soil mechanics question and answers

8/7/2019 Soil Mechanics Question and Answers

1/43

SECTION CIiHORT TYPE QUES1ION it

. A sample with a volume of 45cm3;is filled with a soil sample. When the soil sample is \~:::::: :::duated cylinder, K displaces 250m' of water. mat Is the porosity and ~.

ns. Given, V=45cm3; ! Iof solid. ( V,) Volume of water displaced

Vv=V-V!l=45-25=20cm3 t

1 +W = Heace Wl+W

n=v JV=20 /45=O.444e=V JV.l=20/25=O.8

2. By three phase soli system, derive a relation betWeenr6 ' and W.Ans. Water content (W) = W JW " ,

wSo y",=Y/1+W

3.What are the index properties of soil?Ans. Index properties of soilars

i) Water contentiii) Particle size distributionv) In-situ density

ii) Specific gravityiv) Consistency limitsvi) Density index

4. What do you mean by consistency of soil?Ans. Consistency of soil is nothing but it is the relative ease with which soil can be

deformed.

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2/43

It denotes degree of firmness of the soil which may be termed as soft, firm,stiff or hard

What are the A!!!rl:>_~rg_lmlts which are used for engineering purposes? Define UquidityIndex.s. The Atterberg limits areliquid limits

plastic limits andshrinkage limitLiquid index (IJ) :The liquidity index or water-plasticity ratio is the ratioexpressed as a percentage of the natural water content of the soil minus it'splastic limit to it's plasticity index.

W-W;pI p, --

. What is Shrinkage ratio? What is the relation between shrinkage ratio and mass specificgravity of the soil in its dry state?

ns. Shrinkage ratio:Shrinkage ratio is defined as the ratio of a given volume change expressedas a percentage of dry volurne , "to the corresponding change in watercontent above the shrinkage limit expressed as a percentage of weight of theoven dried soil.Shrinkage ratio = mass specific gravity of the soil (in its dry state)

7. What do you understand by in-situ unit weight of a soil? Name two important methods forthe determination of in-situ unit weight.

Ans. The in-situ unit weight refers to the unit weight of a soil in the undisturbed conditionor of a compacted soil in-place.Two important methods for the determination of the in-situ unit weight are beinggiven:(i) Sand-replacement method.(it ') Core-cutter method.


3/43

What is the difference between the volumetric shrinkage and linear shrinkage?s. Volumetric Shrinkage (Vs )

I'The 'Volumeter Shrinkage' (or Volumetric change Vs) is defined as the decrease in Ii'I '1I 't )I'

the volume of a soil mass, expressed as a percentage of the dry volume of the soilmass, when the water content is reduced from an initial value to the shrinkage limit:

(Vi - Vd) ,V = V x 100'Linear Shrinkage lLs1 d

i. = ( 1 - 3 1 0 0 ) X 1 0 0 's Ys+I00'Linear Shrinkage (Ls)' is defined as the decrease in one dimension of the soil massexpressed as a percentage of the initial dimension, when the water content isreduced from a given value to the shrinkage limit. This is obtained as follows:

. Differentiate between 'residual' and 'transported' soils. In what way does thisknowledge help in soil engineering practice?

ns. Soils which are formed by weathering of rocks may remain in position at the place oforgion. In that case these are 'Residual Soils'. These may get transported from theplace of origin by various agencies such as wind, water, ice, gravity, etc. In this casethese are termed "Transported soil". Residual soils differ very much fromtransported soils in their characteristics and engineering behaviour. The degree ofdisintegration may vary greatly throughout a residual soil mass and hence, only agradual transition into rock is to be expected. But in transported soil high degree ofalteration of particle Shape, size, and texture as also sorting of the grains occursduring transportation and deposition.

10. Distinguish between( i) Texture and Structure of soil.(ii) Silt and Clay.

Ans. Differentiation between Texture and Structure of soil.The 'structure' of a soil may be defined as the manner of arrangement and state ofaggregation of soil grains. But the term 'Texture' refers to the appearance of thesurface of a material, such as a fabric. It is used in a similar sense with regard tosoils. Texture of a soil is reflected largely by the particle size, shape, and gradation.

