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PROBLEMS IN SOIL MECHANICS AI{D FOT]NDATION ENGINEERING A'M'LE'(ndia); [ForB.E.(Civil);M'E'(Civil); Examinationsl U.P.S.C'andotherC-ompetitive DEBASHIS MOITRA Departmentof civil Engineerilq BengalEngineeringCollege DeemedUniversity Howrah' \|\\ ,j r u" \ { i i , DHANpA_rLtcATto hls(p)' LrD.84! F_uB_ FIRSTFLOOR,.6ZI+ IT,INONAS HOUSd DARYAGA+TJ, NEWDELHI-J1OOO2 PHONES:3274073 Note: This book or part thereof may not be reproduced in any form or translated without the written permission of the Author and the Publisher OTHERUSEFT'LBOOKS AdvanceTheoryofStuctures N.C Sinha Concrete Testing Manual MI Gambhir Fundamentalsof Limit Analysis of Structures V.K Manicka Selvam Modern method of Structural Analvsis V.K Manicka Selvam Multistorey Building & Yield Line Analvsis of Slabs V.K Manicka Selvam Energy Methods in Structural Mechanics V.K Maniclca Selvam Analysis of Skucture in Earth Quake Region V.K Manbka Selvam Dock and Harbour Engineering S.P Bindra Foundation Design Manual N.V.Nayak Preface This book is primarily intended for the undergraduatestudentsof Civil Engineering However, it will be helpful also to the diploma-level students, A.M.I.E students,and, in some cases,even to the post-graduatestudentsof Soil Mechanics and Foundation Engineering FirstEdition1993 Reprint : 1998 ( A thorough understandingof the basic principles of a subject like Soil Mechanics calls for lhe solution of a large number of numerical problems In the presentbook a briefinfoduction to the contentsofeach chapterhas been given, which is followed by a number of worked-out examples and quite a few practice problems For a better understandingof the topics and students are required to solve all the problems by themselves.Effort hasbeenmade to explain the basic principles underlying the solution of the problems so tlat the students may develop the habit of having a logical insight into the numerical problems while solving them Commentsand 5rrggestionsregardingthe book, from the studentsaswell as the teachers,will be highly appreciated Price:Rs.60.00 Calcutta, 9, March 1993 Ptfulishedby Prittted at Ish Kapur for Dhanpat Rai Publications (p) Ltd : A.P Of1.sc.t Navecn Shahdara.Delhi- | t(X)32 DEBASHISMOITRA WEIGHT.VOLUME REI.ATIONSHIPS CONTENTS Clwpter Page "{ Weight-VolumeRelationships ,/, Index Propertiesand Soil Classification 24 ,/ I Capillarity and Permeability 49 'g r' lz, Seepageand Flow-nets 81 StessDistribution to7 Consolidation 133 Compaction 165 Nr Shcar Strength 181 '9J/ Earth Pressure 2r3 10 Stability of Slopes ?54 L1 Bearing Capacity ?33 12 Deep Foundations 310 /J ,€ o Matter may exist in naturein threedifferent states,viz., 1.1 Introduction: solid, liquid and gaseous.A soil massin its naturalstatemay consistof all ' three phases.The basic ingredient is the solid grains which form the soil skeleton,while the intermittent void spacesare filled up by either air, or water, or both Thus, a soil massin its natural statemay be considereda three-phase system 1.2 Soil Mass as a Three-phase System : In a soil mass in its natural state,tle three phases,viz., solid, liquid andgas,are completely intermingled with one another However, if one can determine the individual volumes of solid grains, liquid (i.e., water) and gas (i.e., air) presentin a certain volume : -Water- : Fis.1.1 I ofa soil, the entire soil mass can be represelted by a schematicdiagram, as shown in Fig 1.1, where the volume of each constituent part is shown as a fraction of the total volume The cross-scctional area of the soil mass fo taken to be unity, so ttat, the volume of each constituent part is numerically equal to ib beight shown in the diagram Again, the mass of each part may be obtained by multiplying its volume by the corresponding density t Thenotations used inthe diagram are defined below: t V = total volume of the soil mass I I I ' \ Problems in SoilMechonics and Fonndation Engineering particlegin the soil % = volume of solid in the soii = voids of volume V, V- = vslspe of water presentin the voids V, = volurne of air presentin the voids s RelationshiP Weight-Vofume i.e., v t=ixrWva .(1.4) dry soils) to 1007o(for fully The value of s may vary from oVo (for saturatedsoils) " is defined as the ratio of the ("tSp"t it'rc gravity of sotids(G".or G) : It to the mass of an equal volume of mass of a given lrotume of solicl grains water, measuredat the sametemperarure' 17 = total mass of the soil !7" = rnassofthe solid Particles W- = mass of water presentin the voids' The massof air presentin the voids is negligible' Vu=V"+Vn Thus, G =Mny : - 1.c., The fundamental physical properties which 1.3 Basic Defrnitions : below : govern the engineeringperformanceofa soil are defined grains M" = massof anyvolurneVofsolid M = massof water of volume V' then in the C'G'S' systen If this volume V is arbitrarily taken as unity' of solid grains (y') and dersity the to Lqu't M" and M become ** i".iry density of water (1.) respectively' Thus' (i)Voidratio(e):Thevoidratioofasoilisdefinedastheratioofvolurne of voids to the volume of solids' massolunitvolunggllglids- Ts O massof unitvolumeof water Y- and, V =V r+V, 0r, V=Vr+Vo+Vn vu i.e., "=v, where, .(1.1) Thevoidratioisadimensionlessparameter,thenumericalvalueofwhich with increasing degree of compactnessof the soil' decreases -aefineAas the ratio of the volume of voids to the 1i4 f-rsity (n): ttis as a percentage' total volume of the soil mass.It is generallyexpressed i.e., fu= + x rooe,o .(r.2) 1' However' as lhe The void ratio of a soil may be greateror less than a soi| mass,its porosity volume ofvoids is alwayslessthalrthe totalvolume of is always lessthan 100% is defined as (ili) Water content(w) : The water content of a soil mass expressedas always is tne ratlo of the rnassof *.i"t to the massof solids' It a percentage i.e., ,/ w *=frxlooVo "'(1'3) ,/ (s) : The degreeof saturation of a soil mass is 4i{ O"gr"" of saturation of voids It is always defin-eias tf,e ratio of volume'of water ro tbe volume expressedas a Perc€ntage' .(1.s) T"= G'Y' or' as the ratio of the mass of (vi) Mass spectftcgravity (G,,) : It is defined volume of water' measuredat a siven volume of soil to theLiti'of tn equal the sametemPerafure i.e., I ; M Y M* \n .(1.6) where " '(vit\ Y= unitweightof thesoilmass' of,thetotal Butka"nrityl, unit weight(v): It is ogrineo15n;-ratio KN/m ' gm/ccor t^n- or ,o.,, of u soil to its totalr olume.Its unit is w l.e.r, \=T .