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2011 PE Review: Michael C. Hirschi, PhD, PE, D.WRE Professor and Assistant Dean University of Illinois

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Acknowledgements: Rod Huffman, PE Review coordinator Daniel Yoder (2006 presenter) Rabi Mohtar & Majdi Abu Najm (2010 presenters)

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

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Sources Environmental Soil Physics; Hillel; 1998 Hi Essentials of Soil Mechanics & Foundations, 2 nd ed.; McCarthy; 1982; M Soil and Water Conservation Engineering –4 th ed. Schwab, Fangmeier, Elliott, Frevert: S4 –5 th ed. Fangmeier, Elliott, Workman, Huffman, Schwab: S5 Design Hydrology & Sedimentology for Small Catchments; Haan, Barfield, Hayes: H USLE/RUSLE Handout: HO

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Soil Physics & Mechanics Soil classes and particle size distributions Soil water –Content –Potential –Flow Soil strength & mechanics

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Soil classes & particle sizes Hi.61

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Soil classes & particle sizes - 2 ISSS easiest –Sand mm ( μ) – Silt mm (2-20μ) – Clay <0.002mm (<2μ) Soil texture classes based on USDA –Sand mm –Silt mm –Clay <0.002mm

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Soil classes & particle sizes – 3 Texture –EX: 50% sand, 20% silt 30% clay? Hi.64

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Soil classes & particle sizes – 4 Particle size distribution –EX: draw in a sandy clay loam? Hi.65

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Soil water content M t = M s + M w + M a V t = V s + V w + V a –t = total, s = solids, w = water, a = air ρ b = bulk density = M s /V t ≈ g/cc ρ p = particle density = M s /V s ≈ 2.65 g/cc Porosity = (V w + V a ) / V t ≈ 25-60% ρ w = water density = M w /V w = 1.0 g/cc 0

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Soil water content – 2 w w = M w / (M s + M w ) –Water content wet basis w = mass wetness = M w / M s –Water content dry basis θ = V w /V t = V w / (V s + V w + V a ) –Volumetric water content

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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?

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Calc.: soil water content – 2 M w / M s = 0.30 – M w = V w * ρ w – ρ b = M s / V t ; M s = V t * ρ b

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Calc.: soil water content – 3 M w / M s = 0.30 – M w = V w * ρ w – ρ b = M s / V t ; M s = V t * ρ b – M w / M s = (V w * ρ w )/(V t * ρ b ) = (V w / V t )(ρ w / ρ b ) – θ = V w /V t – θ *(ρ w / ρ b ) = 0.3; θ = (ρ b / ρ w ) * 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 Hi.157

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Soil water potential – 2 Cuenca.58

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Soil water potential – 3 S5.337

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Soil water potential – 4 Hi.162

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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

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Calc.: soil water potential - 2 1) pick datum 2) add pressures – Suction – Water depth – Gravity T = z + p + p os – = -86cm – Above water table

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Soil water flow q = A*K*H/L – K = (q*L)/(A*H) K values S4.359; S5.261; H.430 A q L H

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Calc.: soil water flow Given: – Need 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)

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Soil strength and mechanics M , Soil stresses – Normal = F n /A = σ – Shear = F t /A = τ F n = normal force F t = tangential or shear force – As σ ↑ τ to cause failure = τ f ↑ –tan Φ = τ f / σ; Φ = angle of internal friction

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Soil strength and mechanics – 2 M.234

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Calc.: soil strength Given: – Well-graded sand; density 124 lb/ft 3 Find: – Ultimate shear strength 6 ft below surface?

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Calc.: soil strength – 2 From table, for well-graded sand, Φ = o = 33.5 o Normal stress = (124 lb/ft 3 )(6 ft) = 744 lb/ft 2 tan Φ = τ f / σ; τ f = σ * tan Φ = τ f = 744 lb/ft 2 * tan(33.5 o ) = 492 lb/ft 2

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Footing bearing loads M q ult = a 1 *c*N c + a 2 *B*γ 1 *N γ + γ 2 *D f *N q – c = soil cohesion beneath footer –γ 1,, γ 2 = effective soil unit weight above and below footer – B = footer size term –N c, N γ, N q = capacity factors – D f = footing depth below surface q design = q ult / FS Length/width Ba1a1 a2a2 1 (square)Width Width Width Width Width StripWidth CircularRadius

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Footing bearing loads – 2 M.375

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Calc.: footing load Given: – Strip footing 3 ft wide – Wet soil with density of 125 lb/ft 3 – Angle of internal friction = 30 o – Cohesive strength of 400 lb/ft 2 – Use factor of safety of 3 Find: – q design in lb/ft 2

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Calc.: footing load – 2 a1 = 1.0, a2 = 0.5, B = width = 3’ γ 1 = 125/2 = 62.5 lb/ft3; γ 2 = 125 lb/ft3 c = 400 lb/ft2 N c = 30, N γ = 18, N q = 20 q design = 23,700/3 = 7900 lb/ft 2

