I.B Soil Conservation Systems
Rabi H. Mohtar
Professor, Environmental and Natural Resources Engineering
Executive Director, Qatar Environment and Energy Research Institute
mohtar@purdue.edu or rmohtar@qf.org.qa
August 2012
1
Materials To Be Covered
1.Principles of Soil Physics (Mohtar)
2.Sediment Transport (Mohtar)
3.Erosion Control (Mohtar)
4.Soil Mechanics (Khawlie)
5.Slope Stabilization (Khawlie)
This will provide you with an overall review and
not necessarily makes you an expert!
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Sources
1.
2.
3.
Environmental Soil Physics; Hillel; 1998 Hillel (1998)
Essentials of Soil Mechanics & Foundations, 7th ed.; McCarthy;
2007; McCarthy (2007)
Soil and Water Conservation Engineering
a)
b)
4.
5.
6.
7.
8.
4th ed. Schwab, Fangmeier, Elliott, Frevert: Schwab et al (1993)
5th ed. Fangmeier, Elliott, Workman, Huffman, Schwab: Fangmeier et
al (2006)
Design Hydrology & Sedimentology for Small Catchments; Haan,
Barfield, Hayes: Haan et al (1994)
USLE/RUSLE: USDA Agricultural Handbook No. 537 (1978)
Cuenca, R. H. 1989. Irrigation System Design - An Engineering
Approach. Prentice-Hall, Inc., Englewood Cliffs, NJ. 552 pp.
Cuenca (1989).
Ward, Elliot 1995 (Environmental Hydrology, Lewis Publishers).
http://cobweb.ecn.purdue.edu/~abe325/: Mohtar soil and water
resources conservation course.
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Soil Physics & Mechanics
1. Soil classes and particle size
distributions
2. Basics of soil water
a) Water Content
b) Water Potential
c) Water Flow
3. Soil strength & mechanics
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Soil classes & particle sizes
Hillel (1998) page 61
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Soil Classes & Particle Sizes - 2
ISSS classification is easiest
1. Sand 0.02-2.0mm (20-2000μ)
2. Silt 0.002-0.02mm (2-20μ)
3. Clay <0.002mm (<2μ)
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Soil classes & particle sizes – 3
Soil Textural Triangle
Example 1:
Find the soil texture for
this soil:
 50% sand,
 20% silt
Hillel (1998) page 64
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Soil Classes & particle sizes – 4
Particle size
distribution
Example 2
Draw in a
sandy clay
loam?
Hillel (1998) page 65
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Soil Structure and Functionality
Clay
particles
Primary Inter-ped
peds
pore space
Primary
mapping
soil
Primary soil
unit
Soil
mapping type
unit
Geomorphological
unit
Pedostructure, primary peds,
primary particles, are functionally
defined and quantitatively
determined using the shrinkage
and potential curve measurement
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Horizon
Pedostructure
=
=
Vertical porosity
(cracks, fissures)
+
Pedostructure
Mineral
grains
Interpedal porosity
(macro-porosity)
+
Primary peds
and free
mineral grains
Mohtar (2008)
Clay pore space
Primary ped
=
Clay plasma porosity
(micro-porosity)
+
Primay particles
and pedological features
9
Soil Water Content
0
1. Mt = Ms + Mw + Ma
2. Vt = Vs + Vw + Va
a. t = total, s = solids, w = water, a = air
3. ρb = bulk density = Ms/Vt≈ 1.1-1.4 g/cc (why
dry basis?)
4. ρp = particle density = Ms/Vs ≈ 2.65 g/cc
5. Porosity = (Vw + Va) / Vt ≈ 25-60%
6. ρw = water density = Mw/Vw = 1.0 g/cc
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Soil Water Content – 2
 Water content wet basis:
Ww = Mw / (Ms + Mw)
 Water content dry basis:
W = mass wetness = Mw / Ms
 Volumetric water content:
θ = Vw/Vt = Vw / (Vs + Vw + Va)
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Calc.: soil water content
Soil Water
Example 3.
Given:
 Soil with 30% water content dry basis
Find?
 Best guess at equivalent inches of water
in the top foot of soil?
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Calc.: soil water content – 2
1. Mw / Ms = 0.30
a. Mw = Vw * ρw
b. ρb = Ms / Vt; Ms = Vt * ρb
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Calc.: soil water content – 3
 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”
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Soil Water Potential
soil characteristic curve
Hillel (1998) page 157
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Soil Water Potential – 2
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Cuenca (1989) page 58
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Soil Water Management
Ward, Elliot 1995 (Environmental Hydrology, Lewis Publishers)
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Soil Water Potential – 3
Fangmeier et al
(2006) page 337
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Soil water potential – 4
Hillel (1998) page 162
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Calc.: soil water potential
Soil Water Potential
Example 4.
 Given:
Mercury tensiometer




