Flow in mobile boundary channel

The shear stress reduction factor (tractive force ratio) is

$$K = \frac{\tau_{cs}}{\tau_{cb}}$$

the ratio of the critical shear stress on a side slope ($\tau_{cs}$) to the critical shear stress on a level bed ($\tau_{cb}$) for the same material.

Setup

Consider a single sediment particle on a side slope inclined at angle $\theta$ to the horizontal. Let the angle of repose of the material be $\phi$, the submerged weight of the particle be $W_s$ and the area exposed to flow be $a$.

Forces on the particle at incipient motion:

• Weight component down the slope: $W_s\sin\theta$
• Drag (tractive) force along the flow direction: $\tau_s\,a$ (acts along the channel axis, i.e. perpendicular to the first force within the plane of the slope)
• Normal reaction: $W_s\cos\theta$, giving a resisting force $W_s\cos\theta\tan\phi$

   side slope (looking along the channel)
          .
        .   particle
      .  ^  Ws sin(theta)  (down the slope)
    . theta    
 ------------------------ horizontal
 Drag tau_s*a acts into the page (along flow)

Derivation

The two driving forces are at right angles, so their resultant is $\sqrt{(W_s\sin\theta)^2 + (\tau_s a)^2}$. At incipient motion this equals the resisting force:

$$W_s\cos\theta\tan\phi = \sqrt{(W_s\sin\theta)^2 + (\tau_{cs}\,a)^2}$$

Squaring and solving for $\tau_{cs}a$:

$$(\tau_{cs}a)^2 = W_s^2\cos^2\theta\tan^2\phi - W_s^2\sin^2\theta$$ $$\tau_{cs}\,a = W_s\cos\theta\tan\phi\sqrt{1 - \frac{\tan^2\theta}{\tan^2\phi}}$$

On a level bed ($\theta = 0$) the particle is held only by friction against the drag:

$$\tau_{cb}\,a = W_s\tan\phi$$

Dividing:

$$K = \frac{\tau_{cs}}{\tau_{cb}} = \cos\theta\sqrt{1 - \frac{\tan^2\theta}{\tan^2\phi}}$$

Simplifying with $\cos^2\theta\left(1 - \dfrac{\tan^2\theta}{\tan^2\phi}\right) = \cos^2\theta - \dfrac{\sin^2\theta}{\tan^2\phi}$ and using $\cos^2\theta = 1 - \sin^2\theta$:

$$\boxed{K = \sqrt{1 - \frac{\sin^2\theta}{\sin^2\phi}}}$$

Remarks: $K < 1$ always, because particles on a slope are helped to move by gravity. $K = 0$ when $\theta = \phi$ (the slope itself is at the angle of repose), so the side slope must be flatter than the angle of repose.

Shields diagram

Shields (1936) showed by experiments that the start of motion of non-cohesive sediment depends on two dimensionless numbers:

• Shields parameter (dimensionless critical shear stress): $\tau_* = \dfrac{\tau_c}{(\gamma_s - \gamma)\,d}$
• Boundary (shear) Reynolds number: $Re_* = \dfrac{u_*\,d}{\nu}$, with $u_* = \sqrt{\tau_0/\rho}$ the shear velocity

where $\tau_c$ is the critical bed shear stress, $\gamma_s$ and $\gamma$ the specific weights of sediment and water, $d$ the grain size and $\nu$ the kinematic viscosity.

 tau*
 0.1 |\
     | \
     |  \                     ___________  0.056
 0.05|   '.                _.-'
     |     '.          _.-'
 0.03|       '-..__..-'      <- minimum
     |    (smooth)  (transition) (rough)
     +------+--------+---------+------------> Re*
           2        10        400
  above curve: bed moves ; below curve: no motion

Important points (based on $Re_*$):

The curve separates the region of no motion (below) from the region of sediment motion (above).

