1. 1. 780 S01].MECHANICS AND FOUNDATIONSR =ZR;.. .(27.3) where R is measured below the high ood level (HFL).' Scour level= ll. F.L.R = H.P.1.. 2R. , . ..(27.4)The grip length is taken as g-R below the scout level according to the code of practice of the Indian Roads Congress and as -}R in Railway practice.This means that the depth of foundation should be at least lR below HFL according to [RC code.and1R below HFL according to Railway practice .it is further recommended that the minimumdepth of embodment below the scour level should not be less than 2.0 m for piers and abutment with arches and 1.2 m for piers and abutments supporting other types of superstructure. According to Terzaghi and Peck.the ultimate bearing capacity can be determined from the following expression: Q, = Q,+ zany,D;.. .(27.5) Q,= til? (I .2 CM + 10, N,+ 0.6yRN, ) . ..(27.6) where N. . Nq.M =Terzaglti's bearing capacity factorsR - radius of well D; =depth of well (deptlt of foundation) 1.= average skin friction 27.4. FORCES ACTING ON A WELL FOUNDATION In addition to the selfweight and buoyancy,a well carries the dead load of thesuper-structure.bearings pier and is liable to the following horizontal forces : (t) braclting and tractivc effort of the moving vehicles. (ii) force on account of resistance of the bearings against movernem due to variation of temperature. (fit) force on account of water current, (iv) wind focres. (v) seismic forces. (vi) earth pressure, (vii) centrifugal forces. The magnitude,direction and point of application of all the above forces can befound under the worst possible combinations and they can be replaced by two horizontal forces.P and Q and a single vertical force W as shown in Fig.27.3.P=Resultant of all horizontal forces in the direction across the pier. Q:Resultant of all horizontal forces in the direction along the pier.W= Resultant of all vertical forces. The analysis is done on the following assumptions (Banerjce and Gangopadhyay.1960):
2. 2. WELL FOUNDATIONS 78!1. The well is acted upon by an uni- directional horizontal force P in a direction across the pier. 2. The well is founded in sandy stratum. 3. The resultant unit pressure on soil at any depth is in simple proportion to horizontal displacement. 4. The ratio between contact pressure and corresponding displacement is independent of the pressure. 5. The co-efficient of vertical subgradc reaction has the same value for every point of surface acted upon by contact pressure. The analysis that follows is that suggested by Banerjee and Gangopadhyay (1960). 27.5. ANALYSIS OF WELL FOUNDATION1. Horizontal soil reactions.When a rigid WC. in sand,813115 moving PBMIICI n(; _ 2-; _3_ poacs ON A wELL_ to its original position.under the action of a horizontal force P.it transforms the soil on one side to passive state of plastic equilibrium and the other side into active state.Assuming that the well movement p.is sufficient to mobilise fully the active and passive earth pressure.the resultant unit pressure at a depth z below the surfaceis given by p,= 1 .2 (K,- X. ) . ..(21.7)where 7 =unit weight of soil K, . K,= co-efficient of passiveand active earth pressure,and dependupon the angle of internal friction ii.and angle of wall friction 5.Let p be the load per unit area of vertical surface of sand and p be the corresponding displacement.Assum- ing that p.is the displacement required to increase the value of resultant unit pressure form zero to p, , we have. p= p|= Yz([(y_]{a)PI PI . ..(27.8) | _______ Kp1zjT+B+_Ka1z_4FIG.21.-8. EFFECT OF WALL MOVEMENT.
3. 3. 784 SOIL MECHANICS AND FOUNDATIONS6. Evaluation of not M,produced by P, M; = ID m E (2-1). ) 4; = ."_'PL(3o -4D.o+D. ) . ..(27.l9) D.D.12D.7. Evaluation of vertical reaction R :Modulus of vertical subgrade reaction is k= %where p a vertical deflection of soil corresponding to vertical reaction P = - p.3/2 5/2 R =2 pd:= 2 kp.dz 0 0or B =up... .(27.20) 8. Evaluation of moment M,produced at the base due to vertical soil reaction pThe rotation of the well is also resisted by a moment M.acting at the base onaccount of the downward deflection of the toe and upward deection of the heel.Fig.27.8 shows the rotation of the base,with displacement p,at the ends.Let p be thedeection at a distance x from the centre 0 of the base. rm n/2M; =2 p. xdx.=2 I: p.xdx :P o o I-x-I 3 32 )5 * /2 B1 =2_[ t. -9-xx = kp 0 B (D, .1Case (ii) Zero.if n*= to. Case (iii) Imaginary or complex if n< co. A: -n= -z-}%11=-Il- Jill-0)": Case (it) gives a value of
4. 11. 800 S011. MECHANICS AND FOUNDATIONSEq.28.1! for case (it) reduces to 3" =0. indicating that for this condition there will be no oscillation.but only a rapid return back to the equilibrium position of the mass [Fig.28.4 (b)].The value of c for this condition is called critical damping c, .c.= 2m % =2-H . ..(28.l2)For case (1).the radical is real ( II > ma) .and c 2 2M. 1 '5 Hence from (0. z= C'e'= C'e1|-ZL: -L) H . ..(28.l3)Eq.28.13 shows that z is not a periodic function of time.Therefore,the motion.when rt of,> 0 is not a vibration,because it can only approach the equilibrium position at t> so.However.the viscous resistance is so pronounced that the weight set in motion from its equilibrium does not vibrate but creeps gradually back to the equilibrium position at time innity [Fig.28.4 (a)].For ease (iii) when nc>. . (blcase ()0 = . . 28.8. Also.the sum or dif- ference of these two solutions mul- tiplied by any constant is also a solution 2zI____C2_'1(). tr_(tz. rj O _ ___ _'_; 'm (28.15 0) """" '- and z1=%[ell~l+el1.l) . ..(2s.15 b)where Cl and C1 are comtants.Substituting the values ofA and 1" mm Eq 28% and I no 2214 nus DISPLACEMENT cuaves r-on $imP1if)'in8- We set ' ' ' oawan vmrwnous.z. =:C. e"sin out and zz'~Cze""cos amt Summation of z,+22 =z renders the general solution of Eq.28.8 in the following form.
