Design of Rectangular Roadside Drains: Step-by-Step Design Example (IS 456 and IS 3370)

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Roadside drains carry stormwater away from carriageways and protect the pavement from water damage. The most common choice for urban roads and highways is the RCC rectangular drain, which is easy to build and handles a wide range of discharges.

Structurally, a roadside drain is a small U-shaped channel that has to resist earth pressure, vehicle surcharge and water pressure. In this example, the drain is also designed as liquid-retaining, so that water does not leak into the subgrade and weaken the road base. The strength design follows the Limit State Method of IS 456:2000, and the reinforcement is detailed to the minimum steel and crack control requirements of IS 3370 (Part 2).

Problem statement: Design of rectangular roadside drain with a clear width of 1.0 m and a clear depth of 1.2 m. The wall thickness is 175 mm, and the base slab thickness is 150 mm.

Data and Assumptions for the Design of Rectangular Roadside Drain

The top of the drain is open, so each wall behaves as a cantilever fixed at the base slab. The coefficient Ka is the Rankine active earth pressure coefficient (Rankine, 1857). The 20 kN/m² surcharge is taken as about 1.2 m of equivalent earth fill, in line with IRC practice (Indian Roads Congress [IRC], 2017)

Design cases

1. Case 1: Drain empty. Earth pressure and surcharge act from outside. This governs the outer (earth) face of the wall.
2. Case 2: Drain full. Water pressure acts from inside with no earth support outside. This governs the inner face of the wall.

2. Case 1: Wall Design (Drain Empty)

Lateral pressure

• At top: p₁ = Ka × q = (1/3) × 20 = 6.67 kN/m²
• At bottom: p₂ = p₁ + Ka × γ × h = 6.67 + (1/3)(18)(1.2) = 13.87 kN/m²

Bending moment and shear force at the base of the wall (per metre run)

The pressure diagram is a rectangle (surcharge) plus a triangle (soil):

• M = (6.67 × 1.2²)/2 + (7.2 × 1.2²)/6 = 4.80 + 1.73 = 6.53 kNm
• V = 6.67 × 1.2 + ½ × 7.2 × 1.2 = 12.32 kN

Factored values (load factor 1.5):

• Mu = 1.5 × 6.53 = 9.8 kNm
• Vu = 1.5 × 12.32 = 18.5 kN

Effective depth

d = 175 − 40 − 10/2 ≈ 130 mm (taken as 129 mm to be conservative)

Limiting moment check

Mu,lim = 0.133 fck b d² (BIS, 2000) = 0.133 × 25 × 1000 × 129² = 55.3 kNm

Since 9.8 kNm < 55.3 kNm, the section is under-reinforced.

Steel required

Using Ast = (0.5 fck/fy) [1 − √(1 − 4.6 Mu/(fck b d²))] b d:

• Mu/(b d²) = 0.588 N/mm²
• Ast = 0.025 × 0.0557 × 1000 × 129 ≈ 180 mm²/m

Minimum steel (0.12% of gross area) = 0.0012 × 1000 × 175 = 210 mm²/m, which governs.

Strength requires only 180 mm²/m, but a liquid-retaining wall needs a minimum steel percentage of about 0.3% of the section under IS 3370 (Part 2)(BIS, 2021), which governs here.

Check for shear

• τv = Vu/(b d) = 18,500/(1000 × 129) = 0.14 N/mm²
• For pt = 0.41% and M25, τc ≈ 0.44 N/mm²
• Applying the slab factor k = 1.25 for a 175 mm section, k·τc ≈ 0.55 N/mm²

Since τv < k·τc, the wall is safe in shear and no shear reinforcement is required.

Check for deflection (cantilever)

• Basic l/d for a cantilever = 7
• Steel stress fs = 0.58 × 500 × (180/524) ≈ 100 N/mm²
• Modification factor ≈ 2.0 (the maximum allowed)
• Permissible l/d = 7 × 2.0 ≈ 14
• Actual l/d = 1200/129 = 9.3

Since 9.3 < 14, the deflection check is satisfied.

3. Case 2: Wall Design (Drain Full of Water)

• Maximum moment at the base: M = γw h³/6 = 10 × 1.2³/6 = 2.88 kNm
• Factored: Mu = 1.5 × 2.88 = 4.3 kNm
• Ast required ≈ 85 mm²/m, so minimum steel governs

Horizontal Wall Reinforcement

As per IS 456:2000, Clause 32.5(c), the minimum horizontal reinforcement in a wall is 0.20% of the gross concrete area for deformed bars not exceeding 16 mm in diameter.

