Problems In Soil Mechanics And Foundation Engineering 1aq3k

< 1 8 0 o Toe Drsin

Fig'4.5 ED is an equipotential line 4.E.2 Conections at Entry and Exit Poinls: and the phreatic line is a flow line. These t'wo should meet each other at right angles. This necessitatesthe correclion at the entry point, which should be drawn by hand. The phreaticline should meet the down streamfaceBCtangentially. This necessitatesthe correction at the exit point. The basic parabola intersectsBC = La, at M. But the phreatic liire should meetBC at N' I-et CN = a and NM the of angle, slope p the on depends + Aa) of Ml(a ie The magnitu downstream face. Its value may be obtained from Fig' 4'6'

I = 11 8E00oD r o i n o g eB t a n k e t

Fig.4.7

It shouldalsobe notectthat for an earthdamhaving a loe drain, chimney drain or horizontaldrainageblanket,thebeginningpoint ofthatdrain on filter, and not the bottornconter of dowustreamface,shouldbe takel into while plotting lhe basic parabola.

EXAMPLES earthdam,30 rn higlr, hasa free board Problem$r1- A hornogeneous and of 1.5 rn. A flownet was constructed the following resultswere noted: No. of PotentialdroPs = 12 No. of tlow channels = J

Problems in Soil Mechanics ond Fottndation Engineering

90

Seepage and Flownets

9l

The darn has a 18 rn long horizotttal t'ilter at its dowrutream ettd. Calculatethe seepageloss acrossthe dam per day if the width of th,edarnbe 200 m and the co-efficientof permeabilityof the soil be 3.55 x 10-' cmlsec. Solution: Using eqn. (4.4), the quantity of seepageloss acrossuuit width of the dam is. N"

(I -- k.H.#

I\ aI

k = 3.55 x 1.0-actnlsec

Here,

_ (3.ssx 10-4)(s64oo)n/cJay = O.3067n/day 100

Nf=3'

Na=12

As tbe downstrearn end is provided with a long horizontal tllter, the downstreamside shouldbe drv.

q =

H = 3 0 - 1 . 5 =28.5 m (0.3067)(28.s)(3)

t2

= 2.185 ^t

14

Total quantityof seeqagelossper day acrossthe ertire width of the darn = (2.185)(200) =' 437 n'. Problerryf.2 : A single row of sbeetpiles is driven upto a depth of 4 rn in a bed of cleadsandhavinga co-efficientof penneabilityof 0.002cin/sec. An impermeableleyer of very stiff clay exists at a depth of 10 m below the G.L. The sheetpile wall has to retainwater upto 4 rn above G.L. The heigltt of water level on the dowrutream side is 0.5 rn. Construct lhe flownet and determinequantity of seepageloss consideringunit width of the sheetpiles. Solution : The flownet is given in Fig. 4.8. Using eqn. (4.4),

N= K H , H Herc,

_ (0.002)(86400) m/day = 1.728 n/daY k=4.002 crn./sec 100 H = 4 - 0 . 5 = 3 . 5m Nf=7,

Na=lZ

(3.s)(7) = _ (1.728) 3.53,n3/d^y t2

I m p e r v i o uLsa v e r Fig.4.8 Problern 43-I-With referenceto Fig. 4.8, determine the following: (i) The piezometric headsat the points A, B, C, D andE. (ii) The exit gradient (iii) Factorof saftty againstpiping. Given, G =2.67, e = 0.95. Solution: (i) Initial piezornetricheadat the groundlevel on upstreamside = 4 m. Headdifference 4 - O.5 = 0.2977m Head drop AH = ofhead No. drops 12 Now, numberof headdropsupto the pointA = 3 .'. Head loss at A = (3)(0.2914 = 0.875m. Residualheadat A = Initial head-head loss = 4 - 0 . 8 7 5 = 3 . 1 2 5n . Similarly, the piezometricheadat B,C and D are computed. Piezorrretriclreadat B = 4 - (5) (0.2917\ = 2.542 n.

ar C = 4 - (10)(0.2917)= 1.083rn

Problems in SoiIMechanics and Foundation Engineering

92

93

Seepage and Flownets a t D = 4 - ( 1 0 ) ( 0 . 2 9 1 7 )= 1 . 0 8 3 m The point -E lies in between the 5th and 6th flow lines. Hence, the piezometrir:head at E shouldbe obtainedby linear interpolation. Averageno. of headdrop at

n =

7 !.,8

_ (6.5 x 10-5)(86400)m/day = 0.05616m/day 100

= l.S

- ( 7 . 5 ) ( 4 . 2 9 1 7 )= 1 . 8 1 2 m .

P i e z o m e t r i c h e a d a tE = 4

k = 6.5 x l0-5 cm./sec

I/=18m Usingeqn(4.4),

(ii) In order to find out the exit grad,ient,the smallestflow eletneutuear the downstreamend (i.e.,the one adjacentto the sheetpile well, markedwith hatc:hlines) is to be considered.Averagelength of this element= 1.1 m.

= 0.265

i" = + = ry Exitgraclie*t,

gradient is givenby, (iii) Thc criticalhydraulic . G-t t'=Ti

= 2 . 6 7 - t = u^ ' ^6 -) -t r 1.0.95

... Facrorof saferyagaircrpiping=

= ;

*#

= 3.23

Probfern 4.4r2/'Constructthe flownet and detennilrethe quantity of weir shownin Fig. 4.9. Given,k = seepage lgssirFr6'/dayfor thecollcrete 6.5 x 10-)cm/sec.

_ (o.os61gx18)(4) = 0.311*3/duy 13 Problerrylr*' A concreteweir of 15 m length has to retain water upto 5 rn above G.L. The cross-sectionof the weir is shown in Fig. 4.10. The foundation soil consistsof a 12.5 thick stratumof sand having & = 0.015 cm/sec.In orderto reducethe seepageloss,a 5 m deepverticalsheetpile cul off wall is placed at the bottom of the upstream face of tbe weir. Draw a flownet and determinethe quantity of seepageloss that will occur in one day, if tbe widtlt of the weir be 55 rn. Also determinelhe factor of safety against piping if the soil has G = 2.65and e = 1.08. Solution : The flownet is given in Fig. 4.10. the number of flow channelsis found to be 5, while the number of head drops = 16. ,= k = 0.015 cm./se<

(0'015) (86400) m/day = 12.96 m/day ff

Iy'=5rn

Fig.4.9 Solution

:

Fig. 4.9 shows the flownet. From the figure we get,

No. of flow channels,

Nf = a'

No. of head drops,

Na = ll.

Fig.4.10

94

Problems in Soil Mechanics and Foundation Engineering

95

Seepageand Flownets

This is the quantity of seepage loss across unit widtlt of the weir. Considering the entire width of 55 m (on a plane perpendicular to that of the paper), total quantity of seepageloss per diy = (n.?5)(55) = 1113.75 rn3.

35m

Again, the averagelength of flre smallest flow elernent adjacent to the weir = 1.2 m.

Exit gradien t

i. --

+

=

H N s x l

c6t,

= o'?-6

Critical hydraulic gradient,

- t ^_^ = u't9 i , = 7 #= 2.65 1 - 1^08

=f f i = 3 . 0 4

Factor of safety agahut piping ,/ Prcblem.4y'. A concrete weir of 52.5 m length is founded ar a depth of 2m in a defosit of fine sand for which the co-efficient of perrneabilityi! the horizontal and ver{icul directions are 1.5 x 10-' cm/sec and 6.7 x 104 crnlsec respectively. The sand is underlain by a rock layer at a depth of37 rn below G.L. The high flood level on the upstream side is 18 m and the downstream side has a frbe standingwater table upto 1.5 m above G.L. Draw the flowret and determine the quantity of seepageloss across unit width of tbe weir.

T r a n s f o r m eS d ection Fig.4.11(a)

Solution: As the co-efficieuts of permeability of the soil in the horizontal and vertical directions are different. the
c -frkh =

- 0.668

Length of the weir in the transformed section = (0.658) (52.5) = 35.07 rn o 35 m.

I

I

Fig. 4.11 (a) shows the toansformedsection.The flownet is drawn in the usual manner. In order to obtain thc true flownet, the transformed sectioh, along with the flownet already drawn, will have to be retransformed in such a way tlat all ve*ical dimensions will remain unchangcdbut all horizontal dimensions will be divided by a constant factor of 0.668. In order to retransforiu the flownet, the location of all grid points (i.e., the intersection betu.een flow lines and equipotential lines) should be

Section Origionol Fig.a.l1 (b) nninlc should shnlld then be he-located lc on the retransfom ned determined. These grid pohrts section. ing thesc points in the appropriate order will give the true fl ownet in which all flow elernents will be rectangular.

phreatic iine. The rem:r:fling portion of the basic parabcla is shown wit& a llrokcnlinc.

Fig. 4.11 (b) showsthe retransformedsectionand the true flownel. The dischargequantity is given by, Nt

Nr

F

e = k H ' f r =1 ' l k 6 x k " x H ' 7 U

-+

22n

1

kn = l.s " 10-3cm//sec= 1.296m/day

Now,

97

Seepageand Flo*'neti


96

I

k, = 6.7 x 10-a curlsec= 0.579m/day

30

Il=18-1.5=16.5m Nt = q.a, ff; = 10

*-{"-

I )

(4.6)/(10) q = ,/(Lzs6)(usze)(16.s) = 6.57 ^3/duy It maybr notedherethat,if we areto find out only the dischargequatltity, the true flownet for the retransformedsection is not required to be drawn, as the flownet for the transformed section ciln selve the purpose' However, for the determinationofexitgradient, hydrostatiressureheads,uplift pressures etc.. the actual flownet has to be drawn.

Fig.4.t2 'i;r

ii+m has a top widfh of 28 nt and a :,lrien'-#.9".t A +,-;.ruhigh e.arttrr f-aeeshave.equal slope of frn. i'hr illlstream arlcirfu.n;rilstrearn l'rt:cir<,ar
Problern 4.,7: The cross-sectionof an earthdarn is shown illFig.4.I2. 3 Draw the phrci'tic line. Solution: Horizontal projection.Lof the wetted portion of the upstream face = 27 m. Hence the first point P of the basic parabola is given by, EP = 0.3I = (0.3X27) = 8.1 m. With P as centreand PC radius,draw an arc CM to intersectthe extensionof tle water surfaceline atM. DrawMlV I DC. MN is the directrix of the basic parabola.The mid point Q of CN is another point in thc basic parabola.l,ocate the points YyY2, Y3, ...... , such that they are

E = 2L , :(.:4, ;rr 3U"; = liiC in. -l'he

equidistantfrom the directrix and the focus. P, YyY2,Y3, ...... , aru)Q to obtain the basic parabola. In order to make lhe correction at the entry point, draw a smooth curye from E to meet the basic parabola tangentially. For the correction at exit, the location of the outcrop point is required. Refening to Fig. 4.6, for slope angle An

8 = 45'. ----

a+La

By measurement,

= 0.34.

crcrss-sectionirf lhe dant is shown in Fig, 4.tr3"

f_28mJ tl

T I

a + A,a = 19m. 62 = (0.34) (19) = 6.46 m.

The distanceCC' is laid off suchthat, C C'= 6.46 m. C' is the true qutcrop point. Draw another smooth curve to mcet tangentially the basiarabola at one end and the downstrearn face at C'. The cuwe EC' is then:the required

L.

Fig..i.t3

Problems in SoiIMechanics and Fottndotion Engineering

98

Scagc and Flownets The basic parabolaand the phreaticlitre are drawn in the usual Inanller and the flownet is sketched.Frorn Fig. 4.13 we obtaitt, Nf = 3.2

H = 44 - 4 = 40m.

(0'3)-69)(24) = o.432 n,/day. k = 0.3 rmn./min-

and,

Scageloss ac:ross the entiredam = (39.168)(175)= 6854.4,n3/d^y.

Na=17 Again,

99

Using eqn. (4.4), the quantity of seepageloss acrossunit rvidth,

Problern;pltr Fig. 4.15 shows the cross-seclionof au earth dam -toe cousistirtg of filter at the clownstrearnend. Draw the flow-net auil dtrlerminethe quantityof seepageloss per day acrossunit width of the darrr. The darnis foundedon an impewiousbaseand the rnaterialof the dam has a co-efticie{t of permeabilityof 3.28 x 10-r crn/sec.Explain the procedureof obtainingthe flownet.

@.42)@-G-A- == r 3.zs ' L r r ^3/d^y. l

Solution : Fig. 4.15 slrows the given cross-section,atongwith the l'ltlwnet.Ttrc procedureis briefly cxplainedbelow:

A 20 rn high dam has a top width of 8 rn, a botlorn width Prohlem {rX of 90 nr and a free board of 3 m. The dam is made of coarsesand having a co-efficient of permeability of 0.01 cm/sec.A 22.5 m long horizontal drainage blanket is placednearthe dowrutreamend of the dam. Draw tlre flownet and dcterminethe quantity of seepageloss if the widtb of the danl be 175 m

(a) Locoting thephreatic line : (i) LocatethepointDin theusualmarurer(ED=0.31= 0.3 x 37= 11.1 m). (ii) The bcttom letl hand comer of the toe is taken as ilre focus of the basic parabola. (iii) Draw the directrix and lotute the point p of the basic parabola. (iv) Locate a numbcr of points whic.h are cquidistant from the direc:trix and the focus. (v) thesepointswith a smoothcurve to obtain the basic parabola. (vi) Make the necessarycorrectionsat the entry and the e.xitpoints.

0 =

Solution :

U

The flownet is shown in Fig. 4.14.

9 0 m- Fig.4.14

Lsl loss for unil widt[ Using eqn. (4.4), the quantity of seepn.3qe

\ Q = k H ' Ntr Here,

k = 0.01 cm./sec = 8.64 m . / d a y . H - 2 0 - 3 = 1 7 m . Nf-4,

Na=15

,_r _(8.64) _ T _(r7) _ (4)

39.168^3/d^y.

s' f=_

L Toe 28m_

Fig.4.l5

(b) Construction of the flownet : (i) Draw a vertical line on tlre rigbt hand side of the downstream face. Divide the vertical diStancebetweentbe water level and the outcrop point into any number of equal parts of the length A f/. (ii) Draw horizontal lines from each of these points and locate their points of intersec.tionwith the phreatic line.

solution : The co-efticielt of per-rn:ability of the rnatciial of the dam is is 10 times thet of the foundationsoil; Hencethc flowilet ill the earthdarn be It c-'an as if it were placedon an inpervious foundation' first cronstfuc:ted seeniiorn Fig. 4.16 tSatthe secoudtIOw line hits the intertacebetWeenthe darnaud the tbuldation soil. Now corne.downto the foundationsoil attd cornpletethe tlownet. Thc sccoM flow liue aswell asalllhe equipotentialliileswill enterthe tbundation ,oil ,nO ,ritt undergoa deviationat ihe interface.In thc soil stralum,the flownet is drawn as orthogonallysquared,asusual.However,a.sthe /c-valueof the soil is only 1/l0th oflhat of the rnaterialof the darn,eachtlow channelin the soil stratumcrnie-sa dischargeequalto 1/10thof the dischargec:arriedby a tlow channel in the darn. consequcntly,the tlow chartttelsin the soil are

(iii) Draw ;l nrtrnberolequipotentiailiucs.frorneachof ttrlsc inlcrseciion points.This "rrlll erisurethat the headdrops aie equal' rhe flow lines and equipotenriallines (iu) t ' braw rh. flow lines. Adjust becomesorthogonally squarcd' flowtret thg until agairi and again Using eqn. (4.4),

Here,

101

Seepageail Flownets

Problents in SoilMeclran;icstnd Fouitdtition Ettgiti':xring

100

t' = 3 28 * l0-3 ctn/scc = 2'834 111/day' + 1L = ll 1r.i. A'/ - :'i

nrarkedas2.l ,2.2elc.

Nt -7

From the figure wc gct,

2 (2.834) (23\ (2'3) -= 11 A z1'a2 n'/day' q = -v?'-:i-:-

lotal nulnberof tlow clrannels, Nr = 2.25

I

Na = lo'

numberof headdroPs,

it:, lt Frc[,!,';.:4.11. i ?C In big,lie'anl]dal:ihavitiga toBwidrlr o| 2i} t , . . , . , ' i . 1 : t t i . f f i 4 q : 1 1e l l d ? f r * c b r r a E i ! + f , ! i ; i r f l r : : : . r ' : ! t d s * . o 6 0 n ' . t b i c k

= 0.003 cm,/sec = ?.592 m/daY.

k for clam-material g = 3 O- 3 = 2 7 m .

'ilj! rrr[jr'!r ]., :rnii'lrrai:l bJ'iti-,.i,'r',1er*teableSll:ill Alre i.rj.,1;r;,'-.-!.14;, ii.i,.;.11;,1 tirc Ca|l and{re lou:l'J:'itiali ,:f co-eilii:ii,Bls ;r;:lil*:iilriiify o1 l::, :iiier:ial of Diaw lhe flowner and relptr:t'ive.ly, rrrd crn,,!cs:. 0.f}fxl3 citijsec soi! are 0.{}ti3 m-iriaY' in ioss oi scepage quantity the rietermine

(27\(2.?5J ,. -5 = rr'i m3/day.

(2.592\ q = ----l0

problem yY' Fig. 4.17 shows the cross-sectiottof an eartlr daln of tbunded on a fervious stratum of 60 n thickness. The c--o-effic:icrtt of the while that x cm/sec 104 1.6 qf is dam the material tbe o[ pcnneability itrundation soil is 1.6 x l0-1 crn/sec.Draw the flownet and detennine the quantity of seepageloss in m'lday. solution : Here the tbundation soil is 10 times more Pe.nneablethan thc lnaterial of the darn. Heuce Inore emphasiswill be given ort seepage throughthis soil. Draw a kial flownet in thc. foundatiou soil neglecting the earth dam' Extend all equipotential lines frorn tlie foundation soil into the dam. These lines shouldo.ui.t" from the inlerface,but tlris shouldbe doneonly by hand nnd eqn.(4.11) neednotbe considered.Now draw the flow lines in the dam scction aud try to rnake the flow ilet orthogonally squared. This rnay especiallytbr the last flow line in tbe dam, necessitate"".tuiu readjustments, which should enter intrr the foundation soil. AII previously drawn flow lines in tbe foundation soil may have to be lowered. The final flownet is shown in

-tii;

Fig.4.17.

4 16

li

il

t02

Problems in Soil Mechttnics ttnd Foundation Engineering

Seepageand Flownets

103

120m

-+ 4.sf 7-sr-

2 0 m| _ ''r

i

/

)b:t ,/ 0.2- 7

7:;? k2=10k1

rs, --J

55m Fig.4.18

Fig.4.l7 As tlre co-efficient of permeabilityof the darn rnaterialis 1/10tb of that of foundationsoil, a flow channelin the dam is equivalentto 1/10thof a flow c'.hannelin the foundation soil. Consequently,the flow channels in the daru are rnarkedas 0.1,0.2,and 0.3. No. of full tlow channelsin the soil skaturn = 2.5. Hence, Nf = 0.3 + 2.5 = 2.8.

rvhich the averagelength in the direction of flow is 3 times (sinc.ek2/\ = 3) lhc avsssgswidth. In orderto copewith this condition,the trial phreaticline drawn in zone II may have to be either raisedor lowered and the flownet shouldbc completedby trial and error. Usingcqn.(4.4), N, q = l c H .# .

da=9 . ,1 1 =3 6 m &&for foundationsoil

lrd

= 1.6 x 10-3cm/sec= 1.382m/day.

(1.382)(36) (2.8) = ra4 r ' t4 1 * 3 / d u Y ' n=-, ,/ Problem 4.lQr/ Fig. 4.18 shows the cross-sectionof a zoned earth dam consisting of two zones. Zone I adjacent to the upstrearnface has t = 0.001 cm/sec while zone II adjacent to the downstream face has /c = 0.fi)3 cm/sec. Draw the flownet and determine the quantity of seepageloss in.37d"y. .

Solution : The material of zone II is 3 times more permeable than that of zone I. Draw the phreatic line in zone I arbitrarily. From the interface between the two zones,the phreatic line should deviate downwards, as water can flow more easily in zone II. Draw this deviated phreatic line arbitrarily. Now draw a vertical line and divide it into any number of equal parts of length AIf. From each ofthese points draw horizontal lines to intersect the phreatic linc. All these intersection points are springing points of the equipotential lines. In zone I, draw the flownet as orthogonally squared.Each t'low line enterszone II after undergoing a deviation at the interface. However, lhc flow elements in zone II will not be squaresbut rectangles, for each of

Now,NJ = 2.J, Nt = 6, H = 37n. k ftrr zone I = 0.001 cm/sec. = 0.864 m/day.

, = @P@

= 832 ^3/d^y.

It nraybe notedthatthemethodsemployedinProblem4.11through4.13, ht)wcvcr crude they may seem to be, will yield results which are within r.l0%, of the results otrtainedby a rnore accurateand vigorous solution.

EXERCISE4 4.1 On a waterlrout,a sheetpile wall of 8 m height is embeddedinto tht: soil upto 6 m bclow G.L. The free board is 1 m while water on the dowrrstreamside standsupto 2 m aboveG.L. The foundation soil consistsof a 15 m thick sand slratum (t = 0.009 crn/sec)which is underlain by an inrpcrviouslayer. Draw the fiownet and determinethe quantity of seepage Iossncrossa I m wide sectionof the sheetpile.

lO4

Probtems k Soil Meclmnics and Fottndation Engineering

Fig. 4. 19 shows tbe cross-section of a concrpte weir. The 4.2 3 foundation soil has a co-efficientof peimeability of 1.25 x 1.0 cm/sec,avoid ratio of 0.88 and a specific gravity of solids 2.65. (i) Sketch the flownel liij O.termine the quantity of seepageloss in *3/dry (iiD Find out the factor of safetyagainstpiping.

105

Seepageand Flowmets

4.5 Draw the phreatic lines for the ctamsectionsshown in Fig. 421(a) through(d):

Drainoge Btqnket

(b)

(o)

C hi mn e y Oroin

t ,2m

"Tffi 'f-6

30m

-|

o.st

|

a3 (a) If a sheet pile cut-off wall of 3 tn depth is introduced in the upstream end of the weir shown in Fig. 4.19, determine the percent reduction in the quantity of seepageloss. (b) If in addition to tlis, another sheet pile of 2 m depth is placed at the downstream end, how will the seepagequantity change?

i00m -------{ (d)

(c)

Fig.4.19

_6fuN -r N__aom J

Fig.4'21 4.6 Sketch the flownets and detennine tbe quantity of seepageloss the aqrossunir width of the earthdarnsslrowu in Fig. a.22(a) through (c)' All = dams are foundedon impervioussoils.Take & 0.002 cmlsec' ^

rJ1^n L-

L .3m

lzom

4.4 Fig. 4.2Oshows the cross section of a concreteweir founded in an anisotropic soil mass.The co-efficient of permeability in the horizontal and verticat directions are respectively 5 x 10* mm/sec and 1.25 x lO--rnm/sec. Sketch the flownetand ditermini the quantity of seepageloss in m3/hr'

i

i

J--'!

!

II

2m

T

I

f_ro.sm__]

16m___l

Fig.4.20

|

i-- 60m-{ 1s0m---------J (c) Fig.4.22

II II

I

II

I

"rd

106

Problems in Soil Meclwnics ond Foundation Engineering

4.7 construct the flow'et for the zone earth darn shown ,tFig.4.23. Hence cornputethe total quantityof seepageloss if the width of the re-servoir bc 125 m.

f

5 =0 ' 0 0 2 cm/ s e c

k =0 , 0 1 rm/seE

asm---J--ism

---J

Fig.4.23 4.8 A'earth darnhaving an overall height of Lg rn, a top width of 10 m and a free board of 1.5 m is founded on layer of clean sand having a co-efficient of permeabilityof 0.01 cm/sec.A rock layer existsat a depth of 45 rn below the ground level. The earth-fill in the dam has a co-efficient of penneabilityof 0.00.2crn/sec.Draw the {low net and detenninethe quantity of seageloss in m'/day. '

4.9 Solve Probrem4.8 assumingthat the co-efficient of permeabiliry of the materialof the dam and the founctationsoil are l5 m/daylnd 5 m/dav respectively.

STRESSDISTRIBUTION S.l. Introduction: The applicationof an exte'ral load on a soil mass r(:sults in an additional vertic:al stress (i.e., iu addition to the existing ovcrburdenpressure)at any point in the soil. The rnagnitudeof this stress dccrt:ades with increasingdepth anclincreasingradial distancefrom the line o l ' a c t i o uo f t h el o a d . The stressconditionsin a soil massdueto externalloadsc-anbe analysed by thc theory of elasticity,assurningthe soil to be a perfectlyelasticmaterial which obeysHooke's law of proportionalitybetweenstressand strain. 5.2 c)verburden Pressure: The overburdenpressureat any point in a soil nrassis defined as the initial vertical stressdue.to the self weight of the soil nrass,and can be obtainedfrom

o-' = Y'

...(s.1)

wltcre, Y= uuitweightof theSoil z = depthbelowgroundlevel. 5.3 Stress Increment: Thc stressincrenrent,Ao-, at any point nray be defilcd aslhe increasein verticalstressdueto the applicationofexternal loacl on thc soil ntcss. Thc total stress,o- , at ally point 4 at'terthe applicationof au extenral krad,is given by,

...(5.2) s.4. Boussinesq'sEquation: Boussinesq'smethoclof determinationof slrcssincrementdue to an extemalload is basedon the assumptionsthat the soil mass is elastic, hornogeiret'rus, isotropic and serni-infinite while the cxlcrnal load is c-oncentratcd at a point and is appliedon the ground surface. With referenceto Fig. 5.1, at a depth z below the ground level and at a radial distancer from the line of actionof the load p, the verticul stressatr. is given by

108

ProLrlemsin Soil Mechanics and Fourdation Engineering

109

StressDistrihution C} A o , = K"o .z' +i

and

...{s8)

5.5. lYestergaard's f,'4uation: Westergaardassurne
...(s.e)

p = Poisuon'sratio of tfte soil. It lbr a given soil, p = 0 , 9qn.(5.9) reducesto Ac-.=

7 Fig.5.1

...(s.3)

A oU A . =:9 = l # 1, 1st2 lt ; T\ )' /' lJ

r

r

13i2

,..(-{.x0)

ij!{ il;*;i !:: ::'j,i:.i; 5 ti. 3 : I *isp';rrie:: ,1!*ti:r:ri : il' +n api.,;:r,.;i!11,;,ir': it iusrssuni':ii iiiri iir.,',:ii+{:t -*ian exlernal l*acl is ilisi',,,'r,:ri ;'ii;1'-t*iraigirt iiri*r.; irrclirredat 2l' : 1Il.

P

e+ Ao - Z- = f il q R '

a

tt; ' i*---J--:---l . xEt . r(;)"j l, '?'!;js

= 17;77

or'

ftt - zfi/(Z zv)+l/izfzn

i7

whc.rc,

r = {/7Q R = {7;7

or'

{6/e-rtl)

a

...(5.4)

trtiith re'l!:isllr.' 9irFig. 5.3, i.,::i.lil1r5i!3;1.r;l;'.1rif : ,1:i)grpliii: r::tf r,rt, S i

(i BL. Ar a depthe, thisloadis distributedo?cran area(I + z) (B + z). Hence, s.he.ss intensit-vat tbis level. . = Au'

oBL

&6tE;4

...(5.11)

..(s.s)

A o -= K r ' q z-

whcre, KB

is called Boussinesq'sinfluencefactor and is given by,

3 f 1 1 xa = 2n'l;/rf"l

l'

'lzl

s n

'(s'6)

I

The tangentialstressA r.. and,0" ,loru,trlr.r', o o, at rhesamepointare givenby NEr, =

KBT

,..(s.7)

r -..{*r3 {l rrJ-

l

1

1 Ij

Fis.5.2

A

110

Stess Distibution

Problems in Soil Meclunics and Foundction Engineering

(3) Determine the radii of the circles frorn

5.7. kobar: If the vertical stressintensitiesat various points in a soil mass due to an external load are computed, and the points having equal stress intensities are ed by a smooth curye, a number of stresscontours, each having the shapeofa bulb, are obtained.Theseare called isobars.The zone in the soil rnassboundedby an isobaris called a pressurebulb. An infinite number of isobars can bc drawn. The zone contained by the isobar corresponding to a stressintensity v.,hichis equal to lOVoof the applied stress is taken to be the zone within which appreciableeffecls of the applied stress may oqcur. This zone is termed as the zone of influence.

. t 1 l - t ) where,

'

whcre,

...(s.16)

m = total numberof radial linesto be drawn.

(6) Draw the radial lines with tbe deflection anglesthus calculated, With the help of the Newmark's chart (Fig. 5.15) the stressintensity at nrrypoint dueto a uniforrnlyloadedareaofanygiven shapecanbedetermined ns follows: (i) Adopt a drawing scalesuchthat the depthat which the stressintersity is required is representedby the numerical value of z on the basis of which Ntrwmark'schart is drawn. '(ii) Draw lle plan of the toadedareaon a tracing paperwith this drawing rcxlc. l.ocate the point P below which the stressis required. (iii) Place the tracing paper on the chart in such a way lhat the point P on thc tracing paper coincideswith the centreof the circles. (iv) Count the number of elementscovered,fully or partly, by the plan of the area. (v) Calculate A o" as:

Ao, = 4l(pirr + e/z).N2+ e/3]gf

...(5.13)

whcrc,

ill

...(s.17)

= number of elementsfrrlly covered.

N2 = number of elernentshalf covered.

II

1y'3= number of elementsof which one-third is covered.

L

...(s.14)

t/ = influence factor

c = radiusof the outermmtcircle

= l/no,of

L a, = stressat a depth z due to the entire loaded area. (2) Selectan arbitraryvalue of z (say,z = 5 cm).

r; = radiusofthat circle.

m

ln order to preparethe chart, proceedas follows: (1) Select the number of elementsinwhich the chart should be divided, and determine the influence value for each element. For example, if 10 concentric circles and 20 radial lines are drawn, tht: number of elements = (20) (10) = 200. When any one of these 200 elernents is loaded,the shessintensityat a depthz is given by,

where,

...(s.1s)

n = total nurnberof circles to be drawn ni = number of tbe circle whose radius is required

0 = $q-

...(s.r2)

or, = 0'005 Ao"

2 =n - ; n

(.5) Deterrninethe deflectionangleof the radial lines from:

5.9. Newrnark's Chart: The stress intensity at any point due to a uniformly loaded area of any shape can be determinedwith the help.of Newmark's influence chart. It consists of a series of concentric circles of various radii and a seriesofradial lines drawn at regular angular intcrvals. The total area of the chart is thus divided into a number of elernents.The elementsmay havedifferentsize.s,buteach ofthem, whenloadedwith a given stressintensity, will give rise to the samevertical stressat a given point.

ort

3 / * l Ir:\tl

(4) Draw all the conceutriccircles.

,ffi'"1

o,"=o#=#['

)

l1+ l-l I \'/ ) [

Str"rs due to a Uniformly Lcaded Circular Area: Frorn {,8/ Boussincsq's equation it can be proved that the stressintensity at a depth z below the centre of a circular area of radius 4 which carries a uniformly distributedload q, is given by, Lor=q

1i1

I

elementsof the chart.

t. 10. StressDue to Vertical Linear I-oad: The load resulting from a long bul nanow wall, or a railway track, is an example of a vcrtical linear load

Ll?

Probiems in Sai!Mechanics trnd Foundation Engineering StressDistributian

tL3

(Fig. 5.3 a). The vrriical strcssat a depth z andat a.radial dirtaicr of ,- &+m the line of action of such a load of intensily f tlm, is given by,

_T

rc==;ffry=#l,tb=]'

...(s.18) q

-oc

\ \

/unit qreq

- --

I

l l

\.nzl

d,1 {1 --.q

. i \

\i

l

7 N I

J

\ B. \-1 -..o(-r. i -y_ l!

Yt'{

'.. -.: {

P ( y , z)

z Fig.5.4

5.13. Embankment Loading: Stressesin subsoils due to embankment loadings can be computed using eqn. (5.20). L€t it be required to comput€ the maximum vertical stressintensity at a depth z below an embankmenthaving a top width 2b, abasewidth 2(a+ b) and a maximum intensity q t/m. The solution can be obtained in the following steps; 1. The embankmentis divided into two equal parts as shown in Fig. 5.5. Two symmebical Eapezoidalloadings are obtained.

A a, =

q

i{n

...(s.1e)

+ s i nc r c o s( c r * . 2 p l)

ITITil] , I'" L-"-l oJ i-"J-o-J F,i

5.1?" r'i*stirtr!StFeii:i li , 1.,'fr!,iiio::li+ z'Y.-+*{a,:Ii,,::u.rr:r:li:i:'.r,:i r , t l ; g i : 1 . ' r . 1 , i.C i i 6 i 1 ; , ' . , : i : e d t i ' ' " . . ,r ' , i a t * r l l r ; r - ' ; i a l i i , i r r $t ! ' . . . : r ' l ' c 1 .+qi . . . r : rl--r',. loadings. : . 1 ; . . ;: r . ' ! ' : .

.

l I

+

$r

a

q -,

t ,it..

i l 1

-ra_r.r .' fq i-'"'i

I

f

I I

"./2!l

ql-t

Fig.5.5

t b\

-{

P

I

r

I

\ \

.

'{l

\--

b

c f eI r n1 ill t-.L-J

.

I

t I

ss tr't

1iP

i j

1i4

Problerns in Soil Mcclmnics and Founcletion Engineering

StressDistribution 2. Thc trapt:zoidABCDE is cxtcnded to fonn L AFD by adding an imaginarylriangular load BFC. 3. Stress intensity at the point P (lying below the ceutre of the enrbankmcni\due tr the triangular loadsAFD andBFC are obtained.The differencebehveenthesetwo givesthestressintensitydue to half theembankment. Hence for the entire ernbankrnentthis intensity has to be doubled. 5.14. Influence Line for Vertical StressIntensity: The combinedstress interuity at any poitt due to a systernof extemal loads can be detennined usrng the concept of influence line diagrams.The melhod of drawing an influencelinc is explaincdbclou,: 1. Considr:ra unit load aptriliedon the grouud surfaceat O. 2. Considera horizottal plaueMrV at a depth zbelow G.L. .3. Using Boussinesq'sequationdeterminethe vertical stressintensities at various points on Mif due lo the unit load. 4. Choosea vector scaleand lay off the correslxrndingordinatesat the rrspeclivt: points to represenlthr computedslresses. 5. theseordinateswith a smoothcurve. This is the influenceline.

115

EX.A,MPLES ,/ Problem6/J. A concentratedload of 40 kN is applied vertically on a horizontal ground surface. Detennine the vertical stress intensities at the followingpoins: (i) At a depth of 2 m below rbe poinr of applicarion of the load. (ii) At a depth of L rn and at a radial distanceof 3 m from the line of action of the load. (iii) At a depth of 3 m and at a radial distanceof 1 m from the line of action of the load. Solution:

We have from eqn.(5.4),

3Q ao, = ?n? (i) Here,

qt'l

Q = 4OkN, z = 2 n r/z=A

|

1s"

and r=O

Henceeqn(5.4)gives (3)(40] tQ" = Lo, = = 4.'17kN./rn2 (2n) (2') 2nt (ii) Inthiscase, O=40kN, z=lm, r=3m

Ao,= ffi (iii)Here,

O=40kN,

t-tt]'n

z=3.m,

,,=t*

(3)(4ol -Lo" l-l--ft" '/ = (?-n)Q)z lr * lrttl2l

Applying Maxwell's reciprocal theorem,the stressat any pointdue to an externalload can now be obtainedas follows: (i) For any given load P acting at O the sfressintensity ar any pointx is obtained by multiplying the ordinate of the influence line at X by the load P. (ii) For any given load y' acting at X, the stress intensity below O is obtained by multiptying the ordinate of the influence line atX by the load P'.

=o.o6kl.r,/m2 = r.63kN,zm2

Problem 5.d e,reaengularfooting,Z m x 3 m in size,hasto carry a uniformly distributedload of 100 kN/mz.plot rhe distributionof vertical stess intensityon a horizontalplaneat a depthof 2 m below the baseof footingby: (i) Boussinesq's method. (ii) 2 : 1 dispersionmerhod. .s9!ution: (i) Boussinesq'smethd. Thc uniformry distriburedroad carriedby thefootingis to beconsidered asaconcentratea toaaactingthrough the centreof gravity of the footing. I = q BL = (100)(2) (3) = 6m kN.

r1,7

Stess Distibutian

tt6


Using eqn. (5.4), the stress intensity at a depth of 2 m and at a radial distancer from the line of action of Q is given by:

La,=#l;4'" = 1rr.ut) [--+-l''

l1 + (r/zYl

A nurnber of points are chosenon the given plane and the sfress intensity at eachpoint is calculated.Thesearetabulatedbelow:

r/z

0

0

t 1.0

t 0.5

t 2.0 -r 3.0 * 4.0

t 1.0

*. Z0

= 5.0

x.2.5

< /', 1 |l--------------;l 1 - ' ' StessAintensiry s,

lr + (r/z)'I

t 1.5

1.0 0.572 a.n7 4.052 0.018 0.007

^

L

E O U S S IENS O , S METHOD

DISPERSION METHOD _\_ I

f t z . o ei

I

ll12.66

&Nlm2 ) 71.62 4.97

Fig.5.7

12.68 3.72 t.28 0.50

(ii) 2:1 dispersion methodt Using eqn. (5.11),

(100)(2)/?\

sB =-----*=30kr\,,m', (2+21(3+2) Q + z)(B + z) The distribution of stressintensitieson the given plane, as obtained from Boussinesq's equation and 2 : 1 metlod, are shown in Fig.5.7. -A o- z, =

I

f-l-fT-[fTT-l = I I I i r i I t i 11 100kN / m2 ^r*_

,o.r*

= (3)(600)-t-l-=lt" (2)(3.14)(2') 11* {rl-l [ \'/l

Radialdistance r (n)

i Q= 600kN

The following differences betweenlhe two stressdistibution diatryams arc to be noted : ii) The maximum stressintensity obtained from Eloussinesq'sequation is 71.62 kN/m', which is nearly 2.4 times the maximurir sEessintensity given by2:1method. (ii) In Boussinesq's method, the maximum stress intensily occurs ditectly below the point of application of the load. The stress inrcnsity decreasesrapidly with increasing radial distance, but except at an inf,mite distance, it never becomeszero. Whereas,the stressiniensiry given by 2 : 1 method remains constant over a radial disganceof 2.5 m on either side of the cente line of &e loaded area. and suddenlv becomesnon-existent bevond rhat lirnit.

method (iii) Evidentty,thepressure distributiongivenby Boussinesq's is morelogicalandshguldbe usedin ordinaryfield problems. ,a

Problem 88t' A concentratedvertical load of 200 t is applied on the surface of a semi-infinite soil mass. PIot the disnibution of vertical stress intensity on a vertical plane situatedat a distanceof (i) 3 m (ii)5 m, from the line of action of the load, Solution: In Fig. 5.8, let P be the point of applicationof the load. kt Y1Y1 and Y2Y2 be the given planes,locatedat radial distancesof 3 rn and 5 m respectively from P. Using eqn. (5.4), the stressintensity at a depth z and radial distance r from the line of actionof a?N t load is,

(i) Lo,=ffi[;io1'"= r lr;*1"' 95.49 )

The stressintensitiesat various points or the planesY_1Y1and Y2Y2may now be computed ftom eqn. (i). The resultsare shown in the following table.

l

I

1 l

,i

II j

'l

l--*.

A

I


118

No.of point

Deptlt (m)

Plane Y2 Y2

PlaneYlYl

r/z

L, o, (t/m2)

r/t

A,o, (t/m21 0.004

r.254

r0.00 5.00 3.33 2.50

t.2

r.643

2.{JO

a.273

3.0

1.0

1.875

1.67

0.380

4.0

0.75

1.955

r.25

0.568

5.0

0.60

t.770

1.00

0.685

9

6.0

0.50

1.518

1.20

0.28-s

10

8.0

0.375

t.473

1.60

0.062

1

0.5

6.0

0.06

2

1.0

3.0

0.302

3

1.5

2.0

0.159

4

2.4

1.5

5

2.5

6 x

0.028 0.083

StressDistribution

consistsof a 6 rn thick clay laycrwhiclris uudcrlairr lrv a l;rvtxof denscsand. Plot lhe dislributionsof: (i) Overburdcnpressur(' (ii) stressincrementdue lo footingloads (iii) grossstressintensity on a horizontalplanethroughtlre middle of the clay layu.'Ihe propr:rtiesof the clay are as follows: G =2.70. e =0.55. w=0,)'t Assunrethe footingslo be frrundedal lhc grtlutxl lcvt ;.

