ (between the two) mean-sectional = 1399.20 m³ (the largest) spread from the smallest to the largest = 0.73% of the prismoidal value

Mid-sectional and mean-sectional bracket the true volume from below and above; the prismoidal formula is Simpson's rule applied to the cross-sectional area and always lies between them. If a question asks you to "compare the methods", that bracketing is the answer — and the worked table above demonstrates it.

Hilly ground — the two-level section

In hilly country the ground itself slopes across the road, so the two side slopes are no longer of the same length and the simple trapezium no longer applies. The section is then a two-level section, and its area is

s·b² + r² ( 2·b·d + s·d² ) A = ------------------------------------ r² − s² b = HALF the formation width, B/2 d = centre height of banking, or centre depth of cutting s = side slope of the road, s horizontal to 1 vertical 1 in r = cross slope of the ground, 1 vertical in r horizontal on level ground r → ∞ and this reduces to A = B·d + s·d²

Do not use A = (B·d + s·d²) × r²/(r² − s²). That expression omits the s·b² term and under-estimates the area — by about 2% in the worked example below, and by more as the formation widens. The full form above is the one that reproduces the tabulated areas exactly, and it is the form Chakraborti derives in §8-7 as s·b² + 2b·r²·H + r²·s·H², all divided by (r² − s²).

A road on a cross slope is called a two-level section, because the ground meets the two side slopes at two different levels. Where the transverse and longitudinal slopes are uniform between two cross-sections, that reach is a two-level section and the formula above applies to it directly.

Worked example — road on a cross slope

Formation width B = 7.00 m, so b = B/2 = 3.50 m · side slope 2 : 1
Ground cross slope tan θ = 0.1, i.e. 1 vertical in 10 horizontal · cross-sections 20 m apart
at the first chainage, d = 1.20 m : A = [ 2 × 3.50² + 10² ( 2 × 3.50 × 1.20 + 2 × 1.20² ) ] / ( 10² − 2² ) = [ 24.50 + 100 ( 8.40 + 2.88 ) ] / 96 = [ 24.50 + 1128.0 ] / 96 = 12.005 m²

Earthwork on a cross slope — two-level section, mean sectional area method

ChainageHeight of bank d (m)Sectional area (m²)Mean area (m²)Distance (m)Quantity (m³)
51.2012.005———
62.0523.95817.98220359.64
71.9021.63022.79420455.89
81.3013.25517.44320348.85
91.2012.00512.63020252.60
102.7035.13023.56820471.35
Total quantity1888.33

Cutting and filling must still be worked out separately, because earth cut on the hill side is used to build up the valley side. The estimate then shows how much material must be borrowed or wasted — which is exactly what a mass haul diagram answers.

The zero point — where cutting changes to filling

Along a hill road the formation crosses the ground surface somewhere between two chainages. That crossing is the zero point (or balance point): the section there is neither cutting nor filling. It must be located, because the cutting volume is measured up to it and the filling volume from it.

distance of the zero point from the first chainage d x = L × --------- d = depth of cutting at the first chainage d + h h = height of filling at the next chainage chainage of the zero point = chainage of the first section + x

Two quantities at the zero point are found by averaging, because the section there is transitional:

cross slope, harmonic mean : r₀ = 2 r₁ r₂ / ( r₁ + r₂ ) formation width : b₀ = ( B₁ + B₂ ) / 2 r₁, r₂ = ground cross slopes (1 in r) at the two chainages B₁ = formation width in cutting , B₂ = formation width in banking

Worked illustration. Cutting 0.70 m deep at chainage 30 m, filling 0.80 m high at chainage 60 m; cross slopes 1 in 12 and 1 in 10; formation 8 m in cutting and 10 m in banking:

x = 30 × 0.70 / (0.70 + 0.80) = 14.00 m zero point is at chainage 30 + 14.00 = 44.00 m r₀ = 2 × 12 × 10 / (12 + 10) = 240/22 = 10.909 b₀ = (8 + 10)/2 = 9.00 m

Note that the cross slope is averaged harmonically, not arithmetically. The arithmetic mean of 12 and 10 would be 11.00; the harmonic mean is 10.909. A slope is a rate, and rates average harmonically — the same reason an average speed over equal distances is a harmonic mean.