SG = 13.6
Situation as shown
Find:
1. Total potential at point C
2. Is point C above or below
the current water table?
Cuenca, (1989) page 64
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Calc.: soil water potential - 2
1. Pick datum
2. Add pressures
a. Suction
b. Water depth
c. Gravity
3. T = z + p + pos
a. z = + 80 cm
b. p = ?
c. T = -86cm
d. Point C is above water table. Why?
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Soil Water Flow
 q = A*K*H/L
 K = (q*L)/(A*H)
 K values
A
H
L
Fangmeier et al (2006) page 261; Schwab et al
(1993) page 359; Haan et al (1994) page 430
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Calc.: soil water flow
Darcy Law Application
Example 5.
 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?
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Calc.: soil water flow – 2
q = A*K*H/L; A = (q*L)/(K*H)
 50000 gal  1 ft  day  1  1 ft 3 
  278 ft 2
 


A  
 day  1  8 ft  3 ft  7.481gal 
d
4A

 18 ft
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Soil erosion and sediment yield
Hillslope erosion
Channel system erosion
Sediment delivery to streams
Sediment transport in streams
Slope stability
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Hillslope soil erosion
 Background
 Detachment
Raindrop impact
 By turbulent overland flow
 Runoff

 Transport downslope

By runoff
Schwab et al (1993) pp:91-111; Fangmeier et al (2006) pp:134-156; Haan et al
(1994) pp:238-285
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Hillslope Soil Erosion Background
At the top of the slope
 Detachment by raindrop impact
 Transport by shallow sheet flow
 Sheet erosion
USDA-NRCS
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Hillslope soil erosion background - 2
 Lower on slope
 Small flow
concentrations
 Start to cut small
channels
 Rills



Roughly parallel
Head straight downslope
Random formation
USDA-NRCS
 Flow from sheet areas
between rills
 Sheet and rill erosion
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Hillslope Soil Erosion Background - 3
 Bottom of hillslope
 Ends at
concentrated flow
channel
 Low area in
macrotopography
 “ephemeral
gullies”
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USDA-NRCS
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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
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USLE/RUSLE
 A = R * K * LS * C * P
 A = average annual soil erosion (T/A/Y)
 R = rainfall erosivity (long empirical units)
 K = soil erodibility (long empirical units)
 R * K gives units of T/A/Y
 LS = topographic factor (dimensionless, 0-1)
 C = cover-management (dimensionless, 0-1)
 P = conservation practice (dimensionless, 0-1)
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USLE/RUSLE – background
 Empirical approach been in use since 1960
 >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
 Sub-factors in later versions
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USLE/RUSLE – approach
 Lookup
 Maps, tables, figures
 Databases
 Process-based calculations
 Show changes over time
 Where don’t have good data
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R factor – rainfall erosivity
 Maps
Haan et al (1994) pp:251; Haan et al (1994) Appendix 8A; Schwab et al
(1993) 99(SI); Fangmeier et al (2006) pp:143(SI); USDA (1978) pp:1-5
 R(customary SI) = 17.02 * R(customary US)
S4
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K factor – soil erodibility
 Soil surveys, NASIS, Haan et al (1994) 261-262; USDA 6
 Erodibility nomograph: Haan et al (1994) 255; Schwab et al
(1993) 101; Fangmeier et al (2006) pp144; USDA (1978) pp: 7
 No short-term OM
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LS – topography factor
 New tables & figures
 Haan et al (1994) 264; USDA (1978) 8
 Know susceptibility to rilling
 High for highly disturbed soils
 Low for consolidated soils
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C – cover-management factor
 Part of normal management scheme
 Lookup: Schwab (1993) 102; Fangmeier et al (2006) pp: 146;
Haan et al (1994) 266; Hillel (1998) Appendix 8; USDA
(1978) 9
 It Changes over time
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C – cover-management factor - 2
 Subfactor approach (RUSLE)
 C = PLU * CC * SC * SR * SM; all 0-1

