5 Channel and Slope Stability

8/16/2019 5 Channel and Slope Stability

1/20

1

Module 5: Channel and SlopeStability

Robert PittDepartment of Civil, Construction and

Environmental EngineeringUniversity of Alabama

Tuscaloosa, AL

Massive Streambank Failure at New Outfall in a Suburban Area

Some potentialsolutions tostabilizestreambanks.

Channel and Slope Protection

• Upslope diversions• Channel Protection

– Channel liners – Check dams

• Slope Protection – Roughening surface – Erosion control mats


2/20

2

SlopeDiversions

Channel Lining


3/20

3

Check Dams

Roughening Slopes while Compacting Ground


4/204

Gabions for Slope Protection

Slope Protection Using Different Materials

Rock and Asphalt for Shaded Areas

Coir Logs


5/205

Channel Design Based on AllowableVelocity and Shear Stress

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

• Shear stress is calculated based on water depth andchannel 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 allowablevelocity because its sediment carrying capacity isalready reduced).

Example Allowable Velocityand Shear Stress Values

Silty Water (on siteand downslope)

Clear Water(diversions)

Manning’sroughness

LinerTexture

1.105.500.915.000.035Cobbles

0.325.000.0752.500.020Finegravel

0.153.500.0752.500.020Firmloam

0.0752.500.0271.500.020Finesand

τo (lb/ft 2)V (ft/sec)τo (lb/ft 2)V (ft/sec)n

Boundary Shear Stress (tractive force):

RS o γ τ =

RS o γ τ =

γ = specific weight of water (62.4 lbs/ft 3)R = hydraulic radius (flow depth is used for maximum shear stresscalculations; for sheetflow conditions, the depth is equal to thehydraulic radius)

S = channel slope Chow 1959


6/206

Chow 1959

Example plot ofallowable shearstress (unittractive force)for cohesivematerials.

COE 1994

Example Allowable Tractive Forces

• Sheetflow: ¼ inch deep on a 10 % slope: – (62.4 lb/ft 3)(0.021 ft)(0.10) = 0.13 lb/ft 2

– >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/ft 3)(0.021 ft)(0.02) = 0.026 lb/ft 2

– > 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/ft 3)(1.5 ft)(0.05) = 4.7 lb/ft 2

– no natural lining material safe

• Channel Flow: 18 inches deep on a 1 %slope: – (62.4 lb/ft 3)(1.5 ft)(0.01) = 0.93 lb/ft 2

– >70 mm noncohesive material OK


7/207

Channel Design Steps for MaximumPermissible Velocity/Allowable Shear

Stress Method1. Estimate Manning’s roughness (n), the

channel slope (S), and the maximum permissible velocity (V) for the channel.

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

5.1

5.049.1 ⎥⎦⎤

⎢⎣

⎡=S

Vn R


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

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

V Q

A =

4. Calculate the corresponding wetted perimeter (P):

R A

P =


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

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

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

5.032

49.1 S nQ

AR =

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


8/208

Example Design of Stable Channel• Noncolloidal alluvial silts channel lining,

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/ft 2

• The previously calculated peak discharge(Q) = 13 ft 3/sec• The channel slope (S) = 1%, or 0.01

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

5.1

5.049.1 ⎥⎦⎤

⎢⎣

⎡=S

Vn R

( )( )

.32.001.049.1020.05.3 5.1

5.0 ft =⎥⎦

⎤⎢⎣

⎡=

The required cross-sectional area:

V Q

A = 27.35.3

13 ft ==

Therefore, AR 2/3 = (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,

Channel options that meet allowablevelocity 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

Safetyfactor forshear

Maxshearstress

Normaldepth y

y/bAR 2/3/b8/3Bottomwidth b

Sideslope z

• As the channel becomes wide, the side slope haslittle effect on the normal depth and therefore on theshear stress.

• Even though all these channels meet the permissible

channel velocity, only those approaching 25 feetwide also meet the allowable shear stress.• Since the allowable shear stress is 0.15 lb/ft 2, the

normal depth must be less than 0.24 ft (about 3inches), requiring a relatively wide channel.

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


9/209

Channel Design using Reinforced Liners

• If a channel will have intermittent flows, it iscommon to use vegetated liners to increasechannel stability.

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

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

vegetation

Sod placed along a

drainage bottom,with grass seedalong the edges

Seeding alongmedian strip swaleof highway project

• barley ( Hordeum vulgare L .), noted for itsearly fall growth;

• oats ( Avena) sativa L .), in areas of mildwinters;

• mixtures of wheat, oats, barley, and rye(Secale cereale L .);

• field bromegrass ( Bromus spp .); and• ryegrasses ( Lolium spp .).