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Differentiation between Silt and Clay.Depending on sizeSilt 0.075 mm to 0.002 mmClay Less than 0.002 mmDepending on specific gravityClay 2.44 - 2.92Silt 2.68 - 2.72

Define:(i) Degree of saturation(ii) Submerged density.

ns. 'Degree of saturation' of a soil mass is defined as the ratio of the volume of water inthe voids to the volume of voids. It is designated by the letter symbol S and 'iscommonly expressed as a percentage:

Submerged (Buoyant) Unit WeightThe 'Submerged unit weight' or 'Buoyant unit weight' of a soil is its unit weight inthe submerged condition. In other words, it is the submerged weight of soil solids(Ws)sub per unit of total volume, Vof the soil. It is denoted by the letter symbol 1':l'= (W,)subV

12. Derive the formula between soil moisture content (w), degree of saturation (S ) , specificgravity (G), and void ratio (e).

Ans.By definition.

..

w =Wr J W s .as fraction; S = V ' / v v > as fraction; e = VJ V sS. = V , i V sw = W fW = Vw, "(s

W If Vs.'Y sw_G = S_e

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3. Derive the relation between void ratio (e), specific gravity of particles (G) and moisturecontent at full saturation (w).By definition,

..w = W'/w" as fraction; S = VJ V V ) as fraction; e = V;V ,s = Vr /Vsw=WIW=Vw Y s

W S V, .Y ,w _ G = S _ e

For saturated condition, S = 1.HrI;.I\ i

14.A fine grained soli is found to have a liquid limit of 90% and a plasticity Index of SO.The Il: , Wsat = eJG or e =Wsat.G

natural water conte ntis 28%. Determine the liquidity index and indicate the probableconsistency of the natural soli.

Ans. Liquid limit, wL = 90%Plasticity index, lp = 50lp = wL-wp:.Plastic limit, wp = wL - lp = 90 - 50 = 40%The natural water content, w = 28%Liquidity index, IL = W-Wp = 0.24 (negative)I pSince the liquidity index is negative, the soil is in the semi-solid state of consistencyand is stiff; this fact can be inferred directly from the observation that the naturalmoisture content is less than the plastic limit of the soil.

15. A clay sample has void ratio of O.SOn the dry condition. The grain specific gravity hasbeen determined as 2.72.What will be the shrinkage limit of this clay?

Ans. The void ratio in the dry condition also will be the void ratio of the soil even at theshrinkage limit: but the soil has to be saturated at this limit.For a saturated soil,e=wG

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orw= e/GWs = e/G = 0.50/2.72 = 18.4%,Hence the shrinkage limit for this soli Is 18.4%.

. The liquid limit and plastic limit of a clay are 100% and 25%, respectively. From ahydrometer analysis it has been found that the clay soli consists of 50% of particlessmaller than 0.002 mm. Indicate the activity classification of this clay and the probabletype of clay mineral.

Liquid limit, wL= 100%Plastic limit, wp = 25%Plasticity Index, Ip = (wL - wp)= (100 - 25)% = 75%Percentage of clay-size particles = 50Activity, A =~where c Is the percentage of clay-size particles.A = 75/50 = 1.50Since the activity is greater than 1.25, the clay may be classified as being active.The probable clay mineral is montmorillonite.


7/43

SECTION 0r

NG TYPE QUESTIONSThe Attei'berg limits of a clay soil are: Liquid limit = 75%; Plastic limit = 45%; andShrinkage limit = 25%. If a sample of this soil has a volume of 30 em" at the liquid limitand a volume 16.6cm3 at the shrinkage limit, determine the specific gravity of solids,shrinkage ratio, andvolumetric shrinkage.

T ?~~;~~~~W d T - - - - - - : : : . : : . : : - - - - - - . : : : : . : : . - : -T . ~ ~ ; ~. .:'.-.!\.'~~..- T~~~~~~i~~_f~~~ r ' : ' 1 3 ~~ { . )~ ~I rw ' \. rw ' V ' I W f '(a) At liqJid limit (b) At shrinkage limit (e) Dry stateDifference,jn the volume ofwnteratLL and SL = (30 -16.6) cm3= 13.4 c :m 3Corresponding di t feren.ce inweight ofwnter :: 0 . .134 NB u t this is (O.76~O.25) W if oro.so W ":. 0.50 W tI= 0.184W tI = 0 .268 NWeigh t of water at SL = 0 .25 W d = 0 .25 x 0 .268 = 0 .0 67 N, ', Vo lume ofwater at SL = 6.7 c uJ

V.= Total vo lum e at 8L - volum e ofw ater at SL.= (16.6 -6.7) c~ = 9.9 em'.

W =0268Nf n2681 ,.= ""'i.9 Nlem8 = 0.027 N/em B = 27 kN/m B:. Sp e c if ic gravity o f so l ids = "III = 27 =2.71'1..,. 9.81

, V o lum e of so lid s ,

Weight of so l ids ,.,.

