Title | Lecture notes, lecture all |
---|---|
Course | Geotechnical Engineering I |
Institution | Memorial University of Newfoundland |
Pages | 200 |
File Size | 14.8 MB |
File Type | |
Total Downloads | 21 |
Total Views | 186 |
Origin of soils.pdf
Compressibility of soil.pdf
In situ stresses.pdf
Mechanical analysis.pdf
Permeability.pdf
Plasticity and structure.pdf
Seepage.pdf
Soil classification.pdf
Soil compaction.pdf
Stresses in a soil mass.pdf
ENGI 4723 – Geotechnical Engineering I
Instructor: Dr. Bipul Hawlader
Origin of Soil and Grain Size [Chapter 2] Soil formation: Soil is formed by the process of Weathering of rock. Weathering: disintegration and decomposition of rocks and minerals at or near the earth surface.
Two types of weathering: a) 1. 2. 3. 4.
Mechanical Weathering: Temperature change Freezing and thawing Splitting action of plant roots Abrasive movement (mass movement of by means of wind, water or ice may cause an erosion and disintegration of rocks)
ENGI 4723 – Geotechnical Engineering I
Instructor: Dr. Bipul Hawlader
b) Chemical weathering 1. Oxidation (oxygen ions combine with some minerals in the rock which finally decomposes as rusting of steel) 2. Carbonation (carbon dioxide and water from carbonic acid, which decompose the mineral containing iron, sodium or calcium) 3. Hydration (water convert some mineral into new mineral) 4. Vegetation (decaying vegetation produces organic acid, carbon dioxide or oxygen, which mixed with water penetrate through the rock and change the chemical contents. For example, silcate change to silica in this manner.
Two types of soils: a) Residual soils: Weathered soil remains in place. Engineering behaviour is different from transported soil. One important characteristics is that the size gradation. Fine grained soil near the surface and grain size increases with depth b) Transported soils: Glacial soils: formed by transportation and deposition of glaciers Alluvial soils: Transported by running water and deposited along stream Lacustrine soils: formed by deposition in quiet lake Marine soil: formed by deposition in the seas Aeoline soils: transported and deposited by wind Colluvial soils: formed by the movement of the soil from its original place by gravity, such as during landslide.
ENGI 4723 – Geotechnical Engineering I
Instructor: Dr. Bipul Hawlader
Typical residual soil profiles
ENGI 4723 – Geotechnical Engineering I
Instructor: Dr. Bipul Hawlader
Soil Particle Size [§2.4] Gravel: pieces of rocks with quartz and feldspar and other minerals Sand: Mainly quartz and feldspar Silts: (microscopic) fine quartz grain & flake shaped particles of micaceous minerals Clay: Kaolinite, Illite, Montmorillonite [§.5].
Notes: 1. Sometimes particles 0 Pw < P
= + u
One‐DimensionalLaboratoryConsolidationTest[11.5] FirstsuggestedbyTerzaghi Performedinaconsolidometer(sometimesknownasOedometer) Soilspecimenplacedinametalring,withtwoporousstones(onetopandone bottom) Specimensize:64mm(2.5inch)dia,25mm(1inch)thick Loadsareappliedusingaleverarm Compressionismeasuredusingadialguage Specimenskeptiswater Eachloadiskept24hours,afterthatloaddoubled Attheenddryunitweightofthespecimeniscalculated
Stage I: Initial compression, mostly by preloading
Stage II: Primary consolidation, excess pore water pressure gradually transferred to effective stress because of expulsion of water
Stage III: Secondary consolidation, after completion of excess pore water dissipation, deformation takes place because of plastic rearrangement of soil fabric
VoidRatio‐PressurePlots[11.6]
Calculationofvoidratio:
1. Calculatetheheightofthesolid,Hs,inthesoilspecimenFig.11.9 Hs
Ws AG s w
Where,Ws=dryweightofthespecimen A=areaofthespecimen G=Specificgravityofsoilsolids w=Unitweightofwater 2. CalculatetheinitialheightofthevoidsHv Hv H Hs H=initialheightofthespecimen 3. Calculated initial void ratio of the specimen e0