0.r68

The pressuredistribution diagramsare shown in Fig. 5.8.

Solutiern: (i) Overburdenpressure:Using cqn. (.i i , :i ,' .irr.rrburdt'u prr:ssurcal llrc nriddlc of the c{ay layer (i.tr.,I = 6,') = t ,', givct by, o.^ = YZ

t'r',, _ ( 2 . 7 0 ) {ot r .,, = |+t, I**l'=l.74tzrn' . L , = ( 1 . 7 4 ) ( 3 . 0=) 5 . 2 2t , n l ,

Now,

The intensityof this;rrcssun:ovcr lht'lrr ri.'orrlalplanc.YYtirroughthe nriddleof clay is conslant. (ii) In order lo dctt:nttitttr fhc strcssillr'ir't'rr'tlldur: l
Verlical stressintensity due to : Pt (1004 r/z (n) a

I

2 J

)

yr FE.s.8

y2

" . / The footingsof threeadjacentcolumrsof a building lie Problem 5/ on the sames{raightline and cary grosstoadsof 100 t, 150 t and 120 t distancebetweenthe t'irst and second respectively.The centre-to-centre footingis 4 m while thatbetweenthesecondandthethird is 3.5rn.Tbesubsoil

119

6 7 8 I l0 1l t2 l3

-l

Pz (Iso t) r/z

aozl (t/n2)

(n)

-o.6? 4 -o.33 4.08 -5

0 531 I 0.33 4.08 2 0.67 2 . r 2 3 1.00 0.94 4 1.33 0.4r 5 1.67 0.19 6 z.M 0.09 7 2.33 0.05 t-) 2.50 0.04 8.5 2.83 0.02 9.5 3 . 1 1 0.01 0

A

-3 -l 0

I

Pt (l)) t) Ao:2

(tln")

(n)

Att

l+ A,,

'u' ao,, I l*A'r', 2)l{r',r:I !ttt^21lttt^21

+--

-2.00 0.14 -95 -3,1; I g.g2 ) ' ) A 3lsrl 'r I s ',' -1.67 o.29 -a.5 -2.8-1| 0.03 4.4(l -r33 0.62 1 < -2.5( I 0.04 5.97 -1.00 t . 4 l -4.5 - 2 . r i | 0 . 0 8 5.57 / |)-' il<)' :' 4.67 3 . 1 7 -5.5 - l 8 r | 0 . 1 6 5.4-s t-*-o.33 6 . l l -45 -i.5(l | (,.33 7.38 r | 5.2.: 0 7.96 -3.5 -1.fi | 034 9 . 1r 1I < : '

1il: ;i ; : : ,l;;; ,l ; ; ,

3.5

0.33 6 . 1 1 J < -o.rJ-r | 1,72 8.02 o.67 3 . t 7 -1.5 -{..5(}3.64 7.76 I 1.00 t . 4 l -o.5 -o.r7 I 5.e3 7.:t() ''-| 't t . t 7 0.93 0 0 i 6.37 .3.1 ' 5.22

4.5

r.50 0.42

1

).)

1.83 0.20

I

I

;

|

t s.22 5.36 I 0.r'. i 1.5-l 2:7lI I 5 ) ? 0 . - 1 3i 4 . 9 2 i

Problems in Soil Meclrcnics and Foundation Engineering

t20

t21

StressDistributian

0.0764 (3)(1)_ A ov-:=A l _ _ J . 1 t " = ,-______3n e r , ) ( zl.rr_z4 1 (1 + 0.16f) 1 L tz'sY \

Lro=,*ffih?l = and,

,

1

0.0306r

,

5.22ttn? X

0 v e r b u r d e nS t r e s s

The computed values of A o, and A t* at various points on the given horizontal plane are shown below:

No.

r (m)

a6t

(tlm21

1. 2.

S t r e s sI n c r e m e n t

5.

(\l

rP o\

o GI

o\

r\,

l--

\l)

o

o

o

.9

(\

fn ''1 -t

*t q{ tn

cO Q GI

\D

({

rJ1

*

ct ql o

r.rl

s

(2.srJ

[

ct52

(t + 0.t6r)

4. 5. 6. 7.

G r o s sS t r e s sI n t e n s i t Y V { e c t o rs c q t e '1: c m= 1 0t / m 2}

0 r0.25 r 0.50 r 1.0 r 1.5 r2.O * 3.0

No.

Lro

r (n)

ho,

@n?1

0 0.0764 0.0745 r 0.0075 0.0693 t 0.0139 0.05n t 0.0211 0.0354 rO.0213 4.0222 * 0.0178 0.0G2 * 0.0099

L 9. 10. 11. 12. t3.

t 4.0 t 5.0 t 6.0 x.7.0 t 8.0

* 9.0

0.0032 t 0.0051 0.0014 x.0.0027 0.0006 tO.0015 0.0003 * 0.0009 0.0002 * 0.0006 0.m01 * 0.0004

The distribution of A o" and Lr.-arc shown in Fig' 5.10'

Fig.5.9

(iii'STotalstess.Tbetotalstressatanypointis tlie sumofthe over-burdert pr"r*r" andthestressincrementdueto footingloads.Tbesearetabulatedin the lastcolumnof the abovetable. The distributionofoverburdenplessurc,stess incrementandglossstress intensityareshorvnin Fig' 5.9. prcblem { Draw thc influencelines for thevertical andshearstress of 2.5 m below the groundlevel dueto a unit vertical a depth at interuities on the glound surface. applied loads concentrated Solutlon: We havefrom eqns.(5.6) and (5.7),

-4

0 0 . 5 t | . 5 2

c\l

I

I n f l u e n c tei n e fa verticolstress

La,-#l-+if"' Lao'#lTi7l'"

and Here,

-5

Q=Lt"

z=25m

Lto

ftln?1 ftn?)

l n f l u e n c el i n e f o r s h e o r s t r e s s ( s c o l e: 1 c m= O ' O 3 i / m)2 Fig.5.10

122

Psoblems.in SoilMechtnics and Foundation Engineering

/ P.oblem f.6. usi'g the i'fluence ri*e plotted in Fig. 5.r0 plot the distribution of vertical stress intensity on a horizontal plane through the rniddle of a 5 rn thick clay layer dueto theloadingschemeshown in Fig. 5.t t. Solution: At any point on the given plane, the combined stress rntensity can be obtained by summing up the stress intensities due to the iudividual loads,whiclr, in tum, rnaybe deterrninedby the processexplained in Art. 5.14.

StressDistibution

Prnblem Vrt ft is proposedto constructa sftip footing of 1.5 m width to carry a load
Vertical stress itiletu;ty ilue to .

Dist.

Dist.

fro the origin (m)

(m)

0

-2

I

-l

0 t

I

4 5 6 7 8 9 10 ll

3 5 6 7 8 I

Pz (80 0

Ordinate of ILD

A Ozr

(t/^:)

(t/m2)

0.0222 1 . 1l 0 0.0527 2.635 0.0764-3.,-20 0.0527 2.635 0.Q222 l l 1 0 0.tK)82 0 . 4 1 0 0.0032 0.160 0.00!4 0.070 o.(xn6 0.030 0.m03 0.015 0.0002 0.010 0.0001 0.005

Dist.

L azz

(tl^21

(t/m-)

(m)

-5 4 -3 -2 -l 0 I 2 3 f,

6

Refore the construction ofthe footing, stressintensity at a depth of3 m below the centre line of the footing is given by,

Pt (65

Ordinateof II.D

0.0014 0 . 1 1 2 0.0032 0.256 0.0082 0.656 o.0222 t.776 0.0527 4.216 o.07u 6.112 0.0527 4:216 0.0222 t.776 0.00a2 0.656 0.0032 o.256 0.0014 0 . 1 r 2 0.0006 0.048

Dist.

(n)

-9 -8 -6 -5 4 -3 -2 -l

0 t

2

The butk densityof the sand,

G + se 2.68 t (0.10X0.6s)(1.0) = r.66t/m3 t = 1 - ; Z ' Y . - ., _ 1+0165

The following table shows the computed values.

Pt (s0 t)

Ordinate of il,D (t/m2)

a oz3

Qzo=lz

! Au,

= ( 1 . 6 6 ) ( 3 . 0 )= 4 . 9 8 t / m ? . (t/^2)

0.0001 0.006 0.0002 0.013 0.00c1 0.019 0.0006 0.039 0.0014 0.091 0.0032 0.208 0.0082 0.533 0.0222 r.433 o.0527 3.426 0.o7u 4.996 o.0527 3.426 0.0222 1.443

(tln2)

r.23 2.91 4.40 4.45 5.42 6.73 4.91 3.29 4.tl 5.24 355 1.50

Skess increment at the sarnelevel due to the construction of the footine may be determinedusing eqn.(5.19).

A o , = 1 , r o * s i n c c o s ( c r+ 2 p ) l 1l' The maximum stressintensitywill occur directly below the centreline of the strip load. With referenceto Fig.5.3 (b).

u=ran-rf+) = o'245tad' I

p =0

and

1Z 1l

l

o

i

1

t

A, = Oro+ AO,

laor

t

2

10.245+ sin (0.245)cos(0.245)l

Hence, lotal stre.ssintensity after the constructionof the footing,

Isot -

\ J, ' /

= 1.83t/rYr2

The distributirxrof vertical stressintensityis slrown in Fig. 5.11.

-

1"23

X{m)

t

,

s

//

6

-4.98+1.83 =6.81t/m2

Prcblemf$ Two long boundary walls of small width run parallel ro each other at aYistanceof 3 m apart.The self-weights of the walls are 25 and 15 kN/m. Plot thc distribution of vcrtical stess intensity due to the walls on a horizontal plane 3 m below the ground level.

i

Z

L S c o l e : 1 c=mJ t / m 2 F i g .5 . 1 1

Solution: The point of applicationof the 25 kN/m linear load is chosen as the origin. The ground surfaceserv€sas tlre l/-axis while tie verticalaxis through the origin is the Z-axis. Mif is the plane under consideration.

124

Probiems in Soil Mechanics and Foundation Engineu ing

t25

StressDistrii;tttion Usirrgeqn.(5.18),the verticalstressintensitydue to a line toad q is givc.n by:

A-

Llu-

=

2q I I f xtz it + \rtzl.l

.+.l------;l

The stressintensitiesat variouspointson the plane arecoinputedand are presentedbelow in a tabular form: No. Dist.

from origin

Stressdue to 25 kNlm lood

r (m)

r/z

Ao:r

'Stress due to 1,5N/M Ioad

r (m)

r/z

(kNlm2)

.)

a

-1

-1

J.

0

0

4.

1

1

2

)

0.67

6.

3

J

1.00

1

A

4

1.33

)

5

r.67

0.37

1. 2.

8.

Loz2 (kNlm2)

Totai ,9tress Lor, + Loz2

2.53

-)

-1.67

4.3r 5.31

*1.33 0.42 -"1,00 4.79

4.73

0

4 -3

4.31

-z

4.67

1.52

5.83

2.53

-1

4.33

2.56

5.09

1.33

0

0

3.18

4.51

0.69

1

0.33

2.56

3.7

2

0.67

1.52

1.89

0.33

t-'ythe ;;r;itrtsr{, 6 tiid C respective.ly'

(wlm2)

4.67 -o.33

a.z2

t Prcbiem 5.9. A long flexible strip footing of 2'5 m yiqt!-h-1uiog run. kN/m 80 of load smooth i;.irir, is subjr+-:irriio a uniformly distributed Detenuiuc the vertical stressitttettsitiesat a depthof 2 rn below: {i) ce.ntreline of the footing (ii) side fac:eof the footing line of the tboti.g at a dista'ce of 3 m liiii a li'e paraliel to the ce'tre liorn it. in Fig' Solutlon: The r:ross-sectionof the given tboting is presented ..*hich repre$ented the siressesareto be determinedare :i.12.1"Lciocations;il

2.75 6.10

The distributionof vertical srressis shown in Fis. 5.12.

2 5k N/ m

I 5 k N/ n r 5ggls:1:50 Fig.5.13 By measurtmertt o. * 64' = l.1L? radian *0'558 radian l} * *J2' =

{i'; Foittt A:

usi-cgeipt. i5.19), ai!

A 6 , r ; r = : ' ' - i 1 . 1 1 7+ s i n{ 1 . i 1 ? }c o s{ 1 . 1 1 7+ :i.

Fig.5.12

= 51.3:1kN/n::. (ii):1'n;r*t8; Here, u = 51.5" = 0.899radian'

F=0"

2 (- o.sssill

Problems in Soil Meclmnics and Foundation Engbreering RN

A o, 16)=

;

StressDistribution

I 0.899 + sin (0.899)cos(0.899 + 0) l

Now,

= 23.96 kN,/m'. s, = 26" - 0.454 radian.

Of,

F = 36.5" = 0.637 radian.

Lo,(r)=

if

Usingeqn.(5.20),stressintensitydueto triangularloadingis givenby,

A . 4 5 4 +s i n ( 0 . 4 5 4 ) c o s ( 0 + . 4Z5.4. x 0 . 6 3 7 ) l

q { y A o-, = - l i ' a /J.r\D

= 9.81 kN,/m2.

For

Problem 5.10. An embankment of trapezoidal cross-sectionis to be constructed by compacting a soil at its oMC (l6vo) and the corresponding 12-"* (1.55 Vm'). The top and bottom widths of the embankment should be 10 m and 22 m respcctively while its height should be 3 m. The foundation soil consis6 of a24m thick layer of soft clay. Detennine the maximum stress intensity at the middle of the clay layer due to embankment loading. Solution: Bulk density of the compactedsoil,

oo", = For

LBFC, .

oo,, =

Hence, the rnaximum surchrge intensity of the trapezoidal loading = (1.80X3.0)=5.4t/m'.

e = 4.5Vm, a. = 2l.5' = 0.497rad., B = g y/b=2 AS

fitz

* 0.497- o) = o.ltt/m2

1

c.T

\

c \

If 10concenbiccirclesof aipropriateradii aredrawrqtre areaunderthe chartwill be dividedinto 10parts.It will thenbe frrrtherdividedinto smaller segments by drawinga numberof radiallines.Evidently,thenumberof radial . z f f i = ttnesto be drawn= 20. lb-

l

l

A

--_ 611

---t. -\ I

-

lp

2E.s'}'l oP I

L i n e q rs c q l e: 1 : 7 5 ,V e c t o rs c o l e : 1 c m= 0 . 2 t l n ? Fig.5.14

l

N-i-o'*5'2ou

r--

5 ' 4 ^ --

L--

" 1.065- 0) = 3.36t/^2.

Aoz = L,or, - O arz- 3.36 - 0,71- 2.65t/m2. -/ Problen lJf/ Draw a Newmark's influence chart on the basis of Boussinesq's equition,fdr an influencefactorof 0.005. Solution: Dercrminethe total numbcr of divisions in the proposed charton thebasisofthe giveninflucncefactor.

F

0

fitz

.

Fig. 5. 14 shows the embankment cross -sectionABC DE The imaginary triangle BFC is added to complete the triangle /-ED.

E

oo

.'. Netverticalstess intensityat themiddleof clay layer,

- 1.55(1 + 0.16) = l.80/m3-

-

- s i n- 2' \) p l .

e = 9.9t,/m, a = 61o = 1.065rad.,0 = 0 y = ?-b, i.e., y/b = 2.

LA.FD,

y = yd(I + w)

llmt I

FC = BC BE AE, . BC (5.4)r' s) BE nA = = =-J= 4.5t/m. nC ln FD = FC + BE = 4.5 + 5.4 = 9.9t/m.

I I i

3!0' = The anglcbctweentwo consecutive radiatlines= tg. 20 - Now, usingeqn.(5.72),theverticalstressat a depthz belowthe cente of a circularareaof radiusr, carryinga uniformlydistributealoadg is given bv:

l

Frr.iblcmsin Soil lr{edtttnics and Foundatian Engineering

128

r

_ Ji-__L_ --

Ao-= all

" I

L29

3'2 I

I r * 1 ' z ' t 3 iJ

t Ao. _ = = 1 _ q

0f'

I

-

StressDistibution

[ J---

7/)

1

l"'" " I

lr + (r/z)"I ", /?

ao. ) a f

{l

= l r'

0fr

l + (r/z)2

"'-

-) /7

! + (riz)z

of'

=i 1 I

a o, \, -

t

, l i

/*-Ao.\-'r,.'r n"!( -'. - = Vll*-, ' . 4 I1 ,

or,

..'(s'21)

A o" /q (tangng lr*m ti,00 tc i't]* at lhe rate of 0.L0)the correspi'l:t1ii,g.,ri":,;s of riz can lrc dcterruined froin eqn' (5.21). Consequently,lhe r:illu* of I csu be obtained!f :: ie knor*'n.Wklle drawing tle chart, we u,ill arbitrariiy lake z = 2.5 cm. The cc,:aput*ti:ns arc given below in a tabvlar f'orm: Fer Ciifsrent values of

fr Â "2-|-;T;F"',"1-^".T;;i ;-l '

Circ ,t{o

i,

!

I

!1cn1lpmtlN ; i o_r_* . _J__*_l__"*_ l{c@l(cm)

i^_f__1 0 . 0 1 0 . 0 0 0?1. . 5l O . l , t ' i { ; j ' L } . 5j 0 9 i 8 : 2 . 5 l ; . 1 { Ji

I i * . r i o r t or .i i , , u s l? jl oa. .s7i r i r o.i r s ! : . ; * ' r . " * s:x. :j i r o : l t . s o la i*.t lc.CIri:5 I I

E i o.+ii .r.l*i z.s i, 't.lt | :.' i r.--0 i 0., l+.srs | : : i i ; . - 5 e i i1r 0. r l i * l o':'t_i_?"?6ili_i .olo.0rtl i 2 . tl *

_i

l_ll--_

i_

i

I i

I

i._ | _- J I

{ine ccncentrlc circles are drawn wrt} the r:cii sirowri-is rhe table. ,t!, rer of r*dinl lincs ar+ lheir drawn frtw lhe centre al ';;:ual deflectial:r"rf l8o. Ibe resulting Nernna;k'ri chart is sbr:wn in Fig. 5.15

-ls*-

Influencevalue= 0.005 Fig.5.15

PrcblemSs// Using the Newmark's chart preparedin problem 5.11, determine the vertical stressintensity at a depth of 2 m below the pointA of tbe raft fo"otingshown in Fig. 5.16. The uniformly distributed load on he raft is 8.5 Vrn'. Solution: In Problem 5.11 the Newmark's chart was prepared for z = 2.5 crn. In order to usethis chart for the rnmputation of vertical stressintensity at a depth of 2 m below any loaded arca, the plan of the area is to be drawn in sucha scalethata distanceof 2.5 crn in the drawing may representan actual distanceof 2 m.

130

Problems in Soil Mechanics and Fottndation Engineering

+-

--T

I

4m.

I

II

14m

II

15m

-f

II

J-

6m

_L

l--,0'-f r'--'l' Fig.5.l6

Hence, the requirid linear scalett,

3#

= 1 : 80.

The plan of the given raft footing is drawn on a tracing paper to a scale of I : 80. The pointA, which is the point of intersection of the two diagonals, is located. The tracing paper is then held on the Newmark's chart in such a way that the point A coincides with the centl€"of the chart. During such

th"_gl,g$l, glj!.gggcsilLp9!-er tqiEqte!er!4l=_ superpositioniog, In Fig.5.15,thep1-Iiofr-tegivenhreais drawnwith brokenlines.The nurnber of elements of Newmark's chart covered by the plan area is now counted. nt - no. of elements fully covered by the area = 39 n2 = no. of elernents1/2 of which is covered by the area = 15 43 = no. of elements 1/3 of which is covered by the area = 4 Tbe stressintensity at a depth of 2 m belowA is then given by, ' L a " = l x n x q

= (0.005) (39 + 15/2 + aft)$.s) o

= 2.A3Un?

EXERCISE 5 load of 50 t is applied on the ground concentrated 5.1 A vertical at a point 3 m belowtheground intensity stress vertical the surface.Compute of theloadby: of action line the from away m and 2 level

!*

StressDbtributi tt,

131

(i) Boussinesq'smethod (ii) Westergaard'smethod, assumingF = 0

[Ans.(i) 1.058t/n?1i;10.681Vrn2] 5.2 A 2 rn x 2 rn squarefooting carriesa grossload of 550 kN. The tboting is foundedat a depthof lT5 m below G.L. The subsoilconsistsof a 2 m thick layer of sand having a unit weight of 1g kN/m3. The sand layer is underlainby a 4 m thicklayerof softclayhavinga unitweightof lj.2kN/m3. compute the maximurnvertical stressat the rniddle of the clay layer betbre and after the constructionof the footing.Use Boussinesq'seguation . tAns. 70.4 kN/m' ; 80.9 kN/m,l 5.3 A vertical concentratedloadof 100 t is appliedat the groundlevel over a 6 m thick clay stratumwhich is underlainby a deepsandstraturn.plot the distributionof verticatstressintensityon a horizontalplaneat thc middlc of the clay layer by : (i) Boussinesq'smelhod. (ii) Westergaard'smetbod. (iii) 2; I disJrersion merhod. Given, Poisson's ratio, p - 0. 5.4 Plot tlre distribution of vertical stressintensity on a vertical plane due to a vertical concentratedload of750 kN appliedon the ground surface at a lateraldistanceof3 m from the given plane. 5.5 Draw the isobarsfor?Svo and10%stressintensitiesdue to a footins of 2.5 m x 2.5 m, carrying a uniformly clistributed load of l0 Vm2. UsI Boussinesq'smethod. 5.s A building is edby a raft t'ooti'g of 15 m x 1g rn pra* area. The gross load of the building, including tbe self weight of the ratt, is 405ffi kN. Plot the variation of vertic:alstressintensities with depth (z) below the ground level, taking 0.5 m < z s 5.0 m, at equal intervalsof 0.5 rn. Use 2:ldispersionmethod. s.7 Two adjacent footings of building, placed at a centre-to-ccntre distance of 4.5 m, have to carry gross loads of 750 kN each. using Boussinesq'stheory,plot the distribution ofvertical stress intensity at a deprh of 3 m below the baseof the footings. 5.E Three consecutivefootings of a building are carrying gross loads of80 t, 120 t and 110 t respectively.The centre-to-centredistance between the first and secondfooting is 3.0 m, while that between thc secondand third footing is 4.0 m. All tbe footings are foundedat 1.5 m below G.L. Determine tbe maximum vertical stressintensity due to the focting loads at a depth ôf 3.5 rn below G.L. t A,ns. 15.06 r/m2 ]

1i2


5.9 (a) Draw the influence line for the vertical stressintensity at a depth of 2.0 m bclow the point of applicationof a unit load. (b) Solve Problem 8 using the influence line thus drawn. 5.10 A strip footi4gof 2 m width carries a uniform load of 8 V*2. Tn. tboting is placed on the ground level over a homogeneous deposit of clay having the following p;operties : G = 2 . 7 2 , e = 0 . 7 8 , w = l Z % o, Determine the initial and final overburden pressure at a depth of 3 m below the centreof the footing. I i\ns.5.a9 Urt ;7lVr#'l 5.11 Two long boundary walls run parallel to each other at a centre-to-centredistanc-eof 1.5 m apart.The width and height of the firstwall are ?50 rnrn and 20fi) nrn respectively, while those of the second are respectively 125 mm and 3000 mm. Plot the distribution of vertical stress intensity due to the walls on a horizontal plane,2 m below G.L. The walls have negligible depth of foundation and are made of brick masonry having a unit wegnt of 1920 kdto3. 5.12 Fig. 5.16. shows the cross-section of an earth dam. The unit weisht of the carth-fill is 1.85 t/*3. Dlt.rmine lhe maximum stress intensity at a depth of 5 m below the baseof the dam. t

5.13 Draw a Newmark's +-influence chad, on the basis of Boussinesq's equation,for a n influencefactorof 0.00556.

J z o mI

6.1 Introduction, (Wn"n an extemalstaticload is applied on a saturated soil mass, an excess pore wat€r pressure is developed. As water is incornpressible for the low stress ranges commonly encountered in foundation problems,this porewater now tries to escapefrom the void spaces. Such expulsion of water results in a decreasein the void ratio and, consequently, a reduction in the volume of tlre soil mass.This processis known as consolidation.I

6.2 Definitions: The following are frequently used to express,the compressibilitycharacteristics of soils:

I

Fig.5.l7

J

|

-T

3m I

I 12n

I

_T 3m

-'f

-lr,]r-- - tzn-43n:r-

CONSOLIDATION

Consolidltion is essentiallya time-dependentprocess.In coarse-grained soils having high co-efficient of permeability the pore water escapesvery rapidly. The time-dependent volume change of the soil mass, therefore, occnrs only in less permeabletine-grained soils like clay and silt.

95m----J

5.14 The plan of a raft footing ing a multistoriedbuilding is shown in Fig. 5.18.The,raftcarriesa u.d.l. of 15 Vm'. Using the Newmark'schartgivenin Flg. 5.14, determinethe vertical stressintensityat a depthof 3 m below pointA.

6

(i) Co-efftcient of compressibility (a): void ratio per unit changeirr pressure. t . e. ,

:ae

a"=

Lp

It is defined as the changein

eo-e 6-

oo

where, e6 and e are the void ratios of a soil under the initial and final vertical stressestr6 and o respectively. (ii) Co-efficient of volutr,e change or volume compressibility (mu): It is definedasthechangeinvolume ofa soil massperunitof itsoriginal volume due to unit change in pressure. l'e't

L V l *, " -V.G

...(6.2)

Fig. 6.1 shows a soil masshaving an initial void ratio eg . If the volume of solids be unity, then volume of voids is given by,

V u = e O . V "* € 0 .1 = f o .'. Total volume. Fig.5.18

...(6.1)

vn=vr+v,

Consolidation

Problems in Soil Mechanics and Foundotion Engineering

134

But,

AC BC A C = % - e = L , e

and

BC =logr - logropo

= 1 + e o to e due to increasein pressure,then Ifthe void ratio uow decreases V t = l + e ' LV

or, changeinvolume

= Vo - Vt

eo-e

=

= ...fu.ol logropfogropo =:+ togroq/p' \ I -eoffiffi[i iidilma} alsobe determined fromrhe

= | + eo - (1 + e) = e 0 - e

The value following ernpirical formulae:

= Ae.

For normally consolidatedclays (sensitivitys 4 ), For rernouldedclays,

, Ae

T1.

l+

wherc,

ui9h% C, = 0.007 (w1 -

...(6.sa) ...(6.sb)

10)

wt = liquid li;,&(n--

--woref----

( r nedl

I

Fig.6.1

; o

^,=#;0.*

G, !

'o

L e 1 m'=Td r-"0 av ffl- - =

...(6.3)

a-

L + e g

Thc unit of both a, and m,rs atlt
V i r g i nC o m p r e s s i o n

o

From eqn. (6.2) we get,

of'

R e co m p r e s s

C" - gradientofA.B - tanO

0.5 0'3 0.40.5 0.7 |

3

4 5 6789tO

t ogt' p ----{Fig.6.2

6.3 Terzaghi's Theory of One-dimensional Consolidation: The process of consolidation is closely related to the expulsion of pore water and dissipation of pore pressure. Terzhghi, in his theory of one-dimensional consolidation, investigated the relationship between the rate of change of. excess pore pressure and tle degree of consolidation, and dcduced the following d ifferential equation:

Problems in Soil Mechanics and Fotrndation Engineering

136

6tt

6l=u, where,

a2u ---; al

...(6.6)

u stand for the excesspore pressureat a depth z t stands for the time elapsedafter the application of the load' C, = Co-efficient of consolidation,whici is defined as:

F:----T-l !L.. = -

where,

...(6.7)

|

Hu\n I |i " Y f-E--co-etficient ofpermeability, cm,/sec

(i) Estimating the probable consolidation settlement of a proposed struclure to be constructedon this soil. (ii) To determinethe time-rate of setflement. The sample is placed in an oedometerbetween two porous stones and arrangementsare made to keep the samplesaturatedthroughout the test. The loading intensities are generally applied in the following order: 0.25,0.5,1 A,2.0,4.0,8,0and 16.0kg&rn'. The vertical deformationsof the sampleunder each loading intensity are measuredwith the help of a dial gauge. The readings are taken at elapsed times of : ,. 0.25, 0.5, L,2,4,8, 15, 30, ffi, 120,240 arnd1440minutes.

Tw = unit weight of water, gn/cE zy = coefficient ofvolume change, t^2 /g^

From the results of the test, the following three curves are drawn: (i) e vs. logrOP curve, to determinethe value af. C.

The unit of C, is "*2/r.". Equation (6.6) is a second order differential equation, the solution of which rrravbe obtained in the form,

u=f(T") where,

where,

Cu't _F

(ii) Dial reading vs. logls t curve ] (iii) Dial reading vs. Vt curve j

In order to plot the e vs. log16p curve, the void ratio of the sample at the end of each load increment has to be detennined from the corresponding dial reading. This can be done by either of the following methods.

...(6.e)

(a) Height of solids method.' After the corrpletion of the test, the sample is taken out from the oedometer,dried in oven and its dry weight 177 is detennined.

I = time required fotU%o consolidation, sec

wd

ft = maximumlengthof drainagepath, un. In caseof double drainagecondition (i.e., when a clay layer of thickness ÊIlies befween two penneablc layen at top and bottom) the maximum length of drainage patb ft = Hlz,whercas in caseof single-drainagecondition (i.e., when the clay lies between a permeableand an impermeable layer), h = H' The time factor ?i, is a dimensionless quantity, the value of which depends on the degree of consolidation taken place at a given time, and not orrthe propertie.sof the soil. Terzaghi suggestedthe following equations for

for

ocu<S3vo,

and, height of solids in the sample,

h,=Yi=h where,

< U < t0070, T, = 1.781 - 0'933log (100 - U)' 53%o

Let e be the void ralio correspondingto a height h of the sarnple.

"'(6.10)

6.4 Laboratory Consolidation Test: f h order to determine the cornpressibility characteristicsofa clay deposit,'laboratoryconsolidationtests are to be performed on representativesamplesof the clay collected from the site. A knowledge of such characteristicsis required for:

vu v-v" A.h-A.hs = = A.lrrv" % h-h,

ofr

'(6'llrlr/

...(6.12)

A = cross-sectionalareaof the sample.

"=

t , = 4n\l fuo\ 'o ) / V

W7

""=T=Gy*

Now, volume of solids,

the determination of Iu :

For

,o 0",".*ine,the value of c,

...(6.8)

U = degreeofconsolidation.

I" = Time factor =

t37

Consolidution

e =

...(6.13)

hs

Thus if the value of ft is known at any tirne during the test, the correspondingvoid ratio can be determincd.The value of ft may be obtained from:

h =H - R.C

r 138

Problems in Soil Mechanics and Foundation Engineering Consolidotion

where,

fI = initial height of sarnple. R = dial reading.

a, =

or,

C = dial gauge constant (b) cfurnge in void ratio method: With referenceto Fig. 6.3, let, Vt = initial volurne of &e sample. V2 = volume of the sample at the end of compression under a loadirg internity p AV = changein volurne = Vt - V2

139

g-t

=1]34

...(6.r4)

h

Thus, knowing the values of e1 and ft, ihe change in void ratio at any given instant can be determinedif t[e.conesponding value of A ft is known. 6.5 Determination of C, : For a given soil, the value of C, is not constant but depends on the magnitude of the applied skess. In order io determine the degree of consolidation of clay layer under an external load, it is required to determinethe initial and final pressures( o , and o. + Ao, respectively ) on the soil. If, for example,the initial and final pressurebeforeand after the application of external load be 1 kg/cm' andZ xglr:;rf , thenthe value of Cumust be obtained {rom this particular range of loading il the consolidation test. The value of Cumay be determinedfrom either of the following metlods : (a) Squareroot of time fitting method: Cu =

--:--Wof -::: er

(T,)ssx rt tq

...(6.ls)

a

(b) Logarithm of fime fitting method: 6.6 Computation of Settlement: given by,

So l i d

,,

= 9)#!

The total setflement,S, of a footing is

S=S;+S.+S" where,

Fig.6.3 Now

V1 = h1.A and VZ = luZ.A

-

hA h.A = W -=v

LV

lq - loz

Lh h, =E

then,

Similarly,

V2=Vr(l+e2)

{ . ' . '" t =

L V = %(1 + er) - V"(l + e) = V"(et Therefore,

olr

M

h -M= h

v"/v"l

6.6.1 Immediate Sctlement: The immediate settlementdue to a vertical concentratedload Q at a depth z and radial distancer is given by,

...(6.18)

ez)

The immediate settlementdue to a uniformly loaded area is givel by,

-

s ; = q B . $ - v h. r ,

V"(e1 e2) LV =T =Z E+er) where, Le 1 +e 1

...(6.r7)

Si = immediate settlement .Sc = primary consolidation settlement Ss = socondaryconsolidation settlement.

The secondaryconsolidation setflementis of irnportanceonly in caseof highly organic soils and peats.

I f e l a n d e 2 b ethe void ratio conesponding to volurneV1 and V2,

Vt = V, + % = e1V" t V" = V"(1 + e1)

...(6.16)

q I F E

= = =

intensity of pressure leastlateraldimensionofloadcd area Poisson'sratio of soil modulus of elasticity of soil

....(6.1e)

140

Problems in Soil Meclnnics and Foundation Engineering

1/ = Influence factor, the value of which depends on: Type of rhe footing (i.e., wherher ir is rigid or flexible) Q (ii) Shapeof rhe footing (iii) The location of the pointberow which settrement is required (i.e., the centre, comer or any other point of the footing) (iv) Length ro breadrh ratio of the footing. be obtained from Tabte 6.1, wtrile Table 6.2 gives .r^ ÎO^:,1,1t-". ?t^!tt"u! rtreetasttc properties ofvarious soils.

Table 6,1 : Influencefactorsof various footings

Consolidation

L4l

6'6'2' corcoridation settreme_nt: Fig. 6.1 representsa soir sampre subjectedto an initial stresspg.I*r eg u.in" uoro ratioof thesoir.Due to a stressincrementa,p,'hevoid ratioreducesto e . Thechargein void ratio, Ae =.c0 - € Again, let /16 andi{ be theinitiarancrfinarthicknesses of lhe soirmass. MI=Ho_Ht Now, by dcfinition,

,,=+'+

Inf'luencefacnr

v a p For a laterally confined soll, area ofcross-sectio'.4

Square Circular Rectangular: L/B = 1.5 L/B= 2.0 UB = 2.5 L/B = 5.0 L/B = 10.0

LV t.t2 1.00

0.56 o.64

0.95 0.85

0.82 o.79

1.36

0.68 0.76 1.05 r.27 r.69

r.20

1.06

1.30 1.83 2.20 2.96

LzA

r.52 2.10 2.54 3.38

1.70

2.ro

v =WT

1 " = A$ H

cr= Table 6.2 : Elastic properties of various soils or,

1..Coarsesand (p = 0.15)

$[i.^r (tr= 0.2) $[i"^r $[/"^r $[i"^r 0.3s)

Medium

Fine sand (tt = 0.25) Sandysilt = 0.30 to

0.61to 0.70

43 45200

38 32qO

is constant.

= LEHI L,p

...(6.20\ The change in thictness-of rhe soir rnass, andhencr the consolidation settlement,canbedeterminedfrom eqn. 6.mi, Again, by definition.

3.4

0.41to 0.50

LH.A

Le

l%rr/po

Le = c..logr;

Assumingunit vorumeof sorids,the initiarand finarvorurneof thesoir are,

v o =l + e 0 ,

N 45200

38 39300

35 3240/0

or,

38 366A0

36 27600

32 23500

But

36 13800

34 1t700

30 10000

Hence,

a n d V 1= l + e ,

AV= 1 + e g - ( 1 + e o ) = e o - e = L , e AV Le

vo

i.A

AH

LV

:

6

E

...{6.2r)

142

Problcms in Soil Meclmnics and Foundation Engineering

or,

L,H= Ht,fu.torro P o + L P Po

n

s.,

...(6.22)

i-1

EXAMPLES Prcblem e/ enonnally consolidatedclay stratumof 3 m thickness has two permeablelayersat its top andbottom.The liquid limit and the initial void ratio of the clay are 36Vo and 0.82 respectively,wlrile the initial overburden presdureat the rniddle of clay layer is Zkg/an'. Due to the construction of a new building this pressure increasesby 1.5 kg/cmz. Compute the probableconsolidationsettlemett of the building. Solution:

Using eqn. (6.21),consolidationsettlernentof the building, S. = H

C. i-

'logrn ",

Po+& p0

Again, using eqn. (6.5a)

Cc = o.oo9(w7- 1o) = 0.009(-36- 10) = A.234 H=3m=300cm ea = o'82

Po = Zkg/crt

...(6.2r)

In eqn. (6.21)pg andp1 representthe averageinitial and final pressure acting over the thickness flg of the soil. Wrile cornputing the probable consolidationsettlementof a clay stratun, generally it is assumedthat, the average stresses are those acting at the mid-height of the clay stratum. However, this assumption is not correcl because,as we have seenin chapter 5, the stressintensitydue to an externalloaddoesnot vary linearly with depth. If thethicknessof the clay stratumis substantiallyhigh, this leadsto erroneous results. In order to determine accurately the probable consolidation settlement of a clay layer of finite thickness, the following steps should be followed, (i) Divide the given clay layer into a number of sub-layersof small thickness. (ii) Determine the effective overburdenpresssureand stressincrement at the mid-height of eachsub-layer. (iii) Compute the consolidation settlement of each subJayer using either eqn. (6.20) or eqn.(6.21). (iv) Ihe probable sefflement of the clay stratum is then obtained by summing up the settlementsof all subJayers,i.e.,

S.=

143

(tutsolidotion

Ap = t.s kg/ctt

' rosro [ti"]"-

s. = (3oo)

= 9'37 cm

A 3 m thick saturatedclay layer is overlain by a 4 m l'r..tblem 6;.{ tlrick sandlayer and is unde-rlainbyrock.The unitweight of the sandandclay tut 1.72t/rn3and 1.85 V# respectively'The clay has a liquid lirnritof 53Eo load of200 t is appliedon the ground nrrda voiclratio of0.65. A concentrated settlementof the clay, consolidation probable the surlhct:.compute (i) consideringthe entireclay layer of equal thickness. liij oivloing the clay layer into three sub-layers Solution: CompJessionindex of clay,

(wl - 1o) c" = o.ooq = 0.009(53 - 10) = 0.387. H,=3rn=3(X)crn eo = 0'65 of claylayer' at rnid-depth pressure (i) Initialoverburden . (1.85)= 9.65tln'= 0.965kg/cttf po = (4.0)(1.72) + (3.012) eqn.,themaximumstressintensityat themiddleof FrornBoussinesq's claylayer,

A' e _

(3)(2m) - =3.r6t/n?=asrckp/q} (4+ 1.5)' (2)(3.14)

^ = (300)(0.387) r. T_fTIil.

togto

0.965+ 0.316 = 8.66cm tg6--

(ii) In this casethe clay layer is divided into threesub-layersofthickness I m each,as shown in Fig. (6.4). Thc consolidation settlementgf each sub-layer is estimatedbelow: Sub-hyer I: Depth of the niddle of layer 1 below G.L = +'O+ f

= 4'5 m

p61 = (a) G.72) + (0.5)(1.85) = 7.805t/mz = O'7l7kg/cnf

6,,, =

(3)(200) - = 4.7tr/m2= a.4lrkg/rmz (2)(3.r4')(4.s)'

Problems in SoilMechanics and Foundation Engineering

144

('tmsolidotion

145

I'robleu-5r3:- A 3 rn thick layer of silty clay is sandwitchedbetween lwo Iayersof densesand.Theêffectiveoverburdenpressureat the ceutreof thc silty clay layer is 2 kg/cm" . How€ver,due to the constructionof a raft to 4 kg/on' . lirurrdation,this pressureirtcreases Laboratorycousolidationtestwas performedon a 2.5 cm thick sarnple ol' thc silty c:lay.Under applied stressesof 2 kglcnrz and 4 kg/c,m2 the conrpressionsof the sample were found to be 0.26 cm and 0.38 cm rcspcctively.Cornputethe probableconsolidationsettlernentof the raft. Solution :

Using eqn.{6.20), = S, frr'Ho'LP

wltt'rc,

--iltA-p-l--

Su b- toyer I

= 2.5 cm for the soil sampleand 300 cm for the soil in-siru. 4p = changein effective pressure

3m 1 m - - - LlJ e.z-- -{ ILA!3.- S u b - t o y e r

=4-2=Zk,/cnz.

