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Channel and Slope Stability

Robert PittDepartment of Civil and Environmental Engineering

University of AlabamaTuscaloosa, AL

Channel and Slope Protection

• Upslope diversions• Channel Protection

– Channel liners– Check dams

• Slope Protection– Roughening surface– Erosion control mats

Slope Diversions

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Channel Lining

Check Dams

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Roughening Slopes while Compacting Ground

Gabions for Slope Protection

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Slope Protection Using Different Materials

Rock and Asphalt for Shaded Areas

Coir Logs

Channel Design Based on Allowable Velocity and Shear Stress

• The concepts of allowable velocity and allowable shear stress are closely linked.

• Shear stress is calculated based on water depth and channel slope.

• Velocity is affected by both slope and depth.• Most allowable velocity charts also include slope

categories for the different liner materials.• Some allowable velocity charts also consider silt

content (water carrying silt has a higher allowable velocity because its sediment carrying capacity is already reduced).

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Example Allowable Velocity TableSilty Water (on site

and downslope)Clear Water (diversions)

1.105.500.915.000.035Cobbles

0.325.000.0752.500.020Fine gravel

0.153.500.0752.500.020Firm loam

0.0752.500.0271.500.020Fine sand

τo (lb/ft2)V (ft/sec)τo (lb/ft2)V (ft/sec)n

Boundary Shear Stress (tractive force):RSo γτ =

RSo γτ =

γ = specific weight of water (62.4 lbs/ft3)R = hydraulic radius (flow depth is used for maximum shear stress calculations; for sheetflow conditions, the depth is equal to the hydraulic radius)S = channel slope Chow 1959

Chow 1959

Example plot of allowable shear stress (unit tractive force) for cohesive materials.

COE 1994

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Example Allowable Tractive Forces• Sheetflow: ¼ inch deep on a 10 % slope:

– (62.4 lb/ft3)(0.021 ft)(0.10) = 0.13 lb/ft2

– >3.5 mm particles OK, for high slit content flow– >5 mm particles OK, for clear flow– Moderately compacted cohesive clays OK

• Sheetflow: ¼ inch deep on a 2 % slope:– (62.4 lb/ft3)(0.021 ft)(0.02) = 0.026 lb/ft2

– > 0.1 mm particles OK, for all flows– Ok for all loose and more compacted cohesive

clays

Example Allowable Tractive Forces (cont.)

• Channel Flow: 18 inches deep on a 5 % slope:– (62.4 lb/ft3)(1.5 ft)(0.05) = 4.7 lb/ft2

– no natural lining material safe• Channel Flow: 18 inches deep on a 1 %

slope:– (62.4 lb/ft3)(1.5 ft)(0.01) = 0.93 lb/ft2

– >70 mm noncohesive material OK

Channel Design Steps for Maximum Permissible Velocity/Allowable Shear

Stress Method

1. Estimate Manning’s roughness (n), the channel slope (S), and the maximum permissible velocity (V) for the channel.

2. Calculate the hydraulic radius (R) using Manning’s equation for these conditions.

5.1

5.049.1 ⎥⎦⎤

⎢⎣⎡=

SVnR

Channel Design Steps for Maximum Permissible Velocity/Allowable Shear

Stress Method (cont.)3. Calculate the required cross-sectional area

(A) using the continuity equation and the design storm peak flow rate (Q):

VQA =

4. Calculate the corresponding wetted perimeter(P):

RAP =

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Channel Design Steps for Maximum Permissible Velocity/Allowable Shear

Stress Method (cont.)5. Calculate an appropriate channel base width (b)

and depth (y) corresponding to a specific channel geometry (usually a trapezoidal channel having a slide slope of z:1).

5b. Chow’s nomograph can be used to significantly shorted the calculation effort by using the following form of Manning’s equation:

5.032

49.1 SnQAR =

Chow’s (1959) nomograph to determine normal depth for different channel geometries and flows:

Example Design of Stable Channel

• Noncolloidal alluvial silts, water transporting colloidal silts:– Manning’s roughness (n) = 0.020– Maximum permissible velocity (V) = 3.5 ft/sec– Allowable shear stress is 0.15 lb/ft2

• The previously calculated peak discharge (Q) = 13 ft3/sec

• The channel slope (S) = 1%, or 0.01

The hydraulic radius (R) using Manning’s equation:5.1

5.049.1 ⎥⎦⎤

⎢⎣⎡=

SVnR

( )( )

.32.001.049.1020.05.3

5.1

5.0 ft=⎥⎦

⎤⎢⎣

⎡=

The required cross-sectional area:

VQA = 27.3

5.313 ft==

Therefore, AR2/3 = (3.7)(0.32)2/3 = 1.7and the wetter perimeter = A/R = 3.7/0.32 = 12 ft

There are many channel options available,

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Channel options that meet allowable velocity criterion:

1.250.120.200.0080.00032250.5

0.940.160.260.0170.0012150.5

1.250.120.200.0080.00032251

1.250.120.200.0080.00032254

0.940.160.26 ft0.0170.0012154

Safety factor for shear

Max shear stress

Normal depth y

y/bAR2/3/b8/3Bottom width b

Side slope z

• As the channel becomes wide, the side slope has little effect on the normal depth and therefore on the shear stress.

• Even though all these channels meet the permissible channel velocity, only those approaching 25 feet wide also meet the allowable shear stress.

• Since the allowable shear stress is 0.15 lb/ft2, the normal depth must be less than 0.24 ft (about 3 inches), requiring a relatively wide channel.

