With ground at 0.00, what are the levels of the bed and the bank crest?

Bed at -3.59 m and crest at +3.00 m

Bed at +3.59 m and crest at -3.00 m

Bed at -3.00 m and crest at +3.59 m

2072 Asoj · 3+2 marks

Balancing depth and canal cross-section

Short answer

Equate cutting area (8+1.5d)*d to filling area 2*3*(2+2*3)=48 m². Solve quadratic 1.5d² + 8d - 48 = 0. Balancing depth = 3.59 m.

An irrigation channel has a bottom width 8 m and side slopes of 1.5H:1V in cutting and 2H:1V in filling. The width of the crest of bank is 2 m and its height above the ground level is 3 m. Compute the balancing depth and draw a neat x-section of the canal illustrating the various dimensions and level it.

Given: $B = 8$ m, $n_1 = 1.5$ (cutting), $n_2 = 2$ (filling), crest width $b = 2$ m, bank height above ground $h = 3$ m. Banks are on both sides with 2:1 slopes on both faces; no berm.

Filling area:

A_f = 2h\,(b + n_2h) = 2\times3\times(2+2\times3) = 48\ \text{m}^2

Cutting area: $A_c = (8 + 1.5d)\,d$

Equate:

1.5d^2 + 8d - 48 = 0\;\Rightarrow\; d = \frac{-8+\sqrt{64+4\times1.5\times48}}{3} = \frac{-8+\sqrt{352}}{3} = 3.587\ \text{m}

Answer: balancing depth = 3.59 m.

Cross-section (level with the ground at 0.00):

      2.0 m                          2.0 m
   |---|                          |---|
 +3.00 ____                       ____ +3.00
      /    \                     /    \
 2:1 /      \  ground 0.00       /      \ 2:1
 ---/        \--             --/        \---
                \ 1.5:1  1.5:1/
                 \            /   depth 3.59 m
                  \__________/ -3.59
                     8.0 m bed

Bed level is 3.59 m below ground; crest of each bank is 3.0 m above ground; bed width 8 m; cutting slopes 1.5H:1V; bank slopes 2H:1V; crest width 2 m.

✓ Answer checked by Claude Opus 5.5 (high effort)

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