II

1m

*--illo:--JrL^-P;-

mu = co-etficieut of volume changefor the pressurerangeof 2kglcm'to 4 kg/cm'

Sub-toyer III

l'or lht:laboratorytest:

Rock

= 2.5cm Tlricknessundera pressureof 2kglctnz =2.5 -0.26 = 2.24sn. Thicknessuudera pressureof 4 kg/on2 = 2.5- 0.38= 2.12crn. lrritialthicknessof the sarnple

Fig.6.4

(100)(0.387)'to8to 0.781+ 0.471 = - -l-] ,s = 4.81cm. r", gq66 -tt Sub-layerII: Depthof middle = 5.5m' .

pm = (4) (1.72) + (1.5)(1.85) = 9.655t/t# Mc =

(3) (200) = 3.157t/ri

(2)(3.14) (s.sr

IlO = ittitial thickness

= O.965kg/artz

.'. Changein thicknesswh"enthe pressu.re =2.24-2J2= 0.l2c.lrn. irrt'reasesfrorn 2 kg/cm" to 4 kg/on" Frorn cqn. (6.20)

0.t2 = (m")(2.s)(2.0) = o.3l6kg/cnl

0.965+ 0.316 " = - (100)(0.387).loglo -2.89cm J", 0.965 1 + 05,5 Depthof rniddle = 6.5 m Sub-layerIII: pw = (41(1.72)+ (2.5)(1.35)= 11'5t/m2 = r-lllq/cmz ar._i@\=2.26cm

(6.sr (2)(3.14)

setllement = 4'81 + 2,89 + 2.26 .'. Totalcstimated = 9.96cm.

or,

m, = 0.024. z/kg.

Again, using eqn. (6.2q, ilre consolidationsetflernentof the silty clay Inyt'r, sc = @.024)(300) (2) = 14.4cm l'nrble4a$,4,.-' Due to the constructionof a new strucfure the average vcrfit'a"lpressureat.the centreof a2.5 m thick clay layer increasesfrom 1 kg/r'rn" to 2 kg/cm". A laboratory consolidation test was performed on a 2 crrrthit:k undisturbedsampleof the clay. Under appliedstressesof 1 kglon' l rrd 2 kg/cm" the equil ibrium thicknessesof the samplewere found to be 1.76 r'ru lrrd 1.63crn respectively.On reinovingthe stresscompletely,the thick-

Problems in SoilMeclmnics snd Fottndation Engineertng

t46

to 1.88c:m.Tht: final moisturecontentand the spec-ificgravity nessinc--reased samplewere found lo be297o and2.7l respectively.Conlpute of the of solids settlementof the structure' consolidatiott probable the Irt e; and Il;be the tinal void ratio and thicknessof the solution: sarnple.

Le LH l + e r H l =

Again,

where, and, ol, Here,

.'.

Rcquired cousolidatiou settlemeutof the clay layer in the field

,5. = rn,,llg Ao = (0.08)(2s0) (2 - 1) = 20 cm. llcnct:, lhc rt:quiredsettlementof tlie structure = 20 cm

Solution :

AH = changein thicluressdue a giveu stress Ae = correspondingchangein void ratio'

when

AH A e = ( t + e 7 ) . u .

I lcrr',

"I

o = Z.}kg/cm"

Agrin, usingequ.(6.7),

c '. , = k

frr\n

NI = 1'88- 136=

lf trc,

m , - 0 . 0 6 5" n 2 / k g - 0.065 t 10-3 "rn2/grn

0'12trm

= 6.5 x l}-s anz/gn

Le = (0.95)(0.12) = 9'114

Hence,void ratio at

k - 3.3 x 10-4 cmlsec.

o = 1.0kg/cm2 = 0'786 - 0'114 = 0.672.

y- - 1 gm/cc.

Letm,,betheaveragevalueoftheco-efficientofvolunrechangeinthe pressurerange of 1'0 to 2.0 kg/on-. We have frorn eqn. (6'2), mv

Le

l+eg

.1=0.065"*z/kg

= ef - Le = 0.786 - 0'238 = 0'548'

l'0kg/cn?'

- 0.73= 0.r2

mv

"'(i)

o = 2.okg/cn? , NI = 1.88 - 1.63 = 0'25crn 6s = (0.95)(0.25) = 9.239

o=

1. L,p

N t = z - | = Ilf./cmz

Ae = 0.95AH

Again,when '

L,e l+es

eo = 0.85, Ae = 0.85

ef = O.lgO, and H1 = 1.88cm.

Hence,void ratio at

Usingeqn.(6.2), frv=

Ae= (1 + 0.786) #t of,

t47

l)rrrblenr *Y In a laboratoryconsolidationtest,the void ratio of the silnrlllc rcducedtiorn 0.85 to 0.73 as the pressurewas increasedfrom 1 to 2 kg/t'rnr. If the co-efficientof penneabilityof tbe soil be 3.3 x 10{ crn/sec, r l cl cr r ni n c : (i) co-efficicntof volume change (ii) co-elficicnt of coruolidation.

wG -= -(0.29)(2.71) = o.786 "- t _ = (1) ,

Then,

('on.rolido lion

1 Ao

(0.672- 0.548) . _ = = J- ^1.0)= o . o 8 r 2 / k g ' (2.0 (1 + 0.548)

cr=

3.3 x 10-a

(6.sx to-5)(t)

= 5.07" 2/""".

g6 rn thick clay layeris drainedat botq lop and bottom. f'nrblcnr d' rrr clli<'icnt of consolidationof the soil is 5 x 1oe cn(/sec. Determine llrr tirrrr rcquircd for 507oconsolidationof the layer due to an externalload.

'f'lrc

Solullon :

Usingequ.(6.9),

148


T

Consolidation

,,=sffq= 375crn

cr't

-

-

149

a

If

Tr.t?

ot,

I =

Tr, =

For 50% consolidation,

'=

C,

n(u\2

Al-trrl

H For double drainagecondition, h = 2

600 2

r u / 5 0 \ 2=

a[.'*/

= ffi0^' "

300 cm .

(0.197)(300)2 , s e c =3.546 x 107sec 5 x l0-' 3.546x 107 . oays = 410 days. s6400 Problem fy Araft footing is to be constructedon a 7.5 qrn thick clay layer which lie3 betweentwo sandlayers.In order to predict the time rate of settlernentof the building , a 2.5 crn thick undisturbed sarnpleof the soil was tested in the laboratory under double drainagecondition. The sample was found to have undergone5O%corsolidation in 12.5 minutes.Determinethe time required for 5O%settlernentof the building. Solution :

= 782 days = 2 years 1 month and 22 days. Problem QV't"a laboratory consolidationtest, a 2.5 crn thick sample of'<'lay reached60% cnnsolidatiron in 17 minutes under double drainage corrdition. Determine the time required for 6OVoconsolidation of a layer of lhis soil in the field under the following conditions: (i) when a 3 m thick layer of the given soil is sandwitched between lwo sandlayers. (ii) when a 5 m thick layer of the soil is overlainby a sandlayer and rundt:rlainby a deeplayer of intact shale. Solution :

Using eqn. (6.11), the time factor for 60% consolidation T, = 1.781 - 0.933 logls (100 - 60)

= 0.?a6 Again, using eqn. (6.9)

,'r' -- c ' ' t

We hdve from eqn. (6.9),

T,,

C r x t =

a

t

of,

C, =

Tr-t?

In thelaboratorytest, 7, = time factor for 50% consolidation = 0.197

h =

12.5min. H 2

2.5 = l.?5 cm 2

In case of the actual building, T' = 0'197

' or, C, =

Tr,' h2

It2

h'

t =

c,

a

(0.197)(375)' - mtn = ---0.0246

0.197.

5 x 10-a"nl2,/r.".

and,

T,' h2

In thr: laboratory test,

t = 17min. h = 2.5/2 = 1.25cm

(0.2s6) (r.?52= vv = ---di-

a

0.A263cm'/min

( i ) Here the soil layer is drained at both top and bottom

H _ ( 3 )(2) (1oo)=15ocm (0'285X150f = z44nxamin= r70 days ,' 0.0263 ( i i ) In this casethe soil layer is drained at top only H=5m=500cm

Problems in Soil Meclmnics and Foundotion Engineering

150

(0'4q)-L5-00)z = 2 1 l 8 6 3 n n i r t = 1 t t 8 8 r l a y s= 5 . 1 7 y e a r s . ', 0.0263 ,/ The consolidatiottsettletnentof a new structuretbunded Problem Q/ on a 5 m thick layer is estimatedas 6.5 cm. The structurewas found to have settledby 1.6 cm in 6 monthsafter the completionof couslruction.If the c{ay layer is underlainby rock and overlainby a layer ofcoarse sattd,detennine: (D tle timc required for 50% consolidationto occut (ii) the amount of settlementwhich will take place in the next six montls. Degreeof consolidationoc:curredin the first six months 1..6 = x Lo07o = 24.627a ;;

151

Consolidation *rfl -L

= 0'0974'

,, = 35.22o/o u == v&q*e-e.ozo t 3.14

()f,

ll x be the amouutof settlernent.then

U=4;xr00 o.)

or,

'=T#

=z.z()cn

Solution :

Time factor for

U = 24.62Vo T, = (x/4) (24.62/10q2 = 0.048.

As single drainagecondition is prevailing, lt = 5 tn.

Solution : h=H=2.5crn

Using eqn. (6.9),

(o#8,)(t2) tu = -1o;1ru,1 =

Problem &J${ndisturbed sampleswere collec-.tedfrom a 3 m thick t'lay stratumwhich lies betweentwo sandstrata.A laboratoryconsolidation It:sl was performed on a 2.5 crn thick sarnpleof the clay. During the test,water was allowedto drain out only throughthe top of the sample.The time required t
6.67x 1o-3^z/d^y.

(i)For50%consolidation, time factor,

7,,= O.197.

Again, for U = SUVo,wehave Tr= 0.t97. Usingeqn.(6.9),

'' = gfft

Using eqn. (6.9), T,x h2

(0.1e7)(s2)

C,

6.67 x lO-'

= 738.4days

As the samplewas testedunder single drainagecondition,

= o'o3s "'n2'l*i"' = 1.781- 0.933 logls (100- 60) = 0.286

Now, tbr 607oconsolidation,

= 2yearsand8.4days. (ii) L,et U be the degree of consolidation that will take place in the next six rnonths, i.e. at the end of 1 year since the completion of construction. We have already found that the time required for 5O% consolidation is 2 years and 8.4 days. Thus, degreeofconsolidation occurredin 1 year must be less than 50Vo. The corresponding time factor may be determined using eqn. (6.10), T" = (n/4)(U/LOC)' =

"r.P 40000

Again, using eqn. (6.9),

(6.67x 1g-3)(36s)= o.oe74. Tr= (5")

For a double drainagecondition,

t,t = H = ; 3n i

= l50cm.

Tu*'h'

t6o =

-7:

(0.286) (1so)2 (0.03s) = 183857 min = 127.7days o 128days For 90% consolidation,

T,x = l'781- 0.933 logls (100- 90) = 0.848.

r52


153

at 2 rn below the ground lev-el.The unit weight of sandabove and below water iable are 1.90 and 2.10 t/m'. Tbe propertiesof tbe clay are as follows;

,u*.*

tco =

Consolidotion

C,

Initial void r'atio= 0.72

= 545143 min

*"*tP

= 379 days. Pnoblem 6.1I. A flexible footing of 2 m x 2msize carries a total load 490 kN, inclusive of its self-weighlThe footing restson a sandlayer having of a modulus of elasticity of 400fi) kN/m' and a Poisson'sratio of 0.38. Estimate the probable settlement below the cenlre and below any orie corner of the

footing.

Here,

f iquid limit = 427o co-efficient of consolidat ion = 2.2 x I 0-3 cm2/sec. Dclennine: (i) Probablesettlemenrof the raft. (ii) The time required to undergo a settlementof 5 cm. Solution : (i) The soil profile is shown in Fig. 6.5. The clay layer is divided into three sub-layersof thickness 2 m each. The settlement of each rub-layer rnay now be computed using eqn. (6.21),

Solution: We have, from eqn. (6.19),

s ; =- R 0

specificgravity of solids = 2.71

- t') r'I'

.,*, Pt L , H = H- n ' r Q -'opo' l+e6

E

intensity of loading

The computation of settlementfor the first sub-layer is shown below :

= g99]- = 02.5kN./m2 (2) (2)

q

= (0.00e)(42 - 10) = 0.288

eo = A'72

B = 2 m

Ho = 2m = 2l)0cm.

p = 0.38, E = 40000kN./m2 The influencefactor//may be obtainedfrom table6.1., /y(corner) = 0.56

Dthofmiddleofthe sub.layerbelowG.L. = 8 + 2D =9 m

11.6m_J

/1(centre)= L.12.

Roft

Immediatesettlementbelowthecentre,

- @# si(centre)

(I=1'90 t/ m3)

'0.r2)

t Y =2 . ' t 0 t / m 3 )

* 0.59 crn

Sond

Immediate settlement below the corner

S;1--e4-

ff

= o.295crn.

Problen 6.12. A 6 m thick clay stratumis overlainby a 8 m thick sua$m of coarsesandand is underlainby an impermeableshale.A raft footing, ingthe columnsof a building, is to be foundedat a depthof 1.2 rn below groundlevel. The sizeof the raft f 8.5 m x 13.6m, and it is loadcduniformlywith a stressintensityof 9.ztlm' . Thewatertableis located

6m >

I

zlzz>

Ctoy ( e e =g , 7,26 = 2 . 7 1 , r o L = L 2 a /Cov = 2 . 2 x t 0 - 3 c r l l s e c l >7 t>>7>rr>zt>

r2l=t>z>7ztz

I m p er v i o u s S h o l e Fig.6.5

,>z

,2

154

Problems in Soil Meclunics and Foundqtian Engineering

Initial effective overburden stressat a depth of 9 rn below G.L. = stressdue to sand above water table + stressdue to sand below water table + stressdue to clav /b = y ft1 + yru6h2 + y.1^,h3.

= 0.317 As single drainagecondition prevailsat site, h = H = 6m = 600crn. tlsing eqn.(6.9),

Here, ulit weights of sand are : y = I'g t/m3 and,

7,,-fi

t=

c,

'kat = 2.1t/m3

- (0.317)(6od-) = 51872327 sec (2.2 x l0-'\ = 60,0days

Ysub= Ysar- Y. = 2.1 - | = l'lt/m3 Again,

155

Consolitlalion

Iclay

-

G + e l + e

t w

(2.7r + 0.72)

l'

(l) = 2.0 t/m3 - 2) + (2.o - 1.0) (1)

= !L.4t/rn2 = l.l4kg/cmz Again, depth of middle of this sub-layer below the base of footing =9-1.5=7.5m. Using the 2 : 1 dispersion method,

Problem 613. The constructionof a rnultistoreyedbuildittg startedin January 1989t'andwas cornpletedin June 1990. The total consolidation scttlcmentof the building was estimatedto be 8 cn. The averagesettlement of tlrebuilding was lneasuredin Decernber1991andwas found to be 2.2 cm. Cornputethe probablesettletnentof the building in January2001. Solution : Lrt C, be the cs'efficient of consolidationof the soil in the appropriate pressurerange, and Il be the effective length ofdrainage path. Time elapsedfrom June 1990 to December l99I = 1.5 years Degreeofconsolidationoccurredin 1.5 years.

u - - Q'2)iroo)vo= z7.5vo (8)

(e'2) (8's) (13.6) !!6o ' = ----9 (B + z)(L + z) (8.5 + 7.5)(13.6+ 7.5)

T, = (x/4) (27.5/rm)2 = 0.059

= 3.15t/m2 = A31^5kg/cnl2 pr = po + Lp = 1.14 + 0.315 = 1.455kg/ctn2

^-, ^ _) ( r . 4 s s \= r'55cm' A r / _ ( 2 q g ) . ( 0 . 2'rogro 8, 8 (r + oJD r-14 I ,J

Similarly, settlementsof the secondand third sub-layers are found to be 2.54 cm and 1.86 cm respectively.Hence,the total settlementof the raft = (3.55 + 2.54 + 1.86)cm = 7.95 cm. (ii) The degree of consolidation correspondingto a settlementof 5 cm, (5)(lm) f f ='6 .^ = 62'8970 u Using eqn. (6.11), the correspondingtime factor is, T, = 1.78t - 0.933 togls (100 - 62.89)

But,

T, = C,

O l t

T,

0.059

H'

7=

C,, .= H'

0.039

. =

ort

Cu't _F

ls

...(D

Again, time elapsedfrom June1990to January2O0l = 10.5 years. Let U be the correspondingdegreeofconsolidation. Assuming U > 53Vo, But,

T, = 1.78I- 0.933logls (f 00 - Lr)

r, =

?;=

c.. I (o.o3e) (10.s) '..; = o.o3el

= 0.4095.

H

'

l

t55


t57

Consolidation

1.781- 0.933 lo916(100 - U) = 0.40995 fog16(100 - u) =

of'

(1'781-: -q'-4095) = 1.47 (0.e33)

NI=H #a

or,

- (2)(1oo)(q.g9s) = 8.44crn (1 + 1.25)

Taking antilog of both sides we get,

100-U=29.51 U= 100-29.51=70.497o

or,

(iii) In the pressurerangeof 2to 4kg/crr2.

^"=#^.b

Amoult of consolidation settJementin January 2001 (8) (70.49) _ < ai _ -(1oo) = = )'o4 cm' Probtem {a! e/ e,2 m thick layer of saturatedclay lies in between two penneable lEdrt'The clay has the following properties : liquid lifrit = 45Vo co-efficient of permeability = 2.8x 10-7 crn/scc initial void latio = 1.25 The iqitial effective overburden pressureat the middle of the clay layer is 2kg/at ,and is likety to increasei axg',t.2 due ro rhe "o*t,u"iilo or, new building. Determine : (i) the final void ratio of the clay. (ii) settlementof the proposedbuilding. (iii) time required for SOVoconsolidation. Solution :

(i) Compressionindex, C, = (0.009) (45 -10) = 0.315. But, by definition,

cr=

0ft

k

Po+Lp , loglo po-

A e = L" fo$19

ps + L,P p,

L e = (0.31s)rogls{(2 + 2)/(2)} = 0.095 .'. Final void ratio = e6 - L,e = l.?5 - 0.095 = 1.155 (ii) Let A/1be theconsolidationsettlementof the clay layer. LH- _ Le _ H 1+tu

=

= o.o2tcm2lkg. (2) (1 + 1.25) + C,' = Jfrr\n

Usingcqn.(6.7),

k = 2.8 x 10-7cm,/sec.

Hcrc,

m v = 0 . 0 2" 1z / u ' g ln = lgm/cc = I x 1o-3t'g/cc. (2'8 x 1o-')- = 0.0133orr2/r"" (0.021x 10-') . w€ have, Tu= O.197 consolidation, For 5OVo Usingeqn.(6.9), a.. "

T,,.h2 (0.197)QN/z)z sec' ' = -T = -afi133; = 1.71days. testwas performedon a 2 Problem 6.L5. A laboratoryconsolidation cm thicksampleof a siltyclay,andthefollowingresultswcreobtained; Pressure (kglcm2)

Final dial gauge reading (mm)

Pressure (kslcnl)

Final dial gauge reading (mm)

o 0.25

5.590 5.234

2.00

3.9U

4.00

3.515

0.50

4.gffi

8.00

2.785

1.00

4.6M

0

5;224

The final moisture content of the sample after swelling was fouttd to be 32.57o.The specificgravity of solids = 2!lO.

159

Problems in SoilMechanics and Fottndation Engineering

158

Consolidation

table',"t Ot:t Note that, in column5 of the above. ::t1"1tt:^?:::rT":

(i) Plotthee vs. logP curue' (ii) Detennile the cornpressiolindex and the co-eft'icientof volutuc changeof the soil. solution: Inordertoplotthe e vs. logp curve,thefinalvoidratios at the end of eachpressureincrementare to be deterrnine.d'

: 9,tl2 I:bvsubtracting * noiff"T:;ilJl""ail t""i-'*"a*':ri''tiEdetermined :ll. l:1":: :il":ffiffi ffi;;;;;"*'1ry: ofAeo:*':l1 Tri

;;i;il.6roPriaie values ratiocorrespondi'g to8kg/crn ;"id ;:ffi i'ipd;il;;il = ef - Ae = 0'878 - 0'233 = 0'645' ) to 4 kg/onandthatcorresponding

The final void ratio of the sampleat the end of swelling .

e=*G

s

_ ( 0 . 3 2 5 ) ( 2 . 7 0= , )0 . 8 7 t . (1)

= 0'645 - (-0'070) =0'715 c' and the co-effic:ient of volume The co-efficient of compressibility' ' in col' eachpressurerauge' and are slrowr change,tz, are then comput;d for 8' c'ol' in are siven Ct values of 6 and 7 rcspectively' fn"'"o'nput"d scale).hav:

The thicknessof the sarnpleat this stage, H = H o - M l

."0 lp (in log Ttn Thevaluesor rc 1lo"i'ritirr.tvt".r"j on a semi-loggraphpaper'to i-axis respectively ana Y tne dr;;;"; Tnf ls sbown in Fig' 6'6

= 2.A - (0.5590 - O'5224)cm = 1..9634cm.

obtain the e vs. loglg P curve'

Now, we have,

N] H

Le l + e

MI ,. . + e) O" = T(1

or,

0.90 0.Es

Substitutingthe final valuesof e andH, we gel,

Ar/ Le = tH tt iO:#t' = 0.e565

...(i)

lo. I

The c.hangein void ratio, and hencethe final void ratio after each load incrernent,arellow detenninedby puttingthe correspondingvaluesof AH itt eqn. (i). The cornputedvalues are showttbelow in a tabular form : Pressure ranSe

Pressure itrcre-

I ncrease

in

ment

(kg/cn2)

&

NT

(kglcm2)

(cm)

Change in void ratto Le

Equilibrium void ratio

4.00 to 8.00

(5) (4) (3) -o.034 -o.0356 0.879 +0 .25 +0.25 4.O274 -o.026 0.863 +0.50 -o.0356 4.034 0.829 +1.00 4.0640 -o.061 0.768 +2.00 -{.0549 -o.0s3 0.715 +4.00 -{.0730 -o.070 0.645

8.0Oto 0

-8.00

(1) 0 to 0.25 0.25to 0.50 0.50to 1.0O 1.0Olo 2.00 2.00 to 4.fi)

/t\

mv 'f/ Ae\ a u \ 4oJ = 1 . -

C, L

o

;0. o L

E'

6 0 e

'"g. k/

l | 7 8910

1c^2/k91 (cn2/kg) (6) 0.136 0.104 0.068 0.061 0.027 0.015

(1) o.072 0.055 0.036 0.019 0.014 0.008

toglgp*

(8)

Fig' 6'6

0.086 0.113 0.203 0.t76 0.199

befoundedin a 3 m thick layerof Problem 6.16. A raft footingis to sandlayel' The initial over underlainiV t nign-vpermeable "lry;;;;l; is likely orilt "fty layeris 2'0 kg/cmrandthis at thece-nire burdenpressure thick cm 2'5 A of the raft' to increaseto 4.0 kg/cm?'l;t to 'ht construction underdoubledrainage sarnpleof this soit is iesteain a consolidometet

+0.2439 +O.233 0.878

i l

160


conditio.. The followi'g data.wereobtainedwhe'the pressure on the sampre was increasedfrom 2to 4kglan" :

161

Consolidation

2000 1900 1800

0

I972

16.00

t727

0.25

L92L

36.00

1642

1.00

1870

64.00

1555

2.25

1848

100.00

I49r

4.00

1813

144.N

L449

9.00

1769

The dial gaugeconstantis, L division = 0.002 rnin. (a) Determine the co-efficient of consoridationof the soir by the square root of time fitting method. (b) Estirnate the time required for 50vo and g}Toconsolidatironto occur in the field.

1700 1600 1500 1400 13C0

tr

6

2' A number of points, each represe'ting a certain dial reading a'd the corresponding value of vl are obtained. A smooth curve is drawrithrough them.

12

1L

16

18

Fig' 6'7

y'tco= 1l'o /so = 134'56min

From Fig.6.7,

Solution: (a) The procedurefor obtaining the value of C,,by the square root of time fitting method is explained below I 1. The valucs of the squareroots of various time intervarsat which the dial readingswere taken are plotted along thex-axis while the corresponding dial readings are plotted along the y-axis ofan ordinary g.aph paper.

10

I

T,(n).hz

.-u= _6_

Now, Here,

Ir (m) = 0.84[t,

h = 2.5/2 = l.?-Scrrr

(- - (0.848)(1.25)2-

" = llffiI&r

= L64 x

10-a.,.,2./r..

3._ The straight portion of the curve is projected to intersect the y-axis _ at R". This is taken as the initial reading.

(b) The time required for 507oand90Voconsolidationto occur in the field may be obtained using eqn. (6.9).

4. A point P is arbitrarily chosenon the curve. 5. Frorn P, a horizontal rine pe is drawn to intersect the y-axis ar e. Lx.l, PQ = 4.

(3oo)2 = 1.081x tossec , r o = W =(0.1971 1.64x 10-

6. The point R is chosenon projecredpe, such thal, pR = 0.15a. 7, R6 and R are ed. The line R6 is then projected to intenect the curve al S. The dial reading corresponding to s representsx)vo consoridation. Let t96be the corresponding time required.

= L?Sldays = 3 years 5 months and 6 days.

t* =

T,(gq-xh2 C, 5386days

= 4.654x r'ssec = (p.q{8)(iqql2 1.64 x l}-a 14years9 monthsand6 days.

162

Problems in Soil Meclnnics and Foundation Engineering EXERCISE6

6.1. Estirnate the consolidation settlement of a 2 rn thick clay layer which is overlain by a layer of sandand underlainby a deep layer of intact shale, ifthe construction ofa new footing itrcreasesthe averageover-burden pressure by 50o/o.Tlre initial overburden pressure was 2 kg,/on'. Giveu, co-efficient of volume c-hange= 0.023 cm'/kg. [4.6 cml

Consolidation

163

6J. Estimate the immediate settlement below the centre of a 15 m x 25 m flexible raft footing carrying a gross pressureof t}t/r*.The raft rests on a sand stratum having a modulus of elasticity of 4080 t/m' ard a Poisson's ratio of 0.25. The influence factors are as follows :

when LlB = 15, If = 1.36 when LlB = 2.A, If = 1.52

[5.8acm]

6.2. An 8 rn thick clay stralurn lies between a 10 m thick sand stJatum at top and a reck layer at bottom. The unit weight of sand is 1.75 t/m3. The clay stratum has the following properties: L.L. = 42c/o, w = 28.5o/o, G = 2.72

6.8. A footing is to be constructedin 3 homogeneousbed of clay having an overall thickness of 3 m. Thc clay layer is underlain by rock and overlain by a sand layer. If the co-efficient of consolidation of clay be 9.5 x 10-a cm-/sec, find out the time required for 907oconsolidation. [930 days]

A raft footing of 15 rn x L5 rn area and carrying a unifonnly distributed load of 20 Vm" is proposed to be constructedat the site. Determine the probable consolidation settlementof the footing. The clay layer should be divided into four sub.layers of cqual tlickness and the stress increments may be computedby 2: I dispersionmethod.

6.9. The total consolidation settlernentof a building founded on a 5 rn thick silty clay layer, drained at both ends, is estimated to be 6.8 cm. The building is tbund to have undergonea sefilementof 2.5 cm in 3 months. The initial void ratio and the co-efficicnt of permeability of tlie soil are 0.88 and L2x lAa crn/secrespectively.Determinethe co-efiicient of compressibility of the soil. [0.265 cm"/kgl

6 . 3 . A 3 m x 3 m s q u a r cf o o t i n g , c a r r y i n g a g r o s s l o a d o 1f 2 5 t , h a s been constructed over a 5 rn thick sand layer whicb is underlain tirst by a 6 m thick layer of soft clay and then a layer of irnpermeableshale. Compute the consolidation settlement of the footing by considering the clay layer (i) as a whole (ii) divided into three layers of equal thickness. Given, unit weight of sand = 1.8 gm/cc compression index of clay = O.42 water content of clay =32Vo specificgravity of clay particles=2.7 KD 3.6 an (ii) a.29 cml 6.4. During a laboratory consolidation test, the void ratio of a soil sample decreasedfrom 1,2 to 1.05 when the pressureon it increasedfrom 2 to 4kglcn' . Determine the co-efficient of compressibility and tbe co-efficient of volurne change of the soil. Will thesevalues remain the sarne if pressure cm2ltg] increasesfrom ito 8 kg/crn2. [0.075 .t?/rg;o.o:+ 6.5. A consolidation test was performed on a samplc of saturatcd clay in the laboratory. Thc liquid limit and the initial void ratio of thc soil were 487o ard 0.96 respectively. What will be the final voîdratio of the soil if the pressur€ is increased from 0.25 kg/*rz to 1.0 kg/cnr2 ? [0.721 6.6. Sample of a silty clay wassubjected to a laboratory oedometertest. Under a veitical pressureof Zkglun'the equilibrium void ratio was found to be 1.05. On increasing the pressureto 3 kg/crn2, the final cquilibrium void ratiq reduced to 0.93. If the co-efficient of permeability of the soil be 1.2 x 104 cm/sec, detennine the co-efficient of consolidation in t# rc^y.

[r.Ttx toa rt/aay1

6.10. A building is to be edby a raft footing laid in a 3 m thick bed of clay, which lies betueen two penneablelayers.A 2.5 cm thick sarnple of the soil is found to have undergone50% c
of theclay layerswhichwill takeplacein 1 year. [2?.SVol 6.12. An isolatedfootingof 2 rn x 2 m plan areais constructed overa saturatedsandyclay sfiatumof 5 m thicftness.The soil has the following properties. . p=0.36, Cc=0.3, w=35Vo, G=2.69. E=3WkN/m', Estimatcthc probablesettlcmentof the footingif it carriesa grossload of 225kN. 112.57clnl 6.13. A 5 m thick layer of normally consolidatedclay sa newly @nstru,ctedbuilding. The weight of sand overlying the clay layer is 660 grn/cm' while the new construction increasesthe stressat the middle of the ilay layer by 450 grn/r:n2. Compute rheprobableconsotidafionsettlementof rhe building. Given,

LL =397a, G =2,7, w=457o.

[439 cm]

t64

Problems in Soil Mechanics and Foundatian Engineering

6.14. The total consolidation seftlement of a.clay layer due to an imposed load is estimated to be 8.5 cm. A setflement'of 2 crtr tooL place in 15 days. Determine the time required tor 5O7oandgUVoconsolidation. [68 days; 292 daysl 6.15. The results of a consolidation test are shown below :

(kglcm2) Pressure

7 COMPACTION

The sample had an initial height of 2 cm and an initial mass of ll2,.O4 gm. After the completion of the test the oven-driedsample was found io weigh 81.39 grn. The specific gravity of solids was 2.71 and the dial gauge consrant was: 1 divn. = 0.02 mm. (a) Determine the equilibrium void ratio of the sample after each toad incremeut. (b) Determine the values of co-efficient of compressibility and co-efficient ofvolume change for various pressureranges. 6.16, An undisturbed sample of saturatedclay, collected from a depth of 5 m below G,L., was subjectedto a laboratory consolidation test. The initial diarneter and thickness of the sample were 7.5 crn and 2 cm respectively. The mass of the sarnple in the wet and dry s[ateswcre L75.2 gm and 138.8 gm respectively. The final cornpressions under various pressures are shown below:

Plot the e vs logl6 p curve and check whether the soil is overconsolidated. If so,determinethepreconsolidation pressure. Given,G = 2.67. ?

i

7.1 Introduction: Constuction of structureson weak soils (e.g., soft clay, loose sand, etc) sometimesrequires "stabilisation" of the soil mass, i.e., an artificial improvement of ib engineer.ingproperties, There are various methods of soil stabilisation, the most common one being the mechanical stabilisation,and the simplest technique of mechanical stabilisation is compaction A soil mass can be compacted by cither a dynamic process or a static one. In the dynamic metlod the soil is compactedby repeatedapplications of a dead load, while in tbe static method compaction is done by a steadily increasing static load. Generally, the dynamic merlod gives better results in coarse-grainedsoils and tle static compaction is suitable for less permeable fine-graincd soils. 72 Moisture-densityRelationships: Whilecompactinga soil inthe field, it is always desirableto compact the soil in such a way tlat its dly density is maximum. If a given soil is compactedunder a specified compactive effort, its dry dcnsity will be the maximum at a certain moisture content, known as the optimum moisture contenL Hence, before compacting a soil in the field, its optimum moisture content and the corresponding dry density must be determined in the laboratory. The test employed for this purpose is called StandardProctor Test. 73 Standard koctor Test In this test, samples of the given soil are prepared at various moisture contents and are compacted in t cylindrical mould, 127.3 mm high and having an internal diameter of 100 mm. The eample is compacted in tfuee layers of equal height, each layer being oubjectedto 25 blows of a compaction rammer having a self-weight of.2600 gm and a heigbt of free fall of 310 mm. Samples are cornpacted in the mould at increasing moisture contents. After each test, weight of the sample compacted is detcrmined and ie bulk and dry densitiesare cornputed. A curve is then plotted to show the variation ofdry density with moisture content (Fig. 7.1). The curve is usually parabolic in shape.Initially thc dry

Problems in Soil Mechanics and Foutdation Engineering

166 2.O

l

ry

U

:

I

t'8

> l 7

l

S t o n tl ord Pr octer I Test

-)

'ld -

We have,

(

E

3

At any given moisture content, the dry density of a soil in the fully saturatedcondition can be derived as follows:

I A A S H 0T e t

t.9

rd= ##c \

15

"'(7'1)

From eqn (7.1) it is evident that, for a given soil, an increasein moisture content will always result in a decreasein y4. Hence the zero air voids line is always a steadily descendfurgline.

r.5

t0

wG = e

F o r a f u l l y s a t u r a t e d s o i ls, = 1 ,

/

L

o

G,{n

l + e

wG=se

and

IA

3 rs

167

Compaction

20

25 2'0

W o t e rt o n t e n t( 7 " ) Fig.7.1

density increaseswith increasingmoisture content, until a certain peak value is reached. Further increase in moisture content results in a decreasein the dry dersity. The moisture content representedby the peak of the cune is the optimum moisture content (OMC) and the corresponding dry density is the maximum dry density of the soil under that particular compactive efforl For heavier field compaction, the moisture-density relationship can be investigated by the modified AASHO test. The test procedure is similar to that of Proctor test except tbat a heavier rammer (weight = 4900 gm, free fall = 450 mm)ris used and the soil is compactedin 5 layers. Under heaviercompaction,the moisture-densitycurve (FigJ.l.) is shifted upwards and simultaneously moves to the left, resulting in a lower OMC but a greater y;.o. . 7.4 Zero Air Voids Line: Compaction is achievedby the expulsionbf air from the voids. However, as the external load acts for a very short time, it is nearly impossible to drive out all the air from the voids. Thus, during compaction, a soil is not fully saturated.If the remaining air could be driven out, its void ratio would have been reducedand consequently,its dry density would have increased.The zero air voids line (FigJ.2) is a theoretical curve which representsthe relationship between water content and dry density of the soil when it is l00To saturated.

1.9

,\

U U

E

I.E

I = c

1,7

^ 807' Soturation |t ,l [i n e

- Z er r i o i r v ' o i d sl i n e I ( SoturotionlineI \ 10.0cl'

N

q,

\

?'.. L

o

\ t.5 1.4

l0

l2

l4

l5

l8

20

22

WuterContent {7") Fig.1.2 75

Califomia

Bearing Ratio (CBR):

The California bearing ratio test is

of immense importance in the field of highway engineering.The CBR value_

n8{fg3i!s1-prob&le The California bearing ratio is defined as the ratio of the force per unit area required to drive a cylindrical plunger of 50 mm diameter at.the rate of

168

Probiemsin SoilMeclnnics and FoundationEngineering

169

Compaction

1.25 mm/min into a soil massto that required to drive the same plunger at the same rate into a standardsarnpleof crushedstone.

Thus,

cBR=###"

xroovo

...(7.2)

or Y L

The test is performed by first compacting the given soil in tle AASHO mould at the specified compactive effort as stated in Art. 7.3. The sample is compacted upto a height of. 127 rnm at the particular moisture content and density at which the CBR value is required. The plunger is then driven into the soil under a steadily increasingstatic load. The settlement of the plunger is measured wifl the help of a dial gauge while the corresponding load is obtained fron the proving ring, From the results a load-settlement curve is plotted and the test loads for 25 mm and 5.0 mm penetrationare determined. The values of unit standardloads coresponding to thesetwo penetrationsare 70 kgcr# and 105 kg*? respectively. Therefore, the CBR-values at 2.5 mm and 5.0 mm penetrationscan be determined.

C u r v eB

o gr c

600

CL

400

= c o

orrected5mm Penefrotion

t

t

ct o J

200

Corre cfed2.5mm (mm) Penetrqtion \F\ Shift of origin Fig.7.3

Generally, the CBR valu e at 2.5 mrn penetation should be greater than that at 5.0 mm penetration. In that case,the former value is acceptedas the CBR value for design purposcs. If the CBR value correspondingto 5 rnm penetrationexceedsthat for 2.5 mm penetation, tle test should be repeated,However, if identical results are obtained once again, the CBR value for 5 mm penebation should be used.

EXAMPL'ES Proctortestareshownbelow-: Problem7.1.Theresultsof a laboratorv

7.5,1 Correction to the cume : The load-penetrationcurve should always be convex upwards (curve A in Fig.7.3). However, tiue to surface irregularities, tlre initial portion of the curve is sometimes concave upwards (curve B in Fig.7.3). The cuwe lhen. must be corrected in the following manner:

No. ofTesl

(i) The straigbt portion of curve I is projected backwarclsto meet the X-axis at O'. (ii) The origin O is shifted to O'. (iii) Subsequently,all penetrationsare measuredfrom the new origin O'. Thus, the points corresponding to 2.5 mm and 5.0 mm penelration should be shifted towards the right by an amount equal to the shift of origin. In order to simulate the worst possible field conditions, the CBR test is sometimes performed on seakedsamples.After compacting the sample in the mould" the sample is kept submergedin water for a period of 4 days, after which the sample becomesalmost saturated.The CBR test is then performed on this soakedsample.

I

.'

a

Wt. of mouldandsoil (kg.)

3.526 3.711 3.797

Water content (%)

8.33 10.40 12.23

4

5

6

3.906 3.924 3.882 L6.20

\7.92

20.39

The mould is 12.7cm high andhasan internaldiameterof 10 cm. The weightof the emptymouldis 1.89kg. (i) Plot the moisturecontentvs. dry densitycurve and determinethe optimummoisturecontentandthemaximumdry density. (ii) Plot tbezeroair void curveandtheL0Voair void curve. Given,G =2.68. Solution:

Volumeof themould= (x/4) (12.7)(L02)cc = 997 cr = 3.526 - 1.89

In the first test, weight of soil

= L.636k.g = 16369m w Bulk density,y =

1636 =

v= 9n

l.64gm/cc,

l7O


and,drydensity, \d - *-

f#*S::

2-2

= 1.515 gm/cc.

The dry density y2"of the soil correspondingto the zero a ir void condition may be obtained from eqn. (7.1).

-o

,/

*F-

\

I

);---

h.

I

/

1.1

I

1.2

louc='ts.z.

1s

15

ro

zo

22

Wotertontent {7.)

Table 7.1 a a

5

4

)

6

Wt. of mould and soil (kg.)

3.526

Water content (7o)

8.33 10.40

Wt. of soil (gm)

1636

t82t

t907

Bulk density (gm/c-c)

1.641

1.826

1 . 9 1 3 2.V22

2.UO 1.998

Dry density,ya(gm/cc)

1 . 5 1 5 t.654

r.705 t.740

1.730 1.660

Dry density for zero air void, p, (gmlcc)

2.I90

2.018

1.810

Dry density for lOVo air void, y9o (gm/cc)

2.I47

3.7t1 3.797 3.906

12.23 16.20 L7.92 20.39 2016

1.869

Frg.7.4 Problem T.|/.-=T,\eoptimum moisture content of a soil is 16.5% and its maximum dry density is 157 gnlec. The specific gravity of solids is 2.65. Determine: (i) the degreeof saturation and percentageof air voids of the soil at OMC. (ii) the theoretical dry density at OMC correspondingto zero air voids.

3.v24 3.882 ?n34

Solution: (i) When the sorl is at OMC, it hasa moisturc content of 16.57o and a dry densityof 1.57Bm/cc.