Vv H v A H v Vs H s A H s
4. If the deformation is H1 for first load increment1,thechangein thevoidratio e1iscalculatedas e1
H 1 Hs
5. Calculatethenewvoidratioe1andstrain1as
e1 e0 e1
1
e1 (1 e0 )
Note:forthenextloadfirstloadincrement2(cumulativestress)causeadditional settlementH2
Example: Heightofthesample=25mm Diameterofthesample=63.5mm Dryweightofthesample=0.586N=0.586/1000kN SpecificgravityGs=2.65 0.586 / 1000 Ws 0.00712 m 0.712 cm AGs w 63.5 2 2.65 9.81 4 1000
Heightofsolid H s
Initialvoidratio e0
H v H H s 2.5 0.712 2.5 Hs Hs 0.712
Appliedstress (kPa)
Heightof specimen(cm)
e=H/Hs
Currentvoid ratio
10 20
2.5 2.478
2.500 2.470
40 80 160 320 640 1280
2.457 2.357 2.142 1.927 1.712 1.497
(2.5‐2.478)/0.712=0.030 (2.478‐2.457)/0.712 =0.030 0.140 0.302 0.302 0.302 0.302
2.440 2.299 1.997 1.695 1.393 1.091
2.60
Void ratio (e)
2.20
1.80
1.40
1.00 1
10
100
Applied stress (kPa)
1000
10000
NormallyConsolidatedandOverconsolidatedClays[11.7] Normallyconsolidated:Presenteffective overburdenpressureisthemaximum effectiveoverburdenpressurethesoilwas subjected(=’c) Overconsolidated:Presenteffective overburdenpressureislessthanthe effectiveoverburdenpressuresoil experiencedinthepast(’v
Page7of21
Factorsaffectingthepermeabilityofsoil 1)Soiltype
2)Thepropertiesofporefluid(viscosity)
k
w K
w =unitweightofwater
(kN/m3) =viscosityofwater(kg/m.s) 2 K =Absolutepermeability(m ) k=permeability(m/s)
kT 0C k 20 0 C
20 T
0C
0
C
3)Thevoidratioofthesoil(willbediscussedlater) 4)Theshapeandarrangementofpore‐verydifficulttodescribemathematically 5)Degreeofsaturation‐increaseindegreeofsaturationincreasesthe permeability Page8of21
LaboratoryDeterminationofHydraulicConductivity(§7.5) a)Constantheadpermeability ‐isusedforcoarse‐grainedsoils
Q Avt A( ki )t i h/L h Q A k t L
k
QL Aht
Q=volumeofwatercollected A= cross sectional area of the specimen t=durationofwatercollection L=lengthofspecimen h=hydraulichead Example‐2:Duringaconstantheadpermeabilitytestonasoilsample260mlof waterwascollectedin2minutes.Ifthelengthofthesampleis10cm,diameter4 cm,andthemaintainedhead20cm,whatisthecoefficientofpermeability?
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b)Fallingheadpermeability ‐isusedforfine‐grainedsoils dh h A a dt L
qk
dt t
aL dh Ak h
dt
0
t
aL Ak
h2
dh h h1
h aL h1 aL ln log10 1 2.303 Ak h2 Ak h2 x x ln 2.303 log10
k 2.303
h aL log 10 1 At h2
q=flowrate A=crosssectionalarea ofthespecimen a=crosssectionalareaofstandpipe L=lengthofspecimen Example‐3: In a falling head permeability test, the hydraulic head dropped from 90 cm to 40 cm in 20 minutes. The cross‐sectional area of the standpipe was 1 cm2.Thesamplewasof4cmdiameterandhadalengthof18cm.Determinethe coefficientofpermeability.
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DirectionalVariationofPermeability(§7.8
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EquivalentHydraulicConductivityinStratifiedSoil(§7.9) Anexcellentexampleofnaturallydepositedlayeredsoilisvarvedsoil.
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Horizontalflow Totalflowthroughunitwidthinunittimeis . 1. . 1. . 1. . 1. ⋯ . . . 1. v=dischargevelocity v1,v2,v3,..vn=dischargevelocityineachlayer ⋯ . ⋯ ⋯ /
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Verticalflow
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Example‐4: A sandy soil with k=3x10‐2 cm/s contains a series of 5 mm thick horizontalsiltlayersspaced300mmoncenter.Thesiltlayershavek=3x10‐6cm/s. Compute equivalent coefficient of permeability in the horizontal and vertical directions. Whendrillingaboreholethroughthissoil layer,howeasywouldit betomiss the siltlayers?Ifyoumissit,howmuch effectwouldtheignoranceofsiltlayerhave oncomputationofwaterflowintheverticalandhorizontaldirections.
