Detailed Estimate

Given data and assumptions

• Formation width $10$ m throughout; banking slope $2:1$, cutting slope $1:1$; ground level across the road.
• Formation level at 180 m is 1197.50 m; level from 120 m to 180 m; falling 1 in 120 from 90 to 120 m, so level at 90 m = 1197.50 + 30/120 = 1197.75 m; rising 1 in 90 from 180 to 210 m, so level at 210 m = 1197.833 m.

Method

Level (plain-ground) sections are used. With centre height $h$, formation width $B$ and side slope $s:1$:

$$A = (B + s\,h)\,h$$

Volume between two sections = mean area × distance; where a section changes from cutting to filling, the length of each part is found by proportion of the depths and the volume of each part = ½ × end area × length.

Sectional areas

Volumes

Answer: Earthwork in cutting = 1,745.94 m³; earthwork in filling = 187.07 m³.

Assumptions from the figure

• Length of decking $= 6.00 + 2 \times 0.45 = 6.90$ m (clear span plus bearing at both ends).
• Overhang of slab beyond outer ribs $= (7.70 - 3 \times 0.35 - 2 \times 2.25)/2 = 1.075$ m (matches the 107.5 cm in the figure).
• Fillets are taken as triangles of 15 × 15 cm at both sides of each rib (6 in total). The railing posts are at 1.36 m c/c: with 6.90 m length this gives 6 posts per side, 12 in all.
• The 20 cm slab is taken from the figure; the kerb is 20 × 20 cm above the slab, and the wearing coat (10 cm) is laid between the kerbs over the 7.0 m roadway.

(i) RCC work (1:2:4), excluding steel

(ii) Cement concrete (1:2:4) in wearing coat

$V = 7.00 \times 6.90 \times 0.10 = 4.830$ m³

Answer: RCC = 17.402 m³; wearing coat = 4.830 m³.

Materials for information (1:2:4, dry factor 1.54): RCC needs 110.3 bags cement, 7.66 m³ sand, 15.31 m³ aggregate; wearing coat needs 30.6 bags, 2.13 m³, 4.25 m³.

Given data and assumptions

• Formation width: cutting $8$ m, banking $10$ m; side slopes: cutting $1.5:1$, banking $2:1$ (H:V).
• Sections at 0+020, 0+070 and 0+140 m, so the distances are 50 m and 70 m.

Method

Ground has a transverse (cross) slope $n:1$ (H:V), so each section is a side-hill section. With centre height $h$ (cutting +, filling −) above/below formation level, formation width $B$ and side slope $s:1$:

$$\begin{aligned} x_u &= \frac{n\,(B/2 + s h)}{n - s},\quad d_u = h + \frac{x_u}{n}\\ x_d &= \frac{n\,(B/2 + s h)}{n + s},\quad d_d = h - \frac{x_d}{n}\\ A &= \tfrac12 B h + \tfrac12\left[\tfrac{B}{2}(d_u + d_d) + h\,(x_u + x_d - B)\right] \end{aligned}$$

Here $x_u, x_d$ are the horizontal distances of the up-hill and down-hill toes from the centre line and $d_u, d_d$ the heights of the ground above formation at the toes (formulas for cutting). For a section wholly in filling the roles swap: the down-hill toe is the far one, $x_d = n(B/2 + s h)/(n - s)$ and $x_u = n(B/2 + s h)/(n + s)$, with $h$ the filling height and $s$ the filling slope. Where the ground line crosses the formation level inside the road width (part cutting, part filling), the cutting and filling areas are found separately from the co-ordinates of the section (toe points from the same intersection equations). Volume between sections = mean area × distance, taken separately for cutting and filling.

Sectional areas

Working for one section:

Take chainage 0+020: centre cutting depth $h = 0.60$ m, $n = 10$, $B = 8.0$ m.

• Ground height above formation at the uphill edge $= 1.000$ m, at the downhill edge $= 0.200$ m.
• Toe of the side slope on the uphill side at $x = 5.765$ m from the centre line (ground above formation by 1.176 m); downhill toe at $x = 4.261$ m (ground above formation by 0.174 m).
• Area from the co-ordinates of the polygon: cutting $= 5.708$ m², filling $= 0.000$ m².

Volumes

Answer: Earthwork in cutting = 533.12 m³; earthwork in filling = 346.58 m³.

Given data and assumptions

• Formation width: cutting $8$ m, banking $10$ m; side slopes: cutting $1.5:1$, banking $2:1$ (H:V).
• Sections 30 m apart. Cutting at 0 m and 30 m, filling at 60 m and 90 m.