Shrinkage ratio, R = Wd = 26.8 = 161 'V 4 1 6 .6 .,Volumetric shrinkage. V.;::R(wi - 10.):;:R(wz, - w.).here

= 1_61 (75 - 25) ::::80.5%.


8/43

Name any two methods for determining the liquid limit of the soil sample. Describe theCasagrande method for determining the liquid limit in the laboratory.

i) Static Cone Penetrometerii) Casagrande's Liquid Limit DeviceCasagrande's Liquid Limit measurement:The liquid limit is determined in the laboratory with the aid of the standard mechanicalliquid limit device, designed by Arthur Casagrande and adopted by the 151.Theapparatus required are the mechanical liquid limit device, grooving tool, porcelainevaporating dish, flat glass plate, spatula, palette knives, balance, oven wash bottle withdistilled water and containers. The soil sample should pass 425-lJ IS Sieve. A sample ofabout 1.20 N should be taken. Two types of grooving tools-Type A (Casagrande type)and Type B (ASTMtype)-are used depending upon the nature of the soil. The camraises the brass cup to a specified height of 1 cm from where the cup drops upon theblock exerting a blow on the latter. The cranking is to be performed at a specified rate oftwo rotations per second. The grooving tool is meant to cut a standard groove in the soilsample just prior to giving blows. Air-dried soil sample of 1.20 N passing 425-lJ 1.5.Sieve is taken and is mixed with water and kneaded for achieving uniformity. The mixingtime is specified as 5 to 10 min. by some authorities. The soil paste is placed in theliquid limit cup, and levelled off with the help of the spatula. A clean and sharp groove iscut in the middle by means of a grooving tool. The crank is rotated at about 2revolutions per second and the number of blows required to make the halves of the soilpaste separated by the groove meet for a length of about 12 mm is counted. The watercontent is determined from a small quantity of the soil paste. This operation is repeateda few more times at different consistencies or moisture contents. The soli samplesshould be prepared at such consistencies that the number of blows or shocks requiredto close the groove will be less and more than 25. The relationship between the numberof blows and corresponding moisture contents thus obtained are plotted on semi-logarithmic graph paper, with the logarithm of the number of blows on the x-axis, andthe moisture contents on the y-axis. The graph thus obtained, i.e., the best fit straightline, is referred to as the 'Flow-graph' or 'Flow curve'. The moisture contentcorresponding to 25 blows from the flow curve is taken as the liquid limit of the soil.

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9/43

Explain the soil phase diagram by representing and giving the engineering significanceof porosity, void ratio, degree of saturation, percent air voids, air content, water content,bulk unit weight, water content, unit weight of water, unit weight of solids.

Va = Volume of airV",= Volume of'waterV ;. =Volume of voidsV, = Volume of solidsV = Tobl volume o r soil mass

Wo = Weicht of' air (necticible O r zero)W~ =Weicht of watel'W" =Weicht O r material occupyinc void spaceW, = Weight of'solidsW = ToUll weiCht of solid ma.ssWII=wll 1

Soil-phase d ia gram (vo lum e s a n d weights o f phases)Porofity'Porosi ty ' ar a soil m ass is the ratio of the volume ofwids to the to ta l volum e of the sail m a s s .Itis d e n o t e d by the l e t t e r symboJ n an d is commonlyexpres&ed as a p e r a m . t a g e :

n ..~)(lO)VHere .V()ldRatio'Void ratio' of a soil muss is d e f i n e d os t h e rat io ofthevo!ume olvoids to the volume of so lid s inthe s oil mos s. I t is d e D O t e d by the le t te r symbol e a n d iigenerfllly expressed as (l d e d m a lunction ;

V.c=-E.V .Here V" =Vo + Vw'Void ratio' i& u se d mo re than 'Poroaity' in s oil me chani~ to tharat:tense the naturalstate of soil. This is . fo r the reason tha t . in void rat iO. tbe denominator, V" or volume of'salida,

is supposed to be re la tiv ely c on stan t u nd er the applic atio n o f p re ssure , while the numerator.V tthe v o l l ! J ) 1 e of voids nloneehQDgeG ; however , intheenSe of porosity, both the numeratorV"and the d enom in ato r V c h a n g e upon applieation ofpmsure .