Temporary Vegetation Plantings

Bermudagrass is most widely used permanent grassin the sourthern US (for permanent plantings)


10/2010

• Channel matting failure is based on soil loss (usuallymaximum 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 channelmat:

( )2

1 ⎟ ⎠ ⎞

⎜⎝ ⎛ −=

nn

C DS s f e γ τ

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

50sudangrass0.50150common lespedeza0.50300lespedeza sericea0.50500alfalfa0.50250yellow bluestem0.50

350weeping lovegrass0.50200grass mixture0.75350blue grama0.87350kentucky bluegrass0.87400buffalograss0.87500centipedegrass0.90500bermudagrass0.90

Reference stemdensity (M),stem/ft 2

Covers TestedCover Factor (C f )(good uniformstands)

Allowableeffective stress fornoncohesive soils

Soil grainroughness fornoncohesive soils


11/2011

Standard VR-n curves for deep channel flows

USDA Stillwater, OK, tests

Indoor Channel Trendlines in Comparision to Stillwater Curves

0.00

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0.80

0.90

1.00

1.00E-04 1.00E-03 1.00E-02 1.00E-01 1.00E+00

VR (m^2/sec)

M a n n i n g ' s " n "

Outdoor Swale Data

Retardance Classes ( A - E)

A

B

C

D

E

Bluegrass

Centipede

Zoysia

VR-n curve for different grasses, showing results forshallow flows; Univ. of Alabama tests (Kirby 2003)

(multiply ft 2/sec by 0.092 to obtain m 2/sec)

Example Problem for the Selection of RoughnessCoefficient for Grass-Lined Channels

Determine the roughness value for a 10-year design storm of70 ft 3/sec (2 m 3/sec) in a grass-lined drainage channel havinga slope of 0.05 ft/ft and a 4 foot (1.2 m) bottom width and 1:1side slopes. The grass cover is expected to be in retardancegroup D.

Long-term design, based on vegetated channel stability:

•use Q peak = Q 10year = 70 ft 3/s (2 m 3/s) initially assume that n vegetated = 0.05

Determine the normal depth of flow:

( )

( )51.10

05.049.1

7005.0

49.15.05.0

32

=== cfs

S

nQ AR

AR2/3/b8/3 = 10.51/40.32 = 0.26

AR 2/3 /b 8/3 = 10.51/40.32 = 0.26

With a 1:1 side slope trapezoidal channel, the ratio of y/b is0.43, and the depth is 4(0.43) = 1.7 ft.

The cross-sectional area = 9.7 ft 2

The velocity = (70 ft 3/sec)/(9.7 ft 2) = 7.2 ft/sec

P = 8.8 ftR = 9.7 ft 2/8.8 ft = 1.1 ftVR is therefore = (7.2 ft/sec)(1.1 ft) = 7.9 ft 2/secFrom the VR-n curve, the “new” n is therefore 0.032

for retardance class D grass.

The normal depth must be re-calculated:

( )

( )72.6

05.049.1

70032.0

49.15.05.0

32

=== cfs

S

nQ AR


12/2012

AR 2/3 /b 8/3 = 6.72/40.32 = 0.17With a 1:1 side slope trapezoidal channel, the ratio of y/b is

0.34, and the depth is 4(0.34) = 1.4 ft.

The cross-sectional area = 7.6 ft 2

The velocity = (70 ft 3/sec)/(7.6 ft 2) = 9.2 ft/secP = 8.0 ftR = 7.6 ft 2/8.0 ft = 0.95 ftVR is therefore = (9.2 ft/sec)(0.95 ft) = 8.7 ft 2/secFrom the VR-n curve, the revised value of n is still close to 0.032

The maximum shear stress (using normal depth instead of

hydraulic radius) is therefore:

γ DS= (62.4 lb/ft 3) (1.4 ft) 0.05 ft/ft) = 4.4 lb/ft 2

This is a relatively large value for shear stress, requiringreinforcement.

Example Specifications for Erosion ControlBlanket (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


13/2013

Example of a permanent channel design, with a liner:

( )( )

( )38.1

08.049.1

2902.05.0

3

2

== AR

One possible solution:

A = [(7.64+5)/2] (0.44) = 2.78 ft2

V = Q/A = 29 ft 3/sec/2.78 ft 2 = 10.4 ft/sec

R = A/PP = 5 + 2(3.16)(0.44) = 7.78 ft.R = A/P = 2.78 ft2/7.78 ft. = 0.36 ft.τ = γRS = (62.4lb/ft 3)(0.36 ft.)(0.08) = 1.8 lb/ft 2

which is relatively large (The permissible shear stress for theunderlying soil is 0.08 lb/ft 2)

n = 0.02

Q = 29 CFSS = 8% (0.08)