1992 Now,we have,

0r' 1.964

1.808

L.747

t.67

optimum rnoisture content = l5.2Vo and rnaximum dry density = 1.76 gm/c.c The zero air void line and the lovo air void line also are shown in Fig.

fd = * !'5't =

t.733

The compaction curve is shown in Fig. 7.4. From the curve we find,

7.4.

ta c o,

o

similarly, the dry densitiescorresponding to theactualproctortest,the zero air void condition,andthe lovo afuvoidconditionarecomputedfor the remainingfive tests.Table7.1 showstheresultsin a tabularform.

2.M6

d6or= 1.752 qm/cc

>\

(2.69)(t nt v, , 4 e o= t + 0 . 2 f r = z l 4 7 g m / c n

2.86

I z Z e r oa i r v o i d s

.--q-:<

1 0/o : oir voids z

U U

=t€

"_g_-(%|fa@ _0.248

1

9)--

2.O

g

(2.68) I * (0.0833)

Again, when the soil has IO% ait void, its degree of saturation isgOVo.

No. of test

{--

>E 1 . 8

G ' b -, , , ! ? 1 % l : 9 1 = .= Z . r e g m / c c

Y* " GG

17l

Compaction

oft

(2'65)(1'0) l + e

r + e= f f i = t . e a e = 0.688

Again, se = w G, or, ,f = n G e

(0.16s) (2.6s)= 0.635 = 63.5Vo (0.68s) Hence,therequireddegreeof saturationis 63.5%o of andthepercentage air void is (100-63.5)Vo= 36.5Vo (ii) At zeroair void thesoil is fully saturated, i.e.,s = 1.

I7Z

Problems k Soil Mechanics and Foundation Engineering

" = 8-

s

(2.6s)= - (0.165) 0.437

existing soil is less than its OMC. Hence,a certainamountof water is to be added to the soil prior to compaction' Now, 14*= 1.66 gnlec = l -66t/n3

l

(2.6s:" ^' =1.844gm/cc ta=ffi, Therefore, the theoretical dry density at OMC for zero air void = 1.844 gmlcc. During tie construction of an embankment, the density Probten.@. attained by field compaction was investigatedby the sandjar method. A test pit was excavatedin the newly compactedsoil and was filled up by pouring sand. The following were the observations:

Thus,for every1fi) rn3of finishedembankmen!the weightof dry soil requiredis, W7 = "tit*, 'V = (1.66)(100)t = 1661.

'w*\ W n = w . W|| d * = ' f r' J; t

Bulk density of sand = t.52gm/cc

l-

Moisture content of embankmentsoil = 167o Detennine tle dry density of the compactedsoil.

thepit,

sorurion: to"'";'="i: L)z \ Volume of the pit = 1550 cc.

Wt

B u t , 1 =7 i , o r , W 4 = \ d . V

And theweightof wateris,

Weight of soil excavatedfrom pit = 2883 gm Weight of sand required to fill the pit = 2356 gm

ffr

173

Compaction

= 37.35t = (0.22s)(166) content soilis 1'78t/# anditsmoisture of theexisting Thebulkdensity is9%6. Dry densityoftheexistingsoil, y4 = #;

J:: or'

But, weight of the soil excavatedfrom the pit = 2883 gm

thesoil, t = .'. In-situbulkdensityof #i

= 1.86gm/cc

And, in-situ dry densitv of the soil,

'P=tfl=I-#G=r'66gm/*' ,a

It is required to construct an embankmentby compacting a soil Prcblerudl excavatedfroln nearby borrow arcas.The optimum moisfure content and the corresponding dry density of this soil were determined in the laboratory and vrere found ta be 22.57oand 1.66 gm/cc respectively. However, the natural moisture content and bulk density of the soil were 9Vo and 1.78 gm/cr respectively. Find out the quantity of soil to be excavatedand the quantity of water to be added to it, for every !.00 m'of finished embankment. Solution: The embankrnentshouid be constructedby compacting the soil cbtained from borrow area at the optimum moisfure content and the corresponding rnaximum dry density. But the natural moisture content of the

Yr=1|fu

-r'633t/nf

The volume of soil, V6 to be obtainedfrom borrow areain order to obtain 166 t ofdry soil is,

, r = Y - #l.oJJ = 1 0 1 . 6 5 m 3 ld Weight of water availablefrom tbis soit, Wnr= Wd. w5 = (166)(0.0) = 14.94t .'. Quantity of water to be added=(37.35 -t4.94>t

tobeadded= ofwater Volume ##Hi# But,densitYofwater,Y,o = 1 gm/cc = 10-6 t/cc

= (1000) 1ro-6;tzfit = l0-3 tlir

174

Problemsin SoitMechanicsandFoundationEngineering

.'. Volumeofwarertobeadded= {22.4}) - 2L4l|litre. (10-') Thus,101.65m3of soil is to beexcavated from theborrowpi ta,.dz24ra litre ofwater is to be addedto it. Pnoblem?F. ett embankmentwas constructedby compacting a soir at a moisturecontentof ISSVoând a dry densityof t.ilTgmTcc.ti"tn. ,p."ln" gravityofsoil solidsbe2.d8,derenninethevoid ratioan-cl degreeof saturation of the embankment soilSolution:

Here,

In the loosest state, bulk densitY = dry density, yd_, =

( 3 3 6 3 . 6 - 2 1 -i 00)_12 l'J39 gm/cc (r43^3r) = =(t''?n)r. = 1.2o6gm/cc. ------Y( 1+ 0 . 1 1 ) l+w

8857.4-2100I b u l k d esni.t V - f t * f f = l . f i 6 2 g m / c c

la=l.72gm/cc,G=2.68

y,r* = drydensity,

- 1.505gm/cc Y/* ld - ldô Relativedeusiw. R^ = x l00vo . -

se = wG, or " = Ag e (9.155)-g{s} " = - 0.744 = 74.4vo (0.55u1

Again,

Yd

Weightof mould+ soil in thelooseststate= 3363.6grn Weigbtof mould+ soil in rhedensest state= 3g57.4gm

$"n

Y/r""

Ydro

={i#i{i#+#i 0oo)'/.

Problem7.6.rnorderto determine thererativedensityof a sandsampre, its naturalmoistureconte.ntandburk densityweredetermined in the fiero and were found tobe 770and1.61grn/ccrespectivery. Sampresof this soirwere thencornpactedin a procror'smourdof i/30 cft capaciiy, at the toosestand thedensest states. The followingdatawereobtained: Weightofemprymould= 2100gm

Solution: Volunreof thc mould =

= r.677gm/cc

ln-sirudrydensity, Va= ffi

e = 0.558.

Moisturecontentof thesarnpleusedin lests= llo/a Determinetherelativedensityof thesandandcommenton

#Tfrb

In-situ bulk densityof the soil = 181 gm/ccand its naturalmoisture cortent = 7%b

t+"-ffi

ot,

=udffiul cc=e43.8ecc

ln the denseststate,

We have,p =

_ (a*ga 1.72 ort

t75

Compaction

its type.

= 70.74 7o / Problernl]l It is required to construct an embankment having a total volume of 64000 cu.m. The required soil is to be collected from borrow pits. It was found that tbe cxisting soil has a moisture content of l4Vo, void ratio of 0.63 and specific gravity of solids of 268. I-aboratory tests indicate that the OMC and maximum dry density of the soil arc l9.5Vo and 1.72 gm/cc respectively. The soil is to be carried from the borrow pit to the construction site by trucks having averagenet canying capacity of 5.5 t. Determine the total number of trips the trucks havc to rnake for constructing the entire embankment.Also find out the quantity of water to be addedto the borrowed soil before compaction. Solution:

In-sirudrydensity of thesoil, ,u = fP l +

e

176


= g'6CIi1'P= L64sm/cr (1 + 0.63)

Compaction

177

Solutiron: When the rock presentin the fill is compactedto the densest state, its dry unit weight is given by,

= 1.64t/m3 .'. In-situbulk density,\ = \a(1 + w) = (1.64,(1 + 0.1a) = L87 t/m3 Now, in 1 n3 of borrowedsoil, quantityof dry soil presentis 1.64t, and quantityof waterpresent= (0.14)(1.64) Wn = w . W6l t... = O.23t while constructingrheembankment, this soil hasto be comDacted at a moisturecontentof l9,5Voandat a dry densityof 172 t/m3. For I m3of finishedembankrnent, dry soil required= 1.72t,

G'{n -i-; (2.s6) (t.o) _ = = 1.48gm/cc. Y/.- = | . " _ o;i

Let us now consider 1 gm of the given fill. According to the question, the weight of rock and soil present in the fill are 0.8 gm and 0.2 gm respectively. \_ Now.volumeof0.8gmofrock

andwaterrequired = (0.195)(1.72\t = 0.335t. quantityof dry soil required= (1.72)(6.t000)= 1,10,0g0t

.'. Total volurneof excavationrequiredto be made=

*P

.'. No. of trip to be made- l'2!'!L8 = 22.812.5- 22,g22 5.5 Weight of waterobtainedfran 67,122m3of borrowedsoil = (67,122)(0.23)r = 15,438r 'Weight of water

finatly required= 2!,4N L

""Quantitv' *".:il;,#'_',o,:HrTlffiX'l Problem 7.8. The rock content in a filr is 8TToby dry weight. The rock can be compactedto a rninimumvoid ratio of 0.73.The maximum dry unit weight to whic.h the soil fraction can be compacted is 1.63 gm/cc. wbat is the maximurn dry density to which the fill can be compactid ? Given, specific gravity of the rock = 2.56.

= 0.54cc.

#cc #

= 0.123 cc:.

Total volurne of 1 gm of fill

= 0.54 + 0.123 = 0.663 cc.

Dry unitweight of rhe filt

= =

W L - 1.508gm/cc. 0J63

Problem 7.9. The results of a laboratorv CBR test are shown below :

= 6-7,122 n3

Grossweightof thissoil = (1.84 (67,lZZ\t = 1,25*51g r

=

and, volume of 0.2 gm of dry soil =

.'. For the entireembankmentof 64000m3 and, quantityof waterreguired= (0.335)(64000)= ll,llg 1 - As the in-situdry densityof existingsoil is 1.64t/m3,everycubicmetre of excavation will produce1.64t of dry soil.

1d^, = l-63 gm/cc.

For the soil.

3

No. of tesl Penetratron

0

0.5

Lmd (kg)

0

19.8 < t <

(-.)

1.0

4

1.5

5

6

)n

?.5

3.0

8

9

4.0

5.0

93.7 t3zl 171"9 20"t.o288.8

a)7 )

t0 't.5

II

L:

10.0 L?.5

401.7431.8458.3

Determine the CBR value of the soil. Given, unit standardloads for 2.5 mm and 5.0 mm penefafions are 70 and 105 kg/cm" respectively. Solution: Fig.7.5 shows the load vs. penetraiioncurve. As the curve is initially concaveupwards, an initial zero correction is required. The shaight portion of the curve is projectedbackwardsto intersect the X-axis at O , which then becomesthe new origin. Consequently,all points on the penetration axis are shifted to the nght by an equal amounl From Fig. 7.5 we obtain, test load for corrected2.5 mm penetration= 200 kg. and,

test load for corrected5.0 mm penetration= 332k9.

Area of CBR plunger

= * fS.Ol'"*2 - 19.635cmz 4 '

178

Problems k Soil Mechanics andFoundation Engineering

.'.Unittestload forL|mmpeneftarion = ffi. and,unittestloadfor 5.0mmpenetrat ion- I lv.oJJ

-t6.- *

CBR value for 2.5 mm penetration = 10.19 CBRvalue for5.0mm penetration= ' I

= 10.19 kg/cn| = 16.91ks7cm|

(i) Plot the water content vs. dry density relationshipand determine the optimum moisture content and the correspondingmaximum dry density of the soii. (ii) If the specific gravity of soil solids be 2.70, pl
1.007o= 14.6%

x l(fi%o = 16.lLo

105

Thus, CBR value for 5.0 mm penetrationis greaterthan that for 2.5 mm penetration"Therefore, the cBR test has to be repeatedand if similar results are obtained once again, then the cBR value of ie.tvo should be accepted. 500 400 ET

)<

7.2.The resultsof a standardProctortest are shownbelow. Water C-ontent(%)

11.6

14.9

t7.7

20.r

22.s

Wt. of soil and mould (gm) 3263.4 3523.28 3734.8 3852.9 3832.7 3765.1

The height and intemal diameterof the mould are 12.6cm and 10.1 cm respectively. The ernpty mould weighs 1950 grn. Plot the compaction curve and determine ttre optimum moisture contentand the correspondingdry and bulk densitiesof the soil. AIso plot the zcro air void line and tbe SOqosaturationline. Given, specific gravity of solids = 2.69. [Ans: OMC = l7Vo,y7 ='1.6gm/cc, 7.3. Tlre in-situ dBnsity of a soil mass is being determined by the'''ifure cutter method. The height and internal diameterof the core are 12.7 cm and L0 cm respectively and is weight, wheu empty, is 1847 gm. When the core is filled with soil, it weighs 3674 gm.If the specificgravity of solids be 2.67 and the degree of saturation of the soil be 63%, determine the in-situ dry densityof the soil. The in-situvoid ratio of the soil is found to be 0.85.[Atts. 1.526gm/ccl 7.4. An embankmentof hapezoidalcross-scctionis to bc constructed for a 2 knr long highway. The embankmentshould have a height af 2.2 m and a top width of 10 rn. The sides of the embankmerttare to be sloped at 2H : 1' V. The soil obtained from the borrow area is tested in the laboratory and is found to have the following properties: Natural moisture content = l2Vo

c,

EN

= crc o EI E CI

-,

P e n e t r o t i o(nm m l Fig.7.5

= 1.8{./m3 In-situbulk density Optirnummoisturecontent= 19%

E)GRCISET 7.1. The following are the results of a proctor compaction test performed on a soil sample"

Bulk Density (gm/cc)

7.8

y = 1.87gm/ccl

L

WaterContent(%)

179

Compaction

9.2 I.524

L2.7 1.749

15:5 1.949

18.3

2.U9

2A.2 2.4t9

Drv derrsitvat OMC

= t.65thf

Determine the quantity of soil to be excavatedand the quantity of water the embankment.[Ans:65055 m'; 7318 to-beaddedto it beforec<mstructing

.31 7.5, Determine the magnitudesof compactive effort imparted to a soil during: (i) StandardProctor Test

! i

i I

I l1 i1

,l jt

';

Il )

Problems in Soil Meclunics ond Fottndatian Eng,ineering

180

Modified AASHO Test' x no' of [Hina: Compactive effort = Wt. of rammer x height of fall x no. oflayersl blowslayer 7.6.The speciticgravity of solidsof a soil is 2'65. Detenninetbe quantity of dry soil audwater requiredto c0lnpactthe soil iu a Proctor rnould having D = 10 cm and H = 12j cm, at a void ratit'rof 0.6 and at a moisture content of 207o.[Ans: 1652 gm; 330 crc] 7.7.T\ree identical triaxial test samplesof 7.5 cm height and 3.75 cm cliameterare to be preparedat a moisture content of 15Voand a dry density of 1.48 gm/cc. Determine the total quantity of oven-dried soil and volume of water requiredfor the purpose. [Ans: 367.8grn, 55.2 cc] 7.t. Determine the CBR value of a given soil from the following data obtained from a laboratory CBR test : (ii)

Load (kg)

0

19.8

50.1

81.8

Penetration (mm)

o

0.5

1,0

r.5

120.0 170.1 421.7 605.9 699.3 662.8 '7.5 lo.0 r2.5 5.0 2.5 2.O

Plot the load-penetration curve and determine the CBR value of the soil' Conrment on the test results. fAns.23.7%l

8 SHEAR STRENGTH When an external load is applied on a soil mass, 8.1 Introduction: sbearing stressesafe induced in it. Ii the shearstressdeveloped on any plane in the soil exceedsa certain limiting value, failure of the soil occurs. The maxirnurn shear stresswhich a given soil can withstand is called its shear strength. The factors goveming the shearstrengthof a soil are : (i) internal frit:tion, i.e., the resistancedue to particle interlocking (ii) cohesion, i.e., the resistancedue to the internal structural bond which tends to hold the particles together. According to Coulomb's law, the shearstrength,r, of a soil is given by: 'E = c + otan ...(8.1) 0 where,

o = normal stressacting on the soil c = cohesion

0 = angle ofinternal friction The factors c and S are called the shearparametersof a soil. When expressedgraphically, eqn. (8'1) can be representedby a straight line called the failure envelope;The general form of failure envelope for a cohesionless,a cohesiveand a c - 0 soil are shown in Fig. 8.1 (o.),(b) and (c) respectively.

T {

I

T

c

J(bt Fig.8.1.

L82

Problems in Soil Meclnnics and Foundation Engineering

The shearpararnetersofany soil dependuot only on the nature ofthe soil but also on such factors like rnoisturecontent and loading conditions. At very low moi_sfurecontent a cohesivesoil may developa certain amount of internal friction. Likewise at high rnoisture conlentsa cohesionlesssoil may show the signs ofhaving an apparentcohesion. 8.2 Mohr's circle of stress: This is a graphical representationof tle stress conditions in a soil masswhich enablesone to find out the stressesdeveloped on any plane within the soil due to an external loading system. In a stressedmaterial, a plane which is subjectedto only a normal stress, but no shear stress, is called a principal plane. Through any point in the material, two such planes exist. These planes are called the major and the miror principal planes, and are ortlogonal to each other. If lhe principal stresses,01 and g,3,?te known, the normal.stess o and shear stressr on a plane inclined at an angle 0 to the major principal plane is given by,

and,

"=Y.rycos2o

...(8.2)

" = 9l-:-or z

...(8.3)

,in 29

183

Shear Stength

8.3 Pole: The conceptof the pole, or the origin of the planes,is very usefrtl in such problems where the locations of the principal planes are not known. Consider the soil element subjectedto a system ofexternal stressesas shown in Fig. 8.3. It is required to determine the normal and shear stresses acting on the planeAA, inclined at an angle 0 to the horizontal. Considering the free body diagram of the element it can be proved that the element can be in equilibrium only if, T, = ayx. The procedure for drawing the Mohr Circle and locating the pole are as follows: (i) Choose tle co-ordinate axesand selecta vector scale. (ii) Locate the points A and B such that they representthe stresseson tbe horizontal and vertical boundariesr€spectively,of the element. (iii) AB. It intersectsthe o-axis at C. (iv) With C as centre and CA = CB as radius, draw the Mohr circle. (") The point A representsthe stress conditions on the horizontal plane. From A, draw a straight line parallel to this plane. It intersects the circumference at P. Again, if from .B a line is drawn parallel to the vertical plane (since the point B representsthe stressesacting on this plane), it will intersect the circle at tle samepoint P" This is the pole of Mohr's circle.

Equations (8.2) and (8.3) can be represented by a Mohr Circle, as illustrated in Fig. 8.2. The co'ordinates of any point on the circumference of the circle give the stressconditions on a particular plane representedby that point ^ +?

T (-r rTtyl ^ -

!y, .'A

xv'

M

Try

(0,0)L/

QlqT) A Txy

cr

-t- llqjor hincipol Shess

"i+"3

Io

vlrv

"i-"3

,-

T Fig.8.3

2

-t- FliRorhincipol

Strrss

a,

(-,Tl \

c\

e

-T Fi8.8.2

(uD From the pole P draw a line parallelto the planeon which the shessesarerequired.This line intersectsthecircleatQ.T\e co-ordinates of give the nonnal and shear stresses on given planeAA. the Q Thus,thepolemay be definedasa particularpoint on thc Mohr's circte suchthat, if a line is drawnfrom this point makingit parallelto any given

184


plane within the soil mass,lhen, the co-ordinatesof the point of intersection of this line with the circle will representthe stressesacting on that plane. 8.3.1 Sign convention' The following sign conventions are rormallv followed for plotting the stressco-ordinates: Normal stress: Compressive stressesar€ taken as positive and tensile stressesas negative. However, soils can with stand only compression and not tension. Hence the normal stresson any plane of a soil element which is in static equilibriurn is always positive. Shcirrstrcss: The sign of a shearstressis determinedon the basis of the direction of its moment about any arbitrary point inside the soil mass. If tbe moment acts in the anticlockrvisedirection, the shearstressis positive, whereas if it acts in the clockwise direction. the shear stress is negative. 8.3.2 Ltrcation oJ the failure plane : Fig. 8.4 representsa soil sample subjected toa rnajor principal stresso1 and a minor principal stress03. As the sample is on the verge of failure, the Mohr circle has touched the failure envelope at P. Evidently, the pole of the Mohr circle is at A. The highest point on the circumference of the Mohr circle is the crown R. The lineAR is inclined to the o-axis at 45'. The corresponding plane in the soil is MN, which is the plane subjected to the maximum shear stress, r,no. Howev er, theIntential failure plane in the soil is not MN, but theplane represented by the poiru P, becausethe stressco-ordinates given by P are

,B / , N

ShearStength

185

such that coulomb's equation is satisfied as the point P lies on the failure envelope. In order to determine the location of this plane, PA and PC. Now,

LPCB=LPAC+/-APC AC = PC,

As

LAPC = LPAC = a

LrcB=ct+ Again,since InAPDG.

q = 2 a ,

DF llOB , LPGF = LPCB = 2c. LPGF = LPDG + LDPG 2 a = Q + 9 0 " [ ,. . P G L DE, .'. LDPG = 90" I

or,

a = 45' + g/2

or,

...(8.4)

In Fig. 8.4, the planeBB, drawn at (45' + Q/2) to the majorprincipal plane,representsthe failure plane. It r:anbe proved that, at failure the relationshipbetweenthe two principal stressesis given by, or = 03 :filoz 145" + Q /2) +'zr'tan@s" + g /2) ...(8.5) o1 = o3.f{6 + \ctlfia

or, where,

" ffO = flo'w value = tan (45" + 0 /2)

...(8.6)

...(8.7)

8.4 Determination of Shear Strength: The following testsare employed tbr the evaluationofthe shearstrengthofa soil : A.

l:boratory tests : 1. Direcl ShearTest 2. Triaxial CompressionTest 3. UnconfinedCompressionTest.

Field Test : 1. Vane ShearTest For a detaileddescriptionof the testprocedures,the readeris refened to any standardtextbook of Soil Mer:hanics.Only the essentialpoinf,sregarding the computationof shearstrengthwill be highlightedhere. B.

-3

H, ,r8/

J

A e

Fig.B.;

8.4.1 Direct Shcar Test: In this test,soil samplcscornpactedat known densitiesand rnoisturecoutentsin a shearbox of 6 crn x 6 cm size,which can be split inlo two halvcs,is shcareCby applyinga graduallyincreasing!ateral load. Three identical sarnplesof a soil are testedunder ditferent vertical compressive stressesand the corresponding shear stressesat failure are determined.A graph is then plotted betweennormal stressand shear qtess. Resultsof eachtest are represented by a singlepoint. Three points obtained from the lhree testsareed by a straightline which is thc failure envelope for the given soil. The siope of this line gives the angle of internal friction, while the interceptfrom the r-axis gives the vaiue of cohesionof the soil.

186

Problemsk SoilMechanicsand FoundationEngineering

8,4.2 Triaxial Cunpressian Test: In lhis test, cylindrical soil specimens of 3.8 cm diameterand 7.6 cm height,enclosedin an impermeablerubber rnembrane,areplacedinsidetheniaxial cell.An all-roundcell pressure,o3, is appliedon the sample.Simultaneously, a gradually increasingvertical stressis applieduntil eithertle samplefails, or its axial stain exceeds2A%. Stressvs. straincurvesareplottedto determinethe normalstressat failure. This stressis calledthedeviatorstress,od. The majorprincipalstress,o1, is obtainedfrom the following relation(referFig. 8.5) : O1=O3*O4 ...(8.8) oi= "3*-d

V r t L V

A'= - nt i where,

[o)

(b)

(c]

Fig.8.5 Three samplesofa soil are testedunder different cell pressures.From the results, three Mohr circles are construcied, and a common tangent is drawn to them. This is the failure envelope. The normal stressat any point during the test is determined by dividing the normal load obtained from the reading of the proving ring by the cross-sectionalarea of the sample. Due to the bulging of the sample during shear, the cross-sectional area should be modified using the following equation : Ac = As/(t - e) ...(8.e) where, Ac = coffected area A0 = initial area where,

e = axial sf]ain = M/L A L = axial compression L = initial length

In the drained tiaxial tests,the volume of the sample may changeduring the test due to expulsion or absorption of water. In that case, fhe corrected area should be detennined from :

...(8.10)

Vt = initial volume of the specimen A 7 = changein volume due to drainage. Ir = initial length of the sPecimen AI = changein length of the specimen

8.4.3 I) nconJircedCompression Test : This is a special caseof triaxial test in which o3 = 0. We have, from eqn. (8.5) As

q3ta6 r1 =

187

Shear Strength

+ Q/Z) ar = o3vn4 (45' + 6/21 + ktang," 03 = 0, for an unconfined cotnpressiontest, or = 2c tan(45' + Q/2)

...(8.11)

A number of tests on identical specimenswill give the same value of o1. Thus, only one equation is availablewhile two unknowns, viz., c and f , are involved. Hence, eqn. (8.11) cannot be solved without having a prior knowledge ofany one ofthe unknowns. P Due to this reason,the unconfined compression test is employed to determine the shearparametersof purely cohesive soils only. For such soils, S = 0", and hence, ol=2clan45" =2c TorqueRod The vertical stress o1 at failure, known as the unconfined compressivestrength and denoted by q* is obtained by dividing the normal load at failure by the correctedarea,asgiven by eqn. (8.9) Tlus,

or,

Vones

qu = 2c

eu c=T

...(8.12)

8.4.4 Vane Sheqr Test.' This is a field test used for the direct deterrninationof the shear strength of a soil. Generally this test is conductedin soft clav situated at a lreat deoth- samolFdrEf,ich are

airfi.ffil The apparatusconsistsof fourmetal blades, called vanes,mounted on a steel rod, as shown in Fig. 8.6. The device is pushed slowly upto the desired depth

I.

r/

f

I

l- o-l Fig.8.6

Vanes

188


189

Shear Stength and is rotated at a uniform speedby applying a torque through the torque rod. The amount of torque applied is recorded on a dial fitted to the rod. Failure occurswhen the vane can be rotatedwithout any further increasein the torque. For a cohesive soil, Q = 0. Hence coulomb's equation reducesto : S = C

Thus, for a cohesive soil, the shearsrrengthis equal to its cohesion.In a vane

shear test, thecohesion, r"I:y

*.ryry*n

bederermined from:

:,"=.d4{A \ I t2 6l I

_ i . - -

'

"(sr3)

+ - l

I : torqueapplied (= p.a) i/ = heightofthevane

where,

D = diameterof tlle vane. t.5 sensitivity: when the shear stressesdeveloped in a soil exceeds its .shearstrength, the soil fails by shear and loses its strength. However, if rhe soil is left in that state for some time, it regains some of its original strength. The sensitivity of a soil is a measureof its capabilify of regaining strength after a disturbance has been causedin the soil. It is expressedas, shear stength in the undisturbed sute o ...(8.14) ' shear strength in the remoulded state

a =

and,

Nature of clay

I t-2

Insensitive f,owsensitive Medium sensitive Sensitive

2-4 4-8 8-L6 >16

Extra sensitive Quick clay

The givenplaneis inclinedat 30' to the majorprincipalstress.But the to the major principal directiouof major principalstressis perpendicular given planeandfte major between the plane.Hencetheangleof inclination principalplaneis, 0=90'-30"=60P (2 - l.l) (2 + t.l') * --T-.cos(2 , /^ .x 60") ar T = 1.55 + (0.45)(cos120') = 1.55 + (0.45)(-l/2)

- r.r\

Q r = #sin

and

sclution: (a) Analytical method: The normal stress,o and the shear stress,r on any plane inclined at 0 to the major principal plane is given by : o ==0--? 1 * o 3 '* 0 -T 1 - o 3

L_--

.^ x 60') (2

- (0.45)(sin120') = 0.39kg/arr2 TlKglcnz]r 0.75

0 t1.33,0.39)

0.50

I 0.25

I

0'39K9

0 t-75

EXA.MPLES

Pnoblern V. A soil sample is subjectedto a major principal stressof 2 kglon- and a rninor principal stressof 1.1 kg/r-' . Determinethe uonnal and shear stressesacting on a plane inclined at 30o to the nrajor principal stress.

cos2o

:

= l.32skg/cmz

f.5 c .

sin 20

z

ot = 2?,g/crf and o3 - t,lkq./cri

Here,

on the basisof the sensitivity,clayey soils are divided in the followine categories: Sensitivity

ol-('3

2.5 { ( K g/ c m 2 l

Fig.8:7

(b) Grophicalmethad:Thegraphicalsolutionis shownin Fig 8.7'The procedure is statedbelow: (i) TheMohr circleis drawnwith at = 2'0kg/cmz andcr3= 1.1kg,/cm2

190

Problems in Soil Meclwnics and Fottndation Engineering

Shear Strengtlt

l m2 y

(ii) From the centre C of tlris circle, CD is drawn at atr augle of 120o ( = 2 0) to the o-axis. This line intersectsthe circle at D. Altenratively, frorn the pointA correspondingto s3, a straight lineAD is drawn at an angle of 600 1= 0) to the o-axis.AD also intenects tbe circle at the samepohrt D.

N(40,'t0)

:16'5kN/m?--+1

(iii) The co-ordinatesof D give the normal and shearstressesacting on the given plane.From Fig. 8.7 we obtain,

t T = 3 ' 6k N /

o = 1.33kg/cmz

30

r = 0.39kg/crnz Problem

Y.

19.

LS S

m. stressesactingon a soil elementdre shown in Fig. 8.8 (a).

10kN/m2

M

(20,-10)

2 0 k Nm /

iot
_T(kN/m2)

to)

Fig.8.8(b)

6. From P, FQ ll ,ttr is drawn to intersectthe circle at p. The solutions to tle three giveu questionsare now obtained as follows :

D Fig.8-8(a)

(i) The points of intersection,R and $ between the circle and the o-axis give the principal stresses.Here,

(i) Detennine the magnitudeand direction of the principal stresses. (ii) Find out the stressesacting on the plaue XX.

ol = 48 kN,/rn2

(iii) If the soil hasa colresionof 5 kN/m2 and an angleof intemal friction of 25o,find out whelher a shearfailure is likely to occur along the planeXX.

ot = l6.2kN/m2

and

Solution:The graphicalsolutionof the problem is presentedin Fig. 8.8 (b). The procedureis as follows:

In order to locate the directions of the principal planes,the points R and S areed to the pole P. Through any pontZ in the soil elelnent, Z -lllPS and Z - 3 ll PR are drawn. The planes Z - t andZ - 3 give the directions of the major and minor principal planes respecrivcly. (ii) The strcsseson,lXare given by the co-ordinatesof Q. From the figure we obtain.

1. Two orthogonal c'o-ordinateaxes and an appropriatevector scale (1 cm = S *NZm2) are chosen. 2" Tbe points M (20, -10) and l{ (40, 10) are chosen to represent the slresseson the planesAB and8C respectively. 3. M andif areed and the mid-point O of MN islocated.

on

4. Witlr O as cenlre and M]{as diameter, the Mohr circle is drawn.

= 16.5kN/m2 andrps=3.6kN/rn2

(iii) The normal stresson.lXis 16.6kN/#. Frsm coulomb's equation, the shear strength of a soil is given by, s - c + otano

5. The pointilf representsthe stresseson the planeA.B.FromM, a straight line ifP is drawn parallel to AB, to intersectthe circle at P. P is the pole.

l

t92

Problems in Soil Mechanics qnd Foundation Engineering Here, c =5 kN/rn2,o = 16.6 kN/m2, 6 = ?5" .'; s = 5 + (16.6)(ran25') = 12.74kN,/m2 > 3.6 kN,zmz Asr

< s, ^failure along)Q( isnotpossible. Problern si.]r4he stressesactingon an elementof erasticsoil mass'areshown in Fig. 8.9 (a). Determinethe nonnal and shearstresseson the planeXX.

193

Slrcor Strengtlt

(ii) FromA (representingthe stressconditionson the plane bc) d:,awAP bc, to intersect the circle at P. This is the pole of the Mohr circle. ll Alternatively, if frorn B, BP ll bc is drawn, it alsowill intersectthe circle at P. (iii) FrornP, drawPQll ,XX.It intenectsthe circle at Q. The co-ordinates of Q give the stresseson the planeXX. From the figure w'eob'.ain,

oxx = 2'35t/mz

5Kg/cm2

rnr = 0'95t/n2

artd

ZKglcnz

Problem S;&;The stressconditionson a soil elementare shown in Fig. 8.i0

(a). 5Kglc nz

X

r

(i) Find out graphicallythe stresseson the plane,4,4. (ii) Draw a freeboclydiagramof thesoil elementandshow thesestresses. (iii) Prove ihat the free body is in equilibriurn.

50kN/m2 (o)

5 5k N / m 2

(a) Fig.8.e

40kN/m2

solution.'Fig. 8'9 (b) slrows the graphical solution. The solution is obtainedin thc lbllowiug sts: (i) Locate the pointsA (2, 0) a^d ^B(5, 0) which represe^trhe pri^cipal stressesacting on the soil element.with AB as diameter.draw the Mohr's circle.

8.61 kN/m2

T ( k N/ m 2 ) A

I

a?

50k N/m2 60kN/m2

a 7 ( K g/ c m 2 )

55kN/m2

q__-L 8.57k N/m2

30

40

cr ( k N/ m 2 )

+ 2 0.95 Fig.3'it-'-

P.

-T(Kglcnzl

tb) Fig.8.e(b)

solution: (i) The graphicalsolutiouof the problernis shown in Fig. 8.10 (b), from which we get, ,\ o = 55 kli/m-


t94

r = 8.67 kN/m2 (ii) The free body diagram of the soil element pqr, bounded by the vertical plaile, the horizontal plane and tbe given plane AA, is shown in Fig. 8.10 (c). (iii) The free body will be in equilibrium if the sum of the components ofall forces acting on it along any two orthogonal axcs separatelybe zero. Let,

P4 = 1unit, p o

l

pr=ffi=6;=2units

g r'e' and-

| ar ' 7 'o -PQ tanh=m-{3n

-

1'732unia

Consideringunit thicknessof theelement, 2 Fx = (- 40) (1) - (8.67)(cos30) (2) + (s5) (cos60")(2) =-40-15+55-0. x Y (- 60)(r.732, + (&67)(sin30")(2) + (5s) (sin60")(2) - -103.92 + 8.67 + 95.25 - 0. Hencethe freebody is in equilibriurn. subsoitata siteconsistsof a 10m thick homogeneous layer Probbm ffie ofdense sandhaving the follouring properties: la - l.62gm/cc' G - 2.68,0 - 35' The nahrralgroundwatertablelies at 2 m belowthegroundsurface. (i) Determinethe shearstrenglhof tbe soil along a horizontalptane througbthe middleof sandlayer. (ii) If duringmonsoon,thewalertablerisestothegroundlevef,how will the shearstrengthalongthesameplanechange? Assumethatthe soil is dry abovewatertable. Solution: Tbe horizontalplaneunderconsiderationis at a depthof 5 m below the G.L. We have,

or, OI,

n{d-

GTn

Shear Stength

195 G + e Ysar= ll;.Yw

Now,

_

2.68+ 0.654,. ^, = -l-a g654 tr'ut

Z.OTgm/cc= 2.02Vm3 (i) The normalstresson thegivenplane., a = ld, 21 + l*6. 22 = (r.62,(2) + (r.02)(3\ = 6.3t/m2 ..'. Sbearstrengthof tbesoil at thisplane, ., = c + otano - 0 + (6.3)(tan35') - 4,41t/n? (ii) In this casetheentiresoil massis submerged. o - ysub.z= (1.02\(5,- 5.tt/m2 = (5.1)(ran35) - 3.57t/mZ -/,, s Problen-td Specimens of a silty sandweresubjectedto the directshear testin thelaboratory,in a shearboxof 6 cm x 6 cm size.Thc normalloadand theconesponding shearforcesat failureareshownbelow : and,

Draw thc failureenvelopeanddeterminetheapparentangleof shearing resistance anfcohesionof thesoil. Solution:Thecross-sectional areaof theshearbox = 6 x 6 = 36 cm2. Therionnalandshearstresses arefirst obtainedusingtherelation, load = SlreSS

area . . These are shown in a tabular form below :

lll'

E?P ' r.62 e = 0.654

IU

?o

30

Shearforce (kg)

9.90

t5.4t

20.88

Normal stress(kg/cm2)

o.28

0.56

0.83

0.275

0,428

0.580

Normal load (kg)

Shearstress(kglcm")

)

Problems in Soil Mechanics and Fonndation Engineering

L96

In Fig. 8.11 the nonnal and shearstressesareplotted along the horizontal and vertical axes respectively. Three points thus obtained are thened by a straightline. This is the failure envelopefor the given soil. The intercept ofthe failure envelopeon^ther-axis representsthe apparent cohesion,which is found to be AJZkg/an'. The apparentangle of shearing resistance is given by the angle of obliquity of the failure envelope to the horizontal, and is found to be 28.5'.

Shear Strength

tgl

be tlre radius whicb must be perpendicularto OQ, since Oe is a tangen/to the circle. Tlrus, in order to locatelhe centreofMohrcircle, draw eC L Oe. eC meets thc o-axis at C, which then, is the ccntre of Mohr circle. (v) With C ascentreand as radius,draw the Mohr circle. It intersects the o-axis atA andB, which, then, representthe minor principal stresso3 and the major principal stresso1 respectively. From Fig. 8.12, we obtain,oe = 1.08kg/5y1rrz,ot = 2.47 kg/cm?.

e\I

E

(vi) Dnw a horizontal lne PQ through Q. It intersectsthe circle at p. This is the pole of the Mohr circle.

8.5

g0'5

q

(vii) PA andPB. Thesetwo lines are parallel to the directions of the planes on which 03 and 01, respectively,act. From the figure we obtain,

3 a.t

o, L +

AAB

v) u 0.2

c =0 . l 2 k g l c n z

a./l

0|

- 32.5' and LPBA = 57.5'

Hence the minor and the rnajor principal planes are inclined to the horizontal at 32.5" and 57.5" respectively. The orientation of the planes are shown in Fig. 8.12 (b).

ct OJ

lo.2 0.4 06 O.g lo N o r m oS l t l e s s , r ( K g l c m 2) Fi8.8.11

,

Problenr 8/ A direct shearte.stwas performedon a sample of dry sand. of 1.5 kg/on", failure occurred when the shear stress Under a nbnnal stre-ss a reached 0.65 kg/cm'. Draw the Mohr circle and the failure envelope. Hence determine the orientation of the principal planes and the magnitude of the principal slresses. Solution: The construction is shown in Fig 8.12. The procedure is as follows: (i) Choosetwo orthogonal co-ordinateaxesand a_suitablevectorscale. The scale chosen in this problem is : I cm = 0.4 kg/cnr2. (ii) Locate the point Q corresponding to o = 1.5 kg/*rz 0.65 kgr'crn?.

and r =

(iii) Since the soil is a dry sand, it should not have any apparentcohesion and tle failure envelope should through the origin, the origin O and the point 8.O8 is the failure envelope. (iv) The point Q representsthe stresseson tle failure plane. But in a direct sheartest, the failure plane is aly3lrs horizontal-.Now, the point p must must touci the iailure envelope. If Q can be ed to the centre of the circle the resulting line will

GI F

; t.2

gr .Y

7oJ o't

Principot Ptane

L

+ t/l L 0 { o

(b)

CJ E t/l

0

2,4q 2.8

NormolS tress(lQ/cm2l (a) Fig.8.12

Prrblem $r(flrree identicalspecimens of a partiallysaturatedclay were subjectedto an unconsolidated undrainedtriaxial test and the following resultswereobtained:

198

,SampleNo.

Crll?ressure(W*\

< l 1= o 3 l V 1+ 2 c ' l N 6

D.eviarorstress (tg/cmJ

1.

0.5

0.80

In caseof tlre first sample,o3 = o.Sl
2.

1.O

o.97

Substitutingin eqn.(8.6) we get,

1.5

1..t3

0.5JV{+Tcfi=t.3

Determinetheshearparameters of tle soil(i) graphically(ii) analytically. solution: In a triaxial testthe cell pressureactsas the minor principal stess,while themajorprincipalstressis thesumof thecell pressureandthe deviatorstressal failure.Tbe valuesof o3 ando1 areshownbelow: SampleNo.

os (kg/cm2)

oa(kglw2)

or (kgcqr2)

0.5

0.80

1.30

2.