(1.7) as the massof soil solids per (viii) Unit weiglt of solids(Yr):It is defined unit volume of solids 1.e., w, Y " =% .(1.8) a soil mass is defiried as the (ix) Dry density (17) : The dry density of volume of ttre soil mass massof soil solids per unit of the total Problems in SoilMechanics and Foundation Engineering ws \d=V i.e., .(1.e) The difference between 1" and y7 should be clearly understood.The dry density of a fufly or partly saturatedsoil is nothing but its bulk density in the dry state.The dry density ofa soil dependson its degreeofcompactness, and hence, on its.void ratio But $e gnit weight of solids depends only on the properties of iie minerals presentin it and is independentof the statein which the soil exists (x) Saturated unit'weight (y.",) : When a soil mass is fully saturated,its bulk density is tenrred as the saturatedunit weight of the soil (xi) Submergeddensity (y.u6): The submergeddensity of a soil massis clefinecl as the subnerged weight of the soil per unit of its total volume 1.4 Functional Relationships : In order to assessthe engineering performanceandbehaviourofa soil, itis requiredto evaluatethefundamental properties enumeratedin fut' 1.3.While some of theseproperties (e'g', w, G, y etc.) can be easily determinedfrom laboratorytests,someothers(e'g', q s, y" etc.) cannot be evaluated directly However, all of these properties are interdependent.Hence, if mathematical relationships between two or mor€ such properties can be developedthen the direct determination of a few of them will lead to the indirect detenninationof the others.Thus, the functional relationships have an important role to play in Soil Mechanics The most important relationshipsare establishedbelow : vu "= v" = Vv + V " , o r , V "- V But, vu vr/v "' e = v - v"= (v:W '.e= = v,/v considerAlternative prool: The samerelationshipsmay alsobe deduced (b)' and (a) 1'2 in Fig' shown ing the schematicdiagrarnof a soil massas (1+e) Fig.1.2 vv We know that, -Vr n [ = +l V r ,= e V r Let us considera soil masshavingtrnitvolume of solids' = T ? ; t'"J e = u e " = i = " n = Again, = +, or vu n'v = Cqnsideringa soil masshavir:ga totalvolume V l, - n' V = l ' n = n , o r , % = V - V , = | v, .e=Vs V, vr/v, vu o r ' ng v J v " = Wm= ys Now, .(1.10) n = T (b) (o) L - n Again, by definition, .(1.11) L + e V v =r - " r-i " n = ' % = l , o t , V r = € ' I = € ' ' Totalvolumeofthesoil, V = V, r V" = | s (i) Relation betweene and n : By definitnn, Weight -Volume Relat ionslriPs -,5 n l - n newion betweene, G, w ands : With referenceto Fig 1.1' , = w% ' t " Vn'\n u * * Problems in Soil Meclnnics and Foundation Engineering G=!, Y" = G'Y Of, ln Vn'ln Ws By definition, tu ,9e G Wn+W, Vr+V, Vn.yn + V".Gyn _ _ t v Vr+V" Vu+V" (1.13) = From eqn weBet,Tsar Iw (1.13)we get,fd = From.eqn _ (s.e + G)/e , = G r s€ , ttt' tw (l+e)/e V = - W v Wn+W" v Wn+W" (1.13) t | (iv) Expressionfor y.", : Olt Y w" wnr w' l or, (V, + GV')/V, - -F;T-q4 tn or, (G+e)/e tw r - r t (l+e\/e G+e tw l+e w" l a = |i , o r , V = j ,ld .(ii) From (i) and (ii) we gel, For a safurat'edsoil, V,n= l/, Vy.y- r V"'G^ln V, + G.V" = ;=-'Y Tsar - -i7,rlV, Vu+1, l+G.(I/el .(i) Y = - - Again, W n + W " vn"(n + %'Y" = W= Bydefinition,ysar i Vr+V, ffi t+l/e ' We know that, G+se r , #.r* = -GTn l;" l+e v' = - v l + e - u u = fi| (vi) Retation between y and y4 : t\r l+l/e +# Foradrysoil,s=0 tw Dividing fle numeratoranddenominator by V, ,we get, VJV, + G.V"/V, s + G/e 1+V"/V, .(1.ls) Eqns.(1.14) and (1.15) may also be derivedfrom,eqn.(1.13) as follows : Forasaturatedsoils=1 Vn + G.V" E Iw G^tn Vn.\n + %.y" Vr+V" v = - = - = - ot, G/e | + l/e \d = TTe 0f, The bulk density ofa three-phasesoil systernis given by, ' Vr,+V" (V, + V")/V, 'w (iii) Relatian between y, G, s and e : = V".G\n %'y" Vr+V, V s.e = w.G W V .(1.14) G.Vs,/Vv V"/V" ' G + e l-Jl'Yw (v) Expression for y1 : vJV, s G/e f Ysar= of' = = vr.Gr" y" G (vr/v,) ' c = VJV" c = Weight -Vo htmeRelationshtps tw - Y= Yd = Wn+W, ( =w' I a Wr\ W " - ' t , = l t * W " /l ' r o - ( l + w ) y a \ v T;-; .(1.16) (vii) Relation between y*5 and y* : A soil is said to be submergedwhen it lies below the ground water table Such a soil is firlly saturated.Now, accordingto Archimcdes' principlc, when Problems in Soil Meclnnics and Foundation Engineering an object is submergedin a liquid, it undergoesan apparentreductionin mass, the amount of such reduction being equal to the rnassof the liquid displaced by the object Consider a soil mass, having a volume V and mass I,Iz,which is fully submerged in water Volume of water displacedby the soil From theconsiderationof degreeof saturation,a soil sample (i) Completely dry (s = 0) -t (ii) firlly saturated(s = 1) Unless otherwisementionedin the problem, a soil sampleshould always be taken to be partially saturated = V(Y."r - Y-) The apparentdensity or submergeddensity of the soil is given by, V(Y."r - Y,r) W' Ysub=V = V Methpd 12'Given ' lT,w, C I ==+Required ' t : [Ta, ' s,A;l' l As e and z are mutually dependent on each other, effectively three unknown parametershave to be determinedfrom the given data Select the appropriate equationswhich may servethis purpose The value of y7 can be determinedfrom : .(r.r7) Y , ' d - l + w Here, Two differentmetlods :ru (iii) partially saturated(0 < s < 1) Apparentmassofthesoil, W' = W - V -,{n = V.ysat - V.,{n Ysub=Ysat-Yw Solution: may be : = V Mass of displacedwater = V \n or, Weight-Volume Relationships ""t"Toyedto solvethe numerical problems in this chapter They are : Method I : Solution using mathematical relationships : This process is somewhat mechanical, one has to mernorise all the equations deduced in fut 1.4 and should select the appropriate equation/s while solving a given problem However, in most of the casesthis method can yield the desiredresult fairly quickly Method II : Solutionfrom first principles : In this method the solution is obtained using only the basic definitions with referenceto a three-phasediagram of the soil massunder consideration This method always allows the student to have an insight into the problem However, in some casesthe solution becomesa little complicated and more time-consuming than method I After going tlrough lhe worked out examples, quite a few of which r'llustratethe use of both of tlese methods,one should be able to realise as to which method of solution suits better to a particular type of problem It may be pointed out that, the methods may also be used in conjunction with one another Problem 1.1 A soil sample has a unit weight of 1.9 gm/cc and a water content of l2%.If the specific gravity of solids be 2.65, determinethe dry density, degree ofsaturation, void ratio and porosity ofthe soil y = unitweightof thesoil = 1.9gm/cc lr = water content = l2%o = 0.t2 \d = T#n= r'6e6gm/cc In order to solve for the other two unknowns,viz., s and e; two equations are required Evidently, the following equationswill serve the purpose : vrG = s€t or re = (0.12)(2.65) = 0.318 Again, or, or, G+se l + e v' = _ l n r.n= f41l@)tr.ol l * e l+e= | ) ' 1.56,or,e=0.56 The expressionof y7 may also be used '{a= of' G'tn y-l s, =(f?P, 1.6s6 OT, 1.696+1.696e=2.65 or, "=ffi=o'56 .