(a) A canal is to be excavated through a soil with c = 20 kN/m², φ = 20°, e = 0·80 and G = 2·70. The side slope is 1 in 1. The…

(a) Unit weights from G = 2.70 and e = 0.80:

• γ_sat = (G + e)γ_w/(1 + e) = 3.50 × 9.81/1.80 = 19.08 kN/m³
• γ′ = (G − 1)γ_w/(1 + e) = 1.70 × 9.81/1.80 = 9.27 kN/m³

A 1 in 1 side slope gives β = 45°. Taylor's stability number is S_n = c/(F_c γ H), with H = 8 m.

Canal running full: the slope is submerged, so the effective (submerged) unit weight acts and the full φ = 20° is mobilised. From the table, S_n = 0.08:

F_c = c/(S_n γ′ H) = 20/(0.08 × 9.27 × 8) = 3.37

Canal rapidly emptied: the soil stays saturated, so the driving weight is γ_sat, but the pore pressures have not yet dissipated, so the frictional resistance still corresponds to the submerged weight. Taylor's approach uses a weighted friction angle

φ_w = (γ′/γ_sat)φ = (9.27/19.08) × 20° = 9.7°.

From the given table, S_n = 0.12 at 10° and 0.14 at 5°; interpolating for 9.7°, S_n ≈ 0.121. Then

F_c = 20/(0.121 × 19.08 × 8) = 1.08 (≈ 1.09 if S_n = 0.12 is read directly).

Rapid drawdown therefore cuts the factor of safety from 3.37 to about 1.1, which is why sudden emptying is the critical case for canal and dam slopes.

(b) Only the soft clay (3 m to 8 m) consolidates; the dense sand and the rock are treated as incompressible.

Clay properties (saturated): e₀ = wG = 0.50 × 2.70 = 1.35; γ_sat = (2.70 + 1.35) × 9.81/2.35 = 16.91 kN/m³; γ′ = 7.10 kN/m³.

Initial effective stress at the middle of the clay (5.5 m depth):

• Sand above the water table: 0.6 × 18.6 = 11.16 kPa
• Sand below the water table: 2.4 × (19.2 − 9.81) = 22.54 kPa
• Upper half of the clay: 2.5 × 7.10 = 17.74 kPa
• σ′₀ = 51.4 kPa

Compression index (Skempton, normally consolidated clay): C_c = 0.009(w_L − 10) = 0.009 × 60 = 0.54.

The load is spread over an extensive area, so Δσ = 50 kPa at every depth. For the 5 m clay layer:

S = [C_c H/(1 + e₀)] log₁₀[(σ′₀ + Δσ)/σ′₀] = (0.54 × 5/2.35) × log₁₀(101.4/51.4) = 1.149 × 0.295 = 0.34 m (about 340 mm).

Dividing the clay into sublayers would give a slightly larger value, because the upper part of the layer has lower σ′₀.

(c) Both weirs draw from the same canal, so they work under one common water level; the rectangular weir crest is set above the V-notch vertex so that the V-notch alone takes the first 340 l/s.

V-notch (90°): Q = (8/15)C_d√(2g) tan 45° H^2.5 = 1.370 H^2.5.

• At the minimum canal flow (425 l/s) the V-notch must pass 340 l/s: H₁ = (0.340/1.370)^0.4 = 0.573 m.
• At the maximum canal flow (680 l/s) it must pass at most 425 l/s: H₂ = (0.425/1.370)^0.4 = 0.626 m (maximum head on the V-notch).

Rectangular weir: Q = (2/3)C_d√(2g) L h^1.5 = 1.713 L h^1.5. It carries 425 − 340 = 85 l/s at the minimum and 680 − 425 = 255 l/s at the maximum, and its head rises by the same 0.626 − 0.573 = 0.053 m as the water level:

• 255/85 = [(h + 0.053)/h]^1.5 = 3, so (h + 0.053)/h = 3^(2/3) = 2.080 and h = 0.050 m at minimum flow.
• Maximum head on the rectangular weir = 0.050 + 0.053 = 0.103 m; its crest is 0.573 − 0.050 = 0.523 m above the V-notch vertex.
• Length: 0.255 = 1.713 × L × 0.103^1.5 = 1.713 × L × 0.0331, so L ≈ 4.51 m.

Check at minimum flow: 1.713 × 4.51 × 0.050^1.5 = 0.085 m³/s, and the V-notch takes 0.340 m³/s, totalling 425 l/s.