Method

Ground has a transverse (cross) slope $n:1$ (H:V), so each section is a side-hill section. With centre height $h$ (cutting +, filling −) above/below formation level, formation width $B$ and side slope $s:1$:

$$\begin{aligned} x_u &= \frac{n\,(B/2 + s h)}{n - s},\quad d_u = h + \frac{x_u}{n}\\ x_d &= \frac{n\,(B/2 + s h)}{n + s},\quad d_d = h - \frac{x_d}{n}\\ A &= \tfrac12 B h + \tfrac12\left[\tfrac{B}{2}(d_u + d_d) + h\,(x_u + x_d - B)\right] \end{aligned}$$

Here $x_u, x_d$ are the horizontal distances of the up-hill and down-hill toes from the centre line and $d_u, d_d$ the heights of the ground above formation at the toes (formulas for cutting). For a section wholly in filling the roles swap: the down-hill toe is the far one, $x_d = n(B/2 + s h)/(n - s)$ and $x_u = n(B/2 + s h)/(n + s)$, with $h$ the filling height and $s$ the filling slope. Where the ground line crosses the formation level inside the road width (part cutting, part filling), the cutting and filling areas are found separately from the co-ordinates of the section (toe points from the same intersection equations). Volume between sections = mean area × distance, taken separately for cutting and filling.

Sectional areas

Working for one section:

Take chainage 0+090: centre filling height $h = 0.85$ m, $n = 10$, $B = 10.0$ m.

• Ground height above formation at the uphill edge $= -0.350$ m, at the downhill edge $= -1.350$ m.
• Toe of the side slope on the uphill side at $x = 5.583$ m from the centre line (ground below formation by 0.292 m); downhill toe at $x = 8.375$ m (ground below formation by 1.688 m).
• Area from the co-ordinates of the polygon: cutting $= 0.000$ m², filling $= 10.880$ m².

Volumes

Answer: Earthwork in cutting = 280.83 m³; earthwork in filling = 460.27 m³.

Given data and assumptions

• Formation width $B = 10$ m; banking slope $2:1$, cutting slope $1:1$. Ground is level across the road.
• Formation levels: 1197.50 m at 60 m; level from 30 m to 90 m; falling 1 in 120 from 0 to 30 m, so the level at 0 m is 1197.50 + 30/120 = 1197.75 m; rising 1 in 90 from 90 to 120 m, so the level at 120 m is 1197.50 + 30/90 = 1197.833 m.

Method

Level (plain-ground) sections are used. With centre height $h$, formation width $B$ and side slope $s:1$:

$$A = (B + s\,h)\,h$$

Volume between two sections = mean area × distance; where a section changes from cutting to filling, the length of each part is found by proportion of the depths and the volume of each part = ½ × end area × length.

Sectional areas

Volumes

Answer: Earthwork in cutting = 1,745.94 m³; earthwork in filling = 187.07 m³.

Given data and assumptions

• Formation width $B = 10$ m; side slope in banking $2:1$, in cutting $1:1$.
• Sections 20 m apart; cutting up to 40 m and filling at 60 m and 80 m.

Method

Ground has a transverse (cross) slope $n:1$ (H:V), so each section is a side-hill section. With centre height $h$ (cutting +, filling −) above/below formation level, formation width $B$ and side slope $s:1$:

$$\begin{aligned} x_u &= \frac{n\,(B/2 + s h)}{n - s},\quad d_u = h + \frac{x_u}{n}\\ x_d &= \frac{n\,(B/2 + s h)}{n + s},\quad d_d = h - \frac{x_d}{n}\\ A &= \tfrac12 B h + \tfrac12\left[\tfrac{B}{2}(d_u + d_d) + h\,(x_u + x_d - B)\right] \end{aligned}$$

Here $x_u, x_d$ are the horizontal distances of the up-hill and down-hill toes from the centre line and $d_u, d_d$ the heights of the ground above formation at the toes (formulas for cutting). For a section wholly in filling the roles swap: the down-hill toe is the far one, $x_d = n(B/2 + s h)/(n - s)$ and $x_u = n(B/2 + s h)/(n + s)$, with $h$ the filling height and $s$ the filling slope. Where the ground line crosses the formation level inside the road width (part cutting, part filling), the cutting and filling areas are found separately from the co-ordinates of the section (toe points from the same intersection equations). Volume between sections = mean area × distance, taken separately for cutting and filling.

Sectional areas

Working for one section:

Take chainage 0+020: centre cutting depth $h = 0.30$ m, $n = 8$, $B = 10.0$ m.

• Ground height above formation at the uphill edge $= 0.925$ m, at the downhill edge $= -0.325$ m.
• Toe of the side slope on the uphill side at $x = 6.057$ m from the centre line (ground above formation by 1.057 m); downhill toe at $x = 5.867$ m (ground below formation by 0.433 m).
• Area from the co-ordinates of the polygon: cutting $= 3.911$ m², filling $= 0.563$ m².

Volumes

Answer: Earthwork in cutting = 256.73 m³; earthwork in filling = 187.23 m³.

