2011 PE Review:
Michael C. Hirschi, PhD, PE, D.WRE
Professor and Assistant Dean
University of Illinois
mch@illinois.edu
Acknowledgements:
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
Hi.61
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 classes & particle sizes – 3
• Texture
– EX: 50% sand,
20% silt
30% clay?
Hi.64
Soil classes & particle sizes – 4
• Particle size distribution
– EX: draw in a sandy clay loam?
Hi.65
Soil water content
0
• 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
• 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
Soil water flow
• q = A*K*H/L
– K = (q*L)/(A*H)
• K values S4.359; S5.261; H.430
A
H
L
q
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)
 50000gal  1 ft  day  1  1 ft 3 
  278 ft 2
 


A  
 day  1  8 ft  3 ft  7.481gal 
d
4A

 18 ft
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
Soil strength and mechanics – 2
M.234
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
Length/width
B
a1
a2
1 (square)
Width
1.2
0.42
2
Width
1.12
0.45
3
Width
1.07
0.46
4
Width
1.05
0.47
6
Width
1.03
0.48
Strip
Width
1.00
0.50
Circular
Radius
1.2
0.60
Footing bearing loads – 2
M.375
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
 400lb 
 62.5lb 
 125lb 
2






qult  (1.0)
(
30
)

(
0
.
5
)(
3
ft
)
(
18
)

(
4
ft
)(
20
)

23
,
700
lb
/
ft
2 
 ft 3 
 ft 3 
ft






• 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
• 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
Soil erosion and sediment yield
•
•
•
•
•
Hillslope erosion
Channel system erosion
Sediment delivery to streams
Sediment transport in streams
Slope stability
Hillslope soil erosion
• S4.91-111; S5.134-156; H.238-285
• Background
– Detachment
• Raindrop impact
• By turbulent overland flow
– Runoff
– Transport downslope
• By runoff
Hillslope soil erosion background
• At the top of the slope
– Detachment by raindrop impact
– Transport by shallow sheet flow
– Sheet erosion
USDANRCS
Hillslope soil erosion background - 2
• 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
USDANRCS
Hillslope soil erosion background - 3
• Bottom of hillslope
– Ends at concentrated flow channel
– Low area in macrotopography
– “ephemeral gullies”
USDA-
Hillslope erosion factors
• 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
• 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
• 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
• Lookup
– Maps, tables, figures
– Databases
• Process-based calculations
– Show changes over time
– Where don’t have good data
R factor – rainfall erosivity
• 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
• 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
• New tables & figures
– H.264; HO.8
• Know susceptibility to rilling
– High for highly disturbed soils
– Low for consolidated soils
C – cover-management factor
• 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
• 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
P – conservation practice factor
• Common practices
– Contouring, stripcropping, terraces
• Change flow patterns or cause deposition
• Lookup tables
– S4.103; S5.146; H.281; HO.10
Calc.: USLE/RUSLE
• 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
•
•
•
•
•
•
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
• 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
• 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
• USLE/RUSLE for hillslopes
– Erosion
– Delivery
• Erosion critical for soil resource
conservation
• Delivery critical for water quality
– Movement through channel system
Sediment delivery – 2
Sediment delivery – 3
• SDR (Sediment Delivery Ratio)
– Hillslope erosion
– Empirical fit for watershed delivery
• Channel erosion/deposition modeling
– Erosion
– Transport
– Deposition
Sediment Delivery Ratio
• 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
• Area-delivery relationship
H.294
Sediment Delivery Ratio – 3
• Relief-length ratio
– Relief = elev change along main branch
– Length = length along main branch
H.294
Sediment Delivery Ratio – 4
• Forest Service Delivery Index Method
H.295
Sediment Delivery Ratio – 5
• 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
Calc: SDR
• Given:
– Flow path length in watershed = 4000ft
– Elevation difference = 115ft
• Find:
– SDR
Calc.: SDR – 2
• R/L = 115/4000 = 0.029
• From figure SDR = 0.45
Channel erosion-deposition modeling
• 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
Channel erosion-deposition modeling
–2
• Large-channel models
– Sediment transport
– Channel morphology
Sediment transport
• Settling (H.204-209)
– Stokes’ Law
•
•
•
•
•

1 d 2g
SG  1
Vs  
18  

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
Vs  2.81 d
2
Calc.: Stokes’ Law settling
• Given:
– ISSS soil particle size classification
• Find:
– Settling velocities of largest sand, silt, and clay
particles
Calc.: Stokes’ Law settling – 2
• 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
Calc.: Stokes’ Law settling
• 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
• Vs = [(1 ft)/(1 hr)](1 hr/3600s) = 2.778*10-4 fps
• d = (Vs/2.81)1/2 = 0.00994mm
Other settling
• Flocculant, hindered, compression
– Order of increasing concentration
– Slower settling with increased conc.
• Settleable solids
– Imhoff cone
H.211
Sediment transport relationships
• 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
Channel morphology
• 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
• Channel-forming discharge
– Greatest impact on channel itself
– Return period ≈ 1.5 years
• Channel roughness
Channel morphology – 3
• Channel regime
– H.405-419
H.416
Application of process knowledge to
control
• 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
• 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
• 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
• 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
• 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
• Reduce transport capacity of flow
• Enhance deposition of sediment
Reduce transport capacity
• 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
• Increase infiltration
Enhance deposition of sediment
• Use flocculant to increase sedimentation
– Usually in sedimentation ponds when other
methods are not adequate
– Expensive…
Slope stability & failure
Possible forms of Failure
McCarthy (1982) page 437-455
McCarthy (1982) page 440
I.B Mohtar
90
Slope stability & failure – 2
• 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
91
Slope stability & failure – 3
• 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
I.B Mohtar
92
Slope stability & failure – 4
For clay soil
For soil with cohesion and internal friction > 0
McCarthy (1982)
page 474
I.B Mohtar
93
Slope stability & failure – 5
• Ns = c / (γ * Hmax)
–c = cohesion force
–γ = soil unit weight
–Hmax = max depth without sliding
I.B Mohtar
94
Calc.: slope stability
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
95
Calc.: slope stability – 2
• Fig. b, φ = 15o, i = 50o
• Hmax = c / (γ * Ns) =
(500 lb/ft2)(1ft3/ 10)(1/ 0.095) =
48 ft
I.B Mohtar
96
Questions?