Application in design

To find the critical shear stress for a given grain size $d$:

1. Assume $\tau_*$ (for coarse material 0.056 to 0.06).
2. Find $\tau_c = \tau_*(\gamma_s - \gamma)d$ and $u_* = \sqrt{\tau_c/\rho}$.
3. Compute $Re_* = u_*d/\nu$, read the correct $\tau_*$ from the diagram and repeat until it agrees.

To check or design a channel:

4. Compute the actual bed shear stress $\tau_0 = \gamma R S_0$ (or $\gamma y S_0$).
5. If $\tau_0 < \tau_c$ the bed is stable; if $\tau_0 > \tau_c$ the bed material moves.
6. For design, choose the bed material (or a stable size $d$) so that $\tau_c \ge \tau_0$. For fully rough flow, the size that just stays stable is $d_{cr} \approx 10\,R S_0$ (with $\gamma_s = 2.65\gamma$).

Incipient motion

Incipient motion is the condition at which the hydrodynamic forces (drag and lift) acting on a bed particle are just equal to the forces resisting its movement (submerged weight and friction). It is the threshold between a stable bed and a moving bed: if the flow increases even slightly, the particle starts to move.

It is described by:

• the critical (threshold) shear stress $\tau_c$: the bed shear stress at which motion starts; or
• the critical velocity $V_c$ at which motion starts.

Quantitatively, for non-cohesive grains, $\tau_c$ is obtained from the Shields diagram, $\tau_c = \tau_*(\gamma_s - \gamma)d$. Motion occurs when the actual bed shear stress $\tau_0 = \gamma R S_0$ exceeds $\tau_c$.

Mobile boundary (alluvial) channel

A mobile boundary channel is an open channel whose bed and banks consist of loose granular material (sand, gravel, silt) that can be eroded, transported and deposited by the flowing water itself. The boundary is therefore deformable: the shape, slope and bed forms of the channel adjust to the flow and sediment load. Natural rivers, and unlined earth canals in sandy soil, are examples.

An alluvial channel is a channel formed in alluvium, that is, in its own deposited sediment. Its geometry is partly determined by the flow, unlike a rigid boundary channel (concrete or rock lined) whose boundary does not change.

Physical meaning of $K$

$K$ is the ratio of the critical shear stress needed to start motion of a particle on the side slope to the critical shear stress needed on the level bed:

$$K = \frac{\tau_{cs}}{\tau_{cb}} = \sqrt{1 - \frac{\sin^2\theta}{\sin^2\phi}} < 1$$

where $\theta$ is the side slope angle and $\phi$ the angle of repose of the material. It tells us by what fraction the bed critical stress must be reduced to get the safe stress for the side slopes.

Why $K$ is needed in design

1. Gravity helps the flow on the sides. A particle on a side slope has a component of its weight down the slope in addition to the drag of the flow. It therefore moves at a lower shear stress than a particle on the flat bed.
2. The sides are the critical part of the section. Without $K$ a designer would size the channel using the bed critical stress only, and the banks would erode first, even though the shear stress on the banks (about $0.75\gamma yS_0$ for trapezoidal channels) is lower than on the bed (about $\gamma yS_0$).
3. It links the side slope to the stability. $K$ decreases as the side slope becomes steeper and falls to zero when $\theta = \phi$. By checking the side-slope stress against $K\tau_c$ the designer chooses a flatter side slope or a different cross-section so that both bed and banks are stable.
4. It gives a uniform safety. Stable-channel design needs the largest of $\tau_b/\tau_{cb}$ and $\tau_s/\tau_{cs}$ to be below 1; $K$ lets both be compared on the same basis.

The two types of channels are designed on different principles.

In short, rigid boundary design is a hydraulic-capacity problem; mobile boundary design is a capacity plus stability problem.

The maximum permissible velocity is the greatest mean velocity that will not erode the channel boundary; the minimum permissible velocity is the smallest that prevents silting and weed growth. The velocity is chosen from tables or empirical formulas for the boundary material.

