(a) A vertical cut, 4·5 m deep, is to be made in a c-φ soil having cohesion = 19·1 kN/m², angle of internal friction = 16° and…

(a) Vertical cut 4.5 m deep in c–φ soil

The vertical cut can theoretically stand unsupported because the required depth of 4.5 m is less than the critical height (about 5.48 m) for an unsupported excavation in this soil.

Given: Depth (H) = 4.5 m, Cohesion (c) = 19.1 kN/m², Friction angle (φ) = 16°, Unit weight (γ) = 18.5 kN/m³.

Calculations: First, determine the Rankine active earth pressure coefficient (Ka): Ka = (1 - sin φ) / (1 + sin φ) = (1 - sin 16°) / (1 + sin 16°) = 0.568. The active earth pressure (pa) at depth z is given by: pa = γzKa - 2c√Ka. The cohesion term is: 2c√Ka = 2 × 19.1 × √0.568 = 28.8 kN/m².

(i) Active earth pressure: At the top (z = 0): pa = 0 - 28.8 = -28.8 kN/m² (Tension). At the bottom (z = 4.5 m): pa = (18.5 × 4.5 × 0.568) - 28.8 = 18.4 kN/m² (Compression).

(ii) Depth of tension cracks (z₀): Tension cracks develop where the active pressure is zero (pa = 0). z₀ = 2c / (γ√Ka) = 38.2 / (18.5 × 0.754) = 2.74 m.

(iii) Maximum depth of excavation (Hc): The critical height for an unsupported cut is: Hc = 4c / (γ√Ka) = 2 × z₀ = 5.48 m. Result: Since the required depth (4.5 m) is less than the critical height (5.48 m), the cut can theoretically stand unsupported, though with a small margin.

(b) Sluice gate jet and hydraulic jump

A hydraulic jump will occur because the supercritical jet must return to the subcritical tailwater depth (2.5 m > y_c = 1.18 m); since the sequent depth of the jet exceeds the tailwater depth, the jump is repelled downstream.

Given: Q = 20 m³/s, Width (B) = 5.0 m, Jet depth (y₁) = 0.40 m, Tailwater depth (y₃) = 2.50 m. Discharge per unit width: q = Q/B = 20/5 = 4.0 m²/s. Velocity in jet: V₁ = q/y₁ = 4.0/0.40 = 10 m/s.

Justification for Jump: Froude number of the jet: Fr₁ = V₁ / √(gy₁) = 10 / √(9.81 × 0.40) = 5.05. Since Fr₁ > 1, the flow is supercritical. Sequent depth (y₂) = (y₁/2) [√(1 + 8Fr₁²) - 1] = (0.40/2) [√(1 + 8(5.05)²) - 1] = 2.65 m. Since y₂ (2.65 m) > y₃ (2.50 m), the tailwater is insufficient and a repelled (free) jump forms downstream, after the jet has slowed to the depth 0.443 m whose sequent depth is 2.50 m.

(i) Energy head loss: ΔE = (y₃ - y₁')³ / (4 y₁' y₃) = (2.50 - 0.443)³ / (4 × 0.443 × 2.50) = 1.96 m.

(ii) Upstream depth: The specific energy of the jet is E₁ = y₁ + V₁²/2g = 0.40 + (10²/19.62) = 5.50 m. Assuming negligible loss across the gate, the upstream specific energy is also 5.50 m. Solving the specific energy equation for the subcritical root: yᵤ + q²/(2gyᵤ²) = 5.50. yᵤ + 0.815/yᵤ² = 5.50. By trial and error, yᵤ ≈ 5.47 m.

(iii) Force on the gate: Using the momentum equation on a control volume from upstream to the jet: F_gate = (1/2)γB(yᵤ² - y₁²) - ρQ(V₁ - Vᵤ). Vᵤ = q/yᵤ = 4.0/5.47 = 0.73 m/s. F_gate = (0.5 × 9.81 × 5 × (5.47² - 0.40²)) - (1000 × 20 × (10 - 0.73)). F_gate = 729.8 kN - 185.4 kN = 544.4 kN.

Applications & Critical Energy: Hydraulic jumps are used for energy dissipation in stilling basins, aeration, and mixing. The critical energy concept cannot be applied to the jump itself because specific energy is not conserved during the transition; instead, the momentum equation is used to relate the depths.

(c) Settlement of the 3 m footing from the plate load test

The settlement of the 3 m × 3 m footing is 10.3 mm, which is well within the permissible limits of Indian Standards.

Given: Plate size (Bp) = 0.60 m, Footing size (Bf) = 3.0 m, Load = 1100 kN. Pressure on footing: q = 1100 / (3 × 3) = 122.2 kN/m².

Calculations: First, interpolate the settlement of the plate (sp) at 122.2 kN/m² from the test data: At 100 kN/m², s = 4.0 mm. At 150 kN/m², s = 7.5 mm. sp = 4.0 + [(122.2 - 100) / (150 - 100)] × (7.5 - 4.0) = 5.55 mm.

For sandy soil, the size correction formula (IS 1888) is: sf = sp × [(Bf(Bp + 0.30)) / (Bp(Bf + 0.30))]² sf = 5.55 × [(3.0 × (0.60 + 0.30)) / (0.60 × (3.0 + 0.30))]² sf = 5.55 × [(2.70) / (1.98)]² = 5.55 × (1.364)² = 10.3 mm.

Comparison with IS Standards: According to IS 1904 and IS 8009, the permissible settlement for an isolated footing on sandy soil is generally 50 mm. Result: The calculated settlement of 10.3 mm is significantly less than the 50 mm limit, making the design safe.