5. 12. MACHINE FOUNDATIONS 80)z =c"" [C.sin o)4t+ C;cos W41] . ..(28.l6) The quantity in the bracket represents the simple harmonic motion of the case ofvibration without damping while e"" is the damping term.Fig.28.4 (c) shows the time displaccment curve for this case.The peroid T of the damped vibration is given by 2::Zn= u)= -VET, -_, -, ' . ..(28.l7)The term (D4 is called the frequency of damped vibrations. J n= J-f; -:: -:-x1-am:0), .- or 04:i 1-4: is.1-[-3) . ..(2s.1s)cm 2:- 2.8.4. FORCED VIBRATIONSForced vibrations of a system are generated and sustained by the application of an extemal periodic movement of the foundation of the system.Forced vibrations constitute the most important type of vibration in machine foundation design.We shall consider the case of forced vibrations with damping.Generally,for oscillating machinery (where the machinery vibrates because an unbalanced rotational force exists).the force can be expressed as a sine or cosine function.such as F.sin cat.The equation of motion for such a case may be writen asmi +c+kz= Fo sin mt . ..(28.l9)or i"+2+'= -sin cut . ..(29.l9 a) m m InThe solution of the above equation may be assumed in the following forms : zn-A001 aot+Bsitt to!.. .(28.20 a) By successive differentiation.we obtain 2 = Am sin tut + Bo) cos or . ..(28.20 (2) 1 = Am cos it Bun sin (or . ..(28.20 c) Substituing into Eq.28.19 (a).we get (Aoicos tot-Bo) sin n)t)+%(-Aw sinmt+Btneos tot)+%(A costar-I-Bsin cut) -:gun cit . ..(28.2l) Equating the co-efcicnt of sin tut to both sides,Bu)--Am+B= E9- . ..(28.22 a) m m m Similarly.equating the coefficients of cos or to both sides., m+--am +5,4.~o . ..(2s.22 1III In Solving Eqs.28.22 (a)and (b) for coefficients A and B.we get
6. 13. 302 SOIL MECHANICS AND FOUNDATIONSA . - -. ..(zs.23 a) a__f"_(. ".; ,9L_.2 . ..(2s.23 b) (k - mar) + (cm) (I: nzur) + (co) Substituting these in Eq.28.20 (a).we get the solution in the fonn to an F0 (k mm) sin to!= j + m28.M 2 (cm):+ (k - ma) 005 M (cu) + (k muf) ( )The equation represents the components due to forced vibrations with the period of7:I .The frequency fo vibrations (in cycles per second) is given by on fE-0) The natural frequency of vibration.as dened earlier.is given by 0). :radiam/ sec . ..(28.25 b) -21-}. -J 1: and j; . 2 - 2 m . ..(28.25 c) Substituting in Eq.28.24. - F0 C0) F0" mm)=2; sin . ..(28.26 a) and I =2; cos . ..(28.26 (1)(cm) + (k - mto) (cm):+ (k - mm) we get 2 - 2; (sin 4 cos nt+oos sin mt)-=2; sin (mt+) . ..(28.27) where the angle 4: is tenned as the phase angle between the exciting force and the motion of vibrating mass. Noting that these terms represent a pair of vectors which must be added to obtain the displacement.the solution for the displacement due to the forced vibratiom of Eq.28.24 becomes2 =Li + 3 F.sin (tor + ) . ..(2s.2s) Substituting the values of A and B.and noting from Eq.28.l2 that c, ..2 M =2: /E =3 . ..(2s.29 a) or k =" . ..(2s.29 b) k mu 2 F0 l ..,we get z= -k- 1 sin (mt + 4) . ..(. .8.30 a) 2 H:(E I (20); ; _ wit} F sin (ml + 4)) or z = "___(23_3o b) 2 L [ I [2 Ct I t + I The maximum deflection z, ,,, ,, is thus given by
7. 14. MACHINE FOUNDATIONS 303Folk z, ... .,_-A =T . ..(28.3l) / (2~*1+lI-(1 J Cr 1;:/5! But %=6, = static deflection of spring Ag_&m_5:5. [zi Lv+ l_[L]2T . ..(28.3l a) C: Inwhere A;= z. .., . =maximum dynamic deflection of the system. Putting -. =5'5": =[. l =magmoalion factor or dyttantfc amplication factor2- =L =r =frequency ratio "and=damping ratio.we get_ I p._; ),+4, ', . ..(28.31 b)Fig.28.5 shows a plot between the magnication factor and the fre-quency ratio r( zi =-:7] for vari- ous values of damping ratio' c iii a 7, )-From Fig.28.5. it is observed that magnication factor suddenly shoots up for the valuts of r between 0.6 to 1.5. Atr =l.resonanceoccurs for an undampod condition.Even for damped conditions.the magni- cation factor (and hence the am- plitude) is maximum at r < l .Thus.these curves show the effect of damp- ing on shifting the frequency for maximum amplication away from the natural foundation frequency.TheMagnication factor ptaim of the designer should be such Freauencvratiot that the frequency ratio f/ }; is either FIG.28.5. AMPLITUDE FREQUENCY R. El. A'nONSlllP less than 0.6 or more than 1.5. How F0 ': FORCED Vmm5'ever.the frequency f of the machine is always constant.and a foundation engineer has to manipulate the natural frequency f,of the machine foundation system by suitably proportioning it.