Rigid boundary (minimum velocity approach)

1. Choose a minimum velocity $V_{min}$ (typically 0.6 to 0.9 m/s for lined canals) to avoid sedimentation and weeds.
2. Choose side slope $z$, Manning's $n$ (from the lining) and the bed slope $S_0$.
3. Required area: $A = Q/V$ with $V \ge V_{min}$.
4. Manning's equation gives the hydraulic radius: $R = \left(\dfrac{nV}{\sqrt{S_0}}\right)^{3/2}$, and the wetted perimeter $P = A/R$.
5. Solve for depth and width from $A = (b + zy)y$ and $P = b + 2y\sqrt{1 + z^2}$.
6. Check that the velocity is below the permissible limit of the lining, and add freeboard.

Mobile boundary (maximum velocity approach)

1. Select the maximum permissible velocity $V_{max}$ for the material (for example Fortier and Scobey's values: fine sand about 0.5 m/s, coarse gravel about 1.2 to 1.5 m/s, hard clay higher) or use Kennedy's critical velocity

$$V_0 = 0.55\,m\,y^{0.64}\ \text{(m/s, } y \text{ in m)}$$

where $m$ is the critical velocity ratio. 2. Choose the side slope $z$ (as steep as the soil permits) and Manning's $n$. 3. Area: $A = Q/V_{max}$. 4. Hydraulic radius from Manning's equation: $R = \left(\dfrac{nV_{max}}{\sqrt{S_0}}\right)^{3/2}$ (if $S_0$ is not fixed, it is found from $S_0 = \dfrac{n^2V^2}{R^{4/3}}$ for a chosen $R$). 5. Solve the geometry for $b$ and $y$: $P = A/R$, $A = (b + zy)y$, $P = b + 2y\sqrt{1 + z^2}$. 6. Check that $V \le V_{max}$ and $V \ge V_{min}$, and adjust $z$, $b$ and $y$ if not.

Limitation: the method uses only the mean velocity and ignores the side-slope effect and the shear stress distribution, so the tractive force method is preferred for important designs.

The tractive force method (Lane, USBR) designs a mobile boundary channel by ensuring the shear stress (tractive force) of the flowing water on the bed and banks is less than the critical shear stress of the boundary material.

Basic ideas

• Tractive force on the boundary: $\tau_0 = \gamma R S_0$ (average). For a trapezoidal channel the maximum values are about $\tau_b \approx \gamma y S_0$ on the bed and $\tau_s \approx 0.75\gamma y S_0$ on the sides.
• Critical shear stress on a level bed, $\tau_c$: from the Shields diagram, $\tau_c = \tau_*(\gamma_s - \gamma)d$ (about $0.06(\gamma_s - \gamma)d$ for coarse material), or from Lane's charts for a given $d_{75}$.
• On the side slope, the critical shear stress is reduced to $\tau_{cs} = K\tau_c$, with $K = \sqrt{1 - \sin^2\theta/\sin^2\phi}$.

Design procedure

1. From soil data find the grain size $d$ (or $d_{75}$) and the angle of repose $\phi$.
2. Select the side slope $z$ (with $\theta = \tan^{-1}(1/z) < \phi$) and compute $K$.
3. Find $\tau_c$ on the level bed; apply a reduction for sinuosity or a safety factor if required.
4. Find the permissible depth from the side-slope condition and the bed condition:
   • bed: $\gamma y S_0 \le \tau_c$
   • sides: $0.75\gamma y S_0 \le K\tau_c$ and take the smaller $y$.
5. With that $y$, use Manning's equation ($n$ from the material) to find the base width $b$ that carries $Q$:

$$Q = \frac{1}{n}A R^{2/3}S_0^{1/2},\qquad A = (b + zy)y$$

6. Check the Froude number (the flow should be subcritical) and add freeboard. If $b/y$ is far from the assumed range, repeat with a modified side slope.

