(a) The discharge Q over a small rectangular weir depends on head H over the weir, the weir height P, gravity g, width of the…

(a) Dimensionless form of the weir discharge equation (Buckingham π method)

The discharge over the weir is a function of six independent variables: Q = f(H, P, g, L, ρ, μ), i.e. f₁(Q, H, P, g, L, ρ, μ) = 0

Dimensions of the variables (M, L, T):

• Discharge: Symbol — Q; Dimensions — L³T⁻¹
• Head over the weir: Symbol — H; Dimensions — L
• Height of the weir: Symbol — P; Dimensions — L
• Gravitational acceleration: Symbol — g; Dimensions — LT⁻²
• Width of the weir: Symbol — L; Dimensions — L
• Density: Symbol — ρ; Dimensions — ML⁻³
• Dynamic viscosity: Symbol — μ; Dimensions — ML⁻¹T⁻¹

Number of π terms. There are n = 7 variables and m = 3 fundamental dimensions, so the number of π terms is n − m = 4.

Repeating variables. Choose H (geometry), g (flow/kinematics) and ρ (fluid). Together they contain M, L and T, and none of them can be formed from the others.

π₁ = Q Hᵃ gᵇ ρᶜ M⁰L⁰T⁰ = (L³T⁻¹)(L)ᵃ(LT⁻²)ᵇ(ML⁻³)ᶜ

• M: c = 0
• T: −1 − 2b = 0, so b = −½
• L: 3 + a + b − 3c = 0, so a = −5/2
• π₁ = Q/(H^(5/2) g^(1/2))

π₂ = P Hᵃ gᵇ ρᶜ. Here c = 0 and b = 0, so a = −1: π₂ = P/H

π₃ = L Hᵃ gᵇ ρᶜ. Similarly, π₃ = L/H

π₄ = μ Hᵃ gᵇ ρᶜ

• M: 1 + c = 0, so c = −1
• T: −1 − 2b = 0, so b = −½
• L: −1 + a + b − 3c = 0, so a = −3/2
• π₄ = μ/(ρ H^(3/2) g^(1/2))

Result. Q/(√g H^(5/2)) = φ[P/H, L/H, μ/(ρ H^(3/2) √g)] or Q = √g H^(5/2) · φ(P/H, L/H, μ/(ρ√g H^(3/2)))

Check. Each group is dimensionless. π₄ is the inverse of a Reynolds number built on the velocity √(gH) and length H. It accounts for viscous effects, which matter only at very small heads. Because Q is proportional to L for a wide weir, the result is usually written Q = C_d L √(2g) H^(3/2), with C_d = φ(P/H, Re, …). This is the familiar weir equation.

(b) Smallest contracted width that does not disturb the upstream flow

Upstream section. B₁ = 4.0 m, Q = 5.0 m³/s, y₁ = 1.0 m.

• q₁ = 5/4 = 1.25 m²/s and V₁ = 1.25 m/s
• Froude number: F₁ = V₁/√(gy₁) = 1.25/√(9.81 × 1.0) = 0.399. The approach flow is subcritical.
• Specific energy: E₁ = y₁ + V₁²/2g = 1.0 + 1.25²/(2 × 9.81) = 1.0796 m

Principle. As the channel is narrowed, the discharge per unit width q = Q/B increases. With no energy loss and a horizontal bed, the specific energy stays E₁. For subcritical flow the depth falls in the contraction, and the minimum specific energy needed to pass q, E_min = (3/2)(q²/g)^(1/3), rises. The contraction can be narrowed until E_min just equals the available E₁, at which point critical depth occurs at the throat. Any narrower and E₁ would be too small to pass the flow: the flow would choke, and the upstream depth would rise (afflux). The upstream conditions would then be affected.

Critical condition at the throat.