Given data and assumptions

• Formation width $B = 10$ m; banking slope $2:1$, cutting slope $1:1$. Ground is level across the road. Stations are at 90, 120, 150, 190 and 210 m (spacing 30, 30, 40, 20 m).
• Formation level at 150 m is 1197.50 m; level from 120 m to 180 m; falling 1 in 120 from 90 to 120 m, so level at 90 m = 1197.75 m; rising 1 in 90 from 180 m, so level at 190 m = 1197.50 + 10/90 = 1197.611 m and at 210 m = 1197.833 m.

Method

Level (plain-ground) sections are used. With centre height $h$, formation width $B$ and side slope $s:1$:

$$A = (B + s\,h)\,h$$

Volume between two sections = mean area × distance; where a section changes from cutting to filling, the length of each part is found by proportion of the depths and the volume of each part = ½ × end area × length.

Sectional areas

Volumes

Answer: Earthwork in cutting = 1,786.15 m³; earthwork in filling = 187.07 m³.

Given data and assumptions

• Formation width $B = 10$ m; side slope in banking $2:1$, in cutting $1:1$.
• Sections 20 m apart: cutting at 0, 20, 40 m and filling at 60, 80 m.

Method

Ground has a transverse (cross) slope $n:1$ (H:V), so each section is a side-hill section. With centre height $h$ (cutting +, filling −) above/below formation level, formation width $B$ and side slope $s:1$:

$$\begin{aligned} x_u &= \frac{n\,(B/2 + s h)}{n - s},\quad d_u = h + \frac{x_u}{n}\\ x_d &= \frac{n\,(B/2 + s h)}{n + s},\quad d_d = h - \frac{x_d}{n}\\ A &= \tfrac12 B h + \tfrac12\left[\tfrac{B}{2}(d_u + d_d) + h\,(x_u + x_d - B)\right] \end{aligned}$$

Here $x_u, x_d$ are the horizontal distances of the up-hill and down-hill toes from the centre line and $d_u, d_d$ the heights of the ground above formation at the toes (formulas for cutting). For a section wholly in filling the roles swap: the down-hill toe is the far one, $x_d = n(B/2 + s h)/(n - s)$ and $x_u = n(B/2 + s h)/(n + s)$, with $h$ the filling height and $s$ the filling slope. Where the ground line crosses the formation level inside the road width (part cutting, part filling), the cutting and filling areas are found separately from the co-ordinates of the section (toe points from the same intersection equations). Volume between sections = mean area × distance, taken separately for cutting and filling.

Sectional areas

Working for one section:

Take chainage 0+020: centre cutting depth $h = 0.70$ m, $n = 12$, $B = 10.0$ m.

• Ground height above formation at the uphill edge $= 1.117$ m, at the downhill edge $= 0.283$ m.
• Toe of the side slope on the uphill side at $x = 6.218$ m from the centre line (ground above formation by 1.218 m); downhill toe at $x = 5.262$ m (ground above formation by 0.262 m).
• Area from the co-ordinates of the polygon: cutting $= 7.717$ m², filling $= 0.000$ m².

Volumes

Answer: Earthwork in cutting = 330.59 m³; earthwork in filling = 162.29 m³.

Given data and assumptions

• Bed width $4$ m; FSD $1$ m; free board $0.45$ m, so $D = 1.45$ m; top width of bank $3$ m (left) and $1.5$ m (right).
• Side slope in cutting $1:1$, in banking $1.5:1$ (inner and outer faces of banks).
• Bed level 109.52 m at chainage 90, rising at 1 in 300 with chainage (assumed direction): 109.52, 109.62, 109.72, 109.82, 109.92, 109.953 m.

Method

Total depth of canal section $D = \text{FSD} + \text{free board} = 1.00 + 0.45 = 1.45$ m. At each station the bed level is compared with the ground level; $e =$ GL − bed level.

• If $e \ge D$: full cutting, area $A_c = (b + s_c e)\,e$ (no banks needed).
• If $0 < e < D$: part cutting, part banking. Cutting area $A_c = (b + s_c e)e$; each bank stands $h_b = D - e$ above ground, so bank area (both banks) $A_b = (t_L + t_R)h_b + (s_i + s_o)h_b^2$ with $s_i, s_o$ the inner and outer bank slopes.
• If $e \le 0$: full banking, bed above ground. Bank top is $H = D - e$ above ground. Gross embankment $= (T + s_o H)H$ with $T = t_L + t_R + b + 2 s_i D$, and the canal space $(b + s_i D)D$ is deducted.

Volume between stations = mean area × distance (separately for cutting and banking).

Sectional areas

Volumes

Permanent land

Width of land at a section = distance between the outer toes of the banks (banking) or canal top width plus the two bank/berm widths (cutting).

Area of land $=$ mean width × length, summed over the reaches $= 2,330.70$ m² $= 0.2331$ ha.

