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### 2011 PE Review:

Michael C. Hirschi, PhD, PE, D.WRE

Professor and Assistant Dean

University of Illinois

Rod Huffman, PE Review coordinator

Daniel Yoder (2006 presenter)

Rabi Mohtar & Majdi Abu Najm (2010 presenters)

Topics

- Principles of Soil Physics (some duplication with Management session)
- Soil Strength & Mechanics
- Sediment Transport
- Erosion Control
- Slope Stabilization

Sources

- Environmental Soil Physics; Hillel; 1998 Hi
- Essentials of Soil Mechanics & Foundations, 2nd ed.; McCarthy; 1982; M
- Soil and Water Conservation Engineering
- 4th ed. Schwab, Fangmeier, Elliott, Frevert: S4
- 5th ed. Fangmeier, Elliott, Workman, Huffman, Schwab: S5

- Design Hydrology & Sedimentology for Small Catchments; Haan, Barfield, Hayes: H
- USLE/RUSLE Handout: HO

Soil Physics & Mechanics

- Soil classes and particle size distributions
- Soil water
- Content
- Potential
- Flow

- Soil strength & mechanics

Soil classes & particle sizes - 2

- ISSS easiest
- Sand 0.02-2.0mm (20-2000μ)
- Silt 0.002-0.02mm (2-20μ)
- Clay <0.002mm (<2μ)

- Soil texture classes based on USDA
- Sand 0.05-2.0mm
- Silt 0.002-0.05mm
- Clay <0.002mm

Soil water content

- Mt = Ms + Mw + Ma
- Vt = Vs + Vw + Va
- t = total, s = solids, w = water, a = air

- ρb = bulk density = Ms/Vt≈ 1.1-1.4 g/cc
- ρp = particle density = Ms/Vs ≈ 2.65 g/cc
- Porosity = (Vw + Va) / Vt ≈ 25-60%
- ρw = water density = Mw/Vw = 1.0 g/cc

Soil water content – 2

- ww = Mw / (Ms + Mw)
- Water content wet basis

- w = mass wetness = Mw / Ms
- Water content dry basis

- θ = Vw/Vt = Vw / (Vs + Vw + Va)
- Volumetric water content

Calc.: soil water content

- Given:
- Soil with 30% water content dry basis

- Find:
- Best guess at equivalent inches of water in the top foot of soil?

Calc.: soil water content – 2

- Mw / Ms = 0.30
- Mw = Vw * ρw
- ρb = Ms / Vt; Ms = Vt * ρb

Calc.: soil water content – 3

- Mw / Ms = 0.30
- Mw = Vw * ρw
- ρb = Ms / Vt; Ms = Vt * ρb
- Mw / Ms = (Vw * ρw)/(Vt * ρb) = (Vw / Vt)(ρw / ρb)
- θ = Vw/Vt
- θ *(ρw / ρb) = 0.3; θ = (ρb / ρw) * 0.3
- θ = 0.3 *(1.3/1.0) = 0.39
- 0.39 * 1 ft * 12”/ft = 4.7”

Soil water potential

Hi.157

Soil water potential – 2

Cuenca.58

Soil water potential – 3

S5.337

Soil water potential – 4

Hi.162

Calc.: soil water potential

- Given:
- Mercury tensiometer
- SG = 13.6

- Situation as shown

- Mercury tensiometer
- Find:
- Total potential at C
- Above or below water table?

Cuenca.64

Calc.: soil water potential - 2

- 1) pick datum
- 2) add pressures
- Suction
- Water depth
- Gravity

- T = z + p + pos
- = -86cm
- Above water table

Calc.: soil water flow

- Given:
- Need 50000 gpd through a 1-ft thick sand filter with K = 8 ft/d, and a total driving head of 3 ft

- Find :
- Required diameter for circular tank?

Calc.: soil water flow – 2

- q = A*K*H/L; A = (q*L)/(K*H)

Soil strength and mechanics

- M.233-237,373-379
- Soil stresses
- Normal = Fn/A = σ
- Shear = Ft/A = τ
- Fn = normal force
- Ft = tangential or shear force

- As σ↑ τ to cause failure = τf↑
- tan Φ = τf / σ; Φ = angle of internal friction

Calc.: soil strength

- Given:
- Well-graded sand; density 124 lb/ft3

- Find:
- Ultimate shear strength 6 ft below surface?