Permanent C350 liner, 5 ft bottom width, z=3 side slope, andphase 3 vegetation plant stage (mature)

A = [(5+9.2)/2] (0.7) = 4.97 ft2

P = 5 + 2(1.21) = 7.42 ftR = A/P = 4.97/7.42 = 0.67τ = γ DS = (62.4lb/ft 3)(0.70 ft.)(0.08) = 3.49 lb/ft 2

(design case using normal depth)V = Q/A = 29 ft 3/sec/4.97 ft 2 = 5.8 ft/sec

( ) ( ) 22

22

/048.0049.0016.0

87.01/49.31 ft lb ft lbn

nC DS s f e =⎟ ⎠

⎞⎜⎝ ⎛ −=⎟

⎠ ⎞

⎜⎝ ⎛ −= γ τ

n s = 0.016; C f = 0.87 phase 3Effective shear stress on underlying soil:

Another alternative design with a liner:

The permissible shear stress for the underlying soil is 0.08 lb/ft 2

Channel Design using Concrete and Riprap Liner Materials

“A lined swale is a constructedchannel with a permanent liningdesigned to carry concentrated runoffto a stable outlet. This practiceapplies where grass swales areunsuitable because of conditions

such as steep channel grades, prolonged flow areas, soils that aretoo erodible or not suitable tosupport vegetation or insufficientspace is available” – AL Handbook

Even concrete-lined channels may fail

Alabama Handbook

Capacity graph forconcrete flumes, depthof flow = 0.50 feet


14/20

14

Riprap-lined Swale“A riprap-lined swale is anatural or constructed channelwith an erosion-resistant rocklining designed to carryconcentrated runoff to a stableoutlet. This practice applieswhere grass swales areunsuitable because ofconditions such as steepchannel grades, prolongedflow areas, soils that are tooerodible or not suitable tosupport vegetation orinsufficient space.” –

Alabama Handbook

Stable rock sizes, for rock lined swales having gradients between2 percent and 40 percent should be determined using the

following formulas from Design of Rock Chutes by Robinson,Rice, and Kadavy.

For swale slopes between 2% and 10%:d50 = [q (S) 1.5/4.75x10 -3]1/1.89

For swale slopes between 10% and 40%:d50 = [q (S) 0.58 /3.93x10 -2]1/1.89

d50 = Particle size for which 50 % of the sample is finer, inchS = Bed slope, ft/ftq = Unit discharge, ft 3/s/ft (Total discharge/Bottom width)

3.310.06.120000

2.57.64.58000

2.36.94.06000

2.06.03.64000

1.85.42.752000

1.54.72.615001.253.72.21000

1.03.01.9500

0.92.61.6300

0.672.01.3150

0.61.751.1100

0.51.40.850

Rectangular Shape LengthWidth, Height (feet)

Mean SphericalDiameter (feet)

Weight (lbs)

Size of Riprap Stones

2000-1000200--5-100050050--4500-200-25-3200-80--102100-50--101d 90d 75d 50d 25d 15d 10

Weight (lbs.)Class

Graded Riprap


15/20

15

Check Dams“Check dams are small barriers or damsconstructed across a swale, drainage ditchor areas of concentrated flow. Check damsare to prevent or reduce erosion bylessening the gradient of the flow channelwhich reduces the velocity of storm waterflows. Some sediment will be trappedupstream from the check dams, but itsvolume will be insignificant and should not

be considered in off-site sedimentreduction.” – Alabama Handbook

Flow Rates through Stone Check Dams

( )[ ]5.0

2

3/2

5.2/ L D L

W hQ

++=

Q = Outflow through the stone check dam (cfs)h = Ponding depth behind the check dam (ft)W = Width of the check dam (ft), not to be confused with thehorizontal flow path length through the check damL = Horizontal flow path length through the check dam (ft)D = Average rock diameter in the check dam (ft)

h = 3 ft W = 15 ft,L = 3 ft D = 9 inches = 0.75 ft

( )[ ]( ) ( )

( ) ( )[ ]sec/9.7

35.275.0/3

153

5.2/3

5.02

3/2

5.02

3/2

ft ft ft ft

ft ft

L D L

W hQ =

++=

++=

Example Design for Reinforced Grass-Lined Channelswith Check Dams and Level Spreader Pads

A new industrial site in Huntsville, AL, has several 2-acreindividual building sites. Each of the sites will be served witha grass-lined channel that will carry site water to a largerswale system. The slopes of the channels vary from about 1 to6.5%. The calculated peak flow from each construction sitewas calculated to be 16 ft 3/sec (corresponding to theHuntsville, AL, 25 yr design storm of 6.3 inches for 24 hours).

A grass-lined channel is to be designed for each site. The bareseed bed is assumed to have a hydraulic roughness of about0.016.