• Current practice is to design channel liners based on shear stress and not on allowable velocity, as it does a better job in predicting liner stability.

Channel Design using Reinforced Liners• If a channel will have intermittent flows, it is

common to use vegetated liners to increase channel stability.

• If channel will have perennial flows, then mechanical liners must be used.

• Reinforced turf mat liner design should examine three phases:– Original channel in unvegetated condition– Channel in partially vegetated condition– Channel in permanent condition with established

vegetation

• Channel matting failure is based on soil loss (usually maximum of 0.5 inch; greater amounts hinder the establishment of vegetation.

• Basic shear stress formula can be modified to predict the shear stress applied to the soil beneath a channel mat:

( )2

1 ⎟⎠⎞

⎜⎝⎛−=

nnCDS s

fe γτ

τe = effective shear stress exerted on soil beneath vegetationγ = specific weight of water (62.4 lbs/ft3)D = the maximum flow depth in the cross section (ft)S = hydraulic slope (ft/ft)Cf = vegetal cover factor (this factor is 0 for an unlined channel)ns = roughness coefficient of underlying soiln = roughness coefficient of vegetal components

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Example Specifications for Erosion Control Blanket (NAG S150BN Straw, 10 month life)

Max. permissible shear stress: 1.85 lb/ft2

0.1000.020>50 ft

0.021> 2 ft0.0700.01020 to 50

0.055 to 0.0210.50 to 2 ft0.0390.00014< 20 ft

0.055<0.50 ft3:1 to 2:1<3:1Slope length (ft)

Manning’s n (unvegetated)

Flow depth (ft)

Slope Gradients (S)

Channel Roughness Coefficients (n)

RUSLE Conservation Coefficients (C)

Example for Matted Channel Liner• Consider the following example:

– Calculated max. shear stress: 2.83 lb/ft2, requiring a NAG P300 permanent mat.

– ns for the soil is 0.016– n for the vegetated mat is 0.042– Cf for the vegetated mat is 0.87– The permissible shear stress for the underlying soil is

0.08 lb/ft2

( ) 053.0042.0016.087.0183.2

2

=⎟⎠⎞

⎜⎝⎛−=eτ

The safety factor is therefore 0.08/0.053 = 1.5 and the channel lining system is expected to be stable.

Slope Stability Applied to Erosion Control• The basic shear stress calculations can be applied to slopes, using

the flow depth of the sheetflow• Sheetflow flow depth can be calculated using the Manning’s

equation:53

5.049.1⎟⎠⎞

⎜⎝⎛=

sqny

y is the flow depth (in feet), q is the unit width flow rate (Q/W, the total flow rate, in ft3/sec,

divided by the slope width, in ft.)n is the sheet flow roughness coefficient for the slope surface, and s is the slope (as a fraction)

The basic shear stress equation can be used to calculate the maximum shear stress expected on a slope:

ySo γτ =

Slope Stability Example• Design storm peak flow rate (Q) = 2.2 ft3/sec• Slope width (W) = 200 ft• Therefore the unit width peak flow = Q/W = 2.2 ft3/sec/200 ft =

0.11 ft2/sec• Slope roughness (n) = 0.24 (vegetated with dense grass; would

be only about 0.055 for an erosion control mat before vegetationestablishment, using the established vegetation condition results in deeper water and therefore a worst case shear stress condition).

( )( )

fty 033.025.049.1

24.0011.0 53

5.0 =⎟⎟⎠

⎞⎜⎜⎝

⎛=

( )( )( ) 51.025.0033.04.62 ==oτ

The corresponding maximum shear stress would therefore be:

(about 0.4 inches)

lb/ft2

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• For an ordinary firm loan soil, the Manning’s roughness is 0.020 and the allowable shear stress is 0.15 lb/ft2.

• Without a protective mat, the calculated maximum shear stress is substantially greater than the allowable shear stress for the soil.

• The effective shear stress underneath the mat would be:

( )2

1 ⎟⎠⎞

⎜⎝⎛−=

nnC s

foe ττ ( ) 067.0055.0020.00151.0

2

=⎟⎠⎞

⎜⎝⎛−= lb/ft2

The safety factor would be about 1.5/0.067 = 2.2Any mat with a Manning’s n greater than about 0.037 would be adequate for this example.

Checking Erosion Yield of Protected Slope• The final erosion control mat selection must be based on

the expected erosion rate for the protected slope.R = 350 (Birmingham, AL conditions)K = 0.28LS for slope length of 300 ft and slope of 25% = 10.81200 ft by 300 ft slope would have an area of 1.4 acres

• For a bare slope (C = 1):Soil loss = (350)(0.28)(10.81)(1) = 1060 tons/acre/yr

• For a protected slope (C = 0.19, and n = 0.055 for a NAG S75 mat):Soil loss = (350)(0.28)(10.81)(0.19) = 201 tons/acre/yr

Checking Erosion Yield of Protected Slope (cont.)

• The unprotected bare slope would lose about 6.3 inches of soil per year, while the protected slope would lose about 1.2 inches per year.

• The USDA uses a maximum loss rate of 0.5 inches per year to allowable plants to survive. Others have proposed a limit of 0.25 inches per year. This is about 42 tons/acre/year (still about 10X the typical USDA limit for agricultural operations).

• The maximum C value for this slope would therefore be about 0.039, requiring the selection of a more substantial erosion control mat.

• The minimum roughness n for this slope is 0.037, based on the previous shear stress calculations.