Critical shear stress for $Re_* > 400$

From the Shields diagram, the dimensionless critical shear stress is

$$\tau_* = \frac{\tau_c}{(\gamma_s - \gamma)\,d}$$

For $Re_* = \dfrac{u_*d}{\nu} > 400$ (fully rough, fully developed turbulent flow) the viscous sublayer is negligible and $\tau_*$ becomes independent of $Re_*$ and equal to a constant:

$$\tau_* \approx 0.056 \approx 0.06$$

Therefore the critical shear stress is

$$\boxed{\tau_c = 0.06\,(\gamma_s - \gamma)\,d}$$

Critical diameter

The average boundary shear stress in uniform flow is

$$\tau_0 = \gamma R S_0$$

At the limit of stability $\tau_0 = \tau_c$ for the critical diameter $d_{cr}$:

$$\gamma R S_0 = 0.06\,(\gamma_s - \gamma)\,d_{cr}$$ $$d_{cr} = \frac{\gamma R S_0}{0.06\,(\gamma_s - \gamma)} = \frac{R S_0}{0.06\,(s - 1)}$$

where $s = \gamma_s/\gamma$ is the specific gravity of the sediment. For quartz sand and gravel, $s = 2.65$, so $s - 1 = 1.65$:

$$d_{cr} = \frac{R S_0}{0.06 \times 1.65} = 10.1\,R S_0$$ $$\boxed{d_{cr} \approx 10\,R\,S_0}$$

with $d_{cr}$, $R$ in the same units (for example metres). Grains larger than $d_{cr}$ remain at rest; smaller grains move.

Flow over an alluvial (sand) bed forms regular bed features called bed forms. As the flow strength increases the sequence is: plane bed (no motion), ripples, dunes, transition (washed-out dunes), plane bed with motion, then antidunes.

Ripples

• Small, triangular bed forms with a gentle upstream slope and a steep downstream (lee) slope.
• Wavelength is small (typically less than 0.6 m) and height is a few centimetres (less than about 0.06 m); the size depends on the grain size and not on the depth of flow.
• They occur in fine sand (grain size below about 0.6 mm) at low shear stress and low Froude number, just above incipient motion.
• Grains are rolled up the gentle slope, fall into the separation zone behind the crest and are deposited on the lee face, so the ripples migrate downstream.
• The water surface is almost undisturbed. They add form roughness.

Dunes

• Larger bed forms of similar triangular shape with a long gentle upstream slope and steep lee face, but with wavelength and height related to the depth of flow (wavelength about 5 to 7 times the depth, height up to about one third of the depth).
• They occur at higher flow strength than ripples, in sand of any size, with subcritical flow ($Fr < 1$).
• Flow separates at the crest and a strong eddy forms in the lee. The water surface is out of phase with the bed (the water surface is lowest above the dune crest).
• They move downstream at a slow rate and give a large form resistance (high roughness and a large increase in flow depth).

 Ripples:   /\/\/\/\/\/\    small, regular
 Dunes:    __/\___/\___/\   large, depth-related; flow separation
 flow ---->

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

$$d_{min} \approx 0.64\ \text{mm}$$

Checks at the limit: $\tau_c(0.643) = 0.318\ \text{N/m}^2$.

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.

Given: $Q = 15\ \text{m}^3/\text{s}$, $S_0 = 0.0015$, 50% of particles $\ge 5$ cm ($d_{50} = 0.05$ m), trapezoidal, $\phi = 34^\circ$, ratio of bed shear to critical shear $K_2 = 0.7$.

Assumptions: $\tau_c = 0.06(\gamma_s - \gamma)d$ with $\gamma_s = 2.65\gamma$; the design bed shear is limited to $K_2\tau_c$; maximum shear on bed $\gamma y S_0$ and on sides $0.75\gamma y S_0$; Strickler's $n = 0.047\,d^{1/6} = 0.0285$.