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Soil compaction and density Soil compaction – Greater strength – Dependent on water content Proctor test – Pack soil into mold with pounding – Modified Proctor > ft-lbs

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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

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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)

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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

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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

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

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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…

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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

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Multi-stage erosion

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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 S ; S ; H Background – Detachment Raindrop impact By turbulent overland flow – Runoff – Transport downslope By runoff

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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 – Flow from sheet areas between rills – Sheet and rill erosion USDA- NRCS

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Hillslope soil erosion background - 3 Bottom of hillslope – Ends at concentrated flow channel – Low area in macrotopography – “ephemeral gullies” 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 (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)

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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

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

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K factor – soil erodibility Soil surveys, NASIS, H ; HO.6 Erodibility nomograph – H.255; S4.101; S5.144; HO.7 – No short-term OM

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LS – topography factor New tables & figures – H.264; HO.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: S4.102; S5.146; H.266; H.Appendix 8; HO.9 Change 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 – roots, buried biomass, soil consolidation CC = canopy cover; % cover & fall height SC = exp(-b * % cover) – B = 0.05 if rills dominant; typical; 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, stripcropping, terraces Change flow patterns or cause deposition Lookup tables – S4.103; S5.146; H.281; HO.10

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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

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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 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 = SC 1 = exp( * 40%) = SC 2 = exp( * 80%) = 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 H 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 H.294

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Sediment Delivery Ratio – 3 Relief-length ratio – Relief = elev change along main branch – Length = length along main branch H.294

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Sediment Delivery Ratio – 4 Forest Service Delivery Index Method H.295

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Sediment Delivery Ratio – 5 MUSLE (H.298) – Y = 95(Q * q p ) 0.56 (K a )(LS a )(C a )(P a ) 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 * q p ) 0.56 /(R * area) R = storm erosivity in US units – Routing for channel delivery

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Calc: SDR Given: – Flow path length in watershed = 4000ft – Elevation difference = 115ft Find: – SDR

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Calc.: SDR – 2 R/L = 115/4000 = From figure SDR = 0.45

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Channel erosion-deposition modeling Process-based small channel models – Foster-Lane model H 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 (H ) – Stokes’ Law V s = 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 68 o F d in mm, V s in fps

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Calc.: Stokes’ Law settling 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 – V s,clay = 1.12*10 -4 fps = 0.04 ft/hr = 0.97 ft/day – V s,silt = 0.11 fps = 405 ft/hr = 1.83 mi/day – V s,sand = fps = 7.66 mph = 184 mi/day

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Calc.: Stokes’ Law settling 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 V s = [(1 ft)/(1 hr)](1 hr/3600s) = 2.778*10 -4 fps d = (V s /2.81) 1/2 = mm

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Other settling Flocculant, hindered, compression – Order of increasing concentration – Slower settling with increased conc. Settleable solids – Imhoff cone H.211

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Sediment transport relationships H Bedload – Einstein equation Suspended load Total load – Yalin total load equation V avg, d, SG, r, т, т c – Yang stream power relationship Stream power instead of critical shear

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Channel morphology H Channel cross-section – Width B, depth D, avg. vel. U, suspended sediment load q s, flow rate q – B = C a * q a, D = C b * q b, U = C c * q c, q s = C d * q d – C a, C b, C c, C d, a, b, c, d = fit coefficients & exponents

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Channel morphology – 2 Channel-forming discharge – Greatest impact on channel itself – Return period ≈ 1.5 years Channel roughness

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Channel morphology – 3 Channel regime – H H.416

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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

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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”)

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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

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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

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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

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Control of Sediment in Runoff Reduce transport capacity of flow Enhance deposition of sediment

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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

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Enhance deposition of sediment Use flocculant to increase sedimentation – Usually in sedimentation ponds when other methods are not adequate – Expensive…

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90 Slope stability & failure Possible forms of Failure McCarthy (1982) page 440 McCarthy (1982) page I.B Mohtar

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91 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

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92 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

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93 Slope stability & failure – 4 For clay soil For soil with cohesion and internal friction > 0 McCarthy (1982) page 474 I.B Mohtar

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94 Slope stability & failure – 5 N s = c / (γ * H max ) – c = cohesion force – γ = soil unit weight – H max = max depth without sliding I.B Mohtar

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95 Calc.: slope stability Example 8: Given: – Cohesion strength = 500 lb/ft 2 – Unit weight = 110 lb/ft 3 – Slope steepness = 50 o – Internal friction angle = 15 o Find: – Max. slope height I.B Mohtar

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96 Calc.: slope stability – 2 Fig. b, φ = 15 o, i = 50 o H max = c / (γ * N s ) = (500 lb/ft 2 )(1ft 3 / 10)(1/ 0.095) = 48 ft I.B Mohtar

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