Answer: Earthwork in cutting = 27.77 m³; earthwork in banking = 2,316.23 m³; total = 2,344.00 m³. Permanent land = 2,330.70 m².

Given data and assumptions

• Bed width $4$ m; FSD $1$ m; free board $0.45$ m, so $D = 1.45$ m; top width of bank $3$ m (left) and $1.5$ m (right).
• Side slope in cutting $1:1$, in banking $1.5:1$ (inner and outer faces of banks).
• The printed bed level 109.52 m at chainage 800 is the level just upstream of the 0.50 m fall; the reach considered starts after the fall, so bed at 800 is 109.02 m, rising 1 in 250 with chainage (same direction as the companion problem): 109.02, 109.22, 109.42, 109.62, 109.82, 110.02 m.

Method

Total depth of canal section $D = \text{FSD} + \text{free board} = 1.00 + 0.45 = 1.45$ m. At each station the bed level is compared with the ground level; $e =$ GL − bed level.

• If $e \ge D$: full cutting, area $A_c = (b + s_c e)\,e$ (no banks needed).
• If $0 < e < D$: part cutting, part banking. Cutting area $A_c = (b + s_c e)e$; each bank stands $h_b = D - e$ above ground, so bank area (both banks) $A_b = (t_L + t_R)h_b + (s_i + s_o)h_b^2$ with $s_i, s_o$ the inner and outer bank slopes.
• If $e \le 0$: full banking, bed above ground. Bank top is $H = D - e$ above ground. Gross embankment $= (T + s_o H)H$ with $T = t_L + t_R + b + 2 s_i D$, and the canal space $(b + s_i D)D$ is deducted.

Volume between stations = mean area × distance (separately for cutting and banking).

Sectional areas

Volumes

Permanent land

Width of land at a section = distance between the outer toes of the banks (banking) or canal top width plus the two bank/berm widths (cutting).

Area of land $=$ mean width × length, summed over the reaches $= 4,298.75$ m² $= 0.4299$ ha.

Answer: Earthwork in cutting = 227.58 m³; earthwork in banking = 3,830.78 m³; total = 4,058.36 m³. Permanent land = 4,298.75 m².

A drawing is not given, so the method is shown on an assumed small aqueduct: a single-span RCC trough carrying an irrigation canal over a drain, resting on two brick abutments with four wing walls. All quantities are by the centre-line / length × breadth × depth method.

Assumed data

• Trough: clear span 6.0 m, bearing 0.3 m each end (length 6.6 m), internal width 2.0 m, depth 1.2 m, walls 0.20 m, bottom slab 0.25 m, so outside width 2.4 m.
• Abutments (2 nos.): length 3.4 m, thickness 0.9 m, height 2.5 m above PCC, in 1:6 brickwork; PCC 1:3:6 bed 4.0 m × 1.5 m × 0.30 m.
• Wing walls (4 nos.): 2.0 m long, mean thickness 0.45 m, mean height 2.0 m; PCC 2.0 m × 0.9 m × 0.30 m.
• Abutment pit depth 1.5 m with 0.3 m working space on each side; wing-wall trench 1.2 m wide, 1.2 m deep.

a) Earthwork in excavation

b) Cement concrete in foundation (1:3:6)

c) Brick work (1:6 cement sand mortar)

d) RCC work (trough, M20)

Reinforcement is then taken separately (about 80–100 kg per m³ of trough concrete as a check).

Answer: Excavation = 40.50 m³; PCC = 5.76 m³; brickwork = 22.50 m³; RCC = 7.13 m³ (for the assumed dimensions; substitute the dimensions of the actual drawing in the same format).

Quantities are taken in the usual order: length × breadth × height/depth, in cubic metres (m³) or square metres (m²), with deductions entered as negatives. Rates are not asked; when costing, apply the current District Rate and DUDBC rate-analysis norms of the district.

No drawing is given. The method is therefore shown on an assumed one-room building with RCC flat roof, using the centre-line method.

Assumed data

• Room 4.5 m × 3.5 m clear; walls 0.23 m (first class brick in 1:6 cement mortar); centre-line length = 2(4.5 + 3.5 + 2 × 0.23) = 16.92 m.
• Trench 0.90 m wide, 1.00 m deep; PCC 1:3:6, 0.20 m thick; two brick footing steps (0.69 m and 0.46 m wide, 0.15 m each); plinth 0.45 m above GL; ceiling height 3.0 m.
• One door 1.0 × 2.1 m and two windows 1.2 × 1.2 m; RCC lintels 0.23 × 0.15 m with 0.2 m bearing each side; roof slab 125 mm; floor PCC 100 mm.
• Wall height from last footing to plinth top = 1.00 − 0.20 − 0.30 + 0.45 = 0.95 m.

a) Earthwork in excavation

b) Cement concrete

c) First class brick work (foundation, plinth and superstructure)

(Lintel bearing is deducted because it is measured under RCC.)

d) RCC work

Answer: Excavation = 15.23 m³; cement concrete = 4.62 m³; brickwork = 16.99 m³; RCC = 2.61 m³ (assumed dimensions; follow the same sequence for the actual drawing).