Step 1: Critical and permissible shear on the bed

$$\tau_c = 0.06(1.65)(9810)(0.05) = 48.56\ \text{N/m}^2,\qquad K_2\tau_c = 0.7(48.56) = 33.99\ \text{N/m}^2$$

Step 2: Side slope. Choose $z = 2$ (the side slope must be flatter than $\phi = 34^\circ$, i.e. $z > 1.48$):

$$\theta = 26.57^\circ,\qquad K = \sqrt{1 - \frac{0.2}{\sin^2 34^\circ}} = \sqrt{1 - \frac{0.2}{0.3127}} = 0.600$$

Step 3: Depth

• Bed: $y \le \dfrac{33.99}{9810 \times 0.0015} = 2.31$ m
• Sides: $y \le \dfrac{0.600 \times 33.99}{0.75 \times 9810 \times 0.0015} = 1.85$ m

The sides govern: $y = 1.85$ m.

Step 4: Base width from Manning's equation ($n = 0.0285$):

$$A = (b + 3.70)(1.85),\quad P = b + 2(1.85)\sqrt5 = b + 8.27$$

Trial $b = 2.14$ m: $A = 10.79\ \text{m}^2$, $P = 10.40$ m, $R = 1.037$ m.

$$Q = \frac{1}{0.0285}(10.79)(1.037)^{2/3}(0.03873) = 15.0\ \text{m}^3/\text{s} \checkmark$$

Step 5: Check

$$V = 1.39\ \text{m/s},\quad T = 2.14 + 4(1.85) = 9.54\ \text{m},\quad Fr = \frac{1.39}{\sqrt{9.81 \times 10.79/9.54}} = 0.42 < 1$$

Answer: Base width $b \approx 2.1$ m, side slope 2H : 1V ($\theta = 26.6^\circ$), depth $y \approx 1.85$ m. The narrow base is a consequence of the high permissible depth; the section may be widened (with lower depth) if a shallower channel is preferred.

Given: $Q = 10\ \text{m}^3/\text{s}$, $S_0 = 0.0016$, 50% of particles $\ge 3$ cm ($d_{50} = 0.03$ m), trapezoidal, $\phi = 34^\circ$, $K_2 = 0.75$ (ratio of permissible bed shear to critical shear).

Assumptions: $\tau_c = 0.06(\gamma_s - \gamma)d$ with $\gamma_s = 2.65\gamma$; maximum shear on bed $\gamma y S_0$, on sides $0.75\gamma y S_0$; Strickler's $n = 0.047(0.03)^{1/6} = 0.0262$.

Step 1: Critical and permissible shear on the bed

$$\tau_c = 0.06(1.65)(9810)(0.03) = 29.14\ \text{N/m}^2,\qquad K_2\tau_c = 0.75(29.14) = 21.85\ \text{N/m}^2$$

Step 2: Side slope. Choose $z = 2$ ($\theta = 26.57^\circ < \phi = 34^\circ$):

$$K = \sqrt{1 - \frac{\sin^2 26.57^\circ}{\sin^2 34^\circ}} = \sqrt{1 - \frac{0.2}{0.3127}} = 0.600$$

Step 3: Depth

• Bed: $y \le \dfrac{21.85}{9810 \times 0.0016} = 1.39$ m
• Sides: $y \le \dfrac{0.600 \times 21.85}{0.75 \times 9810 \times 0.0016} = 1.11$ m

The sides govern: $y = 1.11$ m.

Step 4: Base width from Manning's equation ($n = 0.0262$, $S_0 = 0.0016$):

$$A = (b + 2.23)(1.114),\quad P = b + 2(1.114)\sqrt5 = b + 4.98$$

Trial $b = 4.62$ m: $A = 7.63\ \text{m}^2$, $P = 9.61$ m, $R = 0.795$ m.

$$Q = \frac{1}{0.0262}(7.63)(0.795)^{2/3}(0.04) = 10.0\ \text{m}^3/\text{s} \checkmark$$

Step 5: Check

$$V = 1.31\ \text{m/s},\quad T = 4.62 + 4(1.114) = 9.08\ \text{m},\quad Fr = \frac{1.31}{\sqrt{9.81 \times 7.63/9.08}} = 0.46 < 1$$

Answer: Base width $b \approx 4.6$ m, side slope 2H : 1V, depth $y \approx 1.11$ m (subcritical flow, $V = 1.31$ m/s).