Calc.: soil strength – 2

- From table, for well-graded sand, Φ = 32-35o = 33.5o
- Normal stress = (124 lb/ft3)(6 ft) = 744 lb/ft2
- tan Φ = τf / σ;
τf = σ * tan Φ =

τf = 744 lb/ft2 * tan(33.5o) = 492 lb/ft2

Footing bearing loads

- M.374-379
- qult = a1*c*Nc + a2*B*γ1*Nγ + γ2*Df*Nq
- c = soil cohesion beneath footer
- γ1,, γ2 = effective soil unit weight above and below footer
- B = footer size term
- Nc, Nγ, Nq = capacity factors
- Df = footing depth below surface

- qdesign = qult / FS

Calc.: footing load

- Given:
- Strip footing 3 ft wide
- Wet soil with density of 125 lb/ft3
- Angle of internal friction = 30o
- Cohesive strength of 400 lb/ft2
- Use factor of safety of 3

- Find:
- qdesign in lb/ft2

Calc.: footing load – 2

- a1 = 1.0, a2 = 0.5, B = width = 3’
- γ1 = 125/2 = 62.5 lb/ft3; γ2 = 125 lb/ft3
- c = 400 lb/ft2
- Nc = 30, Nγ = 18, Nq = 20
- qdesign = 23,700/3 = 7900 lb/ft2

Soil compaction and density

- Soil compaction
- Greater strength
- Dependent on water content

- Proctor test
- Pack soil into mold with pounding
- Modified Proctor > 56000 ft-lbs

The Erosion Process

- Soil erosion is a multi-step process:
- Soil particle/aggregate detachment
- Soil particle/aggregate transport
- Soil particle/aggregate deposition

- There must be detachment and transport for erosion to occur
- Deposition (sedimentation) will occur somewhere downstream

A little soils review…

- Soil primary particles:
- Sand, 0.05mm to 2mm, 2.65 g/cc density
- Silt, 0.002mm to 0.05mm, 2.65 g/cc
- Clay, <0.002mm, 2.6 g/cc

- Soil aggregates, chemically/electrically bonded sets of primary particles:
- Large, in the sand range, 1.6 g/cc
- Small, in the large silt range, 1.8 g/cc

- These aggregate sizes are approximately those used in the CREAMS model (USDA-ARS)

Detachment

- There are many sources of force and energy required to detach soil particles & aggregates:
- Raindrop impact
- Shallow surface flow shear
- Concentrated flow shear
- Many more, at larger scales

Transportation

- Many of the same processes contribute force and energy for soil particle & aggregate transport:
- Raindrop impact
- Shallow surface flow
- Concentrated surface flow
- Channelized flow
- Others

Balancing act

- Foster & Meyer (1972) proposed a balance between detachment and transport for flowing water:
- 1 = (transport load/transport capacity) + (detachment load/detachment capacity)

Essentially, if the flow is using all its transport capacity transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

Example transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

- In the 80’s and 90’s there was a successful push to conservation tillage as a method to reduce sediment in lakes and streams
- In many situations, no improvement was seen, but streambank erosion became more of a problem than it was in the past
- I contend that now that the streams are receiving cleaner water (because of less upland erosion), but at similar rates, from farm fields, the stream uses less of its erosive energy to transport load it receives from runoff water, so it has capacity to undercut banks and scour the streambed

Multi-stage erosion transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

Soil erosion and sediment yield transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

- Hillslope erosion
- Channel system erosion
- Sediment delivery to streams
- Sediment transport in streams
- Slope stability

Hillslope soil erosion transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

- S4.91-111; S5.134-156; H.238-285
- Background
- Detachment
- Raindrop impact
- By turbulent overland flow
- Runoff

- Transport downslope
- By runoff

- Detachment

Hillslope soil erosion background transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

- At the top of the slope
- Detachment by raindrop impact
- Transport by shallow sheet flow
- Sheet erosion

USDA-NRCS

Hillslope soil erosion background - 2 transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

- Lower on slope
- Small flow concentrations
- Start to cut small channels
- Rills
- Roughly parallel
- Head straight downslope
- Random formation