(t Problems in SoilMechanics and Foundation Engineermg 10 - = ! = 8= v o 0.56 From (i), e " = Ti; Answer 0.56 = , ;s = n0.36= 36vo Dry density = 1'696 gm/cc' void ratio = 0'56 Degree of saturation = 56'87o,Porosity = 36Vo wn w=-w Now, s wn ; l Void ratio, "=2=ffi=os6 Porosity, =36vo " =+ =ffi x roovo = rrr, r'n volume of 300 Problem It2-'F'nundisturbed specimenof soil has a its weight hours' 24 for 105'C at oven in drying After +66got' "" tJ*.igh reducedto-+sog*.oeterrninethevoidratio,porosity,degreeofsaturation and water conteut Assume G = 2'70' Solution: Wn = o.lZgm 0r, = l'I2gm Totalmassofthesample, W = Wo + W4 rYr " = Volume of solids, v' t ' = Volume of water, W" T wn Weight ofwaterevaporated, Vo=V-(%+V)=0.092cc = ' Volume of voids, V, = Vo t Vn = 0.12+ 0'fp.2 0'2l2cc o.r2 = ' Volume of air, Degree of saturation, t= fr = ffix '1"'n't' t I cir"n,fr wg5 cf+ Required ' " fuid, weight of the dry sample, = o'12 - 0.12cc v' =Yl! t- l'rz = 0.589cc Total volurne of soil, Methodl: After drying itt oven,thewater presentln m€' soti"ffitatts becomescomPletelYdrY W = 498 gln Now, weight of the moist sample, I -W' - 0" ' -3"7 7* c r ,, = = c\"= (2f5)(l) l* ,, =+ =#F = r'6nsm'/cc Dry densitY, Method II: Letusconsidera'specimenofthegivensoilinwhichthemassofsolid 1'3' grains = gm The tnree-phasediagrari of the soil is shown in Fig' 11 Weight-Vol umeReIat ionshiPs r o o % 5o o and the soil Wa = 456 gn' W-='W -Wa= 498 -456 = 42gm' = 0'0921= 9'21% Watercontent,w - Y 456 wd +G'r'u \d=T;e Dry densitY, \d = But wa -456 = r.szpm/cc v 300 G'tn = L5z l + e \h (0.092cc1 Vw (0'12ccl V ( 5E c c ) 1.s2(r + e) = (2.7)(r) +1.52=2.7 L.S?z e 0.78 Void ratio= 0.78 or, or' : -Woter : : - -_:_-_-_-_-_-_-_ _- of, VJ (1'12gml Again,porosity Vs (0.377rc) , - e TT; = From eqn (1.12), t+G= s€t F i g 0.78 = 0.438= 43'8vo ,ft or, , = I9 , LZ Problems in Soil Mechanics and Foundation Engineering Weight-VolumeRelationships Ort (o.oe2r\ (2.7\ "=ff=0.319=3l.9flo Problem !J A saturatedsoil sample,weighing 178 gm, has a volume of 96 cc If the specific gravity of soil solids be 2.67, determinethe void ratio, water content and unit weight of the soil A Method II : With referenceto the three-phasediagrarn shown in Fig' 1.4, V-= :==42cc Volume of solids, v' s - w " - w ' Gln AS6' = 168'8e cc ' 0f' V'=V-V' Ot, Vu = 3N - 168.89= 131.11cc " = vu 131.11 = o'78 = 16s€, ," vu ,=T= s= w= Vn fi Again, 131.11= O = | V o 3Of = Wo W= 42 trfu 2'67-+ | x e\(1'o) = 1'954 r + e ) 1.854+L,854e=2.67+e 0.85k = 0.816 e = 0.955 0r, V=300cc Total volume, tu, = l]|.u But, Y" ,?in Required : y,",={ =y9 =1.854gm/cr v yw = of' Given,W, VEe+ Unit weight of the soil, w Volume of water, Volume of voids, Solution: wn=498-456=42gm Weight of water, 13 = o32= 327o 42 =9'2lc/o ^5t=0'0921 (0L0255) - 0.358= 35,.8vo * =-X - u t Prcblery{ A ftrlly saturatedsoil samplehas d volume of 28 cc The sample was drled in oven and tle weight of the dry soil pat was found to be 48.86 gm Determine the void ratio, moisture content, saturateddensity and dry densityof the soil mass.Given G =2,68 Solution: Given' F % e;l=+ A schematicrepresentationof the given soil is shown in Fig 1,5 Here, total volume V=?3cc Volumeofdrysoil, % = { 1c1 )c Required : T#"c=18.23cn Assuming that there was no changein void ratio during ovcn-drying, volumeofwaterevaporated,Vn= V - % = QA - L8.23)cc 7,'9.77cc w (4 g m l Sotid Void ratio, v, Ws{ m ) = Fig.r.4 Vn - v " v , l r = - = - o11 ffi = o'536 l'.'v"=vnT 309 Bearing Capacity 308 Problems in Soil Meclwnics and Foundation Engineering and the depth of foundation was 1.4 m, The rubsoil consistedof r deepstretum of medium clay (y - 1.8 t/m3) Find out the average - unit cohesion of the [Ars: c = 3.5 tlnzl clay 11.11 The fmting of a column is 1.5 m x 1.5 m in size, and is founded at a depth of 25m belcnvthe ground level The properties of the foundation soil are: c - 0.1 kg/un2, - 15', \ - 1.75 gm/cc' Detennine tlc srfc load the footing can carry with a factor of safety of 2.5, when thc water tablc is at: (t) 0.5 m below the ground level [Ans: (i) 24.99 t (it) 28'29 tl 1iq O.Sm below the base of fmting 11.12 The subsoil at a site consistsof a homogeneousbed of ilormally consolidated soil having the following properties: y = 1.85 t/m3, c = 3.5 t/m?, = 10' AZ m x 3.5 m footing is to be foundedon this soil at a depth of 1'5 m' Detennine the safc load the fcroting can cary with a factor of sat'ety'of 2.5 Use Brinch Hansen's method Given, for $ = 1g', N = 8.34, Nq - 2.47, Ny = 0'47' [Ans:152.44t] by IS: recommended the method 11.13 Redo Problem 11.12 using = l'22' NT = 2'47' Ns 8.35, N 10', 6403-1981 Given, for $ [Ans: 152'0$ tl 11.14 Detennine the factor of safcty against shear failure of a 1.5 rn wide strip footing located at a depth of I m below the ground level in a bed of dense sand having Y = 1.9 t./m3 and = 40", if it canies a uniformly distributed load of 22tpet metre run Use Terzaghi's equatiou Given, for - 64.18, and flr - 95'et' [Ans:2'61] - 40o, N, = 75.32, Nq 11.15 An R.C.C column is subject to a vertical force of 900 kN actittg through its centrc line and a horizontal thrust of 120 kN actingat2.T m above G.L ihe column is supportedby a squarefooting of 2.5 m x 2.5 m size, placed at a depth of 1.2 rn below G.L The foundation soil bas an angle of internal friction of 35' and a bulk density of 18'5 kl'I76'' Assuming a factor of safety of 3.0 detennine the safe load Use: (i) Brinch Hansen's method (Nc - 46'12, Nq = 33'3' /VY - 4[.69) (ii) Recommendation of IS: 64O3- 1981 (N = 46'12, Nq = 33'3, IVY = 48'03) [Ans: (i) 3458 kN (ii) 2687 kNl I1.15 In order to assessthe bearingcapacityof a 2.5 m squarefooting, a plate load test was conducteclat a site with a squareplate of 60 crmx 60 cm size.The tbllouritrgresultswere obtained: 180 | 360 Seulement(mm) 0.82 1.78 720 1080 144A 1800 3.62 5.40 9.30 If the allowablesettlerneutofthetbotingbe 1.5crn,find outtheallowable [Ans:284.4t] load on the footing 3tl Pile Foundotions L2 PILE FOUNDATIONS According to Terzaghi, a foundation is called a de.ep12.L Introduction: Various types of fcrundationif its width is less thin its depth (i.e., DIB > L)' are: deep foundations Pile foundations Well tbundationsor opell caissons' Pier foundations or drilled caissons' thc load of a l2.2 Pile Foundations: Piles are generally used to trhnst'er of piles applications other The structure to a deep-seated,strong soil stratum' are as follows: (i) to compacr a loose soil layer (compaction piles) subject to uplift or overturning forces 1ii) to nori down structures (tension Piles) provide anchorage against borizontal pull applied on earth(iii); retaining structures(anchor piles) vessels (iv) to protect waterfroni structures from the impact of tnarine (fender Piles) (v) L resisioblique compressiveloads (batter piles)' |2.3C|assificationofPilesAccordingtol-oadDispersalCharacterktics: classified into the on the basis of the rnode of load dispersion, piles can be following two categones: but its tip (i) Bearing piles when a pile passesthrough a iveak stratum the pile transfers the p"n"ii.Gilrrtoa stratum of substantialbearing capacity, pile bearing a called pile is a ioad imposed on it to the stronger stratum.Such a pile is extendSdto a considerable depth in a (ii\-Fri9!ion!!9When capacity, it derives ia load carrying capacity from sratumTt poii66riirg -tn.rt on ihe sides of the pile Such a pile is called a the friction-of the soil friction pile of a pile may be 12.4 Bearing Capacity of Piles: The bearing capacity a pile without by defined as the maximum load which can be sustained producing excessivesettlement' The bearing capacity of an individual pile may be determinedby the following methods: (i)bynamic lbrrnula (ii) Staticlormula (iii) Pile load test 12.5 Dynanric Forrnulae: The dynamicformulaearebasedon the conct:pt rhar a-p=il-e-jerj-\,-i;lG5;fiingcapacity from the energy spent in driving il The following dynamic formulaeare most widely used: According to this fonnula, the safe ,IrEfigineering News Formuls: given by: pile is ol'a capacity bearYng Er,u.r =#r a=#+4 .(12.1) where, g = safHoad in kg W = weight of hammer in kg H = fallofhamrnerinctn s = averagepenetrationof the pile in the last n blows in cm For drop hamrners, n = for steam hamnters,n = 2A additional penetratiou of the pile which would have taken placehad therebeenno loss ofenergy in driving the pile For drop hatntners, d = 2.5 cm for steam hammers,g= 0.25 cm' c = Equ (12.1) gives the general fonn of the Engineering News Fonnula The specific fonns of this fonnula for dift'erent types of harnrters are given below: h = (i) For drop hammerI A = -J (s + 2.5) .(r2.2) wh = (ii)ForsingleactingsteamhammertQ OG r ) (W p) h + a (iii) For doubleactingstean hammer:n (s + 0.25) .(12.3) n2.4) where, a = eft'ectiveareaof thepistonin cm2 p = meaneffectivesteampressurein kg/cm2 the Hiby Formula: IS : 2911(Part1) - 1964recommends ;filoatfUa by Hiley: derived expression originally following formulabasedon an ^ t l f t' W ' H ' " t s 9"= ,*u, .(12.s) where, O - ultimate load on pile (kg) W, H, s and c have the samemeaningas in eqn' (12'1) Il I I - et-ficieucYof hamtner' I rla ' efficiencYof hammerblow - thc ratio of energy after impact to the striking energy of the ratlr When ff > eP, when W < eP, \b = w +3P w ;; -eP12 t t b -w- +we-zl -PF l-wl w -r 313 Pile Foundations Problems in SoilMeclmnics and Foundotion Engineering 312 ? The value of F" shouldlie between2and3 12.5 Static Forrnulac: The strtic formulae are based on the concept that the ultimate load bearingcapacity(0,) of a pile is equal to the sum of the total skin friction acting on the surfacearea ofthe ernbeddedportion ofthe pile (p1) and the end bearingresistanceacting on the pile tip (pb), as illuslrated in Fig 12.1 Q"=Q1+Q6 -1?r But, Q1 = Qf Af alndQ6 = qb.Ab Q1 = q''+S + Qa'At "'(12'6) wbere, ql - ,(r2.7) P = weight of the pile alongwith anvil, helmet, etc e = co-efficient or restitution,the value ofwhich may vary between and 0.5, dependingon the driving systemas well as the material of the Pile' In eqn (12.5), C representsthe temporaryelasticcornpressiou,wltich is given by, /-l c,' = t.77Y! A p c, = o.ostQiL Ap .(12.10) , q =3.ssfr where, 4rLrl) Ap = cross-sectionalareaof the pile, cm I = length of pile, m eu The methodt of evaluating Qy and q6are explainedbelow: l Colrcsive Soils: Average unit skin friction, 4y = a c cr = adhesionfactor, which dependson the consistencyof the soil and mav be determinedfrom Table 12.1 Average point bearing resistance According to Skemptou, for deep foundations,lV"= qb = 9c For a pile of diarneter^Band embeddeddepthD' .(r2.r2) , (r2.ts) where c= unitcohesion Qu=scAl+9cA6 The safe load on a pile may be obtainedfrom: V s = \ 9u Q b= c N " I /', Ab = c/s area of the pile at its tiP' Fig.12.1 .(rz.e) aver;r+euniiskin frihion A/ = surfacearea ofthe Pile on which the skin friction acts (12.8) where, Cl,CZand c3 representtheelasticcompressionsof the dollyand picking, the pile and the soil respectively Their values may be obtainedfrom: .(12.14) qO = poittt bearing resistanceofthe Pile tiP where, C=Cr+C2+C3, .(12.13) Ou=X* andAy=nBD .(12.16) '(12.r7) (12.18) Problems in Soil Meclnnics and Foundstion Engineering 314 to: Eqn.(12.14)thercfort:reduces .(r2.re) e,, =-F(B D cJ,+225 x 82 c Table 12.1:AdhesionFactors Pile moterial Consistency Colrcsion (tlm2) Adltesion foctor ct soft nrediurn stiff o -3.75 3.75-7.50 7.50- 15.0 Steel solt ntc:diurn - 3.75 3.75-7.50 - 0.90 0.90- 0.60 0.60- 0.45 1.0-0.80 0.80- 0.50 sriff 7.50- 15.0 < 0.50 = fir u K,tan6 q, ,t i.e., qa= Jq sY = ShaPetactors B = width or diatneterof Pile D = length of Pile For a squareor rcctangularpilc, sy = 0'5 .(12.24) ternr of eqn (12.22) is For piles of snrall dianrcteror width, the sec:orrd practicalptt4loses, all t'or Tltus, tenn the first to negligitrleas cornpared .