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):

        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.

1. Maximum permissible velocity method

The channel is stable if the mean velocity is below the velocity that would erode the boundary.

1. Select the maximum permissible velocity $V_{max}$ for the soil (tables, or Kennedy's $V_0 = 0.55\,m\,y^{0.64}$).
2. Select side slope $z$, bed slope $S_0$ and Manning's $n$.
3. Area $A = Q/V_{max}$; hydraulic radius $R = \left(\dfrac{nV_{max}}{\sqrt{S_0}}\right)^{3/2}$; perimeter $P = A/R$.
4. Solve $A = (b + zy)y$ and $P = b + 2y\sqrt{1 + z^2}$ for $b$ and $y$.

2. Tractive force method

The channel is stable if the shear stress on the boundary is below the critical shear stress of the soil.

1. Find $\tau_c$ for the grain size: $\tau_c = \tau_*(\gamma_s - \gamma)d$ ($\tau_* \approx 0.06$ for coarse material).
2. Select $z$, compute $K = \sqrt{1 - \sin^2\theta/\sin^2\phi}$ with $\tan\theta = 1/z$.
3. Bed: $\gamma y S_0 \le \tau_c$. Sides: $0.75\,\gamma y S_0 \le K\tau_c$. Take the smaller permitted depth $y$.
4. Find $b$ from Manning's equation: $Q = \dfrac{1}{n}AR^{2/3}S_0^{1/2}$.
5. Check the Froude number and add freeboard.

3. Regime theory (Lacey)

For a channel in alluvium that carries sediment, the dimensions follow empirical regime relations. With silt factor $f = 1.76\sqrt{d_{mm}}$:

$$V = \left(\frac{Qf^2}{140}\right)^{1/6},\quad A = \frac{Q}{V},\quad P = 4.75\sqrt{Q},\quad R = \frac{5V^2}{2f},\quad S_0 = \frac{f^{5/3}}{3340\,Q^{1/6}}$$

Then find $b$ and $y$ from the geometry (side slope 0.5H : 1V).

Given: $Q = 15\ \text{m}^3/\text{s}$, gravel $d = 8$ mm (rounded), $S_0 = 0.0009$, side slope 2H : 1V ($z = 2$), $\phi = 32^\circ$.

Assumptions: the Shields value $\tau_* = 0.06$ (fully rough), $\gamma_s = 2.65\gamma$; maximum shear on the bed $\gamma y S_0$ and on the sides $0.75\gamma y S_0$; Manning's $n$ from Strickler's formula $n = 0.047\,d^{1/6}$ ($d$ in m).

Step 1: Critical shear stress on a level bed

$$\tau_c = 0.06(\gamma_s - \gamma)d = 0.06(1.65)(9810)(0.008) = 7.77\ \text{N/m}^2$$

Step 2: Side slope factor

$$\theta = \tan^{-1}\frac{1}{2} = 26.57^\circ,\qquad K = \sqrt{1 - \frac{\sin^2 26.57^\circ}{\sin^2 32^\circ}} = \sqrt{1 - \frac{0.2}{0.2808}} = 0.536$$

Step 3: Limiting depth

• Bed: $\gamma y S_0 \le \tau_c \Rightarrow y \le \dfrac{7.77}{9810 \times 0.0009} = 0.880$ m
• Sides: $0.75\gamma y S_0 \le K\tau_c \Rightarrow y \le \dfrac{0.536 \times 7.77}{0.75 \times 9810 \times 0.0009} = 0.629$ m

The side slope governs: $y = 0.63$ m.