PLU = prior land use



CC = canopy cover; % cover & fall height
SC = exp(-b * % cover)



roots, buried biomass, soil consolidation
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
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P – conservation practice factor
 Common practices
 Contouring, strip cropping, terraces
 Change flow patterns or cause
deposition
 Lookup tables
 Schwab (1993) pp:103; Fangmeier et al (2006) pp:146;
Haan et al (1994) pp: 281; USDA (1978) pp:10
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Calc.: USLE/RUSLE
Example 9:
 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
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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
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Calc: USLE/RUSLE – 2.1
Example 10:
 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?
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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%
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Sediment delivery
 USLE/RUSLE for hillslopes
 Erosion
 Delivery
 Erosion critical for soil resource
conservation
 Delivery critical for water quality
 Movement through channel system
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Sediment delivery – 2
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Sediment delivery – 3
 SDR (Sediment Delivery Ratio)
 Hillslope erosion
 Empirical fit for watershed delivery
 Channel erosion/deposition modeling
 Erosion
 Transport
 Deposition
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Sediment Delivery Ratio
 Haan et al (1994) pp:293-299
 SDR = SY / HE
 SDR = sediment delivery ratio
 SY = sediment yield at watershed exit
 HE = hillslope erosion over watershed
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Sediment Delivery Ratio – 2
 Area-delivery relationship
Haan et al (1994) pp:294
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Sediment Delivery Ratio – 3
 Relief-length ratio
 Relief = elev change along main branch
 Length = length along main branch
Haan et al (1994) page .294
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Sediment Delivery Ratio – 4
 Forest Service Delivery Index Method
Haan et al (1994) pp:295
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Sediment Delivery Ratio – 5
 MUSLE (Haan et al (1994) pp: 298 and 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
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Calc: SDR
Example 11:
 Given:
 Flow path length in watershed = 4000ft
 Elevation difference = 115ft
 Find?
 SDR
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Calc.: SDR – 2
 R/L = 115/4000 = 0.029
 From figure SDR = 0.45
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Channel Erosion-Deposition Modeling
 Process-based small channel models
 Foster-Lane model

Haan et al (1994) pp285-289

Complicated and process-based
 Ephemeral Gully Erosion Model
EGEM
 Fit to Foster-Lane Model results

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Channel Erosion-Deposition Modeling – 2
 Large-channel models
 Sediment transport
 Channel morphology
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Sediment Transport
 Settling (Haan et al (1994) pp: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
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

1 d g
SG  1
Vs  
18  

Vs  2.81 d
2
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Calc.: Stokes’ Law Settling
Example 12:
 Given:
 ISSS soil particle size classification
 Find:
 Settling velocities of largest sand, silt, and clay
particles
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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
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Calc.: Stokes’ Law Settling
Example 13:
 Given:
 Stokes’ Law settling
 Find: particles larger than what size can be
assumed to settle 1 ft in one hour?
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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
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BREAK
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Soil Strength and Mechanics
From McCarthy (1982) pages 233-237,373-379
 Soil stresses
 Normal Stress = Fn/A = σ
 Shear Stress = Ft/A = τ


Fn = normal force
Ft = tangential or shear force
 As normal stress (σ) ↑, sheer stress (τ) to
cause failure (τf)↑ i.e. shear strength ↑
 tan Φ = τf / σ; where Φ = angle of internal
friction
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Soil strength and mechanics – 2
McCarthy (1982) page 234
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Calc.: soil strength
Example 6:
 Given:
 Well-graded sand; density 124 lb/ft3
 Find:
 Ultimate shear strength 6 ft below
surface?
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Calc.: soil strength – 2
 From table 10-1, 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
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Footing bearing loads
qult = a1*c*Nc + a2*B*γ1*Nγ + γ2*Df*Nq
total support=soil cohesiveness+ below footing +soil bearing
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
McCarthy (1982) page 374-379
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Footing bearing loads – 2
McCarthy (1982) page 375
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Calc.: footing load
Example 7:
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
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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
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Soil compaction and density
 Soil compaction
 Greater strength and reduced permeability
 Dependent on water content dry soil cannot be
compacted well
 Proctor test
 Pack soil into mold with pounding at various
moistures. Find soil moisture for maximum
compaction and density.
 Modified Proctor > 56000 ft-lbs of energy
exerted.
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Slope stability & failure
Possible forms of Failure
McCarthy (1982) page 437-455
McCarthy (1982) page 440
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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)
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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
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Slope stability & failure – 4
For clay soil
For soil with cohesion and internal friction > 0
McCarthy (1982) page 474
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Slope stability & failure – 5
Ns = c / (γ * Hmax)
c = cohesion force
γ = soil unit weight
Hmax = max depth without sliding
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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
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Calc.: slope stability – 2
 Fig. b, φ = 15o, i = 50o
 Hmax = c / (γ * Ns) =
(500 lb/ft2)(1ft3/ 10)(1/ 0.095) =
48 ft
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Materials Covered
 Principles of Soil Physics
 Sediment Transport
 Erosion Control
 Soil Mechanics
 Slope Stabilization
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Thank You and Best Luck
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