Quantities are taken in the usual order: length × breadth × height/depth, in cubic metres (m³) or square metres (m²), with deductions entered as negatives. Rates are not asked; when costing, apply the current District Rate and DUDBC rate-analysis norms of the district.

Data. Internal dia = 2.0 m; masonry wall 0.30 m, so outer dia = 2.0 + 2 × 0.30 = 2.60 m. Founding depth below GL = 14.0 m (including the 0.4 m curb); water table at 3.8 m; well rises 0.6 m above GL.

• Outer cross-section $A_o=\frac{\pi}{4}(2.60)^2=5.309$ m²
• Annular area of steining $=\frac{\pi}{4}(2.60^2-2^2)=2.168$ m²

a) Earthwork in excavation (above water table)

Dry excavation from GL to 3.8 m: 5.309 × 3.8 = 20.18 m³.

b) Sinking of well (below water table)

The remaining depth 14.0 − 3.8 = 10.2 m is sunk by dredging under water: 5.309 × 10.2 = 54.15 m³. (If a single item is wanted, total = 74.33 m³.)

c) Brick masonry (1:6) in steining

Height of steining = 14.0 − 0.4 (curb) + 0.6 (above GL) = 14.20 m.

Volume = 2.168 × 14.20 = 30.78 m³.

d) PCC (M25) for RCC curb

Curb ring area is the same, depth 0.4 m: 2.168 × 0.4 = 0.867 m³ (reinforcement is measured separately in kg).

The 0.45 m well cover (1:4:8) is not asked here; it is π/4 × 2.0² × 0.45 = 1.41 m³.

Answer: Excavation 20.18 m³; sinking 54.15 m³; brickwork 30.78 m³; curb concrete 0.867 m³.

Data. Internal dia 2.0 m; wall 0.30 m, so outer dia = 2.60 m. Depth of well below GL = 14.0 m (curb included); well rises 0.6 m above GL; curb 0.4 m; cover 0.45 m thick placed inside the top of the bore.

• Outer area $A_o=5.309$ m²; inner area $A_i=\frac{\pi}{4}\times 2^2=3.142$ m²; ring area $2.168$ m².

a) Sinking of well

$A_o\times 14.0 = 5.309\times 14 =$ 74.33 m³ (of this, 3.8 m is above the water table: 20.18 m³ and 54.15 m³ below it, which is priced separately if required).

b) Brick masonry (1:6)

Height = 14.0 − 0.4 + 0.6 = 14.20 m; volume = 2.168 × 14.20 = 30.78 m³.

c) Sand filling

Filling goes from the top of the curb to the underside of the cover: height = 14.0 + 0.6 − 0.4 − 0.45 = 13.75 m; volume = 3.142 × 13.75 = 43.20 m³.

d) PCC (1:4:8) in well cover

Volume = $A_i\times 0.45=3.142\times 0.45=$ 1.414 m³.

Answer: Sinking 74.33 m³; brickwork 30.78 m³; sand filling 43.20 m³; well cover concrete 1.414 m³.

Data. Internal dia 5.0 m, wall 0.80 m, so outer dia = 5.0 + 2 × 0.8 = 6.60 m. Founding depth = 15.0 m below river bed (dry river, so no water-table complication). Assumptions: the top of the well is at river-bed level and the steining includes the curb (a separate RCC curb would be deducted from the brickwork).

• $A_o=\frac{\pi}{4}(6.60)^2=34.212$ m²; $A_i=\frac{\pi}{4}(5)^2=19.635$ m²; ring area $=14.577$ m².

Answer: Excavation 513.18 m³; brickwork 218.65 m³; bottom plug 29.45 m³ (1:4:8); top seal 19.63 m³ (1:2:4); sand filling 245.44 m³.

Data. Internal dia 4.5 m, wall 0.80 m, so outer dia = 4.5 + 2 × 0.8 = 6.10 m. Founding depth 15.0 m below river bed (dry). Assumptions: top of the well at bed level and the steining includes the curb.

• $A_o=\frac{\pi}{4}(6.10)^2=29.225$ m²; $A_i=\frac{\pi}{4}(4.5)^2=15.904$ m²; ring area $=13.320$ m².

Total 1:4:8 concrete = 27.83 m³.

Answer: Excavation 438.37 m³; brickwork 199.81 m³; CC 1:4:8 = 27.83 m³ (bottom 15.90 + top 11.93); sand filling 210.73 m³.