For the 175 mm thick wall:

For 10 mm diameter bars at 150 mm centre-to-centre:

Since (\(524 > 350\text{ mm}^2/\text{m}\)), the proposed 10 mm diameter horizontal bars at 150 mm c/c satisfy the minimum horizontal reinforcement requirement of IS 456.

The maximum permitted spacing is:

Therefore, provide 10 mm diameter horizontal bars at 150 mm c/c on both faces of the wall. Final crack-width and liquid-retaining requirements should also be checked according to the applicable provisions of IS 3370 and the approved structural drawings.

Crack control check (indicative)

Using service-load moments and an approximate lever arm of 0.9d (about 116 mm):

• Inner face (water): σs ≈ 2.88 × 10⁶/(524 × 116) ≈ 47 N/mm²
• Outer face (earth): σs ≈ 6.53 × 10⁶/(524 × 116) ≈ 107 N/mm²

Both stresses are low, so cracks should stay well within the 0.2 mm limit for liquid-retaining members. For a final design, verify crack width with the IS 456 Annex F or IS 3370 procedure.

4. Design of Base Slab (150 mm Thick)

Vertical loads per metre run

Soil bearing pressure (base width = 1.0 + 2 × 0.175 = 1.35 m)

• Drain empty: 15.56/1.35 ≈ 11.5 kN/m²
• Drain full: 27.56/1.35 ≈ 20.4 kN/m²

Both values are far below the assumed SBC of 100 kN/m², so the foundation pressure is safe.

Bending moment and steel in the base slab

The wall moment is transferred into the slab at the corner. Taking the slab’s critical moment equal to the wall’s base moment is a conservative approach:

• Mu = 9.8 kNm
• d = 150 − 40 − 5 = 105 mm (104 mm used)
• Mu/(b d²) = 0.905 N/mm²
• Ast = 0.025 × 0.087 × 1000 × 104 ≈ 226 mm²/m
• Liquid-retaining minimum (about 0.3% of 150 mm) ≈ 450 mm²/m, which governs

Shear in the base slab

Vu ≈ 6 kN/m gives τv ≈ 0.06 N/mm², far below τc, so the slab is safe in shear.

5. Development Length and Corner Detailing

For M25 concrete and Fe500 (HYSD) bars in tension:

• Design bond stress τbd = 1.4 × 1.6 = 2.24 N/mm²
• Ld = (0.87 × fy × φ)/(4 × τbd) = (0.87 × 500 × 10)/(4 × 2.24) ≈ 485 mm

Also, read: Development Length Formula (IS 456) for Footing, Column and Beam with Examples

6. Summary And Design Details

6.1. Summary of Reinforcement

Provide a 100 mm thick PCC (M10) levelling course below the base slab, projecting 100 mm beyond each face (1550 mm wide overall).

6.2. Sectional Detail Drawing for Design of Rectangular Roadside Drain

7. Practical Design Notes

1. Uplift and buoyancy.
If the water table can rise to ground level, the buoyant force is about 10 × 1.35 × 1.35 ≈ 18.2 kN/m, which exceeds the empty weight of the drain (15.6 kN/m). In such cases, check for flotation or add a heel or extension to the base slab.

2. Watertightness and joints.
Because the drain is designed as liquid-retaining, provide expansion or contraction joints every 6–10 m, with PVC waterstops and a suitable sealant, to accommodate temperature and shrinkage movement without leakage.

3. Reinforcement economy.
This example uses 10 mm @ 150 c/c on every face for simplicity and a single bar size on site. The distribution steel (horizontal in the walls, longitudinal in the slab) is on the conservative side, so it can be checked against the IS 3370 minimum if you want to optimise.

4. Earth pressure condition.
This example used active earth pressure (Ka)(Rankine, 1857). If the walls are restrained by a stiff cover slab or sit against a rigid structure, use at-rest pressure, K₀ = 1 − sin φ = 0.5. This raises the wall moment by about 50%.

5. Cover slabs.
Where the drain is covered with slabs for pedestrian or vehicular access, the walls are propped at the top and behave differently from a cantilever. Redesign the walls as propped members.

Conclusion

A rectangular roadside drain looks simple, but a rational design needs both loading cases, proper corner detailing and checks for shear, deflection, bearing, crack width and uplift. For this 1.0 m × 1.2 m liquid-retaining drain, 10 mm bars at 150 mm c/c on all faces, with a 500 mm anchorage at the corner and a 100 mm PCC base course, satisfy the requirements of IS 456:2000 and IS 3370 (Part 2).

Frequently Asked Questions

Disclaimer: This article is for educational purposes. Actual designs must be based on site-specific data and checked by a qualified engineer.