10/43

vn =;J.x 1 0 0 VA ir c o n t e n t ' o f a s o i l m a s s i s d e f in ed a s t h e ra t io o f t h e v o l u m e o f a i r v o id s t o t h e t o t o l v o lu m eo f v o i d s . I t is d e s i g n a t e d b y t h e le t t e r ~ ' l n b o l 0 t a n d i s c o m m o n ly e x p r e s s e d 8& a p e r e e n t a g e :

Va =..Lx 1 0 0e V a tW a te r ( M o is t u re ) C o n t e n t'W a t e r c o n te n t ' o r 'M o i s t u r e e o n te n t ' o f a s o i l m a s s is d e f i n e d a s th e ra t io o f t h e w e ig h t o fw a t e rt o the w e i g h t o f s o lid s ( d r y W e ig h t ) o f th e s o i l m a s s . It is d e n o t e d b y the le t t e r s ym b o l a t n n d isc o m m o n l y e x p r e s s e d o s a p e r c e n t a g e :

W 0 0w= W x lW .( o rW , )& (W - W , ) x 1 0 0

W c rI n t h e f ie ld o f G e o lo g y . w a t e r e o n t e n t i s d e f i n e d a s t h e r a t io o f w e ig h t o fw o . t e r t o the

t o t a l w e ig h t o f s o i l m a s s ; t h is d i f t ' e r e n e e ha$ t o b e b o r n e in m i n d .F o r the p u r p o s e o f the a b o v e d e f i n i t i o n s . o n l y the f r e e w a t e r in th e p o r e s pa c es o r v o id s.. d--.Jl S C O D S J "IRU.

B ul. (M a ss ) U nit Wel,h t' B u l k unit w e i g h t ' o r 'M a s s unit w e i g h t ' o f a s o i l m a s s i s d e f i n e d a s t h e w e i g h t p e r u n i t v o lu m eo f t b e s o i l m o s s . It i s d e n o t e d by t h e l e t t e r & y m b o l y .

H e n e e , Y = \ V I VH e r e W = W w + W ,

a n d V = V a + Vw + V,T h e t e rm 'd e n s i t y ' is l o o s e l y u s e d f o r ' u n i t w e ig h t ' in s o i l m e e b a n i e s . o J t b o u g h . s t r i t U ys p e a k i n g . density m e a n s the m a s s p e r un i t v o lu m e a n d n o t w e ig h t .


11/43

r ee o f Sa .tu ra tio notsoturat ion ' of a s oil mo s s is def ined es the ratio o f t h e volume ot\ \ 'ater in the voids

the vol\llI1e of 'v 'Dids:I t is designated by th e le tte r s ymbo l S end is common ly exp re s sed as a .rcentagt! ;

S:~}(10D~,Here Vtr = V I I . . . V I I . 'F ' o r a fully s atu rated s oil m a ss . V I I ! = V I / "Therefore. fo r 0 . s a tu ra .t ed so il ntaS$ S = 100% .For a dry soil n\aS$. Vwis :zero.'Iherefore, fo r a perfect ly ~r soUsamp\e S is z ero .In bo th these c on ditio ns . the &a ilis considered to be a two -pha se &)'Stem.The degree o f saturation is b etw een z ero an d 1 0 0% . the s oil m ass b eing said to be ' p a r -ia lly ' s atu ra te d--th e mo st c ommon c on ditio n in nature.

erwnt Air VDidsair \ 'Oids" of 0 . so ilrnas s is deflned a s the ra tio o f the vo lume of air \ ro ids to the total

olum e of the soil mo.s s . I t is denoted by th e J ette r $Y1 tlb oliiG nnd i3 commr:mly expressed Q$ ~ercen. :B u lk (M a ss ) Unff WeIgh t"Bulk unit weight' or 'M os s.unitweighf o f a $0\1 IDUSS is defined us the weight per Ullit volumeof the s oU mos s. It is deno ted by the letter symbol-y.Hence. y = WJV

Herea n d

w = w + W .III tV ~ V G + V"' + V #

The tenn 'density' is loosely U 8 e d for \mit weight ' in so i l meclumies, al though. strictlysp en kin g. d en sity m eans the mass per uni t volume and not\\ 'eight.Unif Weight ofSolidlJ'U n it w e ig ht of solids ' is th e weight of soil so lids per unit volum e of so lids alone . It is a l s Ds ometime s c alle d the ' absolute u m t weight ' of a $Oil , It is d en ote d by the letter $ Y I D b o l 1 ,:

UnifWeighf Q(Wafer'U n it w eig ht o fw nter' is the \\'eig ht per unit vo lume o fwa t er. Itis d en.o ted .by the lette r s ymb ol' Y i l t ;

W'Y . =~W V I "

It should be n oted that the unit \veight of w ater varies in a small range ''\ith tenperature. Ithes a oonvenient value at 4C 1e ,which is the $tandt l l "d tamperat\ll 'e fOt ' this p t . U ' J J O S e . " Io isthe sym bol used to denote the uni t w eight a tw ater at 4~C .