Step 4: Base width. Strickler: $n = 0.047(0.008)^{1/6} = 0.021$. Manning's equation with $y = 0.629$ m, $z = 2$:

$$Q = \frac{1}{n}AR^{2/3}S_0^{1/2}, \quad A = (b + 2 \times 0.629)(0.629)$$

Trial gives $b = 22.5$ m: $A = 14.93\ \text{m}^2$, $P = 25.27$ m, $R = 0.591$ m, and

$$Q = \frac{1}{0.021}(14.93)(0.591)^{2/3}(0.03) = 15.0\ \text{m}^3/\text{s} \checkmark$$

Step 5: Checks

$$V = \frac{15}{14.93} = 1.00\ \text{m/s},\quad Fr = \frac{V}{\sqrt{gA/T}} = 0.41 < 1\ \text{(subcritical)}$$

Answer: Trapezoidal section with $b \approx 22.5$ m, $y \approx 0.63$ m, side slope 2H : 1V, $S_0 = 0.0009$. The channel is wide and shallow because the small gravel cannot withstand more shear stress. Add a freeboard (about 0.3 m).

Given: $Q = 20\ \text{m}^3/\text{s}$, side slope 4H : 2V ($z = 2$), $S_0 = 0.0015$, $D_{75} = 2$ cm (moderately rounded), $n = 0.025$, $\phi = 40^\circ$, slightly sinuous channel.

Assumptions: $\tau_c = 0.06(\gamma_s - \gamma)D_{75}$ with $\gamma_s = 2.65\gamma$; for a slightly sinuous channel $\tau_c$ is reduced to 90% (Lane); maximum shear on bed $= \gamma y S_0$, on sides $= 0.75\gamma y S_0$.

Step 1: Permissible shear on the bed

$$\tau_c = 0.06(1.65)(9810)(0.02) = 19.42\ \text{N/m}^2,\qquad 0.9\,\tau_c = 17.48\ \text{N/m}^2$$

Step 2: Side slope factor

$$\theta = \tan^{-1}\frac{1}{2} = 26.57^\circ,\qquad K = \sqrt{1 - \frac{0.2}{\sin^2 40^\circ}} = \sqrt{1 - \frac{0.2}{0.4132}} = 0.718$$

Permissible side shear $= K(0.9\tau_c) = 12.56\ \text{N/m}^2$.

Step 3: Depth

• Bed: $y \le \dfrac{17.48}{9810 \times 0.0015} = 1.188$ m
• Sides: $y \le \dfrac{12.56}{0.75 \times 9810 \times 0.0015} = 1.138$ m

Side governs: $y = 1.14$ m.

Step 4: Base width. Manning's equation, $n = 0.025$, $S_0 = 0.0015$, $z = 2$, $y = 1.138$ m, $Q = 20$:

$$A = (b + 2.276)(1.138),\quad P = b + 2(1.138)\sqrt5 = b + 5.09$$

Trial $b = 9.7$ m: $A = 13.63\ \text{m}^2$, $P = 14.80$ m, $R = 0.921$ m.

$$Q = \frac{1}{0.025}(13.63)(0.921)^{2/3}(0.0387) = 20.0\ \text{m}^3/\text{s} \checkmark$$

Step 5: Froude number check

$$V = \frac{20}{13.63} = 1.47\ \text{m/s},\quad T = 9.7 + 4(1.138) = 14.25\ \text{m},\quad D = \frac{A}{T} = 0.957\ \text{m}$$ $$Fr = \frac{1.47}{\sqrt{9.81 \times 0.957}} = 0.48 < 1$$

The flow is subcritical, as required.

Answer: $b \approx 9.7$ m, $y \approx 1.14$ m, side slope 2H : 1V (4H : 2V), $V = 1.47$ m/s, $Fr = 0.48$. Provide freeboard in addition.

Given: trapezoidal channel $y = 1.5$ m, $b = 10$ m, $z = 2$; gravel $d_{50} = 60$ mm; $\phi = 37^\circ$; $K_2 = 0.9$.