1.0

o.97

I.97

3.

1..5

L.t7

2.67

1.

199

ShearStength

Problems in Soil Mechanics and Fowtdation Engineering

-

where

ilO = t o? (45" + Q/21

Similarly, for the second and third samples, the following equations are obtained:

and,

Nq+2c4$=1.97 1.5/V6+ k,/q = L63

....(ii) ...(iii)

(i) from(ii) weobtain, subtracting 0.5f0 - 0,67, of, ilO = 1.34 ort

{i) Graphical solution : Three Mohr circtes are constructed and a common tangent is drawn tlrough them (Fig. s.13). The shearparametersare found to be :

0rt olt oft

c = 0.27W*r2 0 = 8.5"

....(i)

f'

ot,

haf (45' + Q/2) = 1.34 tan(45' + i/2) = 1.157 45'+$/2=49.7 {/2 = 4.2" 0 = 8'4'

Substitutingfor f in eqn.(i), (0.s)(1.34)+ (2c) (1.157)= 1.3 ort

c = 0.27W*?

Check: Substitutingtle valuesof c and0 in eqn.(iii), we get, L.H.s. = (1.s)(134) + (2)(0.27)(1.1s7) =2.63=RH.S.

0.75 1-0 12.5 1.5 1.75 2-00 2.25 ?.50 N o r m o lS t r e s s( K g / c m 2 |

Fig.8.l3

(ii)Arulytical solution: Fromcqn (8.6)wc havc,

Problengp
2OO

?ol

Problemsin SoitMechanics and Fonndation Engineering

Shear Strength

and o1 = 6'38 kg/cm2,a Mohr circle is drawn (Fig'

SampleNo.

Ccll pressu;e (kgicml

Deviator stressât failure (kgicrn')

Pore pressureat failure (klcm')

1.

1.0

2.U2

0.41

2.

1.5

z.t8

o.62

J.

2.0

2.37

0.70

With o3 =Zk{an?

8.14). 'sincetlesampleismadeofcoarsesandandsinceitisinthedrystate' esthrough no apparent cohesion will develop and the failure envelope the origin. to the Mohr.circle Iriorder to locate the failure envelope,draw a tangent fromtheorigin.Byrneasurement,theangleofobliquityofthislineis3l'' Hence, the shearParametersare: ' c = O , O= 3 1 ' (ii) We have, from eqn; (8.6)' o1 =o3Nq+2c4$ Asc = 0, ".

solution: The values ofcell pressuresand deviator stressesgiven in the problern aia the total stressvalues.The conespondingeffective stressesmay be obtained from the relation :

o1 = 03 lV6

01 = 03 tan21+5' + Q/2')

or,

Detennine the shearparametersof the soil considering (i) total stresses(ii) effective stressess.

o ' = 6 - l l

"'(i)

(\a

The major and minor principal stresses,consideringthe total stress analysisaswell aseffectivestressanalysis,aretabulatedbelow :

E c'l

Sample No.

= v, v, o,

o3

gg"r"1

ad (kg/cm2)

o1 (- ol + oa)

g'l o'3 ( a s * ) ot 1- -;r) Gc/"-1 (kg/cm") (kg/cm:)

(kg,/cm2)

L

+ tt L E

o, E 3/'l

NormolStress (Kg/cm2) Fig.8'14

Here,o3 =3k{u?,0

= 31'

g'37kg/ctt o1 = (3)[tan(45"+ 3r"/2)f = Deviatorstress'o7 = 01 - 03 - g.37-3=6.37 kg.tt ^ problen g.l {Tl" tollo*ing resultswereobtainedfrorn a laboratorytriaxial test with aX6ngementsfor porepressuremeasuremenls:

1.

1.0

2.U2

3.U2

0.41

0.59

2.59

2.

1.5

2.t8

3.68

0.62

0.88'

3.06

t,

2.0

2.37

4.37

0.70

1.30

3.67

Total stressanolysis:Three Mohr circles are drawn using the three sets of values of o1 and o3. In Fig. 8.15, thesecircles are shown by firm lines. A common tangent is drawn through them, which is the failure envelope for toal stressanalysis.From the figure we obtain. ' c =o.75 Wt"f and O = ?.5 Effective stess analysis:In this casethe Mohr circles are drawn with the three sets of values of o1' and o3'. In Fig. 8-15 the effectivc stresscircles are representedby broken lines. Thc values ofthe correspondingshearstrength paramete$ are, c' = 0.65 kg/un2 arld O' = 13' t


202

Now,

-o--+

O -13o

e q

.Y

203

SIrcar Stengtlt a7 = a1 - 63

: 6.05- 2.5= 3.55us/""? Herrce the required deviator stressat failure is 3.55 kg/cm2. (iD Let the required cell pressurebe xWMr?. 6l=6d,

o

c -

I.

...(ii)

01 =l;68+r

olt

6

_G

lo!'

Substituting for o1 and 03 in eqn. (i), we get

' 0 5

c =0 ' 7 5

t

I

l--.

ntitni''

t.5 nn

2-0 -33

a.o oil

0ft

' N o r m oS l tressllQ/cm2) Fig.8.15

,.' -Problgm S.|{'fle shearstrengthparametersof a given soil are, c 0.26' kglcm" an{ 6 = 21'. Undrainedtriaxial lests are to be carried out on specinensof this soil. Determine: , (f deviator stressat which failure will occur if the cell pressurebe 2.5 kglont. (ii) the cell pressure d-uring the test, if the sample fails when the deviator stressreaches1.68 kg/crn'.

Solution: (D We havefrom eqn.(8.6). 01 -o3'lVq+zc'l$

x = 0.83

ot,

.'. The requiredcell pressureis 0.83 kglonz. Problem $A*Tlie following aretheresultsof a setof drainedtriaxial tests mmdiameterand76mmheight: performedffiree identicalspecimensof3S Sample No.

i i' $ll preslure , (kN/n')

1.

:50

2. 3.

Deviatorload at failure (kN)

Change in Volume (cc)

Axial Deformation (mm)

0.f/11

- 0.9

51,

i00

0.659

- 1.3

7.O

150

0.0956

- 1.6

91

Determine ttre shearparametersof the soil. Solution: The deviator loaJs at failure corresponding to each cell pressurearegiven. In order to determinethe correspondingdeviator stresses, these loads are to be divided by the correctedarea of the sample,which can

For thegivensoil,g = O.26kgc#and Q - /1'

ffo * t"n2(as"+ Q/2) = ut(45" + 21"/2) = 2.1!7.

and 1fr; - AtrI

1.68+x=2.Il7x+O.757 l.Ll7x = 0.923

- 1.455

be obtained from

Hence,eqn (8.6)reducesto : or - 2.117os + (21(0.26)(1.455)

"^c

=vrtLv Lr-LL

ot;

at - 2.1!7 a3 + o.757

When

a3 - 25Woo2

- (n/4)(3.82) (7.6) cc

o1 -(Lll7)(LS)+0.757

= 86.19cc

- 6.05kg/ct&

....(i)

Here,

Vr - Initial volume of the specimen

Lt - 7'6crrr

Problems in SoitMechanics ttnd Fottndation Engineering

zo1

For the first sarnPle,A V = - 0.9 cc and AI = 5.L crn 86.19- 0.9 = 12.03 c-mz = I2.A3 * 10-4 m2 n" =

/r'

problenr SWAv unconfined compression test was performed on an of norrnallyconsolidatedclay, having a diameterof 3.75 sample undisturbed occuned under a vertical compressive load of Failure crn high. cm and 7.5 recordedat failurewas 0.9 cm. A remoulded de formation axial t 16.3kg. The sAnrpleof the same soil failed under a compressiveload of 68.2k9, alrd the correspondingaxial cornpressionwas 1'15 crn-

7f,:ilf

:.

0.0711 = 59.L0kN/rn' 12.03x 10-a

ad=

and,o1 = 03 + oa = 50 + 59'10 = 109'10kN/rn2 for two othersarnplesarecomputedin a similar The majorprincipalstresses below: tabulated are results Thc manner. Sample No. I

2

o3

1rxlm2)

Fa (kN)

LV (*)

AL ("m)

Ac (".2)

od

o1

GN/#)

ltcN/m2)

50

0.0711

- 0.9

5,1

12.O3 5 9 . 1 0

109.10

100

0.085e

- t.3

7.0

t7^36

69.50

169.50

o.tB56

- 1.6

9.1

12.65

75.61

225.61,

15(l

205

Shear Stengtlt

Threc Mohr circles are constxuctedand their common tangent is drawn. Ttris is the failure envelope of the soil (Fig. 8' 16)' By neasurementwe obtain, " c = 25 kN/rn', O = 3.8

Determine tle unconfined compressivestrengthand cohesionof the soil in the undisturbed as well as remoulded state. Also determine the sensitivity of the scil and hence classify it accordingly. Sofution:

(a) Undisturbed state :

Initial areaofcross-sectionofthe sample, As = (ni4) (3,7il2 = 11'04crn2

t = + = ffi = o.r, Axialstrainarfailure, Corrected atea,Ar=

* 11'04 - 1"55cm2 = L.'

-=

T=-d.tz

Normalstressat failure=

= 9.27kg/anz

#

stength,4u = 9'27kg/cri Unconfinedcompressive J

and,cohesion,= + =

F

; t00 o Y

t4 L

= 4.64kg/cn?

stote: (b) RemouMed 15 1.153 € = 1T. 3 -=L

o h s n ct 0 c

:y!_L__ 50 -3t

a. = , 1l#sl = 13'o3cm2 r00

n32nn

150

It Az N o r m oS l t r e s s( K g / c m 2 l Fig.8.16

n'= #.L* = 5'?3ks/cm2 4u 5'23 or' c=;=;=2'62kglcIn'

)

sttengthin theundisturk Scnstttvlty=@t"

246

Prablems in Soil Mechanies and Foundation Engineering 9'n - L1"' 7 ' '7 5.23

As the value of sensitivity lies between I and2,the soil is classified as a low sensitivesoil. Probfem S.+4.;lftf ,9U triaxial test, a soil sample wa^sconsolidated at a cell pressure ot}Yg/cffiz and a back pressureof I ig/cm2A for 24 hours. On the next day, the cell pressurewas increasedto 3 \E/cm'. This resulted in the development of a pore pressure,of 0.08 kgfcrn'. The axial stresswas then gradually increasedto 4.5 kg/crn', which resulted in a failure of the soil. Tlie pore irressurerecordedat failure was 0.5 kg/crn'. Determine Skempton's pore

pressureparameters A andB.

torque head at failure was 417.5 kg-cm. The vane was then rotated very rapidly in order to comptetely remould the soil. It was found that the remoulded soil can be sheared by applying a torque of ?-83.2kg-cm. Determine the shearstrcngthof the soil in the undisturbcd and remoulded."statesand its sensitivity. Solution: Weknow tha! S -

S -

L , u = B [ A o 3 + A ( A o 1- Aor)1, whereAandBare Skernpton'sporepressure parrmctcls. hrthefirstcase,Ao3 = 3 - 2 -

A cr1- 4'5 - t - lsVcrf, Ao3 - 0 0.50- 0 . 0 8 - 8 l O + A ( 3 . s - 0 ) l O.42- 35A8

-- s -

S -

...(D

In the second case,

or,

Orr

T (xl (7.52)(rr.25/2 + 7.5t6) T ttr3.67

state,f = Ctl.Skg-cm, In the undisturbed 4175 -

lkg/em2' Ao1 - I

o . 0 8 + B [ 1+ A ( O - 1 ) l B(l - A) - 0.8

"nfr '?)

Here,If=llJS cmandD= 7.5cm,

Solution:We have

or,

2W

ShearStrength

...(iD

rtt}.67

o37Wr;rr2

In the remouldedstate,T = 733,2kg-cmr

Sensitivit, -W-

1.48

Dividing (i) by (ii), weget,

l-A - 0.08 15 A 0.42 1 - A oft

T-o'6'l

'or,

l - A - 0.67A, or, A - 0.6 SubstitutingthisvalueforA in (i), wc obuin I

oI,

I

I

B(l-0.6)=0.08 0.08

^ = a 7 = u n^ E .z

I .

Problem ttfl6. A vane sheartest was carried but in the field to determine the sbearing strength of a deep-seatcdlayer of soft clay. The varte was 11.5 crn high and 7.5 cm across the blades. The equivalent torguc recorded at the

EXERCISES 8.1. The normal stress€sacti4gon two orthogonalplanesof a soil sampleare250kNlm2and110kN/m{ Findoutthenormalandsbearstresses " ofl a planeinclinedat 60 to thedirectionof themajorprirtcipalsress. [Ans. o = 215 k]-.[/m2,r = 60.6 trtl/m2] t.2. The stressconditionson a soil elementare shownin Fig. 8.17. Dctermine: (D The orientationandmagnitudeof thc principalstesses. (ii) The stresses actingon thehorizontalandthc verticafplanes. [Ans. (i) ot = 2,76kg.t]

at 98.5'witb horizontal; 03 = 0.83 Ug*rz at 8.5'

withhorizontal(ii)o11 - 0J7kg/artz, aH ov = 2.72kg/c.rr?,av= 0.3 Ugl*r2l

-0.3k4/cm2;

J

248

Problems in Soil Mechanics ond Foundotion Engineering Slrcar Strength

lKglcrs?

209

8.5. In problem 8.4, if the water table rises from a great depth to the ground surfaceso that the soil becomesfully saturatedand its natural moisture content increasesto l9%o, how will the shear strength on the given plane change? [Ans. Reducedby 0.85 t/m']

lKglcn?

2Kglcn2

8.6. The stressconditions on an infinitely small soil elementare shown in Fig. 8.19.Find out the magnitudeand directionof the principal stresses. [Ans: o1 = 1'68 kg/cm2 at 12' to the horizontal 03 = 0.47 kg/cn? atl02' to the horizontal]

0'5Kglcr] 015Kg1c62

Fig.8.17 8.3. Fig. 8.18 illustratesthe stressconditionson a soil elemenl (i) Determine the normal and shear stresseson the planeX-X. (ii) Draw a free body diagram of the element bounded by plane X-X and show thesestresses.

1l tn2 X Fig.8.l9

2.5iln2 2 . 5 tt n ?

8.7. The results of a direct sheartest perfonned on a soil sample in a shearbox of 6 c-rnx 6 crn size are given below:

Shearforce at failure (kg.) Fig.8.18

(iii) Prove trrat theu* o'iii',

Plot the failure envelopefor the soil atrd detennine its shear parameters.

i,TJi::?ftg*,r, r =0.65rg.'/,l

8.4. Thesubsoilat a siteconsists of a5 m thickstratumof a cohesionless soil which is underlainby a rock layer.A surcharge of 5 t/m2is placedon the groundlevel.Thepropertiesof thesoil areasfollows: G = 2.68,e = A,7,w --6Vo,S= 30' Determinethe shearstrengthof the soil on a horizontalplane at a depth of 2 m below the G.L. [Ans : 4.8? t4n']

[Ans.c=0,0=33'] 8.8. A given soil has a unit cohesion of 2 vumzand an angleof internal friclion of 28'. Samplesof the soil were testedin the laboratoryi4 a triaxial apparafusunder the undrained c-nndition.Determine : (i)

Deviator stressat failurc when the cell pressureis 1.5 kglcm2.

(ii) The applied cell pressure,if the sample fails under a total vertical pressureof 5.09kg/on2. tAns. (it332 k4.lcmz, (ii) z.s kg/cmzl 8.9. A set of triaxial testswere performed on three samples of a isoil. The cell pressuresand the deviatorstressesat failure are given below:

Problems in Soil Mechsnics and Fottndation Engineering

210

CellPr. (kN/m2)

Deviator stress(kN/m-)

I

2M

690

2

300

855

J

400

1030

Sample.No.

8.f3. A set of triaxial tests were perfonned on three samples of a line-grainedsoil. The height and diameterof eachsamplewere 75 mm and 37.5 mm respectively.The following are the results:

SampleNo.

Cell Pr.

Deviator load

Axial Deformation

(tg)

(".)

$il"fi

Plot Mohr's circles of slressand determinethe apparentcohesionand "1 angle of internal friction. [Ans. c = 112 kN/m' , 0 = 27 8.10. A direct sheartestwas performedin a 6 cm x 6 cm shearbox on a sample of dry, colresiortlesssoil. Under a nonnal load of 40 kg, failure occurred when the sheariug force reached 26.65 kg. Plot the Mohr strengtlr envelopeand detenninetbeangleof slrearingresistanceof the soil. Detennine graphic:allythe rnagnitudeand direction of the principal stressesat failure. = [Ans. $ = 36"i o3 = 0.64 kg/crnz at 27" to Il, o1 2.47 kglcn? al ll7" ro Hl 8.11. Two triaxial tests were perforued on sanples of a moist soil in an unelrainedcondition. The all-round cell pressuresduring thesetwo tests were 2.5 kg/crn' and 4.0 kg/on' afid the sarnplesthiled under deviator stresses of 4.85 kg/crn' and 6.70 kg/crn' respectively. Detennine the apparent cohesiou and the apparent angle of shearing resislanc:eof the soil (i) analytically (ii) grapbicallY. Do you expect to obtain the sarnevalues of the sbear pararnetersif the samples were tested in a drained condition ? Explain your answer with reasons. [Ans; c = 0'59 kg/crn', Q= 22.q'l 8.12. Irboratory triaxial testswere perfornred on three soil sarnplesof 3.8 cm diameter and7.6 on height.The following resultsrvereobtaitted: Cell Pr, (kdcm')

Deviator load at failure (ke)

Changein volume (cc)

Atial Deformation (cm)

1

0.5

45

1.1

0.92

z

1.0

52

L.5

1.15

3

'/ z.o

79.5

L.7

L.22

SampleNo.

ztL

Shear Stengtlt

Plot Mohr's circles and determine the apparent val'ues ot\hear paramelers of the soil. [Ans. c = 1 kg/crn', e = 18.7"]

1

1.45

29.5

0.98

2

2.70

37.9

L.t3

3

?

42.8

1.16

Determinethemissiugvalueof cell pressure in testno. 3. 8.14. The following resullswereobtainedfrorn a set of consolidated uudrainedtestswith arrangernents for porepirssuremeasurernents: TlestNo.

I

2

J

1.0

2.0

3.0

Deviator Stress(kgicm')

L.31.

I.62

1.89

(kg/cmJ Porepressure

0.18

0.42

0.86

Cell Pr.(kg/cmJ

Detennine the shearparametersof the soil, considering (i) total stress (ii) Effective stress. [Ans. (i) c =A.46kglclr2,0 = 6.5' (ii)c' = 0.42kg/crt,

0' = 9.8'l

8.15/ Aa unconfined compressiontest was perfonned on a silty clay samplq/of 4 cm diameter and 8 cm height. The sarnple failed under a compressiveload of 23 kg and the deformation recordedat failure was 1.42 cm. A triaxial test was performedon an identical sampleof the samesoil. The all rouncl cell pressurewas 1 kglon2 and the sample failed under a deviator load of a 39.5kg, The axial deformationrecordedat failurewas 1.L8cm. Find out the apparentvaluesof shearparameters(i) graphicallyand (ii) analyti"J cally. tAns c = O.70kg/cm',$ = 4.5 8.16. 421.5 cn long c:ylindricalsoil sanrplehaving a diameterof 10 crn was subject to an increasingvertical compressiveload. Failure occurred

2t2

Problemsin SoilMeclnnics qnd Foundation Engineering

whentheloadreachedl5lkg,andlhecorrespondingaxialdeformationwas : 2 cm. The sarnplewas made of clay and had the following properties

9

G=2.67,s=O.69,w=26Vo Determine the sndarparametersof the soil

o, = [Ans' 0 = 0 c A'77 kglc11n'i cylindrical 8.L7. An unconfiinedcompressiontest was performed on a sample The 75 mm' of a height soil sample having a dihmeter of 3?.5 mm and at recorded strain axial The kg. 23.5 failed afa vertical cornpressiveload of 53' at inclined to be observed was plane failure was L.16 "* .od the failure the soil' to the horizoiltal Determine the apparentshearparametersof = 16"] [Ans. c = 0.68kg/on2, 0 an 8.L8. A triaxial test was performed on a sample of dry sand having applied 5 "ppui.",6urlue of 36'.Initialiy, a chamberpressureof TglT.ytt Keeping this and the deviator stress was gridually increasedto 3 kg/on'' reduced. Al gradually the. was deviator stressunchanged,the cell pressure will fail? what value of cell pressurethe sample

[Ans. 1.05tg/"*2] to prevent 8.19. Determine the minimum lateral pressure required" The shear kg/on'. 10 of stress vertical a total to failure of a soil subjected "' = 17'5 = kglcm', 0'3 : c given as I parametersof the soil are [Ans.4.94 xgcrtl an undisturbed 8.20. A laboratory vane shear test was performed in 6'3 mm and were vane sample of soft clay. The diameter and height of the of 110 gm torque applied an if .: *- respectively' The sample failed under rapidly' vane the by rotating cm. The ,.*pl" was ttren "ott pi.t"ly disturbed the Determine gm-cm' of'45 torque " The rernouldid soil failed ,tna"t states remoulded an{ undisturbed the in undrained shear strength of the soil respectively; 2'5] and compute its sensltivity.[Aor.0. 55 and}.22kglurr2 in 8.21. If a field vane shear test is performed on the soil mentioned determine above problem,witha vane of 11.3 cmheightand 7.5 crn diameter, states' remoulded and undisturbed the in soil the to fail required the torques

BARTH PRESSURE 9.1. Inhoduction: It is often required to maintain a difference in the elevation level of the ground on the left and right hand sides of a vertical section. Such sihtations call for the construction of an earth-retaining structure,e.g., a retaining wall or a sheet-pilewall. The earthretainedby such a structure exerts a lateral thrust which is of paramount importance in the design of the retaining structure. Depending on the conditions prevailing at the site, tle lateral earth pressuremay be divided into the following three categories: (i) Earth pressuteat rest. (ii) Active earth pressure. (iii) iveearth pressure. 9.2. Earth kessurc at Rest: Fig. 9.1 (a) shows a retaining wall, embedded below the ground level upto a depth D, and rctaining earth upto a height l/. If the wall is perfecfly rigid, no lateral movement of the wall can occur. And hence,no deformation ofthe soil can take place.The lateral pressureexerted by the soil is then called the earth pressureat rest. S o i tW e d gA eBt

Ur', Pa

[Ans. 670.6 kg-cm; 268'2kg-cm]

Fig.9.l

Problems in SoitMechanics and Fonndation Engineering

214

Theconjugaterelationshipbetweenthelateralearthpressureandthe vertical overburden pressureis given by: "'(9'1) ol = Ko'ov, or oh = Ko'\z where

K0 = co-efficient of earth pressureat rest'

Y = unitweightof soil z = depth at which lateral pressureis measured' The value of K6 dependson the properties of the soil and its stresshistory, and is given by:

& =t h where,

.,(e.2)

p = Poisson'sratio of the soil.

9.3. Active and ive Earth Pressupes: In reality, a retaining wall is not rigid, but flexible, i.e', it is free to roiate about its base' In Fig' 9'1(a)' let pl and pg,be the at-restlateral thrusts acting on the back and front faces of the wall respectively. Due to the difference in elevation levels, Po , Po'' Hence, a flexible wall will yield away from the bac$fill. The soil wedgeABC will then tend to slide down along the potential slidilg surface BC. This condition is illustrated inFig.9.1(b). The frictional resistanceFR againstsuch movement will act upward alongBC. Its horizontal componentFs will act in will the opposite direction to that of Pg. Thus the net pressureon the wall decreaie. Such a stateis called the active stateof plastic equilibrium, and the lateral pressureis called fte active earth pressure' simultaneously,thesoil w edgeDEF inftontof the wall getscompressed. The frictional resistanceFn' io this case acts along ED and its horizontal componentFg' actsin the samedirection asthat of P6', Hence the net pressure on the wall increases. Such a state is called the ive state of plastic equilibrium and the lateral pressureis called the iveearth pressule.. The active and iveiarth pressuresare usually computed by either of the two classical ""tth pr"tt.tte theories, viz., Rankine's and Coulomb's theory. g.4. Rankine's Earth Pressure Theory: This theory is based on the following assumPtions: l.Thesoilishomogeneous'semi.infinite'dryandcohesionless. 2. Theback of the wall is perfectly smooth and vertical' 3. Deformation of the wall is sufficient to create a state of plastic equilibrium'

2t5

Earth Pressure

4. On any vertical plane in the soil adjacent to the wall a coiljugate relationship existsbetweenthe lateral earlh pressureand the vertical overburden pressure. This theory was later extendedby other investigatorsto take into cohesivebackfills and walls with batteredbackface. The equation governing the relationship between the major and minor principal stresses,acting on a soil element, is given by,

o 1= o 3 N 4 . k - q

...(e.3)

where,Nq= (45" +"Q/2) 0 = angle ofinternal friction c = cohesion. Let us consider an intinitesimally srnall soil element at a depth Z below the ground level, adjacentto a retaining wall, as shown in Fig. 9.2'

3m

I

J-

1'5m/

J-l

t.toyeySond I = 1 ' 8 5t / m 3 Q=2t" c=1tlm2 0enseSantl Y = 1'95t/m3 @=36o Fig.9.2

o,, = vertical overburdenpressureon lhe elemenL orr = lateral earth pressureon the elemenl According to the fourth assumption stated above, a conjugate relationship exists between,ou and otr . The relationship is similar to the one expressedby eqn. (9.3). However, the exact form ofthe equation dependson thi prevailing conditions, i.e., whether the backfill is in an active state or in a ive state. (i)Active state : In this cas€,
But, o, = 1z

EarthPressure


2t6 and,

or, = active pressureintensity = pr'

f

.'. Eqn. (9.3) gives,

\z = pa'ilq + 2cVF 2c \z n = L - : ro 'l{o vflo

of,

(ii) ive state : Here, o1 = oy, and ot = 6u

...(9.4)

H

/. .t t{d

I

;,ta^

h

toT

l

^

-r.tlt

But, ot, = yz

and,o7, = ivepressureintensity = po

Ht3

_L

-J *otH+(b)

(ol

.'. Eqn. (9.3)gives, PP=YzNq+?*fi

...(e.s)

Fig.9.3 Fig. 9.3 (b) shows the disnibution of active Pressure intensity. The magnitude cf resultant thrust per unit length of wall may be obtained by multiplying the averagepressureintensity by the height of the wall.

9.4.1. Computation af Eorth PressureUsingRsnkine's Theory z (A) Act ive Earth Pressure: soils: (a) Cohesionless soil, c = 0. For a cohesionless .'. Eqn (9.4)reducesto

0+K-yH Average pressureintensity,pou

_ t L = _ n . : = ^ = y r\F. i1; *+ lsinf Q tan" 145" + g/2)

= = .'. Resultantthrust,P4 l,*'rrr' IX,IH'H

Pa = N+ or,

II

where, Ko = co-efficient of active earth pressur" = i;*l*

"'(g'7)

Po = Ko'{ z Ar the top of the wall

(z = A), Po = O

At the base of the wall (z= 14, Po = Ko:v'H

i.*or,

...(e.8)

(ii) F ully Submerged B aclfill:

Eqn. (9.6) and (9.7) can be usedto compute the active earth pressurefor various backfill conditions, as discussedbelow: (i) Dry or Moist Baclfill with Horizotxal Ground Surface: Fig. 9.3 (a) shows a retaining wall ing a homogeneous'backfill of dry or moist soil, uPto a height.FL At any depth z below the top of the wall.

1

It is evihent from eqn. (9.8) that the resultant thrust is given by the area of the pressuredistribution diagram. This thrust acts through the centroid of the hiangleA.BC, i.e.,is applied at a heigbt of Hl3 abovethe baseof the wall.

"'(9'6)

Po = Kalz

=

This condition is shown in Fig. 9.a (a). As the soil is frrlly submerged,its effectiveunitweightis' T, = ysar_ yw t I I

I

At any depth z below the top of the wall, the total active pressureis the sum of pressuresexertedby the soil and water. According to Pascal's law, a fluid exerts equal pressurein all directions at any given depth. Hence, at a depth z, Pa=KaY'z+Ynz

...(e.e)

pressure distributiondiagramis shownin Fig. 9.4 (b) The corresponding (iii) Part ially Submerped Bqcffit t: (a) Backfill havingsimilarpropertiesaboveandbelowwatertable:

I

Problems in SoilMechanics and Fottndation Engineertng

218

219

Earth Pressure

Eqns. (9.10) and (9.11) may be usedto determiile the resultant thrust aild its poirit of application correspondilrgto any pressuredistribution diagram'

KqYhr D

T lB Y1 Pz P r l

Yz

v

(b)

(o)

iPs F_ I

Fig'9.4 Ilr Fig. 9.5 (a), the retainiug wall has to retain earth upto a heightfl. The ground water table is located at a depth ft1 below ground level. The active pressureintensities are given bY: Above ground water table: Po = Koyz (O s z s h) Below ground water table: Po = Koyhl + Koy'z + ynz (A < z s h2, zbeing measuredtfromG.W'T.) Fig. 9.5 (b) shows the correspondingpressuredistribution diagrarn.The resultant active thrust pbr unit run of the wall is given by the entire area of this diagfam. It is easierto detennine the areaby dividing it into a nurnber of triangle and rectangles' In Fig. 9.5 O). Pt = LABP,

P2 = areaof BCED

Ps = LDEF,

P4 = LDFG.

(b)

(o) Fig.9.5

(b) Backfill having ditterent propertiesabove aild below water table: Fig. 9.6 (a) and (b) illustrate this backfill condition and the corresponding pressuredistributiott diagratn.

11-

Kqrlrhr

I I'

ttst

H I

Resultantactive thrust,

q=11h1

I ,

n

Pn=Pt+P2+P3+Pa=)4

...(e.10)

j-1

The point of application ofP4 canbe determinedby taking moments of individual pressureareasabout the baseof the wall' Thus, Pa'l = Pfr + PzJz + P1Y + PaYa

2 't''t

i-l

0f'

v=-;lp,

2

t- 1

f (q)

y'ro **r,n,i*"rtin4 l (b)

Fig.9.6

(iv) Baclcfill with UniformSurch'arge: Fig.9.7 (a) illustratesa retainingwall inga backfill loadedwith a uniforrn surchargeq. The correspondingpressuredistnbution diagram is shownin Fig. 9.7 (b). Frdm the figure it is evidentthat the effect of the

Problems in Soil Mechanics qnd Foundation Engineering

220

surchargeis identical to that of an imaginary backfill having a heightzo placed above G.L., where,

'- " _- q Y-

221

EarthPressure

.,(e.r2)

T

zs=gttl-t,, 9/unitareq

* -)r H

Fig.9.8

I

I

JJ*ou

(o)

tQlH'-J

...(9.14) BC = H (1 + anetan p) onthisimaginaryplaneBC,usingeqn. (iv) Determinetheactivepressure (e.13). (v) For degtheyall, computethe self-weightof the soil wedge ABC andconsideris effecton thestabilityof thewall separately'

(b)

Fig.9'7

(v'SBaclfill with a SlopingSurface, The conditionis shownin Fig. 9.8 (a). The activeearthpressureat any depthz belowthe top of thewall actsin a directionparallelto thesurfaceof the backfill andis givenby: Po = KolH

where, &=cosp #

...(e.13)

(vi)Wall Having an Inclkd Bac$ace: In order to determine the active earth pressure in this case using Rankine's theory, the following stepsshould be followed (Ref. Fig. 9.9) (i) Draw the wall section and the ground line. (ii) Draw a vertical line through the base of the wall to intersect the ground line at c. (iii) Compute the length8C from:

Fig.9.9 (b) Cohesive-ftictional Soils: From eqn. (9.4), the active earth pressureat a depth z is given by, .rz 2c n = L - : re Vffo /{6

'JrrQ r +L{w4 Ipv' Problems in Soit Mechanics {tnd Fottndotion Engineering

222

Atz = 0, pa =

k

or,

v1%

where,

A t z =eH^ ,= f rf t

1 1 + sin$ l-tittq=4

=&

=U,Or, ",

H, _

or,

Pp = Ko\z Kp = co-efficientof iveearJhpressure No=oo'(a5"+$/z)

Let H"bethe depth at which pressureintensity is zero. '{H" ^{H" 2c 2c ^

rt-q

223

Earth.Pressure

Fig. 9.11(a) arld(b) showsa retainingwall subjectedto a ivestate' distributiondiagram' ivepressure andthe corresponding

ZE\q

...(e.1s)

I

, 2 c --{G

]'1b

-46.j r

2clNo

H

y

I

L ,

{o} d l

(b) Cohesive-fr i ct iono I Soi Is:

(b) Fig.9.10

Fig. 9.10 (b) sbows the distribution of active pressure.The negative side of this diagram (i.e., A abc) indicatesthe developrnentof tensiottupto a depth flr. Since soils cannot take tension, rracks will be formedin this zone. Tbe depth .I{, is, therefore, called the zone of tension crack. The resultant lateral thrust is obtained by computing the area of the positive side of the diagram (i.e. L, cde). (B) Pcssive Errrtl, Pressurei (a) Cobeslonlesssoils: soil, eqn.(9.5) reducesto: For a cohesiortless PP = ^lzNf

C- * Soil (c)

Fig.9.11

-r, J.-tH ' N @ / F _l o {o}

I o h e s i o n l e sSso i l (b)

From eqn. (9.5), we havc' P p = Y z N q+ 2 c ' [ $ For the retainingwall sltowu in Fig' 9.1i (a)'

Atz=0,po=Zcfi alz=H, Pe=tHNr+Ufi ThecorresponctingpressuredistributiondiagramisshowninFig.9.ll (c)'

EartftPressure.

224


andthc failurc In the activc statc,thewall rnovcsawayfrom thc backfill frictional down' slides it As wedge C.BCtends to move downwards' (soil-wall wall the of r"sisln"es actupwardalongthebackfacc lifion),anO of rhefrictionalforceFar, the failureplane(soil+o-sol rruion;. In absence backface'Butnow theactivethrustPwouldhavebecnactingnormallyonthe to the normalon the 6 an angle at inclincd is Fx, the resultantP4 of P and

6i+Qtz

backface.Duetosimilarreasons,thcsoilreactionR4willalsobeinclinedat an angle{ to thc normalon thc failurclurfacc' Thesamcargqmcntsleadustothcconclusionthatinaivestatealso to the normalson nr -{n rii arwill b! inclinedat angles0 andf respectively of Pa andR4 lie action of the lines state, ^nafC.However, in thc actlve linesof action the state' ive inthe whcreas belowthc respectivcnormals, them. of P- andR- lie above ""'A of il"b.t oigt.pniol andanalyticalmethodstqr tllaejgrn]i'ation theory' Coulomb's trt"rui ""rtl prorit"iave beenproposedor thebasisof The mostirnPortantmethodsarc: (i) Culmenl.'smethod Graphicalrnethod: (ii) Rebhann'sconstruction

P

B a ) A c t i v eS t o t e

b ) P o s s i v eS t o t e o Fig.9.t2

(i) Trial wedgemethod' Analyticalmethod: may refer to any For detaileddcscriptions 'Soil of thesemethods,the reader of these application the However Mecbanics. standardtext-bookof workcd-out of number a by chapter this in illustated ,n"*noO,havebeen problems. to solve more Some of the special techniquesrequiredto enableus the shapeof in irregularities complexproblemsinvolving externailo;ds, or with' dealt bcen the watt or the groundssrfac€-havedso

9.5. coulomb's Earth Pressure Theory: Instead of analysing the stresses on a soil element, coulomb considered the equilibrium of the failure soil wedge as a whole. The rnajor assumptionsin Coulomb's theory are: (i) The soil is dry, homogeneousand isotropic. (ii) The failure surfaceformed due ro the yielding of the wail is a plane surface. (iii) The failure wedge is a rigid body. (iv) The backface of the wall is rough. (v) The resultant thrust acts on the backface of the wall at one-rhird heigbt and is inclined to the normal on the wall at this poinr at an angle 6 , where,

EXAMPLES ./ prcblem g.{ nS m brghrigid retainingwall hasto rctaina backfill of dry, cohesionlesssoil havingthe following properties:

6 = angleof wall friction. Basedon this theory,the lateralearthpressurecan be determinedby the trial and error method. As the location of the actual failure surface is nor known, a numberof potentialfailure surfac'.es are chosenand the lateralearth pressureis determinedfor eachof tbem.The one for which the lateralthrust reachesa certainextremevalue (rninirnumfor active stateand rnaxirnumfor ivestate) is accepredas the true failure surfar:e,and the corresponding lateral thrust is acceptedas {he active or ivetirrust.as the casemay be. 9.S.l Wallfriction: The conceprof wall fricrion is illustratedin Fig. 9.12 (a) anri (b).

22s \

(i)Plotthedistributionofhteralcarthpressureontbewall. andpoint of appticationof rheresultant tiii ij"t"rrr"e rhc magnftude thrust in thelateral'thrustif the waler table (iii) ' ' Computetbe pcrcente"ng-" risesfrom agpeatdepthto the top of the backfill' I

I

t

t

Solution:(i) Bulk densityof thc dry backfill,

226

Problems in Soil Mechanics and Fotrndetion Engineering

=t,54t/nf.

rd=*=?T#?

As the wall is rigid, the lateral pressureexerted by the backfill is earth pressureat rest. Co-efficient ofearfh pressureat rest,

= 'o'tf='=A56?5 Ko==L 1-p 1-0.36 At the top of the wall (z = 0), po = 0

227'

EarthPressilre

to retaiu Problem 9.2. A retaining wall with a smooth,vertical backfacehaos placed is t/rn" of 5 surcharge unifonn rn. A of +.5 a leignt upto a sandbacktill over the backfill. The witer table is at 2 m below G.L. The specific gravity of solidsand the void ratio of thebacktill are2.68and 0.82 respectively.The soil above the water table has a degree of saturation of lo7o. The angle of internal friction of the soil, both aboveand below water table, is 30'. Detenninethe magnitudeand point of applicationof the resultantactive thruston the wall. Solution: Bulk densityof the soil abovewater table,

At the baseof the wall (z = 5 m), p0 = Ko\z

G + s e y= | +7'l*

(0.s62s) (1.s4) (s.0) = 4.j3/mz The distribution of lateral earthpressureis shown in Fig. 9.13. (ii) Resultantlateral thrust on the wall (consideringunit width), 1

Ps = lKgyH'

4

= (r/2!(0562s)(1.s4)(s.o)z

(0.10)(0.82) -2.68 + (1) = 1'517t/rn3 1+0.82 densityof thesoil belowwatertable, Submerged 2 . 6 8- 1 r , , = 0 . g 2 3 t / m 3 G-1. Ysub=lllY,=11-537(r/ Co-efficientof activeearthpressure, 1-sin30'

K"=fi61F=5'

= 10.83 t per m run The resultant thrust is applied at a height of 5f3 = t.67 m above the base of the wall. (iii) If the water table rises to the top of the backfill, the soil wilt get.fully submcrged.

= +;+' t,"= (##) Ysub

t/nz ttr = o'e65

1

n

|*oy,ubH2

*

|t.tf

= 19.?8t per m run Percentincreasein lateralthrust = f19.28 f i x I 0 A10.83 Vo = 787o,

Ir l at B dueto moistsoil abovewatertable Active pressure

= Kr,tz=

[})

tt.ttt)(2) = r.or ttmz-

= 0.77t/mz . = KaTsubr= fl Q.gz3)(2.5) \'/

l

= (r/2) (s.0f [(0.e6s) (o.s62s) +U

/l)
Active pressureat C due to slbmerged soil

Resultantthrust =

= Ko = dueto surchatge Activepressure I

1

,

Lateral PressureexertedbY water = \wz = (1) (2.5) = 2.5 t/trf . The pressuredistribution diagram is shown in Fig' 9'14' The resultant active thrust is equal to the areaabcde. For convenie this areais divided into a numberof trianglesand rectangles.Considering

5m

I J-

width of thewall,

Fig.9.13

228

Problems in Soil Mechanics and Foundatbn Engineering

229

EarthPressure Solution: Coefficient of active earth pressure,

- ff-o'^ s"or Kd=co e0m

5t ln2

*70-

. coslo"-G-'1{-re

=(coslu=l:

t-

cos10" + Vcos' tO" - cos'32"

= 0.296

thrust,P4 = I *" ,rf Resultant /1\

= []l1o.zro) (s)2 (r.82) Yl = 6.734t/m

(bl Fig.9.14

P1 - ( 1 . 6 7 ) ( 4 . 5- ) 7 . 5 1t P2

= fif 1t.ot1(2) - 1.0tt

\") Ps - (1.01)(2.5) - 2.52 |

!r=4.52=2.?5m

h= 2.5+ 2t3=3.r7m h=

2.512= 1.25m

/r \

Pa

- l|l1o.tt + 25)(2.5)- 4.691 ta= Z.SR = 0.83m \"1 Rcsultantthrust P6 - Pt + P2 + Pg + Pa - 15,13t pcrmrun.