16/20

16

6.41.30.0390.466.5%

5.51.40.0350.425%

4.82.20.0230.283%

3.14.20.0120.141%

Maximumvelocity withmaturevegetation(ft/sec)

Safetyfactor(allowableshear stressof 0.05lb/ft 2)

Unvegetatedmat shearstress, effecton soil (lb/ft 2)

Bare seedbed shearstress(lb/ft 2)

Slope

The seed bed has an allowable shear stress of about 0.05lb/ft 2. The calculated values for unprotected conditions areall much larger. Therefore, a North American Green S75mat was selected, having an allowable shear stress of 1.55lb/ft 2 and a life of 12 months.

The check dams are assumed to be 2 ft high to the maximumover-topping elevation in 3 ft deep channels. In channelswith 5% slopes, the check dams would have to be about 40 ftapart (or less) to ensure that the toes of the upstream checkdams were at the same, or lower, elevations as the overflowsof the downstream dams. Similarly, the check dams in the6.5% sloped channels would have to be no more than 30 ftapart.

Slope Stability


17/20

17

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 ⎟

⎠ ⎞

⎜⎝ ⎛ =

sqn

y

y is the flow depth (in feet),q is the unit width flow rate (Q/W, the total flow rate, in ft 3/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 themaximum shear stress expected on a slope:

yS o γ τ =

Slope Stability Example• Design storm peak flow rate (Q) = 2.2 ft 3/sec• Slope width (W) = 200 ft• Therefore the unit width peak flow = Q/W = 2.2 ft 3/sec/200 ft =

0.11 ft 2/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 resultsin deeper water and therefore a worst case shear stresscondition).

( )( )

ft y 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/ft 2

• For an ordinary firm loam soil, the Manning’sroughness is 0.020 and the allowable shear stressis 0.15 lb/ft2.

• Without a protective mat, the calculatedmaximum shear stress is substantially greaterthan the allowable shear stress for the soil.

• The effective shear stress underneath the matwould be:

( )2

1 ⎟ ⎠ ⎞

⎜⎝ ⎛ −=

nn

C s f oe τ τ ( ) 067.0055.0020.0

0151.02

=⎟ ⎠ ⎞

⎜⎝ ⎛ −= lb/ft 2

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.


18/20

18

Checking Erosion Yield of Protected Slope

• The final erosion control mat selection must be based onthe 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 inchesof soil per year, while the protected slope would loseabout 1.2 inches per year.

• The USDA uses a maximum loss rate of 0.5 inches peryear to allowable plants to survive. Others have

proposed a limit of 0.25 inches per year. This is about42 tons/acre/year (still about 10X the typical USDAlimit for agricultural operations).

• The maximum C value for this slope would therefore beabout 0.039, requiring the selection of a moresubstantial erosion control mat.

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

0.110.060.070.120.150.660.110.190.220.66S ≥ 2:1

0.090.050.070.080.100.300.090.110.180.30S between 3:1to 2:1

0.070.040.0070.010.020.190.070.100.120.19S ≤ 3:1

Length ≥ 50 ft (15 m)

0.100.050.060.0920.1180.450.100.150.170.45S ≥ 2:1

0.060.030.040.0550.070.210.060.790.120.21S between 3:1to 2:1

0.040.020.0030.0050.0100.110.040.510.060.11S ≤ 3:1

Length between 20 and 50 ft (6 to 15 m)

P300

C350C125BN

CS150BN

S150BN

S75BN

C125SC150

S150S75Slope lengthand gradient

North American Green Conservation Factors f or DifferentErosion Control Mats, for Diff erent Slopes and Slope Lengths


19/20

19

RUSLE Cover Factors (C) for Grasses

0.0050.140.14all--Synthetic mats

0.0050.070.07all--Organic andsynthetic blanketsand compositemats

0.020.050.2034 – 502

0.020.050.1726 – 332

0.020.050.1422 – 252

0.020.050.1116 – 202

0.020.050.0711 – 152

0.020.050.06


20/20

20

Chemical Treatment of Exposed Soils

These newly developed materials act by chemically combining

small soil particles into larger discrete particles that are moreeffective in settling in ponds and in channels. Polyacrylamide(PAM) is the most common chemical being sold now.Polyacrylamide used for erosion control should have a negative(anionic) molecular charge.

Homework for Chapter 5

1) Identify several different slope categories on yourconstruction evaluation site and propose suitablecontrol practices for each type. Justify your selectionswith appropriate calculations.

2) Design an appropriate diversion swale, or a maindrainage swale for your evaluation site. Using thepreviously calculated flow rates, select a suitablechannel lining, including the consideration of checkdams. Justify your selections with appropriatecalculations.

Documents

5 Channel and Slope Stability