(12.21) K " = co-efticient of elrth pressure,thevalue of which tnay varv " frorn-0.5 for loosesandto 1.0 for densesand = tiictionangiEof-GE6iiloithepilEl\ilEiC[?Ep€iiGon the a:rgleof internalfriction Q of the soil The value of mav be obtainedfrom Table 12.2 (12.23) The value of Nomay be detenninedby the tbllowing methods: (i) Vesic'smethodiAcc'ordingto Vesic:,tq = 3' averageou.rffi v"z sv = 0'3 tbr a circular pile, e u= , t D N o s o CohesionlessSoi/s: For piles driven in cohesionlesssoils, Qa= , the point bearingrcsistanceis For a purcly cohesionlesssoil, c = Henc:e given by, (r2.22) qb = lDNrso + YBNrst whcrtr, No N, = Bearingcapacityfactors' Timber & Concretc where, 315 Pilc Foundotions and, N q = u ' { t a n o ' t u ng25 " + Q / ) Hence, qr=3QNq "'(12'24) The valuesof Nn for variousvalucsof Q aregiven in Table 12'3' Table 12.3: Bearing Capacity Factors Q @egrees) Table 12.2: Friction Angle Smootb (polished) Rough (rusted) 0.54 0.76 0.64 0.80 Parallelto grain Perpendicularto grain 0.76_ 0.88 0.85 0.89 Smooth (maciein metal form work) Grained (made in tirnber tbrrn work) o.76 0.88 0.80 0.80 Rough (cast on ground) 0.98 0.90 Nq $ (degrees) Nq 1.0 30 9.5 t.z 35 18.7 10 1.6 40 42.5 15 2.2 45 115.0 20 3.3 50 422.4 ,q 5.3 (ii) Berezantsev's methodi According to Berezantstu-tl: Nn values Aependon the D/B ratio of the pile and the angle of internal friction of the soil TheN4value may be obtainedfrorn Fig 12.2' 3r6 Problems in Soil Meclmnics qnd Foundation Engineering I| 317 200 Pile Foundations 150 piles is generally less than the product of capacityof a single pile and the numter of piles in the group In or4er to determine the bearing capacity f a pile group, 9g, a correctionfactorrl, is requiredto be used .(t2.28) Qr - nQut1, I 100 rt z , f;/ 50 whe.rc, n - number of piles in the group Ou - ultirnate bearing capacity of eachpile re rls - efficiencY of the Pile grouP The value of 11,may be obtainedfrom the following empirical formulae: (i) Converse-Lttbtrre' Formula: 20 25 30 35 S(Degrees)* /.0 /.5 ,t = numhr of piles in each row 12.7 Pile Capacity frorn Penetratfun Tests: The pile capacity czn also be detennined from the results of the standard Penetration Test or statie cone Penetration Test performed in the field, using the following equations: (i) Standard Penetration Test: where, .(r2.2e) where, tn = nunrber of rows of pile in the group Fig.12.2 Qr=4NAu+O.OZNA1 , r g = 1* [ W ] .(r2.2s) g = tan-l where, d - diameterof eachpile s = spacingofthePiles (ii) LosAngelesformula: a ,lr o -;#, Oa - ultimate bearing capacity of pile in kg l{ - blow count (witboul overburden correction) | , L^ln - ,r; : -r) + n (m-r\ +,/T@:11@r11 '(12.30) 12.9 Dcsign of a Pile Group: The piles in a group are conttectedto a rigid pile cap so that the group of piles behavesis a unit The group capacitymay te derirmined by rhe efficiency equation (eqn 12.24) A more rational Aa = base area of pile in crn2 A/ = tutfu" area of pile in cmZ fI I L However, for a bored pile, Qu=1.33NA6+0.O2NAf .(r2.26) (ii) Statb cone penetration test: Qu = Q"Ab + l,u"o, L .(r2.27) Where, Qc= Co11.e resistanceat tip 12.t Group Action ln Piles: A pile foundation consists of a number of closely spaced piles, known as a pile group Due to the overlapping in the stressedzone ofindividual piles, the beariug capacityofa group offriction r]-f r ll SoftSoit I tt- r -B m S e cA - A Fig.12.3 Problems in Soil Meclranics ond Foundatian Engincering 318 melhod is the rigid block method recontutendedby Terzaghi and Peck According to this method the ultimatebearingof a pile group cqualslhe suru of tlre ultirnate bearing capacity of block occupied by the gronp and the shcaring resistancenrobilised along the perimeter of the group With referenceto Fig 12.3 .(12.31) Qs = Q,BL + DIQB + 2l)s - yDSBL where, Qg eu ultimate bearingcapacityof the pile group 319 Pile Foundotions (ii) The load o1lthe pile group is ett-ectivelytransmittedto the soil at this lower one-thirdpoitrt of pile below this level is ignored' (iii) The presence (iv) The tratrsrnittedload is dispersedas 60" to the horizontal' With referenceto Fig 12.4,the settlementof the group is given by: P= H # " ' r o s r o o6 + Ao o6 .(r2.34) ultirnatebearingcapacityper unit areaof the stressed areaat a depth D1 B, L = width and lengtb of pile group Y = unit weight of soil s = averagcshearingresistanceof soil per unit areabetwt:t:lt ground surfaceand the bottom of pile D/ = depth of enrbedtnentof piles The safeload on the pile group is given by, 0rr=? .(r2.32) The rninirnurnvalue of F" shouldbe takenas 3.0 Tbe aboveequationsarc applicablelo coltesivesoils For crtd hearing piles on hard rock (inespective of the spacing) and ort deusc sand (with spacinggreaterthan timcs pile diamt:tt:r)thegroup capacrityt:qualslht: sum of individualcapacities.i.e., Qr = N'Q, Rock Fig.12.4 .(12.33) of IS : 2911 @art 12.10 Spacing of Piles: As per thc reccrtuurcudatious 1)-1964,the spacingof pilcs n'raybe oblainedfrorn lhr: following gcneral rules: (i) tbr triction piles, s f d ( i i ) t b r e n d b e a r i n g p i l e s p a s s i n g t h r o u g h c o n r p r c s s i b l es o i l , s ]2.5d (iii) tbr endbearingpilespassingthroughcornpressiblesoil but restitg eP I/ = thickness of soil layer which tends to move downwards Using eqn (12.6), Y = unitweightof soil K = co-efficientoflateral pressure (Ko s K s Ko) rlb= EXAMPLES Protrlem tU{ e dmber pile is being drive n with a drop hammer weighing 20 kN and having a free fall of m The total penetrationof the pile in the last five blows is 30 rnm Detenninethe load carrying capacily of 1he pile using the Engineering News formula Here, I,I/ I/ c s = = = w+P e , = W &l.O Q +- $2.5/2 I=42.47t Now, using eqns.(12.9) through (l2.ll), - \ft\@z'!) = o.1o6crn cr = r.779" Ap Solution: Using eqn (l2.lr, o Y = + (o.ss2)(l.s) w * &p = -2.0 = 0.7 7.0+1.5 In order to find out the value of Q, assutneas a first approximation, c = 2.5 cm = friction angle, (6 s Q) , 321 Pile Foundations I x (30)2 wH 6(s + c) ,r=94JJ2,/9)J4J = 0.868 I x (30)2 (3.ss)(42.47)= /, 0.213cm w-3 _= - , weight of hammer = 20 kN height of free fall = I m =" 100 crn' 2.5cm penetrationfor the last blows ave.rage Z 3[l = ? = 6mm = 0.6crn ) O= f f i = x (30)2 C = Ct + C2 + Cs = t.187 cm < 2.5cm ' k N Problem 12.2 Determine the safe load that can be carried by a pile having a grossweight of 1.5 t, using the modified Hiley's formula Given, weight of hamtner = 2.4 t = 91 cIn height of free fall hanmerefficiency =75Vo averagepenetrationunder the last blows = 10 mm = 22 n length of pile = 3iX)rum diatnr'tt:rof pile co-efficicntofrestitution= 0.55 Let = L.397cm Qu= Sot, : c = tt'tat/:l+tot e,=PffiP=56.2st Let Qu=55t,.'