--- title: "Bed motion check and non-moving gravel size" subject: "Hydraulics" chapter: 10 key: "semester_4/hydraulics:sediment-bed-motion-and-minimum-gravel-size" source: https://www.bctengineer.com/q/semester_4/hydraulics/sediment-bed-motion-and-minimum-gravel-size site: BCT Engineering --- # Bed motion check and non-moving gravel size Hydraulics · Chapter 10: Flow in mobile boundary channel - Asked in: 2071 Magh, 6 marks; 2068 Magh, 6 marks (asked 2 times) - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/hydraulics/sediment-bed-motion-and-minimum-gravel-size - Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-10 ## Question A stream has a sediment bed of median size 0.35 mm. The slope of the channel is $1.5\times10^{-4}$. Stream is considered as trapezoidal with base width 3 m and side slope 1.5H:1V. a) If the depth of flow in the channel is 0.25 m, examine whether the bed particles will be in motion or not. b) Calculate minimum size of gravel that will not move in the bed of channel. Use empirical equation of critical shear stress as: $\tau_c\ (N/m^2) = 0.155 + \frac{0.409\,d_{mm}^2}{(1+0.177\,d_{mm}^2)^{1/2}}$. ## Short answer Bed shear stress is 0.318 N/m^2, exceeding critical stress 0.205 N/m^2, so particles move. Setting critical stress equal to bed shear gives the minimum immobile grain size as about 0.64 mm. ## Answer **Given:** $d_{50} = 0.35$ mm, $S_0 = 1.5 \times 10^{-4}$; trapezoidal channel $b = 3$ m, $z = 1.5$ (1.5H : 1V), $y = 0.25$ m. Critical shear stress (empirical, $d$ in mm): $$ \tau_c = 0.155 + \frac{0.409\,d^2}{\sqrt{1 + 0.177\,d^2}}\ \ \text{N/m}^2 $$ #### a) Will the bed particles move? **Hydraulic radius:** $$ A = (3 + 1.5 \times 0.25)(0.25) = 0.844\ \text{m}^2 $$ $$ P = 3 + 2(0.25)\sqrt{1 + 1.5^2} = 3.901\ \text{m},\qquad R = \frac{0.844}{3.901} = 0.216\ \text{m} $$ **Bed shear stress:** $$ \tau_0 = \gamma R S_0 = 9810 \times 0.216 \times 1.5 \times 10^{-4} = 0.318\ \text{N/m}^2 $$ **Critical shear stress for $d = 0.35$ mm:** $$ \tau_c = 0.155 + \frac{0.409(0.1225)}{\sqrt{1 + 0.177(0.1225)}} = 0.155 + \frac{0.0501}{1.0108} = 0.205\ \text{N/m}^2 $$ Since $\tau_0 = 0.318 > \tau_c = 0.205\ \text{N/m}^2$, **the bed particles will be in motion.** #### b) Minimum size that will not move Set $\tau_c = \tau_0 = 0.318$: $$ 0.155 + \frac{0.409\,d^2}{\sqrt{1 + 0.177\,d^2}} = 0.318 \;\Rightarrow\; \frac{0.409\,d^2}{\sqrt{1 + 0.177\,d^2}} = 0.163 $$ Trial: $d = 0.64$ mm gives $\dfrac{0.409(0.4096)}{\sqrt{1.0725}} = 0.1617$; $d = 0.643$ mm gives $0.1630$. Hence $$ d_{min} \approx 0.64\ \text{mm} $$ Check: $\tau_c(0.643) = 0.318\ \text{N/m}^2 = \tau_0$. **Answer: (a) Particles move ($\tau_0 = 0.318 > \tau_c = 0.205\ \text{N/m}^2$). (b) Grains of about 0.64 mm and larger will not move.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the bed shear stress computed in part (a)? A. tau_0 = gamma b S0 B. tau_0 = gamma R S0 with R = A/P of the trapezoid C. tau_0 = gamma y only D. tau_0 = rho V^2 Answer: B. Average boundary shear is gamma R S0 with R = 0.216 m, giving 0.318 N/m^2. (Key point: Bed shear formula) 2. What is the hydraulic radius of the trapezoid (b = 3 m, z = 1.5, y = 0.25 m)? A. 0.844 m B. 3.9 m C. 0.216 m D. 0.25 m Answer: C. A = 0.844 m^2 and P = 3.901 m, so R = 0.216 m. (Key point: Hydraulic radius) 3. The bed shear is 0.318 N/m^2 and tau_c for 0.35 mm is 0.205 N/m^2. What is the conclusion? A. Particles stay at rest since tau_0 > tau_c B. Particles stay at rest since tau_c < 1 C. Cannot decide D. Particles move since tau_0 > tau_c Answer: D. Motion occurs when actual shear exceeds critical shear. (Key point: Decision for 0.35 mm) 4. How is the minimum gravel size that does not move found? A. Set tau_c(d) equal to tau_0 = 0.318 and solve for d by trial, giving about 0.64 mm B. Set d = 0.35 mm C. Use d = 10 R S0 D. Set tau_c = 0.155 Answer: A. The grain size at which tau_c equals the bed shear is the smallest size that stays at rest. 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