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. 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. 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
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.
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
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
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.
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.
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)
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.
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).
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).
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.
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.
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.
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
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
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.
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
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.
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)
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)
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.
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 :