--- title: "Thickness of d/s impervious floor of a fall" subject: "Irrigation and Drainage Engineering" chapter: 7 key: "semester_6/irrigation_and_drainage_engineering:fall-downstream-floor-thickness-q10" source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/fall-downstream-floor-thickness-q10 site: BCT Engineering --- # Thickness of d/s impervious floor of a fall Irrigation and Drainage Engineering · Chapter 7: Regulating Structures - Asked in: 2071 Magh, 12 marks - Video: [Design of sharda fall | Vertical drop structure | Irrigation Numerical | Most important problem](https://www.youtube.com/watch?v=r_VaWVvL3YY) — Sudip Dhakal - 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/fall-downstream-floor-thickness-q10 - Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-7 ## Question Find the thickness of downstream impervious floor for a fall having following data: a) Discharge u/s = d/s = 10 cumecs b) Full supply level u/s = 201.50, d/s = 200.25 c) Drop = 1.25 m d) Bed level u/s = 200.00, d/s = 198.75 e) Bed width u/s = d/s = 9.0 m f) Full supply depth u/s = d/s = 1.50 m g) Bligh's creep coeff. = 8. ## Short answer Rectangular crest at 200.78 m, 9 m long, top width 0.78 m, base 1.84 m. Cistern 4.82 m long, 0.24 m deep, floor level 198.51 m. Total floor 11.32 m; d/s floor thickness 1.26 m reducing to 0.61 m. ## Answer #### Given data and assumptions - $Q = 10$ m³/s, bed width $B = 9$ m, side slope 1:1, u/s FSL $= 201.50$ m, u/s bed $= 200.00$ m, d/s bed $= 198.75$ m. - Full supply depth u/s = d/s $D = 1.5$ m (d/s FSL $= 200.25$ m); drop in water level $H_L = 1.25$ m. - Bligh's coefficient $C = 8$; specific gravity of the masonry/concrete $G = 2.24$ (assumed). - Side slope 1:1 is assumed. The fall is designed as a Sarda-type vertical drop; the crest, cistern and floor length are found first, because the d/s floor thickness depends on them. #### 1. Upstream flow conditions $$ \begin{aligned} A &= (B + mD)D = (9 + 1\times1.5)(1.5) = 15.750\ \text{m}^2 \\ V_a &= Q/A = 0.635\ \text{m/s}, \qquad h_a = \frac{V_a^2}{2g} = 0.021\ \text{m} \end{aligned} $$ #### 2. Crest Since $Q < 14$ m³/s, a **rectangular crest** is used, with crest length equal to the bed width, $L = 9$ m, and $$ Q = 1.835\,L\,H^{3/2}\left(\frac{H}{B_c}\right)^{1/6}, \qquad B_c = 0.55\sqrt{d} $$ where $H$ is the head over the crest (measured from the u/s FSL), $B_c$ is the top width of the crest and $d$ is the height of the crest above the d/s bed. Solving by trial: $$ H = 0.722\ \text{m}, \qquad \text{crest level} = 201.50 - 0.722 = 200.78\ \text{m}, \qquad d = 200.78 - 198.75 = 2.03\ \text{m} $$ Top width $B_c = 0.55\sqrt{2.03} = 0.78$ m. Base width of the crest wall: $$ B_w = \frac{H + d}{\sqrt{G}} = \frac{0.722 + 2.03}{\sqrt{2.24}} = 1.84\ \text{m} $$ #### 3. Cistern Energy head over the crest $E = H + h_a = 0.722 + 0.021 = 0.743$ m. Using the UPIRI formulas: $$ \begin{aligned} L_c &= 5\sqrt{E\,H_L} = 5\sqrt{0.743\times1.25} = 4.82\ \text{m} \\ X &= \tfrac14\,(E\,H_L)^{2/3} = \tfrac14\,(0.928)^{2/3} = 0.238\ \text{m} \end{aligned} $$ Cistern floor level $= 198.75 - 0.238 = 198.51$ m. #### 4. Impervious floor (Bligh's theory) Maximum seepage head occurs with no flow, water up to the crest on the u/s side and the d/s side dry: $H_s = d = 2.03$ m. Required creep length $L = C\,H_s = 8\times2.03 = 16.22$ m. - Cut-offs: u/s $d_1 = D/3 + 0.6 = 1.10$ m; d/s $d_2 = D/2 + 0.6 = 1.35$ m. - Horizontal floor length $b = L - 2(d_1 + d_2) = 16.22 - 4.90 = 11.32$ m. - Minimum d/s floor (from the d/s face of the crest wall) $= 2(D + 1.2) + H_L = 6.65$ m, which is more than $L_c$; adopt $L_d = 6.65$ m. - U/s floor $= b - L_d - B_w = 11.32 - 6.65 - 1.84 = $ **2.83 m**. - Total floor length $= 2.83 + 1.84 + 6.65 = 11.32$ m. #### 5. Floor thickness Creep up to the d/s toe of the crest wall $= 2d_1 + b_u + B_w = 6.87$ m, so the residual uplift head there is $h_1 = H_s\,\dfrac{L - 6.87}{L} = 2.03\times\dfrac{16.22 - 6.87}{16.22} = 1.169$ m (with $L = 16.22$ m, the creep provided). $$ t_1 = \frac{4}{3}\,\frac{h_1}{G - 1} = \frac{4}{3}\times\frac{1.169}{1.24} = 1.26\ \text{m} $$ At the end of the cistern the creep is 11.69 m, $h_2 = 0.567$ m and $t_2 = 0.61$ m. Adopt a d/s floor **1.26 m** thick near the crest wall, reducing to 