Assumptions from the drawing

• Length of decking $= 6.00 + 2 \times 0.45 = 6.90$ m (clear span plus bearing at both ends).
• Overhang of slab beyond outer ribs $= (7.70 - 3 \times 0.35 - 2 \times 2.25)/2 = 1.075$ m (matches the 107.5 cm in the figure).
• Fillets are taken as triangles of 15 × 15 cm at both sides of each rib (6 in total). The railing posts are at 1.36 m c/c: with 6.90 m length this gives 6 posts per side, 12 in all.
• The 20 cm slab is taken from the figure; the kerb is 20 × 20 cm above the slab, and the wearing coat (10 cm) is laid between the kerbs over the 7.0 m roadway.
• Reinforcement as given: beams (ribs and fillets) 2.5%, slab and kerbs 1%, posts 1% of the volume of concrete; density of steel 7850 kg/m³.
• The railing has 4 lines of 40 mm G.I. pipe on each side, running the full length of the decking (assumption; the figure shows "4 nos").

Quantities

Answer: RCC (1:2:4) = 17.402 m³; wearing coat = 4.830 m³; reinforcement = 2,060.7 kg; 40 mm G.I. pipe = 55.2 m.

Assumptions from the figure

• Length of decking $= 6.00 + 2 \times 0.45 = 6.90$ m (clear span plus bearing at both ends).
• Overhang of slab beyond outer ribs $= (7.70 - 3 \times 0.35 - 2 \times 2.25)/2 = 1.075$ m (matches the 107.5 cm in the figure).
• Fillets are taken as triangles of 15 × 15 cm at both sides of each rib (6 in total). The railing posts are at 1.36 m c/c: with 6.90 m length this gives 6 posts per side, 12 in all.
• The 20 cm slab is taken from the figure; the kerb is 20 × 20 cm above the slab, and the wearing coat (10 cm) is laid between the kerbs over the 7.0 m roadway.
• Percentage of steel is not stated; the usual values for T-beam decking are taken: beams 2.5%, slab and kerbs 1%, posts 1%. Density of steel 7850 kg/m³.

(a) Concrete for the RCC work (1:2:4)

(b) Reinforcement

Answer: concrete in RCC members = 17.402 m³; reinforcement = 2,060.7 kg (2.061 tonne).

Given data and assumptions

• Formation width $B = 8$ m in both cutting and filling; side slope in cutting $1.5:1$, in filling $2:1$ (H:V).
• Chainages are 30 m apart; cross slopes are taken as $n:1$ (H:V), e.g. 12:1 means 12 horizontal to 1 vertical.

Method

Ground has a transverse (cross) slope $n:1$ (H:V), so each section is a side-hill section. With centre height $h$ (cutting +, filling −) above/below formation level, formation width $B$ and side slope $s:1$:

$$\begin{aligned} x_u &= \frac{n\,(B/2 + s h)}{n - s},\quad d_u = h + \frac{x_u}{n}\\ x_d &= \frac{n\,(B/2 + s h)}{n + s},\quad d_d = h - \frac{x_d}{n}\\ A &= \tfrac12 B h + \tfrac12\left[\tfrac{B}{2}(d_u + d_d) + h\,(x_u + x_d - B)\right] \end{aligned}$$

Here $x_u, x_d$ are the horizontal distances of the up-hill and down-hill toes from the centre line and $d_u, d_d$ the heights of the ground above formation at the toes (formulas for cutting). For a section wholly in filling the roles swap: the down-hill toe is the far one, $x_d = n(B/2 + s h)/(n - s)$ and $x_u = n(B/2 + s h)/(n + s)$, with $h$ the filling height and $s$ the filling slope. Where the ground line crosses the formation level inside the road width (part cutting, part filling), the cutting and filling areas are found separately from the co-ordinates of the section (toe points from the same intersection equations). Volume between sections = mean area × distance, taken separately for cutting and filling.

Sectional areas

Working for one section:

Take chainage 0+220: centre filling height $h = 0.70$ m, $n = 10$, $B = 8.0$ m.

• Ground height above formation at the uphill edge $= -0.300$ m, at the downhill edge $= -1.100$ m.
• Toe of the side slope on the uphill side at $x = 4.500$ m from the centre line (ground below formation by 0.250 m); downhill toe at $x = 6.750$ m (ground below formation by 1.375 m).
• Area from the co-ordinates of the polygon: cutting $= 0.000$ m², filling $= 7.188$ m².

Volumes

Answer: Earthwork in cutting = 344.80 m³; earthwork in filling = 510.49 m³.

Given data and assumptions

• Formation width $B = 8$ m; side slope in filling $2:1$, in cutting $1.5:1$ (H:V). Ground is taken as level across the road (no cross slope given).
• Formation level at 0 m is 100.00 m. The gradient is read as 1 in 200 upward from 0 to 300 m and 1 in 400 downward from 300 m to 600 m (the printed table shows only the two gradients).