8. 15. MACHINE FOUNDATIONS N5or ,5. =2L " -'53- . ..(2s.34 b)1 it m. ,+m. _2x . +W.where W.= weight of the vibrator.and W,= weight of the apparent soil mass. Unfortunately.the size of the co-vibrating body of soil cannot be determined exactly as yet because it depends on frequency and is influenced by the size of the base area of the vibrator (foundation) and by the elastic properties of the soil (spacing). We shall consider here three methods of determining the natural frequency of foundation soil system :(1) Barlten's method.(2) Balaltrishna Rao's method.(3) Pauw's method. 3.7. BARK]-INS METHOD Barlten suggested the following equation for the natural frequency of systemm. ,=) C . ..(2s.3s a) mwhere C. .= co-efficient of elastic unifomt compression of soil A = -contact area of foundation with soil in -- mass of machine plus foundation.The amplitude of displacement is given by FA = 28.35 b 1 mm} (1 - r) ( ) where A;= Zara =maximum displacement .. .UJ F0 =exciting force ;r 2 frequency ratio = b.PIThe above formulae for natural frequency takes no account of the mass of soil vibrating with the foundations. Barken gave the following equation for the co-efficient of elastic uniform compression of soil.obtained from the solution of theory of elasticity problem concerning the distribution of normal stresses under the base contact area of a rigid plate : E l C =M328.36 in I _ H,W ( ) where E:Young's modulus of soil ;u =Poisson's ratioThus.C.depends not only on elastic constants E and u but also on size of thebase contact area of foundation.The co-cffeicient C.changes in inverse proportion to the square root of the base area of the foundation : C 2337 C" VA . ..( .)2 Table 28.1 gives the recommended value of C,for A =10 m.for various soils.
9. 16. 806 SOIL MECHANICS AND FOUNDATIONSTABLE 28.].RECOMMENDED DESIGN VALUES OF THE CO-EFFICIENT OF ELASTIC UNIFORM COMPRESSION Cu (BARKER.I962)Prnuisxible load on soil under Coeicirnt of elastic nrdforrn action ofstadc load only , o,, ,p, ,, ;0,,cu (kg / .,,3)Weak soils (clays and silty clays with sand.in at plastic state.clayey.and silty sands;also soils of categories II and III with laminae of organic slit and of peat)Soil of medium strength (clays and silty clays with sand.close to the plastic limit:l.5-3.5 sand)Strong soils (clays and silly clays with sands of hard consistency;gravel:and 3.5-5 gnvelly sand.locss and ioessial soils)Rocks Grater than S28.8. BULB OF PRESSURE CONCEPTThe calculations of natural frequency by Barken took no account of the mass of soil vibrating.But tlte work done by DEGEBO indicates that when a vibrating load acts on a soil.a certain mass of soil ranging from 4 to 5 times the vibratory load participates in the vibration.Balakrishna and Nagraj (1960) proposed the bulb of pressure concept of calculating the apparent mass of soil participating in the vibration.According to this.the vibrating mass of soil is assumed to be contained by the boundary of a pressure bulb.For the purpose of simplicity.the load acting on any surface is replaced by an equivalent concentrated load acting at the mass centre of the original area.If y is the unit weight of the soil in lb/ cu.ft. . then aocottiing to the pressure bulb concept.the apparent mass of the soil is the mass enclosed by the pressure bulb of intensity 0, lb/ sq.n.such that0; =I y I . ..(28.38)For example.if the uttit weight of soil is 110 lb/ cu. ft. . the apparent mass of the soil will be the mass of the soil contained by a pressure bulb of intensity 110 lb/ sq.ft.From Boussinesq analysis.the vertical stress or,at a depth 2 and radial distance is r=0 is given by U;= 0.4715 9,. (-Hence I 1 I =0.4775 __ . ..(2s.39)In the above equation H] and W,are known.Hence 2 =diameter of the pressure bulb is known. (I 4 Weight W,of soil= i1r 5 = 4?3 3It should be noted that Eq.28.40 is not dimensionally homogeneous.and is applicable only in F. P.S.units.where W.and W,are in pounds and y is lbs/ cu.ft. . ..(2s.4o)
10. 17. 808 SOIL MECHANICS AND FOUNDATIONS TABLE 28.1. VALUE OF Ho AND [1T B '*= '"""" *'( Dem:sand and gravel 700-1150 I-11.5 Dense sand 400-58 4-5.8 Loose sand 70 -140 0.'ll.4Loose silty sand 58 -115 0.58-1.15Dense silty sand 230-450 2.30--4.60Clay.semi-solid its -23 o. iis -0.23Clay.am plutic o. iis -0.23 0.1 l5 --0.23The spring constant for the truncated pyramid is calculated by first of all determining the surface deformation So given by following innite integral :-52 ____"Z___' B E,(a+az)(b+az)(Ii+z) "'(28'44 0) =31 ___""jor Bbzj;(H_m)(l+m)(s+m) . ..(28.44 b)where l= %(a2b) ;:=5;- ;and m= E; . ..(28.45)The equivalent soil spring constant k in the vertical plane is given by I:= (by denition) L:L In __""": : 33 45 I:ma 0 (I+m)(l+m) (J+m) "'( ' )If the base of foundation is circular,a truncated cone wll be considered in the place of truncated prism.and the above expression will be modied as under : ac 1:4,] _d; _ . ..(2s.47) 1 b 0 (l+m)(. r+m) where m= gbz and : =% ;b= diamcter of the foundation. The values of I:are determined by curves of Fig.28.7 in which k is given by the equationt= pb'2. (for rectangular base) . ..(28.48)and k= iibi (for circular base of dia.b)... .(28.48 a) .at}:a .lb ..For given values of .7 and 3 ratios.-5- is determined from the curves.and thusit is known.Then I:is calculated from Eq.28.38.