12/43

Differentiate, between saturated unit weight and submerged unit weight. Derive theexpression for submerged unit weight of soil.

Soli phase d iag ram show ing a d d it io n a l e q uiva le n ts o n the- w e igh t s id eSa tu ra te d U n it W e i,h tThe 'Saturated u nit weight' is defined as t h e b ulk u nit w eig ht o f the soil mass in the saturatedc o n d i t i o n . This is d e n o t e d by the l e t t e r symbol 'Y m .Submerged (Buoytmt) U ,dt W e tgh tThe 'S ubme rg ed unit \ \ 'eight' or 'B uo yan t u nit w eight' of a so il is its un it w e igh t in the s ub -merg ed conditio n. In o the r words, it, is the submerged ,, 'eight o f sail so li t: l& (W')sub per unit oftotal v o l u m e , V o f the soil. It is denotE d by the letter s ymbo l ' 1 ' ;

t= ( W " ) ' 1 l l IV(W,),lIb is e qu al to the \\'eight o f s o lid s in o 1 r : minu.s t h e "" 'eight o f \" 'n te r d is p la c ed by th e sol ids .Th is le ad s to :

(W ,)s ulI ;; W , - V s' lwSince th e soil is submerged, the voids m ust be full ofwo.ter j the total volume V,then,

m u s t b e equal t o (V I + V 1 1 , ) { W . .)tuh may n o w 'be lW i t t e n a s :(Wt)~tIh=W - Will- V, . l W= - W - V u; Y Ill- V J : i '1 l l= If - i ' w < V w + V$)=W-V lw

Dividing throughout by V , the to ta l volume,(W~)~ :: (.w,'lll'l_V '. IY, 1 [ 1 . 'or r="IMl-1I t may be rw ted that a subm erged soil is invariab ly saturated , while.Q s atu rated s oil

n e e d not be sumberged.


13/43

. Derive the expression the relation between porosity, void ratio, degree of saturation,ercent air voids, air content, and water content.

elatlonshlps Involving Porosity, Void Ratio, Df:'grel'of Saturation,ater Content, Percent Air Voids and Air Content

n = v,. , as 0 . fract ionV= V-V'=l_~ .. l_ lV,V V G yu.,V

n= 1- \VdG ' t l J . ' Vis m ay p ro vide a p rac tic al o .pp ro ac hto the de te rm in ation ofn .

.T h is m n y p rD \ id e a p r o . c t i ~ a l a p p r o a c l l t o th e d e tm n in a tiQ n o f c .~, ~,n:- e=-V V,

1 1 unT V,+~~ V , V~ " 1 (1+e )n = v r r . = .-+-. . l t e + . - -. u 'V v V & , V u ee'f n:--(l+e)c : : n I l 1 - n ) , b y a l g e b r a i c m a n i p u l a t i o n


14/43

T h e s e i n t e r r e l a t i o n s h i p ! b e t w e e n l l t m d e f a c i l i t a t e , c o m p u t O i t t O D o f o n e i f t h e o t h e r is

. . V(I, =...!t V .v '. ,~1n,=-V.. VM=.: =nr V QRQ;: 'M~(

B y d e f i n i t i o n .U J : : tV J W ~ .a s f r a c t i o n ; S ; ; V r iV p a s f r a c t i o n ; c . : V .} V aS ,e = V J V ,

: : W . ID1.. ' = . " ' . ' l 1 ' . l t = - V w ' Y $ ~ t , . Jt!". 'G . , . . S _ iGW 4/,r, " . .. - \'t t l ". . -.1~. it V J . G y w: . w ; G = .S .e( N o i t . T h i s is v a l i d e v e n i f b o t h w a n d S ate ~ Upe iUn tnps ) . F ' O l ' a t u n t t c o n d i t i o n ,

".. .~

Wu.t aelGore -w_.G~ , _ E s . . e - , v u ;. 1 1 ; . v ; - V k f v. u~ v.n -----. _: -_......:......'~. V-V. +V~_. 1+ VII' l+e

v~S.e=Vp.e-S.e e(1-8)no~ - .l+e l+eIt...:;(Q,.jl+'el(l-B) = n(l- S)

.~

But

". .


