---
title: "Design steps of a guide bund"
subject: "Irrigation and Drainage Engineering"
chapter: 6
key: "semester_6/irrigation_and_drainage_engineering:guide-bund-design-steps"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/guide-bund-design-steps
site: BCT Engineering
---

# Design steps of a guide bund

Irrigation and Drainage Engineering · Chapter 6: River Training Works

- Asked in: 2079 Jestha, 5 marks; 2076 Baisakh, 8 marks (asked 2 times)
- Video: [Design of Guide Bank River Training Structure | Irrigation Engineering | IOE TU](https://www.youtube.com/watch?v=3p4xQ418oOk) — Parash Joshi - Civil Construction and Tutor
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/guide-bund-design-steps
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-6

## Question

Write down the design steps of the guide bund (guide bank) with a suitable sketch / example.

## Short answer

Steps: determine Q, silt factor f = 1.76√d; waterway W = 4.75√Q; upstream shank 1–1.5W, downstream 0.2–0.5W; curved heads; top level HFL + 1.5 m, top width 6 m; pitching thickness t = 0.06Q^(1/3); scour depth R = 0.47(Q/f)^(1/3), max 2R at nose; apron width 1.5D' (D' = scour – flood depth), thickness 1.5t launched.

## Answer

The design of a guide bund (guide bank) follows Lacey's regime theory and Garg's recommendations. The steps are listed first and then worked for an example.

#### Design steps
1. Find the design discharge $Q$, HFL, bed level, bed material size $d$.
2. Silt factor $f = 1.76\sqrt{d_{mm}}$.
3. Waterway $W = 4.75\sqrt{Q}$.
4. Length: upstream shank $= 1.0$ to $1.5W$, downstream $= 0.2$ to $0.5W$.
5. Curved heads: radius and sweep angle.
6. Top level $=$ HFL $+ 1.5$ m (free-board); top width 6 m; side slopes 2:1.
7. Pitching thickness $t = 0.06\,Q^{1/3}$.
8. Scour depth $R = 0.47(Q/f)^{1/3}$; maximum scour $2R$ (nose), $1.5R$ (shank) below HFL.
9. Apron: $D' = $ scour $-$ flood depth; width $1.5D'$; thickness $1.5t$ (launched).
10. Draw plan and sections.

#### Example
$Q = 5000$ m³/s, bed material $d = 1$ mm, HFL $= 100.0$ m, bed $= 96.0$ m, so the flood depth is $h = 4.0$ m and the silt factor is $f = 1.76\sqrt{1} = 1.76$.

#### 1. Waterway and length of guide bund
Lacey's regime waterway: $W = 4.75\sqrt{Q} = 4.75\sqrt{5000} = 335.88$ m. Adopt $W \approx 336$ m.
Length of guide bund (Garg): upstream shank $=1.0$ to $1.5\,W$, downstream tail $=0.2$ to $0.5\,W$. Adopt
$$
\begin{aligned} L_{us} &= 1.25W = 1.25\times 336 = 420\ \text{m} \\ L_{ds} &= 0.25W = 84\ \text{m} \end{aligned}
$$
Curved heads: upstream radius $R_u \approx 0.5W = 168$ m with sweep $120^\circ$ to $145^\circ$ (adopt $120^\circ$); downstream radius $R_d \approx 0.25W = 84$ m with sweep about $45^\circ$.
Arc length of upstream head $= R_u\theta = 168\times 2.094 = 352$ m, so total length on the upstream side $\approx 772$ m.

#### 2. Thickness of stone pitching
$$
t = 0.06\,Q^{1/3} = 0.06\times 5000^{1/3} = 0.06\times 17.10 = 1.03\ \text{m}
$$
Adopt $t \approx 1.1$ m (pitching on the river side slope 2:1, over a 0.15 to 0.3 m graded filter).

#### 3. Scour depth and launching apron
Lacey's normal scour depth:
$$
R = 0.47\left(\frac{Q}{f}\right)^{1/3} = 0.47\left(\frac{5000}{1.76}\right)^{1/3} = 6.66\ \text{m}
$$
Maximum scour below HFL: $2R$ at the upstream nose, $1.5R$ at the shank (straight portion).
The apron must launch to the slope $2H:1V$, so its horizontal width is $1.5D'$ and, when launched, it covers a slope length $\sqrt5\,D'$ with thickness $1.5t$.

| Portion | Scour factor | Scour depth below HFL | Depth below bed $D'$ | Apron length $1.5D'$ |
|---|---|---|---|---|
| Nose (u/s curved head) | 2.0R | 13.31 m | 9.31 m | 13.97 m |
| Shank (straight) | 1.5R | 9.98 m | 5.98 m | 8.98 m |

Thickness: pitching $t = 1.03$ m, so the launched apron thickness is $1.5t = 1.5\times1.03 = 1.54$ m. Equating volumes per metre, the thickness as laid over $1.5D'$ width is $T = \dfrac{1.5t\times\sqrt5 D'}{1.5D'} = \sqrt5\,t = 2.29$ m (about $2.25t$).

**Result:** $W \approx 336$ m, length of upstream bund $= 420$ m, $t = 1.03$ m, shank apron width $= 8.98$ m and nose apron width $= 13.97$ m, launched thickness $= 1.54$ m.

```text
 Plan:  ________ shank ________   bridge
       (  curved head           |=|
        \____________________   |=|
 Section: top 6 m, HFL+1.5, slope 2:1, pitching t, apron
```

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which formulas give the silt factor and waterway?
   A. f = 1.76 sqrt(d mm); W = 4.75 sqrt(Q)
   B. f = 0.47 d; W = Q^(1/3)
   C. f = 4.75 sqrt(d); W = 1.76 sqrt(Q)
   D. f = d; W = Q
   Answer: A. Lacey's silt factor depends on grain size in mm and the regime width is 4.75 sqrt(Q). (Key point: Silt factor and waterway)

2. For Q = 5000 m3/s, what are W and upstream bund length (1.25W)?
   A. W = 336 m; 84 m
   B. W = 168 m; 210 m
   C. W = 672 m; 840 m
   D. W = 336 m; 420 m
   Answer: D. 4.75 sqrt(5000) = 335.9 m, and 1.25 x 336 = 420 m. (Key point: Example waterway and length)

3. In the example (f = 1.76), what is R and what governs scour at the nose and shank?
   A. R = 6.66 m; nose 2R, shank 1.5R below HFL
   B. R = 13.3 m; nose R
   C. R = 6.66 m; nose 1.5R, shank 2R
   D. R = 1.76 m
   Answer: A. R = 0.47 (5000/1.76)^(1/3) = 6.66 m, with 2R at the nose and 1.5R at the shank. (Key point: Scour depth)

4. What are the shank and nose apron widths in the example?
   A. 8.98 m and 13.97 m
   B. 13.97 m and 8.98 m
   C. 4 m and 6 m
   D. 5.98 m and 9.31 m
   Answer: A. Shank D' = 9.98 - 4 = 5.98 m so 1.5D' = 8.98 m; nose D' = 9.31 m so 13.97 m. (Key point: Apron width)

5. Why is the laid apron thickness about 2.25 t while launched is 1.5 t?
   A. The stone laid over width 1.5D' must cover the slope length sqrt5 D' at 1.5t, so T = sqrt5 t
   B. Because of the free-board
   C. Because the stone is lighter
   D. Because 2:1 slope gives 1.5 only
   Answer: A. Equating volumes per metre: T x 1.5D' = 1.5t x sqrt5 D', so T = sqrt5 t, about 2.25t. (Key point: Apron thickness as laid)

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