--- title: "Design procedure of sloping glacis weir bay" subject: "Irrigation and Drainage Engineering" chapter: 5 key: "semester_6/irrigation_and_drainage_engineering:sloping-glacis-weir-bay-design-steps" source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/sloping-glacis-weir-bay-design-steps site: BCT Engineering --- # Design procedure of sloping glacis weir bay Irrigation and Drainage Engineering · Chapter 5: Diversion Headworks - Asked in: 2078 Poush, 10 marks - 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/sloping-glacis-weir-bay-design-steps - Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-5 ## Question Write down the design procedure of sloping glacis weir bay in a typical diversion headworks (crest level, d/s floor level and length, total length of floor, cutoff depth and thickness of floor). ## Short answer Design steps: collect hydraulic data, compute waterway via Lacey's perimeter and discharge intensity, set crest level from flood routing and canal FSL, find scour depth, determine downstream floor level and length using hydraulic jump conjugate depth, fix total floor length and cutoff depths via Bligh/Khosla, calculate floor thickness from uplift pressure, and provide protection aprons. ## Answer Design of the weir bay of a sloping-glacis weir (Garg / Punmia style). **Step 1: Data.** Maximum flood $Q$, HFL, pond level (PL), river bed levels, silt factor $f$, afflux, safe exit gradient. **Step 2: Waterway and discharge intensity.** Total waterway $L$ (from Lacey's perimeter, $P = 4.75\sqrt Q$), less the width of the undersluices and piers, gives the weir bay length. Discharge per metre $q = Q/L$. **Step 3: Crest level.** Fixed from the pond level and head required to pass the design flood over the crest: $Q = \frac{2}{3}C_d\sqrt{2g}\,L\,H^{3/2}$ (crest level $=$ HFL after afflux $-$ $H$); also crest $=$ canal FSL $+$ regulator losses. **Step 4: Regime (scour) depth.** $R = 1.35\,(q^2/f)^{1/3}$. **Step 5: Downstream floor level and length.** - Total energy $E_L = $ (u/s TEL) $-$ (d/s bed or floor level); $H_L = $ (u/s TEL $-$ d/s TEL) and use Blench's curves (or Crump's) to find $E_2$ (the conjugate depth after the jump) for the given $q$ and $H_L$. - d/s floor level $=$ d/s HFL $-$ (d/s depth required for the jump) $= $ TWL $- \,E_2$ (with a safety margin of 0.5-1 m). - Length of the horizontal floor $= 5\,(E_2 - E_1)$ (about $5\times$ height of jump) – the glacis length is included with a slope of 3:1. **Step 6: Total length of floor and cutoffs.** - Total length from Bligh's/Khosla's criterion: Bligh $L = C\,H$ with $H$ = max static head; or by Khosla's exit gradient. - U/s cutoff depth $= 1.0R$-$1.5R$ below u/s HFL; d/s cutoff depth $= 1.5R$-$2R$ below the d/s bed. **Step 7: Thickness of impervious floor.** Check Khosla's pressure at key points and take the thickness $$ t = \frac{4}{3}\cdot\frac{h}{G-1} $$ at the toe of the glacis (max thickness), provide about 0.6 m at the ends. Taper thickness along the floor. The d/s floor under the jump should also resist the dynamic forces. **Step 8: Protection works.** U/s and d/s aprons (blocks and inverted filter) with launching aprons of length $1.5$-$2\times$ scour depth; and check exit gradient $G_E \le$ safe value. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the total waterway first fixed in the design of the weir bay? A. Lacey's perimeter P = 4.75 sqrt(Q), less undersluice and pier widths for weir bay length B. From the exit gradient C. From the floor thickness D. From Bligh's coefficient Answer: A. The waterway comes from Lacey's regime perimeter; deducting undersluices and piers gives the weir bay length. (Key point: Waterway) 2. How is the weir crest level fixed? A. HFL after afflux minus the head H needed to pass the design flood over the crest B. River bed level plus 1 m C. Equal to the d/s floor level D. Pond level plus scour depth Answer: A. H is found from Q = (2/3)Cd sqrt(2g) L H^(3/2), then crest = HFL after afflux - H. (Key point: Crest level) 3. The regime (scour) depth used for cutoffs is: A. R = 1.35 (q^2/f)^(1/3) B. R = 4.75 sqrt(Q) C. R = H/(G-1) D. R = C H Answer: A. Lacey's scour depth R = 1.35 (q^2/f)^(1/3) with q discharge per metre and f silt factor. (Key point: Scour depth) 4. Typical cutoff depths below HFL are: A. U/s 1.5R-2R, d/s 1.0R-1.5R B. U/s 1.0R-1.5R below u/s HFL, d/s 1.5R-2R below the d/s bed C. Both 0.5R D. Both 3R Answer: B. The d/s cutoff takes the exit gradient and deeper scour, so it is deeper than the u/s cutoff. (Key point: Cutoff depths) 5. Which is used for the floor thickness and d/s horizontal floor length? A. t = (4/3) h/(G-1), maximum at glacis toe; length about 5 x jump height B. t = h (G-1); length = 1 x jump height C. t = H/C; length = R D. t = h/(G+1); length = 2R Answer: A. Thickness is 4/3 times unbalanced head over (G-1), greatest at the glacis toe; the floor under the jump is about 5(E2 - E1) long. 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