15/43

. Determine the (i) water content (ii) Dry density (iii) Bulk density (iv) void ratio, (v) degreeof saturation from the following data:

Sample size 3.81 em dia. x 7.62 em ht.Wet weight = 1.668 NOven-dry weight = 1.400 NSpecific gravityWetweight1Oven-dry weight,

Water content,Total volume of soil sample,

Bulk unit weight,

Dry unitweight,Spec i f i c gravity of solids,

. . V oid ratio,Degree of saturation,

= 2.7w= 1.668 NWd = 1.4ll0 N

w = (1668 - ~4[J) x 100% = 19.14%140

V = 1t x (3.81)2 x 7.62 ems4= 86.87cm!l1668 ." ( = WW = = 0.0192 N/cms86.87= 18..84 kNhns

" ( = 18.84 kN/ms = 15.81 kNlm3'Y d = (1+w) (1+0.1914)G=2.701 . = G. 'Y UJ ' Y , w = 9.81 kN/ms

d (1+e)15.81 = 2.7 x 9.81 (1 + e)= 2..7 X 9.81 = 1.675(1+ e) 15.81

e =0.675s= wG = 0.1914x 2.70 = 0.7656 = 76.SSO/(l.e 0.6757. A dry soil has a void ratio of 0.65 and its specific gravity = 2.80.i) What is its unit weight?ii) Water is added to the sample so that its degree of saturation is 60% without any change

in void ratio. Determine the water content and unit weight.

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16/43

The sample is next placed below water. Determine the true unit weight (not considering;buoyancy) if the degree of saturation is 95% and 100% respectively.

( i) Dry Soiloidratio, e = 0.65

specificgravity, G = 2.80weight, _ G .'Y u.._ 2.80 x 9.8 kN l s -1665 kNl S'Y a- - m -. m.(1+e) 165Partial Saturation of the Soil

of saturation, S =60%ince the void ratio remained unchanged, e = 0.65ater content, w = S.e = 0.60 x 0.65 = 0.1393

G 2.80=.13.93%U' . ht (G+ Se) (2.80 + 0.60 x 0.65) 981 kNl Srut wag = (1+ e) . 'Y w = 165 . m

= 18.97 kNlms.(iii) Sample below WaterHigh degree of saturation S = 95%.

Unit ight(G+Se). = (2.80+0.95 x 0.65) 9.81kN/msm wei


17/43

Explain the followings.What do you mean by Speciflcgravity of solids? What is its Unit.? iIThe mass Specific gravity of a fully saturated specimen of clay having a water content of}

~I.]30.5% is 1.96. On oven drying, the mass Specific gravity drops to 1.60. CalculateSpecific gravity of clay.

w=30.5%Gm =1.96rat =Gm 'YIII = 1.96 ' V . . .Gm=1.601., = Gm Y... = 1.6Oyw= 196 = (G+ e n l l lYat . l U l (1+e)1!n Gyw' Y . J = .uu'YIII = (1+e)

e=wGe=0.305G

S pe cific G ravity of Solicls .The 's pec ific gravity of s oil s olid s' is d efined as the ratio of the l1l1it weig ht o f s olid s (a bs olu teunit w eight o f so il) to the unit weight of w ater a t the s ta nd ard temp era tu re (4C). This isdenoted by th e letter symbo l G an d is g iv en by :

G= lL1ftThis is als o know n as IAbsolu te specific gravity' and, infact , more popularly as 'O r .Specific Gravi tY . Since th is is relatively constan t value for a given soil, it enters into numcomputations in the fteld o f s o il m e c l u m i c s .

(ii) Saturated clayWater content.Mass specific gravity,On oven-drying.

For a saturated soil,From ro ,

...(i)

...(ii)

1.96 = (G+0.305G) = 1305G(1+Il305G) (1+0.305G)1.96 + 0.5900 = 1.305G

~960G=-- =2.770.707From (ii).1.60=GI(1+e)G = (1+ 0.305G) 1.6G = 1.6+0.485G

::::) 0.512G= 1.6::::) G = 1.610.512= 3.123The latter part shou ld not have been given (additional and inconsistent data).

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18/43

A 'sample of clay taken from a natural stratum was found to be partially saturated andwhen tested In the lab gave the following results:Specific gravity of soli particle is 2.6;Wet weight of sample = 2.S0N;Dry weight of sample= 210N.Volume of sample=1S0 cm3Compute the degree of saturation.