11. 18. MACHINE FOUNDATIONS 8090.2V . ::: =:1=a _ 20 + ll)Z 'z o. - 2[ l + 2p -(a1+z1)m + (az+z2)m . ..(29.3) where u=Poisson's ratio.9 2. Elude deformation under circular load :Thevertical strains.due to the triaxial " '"2" " load (o, . a, . o,= o, ) under the centre of the ,pavemam plate is given by Hoclct-. 's law :l l e =% (a,- 2ua, ) . ..(29.4) sum where E:modulus of elasticity of the As= Ela8tic strain from subgfadc.2 lo innity Subsituting Eq.29.2 and 29.3 in Eq.HQ 29-1 51-55719 D5*RPgAN 29.4. and integrating between z=2 and UNDER GCUMR LOA ' z= ao,the elastic strain A of the sub-grade is given by . _. _ 2 2 I/2_'_"'_Ll. :.1- A 40 2tt)(a +1) (a! +). /2+(t. |+2tl 1):.. .(29.5) Taking u20.5. the above expression reduces to - 3pc -90 A - . ..(29.6) of A --E-- F3 . ..(29.7) where F} = % hf =Boussincsq settlement factor... .(29.8) {Hm 1Eq.29.6 or 29.7 give the elastic deformation of the subgrade only.The elastic deformations from surface to depth z are not considered since the only siginiftcant deflections are in the sub-grade.If the load is at the surface of the sub-grade (z= O).Eq.29.6 reduces toA =1.s PE!.. .(29.9)Fig.29.3 gives the curves for deflection factor for various values of E and-(: -. The5 ratio corresponding to Eq.29.6 is zero and the curve of 5:0 in Fig.29.3 givesa a the deections according to Eq.29.7 (Foster and Ahlvin.1954).
12. 28. 8280.1SOIL MECHANICS AND FOUNDATlO. 'SDellealon factor F30.15 02 0.3 0.4 0.5 0.6 0.8 1.0 1.5 2.0IIIlI| IIIlI| I|lIIIIlM3} tIII| IItIII| III IIIIIIIIIIII . |60 v .0 A/ .21. I/ IA-III I7/AIIIIIIII-III Ill/ /IIIIII II-IIIIIII'l7W. I.III II-III HIV/ MIIIIIIIIII-III Vertical delledlon (Poisson's ratio=0.5)FIG.29.3. CHARTS FOR VERTICAL DEFLECHONS (FOSIER AND AHLVIN 1954). 3. Burmister analysis :The exible pavements consist of a number of layers the moduli of elascity of which decreases wtih depth.In the previous analysis.the effect of the pavement components was ignored while calculating the deflections.Burmister (1943, 45. 58).took into account the effect of various layers.In the simplest case,the wholestructure may be thought to be made up of two layers : the base course or pavementlayer.and the sub-grade layer.In the analysis of the two layer system.following assumptionsare made :is innite in extent in the lateral direction.but of - nitt:depth.(ii! ) the sub- grade layer is infinite in both horizontal and vertical directions.(iv) both the layers are in continuous contact.and (v) surface layer is free from shearing and normal stress outside the loaded area.The ver- tical stress at any depth z.at the cente of the plate is given by6; =CR .p . ..(29.l0)(i) the layers are homogeneous.isotropic and elastic.(-'0 the surface layerI! .! ~Baee course or pavement layerW?FIG.29.4. BURMISIT-IR TWO-LAYER STRESS INFLUENCE CURVES IBURSMISTER.I958).
13. 29. 830 SOIL MECHANIS AND FOUNDATIONSthe dual wheels are equal to those of a single wheel depends upon the spacing of the wheels.Fig.29.6 shows the inuence of mul- tiple wheel on stresses. At a depth approxi- mately half the face-to~face spacing (d),the wheels cease to act independently.and at a depth equal to twice the centre to centre .to spacing (S),the overlap of messes becomes ne8_ FIG.29.6. INFLUENCE OF MULTIPLE WHEELS ON STRESSES.ligible.An equivalent wheel load can be found either from the equal deection criterion or equal-stress criterion.Based on the equal deflection criterion.the following expression for the equivalent wheel load I,results : Eproximate 33pm . --(.94.2t. r|. aP . ... .. -. In/7. =(F.+ F, ) J?.. .(29.13) where P, =equivalent wheels load1:wheel load of each of the dual tyresF. = settlement factor for equivalent wheel loadF.=_ settlement factor contributed by one tyre of dualsF;= settlement factor contributed by the other tyre. The solution of the problem is accomplished by determining values of P,and F,so that the Eq.29.13 is satised.The load P is known and factors F,and F,canbe known from Fig.29.5 for various values of 5 ratios.The maximum value ofF,+ F,occurs at a small distance from the centre of the tyre.However.for practical problems.the values of F.+F,under the centre line,and under a tyre,need by calculatedand the greatest of the two may be taken to he use!in Eq.29.13. Thus the R. H.S.of Eq.29.13 is known.A number of values of P,are assumed and the values ofxiii.F,are computed on Eq.29.13 is satised. 29.5. DESIGN.METHODSThe flexible pavement design methods can be broadly classied into three distinctgroups :(1') Empirical methods based on soil classication and other factors such as climate and moisture.They include the following: (a) Group index method.(b) Federal aviation agency (U. S.A. ) method.