Method

Level (plain-ground) sections are used. With centre height $h$, formation width $B$ and side slope $s:1$:

$$A = (B + s\,h)\,h$$

Volume between two sections = mean area × distance; where a section changes from cutting to filling, the length of each part is found by proportion of the depths and the volume of each part = ½ × end area × length.

Sectional areas

Volumes

Answer: Earthwork in cutting = 2,410.45 m³; earthwork in filling = 1,875.00 m³.

Given data and assumptions

• Two-lane road: carriageway $7.0$ m with $1.5$ m shoulders on both sides (Nepal Road Standard 2070), so formation width $B = 10.0$ m. The earthwork is taken to subgrade (formation) level.
• Cut slope $1.5:1$, fill slope $2:1$, ground cross slope $9:1$, centre depth of cutting $h = 0.35$ m at every section, so the section is uniform along the 1 km.
• Wearing coat and base course 7.0 m wide; sub-base spread over the full formation width of 10 m.

Earthwork (side-hill section)

At the up-hill edge the ground is $0.35 + 5/9 = 0.906$ m above formation (cutting), and at the down-hill edge it is $0.35 - 5/9 = -0.206$ m (so that side falls below formation and needs filling). The ground cuts the formation level at $9 \times 0.35 = 3.15$ m from the centre line.

• Up-hill toe: $x_u = 6.630$ m from the centre line, ground $1.087$ m above formation
• Down-hill toe (filling slope): $x_d = 5.529$ m, ground $0.264$ m below formation
• Area of cutting $= 4.428$ m²; area of filling $= 0.244$ m² (from the co-ordinates of the section).

$$\begin{aligned} V_{cut} &= 4.428 \times 1000 = 4,428.2\ \text{m}^3\\ V_{fill} &= 0.244 \times 1000 = 244.5\ \text{m}^3 \end{aligned}$$

Pavement quantities for 1 km

Answer: cutting = 4,428.2 m³, filling = 244.5 m³ per km; wearing coat 350 m³, base 1,750 m³, sub-base 3,500 m³. The cost follows by multiplying each quantity by its rate (District Rate) and adding overhead, profit and VAT.

A balanced section is a canal section partly in cutting and partly in banking in which the earth excavated from the bed is exactly enough to build the banks, so there is no surplus spoil and no earth is borrowed. The depth of cutting at which this happens is the balancing depth.

 bank top ______                    ______
         /      \                  /      \
        /  bank  \  F.S.L ~~~~~~~/  bank  \
 G.L ===/          \            /          \===
                    \          /
                     \________/
                      bed width b

Condition: area of cutting below ground = area of the two banks above ground.

$$(b + s\,d)\,d = 2\,(t + s\,h_b)\,h_b,\qquad h_b = D - d$$

where $d$ is the depth of cutting, $D$ = full supply depth + free board, $t$ = top width of one bank and $s$ the side slope.

Example (assumed $b = 5$ m, $s = 1.5$, $t = 2$ m, $D = 1.6$ m): solving gives $d = 0.810$ m, cutting area $= 5.033$ m² and bank area $= 5.033$ m² (equal).

Quantities of road earthwork are found from the longitudinal section and cross-sections.

1. Take cross-sections at regular intervals (e.g. 10–30 m, closer on curves) and at points where the ground changes.
2. Find the depth of cutting or height of filling at the centre line from the ground level and formation level; the formation level follows the design gradients.
3. For each section calculate the area of cutting and filling from the formation width and side slopes ($A = (B + s h)h$ for level ground; side-hill formulae when the ground has a cross slope).
4. Find the volume between consecutive sections by the mean sectional area method ($V = \tfrac{A_1 + A_2}{2} \times L$), or the prismoidal formula for greater accuracy.
5. Add the quantities of pavement layers (area × thickness), drains, culverts and protection works, and put everything in a bill of quantities.

Given data and assumptions

• Formation width of a dedicated four-lane road (NRS 2070): four lanes of 3.5 m, a 1.5 m median and two shoulders of 2.5 m give about 21.0 m; $B = 21$ m is used (check with the standard in use).
• Side slope in embankment $2:1$ and in excavation $1.5:1$ (the question writes 1:2 and 1:1.5, i.e. 1 vertical to 2 and 1.5 horizontal). Cross slopes: "plain" is taken as level ground.
• The printed formation levels are unclear. Formation level at 0+000 is taken as 152.00 m, rising 1 in 200 up to 0+090 (152.45 m, as printed) and then 1 in 300 to 0+180 (152.75 m).