• y_c = (2/3)E₁ = (2/3) × 1.0796 = 0.7198 m
• q_max = √(g y_c³) = √(9.81 × 0.7198³) = 1.9126 m²/s
• B_min = Q/q_max = 5.0/1.9126 = 2.61 m

Check. At the throat, V_c = 5/(2.614 × 0.7198) = 2.657 m/s and F = 2.657/√(9.81 × 0.7198) = 1.00. The specific energy there is y_c + V_c²/2g = 0.7198 + 0.3598 = 1.0796 m = E₁ ✓.

Result. The width can be reduced smoothly from 4.0 m to about 2.61 m without any change in the upstream depth of 1.0 m. At that width the flow passes through critical depth (0.72 m) at the throat. Any contraction narrower than 2.61 m causes choking and backs water up the channel.

(c)(i) Active thrust on the battered retaining wall (Coulomb's theory)

Data. H = 5 m and γ = 15.5 kN/m³. The back face is inclined at θ = 6° to the vertical: the heel projects under the backfill, so the soil rests on the back face. The backfill surface slopes at β = 12°, φ = 32°, c = 0, and δ = 0.5φ = 16°.

Because the wall face is inclined and wall friction acts, Rankine's theory does not apply. Coulomb's wedge theory is used. With α = 90° − 6° = 84° as the angle of the back face with the horizontal, measured on the soil side:

K_a = sin²(α + φ) / { sin²α · sin(α − δ) · [1 + √( sin(φ + δ) sin(φ − β) / (sin(α − δ) sin(α + β)) )]² }

Substituting the values:

• sin²(116°) = 0.8078
• sin²84° = 0.9891
• sin 68° = 0.9272
• sin 48° = 0.7431
• sin 20° = 0.3420
• sin 96° = 0.9945

The square-root term is √(0.7431 × 0.3420/(0.9272 × 0.9945)) = 0.5250. K_a = 0.8078/[0.9891 × 0.9272 × (1.5250)²] = 0.379

Active force. P_a = ½ K_a γ H² = ½ × 0.379 × 15.5 × 5² = 73.4 kN per metre run

A trial-wedge check gives the same maximum thrust, 73.4 kN/m, on a failure plane at about 56.6° to the horizontal. For comparison, a vertical wall would have K_a = 0.324. The battered back face increases the thrust, because part of the soil load bears on the wall.

Direction. The thrust makes the angle δ = 16° with the normal to the back face, on the side that opposes the downward movement of the wedge. The normal is itself 6° below the horizontal, so P_a acts at 6° + 16° = 22° below the horizontal, pointing towards the wall.

• Horizontal component: P_h = 73.4 cos 22° = 68.0 kN/m
• Vertical component: P_v = 73.4 sin 22° = 27.5 kN/m (downward)

Location. With c = 0, the pressure grows linearly with depth, so the distribution is triangular. The resultant therefore acts at H/3 = 1.67 m above the base, which is 1.68 m measured along the back face.

(c)(ii) Effect of an earthquake on lateral earth pressure

During an earthquake, horizontal and vertical ground accelerations act on the soil wedge as inertia forces. In the pseudo-static (Mononobe–Okabe) method, these are the forces k_hW and k_vW, where k_h and k_v are the seismic coefficients. They tilt the resultant body force by ψ = tan⁻¹[k_h/(1 ± k_v)]. The effects are as follows:

• The critical wedge becomes flatter and larger, so the active pressure increases: P_ae = ½γH²(1 ± k_v)K_ae, with K_ae greater than K_a.
• The passive resistance decreases (K_pe is less than K_p). This reduces the resistance in front of the toe.
• The dynamic increment ΔP = P_ae − P_a acts higher than H/3. IS 1893 places it at mid-height, which raises the overturning moment.
• Saturated backfill can develop excess pore pressure or liquefy, and the wall's own inertia adds to the sliding and overturning forces.

Walls in seismic zones therefore need larger factors of safety against sliding and overturning, and free-draining, well-compacted backfill.