14. 30. DESIGN OF I"-LEXIBLE PAVEMENTsoaked as well as unsoaked samples are determined.Both during soaking and penetration test,the specimen is covered with equal surcharge weights to simulate the effect of overlying pavement or the particular layer under construction.Each sur- charge slotted weight.147 mm in diameter with a central hole 53 mm in diameter and weighing 2.5 kg is considered approximately equiva- lent to 6.5 cm of construction.A minimum of two surcharge weights (La 5 ltg surcharge load) is placed on the specimen.Load is applied on the penetration piston so that the penetration is approximately 1.25 mmlmin.The load readings are re- corded at penetrations.O.0.5. 1.0. 1.5, 2.0. 2.5. 3.0, 4.0. 5.0. 7.5. 10 and 12.5 mm.The maximum load and penetration is recorded it it occurs for a penetration of less than 12.5 mm.The load penetration curve is then plotted as shown in Fig.29.8.The curve is mainly convex upwards although theportion of the curve may be concave upwards due to surface irregularities.A cor-833LoadonpbtonlnlqjOorrected zeroFIG.29.8. LOAD PENETRATION CURVFS IN CBR TEST. nation is then applied by drawing a tangent to the curve at the point of greatest slope.The corrected origin will be the point where the tangent meets the abscissa. The CBR values are usually calculated for penetrations of 2.5 mm and 5 mm.Generally the CBR values at 2.5 mm penetration will be greater than that at 5 mm penetration and in such a case the former is to be taken as the CBR value for design purposes.If the CBR value corresponding to a penetration of 5 mm exceeds that for 2.5 mm.the test is repeated.If identical results follow.the bearing ratio correponding to 5 mm penetrationis taken for design. Fig.29.9 gives the design curves for determining the appropriate thickness of construction required above a material with a given CBR.for differem wheel loads and traffic conditions.These design curves for roads have been proposed by the Road Research Laboratory.England. and are also followed in India.
15. 31. DESIGN OF FLEXIBLE PAVEMENT 335the tip of the cone.A Cone bearing ratio (psi)--> correction is,therefore, added to or subtracted 5 ' ' "from all the readings E 3 I IIIIIIIso that the penetration ; ,~ g,under 9 kg becomes 3 1 mmmmsl-HMMIII PM under 36 kg.3,0 4111111: Correction ' C =on - 2 9 1 . ..(29.17) 5 '2WWWEHIIIIIIIIbearing value.the 'thickness of pavement.. lIC ts dctcrrnmed from Fig. 29.10. A minimum FIG.29.10. NORTH DAKOTA DESIGN CURVE (AFTER WISE.I955).thickness of 24 cm is provided for hearing values of 28 kg / cm or more.29.9. BURMISTERS DESIGN METHODBurrrtister's design method is based. on the concept of a two-layer system.consisting of the road surfacing.base course and the sub-base as the top layer of thickness It.and the sub-grade as the bottom layer of innite extent.The displacement of such a system.under a loaded area of radius a with load intensity p is given by Eq.29.11.. _ A - L5 E:F where 15;:modulus of elasticity of the sub-gradeFudeflcetion factor.determined from Fig.29.5.The method consists in selecting various values of the thickness It of the top layer and nding the value of the deflection corresponding to each value of h.from Eq.29.Il.the value of factor F being taken in each case from Fig.29.5. The thickness It corresponding to an arbitrary deection of A=0.2 inch (5 mm) has been recommended by Bunnister as the required thickness of the pavement.Tentative design curves for exibe runway pavements. using 0.2 in.as limiting deformation have been drawn assuming approximate value of modulus of elasticity for various types of sub-grades. 29.10. U. S. NAVY PLATE BEARING TEST METHODThis method is also based on Burtnister's two-layer theory.It consists of the following three steps : Step 1. The thickness In of the base course is calculated on the basis of the twolaycr theory.For this the values of moudulus of elasticity E,and E,for the base course and sub-grade are detcnnined from two plate bearing tests.
16. 32. 836 SOIL MECHANICS AND FOUNDATIONSStep 3. Trial sections are constructed with the pavement thickness equal to It.h and Step 3. Plate bearing tests are run on these trial sections and nal thickness is chosen on the bar.of these tests which produces a deection of 0.2 in.(5 mm). Step 1. In order to use the two layer theory of calculation of required pavement thickness in step 1. it is necessary rst of determine the value of E,by the plate bearing tests on the sub-grade.A 30 inch diameter plate is recommended for this test.In performing the test.it is essential to use a series of staked plates to minimize the bending of the plate upon loading.The load P corresponding to a deection of 0.2 in.is determined from the test.and the modulus of elasticity E,is calculated from Eq.29.12 by taking the plate to be rigid.The deection factor F for this test is equal to urtity since the test on sub-grade soil results in a one layer system.HenceA:0.2 in. =1.1s. _... .(29.1s) 52From this.E,is detennined.After the modulus of elasticity E,is known.a test section consisting of the base course material is built and plate-hearing test is made on this.The test section should be 5 m by 5 m square (or larger) and 15 to 30 cm deep.The load intensity p corresponding to A :0.2 in.is determined from the tests.Knowing E,from the previous test.and A(=0.2 in. ) and p from the present test.the factor Fis calculated from t-: q. 29.12 A:mags 2 A .E2 or F- spa . ..(29.l9)Thus.factor F is known.From Fig.29.5, the value of E,/ E,is found corresponding to the value of F and It/ a ratio. After having known E,and 5/5, (and hence 5. also).the value of F corresponding to a given wheel load intensity p is computed from Eq.29.11 by taking A:0.2 in. A -_ 0.2 in. 1.5 g F . ..(29.2o) 1(Considering the wheel load to be a flexible plate) In this equation A.5,.p and a (radius of the tyre contact area) are known.Knowing F.and E, /E.ratio.the thickness h of the base course is detennined from Fig.29.5.Step 2. In the next step.trial sections are constructed of thickness 1:.h and % Itcalculated in step 1. Each trial section of a given thickness is constructed for three different soil condidtions:one on a typical ll section,another on a typical cut section.and a third at a position on grade.Thus.in all.nine trial sections are built.The sub-grade and base courses are compacted to the densities that will be expected during construction. Step 3. Plate-bearing tests are performed on these trial sections.The data then are used to determine the required pavement thickness which will result in the assumed deection