Method

Ground has a transverse (cross) slope $n:1$ (H:V), so each section is a side-hill section. With centre height $h$ (cutting +, filling −) above/below formation level, formation width $B$ and side slope $s:1$:

$$\begin{aligned} x_u &= \frac{n\,(B/2 + s h)}{n - s},\quad d_u = h + \frac{x_u}{n}\\ x_d &= \frac{n\,(B/2 + s h)}{n + s},\quad d_d = h - \frac{x_d}{n}\\ A &= \tfrac12 B h + \tfrac12\left[\tfrac{B}{2}(d_u + d_d) + h\,(x_u + x_d - B)\right] \end{aligned}$$

Here $x_u, x_d$ are the horizontal distances of the up-hill and down-hill toes from the centre line and $d_u, d_d$ the heights of the ground above formation at the toes (formulas for cutting). For a section wholly in filling the roles swap: the down-hill toe is the far one, $x_d = n(B/2 + s h)/(n - s)$ and $x_u = n(B/2 + s h)/(n + s)$, with $h$ the filling height and $s$ the filling slope. Where the ground line crosses the formation level inside the road width (part cutting, part filling), the cutting and filling areas are found separately from the co-ordinates of the section (toe points from the same intersection equations). Volume between sections = mean area × distance, taken separately for cutting and filling.

Sectional areas

Working for one section:

Take chainage 0+060: centre cutting depth $h = 0.30$ m, $n = 12$, $B = 21.0$ m.

• Ground height above formation at the uphill edge $= 1.175$ m, at the downhill edge $= -0.575$ m.
• Toe of the side slope on the uphill side at $x = 12.514$ m from the centre line (ground above formation by 1.343 m); downhill toe at $x = 11.880$ m (ground below formation by 0.690 m).
• Area from the co-ordinates of the polygon: cutting $= 9.467$ m², filling $= 2.381$ m².

Volumes

Answer: Earthwork in cutting = 1,315.99 m³; earthwork in filling = 610.78 m³.

Given data and assumptions

• Formation width $B = 10$ m; side slope in cutting $1:1$, in filling $2:1$. Ground is level across the road.
• Formation level at 0 m is 101.00 m, rising 1 in 200 up to 120 m (101.60 m) and falling 1 in 120 from 120 m to 180 m (101.10 m). The printed 102.15 at 120 m does not fit a 1 in 200 gradient, so the stated gradients are used.

Method

Level (plain-ground) sections are used. With centre height $h$, formation width $B$ and side slope $s:1$:

$$A = (B + s\,h)\,h$$

Volume between two sections = mean area × distance; where a section changes from cutting to filling, the length of each part is found by proportion of the depths and the volume of each part = ½ × end area × length.

Sectional areas

Volumes

Answer: Earthwork in cutting = 1,623.40 m³; earthwork in filling = 137.02 m³.

Given data and assumptions

• Formation width $B = 10$ m; side slope in cutting $1:1$, in banking $2:1$ (H:V).
• Sections 50 m apart; cutting at 0 and 50 m, filling at 100 and 150 m.

Method

Ground has a transverse (cross) slope $n:1$ (H:V), so each section is a side-hill section. With centre height $h$ (cutting +, filling −) above/below formation level, formation width $B$ and side slope $s:1$:

$$\begin{aligned} x_u &= \frac{n\,(B/2 + s h)}{n - s},\quad d_u = h + \frac{x_u}{n}\\ x_d &= \frac{n\,(B/2 + s h)}{n + s},\quad d_d = h - \frac{x_d}{n}\\ A &= \tfrac12 B h + \tfrac12\left[\tfrac{B}{2}(d_u + d_d) + h\,(x_u + x_d - B)\right] \end{aligned}$$

Here $x_u, x_d$ are the horizontal distances of the up-hill and down-hill toes from the centre line and $d_u, d_d$ the heights of the ground above formation at the toes (formulas for cutting). For a section wholly in filling the roles swap: the down-hill toe is the far one, $x_d = n(B/2 + s h)/(n - s)$ and $x_u = n(B/2 + s h)/(n + s)$, with $h$ the filling height and $s$ the filling slope. Where the ground line crosses the formation level inside the road width (part cutting, part filling), the cutting and filling areas are found separately from the co-ordinates of the section (toe points from the same intersection equations). Volume between sections = mean area × distance, taken separately for cutting and filling.

Sectional areas

Working for one section:

Take chainage 0+000: centre cutting depth $h = 0.50$ m, $n = 12$, $B = 10.0$ m.

• Ground height above formation at the uphill edge $= 0.917$ m, at the downhill edge $= 0.083$ m.
• Toe of the side slope on the uphill side at $x = 6.000$ m from the centre line (ground above formation by 1.000 m); downhill toe at $x = 5.077$ m (ground above formation by 0.077 m).
• Area from the co-ordinates of the polygon: cutting $= 5.462$ m², filling $= 0.000$ m².

Volumes

Answer: Earthwork in cutting = 470.38 m³; earthwork in filling = 412.95 m³.