--- title: "Boundary shear distribution in alluvial channel" subject: "Hydraulics" chapter: 10 key: "semester_4/hydraulics:shear-stress-distribution-alluvial-channel" source: https://www.bctengineer.com/q/semester_4/hydraulics/shear-stress-distribution-alluvial-channel site: BCT Engineering --- # Boundary shear distribution in alluvial channel Hydraulics · Chapter 10: Flow in mobile boundary channel - Asked in: 2072 Magh, 3 marks - Video: [Derive shear stress in Open channel flow. | Hydraulics | 4th Sem | Prashant YT | BE Civil |](https://www.youtube.com/watch?v=_7FGkYz6tyo) — Prashant YT - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/hydraulics/shear-stress-distribution-alluvial-channel - Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-10 ## Question Show the shear stress distribution on the alluvial channel boundary with values. ## Short answer Shear stress is zero at the water surface and corners, reaches about 0.97γyS0 on the bed centre and 0.76γyS0 on the lower third of the side slope. For design, bed stress is taken as γyS0 and side stress as 0.75γyS0. ## Answer In a channel of uniform flow, the average boundary shear stress is $\tau_0 = \gamma R S_0$. It is **not uniform** around the wetted perimeter. It is largest on the bed near the centre line, falls towards the banks, is a maximum again at about the lower third of the side slope, and is zero at the water surface and at the corners (bottom edges) of the section. For a trapezoidal channel with side slope 1.5H : 1V and $b/y \ge 4$ (USBR results, $y$ = depth of flow): | Location | Maximum shear stress | |---|---| | Bed (centre) | $0.97\,\gamma y S_0$ | | Side slope (about the lower third) | $0.76\,\gamma y S_0$ | | Bed of a very wide channel ($b \gg y$) | $\gamma y S_0$ | | Corners and water surface | tends to zero | ```text water surface (tau = 0) \ / \ 0.76 gamma y S0 / side \ (max on side) / \ / \_______________________/ 0 0.97 gamma y S0 0 bed corner (max on bed) corner ``` **Design values.** For simple design, take the maximum shear stress on the bed as $\tau_b \approx \gamma y S_0$ (0.97 rounded up) and on the sides as $\tau_s \approx 0.75\,\gamma y S_0$. For a wide channel ($b \gg y$) the bed shear stress is uniform at $\gamma y S_0$. For example, with $y = 1$ m and $S_0 = 0.001$: $\gamma y S_0 = 9.81\ \text{N/m}^2$, so the bed carries about $9.5\ \text{N/m}^2$ and the side slope about $7.5\ \text{N/m}^2$. The side stress is lower than the bed stress, but the critical stress on the side is also reduced (by the factor $K$), so the banks often govern the design. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is boundary shear distributed around the perimeter of the channel? A. Uniform equal to gamma R S0 everywhere B. Not uniform; zero at the water surface and corners, with maxima on the bed and on the side slope C. Maximum at water surface D. Maximum at corners Answer: B. Shear is highest on the bed centre and again on the lower third of the side, and tends to zero at corners and surface. (Key point: Distribution shape) 2. What are the maximum shear values for a trapezoid of 1.5H:1V with b/y >= 4? A. Bed 0.76 gamma y S0 and side 0.97 gamma y S0 B. Bed 1.5 gamma y S0 and side 1.0 gamma y S0 C. Bed 0.97 gamma y S0 and side 0.76 gamma y S0 D. Both equal gamma y S0 Answer: C. USBR gives 0.97 on the bed and 0.76 on the side, rounded in design to 1.0 and 0.75. (Key point: USBR values) 3. Why do banks often govern even though side shear is lower than bed shear? A. Side material is always finer B. Side shear is higher C. Water is deeper at the sides D. The critical stress on the side is also reduced by K Answer: D. Side stress 0.75 gamma y S0 is compared with K tau_c, which can be much smaller than tau_c. 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