The point of applicationof this thnrstabovethebaseof the wall may be obtaincdfrom eqn.(9.11). .Y -_ -(7.s1)(2.2s)+ (1.01)(3.17)+ (2.s2)(1.2s)t(4.0e) (0.83) 15.13 ?5.64 -

Gfr /

l'76m'

Prublem ?3/ A 5 rn high masonryrctainingwall hasto rctain a backfill of sandysoil fi6vinga unitweightof 1.82gm/ cc andan angleof internalfriction of 32'. The surfaceof the backfill is inclined at an angle of 10' to ihe horizontal.Determinethe rraFitude andpoint of applicationof the active thruston thc wall. ,1"".--

This thrust is inclined at 10" to the horizontal (i.e., acts parallel to the ground surface) and is applied at a heigbt of 5/3 = 1.67 m above the base of the wall. Problen 9.4. A retaining wall with a smooth vertical back has to retain a backfill of cohesionlesssoil upto a height of 5 m above G.L. The soil has a void ratio of 0.83 and the specific gravity of soil solids is 2.68. The water table is located at a depth of 2 m below the top of the backfill. The soil above the watertabl eis2}Vosatutated.The angleof internal frietionof the soil above and below water table are found to be 32' and 28' respectively. Plot the distribution of active earth pressureon the wall and detennine the magnitude and point ofapplication ofthe resultantthrust. Solution: Bulk density of the soil abovewater table, G+se t

t w

l + e

2.8 + (0.2)(0.83)

ffi-(1)

= 1.55 t/m3

densityof thesoil belowwatertable, Submerged

Ysub ?#

t" = ?u-!;sl0) - o'etttnf

Active earth pressureabove water table: Co-efficient of active earth pressura,Ko, -

i#fffffff

-0.307

AtA(z - 0), pa = 0 AJB(z-zm|pB-Kor\zl

- (0.307) (1.55) (2) - o.sSvrt

234


Active pressure below water table: In this casethe upper layer (i.e., the moist soil above water table) should be treated as a uniform surcharge, for which the interuity 4 is equal to the self-weight of the layer.

"

231

Earth Pressure

./ For the retaining wall shown in Fig. 9.16 (a), plot the Problem 95 distribution of active earth pressureand determine the magnitude and point of application of the resultant active thrust.

q = yzr = (1.55)(2) = 3.I\t/mz I - sin 28" K,r=i;;;F=o-361

Now,

AtB(/ = 0), pa = Ko.e = (0.361)(3.10) = t.tTt/n? AIC(z'=

3 m ) , p c = K o " e t K o " , y " o 6 z+' y n z '

= r.1,2+ (0.361)e.92)(3) + (1X3) =1.12+0.99+3

-1o7st_ -1vd41-1m

f-

1.2m

= 5.llVmz The pressuredistributiondiagramis shownin Fig. 9.15(b) /1 \

Now , P=1 l : l @( O.e s)= 0 .e 5 t/n Yl

A L ooseSond (y=1.54tlm3,+= 2f ) L,ooseSond {lro1=1'8t/m39=22o;1 C DenseSsnd (Yo1=2.05t/m31 + = 32o)

D

yrJz+2/3=3.67n

f zooJ"l--2.30-+

lz = 3/2 = 1'5rn

P2=0.12)(3)=3..36ttm

L (q)

"r = fll(0.ee+ 3)(3)= 5.e8t/m!3 = 3/3 = rn

(b)

\.J

= Pt + P2 + P3 ResultantthrustP4 = 10.29t perm run.

(t) t_-p!6)(1.5)+ (s.e8)

Fig.9.16 Solution: Active pressuresexertedby various strata are as follows: StratumI: < " iin ' 7? { -o = 0 . 2 1 0 6 1 + sin25" pe=o

10.29

K- =' "at

= 1..41. m .', The resultantthrustof n.29 t per m run is appliedat 1.41,m above thebaseof thewall.

t/m2' 0.95

I

-

pB = Kor,trHr = (0.406)(1.64)(1.0) = 0.67t/m2 While computilg theactive Statum II: This stratumis frrlly submerged. eadhpressurein this region,shatumI is to be treatedasa unifonn surcharge of intersity q1,where,

3'67mI Now,

= I.64t/nf . Qr = Trq = G.64)(1.0) | - sin22' = 0 . 4 5 5 K""= t*G6 pB = Koz ql = (0.455)(1.64)= O35t/^2. Pc = KozQt + Kory'2H2 + ynH2

(sl

- 0 . 7 s + ( 0 . 4 s 5 ) ( 1 . 8 0- 1 . 0 ) ( 1 . 2 ) + ( 1 . 0 ) ( 1 . 2 ) Fig.9.r5

Probkms in SoilMechanicsandFoutdation Engineering

232

233

Earth Pressure

Hence the resultantactive tlrust of 10.315t per m run is applied at 1.409 m abovethebaseof thewall. I Prcblen-{.5. A retaining wall with a smoothvertical backface has to retain a backfill of c - $ soil upto 5 m above G.L. The surface of the backfill is horizontal and it has the following properties:

= 0.75 + O.M + L.2 = 2.3gt/n? StratumIII: Equivalentsurcharge eZ=ltH1 +y'2H2 - (1.64)(1.0) + (1.80 - 1.0)(1.2)

c = l.5t/m2,0

y = 1 . 8t / r f ,

-.Z.ffiVtt.

= 12'.

pressureon the wall. $)?tot the dishibution of active earth point of application of active thrusl and magnitude the .(iipetermine (flglDeterminc'the depth of the zone of tension cracks. (iv) Detennine the intensity of a fictitious uniform surcharge,which, if placed over the backfill, can preventthe formation oftension cracks. (v) Compute the resultant active thrust after placing the surcharge.

-

sin 32', t |' K",-i;jffi-0.307

Pc=K".82+lnHZ - (0.307)(2.60) + (1.0)(1.2) - 0.80 + !.2 - 2.0OVn?.

Solution: Thewall section is shotrn in Fig. 9.17 (a)

PD-Pc+Ror{gHg+l*Hg

- ZtX) + (0.307)(2.85 ' 1.0)(2.3) + (1.0) (2.3) * ZW + OJ4 +,L3 = 5.04tht. The distibution of activeearthpressureis shown-ihFig.9.16p)

1?fht-

I

Computationof forcesandleverarms: P1 = (0.5)(1.0)(0.67) = 0.335t/m /r = 3'5 + l0/3 = 3.83m Pz = Q.2) (0.?5) = 0.90t/m h = 2.3 + t.2/2 = 2.90m

2'06n

P3 = (0.5)(r.2)(0.44)=0.2641/m h - 2 . 3 + L - 2 / 3= 2 . 7 o m Pa = (0.5)(1.2\(1.2\- 0.72t/m lq = 2.3 + 1.2/3 - 2.70m Ps = (2.3)-(2.0)= 4.6t/m fs = 2-3/2 = 1.15m P6 - (0.5)(n)Q.74) - 0.851t/m % = 2.3/3 - 0.77m P7 = (0.5)(2.3, (2.3) = 2.&5 t/m h = 2.3/3 = 0.77m '

,

s-I

f-I '94m I

J-!

c

'17r/#l(b)

(o)

t

Fig.9.17

Pa,= D Pi = Lo.3tSt/m i-L

For a c - $ soil, the intersity of activeearthpressureat any depthz is givenby:

n

2c o , -vzf r -@

IAaa

i-l

Y='--=ffi=1'409m \Z. P , ;- l

Here,

= 1.525 ff' = oo'(45' + t2'/21 - 1g11251'

= 1.235 4 At the top of thewall (z = 0), and,

Problems in Soil Mechanics and Fonndation Engineering

234

The pressuredistribution diagram after placing the surchargeis shown in Fig. 9.17 (c). The resultantactive thrust in this caseis given by,

At the base of the wall (z = 5 m),

PA = (0.5)(5.9X5)= 14.75t/m, appliedara heighrof 5/3 = 1.67m above the base.

W=3.47t/mz.

The pressuredistribution diagram is shown in Fig' 9.17 (b). The depth of the zone of tension ffack is given by,

Problem LZA'ietaining wall of 5 m height has to retain a stratified backfill as shown in Fig. 9.18 (a). Find out the magnitudeof total active thrust on the wall and locate its point of application.

H"=?:ifi,

Solution: (i) Sandy silt layer:

(2)(t'tlt'zrst 2.06m. r. -

olt

iv{ - tan2(45" + 2o'/z) = z.o4

The resultant active thr,ust is given by the part abc of the pressure distribution diagram.

= 5.!ot/n2 -/ .^ = fl Q.s4)(3.47) \" J

The point of application of P4 is located at L946 = 0.98 m, above the v baseof the wall. at The maximum negative pressure intensity developed the top of the wall = -2.43 t/mz. Evidently, the formation of tension cracks canbe prevented by placing a surcharge q on the backfill which can neutralise this negative pressure, Now, after placing the surcharge4 the vertical shess oy at any depth Z is given by,

ov=q+\z / ) ^= rA

o+vz No

+ - :

2c Vtro

# .- 2 c A t z = 0 , p e =,,0 VFo' But the magnitude of q is such that, at z = 0, pA = O,

ft-#,r=o or,

(r.23s)= 3lt/m2. Q = ?rvN; = (2)(1.s)

atz= H, O^= Again,

235

- :'z t !t='--e)(s) - (2]=(1'l= 5.et/n2 1.525 1.235

= -2.43 = - (?{1'i) t/m2. + r.235 Vilo pr=(#

Earth Pressure

...(i)

166 -

1.438

- _-( -2r) z ( 13. 0E) -_a - rr 't o r ,l / y 1 2 . Pn = PB

= (1.8ilg'e)_ (2](1.i0) = 0.33t/m2. 2.04

7.438

(2)(l'-o-)lt'a3e) = 1.55m H" = (ii) Loose sandlayer:

a,=f-ffi=033 Equivalentsurcharge interuity,41= (1.85)(1.9)= 3.5LVri = t.t7th#. pB = Kozql = (0.33)(3.51) Pc = KozQr+ Koz\2H2= 1.L7+ (0.33)(1'72)(1.0) ' = l.I7 + 0.57 = t.74t/^2. (iii) Densesandlayer:

a,=l-ffi=o'26 Equivalent surcharge intensity,qz=(f .S5)(1.9) = 5.?3r/mz + (1.72X1.0) Pc = Kot,qz= Q.26)(5.23)= 1.36t/mz PD=Kotch+KorlsHl

W

ft

= 1.36 + (0.26)(1.88)(1.6)

Problems in Soil Mechanics and Foundqtion Engineering

236

= 1.36+ 0.78= 2.14th? of forcesandleverarms: Computation

-

f

at 10' to the horizontal. The angle of wall friction is 20'. Determine the total lateral pressureexertedby the backfill, using:

I

T 1.55m

t

,

+-

0'33t/m2

to;il, P1

Dense Sond Y= 1 ' 8 8 t / m 3 0= 36o

I 1.6m

I

Yt )-

D (o)

Solution: (e) Culmann's method: Fig. 9.19 illustrates the solution of the problem by Culmann's method. The procedurpis explained below:

\

LooseSond 't=1'72tlm3,@=30o

1.0m

(a) Culmann's method (b) Rebhann's method.

a139/,htr

SondySitt Y= 1'85t/m3 c = 1.0t/mz Q= 20"

1.9m

I I r.:slo.zs

(i) The backfacc' -B is drawn to a scaleof I : 100. (ii) The ground line AC, S line BC and rp line 8X are drawn. Here,

{ - 9d' - (o + 0) - 90" - (n" + 8') = 62'. (iii) The pointsD1, D2, .'., Dg arc chosenon AC at equal intervalsof 1 m. BDyBD2, ...,BDgareed. (iv) From B, BN L AC is drawn. Its length is measuredand is found to be 4.06 m. F) Alternatively,BN = BA'cns(cr - F) = rr'cosG: (l cos -=

{bt Fig.9.18

P1 = (0.5)(0.35)(0.33) = 0.06r/m, y1 - 2 . 6 + 0 . 3 5 / 3 = 2 . 7 2 m P2 = (1.17)(1.0) - L.t1 t/m, Y 2 - L.6 + 1.0/2 = 2.10m P3 - (0.5)(1.0)(0.57)- 0.29t/m, P4 = (1.36)(1.6) - 2.l8/m,

h = 1.6 + l.O/3 = 1.93m Y 4 = L.6/2 = 0.80m

P5 = (0.5)(1.6)(0.78)= A.62t/m, ys = 1.6/3 = 0.53m

i

237

Earth Pressure

* - 4'32t/m

(v) Considering unit width of the wall, the self-weights of various wedges are computed.For examPle, Weigbt of the wedgeABD1 = W, = *'AD1'

BN '7

= (0.5) (1.0) (4.06) (1.78) = 3.61 t per m' Weight of the wedgc

ABDr = Wz = TWr = (2) (3.61)- 7.22t/m. = 18.05Vm Similarty,Wr = 10.83t/m,W+=14.44tlm,Ws Wo=2l.66tlm,W =?5.27Vm,W3= 28.88t/m. (vi) Using a vectorscaleof 1 crn = 3,61 t/rn, the weightsof various wedgesare plotted alongBC, and the points C1, C2, ..., Cg are

i - 1 n <

i-l

(a.0) cos (10'- - 8') = 4.06 m cos 10'

!---;--*ffi,=1.216m S P- t .

Z. i - l

Problem 9.8. A 4 m high retaining wall has a backface inclined at a positive batter angle of 8'. Thebackfill (1 - 1.78 t/nf , 0 - 30") is inclined upwards

Oii) iHffi',Cr/,rll

atr1. tointenectBDl 8x isdrawn

(viii) Similarly,a numberof lincsaredrawnparallclto thery-lineBXfrom intersectionpoints thc pointsC2,C3,...,Cg,alld thecorrespgnding arelocated' E2, Eg,".,, EgwithBDz,&fu,, ..., BDgrespectively cuweis obtainedby jotnngEy E2,...,Egby a smooth (ix) The pressure curve'

238

Problems in Soil Mechanics and Foundarion Engineering

0\ O. oo

o

N (o ll l-

Earth Pressure

c-! 6 oo lJ.

24O


2.41

EarthPressure CE denotes the (ix) A tangent to this curve ls drawn at E, such that ed and extended is maximum ordinate of the pressurecurve' BE plane' the failure is to intersect the ground line at D' 8D the distance C4 The magnitude of t6 resultant active thrust is given by cm' the length of which is found to be 1'58 P,1 - (1'58) (3'61) - 5'70 t/m' (b)Rebhann'smethod:Fig.g.20illustratesthesolutionbyRebhann's method. The procedure is as follows: (i) The backface is drawn to a scale of 1 : 100' 8C and rp-line BX ate draq'n' lilj fn" ground lineAC, $-line BC as diameter' with
= +'GI'HJ'y - 5'77t/m' = ,-0.r,(2.76,(L35t(1.78)

rough backface-taving a Problern g"9. A gravity retaining wall with a backfill upto 4.5 positive batter angle of 1b", has to ietain a dry, cohesionless : m above G.L. Tbe properties of the backfill are \*llkN/m3,Q-?5 Theropoftbebackfillisslopedupwardsat20.tothehorizontal.Theangle the total active thrust on the of wall friction *"y u* taken as 1i" Determine wall bY Rebhann's construction' by the conventional Solution: This protrlem cannot bc solved at a great distance meet will and f-line Rebhann's rnethod, as te groundJine over Rebhann's modifications certain However, l'.' B is nearly equa! to 4 i' in Fig. presented is solution The problem. the * u,olu* method will enable * 9.21, while the procedure is explained below : of 1 : 80' (i) The backface of the wall, A8, is drawn to e scale

Fig.9.2l (ii) The ground-lineAC Q-line BD and rf-line BX are drawn' Here,Q = ff) - (10' + 15') = 65' (iiD An arbitrarypoint E is taken onBD' (iv) A setni-circle is drawn with BE as diameter' (v) EF lleC is drawn.lt interscctsA-Bat F. (vi) r.G ll BX is drawn. It intersectsBD at G. (vii) GIl L BD is drawn. It intersectsthe circle at-Fl arc II1 is drawn to intersectBD 1"iiij witn B as centre and 8Il radius, an at I. (ix) F/is ed. ( x) N ll F/ is drawn. A"I intersectsAD atJ. (xi) From .[ JK ll BX is drawn to intersectA C at K' an arc KL is drawn to intersect lxiii Witn J as cente and JrK as radius, BD AtL.

242

Problems in Soil Meclmnics and Fotrndation Engineering

(xiii) KL is ed. (xiv) FromK, KM I BD is d,rawn. Now, P4 = weightof thesoil wedgeJKl = l.XU a

.LI -y

= (0.s) (3.0)(3.3) (17) = 84.2 kN/m' Problem 9.10. Determinethe magnitudeof the resultantac:tivethrustexerted by a cohesionlessbackfill on a 4.0 m high retainingwall having a backtace inclined at 10" to tbe vertical. The top surface of the backtjll is inclined to the horizontal at 25", The unit we ight and angle of internal friction of the backfill are 1.8 Vrn2and 25" respect-ively. The angle of wall friction may be taken as 15". Solution: In this case,the Q-line and tbe groundline are parallel to eaclr other (since F = 0 = 25") and will never rneet each other. The linal soil wedge may be construcled anywhere on ilre $-line. With refere.nceto Fig. g.Zz,the procedureis explainedbelow :

c

Earth Pressure

243

(i) The backtaceAB, ground litte AC, S-lirc BD and E-line BX are drawn as usual.Here,q, = 90' - l0' - 15' = 65". (ii) An arbitrary point E is rakenon the O-line. (iiD EF' ll BX is drawn to inrersect.ACat F. (iv) Witb E as ceutre,an arc FG is drawn to interseclBD at G. (v) FG is ed. L EFG is rhepressurelriangle. (vi) FH LBD is drawn. Now, P1 = w-eightof the wedgeEFG =

1 .y ;.GE.FH

= (0.s)(4.28) (3.e0) (1.8) = 15.02r,/m run. Prob[ern 9.11. It is required to constructa gravity retaining wall to retain a sandbackfill upto a heiglrtof 3.5 m. The wall is requiredto have a backface which is batteredat 10'. The surfaceof the packfill has an upward slope of 8' and carriesa uniform surchargeof I tlm?.The unit wciglt and ary;ie of intenralfriction of the backfillare 1.80t/rn'and 33'respectively.Delennine the total active thrust on the wall. solution: As the value of angleof wall friction has not been supplied, ) = (z/3) (33") = 22. we wiil assume,0 = ;0 The unifonn surchargeplaced over the backfill may be repracedby an equivalent soil layer which will exert the samevertical stresson the backfill. The beightof this equivalentsoil layer is, , = q 1= . 0 = tt" U.56m. m i The problem can tlow be solved by Rebhann's rnethod, with a few modificatiorn as explained below : (i) The backfaceAfl ground line AC, g-line BC and rp-line BX are drawn as usual(ii) FrornA, a vertical line AA' is drawn and the diskncer4.41is laid off from it such that,r4A| = he = 0.56 m. (iii) Through Av A1c2 is drawn parallel to the ground line to intersect the g Jine at C2.AB andA1C2 are projectedbackwards to meet at Az. A2B andA2Ct will now be treatedas the modified backfaceand the ground line respectively. The problem is then solved by Rebhann's construction in the usual maruler.With referenceto Fig, 9.23, the resultantactive thrust,

Fig.9.22

Earth Pressure 244

245


s cl !b r

E I Fte.9.23 . Pa = weight of the wedge - FGH = z\ ,,

. GI . y

= (I/2)(273)(2.32)(1.80)= 5.70 tpermrun. backfill upto Prublem 9.12. A retainingwall hasto a homogeneous 3.6 m aboveG.L. A long brick wall, 375 mm thick and 3.5 m high, runs parallelto the retainingwall at a horizontaldistanceof 2.8 m from the top cornerof theback of the wall. Determinethe total lateralthruston thewall, given, y = 1 . 8t / m } , 0 = 3 0 ' , c = 0 , c r = 7 8 ' , F = 6 ' , 6 = 1 5 " . Solution: Unit weight of brick masonry= t92Okgl.3. .'. Self-weigbtof thebrick wall permetrerun = (0.375)(3.s)(1920)=2.52t. linear load on the backfill. The The brick wall actsas a concentrated problem can be solved by Culmann'sgraphicalmethod.The solutiott is procedureis statedbelow : presentedin Fig. 9.24 andthe step-by-step


246

247

Earth Pressure

(i) The backthce A.B, ground lhte AC, Q-line BC and rp-line BX are d r a w n a s u s u a l ' H e r e , r p= c r - 6 = 7 8 ' - 1 5 " = 6 3 " . (ii) A numberof points,Cr C2,...,C5 arechosenon the ground lineand 8C1 through.BC5areed. Thesepoints are chosenin sucha way that the line of aclion of the linear load esthrough one of them. In the presentproblem, thesepoittts are chosenat equal intervals of 1.41m. (iii) Sell'-weightof the trial failure wedgesare now cornputed.

- 6") = It = AB'cos(12' eachwedge, Altirudeof +;ffig' = (3.6) (cos 6')./cos 12" = 3.66 m .'. Self-weight of eachwedge = (ll2\ (1.41) (3.66) (1.8) = 4.64t|m. Vector scalechosen: 1 cm = 4.64t1m. (iv) Lay- otTthe distancesBD pnd BD2 from BC, using the chosenvector scale, to representthe self-weight of ABCI andABC2 respectively. Just after crossingC2,the linear load c.omesinto action, and has to be addedto the self-weightof ABC2and all subsequentsoil wedges' From D2 lay off the distanceD2D2' to representthe linear load.

i.e.,

D2D2'='#

= 0.54m.

-

(v) The distancesD2'D3,DPa and DaD5are laidoff torepresentthe weightof thewedges C2BC3, C3BC4 allrd CaBC5respectively. (vi) From DyD2,D2',...,D5 a numberof lines are drawn parallel to rp-line to intersectBCy BC2, ..., BCs at EyE2,E2',...,E5 respectively. (vii) The pressure cune is drawn. A tangent to this curve is drawn at E2', which is the farthest point from the $-line. BC2 representsthe potential failure plane. By measuremenl,E2'D2' = 1.37 cto. .'. Total lateral thnrst -- (1.37) (4.64) = 6.36 t/m.

?F5

,7

noA \

**_{ l-r l"\ |

?0.\

,'r< | )4-l

Rr\'\^4 bl

o a - \ 1 " '\l \i\l

\-,,\-,,\i \*.

Prr=63'5kN/m

\_, Par=110.1 A?=94'9kN/m kN/mtt=rnU*r.torriaU*r,

Problem 9.13. A 5 m high gravity retaining wall hasto retain a cohesionless backfill (y = tg kN,/m3, 0 - 33') upto a heightofs m.Thebackfaceofthe wall has a positive batter angle of 12', and the ground surfacehas an upward inclination of 15". The angle of wall friction is Z)". Determine the total active thrustby the hial wedge method. Solution: Fig.9.?5 (a) shows the section of the wall, drawn to scale. The ground lineAC and tle S -line BD are drawn.

)Yt1

Fig.9.25


248

Equal distancesAC1 = CtC2 = ... 3 C4C5 = 1.5 m are laid off from AC. The lines 8C1, BC2, ..., BC5 areed. Theseare tle trial failure lines. Let W1,W2, ..., W5 be the self-weigbtsof the wedgesABC1,C1BC2,..., C&CS:Accordingtotheconstruction,l[ = Wz =...= W5 = lf(say).

249

EarthPressure

Altitude of each wedge = BN = 5.15 m.

(v) At E, a tangent XEY is drawn to the curve, making it parallel to AC. The distance of this tangent from DE gives the maximum value ofP4. (vi) From E, draw EC' LAC'. BC', which now represents the potential rupture plane.

(1e)= 73.4kN/m. :.Wr= (rl2)(r.s rel="nofollow">(s.1s)

From Fig. 925weget,Pn

For each wedge, the resultant aclive lhrust acts at an angle 6 = 20", to the normal to the backface of the wall. Again, for the failure planesBCyBC2, ...,8C5, the resultantsoil reactions.R1, R2, .. .,R5 areinclined to the normalson the respectiveplanesat,

o .=33'.

Constructionof Force Triangles : Let us consider the first ftial wedgeA8Cl. The forces acting on it are : (i) Self-weight W1 @oth magnitude and direction are known). (ii) Soil reaction R1 (only the direction is known). (iii) Wall reaction P,4, (this is equal to tle resultant active thrust on the wall and only its direction is known). Let us choose a suitable vector scale. A vertical line a1b1is drawn to represent I4r1in magnitude and direction. From c1, a1c1is drawn parallel to the direction of Pa, while frombl, b1c1isdrawn parallel tofi1. They intersect at c1, and tle force triangle a1b1c1is closed. d1c1now representsthe lateral thrust for the trial failure plane BC1, in magnitude and direction Force triangles for all other kial wedges are constructed in a similar manner. Note that for eachwedge, (i) The magnitude of lTandPn change,but their directions do not. (ii) Both magnitude and direction of,R change. In order to compute the actual value of the resultant thrust and to locate the corresponding potential failure plane, a pressure curve is drawn as follows: (i) At any height aboveAC, DlD5llAC is drawn. ( i i ) A t C y C 2 , . . . , C 5 t h e l i n e s E 1 C 1 , E 2 C 2 , . . . , E 5 Ca5r c d r a w n perpendiculattoAC. (iii) The dist ancesE 1Dy E 2D2, . . ., E 5D5 arelaid off from thosenormals, to representP4 PAr, .. ., Pn, to the chosenvector scale. ,, (iv) The points E1, E2, ...,E5are ed by a smooth curve. This is the pressure cufle.

= 113.5kN,/m.

wall"r. Jf*l:lt"tlun, e.1.Aretaining

vertical backhas , smooth

to retain a sandbackfill having the following properties: y=1.85t/m3,q=3g". (i) Detennine the total active thrust exertedby the backfill on the wall. (ii) Determine the percentchangein active thrust, if the water table rises from a great depth to a height of 2 m above the baseof the wall. I Ans. (i) 4.93 tlm (ii) Increasesby 27.2%b] 9.2. A6 m high earth fill is suppodedby a retainjng wall with a smooth vertical backface and canies a surchargeof 30 kN/m'. The angle of internal friction of the fill soil is 30", while its bulk density is 17.5 kN/m'. Plot the distribution of activeearthpressureon thewall. Also determinethe magnitude and point of application of the resuitantthrust. I Ans. 165 kN/m, appliedat236 m abovebase] 9.3. A vertical retaining wall has to retain a horizontal backfill upto a height of 4 m above G.L. The propertiesof the backfill are : c = 0, 0 = 28', G = 2.68, w = ll7o, s = 55Vo, F = 0.5 If the wall is rigidly held in position, what is the magnitude of active thrust acting on it? I Ans. 15.5t/m ] 9.4. With referenceto Problem 3, determinethe percentagechangesin active thrust if the wall moves : (i) towards the backfill (ii) away from the backfill Assume that, the lateral rnovementof the wall is sufficient to bring about a stateof plastic equilibrium. t Ans. (i) Reducesby 63.97o (ii) Increasesby 17697o I 9.5. A masonryretaining wall, 5.5 m high, retainsa backfill of cohesionless soil, having a horizontal top surface.The soil has an angle of internal friction of 27.5", a void ratio of 0.83, and the specificgravity of solids is 2.65. The water table is located at 2.2 m below the top of tle wall. Above the water table, the averagedegreeof saturationof the soil is loVo,Plotthedistribution

250

Probkms in Soil Mechanics and Foundation Engineering

of active earth pressureand compute the magnitude and point of application of the resultant rhrusr. I Ans 12.56 t/m applied at 1.5g m above the base ] 9.6. A cohesionlessbackfill, retained by a 5 m high retaining wall with a srnooth vertical back, is bounded by a horizontal surface. Tbe water table is at 2 rn below the top of the wall. Above the wa-tertable, the angle of internal friction andbulk densityof the soil are 18 kN/m3 and 30; respeJtively.Below the water table, the bulk density increasesby Lovo while the frictiln angle decreasesby 2ovo. Detennine the resultant active pressureon the wail.

e,7.Aretai'ing wauhaving a smooth vertical rjffjfiTfi:

cohesionless backfill. State, giving reasons, how the active earth pressure exerted by the backfill will change in each of tlre following cases: (a) the backfill becomes saturated due to capillary-water, while the ground water table rernains belgw the baseof the wall. (b) the ground water table rises above the base,but there is no capillary water. (c) the given backfill is replacedby a cohesionlesssoil having : (i) sarneunit weight but greater angle of internal friction. (ii) same angle of internal friction but greater unit weighl (iii) same unit weight and angle of internal friction, but having a small apparent cohesion. 9.8. cornpute the total active thrust and its point of application for the retaining wall shown i'Fig. 9.26. The wall has a smooth backface. I Ans. 3.6 r/m, 0.90 m above the base]

3m

I -r1'5m

CtoyeySond f = 1 . 8 5t / m 3 * =2lro c= ll/ p2 D e n s eS a n d I = 1 . 9 5/fm 3 g=36o Fig.9.26

9.9. A smooth vertical retaining wall has to retain a backfill of cohesionlesssoil uptoa heightof4 m aboveG.L. The properriesofthe backfill are :

y = 19 li.{,/m3, 0 = 36'

251

EarthPressure

(a) Determine the active thrust on the wall if the backfill has a horizontal top surface. (b) Determine the percentchangein the active thrust if, insteadof being horizontal, the backfill is now sloped upwards at an angle of 15" to the by 8.85%l horizontal. I Ans.(a)39.52kN/m.(b) Increases 9.10. A masonry wall has to retain a cohesive backfill having an unconfined compressivestrengthof 4 t/m' and a bulk density of 1,72 gmlcc. The overall height of the wall is 6 m. Determine : (i) the depth upto which tensioncrackswill be extended. (ii) the magnitude and point of application of the active thrust IAns. (i) 2.32m (ii) 11.63kN/rn at 1.23m abovebase] 9.11. With referenceto Problem 9.10, detennine the minimum intensity of a uniform surcharge,which when placedover the backfill, will prevent the formation of tension cracks. I Ans.3.08tAn] 9.12. A5 m high masonry retaining wall with a vertical backfaceretains a horizontal backfill of dry sandhaving T = 20 kN/m3 and 6 = 32'. Compute the resultant active thrust on the wall by : (i) Rankine's theory (ii) Coulomb's theory, using the trial wedge rnethod. Which one of the results is more realistic and whv?

I Ars. (i) 76.75kN/m (ii) 79.3kN/rn,assuming0 =

!t

9.13. An RC.C. retaining wall, having a backfaceinclined to the vertical at 10', has to retain a horizontal backfill of dry sand upto a height of 5.2 m. The soil has a unit weight of 17.5 kN/m' and an angle of internal friction of 28". The angle of friction between soil and concrete may be taken as 18'. Determine the point of application, direction and magnitude of the active thrusl Use the trial wedge method. I Ans. 98 kN/m, at 18" to the normal on the backface] 9.14. Solve Problem9.13 graphically,using : (i) Culmarur's method. (ii) Rebhann'sconstruction. 9.15. A gravity retaining wall has to retain a 6 m high backfill of dry, cohesionlesssoil (1 = 19 kN,/m3, O = 36') having a surchargeangle of 8'. The back of the wall has a positive batter angle of 10". The backfill carries a linear load of 5 t/m, running parallel to the wall, at a distanceof 3 m from the top of the backface, measuredalong the ground. Compute the total active thrust on the wall by Culmann's method.Locate the point of application and

., directionof thisthrust.Assurne,6 = i 0. J

I Ans. 155kN/m]

252


9.15. Qompute the total active thrust exerted by the backfill on the retaining wall system shown in Fig. 9.27. Locate the position of the potential rupfure surface.

C l e o nS o n d (Y=19kN/m3, @ = 3 3 oc, = g 1

Fig.9.27 9'17. T\e backfit praced behi'd a 5 m high masonry retaining walr consists of a partiaily sarurated crayeysilt, havin! rhe follo;ing p-fir,i.r, unir weighr = 18.5 kN/m3 cobesion= 10 kN/mz angle ofinternal friction = 21" angle of wall friction = 12. adhesion between soil and wall = g kNlm2 surchargeangle ofbackfill = 10. The back of the wail is inclined to the horizontar g0'. at Determine the magnitude and direction of the aclive thrust by the triar wedge -"rtoJ. arro determine the depth to which tension cracks wiil be extende-d. 27.5 kN/m run; 1.57m] IAns. A retaining wall, 4.5 m high and having a positive batter _ _^ ?.18. angle of 15" has to retain a cohesionlesstactnfl having a unit *.ight of 1.g5 t/*?;; an a.gle of internar friction of 31'. usi'g Reiharur,s meilod, aetermine rne magnitude of laterar thrust on the walr, ii the surcharge angle of the backfilr is: (i) 10'(ii) 2s. (iii) 31". 9.19. A4 m high retaining wall with a vertical backface was eonstructed to retain a backfifl of loose send with a horizontal top surface nurt.a to tn. top of the wall. raboratory investigations revealed that the sand had the following properties:

EarthPressure

253

0 = 25', G = 2.65, e = L.05, s = 0 The back of the wall is relatively smooth.Compute the total active earth pressureexertedby the backfill using any suitable theory. A few months after construction,the backfill was thoroughly compacted and consequently,its Q -value increasedto 32'. However, the top surface of the backfill was depressedby 80 cm. Determine the percent change in the total active earth pressure. ).20. A 4 m high earth-retaining stucture having a smooth vertical backfacc retains a backfill having the following properties: c = 2l/m2, Q - 22', rl = 1.85t/m2 Plot the distribution of ive pr€ssureon the wall and determine the magnitude and point of application of the total lateral force. I Ans. 56.3 Vm ; 1.61 m above the base ]

255

Stability of Slopes and,

a = o" silB = yz cosBsinp

...(10.3)

Failure will occur ifthe shear stressr exceedsthe shear strength rJ of the soil. The faclor of safety againstsuch failure is given by,

10

X

STABILITY OF SLOPES 10.1 Introduction: A slope in a soil mass is encountered when the elevation of the ground surface gradualy changes from a rower lever to a h.igherone. Such a slope may be either naiurar 6i lirty region) or man-made (in artificially comkucted ernbankrnentor excavations). The soil mass bounded by a slope has a tendency to slide down. The principal factor causirrgsuch a sridi'g fairure is the serf-weight of the soil. However, the failure may be aggravateddue to seepageof water or seismic forces. Every man-rnadesrope has to be properly desig'ed to ascertai. the safety ofthe slope againstsliding failure. various methods are availabre for anarysing tbe stability of slopes. Generally these methods are basedon tbe following assumptions : 1. Any slope stability problern is a two_dirnerrsionalone. 2. The shearparametersof the soil areconstantalong any possible slip _ surface. 3. I* problerns invorvi'g seepage of water, the flow'et ca' be constructed and thc seepageforces can be determined. 10.2 stability of rnfinite sropes: In Fig. 7o.r,x-x represe'ts a. infinire slope which is inclined to the horizontal at an angre p. on any prane yy (YY ll w at a depth z below the ground lever thJ roil prop"rti"i and the overburden pressure are constant. Hence, failure ,my o""ui along a plane parallel lo the slope at some depth. The co'ditions for such a failure may ue analysedby co.sidering the equiribrium of the soir prismA,BCD of width b. Considering unit thickness,volume of the prism V = z b cos F and, weight of the prism, W=yzbcosp Vertical stess on llPdue to the selt'-weight.

",=l=yzcosp

Fig. 10.1.

r = T! \t):f

...(10.1)

...(r0.2)

...(10.4)

olrcsionlesssoi/s: We have frorn Coulomb's equation,

rf = c + otano soil,c = 0, For a cohesionless r/ = otan0 in eqn.(10.4) Substituting ohno F = a

Again, substitutingthe expressionsfor o andr. c o s ' B . t a n O_ t a n O , / F _ 11 = /

''tzcosPsinP= ;;ft

r1t "'(10's)

When S = p, F. = 1. Thus a slope in a cohesionlcsssoil is stable till s F Q, provided tbat no extemal force is present. ((ii)) c - f sor/s: In this case,the factorofsafety againstslopefailure is glverf by, n

This vertical stressrranbe resolvedinto the following two componenfs:

o = o, cosp = yt.or2B

T

c * otano T

) ^ ,' _ c + yzcos'Ftan0 yzcospsinp /

...(r0.6)

257

Stability of Slopes 6

Problems in Soil Mechanics and Foundation Engineering Let H" be the critical heigbt of the slope for which F = 1 (i.e, ,f = r) YfI" *tB

sinp = c * yH, "*28

t nq

4_ffi

or,

Hc-+

Of,

y "or2 p (tanf - tan p) Eqn.(10.7)mayalsobewrifrenas: C

q) Foce

.

- cos"B(tanF - tano)

...(10.8)

- tano) Sn= cos2B(tanB

...(10.e)

:? or'

...(10.7) c) Base Foilure

Fig. 10.2

where, s, is a dimensionlessquantity known as the stability number and is given by :

s'=yt

b) Toe Foi [ure

Foiture

(a)Purely cohesive soils: I.etAB representthe slope whose stability has to be investigated.A trial slip circleASlC is drawn with O as centre and OA = OC = R as radius.

,+' I

...(10.10)

---

If a factor of srfety F, is applied to the cohesion such that the mobilised cohesion at a depth H is, c

cn=q Then,

...(10.11)

^ c r- " - c yE= F;yH

tr

P \!

\ \ B

---

r-o

...(10.12)

From eqns.(10.10)and (10.12),we ger, c

c

y4=+rH or,

F, = + = Fn.

Hence, the factor of safety againstcohesion,.F., is tle sameas the factor of safety with respectto height, Fa . 10.3 stability of Finite slopes: In case of slopes of limited extent, three tlpes of failure may occur.Theseare:facefailure, toe failure and basefailure (Fig. 10.2 a, b and c respectively). Variousmethodsofanalysingthefailureoffinite slopesarediscussedbelow. 10.4 Swedish Circle tleUgAL

assuffiill-i6Efifll

oI a circle.

In this merhod, rhe surface of sliding is

Fig.10.3 Let Wbe the weigbt of the soil rnassA.SICBacting vertically downwards through the centre of gravity and c be the unit c-ohesionof the soil. The self-weight tends to causethe sliding while the shear resistancealong the planeASIC counteractslt. Now, arc le4gth ASIC = R '0 where, Q = IAOC

(expressedin radians)

258

Problems in Soil Meclutnics rrnd Foundotion Engineering

Stobiliry of Slopes

.'. Total shearresistancealong the plane ASIC = R 0 c Restoringrnornelt = shearresistancex lever arrn ot

Mn = nOc x R = R20c

il),r-l>l ! o. d (r.t lL-k

...(10.13)

I ,

Consideringunit thicknessof the soil rnass,

.

I

l

\

r

\

- \ \

\

u V

Q

W = A . I . y = A t where, Y = unit we.ightof the soil A = cross-seclionalareaof the sectorASr CB. The areaA can be detenninedeitherby using a plauirneteror by drawing the figure to a proper sc-aleon a graph paper and counting tht'.number o[ divisions ofthe graph paper coveredby the area. Now, disturt'ringrnornent,MA = W .a where,

Nr

d = lever ann of lIlwith respectto O.

The distancedrnay be detenninedby dividitrg tbe areainto an arbitrary numberof segrnentsof small width, andtakingmLrnlentsof all thesesegrnents about O. Thus, the factor of safety against slope failure,

-<

-l

M* = - = - - a

MD

cRzo {rWd

...(10.14)

A rrurnber of trial slip circles are cboseu and tbe lhctor of sal'ety with respectto each of them is computed.A curve is then plotted to show the variation of factor of safety with various slip circles (the curve F1 F2 Fg in Fig. 10.11).The slip circle correspondingto tbe minimurn factor of sat'etyis identified from this curve. This is the potential slip surface, and the corresponding factor ofsafety is the factor ofsafety againstfailure ofthe slope

AB. (b) Cohesite frirtional soils: With reference to Fig. 1O.4,a trial slip circf eASlC is taken and the sector 1,SlCB is divided into a nunrber of verticzl slices,preferablyofequal width. The forcesacting on eachslice are: (i) Self-weight, 17,of the slice, acting verticalty dounwards through the centre of gravity. ConsiCeringunit thickness of the slice, W=\xboxln

...(10.15)

where, boant-lo represent the average height and length of the slice i respectively. (ii) The cohesive force, C, acting along the arc iu a direction opposilg the probable motion of the sliding soil.