17. 33. DESIGN OF FLEXIBLE P/ 'E. (EN'l' 337of 0.2 in.In making these tests.a plate is employed which has a radius corresponding to the effective tyre radius a. In the above method,the design thickness is the total thickness of base material to sustain a given load at a given deection.and no consideration of the type or depth of wearing surface is given.However.the structural qualities of wearing surface material are always better.and hence a certain thickness of surface material can be substituted for the base course material.This will.in effect.produce an added factor of safety. 29.11. LABORATORY EXPERIMENTSEXPERIMENT 2| :DEl'ERMINA'l'l0N OF CALIFORNIA BEARING RATIOObject and scope.The object of the experiment is to dctcrmirtt:the California Bearing Ratio (C . B.R. ) of . i compacted soil sample in the laboratory.both in soaked as well as unsoalted state.The method also covers the detemimtion of CBR of undisturbed soil sample obtained from the eld. Material and equipment.(t) Cylindrial mould (C. B.R.mould) with inside diameter I50 mm and height I75 mm.provided with a detachable extension collar 50 mm height and a detachable perforated base plate I0 mm thick.(it) Spacer disc.148 mm in diameter and 47.7 mm in height.along with a handle for screwing into the disc to facilitate its removal (iii) steel cutting collar which can fit flush with the mould both outside and imlde.(iv) Metal tammers :(a) weight 2.6 ltg with a drop of 310 mm or (b.weight 4.89 ltg.widt a drop of 450 m,(V) Annular slotted weight weighing 2.5 kg each I47 rrun in diameter with a centre hole 53 mm in diameter.(NI) Penetration piston.50 mm diameter and minimum of [00 mm long.(iii) Extemion measuring apparatus consisting of :(a) perforated plate I48 mm diameter.with :threadul stem in the centre.(b)adjus1able contact head to be screwed over the stem.(6) metal tripod.(viii) Loading device.with a capacity of at least 5000 kg and equipped with a movable head or base that travel:at a uniform rate of 1.25 mm/ min :complete with load indicating device.(ix) Two dial gauges reading to 0.01 m.(it) Sieve 2 4.75 mm and 20 mm [5 Si-eves.(xi) Miscellaneous apparatus,such as a mixing bowl.straight edge.seal .soaking tank or pan.drying oven.water content deter-mirettion tins.lter paper etc. 1'5: Procedure (0 PREPARATION OF TEST SPECIMENI.Remoulded Specimen :Remoulded specimen may be prepared at Proetors maximum dry density and optimum water content or at any other desired density and water content.The material used should pass a 20 mm IS Sieve.Allowance for large material should be made by replacing it hy an equal amount of material which passes a 20 mm IS Sieve but is retained on 4.75 mm IS Sieve.The specimen may be prepared either by dymmie compaction or by static compaction. (a) Dynamic eontpaedoa.Take about 4.5 to 5.5 kg of soil and mix it thoroughly with the desired water.If the sample is to be compacted at optimum water content and the corresponding dry density 1' termined by compaction test (light compaction or heavy compaction) take exact weight of soil required and necessary quantity of water so that the water content of the soil sample is equal to the detennioed optimum water content.Fix the extension collar to the top of the mould and the has:plate to its bottom.Insert the spacer disc over the base (with the central hole of the disc at the lower side).Put a disc of a coarse filter paper on the top of the displacer disc.Compact the mixed soil in the mould using either the light compaction or heavy compaction.For light compaction.compact the soil in 3 equal I: t)1:n.each layer being given 56 blows.uniformly distributed.by the 2.6 It;rammer.For heavy compaction.compact the soil in 5 layers.by giving 56 blows to each layer by the 4.89 kg rammer.Remove the collar and trim off excess soil.Turn the mould upside down and remove the base plate
18. 34. 838 SOIL MECHANICS AND FOUNDKUONSand the displacer disc.Weigh the mould with the cotnpacted soil.so that its bulk density and dty denstity may be detennined.Put filter paper on the top of the compacted soil (collar side) and clamp the perforated base plate on to it. (b) Static compaction.Calculate the weight of wet soil at the required water content to give the desired density when occupying the standard specimen volume in the mould by the following expression:W ' 74 (1 + W) V where W--weight of wet soil ;y4'= desired dry demity wdesircd water content;V: -volume of specimen in the mould = =2250 cm. Take about 4.5 to 5.5 Itg of soil and raise its water content to the desired value w.Take weight W (calculated above) of the mix soil and put it in mould filled with the base plate and filter paper at its bottom.Tamp the soil by hand during pouring.Place a lter paper and the displacer disc on the top of the soil.Keep the mould assembly in any compression machine and compact the soil by pressing the disptacer disc till the level of the disc teaches the top of the mould.Keep the load for some time,and then release.Remove the displacer disc. 2. Undistttrbed specimen :To obtain undisturbed satnplw,attach the cutting edge to the mould and push it gently in the ground.When the mould B sufficiently full of soil.remove it by under-digging.The top and bottom surfaces are then tritnmed at so as to give the required length of the specimen.The density of the soil should be detemtined by weighing the soil with the mould or by any eld method (such as the sand replacement method) on the soil in the vicinity of the spot at which sample is collected. (in) SOAKING OF SPECIMEN AND TEST FOR SWELLINGI.Put a filter paper on the top of the soil and place the ttdjumble stem and perforated plate on the top of the filter paper. 2. Put annular weights to produce :1 sunzharge equal to the weight of the base material and pavement expected in actual comuuction.Each 2.5 lrg weight is equivalent to 7 cm of construction.A minimum of two weights should be put. 3. Immerse the mould assembly and weights etc.in a tank of water allowing free access of water to the top and bottom of the specimen. 4. Mount the tripod of the expansion measuring device on the edge of the mould and note down the initial reading cl dial gauge. 5. Keep the set-up undisturbed for 96 hours (4 days).Note down readings every day against the time of reading.Maintain comutnt water level in tanlt. 6. Take the final reading at!the end of period.remove the tripod and take out the mould.Allow the specimen to drain,for IS minutes.Remove all the free water collected in the mould taking can. - that the surface of the specimen is not disturbed during the process. 7. Remove the weights.perforated plate and top lter paper and weigh the mould with soaitcd soil specimen.(ii! ) PENETRATION TESTI.Place the surcharge weights back on the top of the soalned soil specimen.and place the mould assembly on the pcnetnttion test tnaehine (loading machine). 2. Seat the penetration piston at the centre of the specimen with the smallest possible losd but in no case excess of 4 kg so that full contact is established between the surface of the specimen and the piston. 3. Set the stress and strain dial gauge to zero.Apply the load on the penetration piston so that the penetration rate is approximately L25 mmlmin.Record the load ruding at penetrations of O.0.5.