- Flow from sheet areas between rills
- Sheet and rill erosion

USDA-NRCS

Hillslope soil erosion background - 3 transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

- Bottom of hillslope
- Ends at concentrated flow channel
- Low area in macrotopography
- “ephemeral gullies”

USDA-NRCS

Hillslope erosion factors transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

- Rainfall erosivity
- Intensity
- Total storm energy

- Soil erodibility
- Topography
- Slope length
- Steepness

- Management
- Reduce local erosion
- Change runoff path
- Slow and spread runoff => deposition

USLE/RUSLE transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

- A = R * K * LS * C * P
- A = average annual soil erosion (T/A/Y)
- R = rainfall erosivity (awful units)
- K = soil erodibility (awful units)
- R * K gives units of T/A/Y

- LS = topographic factor (dimensionless)
- C = cover-management (dimensionless)
- P = conservation practice (dimensionless)

USLE/RUSLE – background transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

- Empirical approach
- >10000 plot-years of data
- International use

- Unit Plot basis; LS = C = P = 1
- Near worst-case management

- R from good fit rainfall-erosion
- K from K = A / R
- C and P from studies
- Subfactors in later versions

USLE/RUSLE – approach transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

- Lookup
- Maps, tables, figures
- Databases

- Process-based calculations
- Show changes over time
- Where don’t have good data

R factor – rainfall erosivity transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

- maps
- H.251; H.Appendix 8A; S4.99(SI); S5.143(SI); HO.1-5
- R(customary SI) = 17.02 * R(customary US)

S4

K factor – soil erodibility transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

- Soil surveys, NASIS, H.261-262; HO.6
- Erodibility nomograph
- H.255; S4.101; S5.144; HO.7
- No short-term OM

LS – topography factor transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

- New tables & figures
- H.264; HO.8

- Know susceptibility to rilling
- High for highly disturbed soils
- Low for consolidated soils

C – cover-management factor transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

- Part of normal management scheme
- Lookup: S4.102; S5.146; H.266; H.Appendix 8; HO.9
- Change over time

C – cover-management factor - 2 transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

- Subfactor approach (RUSLE)
- C = PLU * CC * SC * SR * SM; all 0-1
- PLU = prior land use
- roots, buried biomass, soil consolidation

- CC = canopy cover; % cover & fall height
- SC = exp(-b * % cover)
- B = 0.05 if rills dominant; 0.035 typical; 0.025 interrill

- SR = roughness; set by tillage, reduces over time
- SM = soil moisture; used only in NWRR

- PLU = prior land use

- C = PLU * CC * SC * SR * SM; all 0-1

P – conservation practice factor transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

- Common practices
- Contouring, stripcropping, terraces

- Change flow patterns or cause deposition
- Lookup tables
- S4.103; S5.146; H.281; HO.10

Calc.: USLE/RUSLE transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

- Given:
- Materials in handout
- 3-Acre construction site near Chicago
- Straw mulch applied at 4 T/A
- Average 20% slope, 100’ length
- Loamy sand subsoil
- Fill (loose soil)

- Find:
- Erosion rate in T/A/Y

Calc: USLE/RUSLE – 2 transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

- R = 150 (HO.1)
- K = 0.24 (HO.7)
- LS = 4 (HO.8-high rilling)
- C = 0.02 (HO.9)
- P = 1.0
- A = R * K * LS * C * P = 2.9 T/A/Y

Calc: USLE/RUSLE – 2.1 transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

- Given:
- Materials in handout
- 16-A site near Dallas, TX
- Silty clay loam subsoil
- Average 50% slope, 75’ length
- Cut soil

- Find:
- By what percentage will the erosion be reduced if we increase our straw mulch cover from 40% cover to 80% cover?