Specific gravity of so il part ic les ,Wet weight,Volume,Dry weight ,

G=2.60W =2.5UN;V= 1 50 emsWa=2.10N

Dry unit weight.

w = (W- Wd) x 100= (2.5- 2.1) x 1~%Wd, 2.1

= OADx 100% = 19.05%2.10T=W1V= 2.50/150 = 0.0167 N/ems I= 16.38kN/ms_ "'( ::: '16.38 kNlms

Td- (1 +w) (1 + 0.19(5)= 13.76kN/niJ

Water content.

Bulkunitweight.

[AlsO, ..,. =i'=2.1W150 = 0.014 N/cm S = 13.734 kNImS], G -TmTd= (1+ e)2.6 x 9.8113.76=-'--(1+ e)

But

Degree o f saturatian,

(1 + e) = 2_6 x . 9.81 = 1_85413_76e = rf854s= wG :::0.1905 X 2_6 = 0.58e 0.854= 58%


19/43

Stresses within the soli mass is caused by external loads applied to the soil and also byIts own weight." Justify the statement basing upon the concept of Geostatic stress

'jacting upon the soil. q. Geostatic stresses:-, ;" . ,;Stresses within a so il m as s are ~u sed by ex te rna llaads app lied to the so il and alsD by th ~ s e lf . Iwright o f th i ! lSl;)il T h e p a t t e r n o f s t r e 1 5 l 5 e s caused by ~malloads 4 U$Wllly v ery ~mpl i - leated ; the pattem of s tresses eaused by the s e l f " w e iS 1 l t o f the s oil als o c an be complicated. But . 1there is one c ommon s itu atio n in\vhieh the self.weight o f th e s oil g iv es ris e to a very s im ple 1p attem o f s tre ss es -tha t is, when the ground s u rf ac e is harizont aland the n atu re o f the soild oe s n ot v ary s ig nif ic an tly inth e horiz on ta l d ire ctio ns . This s itu atio n e x i s t . s frequently in the l 'case o f sedimentary d ep os its . T he s tre ss es insuch a Situation are referred to a s 'O eo sta tic ,'Stre sse s", '

'Further, inthis s itu atio n, th ere can b e no shear st resses upon vertical and horizontallp la ne s w ith tn the s oil m as s. T herefo re, the vertic al g eo static s tres s mayb e c om p uted s im p lyby cons ide ring the w eight of the soil ab ove that depth.

Ifth e unit weight of the soil is cons tan t with depth,where 0 " 1 7 - = vert ical g e o 1 5 t a t i c stress

" /:: m it w eight o f lSojlz :: d ep th u nd er c on sid eratio n..--~---------_"O"v

Ve rt ic a l ge os ta t ic s tre ss in soilwith h on zo n fa l s ur fa ce

Figure1. What are the methods to determine the water content as per lSI ? How can youdetermine the water content by using (a)Ovendrying method (b) Pycnometer method?

ns. Themethod to determine the water content are:-(i) Over-drying method(ii) Pycnometer method

32


20/43

(iii) Rapidmoisture Tester methodOven-drying Methodemost accurate approach is that of oven-drying the soil sample and is.adopted in theboratory. A.clean container of non-corrodible material is taken and its empty weightlong-with the lid is taken. A small quantity of moist soil is placed in the container, thed is replaced, and the weight is taken. Thelid is taken removed and the container withe soil is placed in a thermostaticallycontrolled oven for 24 hours, the temperatureeing maintained between 105-11o o e . After drying, .the container Is cooled in a

the lid is replaced and the weight is taken. For weighing a balancewith anccuracyof 0.0001N (0.01g) is used.us, the observations are:eight of an empty container with lid =W1eight of container with lid +wet soil = Wleight of container with lid + dry soil =W3ecalculations are as follows:eight of dry soil = W3- W1eight of water in the soil = Wl - W3

r,

..Wt ofwater ..fV"tCII&Pw= . X ..t.lAJ".I'OWt of dry soil

w = (WI - Wa ) X 100%tWa-WI)Water content.,

Sandysoils needonly about four hours of drying, while clays need at least 15hours. Toensure complete drying, 24 hours of oven drying is recommended. A temperature ofmore than 1100emay result in the loss of chemically bound water around clay particlesand hence should not be used. A low value such as 600e is preferred in the case oforganic soils such as peat to prevent oxidation of the organic matter. If gypsum issuspected to be present in the soil, drying at s o o e for longer time is preferred to preventhe loss of water of crystallisation of gypsum. To obtain quick results in the field,sometimes heating on a sand-bath for about one hour is resorted to instead of ovendrying. This is considered to bea crude method since there is no temperature control.PycnometerMethod


21/43

is method _1 e u s e d wben the specific gmvityof solids is known. This is a relat ively q m e kand is oonsidered s u i t a b l e fur coan&grained soils only.