Fig'l0'4 C = c'la whert:, c = unitt:ohesion, lo = averagelengthofslice (iii) I-ateraltlrrusttiorn adjacentslices,El andEp .In simplified analysis it is assurned that, E1 = ER 'Hence the effects of these two forces are neglcctcd. (iv) Soit reactionR ac.rossthe arc:.According to the laws of friction, when the soil is about tc slicle, R will be inclined to the normal at an angle Q. (v) The vertical stresses,V1 and7p , which areequal and oppositeto each other and henceneednot be considered. The weighl W is resolved into a normal componentN and a tangential colnponent L For sorneof the slices Twill enhancethe failure, for the others it will resist the tailure. The algebraicsurn of the normal and tangential conponentsare obtainedfrom :

and, Now,

2 T = 2 ( U rs i n c ) t t/ = E (I4zcoso) Mo=R2T driving lllomellt,

...(10.17) ...(10.18) ...(10.1e)

Problems in Soil Mechanics and Foundotion Engineering

260

and,Restoringmoment, Mn = R[clAI + x/V tan$] = R0 >At = totallengtbofarcASlC But Mn = RIcR0 + trYtanOl

...(10.20)

.'. Factorof safety, M^ - RlcR0 + xNtan0l '- R'T MD F

=

-

Slope (V, H)

Slope angle ($)

...(10.21)

A number of txial slip circles should be consideredand the factor of safety for each should be determined.The one correspondingto the minimum factor of safety is the critical slip surface. 1O5 Method of l-ocating the centre of the Trial slip circle: The number of rrials reguired to find out the critical slip circle can be minitnised by an empirical method proposedby Fellenius. According to _him,the centre of the critical slip circle is located on a straight lne PQ, which can be obtained as follows: (i) Draw the given slopcAB and determine the slope angle-,[ (ii; Oetermine the values of the anglesa1 and crz (Fig. 10'5) from Table 10.1. (iii) FromA, drawAP at at *ngle of c1 to A8.

(iv) From 8, drawBP,making it inclined to the horizon lal at a2' BP and AP intersect atP, which is a point on the desiredline PQ' (v) The other point Q is located at a deptl ,Frbelow the toe of the slope anOit a horizontal distanceof 4.5 II away frorn it. Locate tlris point and PQ' on located will be slip circlc PQ.The centre of the critical

Table 10.1

-

.y't c{_9__t_X_ffeLQ. Vr = zr

or,

261

Stability of Stopes

Valuesof angles cr1

Q.,t

40' 37" 35'

1:1.5

33"48'

29" 28" 26"

L:2

26"36'

25"

35'

1:3

18"24'

25'

35'

1:5

11'1g'

25"

27"

1 : 0.58

60'

1:1

45"

10.8 Friction Circle Method: This methodis basedon the assumptionthat the resuttant forceR on the rupture surfaceis tangential to a circle ofradius y = R sin Q which is concentric with the tial slip circle. Various steps involved ate givenbelow : 1. Draw the given slope to a chosenscale' 2. Seleeta triatslip circle of radius R, the centre of which is located at

o (Fig. 10.6a) 3. ComPuter (= fi sin Q) and draw anothercircle of radius r, with O as the centre. 4. Now considerthe equilibrium of the sliding soil mass under the following forces: (i) Self-weightWof the sedorABCD. 1il; fne cohesiveforce C along the planeADC, the magnitudeand directionof which canbe computedasfollows : Let c be the unit cohesion.\\e arcADC is dividedinto a numberof be themobilisedcohesiveforcesalong l*t C1,'C2,.-..,Cn smallelements. them. The resultantC of theseforcescanbc determinedby drawinga force polygon. Now, themobilisedunit cohesion,c-', is givenby : Fig.10.5

262

Problems in Soil Meclmnics and Foundation Engineering

263

Stability of Slopes

(i) Drawa vertic:alline rtb to representW (Fig. 10.6(b))' (ii) From rr clrawac, making it parallelto the line of action of Fp . (iii) Frorn b drop a perpendicularbd on ac. The line bd now represents' in magnitudeand direction,the cohesiveforc.eCR requiredto maintain the equilibriurn of the soil massABCD along the closen slip circle. 6. Dek:rmine the unit cohesioncrrequired for stability from :

c, =

...( 1{) .26)

i

7. The thctor of sal'etvw.r.t. cohesionis now obtainedliotn : -

F

=

-

acfualcohesiorl requiredcohesion

=

-

c cr

...(10.27)

8. The factor of safely w.r.t. shearstrengthc:anbe obtained as follows: (i) Assurne a certain factor of sat'etywith respectto the angle of intemal triction. Let it be,F6.The mobilisedangleof intenni frictiott is then given by: (q)

(b) Fig. 10.6

.,.(r0.22)

' c

The cohesive force is givel by c'L.. O:

...(10.23)

But, sumrning up the moments of all forces about o and equating to zero, we get, C'Lo'R=C.Lr.a

...(1o.24)

where, a = perpendicular distanceof line of action of c frorn the centre of 'the slip circle.

"= !^

r'= Rsin0n

...(10.2e)

(iii) The lactor of sat'etyw.r.t. cohesionF. is thenobtainedby tbnning anothFr triangle of forcres.CompareF6' and F6. If they are different, go for

where, F. = factor of safety with respectto cohesion.

Lc

...(10.28)

(ii) Draw a uew tiiction circlewith O ascentreand r' asradius,where,

, c ' c ^ = n

, C=C^tr=

tan 0 tanOz= n

...(10.2s)

(iii) The other force is the soil reaction d which is assumedto be trngential io the friction circle. 5. Draw the triangle of forces in the following lnanner :

alother trial. (iv) In this mauner, adjust the radius of the circle until ,FOand F. becorneequalto eachother.This value is theuacceptedasthe factor ofsafety for shearstrengthof the soil w.r.t. the given trial slip circle. 10.9 Taylor's Stability Number: Taylor carried out stability analysis of a heights, slope angles and soil large @rious proposed a simple method by which basis of the results, he properties.On the given can be easily detennined with a finite slope safety of the factor of reasonableaccuracy.Taylor introduccda dimensionlessparameter,called Taylor's Stability Nurnber,which is given by,

,./r,, = F , \

H

The value of S,, rnay be obtained frorn Fig. 10.7.

...(10.30)

Problems in SoitMechanics and Fottndation Engineering

265

Stabilityof SloPes 0'18

0-20

e

0.16 o

% /,

I I 01s

2 #

L' L

(U -o

E a

z a.n > =

/

-ct

+ 0.14 I

/

e. a ! @

4.12

E

E a z

t 5 o +

/

o

0.10

../l

+ t/l

/

0.05

,l 10

/

/,

i

0.06

20 30 /.0 50 60 70 80 ----* -Angte(Degrees) Slope

0.05 ;

90

I

D e p t hF c c t o r , n 6 * Fig.10.8

Fig.10.7

D - Depth of bard stratumbelow toe f/ = Height of sloPeabovetoe' slope angles'-Each Fig. 10.8 consists of a family of curves for various curveconsistsoftwoparts.Theportionsdrawnwitifirmlinesareapplicable while the portions drawn with to field conditions iffustraiea in Fig' 10'9 (a), in Fig' 10'9 (b)' shown broken lines are meant for the conditions shown with broken lines' of cuwes' set third Tbe figure also consiss bf a x of the rupture circle distancc the n represents where for various values of 4 given by' from the toe, as illustated in Fig. 10'9 (a), and is

where, w.r.t cohesion, The stability numbers are obtained for factor of safety unity' as taken initially is Fq friction, w.r.t' while fte factorof safety ThevaluesofS'obtainedfromFig'10'Tareapplicableforslip.circles of limited ingthrough tne ioe. However for slopesmade in cohesivesoils the below es circle slip the critical ieptn i'na "nierlain bya hard stratum, toe.Insuchcases,thevalueofs'shouldbeobtainedfromFig'10'8'Intlis as : figure, the depth factor plotted along the 'r'axis is defined

nc=

D + H H

...(10.31)

n =xf r

I

-l

266


261

Stttbility of Slopes

(b)

{o} Fig. 10.9

EXAMPLES / Problern l0y'J A slopeof infinite extentis rnadein a densesandlayer at an angle of 30'to the horizontal.Detenninelhe factor of safety of the slope againstshearfailure if the angle of internalfriction of the soil be 36'. Solution: With referenceto Fig. I0.I, XX representsthe given slope, while )Yis a plane parallel to it at a depthz. Vertical stresson XIldue to overburden. oz=lz where, Y = unit weight of the soil Normal stresson YY,

o = oz cos2p

Shear stressorr YI,

a = o" c.osF sin 0

Fig,10.10 at P' line PQ represents oue sucli equipotential line, which intersects \Y Hence, the piezonetric head at P is given by PR'

Fromgeotnetryof the tigure, PR = P0 cosF PQ = zcos9 and, PR = zcos20

([] = slope angle)

Shearstrengthof the soil on the plane 17,

F t o wL i n e s

Y

Therefore,neutralpressureal

Now. total vertical stressat P due to overburden, o= = YsatZ

"f = otanQ = o-(:os2ptanQ. But, tactor of safteyagair:stsheartailure, a

',r- - - ' f r

o"cos'F tanO

- t a n0 ozcosBsinB lan0

tan 36' =rr"3tr=I'46' / Problem nd fslope inc:linecl at 16' to the horizontalis to be made in a cohesionlessdeposithaving the following properties: G=2.70,e=0.72,0=35', Detennine the factor of sat'etyof the slope againstshearfailure if water percolatesin a direction parallel to tlre surfaceofthe slope. Solution:

The given conditiousare shown in Fig. 10.10.

YYis a plane locatedat a depthzbelow the slope.As water percolatesin a directio;rparallel ro the slope,all llow lines rnust be parallel to the slope. Therefore.,all equipotentiallines should be perpendicularto the slope. The

1 ^ P = 'l', z cos- p

Total nrlrmal slressat P, Shearstressat P,

o = o; c:os2p = yr",, "or2 p r = o : c o s p s i n 0= Y s a tczo s P s i n P

Effec-tivenormal stressat P = total normal stress- neutral stress or,

o' = ysat', .or2 B - Yrz coszP = Z cos2F (y"ar- y.) = ysuu, "or2 P .

However,tbe shearingstressis entirily intergrannular. Shearslrength of the soil on Ylz, "f = o' tan S = ]116z cos2P tan $ .'. Factorofsafety againstshearfailure, 'is ]*5 z cos2p tan Q -

" = i = l;;"'P't"P

y.ubtan 0 yr", tan p

Problems in Soil Meclunics and Foundatian Engineering

268

Now,

ysar

(2'79-*0J3li1'0)= r./rn3 Il +*e "' nu...= 1.988 (l+0.72)

Ysub

1.983-X=0.988t,/m3

Fs=

(0.988)(tan35')

6trsffi=r'2r

Stability of SloPes

1.55

,

Problern 10.3 A slope of 35' inclination and 6 m vertical height is to be rnade in a purêlycohesive soil having a unit weight of 1-.85Vmt and a cohesion of 6 t/m' . Determine the factor of safety of the slope against sliding failure. Solution : The problem will be solved by the Swedish circle method. The solution is presentedin Fig. 10.11 and the procedure is explained below: (i) The given slopeA3 is drawn to a scale of 1 : 200. (ii) The values of c1 and cr2for p = 35' are determined from Table 10.1 by making linear interpolation between F = 33"48' and B = 45". The following values are obtained : al = 26.2", eZ = 35" (iii) The point Q lying at a depth of .EI= 6 m belowA and at a linear distance of.4.5 H = 2'7 m fromA is located. (iv) FromA and .B, two straigbt linesAPand BP are drawn such that, LPAB = ?.6.2", and LHBF = 35" AP and BP intersect at P. (v) PQ is ed. The centre of the critical slip circle should be located on this line. (vi) PB is measuredand found to be 4. 6 m. On projected Q,two more p o i n t s P ' a n d P " a r e t a k e n s u c h t h a t , P P '= P ' P "

= &

rr 2?'5m

I

= 2.3m.

(vii) Three trial slip circles are drawn with P, P' and P" as centres and PA, F A and F' A respectively as radius. The factor of safety with respect to each circle is determined separately. Fig. 10.11 shows the determination of Fg with respect to the first trial slip circlg having its centre atP. The ptocedure is statedbelow : (i) The area under tle slope and the slip surface is divided into 7 slices. The first 6 slices have a width of 2 m each while the width of the 7tb slice is

2.2m.

J-o, Fig.10.11

distutbittg (ii) Considering unit thickness; the area, weigbt and the sliie are d9t€rmine4'Thes?a4 lablla o*K f*.".h

r--'=

Width lstu" llVo. (n) I

2.0

Averagelenglh (n)

Area

o*t 2.2

Weight Level ' Moment f t x A x l ) arm aboutP (t)

4.07

abovl P (m)

3.7

(t-m) -15.06

210 Slice Widtlt No.

Aterage Iengrlt

(m)

2

2.0

2.0

(m)

2.4+ 4.0 2 *

- - =

Area

Weiglt (yxAxI)

(^2)

(t)

6.2

1 l

1 . f

Typeof soil

gfl=o'ut

9.3

t7.21

tr.J

11.6

2r.6

z.)

49.36

5

2.0

6.3+5.6_.o. -"2

r 1.9

22.01

.1. -t

94.64

6

2.0

wl2=o'"

9.5

17.58

6.3

r10.75

4.29

7.94

8.4

66.1O

ff=r.ns MD

Again, restoring lnornenl, MR Here,

c (kNlm2)

Very sotl claY

r7.5

0

L2

1

4-1

Mediurn claY

18.0

0

35

J

7 -t4

Stiff clay

19.0

0

68

4

14-o

Roclk

probablebase Cotnpute tlre thctor of safetyof the slopewith respectto a radius' m of 13.5 lailure along a slip circle prcibletncan solution: Fig. l}.l2illustrates the given conditions.The be solved by the Swedish c:irclemethod' this 1DGE representsthe slip circle of radius 13.5. O is the centre of circle.Astheslipcircleesthroughthreedifferentlayers,thefailure

- l Y l= 1 ? ' 5k N/ m 3 qI tr = 78 ril/m2

2M = 292.A5t-nt Disturbing Mornent,

Q ( /

0-4

5.16

5.3+ 6.3= 5 ' E 2

y (Wlmi)

I

"

2.0

2.2

No. Depth (m)

-19.50

4

.|

50' to the Problern 10.4 A 10 rn deep cut, with the sidesinclirted at follows: as are conditions horizontal.hasto tretnadeat a sitewlrerethesubsoil

Let,er Moment Qrm about P ebout P (m) (t-m)

T1.47

27L

Stabilityof SloPes

Problcms in SoilMe:hanics and FoundtrtionEnginecring

292.05 t-st

- r Y 2 =1 8k N / m l r g2= 35;1r11s2 ln

c R2.o

c = 2.5t/nt? R = PA = 9.9rn g = LAPD = 102.5'= 1.789radian

[ = 1 9k N / r n 3 ? m C 3 = 1 51 P 7 6 1 2

MR = Q.s) (e.e-)(1.789)= 438.35t-m

= = Facforofsal'ery #; ffi=

1.50 Fig.10.12

In a similar manner,thc factor of safcty of the slopew.r.t. the two other slip circles (having their ceutresat P' and P"1 aredetenninedand are lbund to be 1"55 and 1.66 respectively.A curve representingtlrevariation of factor of safety is then plotted. Tbe minimum factor of safety of the slope, as obtained from this cuwe, is 1.45.Tlie correspondingcritical slip circle will bave its centrelocatedat P^.

of wedgeconsistsof threedifl'erentiones. lrt IV1,W2and W! be the weights O. about th" t[r.. zonesand.r1,.r2and .'3 be the correspondinglever arms FromB,drawBlLHF.Zorrelmaylrowbedividedirttotlretriangle BHI and the rectangleB/FE Weight of I

I

L BHI = (0.s)(17.s)(3.3)(4.0)kN = 11s.skN.

Problems in SoiI Meclmnics and Fottndation Engineering

272

Weight of sector BIFE = (17.5)(6.5)(4.0)kN = 455kN. Wr = 115'5 + 455 = 560'5kN. (115.s)(7.2 - 33/3) * (65)Q2 + 65/2) = 9.?4m. 560^5 to be a parallelograrn, thecentroidof The secondzoneHFGJ is assurned which lies at the intersectionof thediagonals. W2 = (18)(10.6)(3)= 572.4kN 'l

=

Asthes|ipcirclct:sthrclughtlrrccdift.ererrtsoillayers,tlreresisting along the three segtneutsof tbrce cotrsistso[ thc cohesivclbrcesmobilised in the figure' the slip circlc. Thc corrt:spcndinganglesare shown Thcrclbre,lotal rcstoriltglllorllellt = c t R 2 o 1+ c r R ? 0 , + c 3 R : 0 3 = .41(cr01 + cr $ + ca 03)

+ (36)(r4'2)+ (78)(16'2)] = (13.s2) # [(26)(101'5)

x2 = 7.5m (by measurement) In orderto find out W3andx3,zoneIII is dividedinto 7 slices.The area, lever arm and momentof eac.hslice aboutO are determined.Theseare tabulatedbelow:

= 14039.69kN-rn' Factor of safetyalonglhe givt:n slip circlt:

= Slice No.

l.

Widtlt

tm)

3.5

z.

Area (^2)

Weight (kN)

4.55

96.3

7.6

-:731.8

9.45

t79.5

4.6

-825.7

1.4

140.6

z.l

-295.3

15.9

342.1

0.4

120.8

19.5

370.5

3.4

t259.7

Itu.n*6.1)=6.5 =5.3 15.9 *4.5) lrc.t

3A2.1

6.4

L933.4

r7r.a

9.9

1692.9

Average Imgtlt (m)

!{o*r.u)=t.t

l{z.u*t.r)=t.rt 2

A q.

a 1

= lo., *6.e) 5.3 3

6.

J

4

|o.s ro)=zzs

9.0

273

Sttrbilityof StttPt's

Lever Moment 4rm about O ubaut O ( w- m ) (m)

14039.69 =

Tzsollg

1.09

made in a-siltyclay having Problem l0.5 A slopeof l V : 2 H is to be a t r a t r g l e o f i r r t e n r a l r r i c t i o n o f 5 . a r r d a c o h e s i o r r o f 0 . 2 5 k g l o n " . T hthe eurrit of cut is 8 m' Cornpute *"isf, of the soil is 1-85grn/cc, and the depth circle method' faclir o[ saft:tyof the slopeby the Swedish 10'13 (t)' T1" centreof a Solution: The given slope is showuiu Fig' sliding wedge is divided The trial slip c:irrrleis lo"ut a uy Feilenius,method. 4 rn' while slice no' 6 of width equal into 7 slices.The tirst five sliceshave slice is tneasured' each of ftngth average and ? are 2.8 rn wide each'The

/ii'i\

-----'72sF

/ ri\.\/(

l

2 M = 3153.9 W3 x x3 = 3153.9kN - m

N1 T1

Now, total disturbingmoment = Wtxt + WZvZ + W3x3

- (s60.5) (7.5)+ 3rs3.e (9.7$ + (s72,4) = 12906.L9

Ws &=l'ls Tg=0

f,

1"1

Fig.10.13(a)

274

Problems in Soil Mechanics and Foundation Engincering

%(1 6 s)/

In lhis nranncr,lhc nomral ancllangentialcomponentsfor eachslice are detcrmined.Thc rt:sull.s are labulatedbelow :

Pzn't+9)

Slice No.

P8{1.82)

P1

(t.eq)

Ps(1,65 ) \t1.71

||

4 ' 5H (b) Fig.10.13

Average length (m)

Weight (t)

t{ (r)

T (t)

1

q

2.4

17.76

t4.7

-10.1

z

4

6.2

45.8t1

44.7

-9.9

J

4

8.7

64.38

64.4

0

4

4

10.3

76.22

73.8

19.4

5

4

9.65

71.4r

62.3

34.6

6 7

2.8

8.4

43.51

30.1

31,.2

2.8

3.6

18.7

10.1

15.9

c R 0 + I y ' y 't a n q Ir t z r

!7 = width x averagelength x unit weight of soil The selt'-weightis assumedto act along the centralvertical line througlr each slice, and not through its geometrical centre. The norn.raland tangential components/y' and I respec-tively,of the self-weight are deternined by constructinga triangle of tbrce. An easyrnethod of doing this is explained here with reference to lhe slice no. 4. Draw the centralvertical line throughthe rniddleof the slice (shownwith a broken line). Measureits length.Extend the line and lay-off an equal length frour it. This new line now representsthe self-weight Wa in magnitude and direclion. The corresponding vector scale used is 1 cm = width of slice x unitweight of soil = (4) (1.85)= 7.4ln In order to construct the force triangle, tlre cetrtre ofthe slip circle to the mid-point of the bottom of slice. Extend this radial line and drop a perpendicular on it from the terminal point of the line representing IVa. The nonnal component.lf4 and tangential component14 are now detennined frorn the force triangle.

= 8 1 . 1t

It should be noted tlrat.as the width of slices6,ard 7 are70Voof that of lhe olhcr slices,lhc length of the vertical lines representingW7 and W6 are 701 of the.avcragelengthof slice no. 6 and 7 respectively. The tactor of sat'etyof the slopew.r.t. the slip circle underconsideration may now be determinedusing eqn.(10.24) :

The weight of any slice may now bc determinedas :

_

Widtlr (m)

2 W = 3 3 7 . 9 t ,X N = 2 5 5 . 4 t , : r

I t

l

275

Stahility olSlqtas

By rneasurernent, R = 15.8m, and,

= (lffiLI*o.= z.oo7rad. o = 115" (2.s)(15.8)(2.007)+ (2s5.$(tan5') =l.E F- s =

In order to Iocate the critical slip circle, i.e., the slip circle with the rniuirnumthctor of sat'ety,proceedas follows : (i) Measure.the distancePyB,l-e.tit be I. (ii) With P0 as centre fonlr a grid consistingof 9 points such that the length of each side of the grJd= 712. (iii) Draw trial slip circles taking each of these9 points in turn as the ccntre.Computethe faclor of safetyof the slopefor eachslip circle. (iv) Plot the values of F5 thus obtained for each grid point and draw contour lines for different values of F5. The slip circle having the minimum value of.F5 can be detenninedfrom this contour.The conespondingvalue of F5 is the fac-torof safety of the slope.

271

Sttrbility of SloPes Problems in Soil Mechanics und Foundation Engineering

276

The processis illustratedin Fig. 10.13(b). It is found thal tht: slip circkr having the rninimum factor of safely is lhc outr drawn with P,7as lhc ct:nlre. Thus, Fellenius'methodyields an accuratcrcsult in this case.Tlre tactoro[ safety ofthe slope.is found lo be 1.25. Prolrlern 10.5 A 12 rn high ernbankutcuthas sidc'gtopc.s oL I V :2 H. The soil has a unit weight of 1.8 Vmr, cohesionof 1.5 t/m/ and angle of internalfric{ion of 15".Determint:the factorof saft:tyof tht:skrpr:with r(:sp(:ct to any chosenslip circle. Use the tiiction circle rncthocl. Solution: The slopeis drawn in Fig. 10.14.A triat slipcircle AEC is drawnwith a radiusR= 20.5m. The cbordAC ised and ils lcrrgthis lirund to be 32 m. [,etD be the rnid point ofAC. The centre of the slip circ{e P rs td lo D aud PD is r:xltndcd. It intersectsthe slope at .F and the slip circlc at E. Thc rnid-point G of EF nray be taken as the centreof gravity of lhr an:aABCE. Norv,

areaABCE = MBC

Tlrroughtltoirrtofintcrsecticltrof!Vatt
The factor of safety wilh respecrtto cohesion is' 1.5 c ^ t

F,=;=ii=1"34

friction' F0 = r'0' F, = l.34when the factor of safetywith respectto that they 4reequal However,thesetwo factorsof safetyshouldbe so adjusted to one another' Fa = l'20 As a tlrst trial, let

tanQ-=s8=q#

+ Ar(:aADCE

l

)

= :.BH.AC + 1AC . DE z _ t l

"*=?=*t"=r.rtt/mz

0- = 8'36''

or,

)

= :(2.2) (32) + ;(31) (7.4-s) ,

.

J

= 1 9 4 . 1n r '

R s i n@

Consicleringunit width of tbe slope,weight of thc soil wt:dgt: A B C E = ( 1 e 4 . 1 )( 1 ) ( 1 . 8 )r

R s i n@ m

= 349.38t. Now, deflectionangle 6 = 102" = 1.78radian . ' . A r c l e n g t h o f A E C = L = R 0 = ( 2 0 . 5 ) ( 1 . 7 8 )= 3 6 . 4 9 r n . The lever arm /o of the cohesiveforce.with respec to P is given by,

t" = ?.R LC

=

#(20.s)

= 23.38 m

At a distanceof 23.38m from P, draw a line paralleI to the chord AC. This gives the direction of the cohesive force C. Again, tfirouglr G, draw a vertical line to repres€nt the self-weight of the soil wedge Iv. The lines of action of W awJC intersect at p. Now, radius of the friction circlc,

r = R sinO = (20.5)(sin10') = 3.56rn.

Fig.10.14

Problems in Soil Mechtrnics ond Foundttion Engineering

278

The new radius of the friction circle is, r' = R sin S,, = (20.5) (sin 8.36") = 2.9ttm Draw anotherfriction circle with this radius. The direction of F slightly changes.A new tirrce triangle is cottslructed. 'fhe value of C obtainedfrom it is 46.7 t. c 46'7 Mobilised cohesiolt , c m = T = 1 6 , = l . 2 l t / : n 2

Hence the l'a<{orot safety of the given slope for the slip circle under is 1.18. consideration I

Problern 1\7 It is requiredto makea 6 m deepexcavationin a stratunr of soft clay having 'l = 18 kN/m'aud c = 26 kN/rn'. A rock layer exists at a depth of 9 rn below the ground level. Determine the factor of safety of thc slope againstsliding ifthe slopeanglebe 40'. The problem will be solvedby Taylor's method.

Here, lhe depth factor,

,o -

I

= rc

For n4 = 1.5 and F = 40", the value of Taylor's Stability NurnberS, as obtainedfrom Fig. 10.8 is, So = 0.172. But, we have from eqn. (10.30),

sr, = F"=

F r n (H

Using eqn.(10.30)

s"= F.fu' ot' F' = or,

F'-F+

Solution:

Assume that friction and cohesion of safety of the slope againstshearfailure' their ultimatevalues' are nobilised to the sameproportionof of friction (i'e" Fq = 1)' the solution: In case of full mobilisation = L5' and p = 30" as obtained value of Taylor's Stability Nurnber for 6 from Fig. 10.7,is, S,, = 0'046'

= l.l8 F -, = + r.zl

F a c t o r o f s a f e t yw . r . t c o h e s i o n ,

' or, F.

Sny/{

26 = 1.40 (0.172)(18)(6)

Hence, the required factor of safety of the slope = 1.46. It may be further observedfrom Fig. 10.8that, for p = 40"and nd= 1.5, the value of n is approximately 0.7. x = n H = ( 0 . 7 )( 6 ) = 4 . 2 m Hence, lhe critical slip circle will cut the ground level in front of the toe at a distanceof 4.2m4 Problern W.{/ n cutting is to be made in a soil mass having ^ \.{ y - 1.8 t/m3, cA- te tt^2 and Q = 15', with.slide slopes of 30' to tle horizontal, upto a delth of 12 m below the ground level. Determine the factor

279

Stabitityof SloPes

1.6 'E. - (0.046)(1.8)(12) = 1 . 6 1

However,asfrictionwillnotbefullyrnobilised,tlreactualvalueofF. by trials' will be less than this, and is to be found out

'Fo = l'25

LeI

tan 15" = o'2r43 tanQ= ffi

0 = 12'1" Referringto Fig. 10'7,for F = 30" S"=0'075 when 0=10",

or'

when

Q=15",

. ' . w h e nb = 1 2 . 1 ' ,

S'=0'046

s,=0.46.ffi = u.058.

+=offina=r'277-r:5 of safety of the slope Hence, as F. and F6 are nearly equal' the factor may be taken as 1.25' E)GRCISE I,O slope of 35' inclination 10.L. Compute the factor of safety of an infinite madeinasanddeposithavinganang|eofinternalfrictiorrof4o".[Ans.1.2] l0.2.Arrirrfiniteslopeof6mheightand35.inclirrationismadeina : layer of densesand having the following-propcrties G = 2 3 0' w=o7o s = 4 . 5 V r n - ,d = 5 o , e = o ' 8 5 , againstsliding' (a) Determine the factor of safety of the slope

280

Problens in Soil Mecl:onics and Foundation Engineering

(b) How will the fac:tor of safety changc if the slope gets fully submergcd? [Ans.(a) 1.2s(b) 1.e81 10.3. Detennine the fac'torof safetyof the slopeAB with respectto tht: given slip circle shown in Fig. 10.15.The soil hasa unit weigtrt ot ig.5 kNln3 and a cohesionof 42 kN/m'. Use the Swedishcircle rnethocl. [Ans : 1.42] 10.4. A 12 rn deepcut is rnadei'a silty clay with side slopesof 3()".The soil has the tbllowing properties: Y = 1.9gm,/cc:, c=0.25 kg/cm', O = 8".

Stability of Slopes

281

3.tn

%

l-

y=;f;,',';,t,H'=,:;: 6m /

/

15m

\Y-

,/

./

sirt crou.r

y = 1 . 9 f / m c3 = , 1.5t/m2 cb=6o Stiff ttov

6'5m

,/

\=195t/m3,:yr/

12n

Rock Fig.10.16

Fig.10.15 Locate the ccntre of the critical slip circle by Fellenius' method and detennine the factor of safety of the slope against sliding tailure by the Swedislrcircle rnethod. lArrs. 1.45] 10.5.Deterrninethe factor of sat'etyof theslopeshown in Fig. 10.16with respectto the given friction circle by the standardrnethodof slices. 10.5. A 10 m deep cut is to be made in a soil with side slopesof 1 v: 1 .F/.The unit weight of the soil is 1,.8gm/cc and the soil has an uncontjned cornpressive strength of 0.63 kglcm". Deterrnine the factor of sal-etyof the slope againstsliding,

1 ' 9g m / c c 0 . 4K g1 c m 2 -o ) 9m

(i) neglecting tensiou cracks (ii) consider:ngtensioncracks

Fig.10.17 Detenniue the thclor of safety of the side slopes of the canal against

10.7. cornpute tbe factor of sat'etyof the slopeshown i' Fig. 10.17 with respectto the given slip circle by the tiiction circle methocl.

slidirrgby Taylor's nlethod.

10.8. An unlined irrigation ca'al hasa depthof g rn and a side slopesof I : 1. The propertiesof the soil are as follows:

10.9. An ernbankment is constructed with a c-Qsoil havingthe following properties :

c=2.0Vr12, q = 15", y = l.gt,/nr3

c=2.5Vm2, q = IZ', y = 1.851,/m3

[Ans.1.25]

?82

Problems in SoiIMechanics and Fotrndation Engineering

to both The ernbar*rnentrnusthave a ihctor of safetyof 1'5 with respect friction' cohesionand angle of intemal of the (a) What will be the maximum allowable slope if the height embartkrnentbe 12 rn ? of the (b) What will be thc maxitnutn allowable height if the sides enbankment are sloPed Lt45" ?

11

l0.l0.A6rnlriglretnbaltklnenlistobemadewnitlraclayeysoilhaving exists a unit weight of 1.7itlm3 and a cohesionof 3.5 t/mz. A hard stratum if a'gle slope the be wlmt should at a depth;f 3 m below the ground level. 33"] ? 2'0 [Ans' the requiredtactorof safetyJgainstslidingbe l0.ll.DetennirretlrefactorofsafetyoftheslopeshowttirrFig'10'18 with respect to the given slip circle. use the friction circle method.

Y = 1 8k N/ m 3 c = ' lf / m 2

BEARING CAPACITY 11.1 Introduction: Structuresofall typeshaveto rest on the soil existing is transmited to the ingsoil at the site.The load of the superstructure throughstructuralcalledfootingswhich ate to be designedproperly so as to ensure: (i) The shearstressdcvelopedon any plane in the loaded soil mass does not exceedthe shearstrengthof the soil. In other words, shearfailure doesnot occur. (ii) The settlenent of the footing due to the applied load does not excred the tolerablclimit. The bearing capacity of a given footing hasto be detenninedcorsidering both of these factors. The present chapter deals with the detennination of bearingcapacityof a footing from the point of view of shearfailure. ll.2

Fig.10.18

Definition of Related to Bearing Capacity:

l. Grossloading intensity (q): This is the intensity of total pressureat the baseof footing due to the load from the superstructure,self-weight of the footing and the weight of earth fill above the baseof footing. Before the conskuction of a footing, the 2. Net loading intensiry fu): soil at the foundation level is subject to an overburden pressuredue to the self-weight of the soil mass. The net loading intensity is the difference between the gross loading intensify and the overburdenpressure. If a footing is founded in a soil rnasshaving a unit weight y at a depth D below the ground level, then,

e, = e - ,(D

...(11.1)

3. Uhimste bearing capacity (q): This is the minimum intensity of loading at the base of the tbundation which will causea shear failure of the soil. This is rhe minimum net 4. Net ultimate bearing capacity (q,): pressure intensity due to the applied load (i.e., excluding the existing overburdenpressure)which will causea shearfailure.

284

,routeryyA A:tEjc{ya Sunttit-nn{leet!9t,

5 . Net safebearing capocity (q,u): The rninirnurn net pressureintensity at the baseof tbotiilg with respectto a specifiedfactor of sal'etyagainstshear failurc,i.e., Qu,t -f Q"r =

285

Bcaring Copacity

the following expressionfor the ultimate bearins capacity of a footing of width B, placedat a depthD below G'L.:

.,(rr.z)

--J r--B l 9 u l

6. Safe beoring capacity (q"): The maximum gross loading intensitY whicb the tooting will sately carry without the risk of shear tailure, irrespectiveof the rnagnitudeof settletnettt. 't ...(11.3) Thus. Q, = Qr" + D 0l'

e'=!+Yn

H

-+

.,(1r.4)

7. AIIow,abIe bearing cupaciry @): Tlris is the net intensity of loading which the foundation will carry without undergoing settlernentin excess of the permissiblevalue but not t:xceedingthe net safebearing capacity' 11.3 Types of shear Failure: The sheartailure of a soil massing a structure may take place in either of the following modes: (i) General shear failure (ii) Local shearfailure (iii) Punchingsheartailure In dense sands and stiff clays, when the loading intensity exceeds a certain limit, the footing ge.nerallysettlcssuddenly into the soil and well defined slip surfaces are fonned. The shear strength of the soil is fully rnobilised along thesesurfaces.This is calleda generalshearfailure.' In relatively loose sands and in medium clays, the footing settle.s gradually.The failure planesare not so well definedand the sbearstrengthof the soil is not tully rnobilised.No heavingof soil takesplaceabovetheground level. This type offailure is called local shearfailure. In very loose sandsand soft saturatedclays, a footing is often found to virrually sink into the soil. No failure planeis formed at all. Sucira failure is due to the shear thilure along'the vertical face around the perilneterof the baseof the tbotilg. The soil beyond this zoneremainspracticallyunaft'ected' Tltis type of failure is called puttchingsheartailure' The type of shearfailure expectedto occur at a site has a direct bearing on the theoreticalcompulationof bearingt:apacity. 11.4 Terzaghi's Theory: This theory is an extension of the concept originally developedby Prandtl. The mode of general shear failure of a footing is illustratedin Fig. 11.1 (a). Consideringthe critical equilibriurn of the.soil we dge xyz underthe forcesshown in Fig. 1 1.1 (b), Terzaghi derive.d

F i g .l l . l

4n = cN, + YDNo + o.5YBi/i,

...(11.5)

wherc, N, , NqarrdNo arebeariugcapacitylactorswhich dependon the atgle of intt:rnalfriction of the soil' Eqn.(1 1.5) is applicableto geucralshear[ailurc.For local shearfailure, the followlng equatiollis to be used:

whcrc,

Q n = c ' N r ' + Y D N+. 0' . 5y 8 { '

...(11'6)

c' = 1c 5

...(11.7)

anct, Nr', No' and Nr'are the bearingcapacitylactorsobtainedtioln Q', rvhcrrr,

Q'= r an-(l3t." )

...(11.8)

Eqn. (1 1.5) is lncattllbr slrip foolings.Horvt:ver,for squareattdcircular lootings thtr follorvirrg nrodified equalionsshould be uscd,whic* take into the shapethclors: For squarcfootings, Q u = 1 . 3c N . + t D N q + 0 ' 4 Y B N ,

...(11.e)

For circular tbotirtgs, Q n = 1 . 3c N . + Y D I ' ! n + 0 ' 3 Y B l { t

...(11.10)

where, ^B stattdstbr the widtb of a squaretboting or the diailreterof a circular footing. Thrrvaluesof Terzaghi'sbearingcapacityfaclorsaregivcn in Table 11.1.

286

e,, = cNrs,dri,

Table 11.1: Terzaghi's Bearing Capacity Factors

a

N.

Nq

0 5 10 15 20 ?5 30 35 40 45 50

5.7 7.3 9.6 12.9 17.7 N.1, 37.2 57.8 95.7 172.3 347.5

1.0 1.6 2.7 4.4 7.4 12.7 22.5 4r.3 81.3 r73.3 4r5.1

0 0.5 1.2 2.5 5.0 9.7 19.7 42.4 100.4 297.4 1153.2

N.'

N;

NY,

5.7 6.7 8.0 9.7 11.8 14.8 19.0 25.2 34.9 5r.2 81.3

1.0

0.o

r.4

0.2 0.5 0.9 1.7 3.2 5.7 10.1 18.8 37.7 87.1

1.9 2.7 3.9 5.6 8.3 12.6 20.5 34.1 65.6

dc , dq , 4 un the depth factors ic, iq, i, arethe inclinationfactors. The valuesof all thesefactorsare given i4 Tables 11.2through 11.5: Table 11.2 Hansen's Bearing Capacity Factors

0

0'

5'

10" 1 5 '

20"

25'

&

1.0 0.09 0.47 r.42 3.54 8 . 1 118.084n

...(rr.r2)

sy

1.00

1.00

L + 0.2BIL

L + 0.2BlL

L -0.4BtL

Square

1.3

r.2

0.8

Circular

1.3

r.2

0.6

Sa

1.00

dq

dr

| + 435 DlB

Forq = y", dq=r.0

1.0

(for all valuesof g)

ForQ>25', dq=d,

dc

...(11.13)

ultirnatebearingcapacityof a footing 's given by,

(for all valuesof g)

Table ll.5: Inclination Factors for Hansen's Equation

...(11.14)

(iv) For rec:tangular footings:

11.6Brinch ".","#:rr#;;;ffi'#T

95.41240.85 t81.84

Table 11.4: Depth Factors for Hansen's Equation

N-=*6 2a

+O.28/L)N,6

KC

sq

Rectangular

(ii) whenP.E:?j:-

i/.=(1

50"

1.0 r . 5 72.47 3.94 6.40 10.6618.4033.2964.18134.85]18.96

Skemptonsuggestedthe following valuesofy'{. : (i) yhef D = 0 (i.e.,when tlre footing is at the ground level) for slrip footingslM. -!!l*

(iii) when DlB > 2.5: ji.=1.5.xNc(surface)

45"

N,I

Contiuuous

c = cohesion. = Bearing capacityfactor which dependson tle shapeof the t'ooting as well as on the depth of foundation. The ultimate bearingcapac'ityis given by:

!! *YI+P/B) N' (sur,race)

40'

. I 4 6.48 8.34t0.97 14.8320.72 30.r446.1 75.32133.89266.89

It

{;

35'

N.

Shopeof footing

where,

folguglg,aLd-circurar-fioorings'

30'

Table ll.3: Shape Factors for llensen's Equation

...(11.I 1)

qn = cN, + YD

...(11.16)

s., sg, sy are the shapefactors

11.5 Skempton's Equation: This equation is applicable to tbotings tbundedon cohesivesoils.The net ultirnatebearingcapacilyof sucha tboting is given by: Quu=7N.

+ q N o s u d u i u + 0 . 5y B . l / r t y d , r \

q ='{D

where,

1gY

287

Beoring Copacity


l6

...(11.ls) ,' -

J.Brinch Hansen, the { l I

, t

H 2cBL

tq

| -o.s+

ly

i?