19. 35. DESIGN OF FLEXIBLE PAVEMENT 8391.0. L5. 2.0. 2.5. 3.0. 4.0, 5.0. 7.5. I0 and 12.5 mm.Record the maximum load and penetrationif it occurs for a penetration of less then l2.$ mm.4. At the end of the penetration test,detach the mould from the loading equipment.Take about20 to 50 g of soil from the top 3 cm layer of the specimen.and lteep it for water content detennination.Tabulation of observations.The test data and observations are recorded as illustrated in Table29.2. TABLE 29.2. DATA AND OBSERVATION SHI-Z! -."l' I-OR C. B.R.DE11-ZRMlNf'l10l'1. Compaction elnraderlstla (rt) Dynarnie Companion : I .Optimum water content2. Wt.of mould + compacted specimen 3. Wt.of empty mould-l.Wt.of compacted specimen5. Volume of specimen (b) State Compaction : I.Dry density . ..(g/ cm) 2. Moulding water content . ..('. ) 3. Wet.wt.(W) of soil compacted . ... (g)2. Soaking and swelling tut 1. Dry density before soaking 2. Built density before soaking 3. Bulk density after soaking 4. Surcharge weight used during soaking Dial reading (mm)Total -e ion mm
20. 36. 8-l0 SOIL MECHANICS AND FOUNDATIONSCalculations and Tut Results 1. Erpansiotr ratio.The expansion ratio may be calculated as follows : Expansion t: tto=5L-Ii'. x 100 where d/ vml dial gauge reading (mm) ;diuinitinl gauge reading (mm) It intial height of specimen (mm). 2. land penetration.Plot the load penetration curve (Fig.29.8).If the initial portion of the curve is concave upwards.apply the correction by drawing a tangent to the curse at the point of greatest slope. The corrected origin will be the point where the tangent meets the abscissa.Find and record the correctecd load reading corresponding to each penetration. Corresponding to the penetration value at which the C. B.R.is desired.corrected load values are found from the curve and C. B.R.is calculated as follows : C. B.R.=x too P5 Pr= coneetedtcstlo. -id corresponding to the chosen penetrzttion from the load penetration curve.P5-Standard load for the samepcnetrationasfor P:taken from Table 29.I. The C. B.R.values are usually calculated for penetrttion of 2.5 mm and 5 mm.If C. B.R.for 5 cnt exceeds that for 2.5 mm.the test should be repeated.If identical results follow.the C. B.R.corresponding to 5 mm penetration should be taken for design.
21. 37. Design of Rigid Pavement30.].INTRODUCTIONRigid pavements are made up of Portlant cement concrete.and may or may not have a base course between the pavement and the subgrade.Because of its rigidity.and high tensile strength.a rigid pavement tends to distribute the load over a relatively wide area of soil.and a major portion of the structural capacity is supplied by the slab itself.For this reason.rrtinor variations in sub-grade strength have little inuence upon the structural capacity of the pavement.The rigid pavements are used for heavier loads and can be constructed over relatively poor sub-grade such as black cotton or plastic soils.peat.etc. A base under rigid pavements may be used for the following reasons : 1. Prevention of pumping of fine grained soil (Pumping is dened as the ejection of water and sub-grade soils through joints.cracks and along the edges of pavements caused by downward slab movemem by the passage of heavy axle loads over the pavement aer the accumulation of free water on or in the sub-grade). 2. Protection against frost action.3. Provides drainage.4. Controls the shrink and swell of sub-grade. S.it forms a working surface on clays and silts.and thus enables the construction to proceed expeditiously during wet weather. 6. Serves as a levelling course on irregular formatiom.7. Lends some structural capacity to the pavement. 30.2. STRESSES IN RIGID PAVEMENTThe factors affecting stresses in rigid pavement can be placed in four broad categories:l. Strr: sssthiett>rtsu'airu: dI: npuannearl1nnisnnedefmmatiors. 2. Stresses due to the externally applied loads. 3. Stresses due to volume changes of the supporting material,including frost action. 4. Stresses due to continuity of the sub-grade support as affected by permanent deformations of the sub-grade or loss of support through pumping. (341)
22. 38. S42 SOIL MECHANICS AND FOUNDATIONSRelative Stiffness of Slabs :When a load is applied on a slab.it deforms in a saucer shape.and the resistance to deformation depends upon the stiffness of the supporting medium as well as flexural stiffness of the slab.The relative stiffness of the subgrade and the slab is indicated in tentts of radius of relative sttness.dened by Wcstergaard (1927) by the following characteristic equation : Eh 12 (1 u)x,.1/4l=.. .(30.l) where I:radius of relative stiffness (cm) (a linear dimension) E= modulus of elasticity of the pavement (kg/ cm) It:thickness of pavement (cm) is = Poisson's ratio of the pavement k,= modulus of sub-grade reaction (kg/ cm)The modulus of sub-grade reaction is defined as the intensity of pressure on the horizontal surface of a soil mass required to cause a unit settlement of surface : 2 E k.P . ..