Assumptions: $\tau_c = 0.06(\gamma_s - \gamma)d$ with $\gamma_s = 2.65\gamma$; the design shear is $K_2\tau_c$; maximum shear on bed $\gamma y S_0$, on sides $0.75\gamma y S_0$; $n = 0.047\,d^{1/6} = 0.0294$ (Strickler).

Step 1: Critical shear stress

$$\tau_c = 0.06(1.65)(9810)(0.06) = 58.27\ \text{N/m}^2,\qquad K_2\tau_c = 52.44\ \text{N/m}^2$$

Step 2: Side slope factor

$$\theta = \tan^{-1}\frac{1}{2} = 26.57^\circ,\qquad K = \sqrt{1 - \frac{0.2}{\sin^2 37^\circ}} = \sqrt{1 - \frac{0.2}{0.3622}} = 0.669$$

Step 3: Maximum slope from bed and side conditions

• Bed: $\gamma y S_0 \le 52.44 \Rightarrow S_0 \le \dfrac{52.44}{9810 \times 1.5} = 0.00356$
• Sides: $0.75\gamma y S_0 \le K(52.44) \Rightarrow S_0 \le \dfrac{0.669 \times 52.44}{0.75 \times 9810 \times 1.5} = 0.00318$

The side governs: maximum permissible slope $S_0 \approx 0.0032$ (about 1 in 314).

Step 4: Discharge at this slope

$$A = (10 + 2 \times 1.5)(1.5) = 19.5\ \text{m}^2,\quad P = 10 + 2(1.5)\sqrt5 = 16.71\ \text{m},\quad R = 1.167\ \text{m}$$ $$Q = \frac{1}{n}AR^{2/3}S_0^{1/2} = \frac{1}{0.0294}(19.5)(1.167)^{2/3}(0.0564) = 41.4\ \text{m}^3/\text{s}$$

Answer: Maximum permissible slope $\approx 0.0032$; discharge $\approx 41\ \text{m}^3/\text{s}$ without disturbing the stability of the channel.

Given: $Q = 75\ \text{m}^3/\text{s}$, $d = 0.65$ mm. Assume side slope 0.5H : 1V (Lacey's usual value).

Step 1: Silt factor

$$f = 1.76\sqrt{d_{mm}} = 1.76\sqrt{0.65} = 1.419$$

Step 2: Velocity

$$V = \left(\frac{Qf^2}{140}\right)^{1/6} = \left(\frac{75 \times 2.014}{140}\right)^{1/6} = 1.013\ \text{m/s}$$

Step 3: Area, perimeter, hydraulic radius, slope

$$A = \frac{Q}{V} = 74.06\ \text{m}^2,\qquad P = 4.75\sqrt{Q} = 41.14\ \text{m}$$ $$R = \frac{5V^2}{2f} = \frac{5(1.0256)}{2(1.419)} = 1.81\ \text{m}\quad (A/P = 1.80\ \text{m, consistent})$$ $$S = \frac{f^{5/3}}{3340\,Q^{1/6}} = \frac{1.419^{5/3}}{3340(75)^{1/6}} = 2.61 \times 10^{-4}\ (\approx 1\text{ in }3830)$$

Step 4: Bed width and depth. For side slope 0.5H : 1V:

$$A = (b + 0.5y)y = 74.06,\qquad P = b + 2y\sqrt{1 + 0.25} = b + 2.236y = 41.14$$

Substituting $b = 41.14 - 2.236y$:

$$(41.14 - 1.736y)y = 74.06 \;\Rightarrow\; 1.736y^2 - 41.14y + 74.06 = 0 \;\Rightarrow\; y = 1.96\ \text{m}$$ $$b = 41.14 - 2.236(1.963) = 36.75\ \text{m}$$

Answer: $V = 1.01$ m/s, $b \approx 36.7$ m, $y \approx 1.96$ m, side slope 0.5H : 1V, bed slope $\approx 1$ in 3830 (add freeboard).