.c=W=L.537 Q,=ffi=sot Probletns in Soil Mechanics and Foundation Engineering 322 In the third iteration fhe assumedand computed values of Q, are quite close Hence, the ultimate load bearing capacity of the pile is 54 t' Consequently, the safe bearing capacity 1= -Qu =4=rr.ur g' 7." z.) t1 Problem l?^3ftiurRcC pile of 18 m overall length is driven into a dee;r stratumof soft c*f having an unconfinedcompressivestrengthof 3'5 Vm" The diameter of the pile is 30 cm Determine the safe load that can be carried by thc pile with a factor of safety of 3.0 Here, d = arerageoverburdeupressure ,tH = \ ! = a = {!ggq = rzt/m2 For loose sand,Ks = 0.5 The value of may be obtainedfrorn Table l2.2.For a srnoothRCC pile embeddedin dry sand, 6/q = 9.76, or, = (0.76)(25') = 19' ey = $2) (o's) (tan1e") =2.A66tln? Q , = e +e u - A u As the pile is driven into a cohesivesoil, Q f= a ' c L a From eqn (12.L4), Solution: 373 Pile Foundatians Using eqn.(12.24), qt=3eNq The value of adhesion factor cr may be obtained from Table 12.7-Fot a q,, i-5 = 1.75t/m"crmaybetakenas0.95 = softclayhaving, = ; t = (3)(1.6)(1s)(s.3) = 38r.6t/mz Af = xBD = r(0.4t))(15)= 18.85m2 Again,wehave, % = 9c n, = f,az = @/41(0.40\= o.r?sm2 Ab = c/s area of pile tip go = (2.066)(18.8s)+ (381.6)(0.126) ) n = - x l - - / E =O l t2=0.07m' = 38.94 + 48.08 = 87.02t = 87 t U0o/ At = surfaceareaof the Pile = r(0.30)(18) = L6.g6ilf gu = (0.e5)(1.7s) (16.%) + (e) (1.?s)(0.07) = 2f + l.l = 29.3t '.safeload, P" = ? - ff - s.ter Problem 124/Asmooth RCC pile of 40 cm diameter and 15 m length is d;iven into a d/eepsratum of dry, loose sand having a unit weight of 1.6 t/mi and an angle of internal friction of 8" Determine the safe load which can be carried by the pile Given, for Q - 25", Vesic's bearing capacity fectoriVo - 5.3 Solution Using eqn (12.20r, qf=dK"tanE e,=+=Y=2gt f s J concretepile of 400 mrn diameter and havirqg Problem l&.ff{bored is an overall length of 12,5 n embeddedin a saturatedstratum of c - S soil having the following properties: c= k N / m ,Q = " , yr*-18kN/m3 Derermine the safe bearing capacity of the pile Given, for Q = 20", the bearing capacity factors are: N"=26, Nq=10, Nt-4 Assume reasonablevalues for all other factors Solution: For piles embeddedin a c - f soil, 4b = cN" + \'D 0ro - tl + 0.5y'B,lV,, 325 Pile Foundations Engineering Problems in SoilMeclwnics and Fottndation 324 For tlre secondlayer, Qf, = nc, = (1s)(26\ + (18 - 10)(12's)(10 - 1) + (0.5)(1s - 10)(0'40)(4) a For the third layer, the skin triction rnaybe neglected' Again, using equ' (12'24), =(3)(1.85 x + 1'9 x + 1'8 x 2)(9'5) eu = 3qNq Qf=ac+flK"tan6' cr = 0.5, K" = 1, 6/$ = 0.80' = (0.s0)(20') = 16" (125/2) (1'0)(tan16") a1 = (0.s)(15) + (1s 10) Again = 21.84kN,/m2 Af = n(0.4)(12.5)= r5'71m? and, '7, -D = 528.67r/nz Ir ,^-" a r r < 1e u = i ' ( ) ' = m - (0'196) + (4)(4'7r)+ (5?f-67) Q, = Q.4)(7.85) = 18.84+ 18.84+ 103-62 = 141.3| = \(0.40)2 = 0.126m2 -s l4l'3 = 47.rt - 4it n E gu = (2r.s4)(1s,71)+ (r2e6'$ (0'126) ft [o undation i s supported Ot I OU"t-t-t:-ul,::is.tiit * P roblem ly'{'ra 300 rows' The diarneterand lengthof eachpile are in of ti pit", "no-ng"a ft' t: piles the mm and 15 m respectively'The spacingbetween ^t'?,*l having g = 3'2 t/m' and foundation soil consists or a sori clay layer group' y = 1.9 t,/rn3'Detenninethe capacityof the pile of piles: Solution: (i) Consideringindividualaction Q 1= a c = 343.1 + 163.3 = 506.4kN O" = ty = 168.8 kr{ s 168kN at a deplh of 1'5 m tZ.l fl"colurnn of a footing is founded 10 by a number of piles each having a length of b"b;;I.;,Mrupp-,.d given are which the properties of m The subsoil consists of thiee layers' below: rll=6'5m I-ayerlzc=3t/rT?., 1=1'85t4t3' 0=0"' rl=3m \=t'go;/nr3' 0=0" LayerIII c=st/r&, = 30" rI = 15m - 1'80t'/m3' kyertrI i c = o, p-Ul"- = (0.9) (3.2) and, Qf, - oc1 = (0'80) (3't = z'atltt' Af, = n(0'5) (5) = 7'85 m2 [Assuming c = 0'901 = 2'88t/mz' Ar = n(0'30)(15) = l4'I4mz = ? (3'2\ = ?l,3t/m2 Qo = c eu = i(o'302)= o'o?lm2 Determinethesafeloadoneachpileifthediameterofthepilesbe500 25' Assume' adhesion factor mm and the required factor of safety be ct = 0.80 piles in the three layers are Solution: The depth of embedmentof the respectivelY,5 m, m and m' Forthe firstlayer' a and' Af, = n (0'5) (3'0) = 4'71m' = 390 + 900 + 6.4 = 1296.4 Assume, = (0'80) (5\ = t/n- of eac.h Pile, IndividualcaPacitY (14'14) + (28'8)(0'071) = Qff) Q, i = 42'77t = Groupcapacity, Qus = (15)(42'77) 641'55t l> I I I A I Problems in Soil Mechanics and Foundatibn Engineering 326 (ii) Considering group action of piles: Assuming a block failure, the capacity of the pile group may be obtained from eqn (12.31): - YDTBL Qs = QaBL + D/28 + 2L)s With refercnce to Fig 12.3, B = Z(I.2) + 2(0.15) = 2.7 m width of the block, length of the block, deptlr of the block, L = 4{1.2) + 2(0.15) = 5.1 m Df = 15 n qb = 9c = (9) (3.2) = 28.8r/n2 s = q f = c r c = ( ) ( )= 8 t / m ' Qs = Qs.8) (2'7) (5.1) + r5(2 v 2'7 + x 5'1) (2'88) - (1.e)(1s)(2.7)(s.1) 3n PiIe Foundutions = 17757kN a block failure'width (ii) Consideringgroupactionof pile-s:.Assuming B=2(1.25) + 2(0'50/2) ofUloct 641.55 t Hence, the ultirnate bearing capacity of the pile group is 641'55 L Safe bearing capacity w.r.t a factor of safety of 2.5' Q,c=ry=:x6'62t-r,6t' , problem qr{ egroup of 12 piles,eachhavinga diameterof 500 rnm a raft foundation.The piles are arrangedin rows and 30 m long,-supports andspacedatl.?s m c/c.Thepropertiesof thefoundationsoil areasfollows: 'l|kN/m2, y' = 11 kN,/m3, Qu = 0"' Assumingcr = 0.80 andF" = 2.5, determinethe capacityof the pile group Solution: (i) Consideringindividualactionof piles: = Qf = dc = (0'80)(75/2) 30 kl'[/m2 qb = 9c = (9) (75/2) = 337'5kN,/m2 Af = x(0'50) (30) = 47'Dt# m2 eu = X(o.sd)= 0.1e6 Capacityof eachPile, g, = (30)(47.12)+ (337.5)(0.196) = 1479.75kN Groupcapacity= (I2) (1479.75) a, = ry - 5x8kN = S?;i8.ZkN of 40 mm Problern 12.9/ Agroup of 20 piles,eachhaving a diameier The capacitvof m c/c' 1'0 spacing a at + rows ""d i0;i;;; ,i(-^"^ierdin the piles' each pile is :g0 kN Determinethe group capacity of pile group' Solution: Using eqn (12.28), the capacity of the Qc = n' Qu'rls' Here, n = 