How is the total depth D of the canal section found?

D = FSD + free board = 1.5 + 0.5 = 2.0 m

At 1+000 the bed RL (1000.00) is above the ground RL (999.50). How is the section treated?

Full banking, because e = GL - bed is negative

Full cutting, because the bed is a known level

Part cutting, part banking, because e is less than D

Not counted, because the ground is lower than the bed

How are volumes between stations obtained here?

Mean of the end areas x distance, done separately for cutting and banking

Area at the first station x distance, for cutting and banking combined

Sum of cutting and banking areas x distance only

Prismoidal rule applied on total depth D

2082 Baisakh · 8 marks

Channel earthwork, chainage 0+500 to 1+500

Short answer

Using ground vs bed level comparison, partial cutting/banking and full banking sections are identified. Volumes: cutting = 2,047.50 m³, banking = 35,930.00 m³, total = 37,977.50 m³.

Calculate the quantities of earthwork from the following data for channel section from chainage 0+500 to 1+500 m. Bed width = 5 m, side slope in cutting 1H:1V, side slope in banking 1.5H:1V, top width of both banks = 2.5 m, full supply depth = 1.5 m and free board = 0.5 m.

Chainage	RL of Bed (m)	RL of ground (m)
0+500	1000.50	1001.80
0+1000	1000.00	999.50
0+1500	999.50	997.50

Given data and assumptions

Bed width $5$ m; FSD $1.5$ m; free board $0.5$ m, so $D = 2.0$ m; top width of each bank $2.5$ m.
Side slope in cutting $1:1$ ; side slope of banks (inner and outer) $1.5:1$ (H:V).
Chainages 0+500, 1+000 and 1+500 are 500 m apart.

Method

Total depth of canal section $D = \text{FSD} + \text{free board} = 1.50 + 0.50 = 2.00$ 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

Station	GL (m)	Bed RL (m)	$e$ (m)	Condition	Cutting area (m²)	Bank area (m²)
0+500	1,001.800	1,000.500	1.300	part cutting, part banking	8.190	4.970
1+000	999.500	1,000.000	-0.500	full banking	0.000	33.375
1+500	997.500	999.500	-2.000	full banking	0.000	72.000

Volumes

Between	Distance (m)	Cutting (m³)	Banking (m³)
0+500 – 1+000	500.0	2,047.50	9,586.25
1+000 – 1+500	500.0	0.00	26,343.75
Total		2,047.50	35,930.00

Answer: Earthwork in cutting = 2,047.50 m³; earthwork in banking = 35,930.00 m³; total = 37,977.50 m³.

✓ Answer checked by Claude Opus 5.5 (high effort)

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