Presentation

Road earthwork is set out chainage by chainage exactly as in the tables above — chainage, ground level, formation level, depth of cut or fill, area, mean area, length, and volume of cutting and of filling in separate columns — and the two columns are totalled separately so that the balance of earthwork is visible.

Formulas transcribed from Dawadi, estimating earth work §6.3; the worked table is computed here.

5.5 Estimation of pavement work and retaining works in roads

Pavement work

Once the formation is ready, the pavement is measured layer by layer. Each layer is an area × a compacted thickness:

LayerMeasurementUnit
Preparation of subgradeformation width × lengthm²
Granular sub-basewidth × length × compacted thicknessm³
Water bound macadam baseas above, each course separatelym³
Prime coat / tack coatarea sprayedm²
Premix carpet, bituminous concretearea × thicknessm² or m³
Seal coataream²
Shoulderseparately from the carriagewaym³

Always use the compacted thickness for payment. The loose thickness is greater — 120 mm of loose ballast compacts to about 80 mm — and the difference is absorbed in the rate through the norms.

Worked estimate — one kilometre of metalled road

Formation width 10 m · metalled (carriageway) width 5 m
Average height of bank 0.80 m · side slope 2 : 1 (H:V) · depth of borrow pit 0.30 m
Soling 150 mm boulder · sub-base and base each 120 mm loose compacted to 80 mm · two-coat bituminous surface.
Description of itemNo.L (m)B (m)H / D (m)QuantityRemarks
Survey, centre-line fixing and site clearance11,000.00——1,000.00measured per km; right of way 15 m
Permanent land required11,000.0030.00—30,000.00m²
Earthwork in embankment1000 × ( 10 × 0.80 + 2 × 0.80² ) = 1000 × 9.289,280.00m³ — A = B·D + s·D²
Temporary land for borrowing soil9280 / 0.30 borrow-pit depth30,933.3333m²
Turfing on the side slopes, both sides2 × 1000 × 0.80 × √(1 + 2²) = 2000 × 1.78893,577.7088m²
Preparation of subgrade, dressing to camber11,000.005.00—5,000.00m²
Soling coat — 75 mm boulder, 150 mm thick11,000.005.000.15750.00m³
Sub-base — 50 mm ballast, 120 mm loose to 80 mm11,000.005.000.12600.00m³
Base course — 40 mm ballast, 120 mm loose to 80 mm11,000.005.000.12600.00m³
Surface, 1st coat — 12 mm ballast @ 0.015 m³/m²1000 × 5 × 0.01575.00m³
Surface, 1st coat — bitumen @ 2 kg/m²1000 × 5 × 210,000.00kg
Surface, 2nd coat — 8 mm ballast @ 0.009 m³/m²1000 × 5 × 0.00945.00m³
Surface, 2nd coat — bitumen @ 1.2 kg/m²1000 × 5 × 1.26,000.00kg
the three quantities that carry the marks cross-sectional area of the bank A = B·D + s·D² = 10 × 0.80 + 2 × 0.80² = 8.00 + 1.28 = 9.28 m² earthwork = 9.28 × 1000 = 9,280 m³ area of borrow land = volume of earthwork / depth of borrow pit = 9,280 / 0.30 = 30,933.33 m² turfing on the slopes — note the SLOPE length, not the height slope length per metre of bank = D√(1 + s²) = 0.80 × 2.2361 = 1.7889 m area = 2 × 1000 × 1.7889 = 3,577.71 m²

The √(1 + s²) factor appears again here. For a 2:1 slope it is √5 = 2.236, so each sloping face is 2.236 times the bank height, not equal to it. Using the height alone would under-measure the turfing by more than half.