20, g, = 380 kN by either of the The efficiency of the pile group' Ie may be determined following formula: (i) Converse- Labarre Formula: Using eqn' (12'29)' ,rr=,Xlffi Here, m=4, n=5, = 2t'8' o = tan-r4s = arr-rf94q'l \ l'ui - t) + + ( - t) s l - 0.624 = 62.4To ' r 8= , - L'8[( s e ol - o t trl I (ii) I.os Angeles fonnula: Using eqn' (12'30)' ,1"= t - d ;-,*Im{n-1) + n(n-1) + {T@:TJ6:T I Problemsin SoilMeclnnicsandFoundationEngineering 328 y y ( r ) t ( s -+15) ( - )+ y ' + - 1 - X , = ' - ; a0.40 = 0.771= 77.r% The lower valueshouldbe used.Hence,the capacityof the pile group Q8 = Qo) (380)(0'624) = 4742.4kN - 4742kN Problern l?.lf It is proposedto drivea goup of pilesin a bedof loose sandto suppoflvrafl Thi group will consistof 16 piles, eachof 300 mm ani 12m length.Theresultsof standardpenetrationtestsperformed Oiameter pile group with a t/# a'd an effective unit weight of 0.9 Vm3 Design the failure' shear factor of safety of against piles in a squale solution: Let us use 16 Nos of 400 rnm o R.C.C formation Let the spacings be equal to d, s = (3) (0'40) = 1'2 * i.e Let I be the length of each Pile' Now, J = I ' (40 ) = o L r r ? CapacitYof each Pile, L) + (21'6)(0'126) gu = (2.16)(r.2s7 or, L + o'eol Ot = = o'eos the averagevalue of N = n,=f,{lo)Z =7n.86cm2 = ?5650k8 = 25.65t' Asthespacingofpilesisashighas5D,itcanbeassumedthatthereis no overlappingof stressedzones GrouPcaPacitY,Qs = n'Qu = (16)(25.65)t = 4101 ii Prpblem 12.1L A raft foundationhas to be supportedby a group of concreiepiles.Thegrossloadto be carriedby thepile groupis 250t' inclusive of or tnewegnt of the pile cap.The subsoilconsistsof a ?5 m thick stratum of 4.8 strength compressive clayhavinganunconfined normallyJonsolidated Qu = 2.715L+ 2'722 ofeachPile, SafebearingcaPacitY n Qu=4NAt+O.02NAS Af - u(30) (12) = 1130'97cm2 (1130'97)ke Q, = (4) (9) (706.86)+ (0'02)(e) = at = 9c - (9)(2.4\ X.16t/n? Af = nBL = (0.40)nL = l'?57 LttJ 1'5 Estirnate the capacity of the pile group' if the spacing of the piles be m c/c + 10 + + 11 + = o Solution: Average N-value = Here, ,=t=+=2.4t/mz = qf = ac = (0'9)(2'4) = 2'l6t/mz' [Assumingcr 0'901 at the site at various depths are given below: Using eqn (12.6),the capacityof a driven pile, 329 Pile Foundations Actualloadto becarriedby eachpift = # or, = 15'6?5t' 0'905t+0'907=15'625 L = I6.2''lm- 16'5m Checkforgroupaction:Consideringtheshearfailureofablockof dimension, BxLxD, B = L = s + d = ( )+ ' = m D = 16.5m (2'16) ' Capacityof the pile grolrp' Qs = Qr'6) (*) + $o's) (4 @ + 4) - (o.e) (r6.il(4) = 894.24t Sar'ebearingcapacityof thepile group Problems in Soil Meclnnics and Fottndatian Engineering 330 894.24 ^ Q,t=T=298t>250L block Hence the designedg4onpof piles is safe from the considerationof failure ,/ e'rcn footing founded at a depth of 1'5 m below G'L' Pr.oblen n.d a dense in a 19.5 thick stratum of normally consolidated clay underlain by diaand m L2 piles oflength 16 sand layer, is to be supportedby a'groupof carried be to load gross The fonnation rneter 400 mm arrangid in a squari The piles are by the pile group (including the self-weight of pile cap) is 350 t' level The ground rhe at is located rable *uter ;;"""d at r.2 m "/" tn are: soil foundation propertiesofthe w =32Vo, G=2.67, L.L= 4tVo consolidation settlementof the pile group' probable the Estimate Solution:WithrefererrcetoFig.L2.4,theloadfromthepi|egroupis point, assumedto be transmitted to the foundation soil at the lower one-third )"t" tZ = m below the pile cap and 8+ 1'5 = 9'5 m below i.e., at a depth of = G.L The tirickness of the clay layer undergoing consolidation settlement and m m, 3 thickness of sub-layers three 10 m Let us divide this zone into nr resPectivelY The settlementof eachsub-layermay be obtainedfrom: p, = H' f; Now we have, 'tor,oo + w G = se, or' e = eo= v c - Ao wG s and, - 1'00 = 0'90 t/m3 Ysub= 1'90 = (4 + tan30")2= 32.86rt 350 ^ o = for = L s' -6 10.65t/mz (300)(0.27e) , e.e-tlq4l = 14'32cn' r'",= f1 frffi'' losto tr of thesecondsub-layer: Settlement os = (0.90)(1.5 + 8.0 + 3'0 + 3'O/2) - I7'6t/m' ,, A2 = (4 + x 4.5 x tan30")2= 84'57ri oo=*=#h=4.r4t/m2 (300)(0.27e) ,^- 12'61]!-! = 5'57cm losto P",= fr'dffi' tr Settlementof the third sub-laYer: lo = ffi (II - 10) = 0.009(41 - t0) = 0'279 0.0@09 ?#r*=Ti ,'s1 = (B + FI, tan30';2 A3 = (4 + x x tan30')2 = 175'?3m2 ryP=08s4 Ysar= Assurningtheloadtobedispersedalorrgstraight|inesinclinedtothe horizontal at 60", the area over which tf e grossload is distributed at the rniddle of thc first layer, At = Q + 2H/2'tan30') (B + 2H/2'tan30') o6 = (0.90) (1.5 + 8.0 + 6.0 + 4'0/2\ = 15'75t/r] o6 AgaiIr, 33t Pile Foundations (1)= l'eotzm3 Settlenrentof the fint sub-laYer: oo = initial overburdenpressureat the middle of the layer = \' z = (0.90)(1.5 + 8.0 + 3.02) g'st/^z Ditnetrsionsof the block of piles, L = B = s + d = ( \+ = m = r.997t/mz 15.75+ L99J - 3.12cm (400)(0.279)' roglo 'pca 1SJS + 0.g54 = + Totalsetflement,Pc P., * Pc, Pr = 14.32+ 5.5'7+ 3.12 = 23 cm EXERCISE12 of an RCCpile drivcn l2.l Determincthesafeloadcarryingcapac:ity by a drophammerweighing3 t andhavinga freefall of 1.5rn, if theaverage [Ans'20'3t] p.n.nrtion for thelastfiveblowsbe 12mm' 332 Problems in Soil Mechnttics and Fottndation Engineering of 10 rn 12.2 An RCC pile having a diarneterof 400 rnrn and a length free tall of a height with kN, 30 weighi*g natruner is bei.g driven with i Orop recorded been has blows few last the for penetration of 1.2 rn The average co-efficient or as rnrn If the ettlciency of the hammer be 7O% aud the using lnoditied restitution 0.50, detenninettre safe load the pile can carry a factor of = Assume Hiley's fornula Given,unit weight of RCC 24 kN/m'' 200 kN] Ans' I safetyof 3.0 into a 12.3 A22 m lorrg pile having a diameterof 500 mm is driven 5.6 of strength compressive deep straturn of sofl clay having unconfined to a respect with pile the of capacity t/# Detennine the staticload bearing tl 40 tAns' facror of safety of 2.5 =16 friction angle = = 10, Nv = 4' [Ans 279 kN] Nq = ?.6, q Nc 20', for an RCC pile of 500 mm 12.8 Deterrninethe ultimate load capacityof sub-soil conditions are The colutnu' a of diarneter supporting the tboting skt:tchcdin Fig 12.5 Given' = 0'9 adlrcsiontactor tbr soft clay = 0'7 silt claYeY and that t'or for = 30' is 9'5' The water table is factorNu capacity bcaring Vt'sit"s neglected' [Ans' 232 t] lrx nlcrl rtt il gr(raldcpth.Skin