Problems in Soil Mechonics and Foundation Enginecring

288

Validupto: H s V tanb - c5'BL

. . .( 1 1 . 1 7 )

wlrere, H attd V arethe horizontaland vertical courponelrtsof the rcsultant load acting on thc footing. I

= lengtlr of tboting parallelto H.

c+ = cohesiottbetweetttboting and sc,il. 6 = angle of tiiction betwcentbotittg and soil. 11.7 Bearing Capacity Equation as per IS Code: Hansen's bcaring capacrityequationwas lalt:r modified by Vcsic. In IS:6403 - 1981, thc following equalions were proposed,which incorporated Vesic's modificalions: F o r g e n e r asl h e a rl a i l u r c : Q , ,= c N r s r d . i , + q N n s n d n i , + 0 . 5y B N . , s r d r i r W ' for local shearfailure. , -) ; . q,, = /y'.'t. dri, + qNq' tqdri,

...(11.1ti)

289

Bearing Capacity

of the soil. When the soil is fully submerged,the submerged density 1ru6 shouldbe usedin placeof 1. But if thewatertableis at the.baseof the footing, only tbe third term is atfected.The generalbearing capacity equatiou is, theretbre,moditied as: "'(11'20) Qu = cNc + 7D1N.W1 + 0'5yBNrW2 where, W1 and W2are the correction factors. For most soils, 1*6 is nearly equal to half tlre value of y' Hetrce,the correctionfaclon are given by (Referto Fig. 11.2):

Wr = 0.5(I + 4/D) Wr = 0.5 Whenthewatertableis at G.L., = 1-0 I4l1 of footing, base is at the when it and Wz = 0.s (l + z2/B)

.,(rr.2t)

...(rt.22)

Whenthewatertableis at thebaseof footing, W2= 0.5 + 0 . 51 B N r s r d r t r ' v y ' ...(11.1e)

The shape fac:torssc , .egand s, are thrr same as lhose used in Brinch Hansen'sequationand canbe obtainedliorn Tablc 11.3. The dth factorsare given by: d,=l+{t.2(DtB\'$q dn=dr=1tb rQ<1 0 '

d,r = d, = 1 + o'1(D/B)' fr,

rct o ' 10'

t

I

-.t-Fig.112

The inclinationfactorsaregivenby: i, = iq = (l - s/g}f .

.

2

i I^ . = f r - g l t\

a /l

rvhere, NO = ton2(45" + q/2) c( = angle of inclination of the resultatrttbrce oIt the footing. for water table. In eqns.(11;18)and (1 1.19),Iy' = correctionfa
I I.8 Fffect of Water Table on Bearing Capacity: In Terzaghi's bearing capacityequation,the secondand third aredependentorrthe unitweight

\

When it is at a depthI below the baseof fmting, Wz = L.O. Here it is assurnedthat, if the water table is at a depth egual to or greater than B betow the base,tbe bearing capacityremains unaffected. IS : 6403-1981 recsmmendsthc usc of a singtc correction factor Iy' to be use.din rhe ftird rerm of equarions(11.18) and (11.19). The value of ly' is as follows: (a) If the water table is at or below a depth of D + B beneaththe G.L., then ly' = 1' 0.5 O) If it is at a depth D or above, t{ ("i rf O. depth of water table is such tiat, D < D n < (D+B), &e value of ly' sbould be obtainedby linear interpolation.

Problems in SoilMechanics and Fottndstion Engineering

290

291

Bearing CaPacitY

dimensionof the In caseof single eccentricity (Fig' 11' a) the effective 7 e' fotrting in the direition of eccentricityis reducedby ,

ll.gEccentricallyLoadedFootings:Afootingissaidtobeeccentrically thc cre[lreof gravity of loaded if the resultantload on it is appliedaway from theload.Suclrfootingsrnaytredesigrredbyeitlrerofthetbllowirrgrnetlrods: (alMetlwdl:[nthisrnetbodtheloadQofeccetttricityeisreplacedby of magnitudeM = Q'e' un .i*t concentricload Q and a balancingrnorneilt load as well a.sthe Stress distribution digrams due to the concentric: stressintensity of maximum balanc:ingmoment a." plotied (Fig' 11'3)' The. bearingcapacity allowable the than the superlnposeddiagramstroutaU" less

B ' = B - 2 e

i.e.,

A' = L(B - Ze)

Incastlofttoubleeccentricitythe
of the footing.

f--i=

t

*f

1.p

"'(11'23)

"'$l'24)

r-B

I

-I

I

I

I

. l

_I Ptan n

l

st'l

F i g1 . l'a

t

r+-t--_]

-{eFEievot'ion

J_ o/A 0riqinotFooting

Etevotion FootinqUnden Loodinq Eouivo-lent

capacity of a footing 11.10 Bearing Capacity from N'value: The bearing PenetrationTe st Standard frorn obtained N-value ile irom may be.de.ternrL"o carried,out in the field from the following equalions: For striP footings:

T-

+ tf1ow, + o'47rfi nw, enu= a.785(100

J - l

For squaretbotings:

J_

Qnu

M.e/l I Q M e A . I -f' P

ressurcDisltt-lgilg!

Fig.I1.3 this method if a footilg is (b) Methotl II (Meyertwf's method): In plan area ofthe footing is ofthe portion a "*polJaio an eccentric ioto, ooty effective area' the as termed is "o*ia*t"e to be useful' This area

where,

* fi)owt + o.3rlt'fBW2 "'(tr'261 = 0.e43(1ry

N = averagecorrectedblow count' D = depth of footing. f i = width of footittg.

w'r wz = correction fataors[u; Qnu

=

"'(11'25)

r'-:'dr table'

net ultimate bearing capacity in kN/m2'

293

BearingCaPacitY Problems in Soil Mecltnnics ttnd Fottndution Engineering

2gZ

= 186.5 - (1'85)(1"s) = $t't/n]' 11.11 Bceri4 capacity frnrn P]ate Load Test: The bearing capacity of a footing to be placld ori a soil rnass may be assessedfrom the results of a plate load test carried out at the site at the desireddepth. However, the process has got several limitations. The method of computing rhe bearing capacity of a prototype footing from the plate load te$tdata is illustratedin Problem 1l'10' The iettlement of the prototype footing, when founded on granular soils, is given by the following relationship suggestedby Terzaghi and Peck'

I Be@ + 0.3)l2

[email protected].| where,

...(rr.27)

Pp - seftlernentof the Plate' p = Settlement of the prototype footing' Bp = width of Plate' .B = width of the prototype foottng. EXAMPLES

problem 1l.l/ A 2 m wide strip footing is founded at a deptb of 1.5 m the below the gtouM tevel in a hornogeneousbed of dense sand, having following properties: 0 - 36', T = 1.85 1,/rn3. Determine|heultimate,netultimate,nctsafeandsafebearingcapacity of the footing. Given, for E - 36" /Vc - 60, Nq = 42' Nt= 47' Assume a faclor of safetYof 3.0. Solution: As 0 - 36', a general shearfailure is likely to occur' (i) Ultimate bearing caPacitY:

Here,

4u n cN" + TDNo + 0'518/Vt c = 0, y = 1.85Vnt3,o = 1.5m, B =L}m' Nq= 42 and N, - 47 (1.8s)(2.0)(47) 4u - (l.ss) (1.s)(42) + (0,5) -

786.5t/n]' (ii) Net uttimatebearingcapacity: 4*-4n-lD

(iii) Net safebearingeaPacitY: Q^=\

Q n " -- - 1 8 3 ' 7 = 6 l . Z t / m 2 . 3.0

(iv) Safe bearing caPacitY: Qs = Qns+ \Df

= 612; (1'85)(1'5)= 64t/m2' -/ beenfounded Pr.oblemlil$,Asqnre footingof 2'5 m x 2'5 m sizehas densityof bulk a having soil cohesive i in groond level at !.2 mbelow tbe strengthof 5'5 t/rn'' Detenninethe uno]nnt .a comprcssive ;:8 rJ;;;r, of safetyof 2.5, ultimateandsafebearingcapaciiyof thefootingfor a factor by (i) Terzaghi'stheory(ii) Skempton'stheory' Solution: Cohesionof thesoil, 4u 5.5 c^ = T = _- T - =_L a. a< t r t/nt? l soils($ = 0) we have' tlrcory:Forcohesive (i) Terznghi's N. = 5"7, Nq = 1'0, NY = o' Usingeqn.(11.9), qu = (t.3)(2.7s)(s.7)+ (1'8)(1'2)(1'0) = 22.54t/ri Qs

= l}.3lt/rri' (ii) SkemPton's methd: 1 7

Here,

D/B=#=o'4at2'5'

Eqn.(11.13)is aPPlicable. N" = (1 + 0.2D/8, ff.lsrrrace) = 6.N' But for squarefootings, ffc (surtace)

4 = { t . r y a } t u . r o l- 6 ; e

- 186.5 - (1.85)(1.5) = t83.7t/rt

I

295

Bearing CaPacitY Problems in Soil Mechanics and Foandatian Eng'ineering

294

I issubjccttoa Sross Problem LI{4. Asquare footingof 2m x 2msize

soil is I m.Thefgynda{on i. me o.,ptlof ioundation

""r,i;;iffio?Yio

Q^, = cN"

= (2.75)(6'7s)= t8.67t/t# = e, = e* + 7D = 18.67+ (1.3)(1.2) 20'83t/n?

of 1.85Vm' attd an coilsistsof a depositof densesandhavinga bulk densisty of safety againstshear ,ngte or intcrnal friction of 3d. Determine the factor failure. Solution: We havc, for Q = 36"

Qnu

+ \u Q" = 7, s

=

1e 6't

7

+ (1'8)(r'2) = 9.63 t/m2.

Detennine the safe load that can be carried by a square Problem llJ. G'L' The footitts of 2.2V x 2'2 m size, placed at a depth of 1'6 rn below foundition soil has the following properties:

y = 1.65t/m3, s = l!l)/mz, 0 = 20', of safetyof 2.5' GiCEn,for Q = /6', a factor Assume N, = 17.7, Nq = 7'4, ffr = 5'o lV.' = 11'8, Nq' = 3'8' NY' = 1'3 thatthesoil Solution: Thelow valueof unitweightof the soil suggests shearfailure is is in the loosestate,Moreovet,Q = Zff < 28"' Hencea local of the capacity likely to occur.Using eqn.(11.26),the net ultimatebearing

footing-*Tit: !.3c'N"'+ vD(N,- 1) + o.4vBN, Here,

, 2 ^--,,-^2 " = j,=(2/3)(1.2)=0.8r/n} N"' = 11.8, ffq' = 3'8, xy' = 1'3 qn,,= (r.3) (0.8)(11.8)+ (1.65)(1'6)(3'8- 1) + (0.a)(1.6s)(2.2)(1,3) = 12.27+ 7.39 + 1'89

= 2t.55 t/rt The safebearingcapacityof thefooting: 2l='25 (1.65)(1.6)= lr.?tt/mz Qnu + q " = F ; - + ' ,o = ?.j Grosssat'eload to be carriedby fte footing' - es X Areaof footittg - {ll'76)(2'2f - 5a'51'

N"=60, Nq=42,N\=47 as c = 0' Using eqn.(11.9) and noting that the first term vanishes q, = (1'8s) (1'0) (42) + (0'4) (1'8s) (2.0)(47) = tql3t/n]. = 1455 t/mz' uuu = 147'3 (1's5)(1'0) Now, actual bearing pressureat the baseof footing'

But,

o,,

qa =

o _ m 'z . - t 8 o -= 4-J5 "t /'|' T Ae>

qb=

? *,o,

=3'3i. F,= #i4=*-ffi#do carryinga net verticalload of

Problenr tt$/Acoluntn of a building, at ,ffipor,"o by a squarefooting'The footingis to be^placed 125;;;;;;. followittg the having soil bf bed a hornogeneous 1.2 m below G'L. in properties: y=1.82grn/cr,O=30" sizeof the footingrequiredto havea factorof rninimum the Detennine safetyof 2.5 againstshearfailure'UseTerzaghi'sformula' = Solution: Net loadon columnfrom superstructure t25t. Addt07ofor the selfweightof the footing = t25 t r37.5t =137.5t= 138t(saY). Grossload soil bearingcapacityof a squarefootingon a cohesionless safe Now, the is givenby 1 r a , = i f t . s c l U .+ y D ( N n t) + O.aVafrl+ y D

= Frorntable11.1,for 0 = 30', N" = 3'7.2,Nq = 22.5,N., 19.7


296

----a-1'^-'^

1

e, = zs t (1.82)(1.2)(22.s- 1) + (0.4)(1.82)(B) (1e.7)l + (1.82)(1.2) = 18.78+ 5.748 + 2.18 = 2O.96+ 5.748 The safeload that canbe carriedby the footing, Q = Q " x A = (20.e6+ 5.7a4*

/( RC.C. column footing of f.8 m x 1.8 m size is Problem tt/.'en founded at 1.5 m below G.L. The subsoil consistsof a loose deposit of siity sandhaving the following properties: | = 1.75t/n3' q = 20", c = l.lvrt tbe ultimate bearing czpacity of the footing when the ground Determine water table is located at: (ii) 0.6 m below ground level. (i) ground level (iiD 2.0 m below the base of footing (iv) a.0 m below the base of footing.

Given,for Q = 20",fr.'= 1L.8,frr'= 3'8, fry'= 1.3'

= 5 .7 4 f + 2 0 96r t

Solution: Assuming a local shearfailure, the ultimate bearing capacity of a square footing is given by,

5.7483 + 20.96Bz = 138, or,

.

f+3.6582=24.04

eu = 1.3c' N.' + I DNq W1 + a.4y BNi w2

Solvingthe aboveegrrationby trial anderror,we obtain, B=Z.O6m=2.10m(say)

Here,

Hence, the reqyired size of the footing = 2'10 m If the size of the footing in Problem 11'5 has to be Problern [rd. restricted to 1.7Ym x 1.75 m , at what depth the footing should be placed? Solution: The bearing capacity of a footing placed in a cohesionless soil increases with depth. In Problem 11.5, the depth of the footing was specified as 1.2 m. The corresponding size required for ing a gross load of 138 t was found to be 2.45 m x 2'45 m' However, if the size of the footing has to be restricted to 1.75 m x 1.75 m (such restrictions are sometimes necessary for avoiding encroachment on adjacent land) and if the column still hastowithstand the samegrossload,its depthhastobe increased. Let d be the required depth. Now,

q, = I[lo{ivu - rl + Q"=

0f'

, ' = ? , = Q / 3 ) ( 1 . 1=)0 . 7 3 t / m 2 . \ = 1 . 7 5 t / m, l D = 1 . 5 m ,B = 1 . 8 m (1.s)(3.8)wl (11.8)+ (1.75) s, = G3) (0.73) ' + (0.4)(1.7s)(1.8)(1.3)wz

olt

Qu = ll.2 + 9.97W1 + t.64W2 (i) Whenthewatertableis at thegroundlevel, z1 = g. Usingeqn.(ll.2l), Wr = 0.5 (1 + 0) = 0.5.

(r.82)d (22.s- 1) + @.4$.82)(r.7s)(re.1)+ (1.82)d 2.5

Qu = ll.2 + (9.97)(0.5)+ 1.64 = L7.82t/mz

ffi = 17.472d+ 10.039 45.061

(ii) When the water table is at 0.6 m below the ground level, z1 = 0.6m,

tYr = 0.50 + A.6/7'5) = 0'7

Qs=77.472d+10'039

W2 is againnot applicable

= 45.M1t/rt

d=2.fi)m. The footing has to be founded at a depth of 2'00 m belorv G'L'

...(i)

(i.e.,W, = 17. W2is notapplicable

0 . a V B N r ]+ \ D

Again,actualpressure,A" = or,

297

BearingCapacity

Qu = tl.2 + (9.97)(0.7)+ 1.64 = 19.82t/m2 (iii) Whenthewatertableis at 2.0m belowG.L., z2 = 2.0-1.5 = 0.5m Wz = 0.5(1 + O.5/2.0)= 0.68

Problems in soil Mechanics tnd Foundotion Engineering,

2g8

Here 171is not agplicable Qu = lL'Z + 9.97 + (1.64)(0'625) = 22.19 t/mZ (iv) when the ground water table is at 4 m below tbe baseof footing, no corr.ectiondue to ground water table is necessary.In other words, the ultimate bearing capacity is not affected by the ground water table' = 22'81 t/nz Qn = 11.2+9'97 +l'64 It is evident from the aboveresults,that, thebearing capacityofa footing increaseswith increasingdepth of the ground water table' prublern l\$/Twoadjacentcolurn's of a building, canyi^g a vertical oeaclr, footing of 2 rn x 3.5 m cornbi'ed by a are ed torrne load of e lhe maximurn Determine ground level. the below m at 1.2 foundecl size., allowable value of Q if the foundationsoil consistsof a deep,homogeneous

er,

Q=37.5t safeload on eachcolumn = 37.5 t' required tbe Hence, (ii) In this case,Brinch Hansen'sequationis expectedto yield a more reliable result.We have tiom eqn. (11'16)' stdttt 4 , , = c N r s r d r i . - + T D N u s n d n i o+ 0 ' 5 v B N r For Q = 10", referringto table 1L'2, N. = 8.34, Nq = 2.47, Ny = 0.47 The shapefactors,cepth factors and inclination tactors are obtained trorn t a b l e sL L . 3 ,1 1 . 4a n d 1 1 . 5 ' For a rectangularfooting of 2.0 m x 3.5 m size,foundedat a depthof 1.2 m below G.L., we gct, s" = rs = 1 + (0.2)(2.0/3.5\ = l'll4 sY = 1 - (0'4) (2/3'5) = 0'77

d. = | + (0'35)(r'2/2'0) = r'71

strafutn of : (i) Saturatedsilty clay (f = 1'9 t/tn3 , c = 4'6 t:nh (ii) Partially saruratedinorganicsilt (y = 1.e t/n3, $ = 10', c = l'6t/m2) solution: Total c:olumnload to be carried by the combined footing : ( a ) l o a d f r o r n t h e c o l u t n n s= Q + Q = 2 Q =O'ZQ ib) self-weightof footing (say l0% of columnloading) g r o s sl o a d = 2 . 2 Q (i; t' tnis case,as the tbundatio^ soil is purely cohesive,Skempto*'s formula may be applied' Here.

299

BearingCaprcity

when Q = 0", dq = 1.0and when$ = ?5', dq = dc = I'21

Bylinearinterpolation,for $ = lO", dq = 1.0 +

= 1'084 Sincetheloadingis vertical, i, = iq = ty = 1 (2.47)(r'rL4) (1.0)+ (1'84)(1.2) .'. 17,= (1.6)(8.34)(1'114)(r.2r) (0.77)(1'0)(1'0) (2.0) (0.47) (1.084)(1.0) + (0.s)(1.84) = 17.99+ 6.59 + 0.67 = ?5'?5t/m2

DIB = t.22.A =O.6 <2.5

U s i n g e q n s . ( 1 1 . 1 1 ) ,( 1 1 . 1 3 )a n d ( 1 1 . 1 5 ) ,t h e n e t u l t i r n a t eb e a r i n g capacity of a rectangularlooting is given by,

* =='o'l.,',: u, :'-T*]u.,i i 0,,,,,,n ; :?'1,'

= 29.51. t/nz a factorofsafetyof2'5, thesafebearingcapacity, Considering ,o ql

n,=t#=rr.Bt/mz Safe gross load on the footing = Q"'A

= (11.8) (2.0)(3.s)= 82.6t 2.2Q = 82.6

{14;-1!} rtol

Qt

25.25- (l'84)(1'2) + (1.8a)(1.2)= L1.42t/mZ ', ' 2.5

Totalgrossloadon tboting= (11.42)(2.0)(3-5)=79'97t' 2.?Q = 7e.97t or, A = 36.35t Problem N.g/A rectangularfooting of 2.4 m x 3.5 rn size is to be at t.YrnbelowG.i. in a c-g soilhavingthefollowingproperties: coustrucrect Y = 1.75VmT, 6 = ?1", c = l.at/mz' The footing hasto carry a grossverticalload of 70 t, inclusiveof its self-weight.In addition,the columnis subjectto a horizontalload of 11 t

Problems in SoiI Mechanics andFottndation Engineering

300

applied at a height of 3.3 m abovethe baseof the footing. Determine the factor ofsafety ofthe footing againstshearfailure : (i) using Brinch Hansen's method. (ii) As per IS : UIJ3 - 1981. solution: The loading condition of the column and the footing is shown in Fig.11.5. Due to the presenceof the horizontal force, the rcsultant load on is inclined, and the footing becor4eseccentrically loaded. Let e th" ilu-n be this eccentricitY'

301

BearingCapacity (i) Brinch.H ansen'seqn:We havefrom eqn.(11.16)' eurt= cN.s"d.i" + yDNrsndnio + 0'5 yBN,tsy4\' For 0 = 20", N" = 14.83,Nq = 6.4, Ny - 3.54 s" = sq - 1 + (0.2)(2.4/3.5\* 1.137 s' = 1 - (0.4)(2'4/3'5) = 0'7t4 dc = | + (0'35)(l5/L4) For

(E%:P i = zot;'do = t.o *

= L'219

(20)- 1.262

4=l'o'

'1.8 m

i, = | - H/TcBL = 1 - lt/(2 x 1.0 x 2.4 x 3'5) = 0.345

iq- | - o.t+- t - 0IlJlU= o.elr

e = 0'52m

\-it=$.szrf-0.&48

(1.5)(6'4)(1'137) (0.34s) (1.21e) + (1.75) (1.137) q, = (r.o)(14.83) (1.0)(0.84t|) (0.7ro (3.s4\ (L4\ (r.7s) (0.s) (0.e21) (1.262) +

Fig.11.5

= 33.79Vmz

Let R be the resultant soil reaction, applied at P, which can be resolved ? into two components,Ry and Rg. 2H = 0 gives, Rs 31.1t 2V=0gives,

Safe bearirqgcapacity,

o"=ff*to

Rv=74"t

summing up tle moments of all forces about tbe mid-point of the base G,

Q,weget,

RvxPQ=LLxz. ort

,pvo -_( 1 1 L ( 3 ' 3 ) = 0 . 5 2 m 70

Of'

Effectivelengthof thefooting, L' = L - 2 e = 3.5 - (2)(0.52)= 2'46m

3 t ' 1 g l+ L 6 l r - 1 1 . 8 6 -I"

Effectivewidth, B' = B - 2.4m Qfr

+(r.75)(r.s)

o"=ry+2.68 Actual pressuredue to the given loading' ,' 7n Q'-*g=ll'86t/m'

e = O.52m

.'. Effectivearea A' - L' B' = (2#) (2.41= 5.gn?

n"=E2=#U?

F" - 3'37 (ii) As per15:6403-1981:

Problems in Soit lt[echttnicsttnd Foundation Engineertng For

= 14.83, Nq = 6'40, ffv = 5'39 S = 20", i/.

Valtresols',snallcls,aretlresarneasilroseobtainedtbrBrirrclr Hansen'snrethod. VFo = tan (45' + 20"/2) = r'428 Now

d,=1.@!#Jg=1.18 d q = d y =t . U T f @

= l'oe

Bearing CaPacttY (i)Plottheloadvssettltnenlcurvearrrldeterrrrine,theultimatebearing of the Plate. capat:ity ' x 1'5 m' placed (ii) Detenninetheultirnateloadwhich a iboting of 1'5 rn settlernentis allowable the if ixrry will soil, same at 1.2 m below G.L. in the 2 rrnt.

11'6' In ordcr Solution: (i) The load-settlenentcrulveis shown in Fig' were two tengents plate' to
Anglc of inclinationof tht resultantload,

= 37'5 t/m2 {a (plate)= 3'75kg/cntz of the prototypelboting is given (ii) Using eq. (11.27),the settletucltt

cr = tan-lE = tou-tft = s'n:' by'

1

i , = ifî = ( t - *9I0 \-

= o'81i

(r-8'93\-=0.:oo ,' r = l . ' n ,|-v.Jw (1'137) i. Qu= (1.0)(14.83) it.rtt't(l.18)(0.811)+ (I.75)(1.s)(6'4) (0'306) (1'0e) (0'714) (1.0e)(0.811)+ (0.5)(r'7s)(z'4\(s'3e) = 35,72t/nt2

3 s . 7 2- g . 7 s ) ( 1 5 )+ ( 1 . 7 5 ) ( 1 . s ) Sal'ebearing caPacitY, 4s = r"

n"=#

or,

I B (8, + 30.5)l2

P=Ppl4iffio'j

)

= lQ gq1' 2 c n r = 2 0 m m , B = l . 5 m = 1 5 0 c ' m 8o

Here,

t 150(30 + 30.5)12 = 2 A = P" (iso - 3o.t

|-30

20 = P p = 2.809 7 . 1 2m m .

or,

+2'6?5 U

33.095

a t

* 2.6L5= *s o"_.--

or,

F" = 3.Sg

./ problem ll.116. The following rc.sultswere obtainedfrom a plate load dth of 1'2 m test perforne
L E

A "

E

g

'Efr GJ

e12 o

L ttc,

u. j6 I

Settlement (mrn)

|

1?

*' " 18 F i g .I 1 . 6

]

f,.XO9 pp

Problemsin SoilMechanicsand FoundationEngineering

3O4

curvewe get, for a settlemetttof 7.12mrn of From the load-settlement loadon theplate the plate,the corresponding = 3.7okg/artz = 37 t/m2. Ultimateload=J)7\L5) (15) = 83.25t. Preblern lft,ff. Determinethe allowablebearingcapacityof a 2 m x 2 m squarefooting foundcdat a depthof 1.5 rn below the groundlevel in a deepstratumof silty clay havingthe following av€rageproperties: y = 1.8t/rf, c = 3t/m2, 0 = 0o, Cc = 0.89, ea = 0'85 The maxirnurnpermissiblesettlementof the footing is 7.5 crn. The atthesiteis at a depthof 1.0mbelowG'L' of thewatertable highestposition

305

Beoring,Copucity

c , " =f + -++=B.zEt/rn2. s z . ) (ii) Computation of SettlementzAs the underlying soil is saturatedsilty clay, only consolidation scttlcmcnt will take place. The zone of influcnce below the brse of tboting is extendedto a tnaximum depth of twice the width of the firoting, i.e., 4 m below the base.lnFig' ll.7,X-X is a horizontal plane throqgh the rniddle of this consolidating laycr. Norv, initial cffcctive ovcrburdenpressur€onX-X pe - 'l zl + ^{subz2 = ( 1 . 8 )( 1 . 0 ) + ( 1 . 8 - 1 ) ( 0 . 5 + 2 . 0 ) = 3.8t/n2 = 0.38 kg/cmz Usirrg 2 : 1 dispcrsion method, stressincrement al X'X, Lp ' =

1m

Po

Ap

tLl __lJ l

(8'28) (2'{J)(2'0)

= 2.07 t/ntl = 0.207kg/m2

(2.0 + 2.0)'

(assumingthc footing to bc loadectwith 8.28 t^n2). "'. Consolidatiousettlerncnt, pn + Lp C.. :po: p, = H i__ "..togro (0.259) '. ,^0.38 + 0.207 _ (400) logto

oJ8

tr + ossi

Fig.1r.? Solution:

Fig. 11.7 shows the given footing.

(i) Computation of Bearing Cupacity: We have, from Skernpton's equation, Qau = cNc

wtere, Here,

N" n

As the estimated settlernetttis greater than lhe uraximum permissible limit of 7.5 cXr,the allowablebearingcapacityof the footing shouldbe less than 8.28 t/m-. Let, q be the load intensity on the footing which results in a settlemenl ol'just 7.5 crn. Let A p be the stressiltensity onX-Xwhen the footing is loaded witb q t/m-' pn + L p c'-7;:

H'

5 ( r + 0,2D/Br(l + 0.28/L)

Q - 1 . 5m ,

B = L - 2.0m.

-

4nu

l \ (6.9) - 6.9c {31 ZOJt/mz

For a faclor of safety of.2.S,thc nct safe bearing capacity is givenby'

| . "o'

(4o0) (o.2se) 0ft

= 6.e * (0.2)(o.r)) sf r . (o.?P)tt ''v

Nc

= lu')d ctn'

(1 + 0.85)

= ,-t

logro

Po + LP = 7.5 logro

Po

P o + L P=

logro

Oft

p0

0.1339

Po+LP= 1.1433. Po

Problems in Soil Mechanics ond Foundotbn Engineering

306

But the value ofpg atX-Xis constant,and is equal to 0.38 kg/."nz

t+*^1

= L36r2

Solving,we get,

But,

6,p = o.l3T2kg/crt = 1.372t/m2 . sBL = qf

uP =

L'372=

@iAT;d

@- zf

q(*)^ (2 + 2)'

or,

Q=5.49t/m2-5.5t/ri Hence, a loading intensity of 5.5 1,1p2will result in a consolidatiott setflement of 7.5 cm. Therefore, the required allowable bearing capacity of the footing = 5.5 Vrn2. EXERCNSE 11 11.1. Determine the ultimate bearing capacity of the following footings placedat 1.2 m below lhe ground level in a homogeneousdepositof firm soil having y'= 1.8 t/nf , O = 20' and c = l.8t/n2. (i) a strip footing of 2 m width (ii) a squarefooting of 2m x 2 m size (iii) a ci'cular footing of2 m diarneter. given,forE = 2g', N. = 17.'7, Nq ='7.4, ffy = 5.0

tln?l tAns.(i) 56.rrth# (ii) 6s.12r/mz liii; 62.80 11.2. A2.5 m x2.5 m square tboting is founded at a depth of 1.5 m below G.L. in a loose soil deposit having the following properties: ,{ = r.65 t/m3, c = o.2kg/cmz, 0 = 15" Determine: (i) the ultimate bcaring capacity (ii) the net ultimate bearing capacity (iii) the net safe bearing capacity (iv) the safe bearing caPacity. The factor of safety should be taken as 3.0. Given, for Q = 15', N, = 12.9, Nq = 4.4, Nr = L5, N"' = 9.7, Nq' = 2,.7, fly' = 0.9.

[Ans.(i) 24.s8t/n?QDzz.srtl# (lii)7.s0!n? (iv)9.98vrnzl 11.3. A circular footing of 2.5 m diameter rests at 1.3 m below G.L' in a soil mass having an average cohesion of 10 kN/m', an angle of internal

Beuring Capacity

307

fridion of 28" anrt a bulk densityof 18 kN/mr. The water table is locatedat a grcat depth. Dt:lerrnine the safebearing capacity of the footing' As-sumea = 18'8 atd gJreral shear thilure' Given, tbr Q = 28" N. = 32'5' ff, Nv = 15.7.Thc faclorof safetyshouldbe takenas3.0. [AIs. 373.7kN/rn'] 11.4. In Problern11.3,if thewater.tableriseslo the groundlevel due to noodiilg, detcrminethe percentchangeill the sal'ebearifig capaeityof the [Ans: DecreasesbY l8.6Vol lootirtg. I 1.5. A squarefooting of 2.2 m x 2.2 msize is foundedat a d€pthof 1'2 m bclow G.L. in a honogeneousbed ofdry sandhavinga unit weight of 1.95 the Vrn3 aild an angle Of intemat tiictiou of 3d. Determinethe safe load tailure. shear footing .un ..rry with respectto a factorof safetyof 3.0 against Givt:n, forq = 36', N. = 65.4, Nu = 49.4, Ny = 54' 'be foundedin a bed ol' t 1.6. A 2.0 nr wide strip tboting is requiredto an angle of shearing and of 2.0 4ensity Vm' a bulk havi*g sa'el 4cnsc rcsistanceof 35'. Plot thevariationof ultirnatebearingcapacityof the tbotittg = 35"' wilh depth of tbunclatiou, Dp tbr 0 s Dy s 3.0m' Given, for q N. = 58, Ns = 41.5, Ny = 42.q. the safeloada Circularfootingof5 m diameterfouuded I1.7. De.rermine at a depth of 1.0 rn below G.L. can caily. The foundationsoil is a saturated claV lraving att uucoutineclcolnpressivestrengthof 6 t/rn2 and a ulit weight .rf i.fS t/nr5.Assumea faUorof iafety of 2.5.Use Skempton'sandTerzaghi's nrelhods attd cotupare the results. State,givilg reasons'wltich one is more rcliable. t, Skempton'smethod] [Ans: Terzaghi:154-92t,Skenpton: 131.48 I I .8. A strip footing has to carry a grossload of 120 kN per tnetre run. Tht. footing is plat-edat l.?5 m below G.L. in a homogeneoussandstratun. The unit *.ignt and algle of internalfriction of the sandare 19 kN/rn' and 32' respectivily. Detennine the rninimum width of the footing required in = 44, or
I

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 1..25m belcnvthe ground level. The properties of the foundation soil are: c - 0.1 kg/un2, 0 - 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?, 0 = 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 0 = 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] 0 - 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 edby 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 1. Pile foundations 2. Well tbundationsor opell caissons' 3. 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 esthrough 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 = 5 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 6 (s + 2.5)

...(r2.2)

wh

= (ii)ForsingleactingsteamhammertQ OG r ) (W p) h + a (iii) For doubleactingstean hammer:n 6 (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 1 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 1

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 0 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"= 9 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

0 - 3.75 3.75-7.50

1 - 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. 0 = tiictionangiEof-GE6iiloithepilEl\ilEiC[?Ep€iiGon the a:rgleof internalfriction Q of the soil. The value of 6 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

2. CohesionlessSoi/s: For piles driven in cohesionlesssoils,

Qa=

, the point bearingrcsistanceis For a purcly cohesionlesssoil, c = 0. Henc:e given by, ...(r2.22) qb = lDNrso + YBNrst whcrtr, No. N, = Bearingcapacityfactors'

Timber & Concretc

where,

315

Pilc Foundotions

and,

N q = u 3 ' 8 { t a n o ' t u ng25 " + Q / 2 )

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

0

1.0

30

9.5

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 4 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 3 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 3 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 endbearingpilesingthroughcornpressiblesoil but restitg
of lhc layer where, I/ = thic-kness C. = cornpiessionittdex, eo = ilitial void ratio oo = initial stressat the ceutreofthe layer due to pilt:s A o = stressinc:rement

= A% A' = areaover which the load is distributedat the centre of tlre layer. 12.12 Negative Skin Friction: Thc downwarddrag actingott a pile due to the relativemovementof the surrounding soil tnassis c-alledthe negativeskin

Problems in Soil Mcchonics and Foundation Engineering

320

friction. This tcnds to reduc-ethe load canying capacity of the pile. Its magnitude can be dctermined from:

Q6 ' P' r' Lf

(i) for cohesivesoils:

soils: O^f = | 4 Oy K tan6 (ii) for c.ohesionless

Solution: Frorn equ. (12.5),the ultimateload on pile,

1 1 ' wH ' n 5 vu = --rll7l-

...(r2.3s) .,(t2.36)

where, p = perimeterofrhepilc

Here, W = 3.0 t, H = 91 crn, Tln = 757o = 4.75 s=10mm=1-.0cm N o w , e P = ( 0 . 5 s ) ( 1 . s )= 0 . 8 2 5 t

c - averagecohesion ofthe soil

W > 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 1 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.

6 = friction angle, (6 s Q) ,

321

Pile Foundations

I x (30)2 4

wH 6(s + c)

,r=94JJ2,/9)J4J. = 0.868 I x (30)2 4 (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 5 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.

1 0 7 ' 5 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 5 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 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

=

\ ! 7

=

a

= {!ggq = rzt/m2 For loose sand,Ks = 0.5 The value of 6 may be obtainedfrorn Table l2.2.For a srnoothRCC pile embeddedin dry sand, 6/q = 9.76, or, 6 = (0.76)(25') = 19'

ey = $2) (o's) (tan1e") =2.A66tln?

Q , = e 1 . 4 1 +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 = 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'

4

= 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..2 + 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=

1 5 k N . / m 2 ,Q = 2 0 " ,

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 5 + 1'9 x 3 + 1'8 x 2)(9'5) eu = 3qNq

Qf=ac+flK"tan6' cr = 0.5, K" = 1, 6/$ = 0.80' 6 = (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 1

Ir ,^-" a r r < . 1e u = i ' ( 0 . 5 ) ' = 0 . 1 9 6 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 3

4.

gu = (2r.s4)(1s,71)+ (r2e6'$ (0'126)

ft [o undation i s ed Ot I OU"t-t-t:-ul,::is.tiit * P roblem ly'{'ra 300 rows' The diarneterand lengthof eachpile are 3 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 - 1'80t'/m3' 0 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 = 9 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, 3 m and 2 m' Forthe firstlayer'

a

and' Af, = n (0'5) (3'0) = 4'71m'

= 390 + 900 + 6.4 = 1296.4

Assume,

= (0'80) (5\ = 4 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 = ( 0 . 9 ) ( 3 . 2 )= 2 . 8 8 t / m 2 .'. Qs = Qs.8) (2'7) (5.1) + r5(2 v 2'7 + 2 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<, = 3m. lengthofblock, L =3(1'25) + 2(050/2)= 425m depthof block, D/ = 30 *' = 337'5kN/m2 4f = lOkN,/m2,% (3)(4'25)+ 3A(2x3 +2 x 4'25)(30) GroupcapacitY, Qs = (337.5) - (11)(30)(3) (4.2s) < 17757kN = 13145.6kN Hence,groupactiongovernsthecapacityof thepile group' kN. O, = 13145.6

= 678.05 t > 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 3 rows and 30 m long,-s andspacedatl.?s m c/c.Thepropertiesof thefoundationsoil areasfollows: 'l|kN/m2, y' = 11 kN,/m3, Qu 0 = 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(-^"îerdin 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) + + ( 4 - t) s l - 0.624 = 62.4To ' r 8= , - 2 L'8[( s

e ol - 1 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 6 y ( r ) t 4 ( s -+15) ( 4 - 1 )+ y ' 2 1 + - 1 1 6 - X , = .' - ; 1 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 3 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 3 d, s = (3) (0'40) = 1'2 * i.e. Let I be the length of each Pile'

Now,

J

4 = I ' (40 . 4 0 ) 2 = o . L 2 6 r r ? CapacitYof each Pile,

L) + (21'6)(0'126) gu = (2.16)(r.2s7 or,

L + o'eol Ot = 7 = o'eos

the averagevalue of N = 9

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 edby 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. 8 + 10 + 8 + 11 + 9 = 9 . 2 o 9 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 = 3 s + d = 3 ( 1 . 2 )+ 0 ' 4 = 4 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 edby 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 = 8 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 3 thickness of sub-layers three 10 m. Let us divide this zone into 4 nr resPectivelY. The settlementof eachsub-layermay be obtainedfrom:

p, = H' f; Now. we have,

'tor,oo 0 +

w G = se, or' e =

eo= v c -

Ao

wG s

and,

- 1'00 = 0'90 t/m3 Ysub= 1'90

= (4 + 3 tan30")2= 32.86rt 350 ^ o = for = 3 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 + 2 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 + 2 x 8 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 = 3 s + d = 3 ( 1 . 2 \+ 0 . 4 = 4 m

= r.997t/mz

15.75+ L99J - 3.12cm (400)(0.279)' roglo 'pca 1SJS 1 + 0.g54 = + Totalsetflement,Pc P., * Pc, Pr. = 14.32+ 5.5'7+ 3.12 = 23 cm. EXERCISE12 of an RCile 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 9 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 ai 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 ing the tboting skt:tchcdin Fig. 12.5. Given' = 0'9 adlrcsiontactor tbr soft clay = 0'7 silt claYeY and that t'or for 0 = 30' is 9'5' The water table is factorNu capacity bcaring Vt'sit"s neglected' [Ans' 232 t] lrx nlcrl rtt il gr(raldth.Skin friction iir sandmay be

Soft CtoY

l2.4.Aconcretepileof30cndiameterisernbedd-edinastratunrofsoft clay straturnis clay lraving 1 = 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 8 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 6 the safecapacityof = determine 9'5' be 30" bearingcapacity faciorNn for 0 = 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'= 9 kN'/m3, 0 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'

0 = 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 1 y = 1 . Et /5n F ) , c= 6 l l m 2

I

J

2.0m

T

I

i I I

Ssnd (t=1.75t/m3,@=30o) Fig.12.5

6 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 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 = 5 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 edby a raft footing piles 96 of consists raft on a pile foundation. The pile group ing 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 = 2 . 0 t / r n 3 c, = 3 . 6 1 / m 2 , O = 0 ' .

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 8 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 3 againstshearfailure. Given. a = 0.85. 12.13. It is required to drive a group of piles in order to 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 = 2 tlnr3 , 0 = 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 4 D. The adhesionfactor for clay = 0.90. Vcsic's bearing capacity factor .lfu for 0 = 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.

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