Calc: USLE/RUSLE – 2.2 transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

- Only thing different is C
- Only subfactor different is SC

- SC = exp(-b * %cover)
- For consolidated soil, b = 0.025

- SC1 = exp(-0.025 * 40%) = 0.368
- SC2 = exp(-0.025 * 80%) = 0.135
- Reduction = (0.368 – 0.135)/0.368 = 63%

Sediment delivery transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

- USLE/RUSLE for hillslopes
- Erosion
- Delivery

- Erosion critical for soil resource conservation
- Delivery critical for water quality
- Movement through channel system

Sediment delivery – 2 transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

Sediment delivery – 3 transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

- SDR (Sediment Delivery Ratio)
- Hillslope erosion
- Empirical fit for watershed delivery

- Channel erosion/deposition modeling
- Erosion
- Transport
- Deposition

Sediment Delivery Ratio transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

- H.293-299
- SDR = SY / HE
- SDR = sediment delivery ratio
- SY = sediment yield at watershed exit
- HE = hillslope erosion over watershed

Sediment Delivery Ratio – 2 transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

- Area-delivery relationship

H.294

Sediment Delivery Ratio – 3 transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

- Relief-length ratio
- Relief = elev change along main branch
- Length = length along main branch

H.294

Sediment Delivery Ratio – 4 transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

- Forest Service Delivery Index Method

H.295

Sediment Delivery Ratio – 5 transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

- MUSLE (H.298)
- Y = 95(Q * qp)0.56 (Ka)(LSa)(Ca)(Pa)
- Y = storm yield in tons
- Q = storm runoff volume in acre-in
- q = peak runoff rate in cfs
- K, LS, C, P = area=weighted watershed values

- SDR = 95(Q * qp)0.56/(R * area)
- R = storm erosivity in US units

- Routing for channel delivery

- Y = 95(Q * qp)0.56 (Ka)(LSa)(Ca)(Pa)

Calc: SDR transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

- Given:
- Flow path length in watershed = 4000ft
- Elevation difference = 115ft

- Find:
- SDR

Calc.: SDR – 2 transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

- R/L = 115/4000 = 0.029
- From figure SDR = 0.45

Channel erosion-deposition modeling transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

- Process-based small channel models
- Foster-Lane model
- H.285-289
- Complicated and process-based

- Ephemeral Gully Erosion Model
- EGEM
- Fit to Foster-Lane Model results

- Foster-Lane model

Channel erosion-deposition modeling – 2 transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

- Large-channel models
- Sediment transport
- Channel morphology

Sediment transport transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

- Settling (H.204-209)
- Stokes’ Law
- Vs = settling velocity
- d = particle diameter
- g = accel due to gravity
- SG = particle specific gravity
- ν = kinematic viscosity

- Simplified Stokes’ Law
- SG = 2.65
- Quiescent water at 68oF
- d in mm, Vs in fps

- Stokes’ Law

Calc.: Stokes’ Law settling transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

- Given:
- ISSS soil particle size classification

- Find:
- Settling velocities of largest sand, silt, and clay particles

Calc.: Stokes’ Law settling – 2 transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

- ISSS classification
- Largest particles size
- Clay = 0.002mm
- Silt = 0.2mm
- Sand = 2mm

- Vs,clay = 1.12*10-4 fps = 0.04 ft/hr = 0.97 ft/day
- Vs,silt = 0.11 fps = 405 ft/hr = 1.83 mi/day
- Vs,sand = 11.24 fps = 7.66 mph = 184 mi/day

- Largest particles size

Calc.: Stokes’ Law settling transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

- Given:
- Stokes’ Law settling

- Find: particles larger than what size can be assumed to settle 1 ft in one hour?

Calc.: Stokes’ Law settling – 2 transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

- Vs = [(1 ft)/(1 hr)](1 hr/3600s) = 2.778*10-4 fps
- d = (Vs/2.81)1/2 = 0.00994mm

Other settling transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

- Flocculant, hindered, compression
- Order of increasing concentration
- Slower settling with increased conc.

- Settleable solids
- Imhoff cone

H.211

Sediment transport relationships transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

- H.223-233
- Bedload
- Einstein equation

- Suspended load
- Total load
- Yalin total load equation
- Vavg, d, SG, r, т, тc

- Yang stream power relationship
- Stream power instead of critical shear

- Yalin total load equation

Channel morphology transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

- H.391-419
- Channel cross-section
- Width B, depth D, avg. vel. U, suspended sediment load qs, flow rate q
- B = Ca * qa, D = Cb * qb, U = Cc * qc,
qs = Cd * qd

- Ca, Cb, Cc, Cd, a, b, c, d = fit coefficients & exponents

Channel morphology – 2 transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

- Channel-forming discharge
- Greatest impact on channel itself
- Return period ≈ 1.5 years