friction iir sandmay be Soft CtoY l2.4.Aconcretepileof30cndiameterisernbedd-edinastratunrofsoft clay straturnis clay lraving = 1.7 t/rn3, Qu= 4'2 t/mz'Thethickness of lhe g m and the pild penetratesthrough a distance of 1.2 m into the underlying = 36"' Detennine the sat'e straturnof de;rsesand,havilrg Y = 1'85 t'lm3 and Q of 3' safety of load carrying capacityof the pile with a lactor capacity faclor = bearing Vesic's = 36', Given, O O.gOQ and for Q [Ans.32.3 t] Nq=23,c[=1,K"=1 is driven 12.5 A stnoothsteelpile of m length and 400 rnm diameter properties: into a cohesionlesssoil masshaving the following = 30' Y"ar= 1.8t,zrn3' Q = 0'60 Qand Vesic's The water table is locatedat the ground level' If the safecapacityof = determine 9'5' be 30" bearingcapacity faciorNn for = 0'7 Ks Given, [Ans' 12'1 t] of 2.5 rhe pil! with a iactor of sifety at a 12.6 A 12 m long pile having a diameterof 300 mm is cast-ih-situ site where the sub-soilconsistsof the tbllowing strata: = 10kN'/m2 StratumI: thiclness =5 m, Y' = 10kN,/m3, 0=30" c = = kN'/m2 Stratun II: thiclness= 16 m, Y'= kN'/m3, 0', c 60 Detenninethesafeloadonthepilewithafactorofsafetyof2.0.Assume ieasonablevalues for all other data' of 500 mm is 12.7 A 16 m long bored concretepile having a diameter properties following the having silt ernbeddedin a saturatedstratum of sandy 'Yru,= 19'5kN'/m3' c = rlkN'/m2' = 2o' with a factor of Detennilte the safe load canying capacity of the pile safety of 3.0 Given, = O'75 adhesionfactor = 0'85 pressure ofearth co-efficient JJJ PiIe Foundations 10m Sitt Ctoyey y = Et /5n F ) , c= l l m I J 2.0m T I i I I Ssnd (t=1.75t/m3,@=30o) Fig.12.5 rows wi-th a 12.g A pile group consists of 42 piles anan$ed in pile is 22 rn long Each centre-to_centiespacing of 1,5 rn in each direction using: pile the of and 500 mm in diameter.Find oul tbe group capacity (i) Convene-Labane formula (ii) tns Angeles formula' Given, load bearing capacity of each pile = 78 t' q [Ans (i) 2142 t (ii) 2624 12.10 A pile group consistingof 25 piles anangedin a sqlare fonnation are L5 m and is to support a iaft iooting The length and diameterof eachpile soil is 300 mm respectively,wiile their spacingis 85ocmc/c Thg-foynfation Determine y 1'85 VT'' a normally consotiAatedclay having c = t/mt and = F" = 3'g' the safe load bearing ""p""ity of thi pile group' Take cr 0'85 and [Ans' 527 t] placed 12.11 A multistoried building is to be supportedby a raft footing piles 96 of consists raft on a pile foundation The pile group supporting the water c/c'The m of 2'0 of 26'm length and 400 mm diameter,with a spacing table is located near tle ground surfaceand the propertiesof the foundation soil are as follows: 334 Problems in Soil Meclwnics snd Foundation Engineering Y " r = t / r n c, = / m , O = ' The adhesionfactor may be taken as 0.95 Determinethe capacityof the pile group with a factor of safetyof 3.0 12.12 Designa pile groupto supporra raft footing of m x 12 m size and carrying a gross load of 760 t The self weight of the pile cap rnay be assumedas 20o/oof tlre gross load on footing The subsoil consists of a homogeneouslayer of soft clay, extendingto a great depth and having the following properties: y' = 0.85 t/nr3, qu = 5.7 t/m2 Design the pile group with a factor of safety of againstshearfailure Given a = 0.85 12.13 It is required to drive a group of piles in order to support a raft footinqof 10 m x 10 m plan area,and subject to a gross pressureintensity of 15 Vm" The subsoil consists of a 12 m deep layer of soft clay (y = 1.8 t/rn3 , qu = 4.5 Vm2) which is underlainby a densesand layer (y = tlnr3 , = 35') The raft is founded at 1.5 m below G.L In order to utilize the bearing resistanceofthe sand layer, each pile should penetratethrough it at least D The adhesionfactor for clay = 0.90 Vcsic's bearing capacity factor lfu for = 35' is 18.7 Design a suitable pile group with a factor of safety of 2.5 againstsbearfailure Assume that the self weight of pile up = 25Voof pressureintensity on the raft 12.14 A raft footing is founded at a depth of 3.5 m below G.L in a ?A rn thick stratum of soft clay having the following properties: y""1= 2.05 t/m3, C, = g.3 The gross load to be carried by the pile group, including the self weight of the pile cap, is 8O0L The group consistsof 81 piles of 400 mm $, arranged in a square formation, and extended to a depth of 12 m below the pile cap The spacingof the piles is 1.25 m The water table is located at the ground level Cornpute the probable consolidation settlement of the pile group Otherusefultitles: Soni, Gupta and Bhatnagar L A Coursein Electrical Power & S ChakravortY Mukheriee P.K Machines Electrical P.V Gupta Fields Electromagnetic in Course lntroductory ortd Gupta Seth Materials Engineering Electrical in A Course and Gupto Dhar & SYnthesis Analysis Network R.K Gaur Microcomputer and Electronics Digital Venkataraman FundamentalsR & Computer Circuits Digital Pulse, Fundamentalsof Microprocessorsand Microcomputers B' Ram Gupta and Satnant Sub-stationDesiguand EquiPment K.D Shanna (Hindi English) & Engineering Electrical 10 N.C Sinha of Structures Theory Advance 11 Selvant Manicka V.K Structures of Analysis Limit of Fundamentals t2 Selvant V.K Manicka DYnamics Structural Elementary 13 Selvam Manicka V.K analYsis Structural Elementary 14 V.K Manicka Selvant 15 Modern method of StructuralAnalysis V,K Manicka Selvam 16 Rudimentsof Finite ElementMethod L7 Multistory Building & Yield Line Analysis v.K Manickaselvanz of 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Abdttlla Shariff Abdulla Shanff 37 Theory of Machines Practice) 38 Computer Programmingfor Engineers(Theory & M.M Hasan (CoveringDigital & AnalogueComputation) M.M Hasan 39 Principlesof Cobol Programming M.M Hasan ulO Computer Programmingin Basic ... defining tbe gradatiottof the soil: (i) Uniformity Co- efficient: = ^u D g Dto .(2.8) (ii) Co- efficientof Curvature: (Dro)2 "=Dto"Doo .(2.e) Problemsin Soil Mechnni.csand FoundatianEngineering 28... soil massunder consideration This method always allows the student to have an insight into the problem However, in some casesthe solution becomesa little complicated and more time-consuming than... properties control the behaviour of the soil in actual field The most important aggragateproperties are: (i) for cohesionlesssoils: the relative density (ii) for cohesivesoils: the consistency,which