(3o.2)where p =sub-grade reaction.p =vertical deflection 30.3. STRESSES DUE TO WHEEL LOADWestergaard (1926) considered three cases of loading :(1) corner load.(2) load at the edge of the pavemem.and (3) interior load.His original equation for interior loading was later modied by him (Westergaard.1933).The equations for edge and comer loading were modied by Sutherland (1943).The modified equations converted irtto metric units.are given below : om. =0.2750 + , r)h:[4 log [-l; ]+ logic i 12 (1 18)} 54.54 c; (ci| ]2] . ..(3o.3) o. .,, , =o. s29(1+ 054,1)4 log; o[%/ i + log, . ..(3o,4)1 .E 13 i am. .. = %'L 1] . ..(3o. s)where ()trJm'or =maximum tensile stress at the bottom of slab due to loading at theinterior (kg/ cm) (e). m =maximum tensile stress at the bottom of slab due to loading at the edge (kg/ cm) (0')mnt(r :maximum tensile stress at the top of the slab due to loading at corner (kg/ em) P:wheel load (kg)
23. 39. DESIGN OF RIGID PAVEMENT 843It =slab thickness (cm) is =Poisson's ratio for concrete l=radius of relative stiffness (cm)E = modulus of elasticity (kg/ cm)k, = modulus of sub-grade reaction (kg/ cm) b=radius of equivalent distribution of pressure (cm).given by the followingequations :b =hm + it - 0.675 It .when a < 1.724 It . ..(3o.5) b -_- a when a 2 1.724 n . ..(3o.7)a =radius of area of contact between the slab (cm).the area being assumed circular in case of corner and interior loads.and semi circular for edge loads c. . C:= correlation factors to allow for a redistribution of sub-grade reactions ;c. ~5l; cge0.2. Pickett (1951) gave the following semi-empirical formulae for corner loading for protected and unprotected corners : For slab corners protected by load-transfer devices such as bars etc.:_ 3.35 P _ M 1 ("''' 0.925 + 0.22 a/ t_!"30 For unprotected corners :_ / I (o). .,, ... , =33 1- I . ..(3o.9); .= o.9z5+o.22 an]The above two formulae are in FPS units in which P in lbs,c is in lb/ in ,I.a and It are in inches and k.is in lb/ in. The formulae for the stresses in pavements are complicated for direct calculations. Use may be made of the inuence diagrams and charts prepared by Docket (1948) and Pickett (1951). 30.4. STRESSES DUE TO WARPINGThe surface of the slab is subjected to wide range of temperature during the daily cycle.whereas the temperature of the bottom of the slab in contact with the sub-grade of the base remains relatively more constam.This temperature gradient through the slab causes differential expamiott or contraction between the top and bottom of the slab. In the day.top surface expands more than the bottom and the slab assumes a shape convex upwards.The weight of the slab and load transfer devices or friction at joint will restrain free warping and will tend to bend it back into its original shape.and hence compressive stresses at top and tensile at the bottom are created. In the night.the sides and corners warp upwards.and might actually leave the sub-grade.In this position.the weight of the raised portions of the slab tend to bend them down.
24. 40. 344 son.MEOIANICS AND FOUNDATIONS Hence tension at top and I_ - IIIIIIIIIIIIII omp ton in the bottom IS de- $. .., ..: . a. ..n. ..,(1938) gave , IIIIII| r.: ::: =: the following equations for edge 'stress and interior stress due to warping :d- agm g IIIIIIIIIIIIII Mme:JVIIIIMIIIIIIIII ~W % IIIIIIIIIIIIII (. .,. .,. ..,. E E;'[ C;+ -9) 3 . , IIIIIIIIIIIIIII P > IIIMIIIIIIIIII where e,= co-efcieiitagofl le)xpan- 'sion of concrete ( 5x10" 9 F' "3) o 2 4 6 3 l0 12 1.:A t=total differenceintem- vain 0' L hmdh perature (may be taken to be equal ' ' to about l F per 1 cm thickness FIG.30.t.WARPING STRESS C0-EFFICIENTS of Slab) (BRADBURY.I938). E= modulus of elasticity of concretein =Poisson's ratioC= co-efcicnt given by Fig.30.].corresponding toL/ I ratio C, =co-efcient in the desired direction(say.x direction) C,= co-efficient in the perpendicular direction (say.y direction)L:length of the edge L1:free length inx direction (t'. e..the direction in which 0'! /uen'or is sought) L, =free width in y direction. 30.5. SFRESSES DUE TO SUB-GRADE FRICTIONStresses can also be set up in rigid pavements due to tuujfonn temperature change,which cause the slab to contract or expand.If the slab is free to move and there is no friction between the slab and the sub-grade.no stresses will result.However.if friction exists between the slab and sub-grade.restraint results from the friction forces and the slab is stressed.During expansion.the under~side of the slab is subjected to compressive stress.while during contraction tensile stresses are induced due to the sub-grade fricion.Fig.30.2 (b) shows the distribution of frictional stresses.as suggested by Kelley (1939).Test results have shown that fully mobilised frictional resistance f, ,, is realised for a distance I2:-X.but from there to the centre of the slab.the stress distribution is parabolic in shape. The equations for the average co-eicient of sub-grade resistance f are as follows :
25. 41. DESIGN OF RIGID PAVEMEINT 845For x._: u zoamo