--- title: "Tractive force design in coarse alluvium, 20 m³/s" subject: "Hydraulics" chapter: 10 key: "semester_4/hydraulics:tractive-force-trapezoid-20m3s-d75-2cm" source: https://www.bctengineer.com/q/semester_4/hydraulics/tractive-force-trapezoid-20m3s-d75-2cm site: BCT Engineering --- # Tractive force design in coarse alluvium, 20 m³/s Hydraulics · Chapter 10: Flow in mobile boundary channel - Asked in: 2078 Chaitra, 6 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/hydraulics/tractive-force-trapezoid-20m3s-d75-2cm - Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-10 ## Question Using the tractive force method, design a trapezoidal channel (side slope 4H:2V) to carry 20 m³/s through a slightly sinuous channel on a slope of 0.0015. The channel is to be excavated in a coarse alluvium with 75 percentile diameter, $D_{75}$ = 2 cm with the particles on the perimeter of channel moderately rounded. Use Manning's n = 0.025. Check if the flow is subcritical under the uniform flow condition as a supercritical flow is not desired in an excavated channel. Take $\phi$ = 40°. ## Short answer Permissible bed shear 13.57 N/m^2, side factor K=0.718, depth 0.883 m. Manning's equation with n=0.025 gives base width 15.5 m. Flow is subcritical (Fr=0.47). Trapezoidal section: b=15.5 m, y=0.88 m, side slope 4H:2V (2H:1V). ## Answer **Given:** $Q = 20\ \text{m}^3/\text{s}$, side slope 4H : 2V ($z = 2$), $S_0 = 0.0015$, $D_{75} = 2$ cm $= 20$ mm (moderately rounded), $n = 0.025$, $\phi = 40^\circ$, slightly sinuous channel. **Assumptions:** For coarse non-cohesive material, Lane's permissible tractive force on the bed is $\tau_c = 0.4$ lb/ft² per inch of $D_{75}$, i.e. $\tau_c \approx 0.754\,D_{75}$ N/m² with $D_{75}$ in mm. For a slightly sinuous channel, $\tau_c$ is reduced to 90% (Lane). Maximum shear on the bed $= \gamma y S_0$; on the sides $= 0.75\gamma y S_0$. **Step 1: Permissible shear on the bed** $$ \tau_c = 0.754 \times 20 = 15.08\ \text{N/m}^2,\qquad 0.9\,\tau_c = 13.57\ \text{N/m}^2 $$ **Step 2: Side slope factor** $$ \theta = \tan^{-1}\frac{1}{2} = 26.57^\circ,\qquad K = \sqrt{1 - \frac{\sin^2\theta}{\sin^2\phi}} = \sqrt{1 - \frac{0.2}{0.4132}} = 0.718 $$ Permissible side shear $= K(0.9\tau_c) = 0.718 \times 13.57 = 9.75\ \text{N/m}^2$. **Step 3: Depth** - Bed: $y \le \dfrac{13.57}{9810 \times 0.0015} = 0.922$ m - Sides: $y \le \dfrac{9.75}{0.75 \times 9810 \times 0.0015} = 0.883$ m The sides govern: **$y = 0.883$ m**. **Step 4: Base width.** Manning's equation with $n = 0.025$, $S_0 = 0.0015$, $z = 2$, $y = 0.883$ m, $Q = 20$: $$ A = (b + 1.766)(0.883),\quad P = b + 2(0.883)\sqrt5 = b + 3.949 $$ Trial $b = 15.45$ m: $A = 15.20\ \text{m}^2$, $P = 19.40$ m, $R = 0.784$ m. $$ Q = \frac{1}{0.025}(15.20)(0.784)^{2/3}(0.0387) = 20.0\ \text{m}^3/\text{s} \checkmark $$ **Step 5: Froude number check** $$ V = \frac{20}{15.20} = 1.32\ \text{m/s},\quad T = 15.45 + 4(0.883) = 18.98\ \text{m},\quad D = \frac{A}{T} = 0.801\ \text{m} $$ $$ Fr = \frac{1.32}{\sqrt{9.81 \times 0.801}} = 0.47 < 1 $$ The flow is **subcritical**, as required. (With $b/y \approx 17.5$, the channel is wide, so taking the bed shear as $\gamma y S_0$ is justified.) **Answer: $b \approx 15.5$ m, $y \approx 0.88$ m, side slope 2H : 1V (4H : 2V), $V = 1.32$ m/s, $Fr = 0.47$.** Provide freeboard in addition. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the slight sinuosity accounted for? A. Critical shear is reduced to 90% (17.48 N/m^2) B. Increased by 10% C. Slope is halved D. Ignored Answer: A. Lane reduces permissible shear for sinuous channels; 0.9 x 19.42 = 17.48. (Key point: Sinuosity) 2. What is K for z = 2 and phi = 40 degrees? A. About 0.54 B. About 0.72 C. About 1.0 D. About 0.40 Answer: B. K = sqrt(1 - 0.2/0.4132) = 0.718. (Key point: Side factor) 3. Which condition governs the depth and what is y? A. Bed, about 1.19 m B. Sides, about 1.19 m C. Sides, about 1.14 m D. Bed, about 1.14 m Answer: C. Bed allows 1.188 m, sides 1.138 m; the smaller side value governs. (Key point: Depth) 4. Why is the Froude number checked and what is it? A. To find the base width; Fr = 1 B. To find K; Fr = 0.72 C. To find the slope; Fr = 2 D. Supercritical flow is not desired in an excavated channel; Fr = 0.48 is subcritical Answer: D. With V = 1.47 m/s and D = 0.957 m, Fr = 0.48 < 1, so flow is subcritical. 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