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Design steps of a guide bund

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.

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

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 QQ, HFL, bed level, bed material size dd.
  2. Silt factor f=1.76dmmf = 1.76\sqrt{d_{mm}}.
  3. Waterway W=4.75QW = 4.75\sqrt{Q}.
  4. Length: upstream shank =1.0= 1.0 to 1.5W1.5W, downstream =0.2= 0.2 to 0.5W0.5W.
  5. Curved heads: radius and sweep angle.
  6. Top level == HFL +1.5+ 1.5 m (free-board); top width 6 m; side slopes 2:1.
  7. Pitching thickness t=0.06 Q1/3t = 0.06\,Q^{1/3}.
  8. Scour depth R=0.47(Q/f)1/3R = 0.47(Q/f)^{1/3}; maximum scour 2R2R (nose), 1.5R1.5R (shank) below HFL.
  9. Apron: D′=D' = scour −- flood depth; width 1.5D′1.5D'; thickness 1.5t1.5t (launched).
  10. Draw plan and sections.

Example

Q=5000Q = 5000 m³/s, bed material d=1d = 1 mm, HFL =100.0= 100.0 m, bed =96.0= 96.0 m, so the flood depth is h=4.0h = 4.0 m and the silt factor is f=1.761=1.76f = 1.76\sqrt{1} = 1.76.

1. Waterway and length of guide bund

Lacey's regime waterway: W=4.75Q=4.755000=335.88W = 4.75\sqrt{Q} = 4.75\sqrt{5000} = 335.88 m. Adopt W≈336W \approx 336 m. Length of guide bund (Garg): upstream shank =1.0=1.0 to 1.5 W1.5\,W, downstream tail =0.2=0.2 to 0.5 W0.5\,W. Adopt

Lus=1.25W=1.25×336=420 mLds=0.25W=84 m\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 Ru≈0.5W=168R_u \approx 0.5W = 168 m with sweep 120∘120^\circ to 145∘145^\circ (adopt 120∘120^\circ); downstream radius Rd≈0.25W=84R_d \approx 0.25W = 84 m with sweep about 45∘45^\circ. Arc length of upstream head =Ruθ=168×2.094=352= R_u\theta = 168\times 2.094 = 352 m, so total length on the upstream side ≈772\approx 772 m.

2. Thickness of stone pitching

t=0.06 Q1/3=0.06×50001/3=0.06×17.10=1.03 mt = 0.06\,Q^{1/3} = 0.06\times 5000^{1/3} = 0.06\times 17.10 = 1.03\ \text{m}

Adopt t≈1.1t \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(Qf)1/3=0.47(50001.76)1/3=6.66 mR = 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: 2R2R at the upstream nose, 1.5R1.5R at the shank (straight portion). The apron must launch to the slope 2H:1V2H:1V, so its horizontal width is 1.5D′1.5D' and, when launched, it covers a slope length 5 D′\sqrt5\,D' with thickness 1.5t1.5t.

PortionScour factorScour depth below HFLDepth below bed D′D'Apron length 1.5D′1.5D'
Nose (u/s curved head)2.0R13.31 m9.31 m13.97 m
Shank (straight)1.5R9.98 m5.98 m8.98 m

Thickness: pitching t=1.03t = 1.03 m, so the launched apron thickness is 1.5t=1.5×1.03=1.541.5t = 1.5\times1.03 = 1.54 m. Equating volumes per metre, the thickness as laid over 1.5D′1.5D' width is T=1.5t×5D′1.5D′=5 t=2.29T = \dfrac{1.5t\times\sqrt5 D'}{1.5D'} = \sqrt5\,t = 2.29 m (about 2.25t2.25t).

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

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

✓ Answer checked by Claude Opus 5.5 (high effort)

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