Retaining and breast walls

Hill roads need a great deal of retaining work, and it is measured as a stack of prisms exactly like a bridge abutment:

masonry wall with a battered face: volume = length × ½( top width + bottom width ) × height

Worked example — 50 m of stone masonry retaining wall, 0.60 m wide at the top, 1.80 m at the bottom, 4.00 m high, on a 0.30 m PCC bed:

Description of itemNo.L (m)B (m)H / D (m)QuantityRemarks
Earthwork in excavation for the foundation150.002.200.8088.000m³
PCC 1:3:6 levelling course150.002.000.3030.000m³
Stone masonry — mean of 0.60 and 1.80150.001.204.00240.000½(0.60 + 1.80) = 1.20 m
Granular filter behind the wall150.000.504.00100.000m³
Coping, 0.60 × 0.15150.000.600.154.500m³
(read row by row — the items go separately to the abstract)462.500m³

Add, separately: weep holes (number, one per 1.5–2 m² of face), and any drainage channel behind or in front of the wall.

Road estimate after Dawadi, 1 km highway, RCC water tank and building §6.6; every quantity recomputed here.

5.6 Earthwork and lining works in irrigation canals

An irrigation canal is estimated the same way as a road — a trapezoidal section repeated along a chainage — but here the section is a channel rather than an embankment, and it usually has to be lined.

Earthwork in the canal

area of the canal section A = B·D + s·D² B = bed width , D = depth of the channel , s = side slope volume between two sections, by the prismoidal formula: Q = (L/6)( A₁ + 4Amid + A₂ )

Cutting and filling are kept in separate columns, chainage by chainage, exactly as for a road. Where the canal runs partly in cutting and partly on a bank, the bank itself is a further filling item with its own section.

Lining

Lining reduces seepage and prevents scour. It is measured on the wetted perimeter — the surface actually covered:

wetted perimeter P = B + 2D√(1 + s²) area of lining = P × length of canal volume of lining = P × length × thickness of lining

Worked estimate — 500 m reach of a lined canal

Bed width B = 3.00 m · full supply depth D = 1.50 m · side slope 1.5 : 1
Reach length 500 m · PCC lining 100 mm thick on bed and sides.
cross-sectional area A = 3.00 × 1.50 + 1.50 × 1.50² = 4.500 + 3.375 = 7.875 m² wetted perimeter P = 3.00 + 2 × 1.50 × √(1 + 1.50²) = 3.00 + 2 × 1.50 × 1.8028 = 3.00 + 5.408 = 8.408 m
Description of itemNo.L (m)B (m)H / D (m)QuantityRemarks
Clearing and grubbing of the alignment1500.0012.00—6,000.00m² — alignment plus working width
Earthwork in excavation( 3.00 × 1.50 + 1.50 × 1.50² ) × 500 = 7.875 × 5003,937.50m³
Dressing and trimming of the bed and side slopesP × L = 8.408 × 5004,204.1635m² — on the wetted perimeter
PCC lining, 100 mm thick, on bed and sides8.408 × 500 × 0.10420.4163m³
Service road on the bank1500.003.00—1,500.00m²
Turfing of the outer slopes2500.002.70—2,700.00m² — slope length 1.50 × 1.803

The √(1 + s²) is simply the slope length of one side per unit depth — for a 1.5:1 side slope it is √(1 + 2.25) = 1.803, so each sloping face is 1.803 D long, not D. Forgetting it under-measures the lining by about half the side area.

Items to be measured

  1. Clearing and grubbing of the alignment.
  2. Earthwork in excavation, by class of soil; and earthwork in filling for banks, in compacted layers.
  3. Dressing and trimming of the bed and side slopes — m².
  4. PCC or brick lining to bed and sides — m² or m³ on the wetted perimeter.
  5. Construction joints and expansion joints — running metre.
  6. Structures on the canal — head regulator, cross regulator, outlets, falls, aqueducts and culverts, each estimated separately.
  7. Service road on the bank, and turfing of the slopes.

Canal sections after Dawadi, estimating earth work; the hydraulic relations are the standard trapezoidal-channel geometry.

Sources consulted for this chapter

These are the references actually used in writing this chapter. Where a worked example is taken from one of them, the figure or the sheet is recomputed here from the stated dimensions rather than transcribed.