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PermeabilityTestintheFieldbyPumpingfromWells[§7.10]
Hydrolo giccycle
Definitions:
Soilprofileshowingcomp lexnatureofgroundwater
Phreaticzone:Portionbelowgroundwatertable. Aquifer: Some soils, such as sands and gravel, can transmit large quantity of groundwater. Theseareknownasaquifer. Aquicludes:Othersoilssuchasclaytransmitwaterveryslowly,whichiskownasaquicludes. Aquitards:Intermediatesoils,suchassiltysand,canpasswaterslow‐to‐moderate rateandare calledaquitards. Unconfinedaquifer:Upperaquifer.Bottomflowboundaryis definedbyanaquicludewhilethe upperflowboundary(groundwatertable)isfreetoreachitsnaturalstate. Confined aquifer: Lower aquifer(s). Both upper and lower flow boundaries are defined by aquiclude.Mostconfinedaquifersareartesian,which meanswateratthetopofthe aquifer is underpressure. Page16of21
UnconfinedAquifers Lowerflowboundaryisfixedbuttheupperflowboundaryisthegroundwater table,whichisfreetoseekitownlevel. dh q k 2 rh dr r1
dr
2k
h2
q hdh r2 r h2 r 2.303q log10 1 r2 k 2 2 h1 h2
Example‐5:Apumpingwelltestwascarriedoutinasoilbedof15mthickandthe followingmeasurementwererecorded.The rateofpumpingwas10.6x10‐3m3/s; drawdownsintheobservationwellslocatedat15mand30 m fromthecenterof thepumpingwellwere1.6mand1.4m,respectivelyfromtheinitialgroundwater level. The initial groundwater level was located at 1.9 m below the ground surface.Calculatethehydraulicconductivity. Page17of21
ConfinedAquifers Bothlowerandupperflowboundariesarefixed. Watercanenteronlyfromaquiferthickness,H. dh q k 2rH dr
dr 2kH h2 q dh r2 r h2 r1
r 2.303q log10 1 r2 k 2H h1 h2
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FieldInstrumentations
Openstandpipepiezometer
ObservationWell Page19of21
Essential Points - Flow of water through soils is governed by Darcy’s law, which states that the velocity is proportional to the hydraulic gradient (v=ki). - The proportionality constant is the hydraulic conductivity. - The hydraulic conductivity depends on soil type, particle size, pore fluid properties, void ratio, pore size, homogeneity, layering and fissuring, and entrapped gases. - In coarse-grained soils, the hydraulic conductivity is determined using a constant-head test while for fine-grained soils a falling-head test is used. - In the field, a pumping test is used to determine the hydraulic conductivity. - Wellpoints are used at a construction site to lower the groundwater level.
Page20of21
Further study from textbook Das, B.M. & Sobhan, K. (2014) Principles of geotechnical engineering 8th ed 1. Sections:7.1to7.5;7.8to7.10 2. Exampleproblems:7.1to7.5,7.12,7.13, 3. Practiceproblemsfromtextbook:7.1to7.6,7.17
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ENGI 4723 – Geotechnical Engineering I
Instructor: Dr. Bipul Hawlader
PLASTICITY AND STRUCTURE OF SOIL [CHAPTER 4]
The presence of water in fine‐grained soils can significantly affects associated engineeringbehavior,soweneedareferenceindextoclarifytheeffects.
SwedishscientistAtterbeg(1846‐1916)developedthismethod When moisture content is very high, soil behaves like fluid, and when very lowitbehaveslikesolid Soilcanhavefourbasicstates:solid,semi‐solid,plastic,andliquid Definitions: LiquidLimit(LL,wl):Theminimumwatercontentatwhichthe soil willflowunder itsownweight PlasticLimit(LL,wp):Theminimumwatercontentatwhichthesoilcanbe rolled intoathread3mm(1/8")diameterwithoutbreakingup Shrinkage Limit (SL, ws): The maximum water content at which further loss of moisturecontentdoesnotcauseadecreaseinthevolumeofthesoil
LL,PLandSLareknownasAtterbergLimits.
