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Balancing depth: bed width 20 m

Short answer

Equate cut area (20+1.5d)d to fill area 2×4×(3+2×4)=88 m². Solve 1.5d²+20d-88=0 to get d = 3.49 m.

Calculate the balancing depth of an irrigation canal using the following data: Bed width of canal = 20 m; side slope of cutting = 1.5:1 (H:V); side slope of banking = 2:1 (H:V); top width of bank = 3 m; height of embankment = 4 m (bank embankments kept 4 m higher than the ground level (berm)).

Given: B=20B = 20 m, cutting slope n1=1.5n_1 = 1.5 (H:V), banking slope n2=2n_2 = 2 (H:V), top width of bank b=3b = 3 m, height of bank above ground h=4h = 4 m. Both faces of each bank are taken at 2:1, and no berm is considered.

Condition: cutting area = filling area (two banks).

Filling area (both banks):

Af=2h (b+n2h)=2×4×(3+2×4)=88 m2A_f = 2h\,(b + n_2 h) = 2\times4\times(3 + 2\times4) = 88\ \text{m}^2

Cutting area:

Ac=(B+n1d) d=(20+1.5d) dA_c = (B + n_1 d)\,d = (20 + 1.5d)\,d

Equate:

1.5d2+20d−88=01.5d^2 + 20d - 88 = 0 d=−20+202+4×1.5×882×1.5=−20+9283=3.488 md = \frac{-20 + \sqrt{20^2 + 4\times1.5\times88}}{2\times1.5} = \frac{-20+\sqrt{928}}{3} = 3.488\ \text{m}

Check: Ac=(20+1.5×3.488)3.488=88.00A_c = (20+1.5\times3.488)3.488 = 88.00 m2^2, equal to Af=88A_f = 88 m2^2.

Answer: balancing depth = 3.49 m.

✓ Answer checked by Claude Opus 5.5 (high effort)

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