0.61 m beyond the cistern (minimum 0.30 m); the u/s floor is given a nominal 0.30 m. #### Summary | Component | Value | |---|---| | Crest | rectangular, length 9 m, level 200.78 m, head $H$ = 0.722 m | | Crest width (top / base) | 0.78 m / 1.84 m | | Cistern (length / depth) | 4.82 m / 0.24 m, floor level 198.51 m | | Floor (u/s / wall base / d/s) | 2.83 / 1.84 / 6.65 m, total 11.32 m | | Cut-offs (u/s / d/s) | 1.10 m / 1.35 m | | D/s floor thickness | 1.26 m, reducing to 0.61 m | **Answer:** crest level 200.78 m (rectangular crest 9 m long, top width 0.78 m, base 1.84 m); cistern 4.82 m long and 0.24 m deep; impervious floor 11.32 m long (u/s 2.83 m, d/s 6.65 m) with cut-offs of 1.10 m (u/s) and 1.35 m (d/s); d/s floor thickness 1.26 m, reducing to 0.61 m. The required d/s floor thickness is **1.26 m** next to the crest wall, reducing to 0.61 m beyond the cistern. Sketch: ```text FSL u/s ~~~~~~~~~~~| | crest u/s bed ___________|__ FSL d/s ~~~~ | u/s floor | \ drop ~~~~~ d1 | | \_____cistern________ d/s bed | |<-Lc->| d/s floor | d2 |____ floor thickness __| ``` ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which formula gives the thickness of the d/s impervious floor? A. t = h1/(G - 1) B. t = (4/3) h1/(G - 1), where h1 is the uplift (residual) head and G the specific gravity of the floor material C. t = (G - 1)/h1 D. t = 4 h1 G Answer: B. The extra 4/3 factor gives a safety margin; t = 4/3 x 0.829/1.24 = 0.89 m. (Key point: Thickness formula) 2. How is the residual head h1 at the toe of the wall found? A. h1 = H_L x (L_act - creep to that point)/L_act, as head loss is proportional to creep B. h1 = H_L / 2 C. h1 = drop height D. h1 = H_L x creep to that point / L_act Answer: A. Bligh's theory loses head in proportion to creep, so the residual head is the share of H_L left over the rest of the creep path: 0.829 m. (Key point: Residual head at wall toe) 3. How does the floor thickness vary along the d/s floor? A. It is thinnest near the wall and thickest at the end B. It is 0.89 m near the wall and tapers to 0.36 m at the end, since residual head falls with distance, with 0.30 m as a minimum C. It is uniform throughout D. It increases to 2 m at the end Answer: B. h2 at the end of the cistern is only 0.335 m, so the required thickness drops and tapers to 0.36 m. (Key point: Thickness taper) 4. What total creep length does Bligh's theory require for C = 8 and H_L = 1.25 m? A. 10.0 m B. 6.4 m C. 13.7 m D. 1.25 m Answer: A. L = C x H_L = 8 x 1.25 = 10.0 m; the actual creep 13.71 m is safe. (Key point: Required creep length) ## Related in Irrigation and Drainage Engineering - [Sarda fall: Q 12, bed level 103 m](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Asarda-fall-q12-bed103) — Irrigation and Drainage Engineering, chapter 7 - [Canal fall design, Q 12, drop 1.5 m](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Acanal-fall-design-q12-drop15) — Irrigation and Drainage Engineering, chapter 7 - [Balancing depth: bed width 25 m](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Abalancing-depth-bed25) — Irrigation and Drainage Engineering, chapter 3, asked 3× - [Vertical drop crest level and width, Q 30](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Avertical-drop-crest-q30-fsl315) — Irrigation and Drainage Engineering, chapter 7 - [Crest width, cistern length of vertical drop, Q 1.55](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Avertical-drop-q155-drop075) — Irrigation and Drainage Engineering, chapter 7, asked 2× - [Siphon aqueduct: Q 50, bed 32 m](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Asiphon-aqueduct-q50-bed32-hfd400) — Irrigation and Drainage Engineering, chapter 8, asked 2× ## Related in other subjects - [Channel earthwork and land area, 50 cm fall](https://www.bctengineer.com/question.md?key=semester_7%2Festimating_and_costing%3Achannel-land-acquisition-h) — Estimating and Costing, chapter 6, asked 2× - [DO deficit 50 km downstream, 1.9 m³/s stream](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Astream-do-deficit-50km) — Sanitary Engineering, chapter 6, asked 2× - [Contraction with lowered bed, 10 m³/s](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Acontraction-lowered-bed-3-5m-wide-10-m3s) — Hydraulics, chapter 7