Page1of21
ENGI 4723 – Geotechnical Engineering I
Moisture(water) content (Increasing)
Instructor: Dr. Bipul Hawlader
Liquidstate:deformseasily; Consistencyofpeasouptosoftbutter
LiquidLimit(LLorwl)
Engineering
PlasticityIndex (PI)
Plasticstate:deformswithoutcracking; Consistencyofsoftbuttertostiffputty
PlasticLimit(Plorwp)
Semi‐solidstate:deformspermanently; Consistencyofcheese ShrinkageLimit(SLorw s) Solidstate:breaksbeforedeformation Consistencyofhardcandy
Page2of21
ENGI 4723 – Geotechnical Engineering I
Instructor: Dr. Bipul Hawlader
LiquidLimit[Section4.2] 1) Casagrandecup
Twotypesofbase: a)USbase ‐Micarta b)UKbase ‐Rubber LLUS 0.94LLUK
(Norman1958)
Page3of21
ENGI 4723 – Geotechnical Engineering I
Instructor: Dr. Bipul Hawlader
Watercontentatwhich25blows close the grove 0.5 inch (12.7 mm)calledLiquidLimit. Obtainwatercontentfor0.5inch closurebetween15‐35blowsand plottheminasemi‐logplot. Thisisknownasflow curve (line).Slopeof theflowlineisflowindex(positive value). Page4of21
ENGI 4723 – Geotechnical Engineering I
Instructor: Dr. Bipul Hawlader
One‐pointmethod: U.S.CropsofEngineers(1949) N LL w N 25
0.121
N=Numberofblow(20‐30) wN=moisturecontentatNblow CommentsonOne‐pointmethod: UsedwhenonlyoneLLtestisrun Fairlygoodassmall changeinmoisturecontentinvolveswithinN=20toN=30, andverysmallchangein
N 25
0.121
forN=20toN=30. 0.121
ForN=20
N 25
ForN=30
N 25
0.973 0.121
1.022
CanbeusedtochecktheLLtestresult Shearstrength: Cassagrande(1932):eachblowcorrespondstosu=0.1kPa
Therefore,for25blowsatLL,shearstrengthsu(LL)=2.5kPa S u( PL) 100 Su( lL) =250kPa
Page5of21
ENGI 4723 – Geotechnical Engineering I
Instructor: Dr. Bipul Hawlader
Example‐1: The data shown in the table were Numberofblowsfor½“ Water obtainedfromaliquidlimittest. closureofthegroove content(%) 10 55 Determine: Liquid limit (ii) flow index 45 (iii)verifytheresultwithonepointtest 16 43 with N=20, (iv) what is the expected 20 22 42 shearstrengthatLL. 27 41 40 36
Flowline
LL=41.5%
25
Page6of21
ENGI 4723 – Geotechnical Engineering I
Instructor: Dr. Bipul Hawlader
2) Fallconemethod(BS‐1377): ‐PopularinEuropeandAsia ‐Aconeofapexangle30andweight0.78Npenetrate20mmin5sec ‐ Perform at least 4 tests at different water content and determine penetration depth ‐Plottheresults(semi‐logarithmic) ‐Moisturecontentatd=20mmistheLiquidLimit ‐Flowindex(IFC): I FC
w 2 (%) w1(%) log d 2 log d1
Page7of21
ENGI 4723 – Geotechnical Engineering I
Instructor: Dr. Bipul Hawlader
PlasticLimit[Section4.3] ‐ Plastic Limit is the moisture content in percentage at which soil crumbles when rolled into a thread of3.2mm(1/8inch) indiameter ‐ Performed by rolling by hand on a glassplate FallConeMethod ‐Sameas LL testusing fall cone,butthemassof the coneis2.35N(not0.78N asusedinLLtest). Page8of21
ENGI 4723 – Geotechnical Engineering I
Instructor: Dr. Bipul Hawlader
PlasticityIndex(PI): plastic. PI=LL‐PL
Describestherangeofwatercontentoverwhichasoilis
Typical Atterberg Limits for Soils (Budhu 2007)
The Atterberg limits depend on the type of predominant mineral in the soil. If montmorillonite is the predominant mineral, the liquid limit can exceed 100%. Why? Because the bond between the layers in montmorillonite is weak and large amounts of water can easily infiltrate the spaces between the layers. In the case of kaolinite, the layers are held relatively tightly and water cannot easily infiltrate between the layers.
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ENGI 4723 – Geotechnical Engineering I
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