- Channel roughness

Channel morphology – 3 transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

- Channel regime
- H.405-419

H.416

Application of process knowledge to control transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

- Limit individual parts to limit whole
- Limit detachment
- Limit transport

- Enhance deposition strategically
- Where damage is minimal
- Where cleanup is possible

Control of Soil Erosion by Water transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

- Detachment limiting strategies
- Reduce raindrop impact (“Stop the Drops”)
- Reduce runoff
- Reduce detachment capacity of runoff
- Increase soil resistance to erosive forces

- Transport limiting strategies
- Reduce runoff volume
- Reduce runoff transport capacity (“Slow the Flow”)

Example – No-Till transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

- Detachment
- Raindrop impact detachment is very low due to high surface cover percentage
- Flow shear detachment is low due to low velocities caused by tortuous flow path
- Soil is resistant to erosion because of low disturbance

- Transport
- Raindrop transport is limited by surface residue
- Flow transport is limited by increased infiltration, lessening runoff
- Flow transport is further limited by small dams created by surface residue

Example – Mulch on newly seeded area transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

- Detachment
- Raindrop impact detachment is very low due to high surface cover percentage
- Flow shear detachment is low due to low velocities caused by tortuous flow path

- Transport
- Raindrop transport is limited by surface residue
- Flow transport is limited by increased infiltration, lessening runoff
- Flow transport is further limited by small dams created by surface residue

Comparison of no-till vs. mulch transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

- Detachment
- Likely higher with mulch for same surface cover fraction because of higher soil disturbance for seedbed preparation
- Likely higher for no-till following dry years because amount of residue cover is dictated by prior year crop growth

- Transport
- Likely higher for mulch, unless “cut” in because no-till residue is effectively “cut” in during planting, at least for a small area, hopefully across slope
- Likely higher for mulch situation because seedbed prep likely reduced average aggregate diameter

Control of Sediment in Runoff transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

- Reduce transport capacity of flow
- Enhance deposition of sediment

Reduce transport capacity transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

- Slow the flow
- Barriers
- Must let water pass, though slowly
- Must be flow-stable, even after use
- Must be where maintenance is possible

- Reduce slope steepness
- Channel must be of adequate capacity

- Barriers
- Increase infiltration

Enhance deposition of sediment transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

- Use flocculant to increase sedimentation
- Usually in sedimentation ponds when other methods are not adequate
- Expensive…

Slope stability & failure transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…Possible forms of Failure

McCarthy (1982) page 437-455

McCarthy (1982) page 440

I.B Mohtar

Slope stability & failure – 2 transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

- Terms
- β=max. slope angle before sliding
- Φ=angle of internal friction

- Cohesionless soil
- tan(β) = tan(Φ)
- Saturated: tan(β) = (1/2)tan(Φ)

I.B Mohtar

Slope stability & failure – 3 transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

- Cohesive soil
- γ*z*sin(β)*cos(β) = c + σ*tan(Φ)
- z = assumed depth
- c = cohesive force
- σ = effective compressive stress

- Rotational or sliding block

- γ*z*sin(β)*cos(β) = c + σ*tan(Φ)

I.B Mohtar

Slope stability & failure – 4 transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

For clay soil

For soil with cohesion and internal friction > 0

McCarthy (1982) page 474

I.B Mohtar

Slope stability & failure – 5 transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

- Ns = c / (γ * Hmax)
- c = cohesion force
- γ = soil unit weight
- Hmax = max depth without sliding

I.B Mohtar

Calc.: slope stability transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

Example 8:

- Given:
- Cohesion strength = 500 lb/ft2
- Unit weight = 110 lb/ft3
- Slope steepness = 50o
- Internal friction angle = 15o

- Find:
- Max. slope height

I.B Mohtar

Calc.: slope stability – 2 transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

- Fig. b, φ = 15o, i = 50o
- Hmax = c / (γ * Ns) =
(500 lb/ft2)(1ft3/ 10)(1/ 0.095) = 48 ft

I.B Mohtar

Questions? transporting sediment, there’s nothing left to detach more. Likewise, if the flow is detaching new sediment at detachment capacity, there’s no capacity to transport any sediment. Natural systems balance out…

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