At 0+900, GL = 109.55 and bed = 109.42, D = 1.45 m. Which case applies?

Part cutting, part banking, since 0 < e = 0.13 < D

Full cutting, since GL is above the bed

Full banking, since e is smaller than D

Full cutting, since e is below 1 m

FSD + free board = 1.00 + 0.45 = 1.45 m

Asked 2/22
2078 Bhadra · 10 marks

Channel earthwork and land area, 50 cm fall

Short answer

Part cutting, part banking throughout; cutting = 1,155.86 m³, banking = 785.21 m³. Permanent land area from outer bank toes = 3,365.00 m².

Calculate the quantity of earthwork and area of permanent land required for the land acquisition purpose for a portion of a channel from following data: Bed width = 4 m; Free board = 45 cm; Side slope in cutting = 1:1; Side slope in banking = 1.5:1; Full supply depth = 1 m; Top width of bank = 3 m left and 1.5 m right. There is 50 cm fall at chainage 800 m.

Chainage	800	850	900	950	1000	1050
RL ground	109.8	109.7	109.55	109.30	109.25	109.15
RL Bed	109.52	Bed slope 1:250

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 ground falls with chainage, so the canal is taken to flow in the direction of increasing chainage, with the bed falling at 1 in 250. The printed bed level 109.52 m at chainage 800 is the level just upstream of the 0.50 m fall. The reach considered lies downstream of the fall, so the bed at 800 is $109.52 - 0.50 = 109.02$ m, falling $50/250 = 0.20$ m every 50 m: 109.02, 108.82, 108.62, 108.42, 108.22, 108.02 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.

Here every station is in part cutting, part banking, so $A_c = (4 + e)e$ and $A_b = 4.5\,h_b + 3\,h_b^2$ .

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

Sectional areas

Station	GL (m)	Bed RL (m)	$e$ (m)	Condition	Cutting area (m²)	Bank area (m²)
0+800	109.800	109.020	0.780	part cutting, part banking	3.728	4.362
0+850	109.700	108.820	0.880	part cutting, part banking	4.294	3.540
0+900	109.550	108.620	0.930	part cutting, part banking	4.585	3.151
0+950	109.300	108.420	0.880	part cutting, part banking	4.294	3.540
1+000	109.250	108.220	1.030	part cutting, part banking	5.181	2.419
1+050	109.150	108.020	1.130	part cutting, part banking	5.797	1.747

Volumes

Between	Distance (m)	Cutting (m³)	Banking (m³)
0+800 – 0+850	50.0	200.57	197.54
0+850 – 0+900	50.0	221.98	167.27
0+900 – 0+950	50.0	221.98	167.27
0+950 – 1+000	50.0	236.88	148.97
1+000 – 1+050	50.0	274.45	104.16
Total		1,155.86	785.21

Permanent land

Width of land at a section = distance between the outer toes of the two banks = canal width at ground level $(b + 2e)$ + inner and outer slopes of both banks $(4 \times 1.5\,h_b)$ + bank tops $(3 + 1.5)$ .

Station	Land width (m)
0+800	14.080
0+850	13.680
0+900	13.480
0+950	13.680
1+000	13.080
1+050	12.680

Area of land $=$ mean width × length, summed over the reaches $= 694.0 + 679.0 + 679.0 + 669.0 + 644.0 = 3,365.00$ m² $= 0.3365$ ha.

Answer: Earthwork in cutting = 1,155.86 m³; earthwork in banking = 785.21 m³; total = 1,941.07 m³. Permanent land = 3,365.00 m².

✓ Answer checked by Claude Opus 5.5 (high effort)

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