T h e 1 b l l D w i n g are t h e s t e p s involved:( i) T h e w e i g h t o f the e m p ty p y c n o m e te r with it s c a p a n d w a s h e r is found (WI)'(ii) T h l w e h o i l s a m p l e is p1aain t h e p y c n o m e t e r ( u p to a b o u t 1 1 4m1 1 3 o f t h e v o l w n e )

a n d it s w e i g h t is o b ta in e d (W z ) .(iii) T h e p y c n om e te r is g r a d u a l l y f i l l o o W ith \~t.tt. st i r r iDgand m i x i n g t h o r o u g h l y withe . .g l a s s rod. s u c h that w a t e r e c n n e s f l u s h , \ l i t h t h e h o le in t h e c o n i c a l c a p . '!he

p y c n o m e t e r is d r i e d o n t h e o u t s i d e with a c lo t h e n d . it s w e i s h t i s o b t a in e d (W ~ .. ( to ) T h e p y c n o m e t e r is e m p t i e d e n d c l e a n e d tharqh1y; it i s t i l l e d with w a t e r u p t o t h e

h o l e i n the c o n i c a l e a p . a n d it s w e i g h t is D b t a i n e d (W J.T h e w a t e r c o n te n t otthe s o i l s a m p l e m a y b e c a lc U la te d I . f o l lo w s :. w = [ ( W ~ WI) ( G - 1 ) - 1 ] x 1 0 0 %(W a - W.) G . .

, ' , . . T h i s c a n b e e a s i l y d e r i v e d f r om the s c h e m I l t i c p h a s e d i a g r a m s $ h G \ v nIf the s o l i d s f r o i n (itt) a r e r e p l a t e d w i t h \ V Q t e r ~ , ; c : , W. ..o f ( i v ) .V o lu m e o fs o U d s = !t.G

(a}mpty (b ) Pymometer+ w e t (e } P y c n o m e t e r + w e tp y e l l a wi. W , 1 0 1 w t . W 2 s o l +w a t e r w t . w!

.:: (WI- Wl) (0-1)_1(Wa -w.. (Jw = [ W a - W J ( . G - 1 ) _ 1 ] ) ( ~Wa-W . G

.~. " ' ' ' ' ' ' . , " ' ' ' ' ' ' ' . ' ' ' .. . .". .- .~." , , ,~!:::;:;:~~::" " . . ._ - . .- _ -" " . . ._ _ -_ - -_ _ . -_ . . " ' . - . .- - _ . ._ -. _ - - _ . _ - - _ .._ - -_ . " ' . - ..- . - . ._ .. .- - _ .. , - _ .._ . . -. . _ . . _ "'. . . . . . _ ._ . . . _ . _ - . _ - . " , . -_ . _ - _ ..- . - .._ -.. . _ . _ - - - - - _ . -(If) Pycnameter +w a t e r w t . w 4


22/43


23/43

. G. lU Ilcf = 1.60.111 1 = (l+e)e=wGe=0.S05G

For a saturated soil...From (i),

1.96 = (G+0.305G) = .LS05G(1+0.3050) (1+0.s05G)1.96 + 0.598G = 1.305G1.960G=- =2.'7'70.707From (ii).

1.60 = GI(l +e)G = (1 +0.3050) 1.6G = 1.6 +0.485G

0.512G=1.6G = 1.610.512 =3.123

1. In a specific gravity test, the weight of the dry ~oil taken is 0.66 N. The weight of thepyknometer filled with this soil and water is 6.7,56N. The weight of the pyknometer full of 1water is 6.3395 N. The temperature of the test is 30C. Determine the grain specific!gravity, taking the specific'gravity of water at 30C as 0.99568.Applying the necessary temperature correction, report the value of G which would beobtained if the test were conducted at .4Cand also at 27C~-The specific gravity valuesof water at 4C and 27C are respectively 1 and 0,.99654.

Weight of dry soil taken,Weight of pyknometer +soil +water (W3)Weight of pyknometer +Water (WJTemperature of the Test (T)Specific gravity of water at BOC (GrDozoBy Equation

Wa=O.66N=6.756 N=6.3395N= 30C=0.99568

Wa .GUTG=----~-W,,-(Wa-W.)

= 0.66 x 0.99568 = 2.69876 __2.'700.66 - (6.756 - 6.3395)36


24/43

If the test were conducted at 4C, G I L T = 1W".1 _ 0.66 x 1

G= W,,-(Ws-W,,) - 0.66-(6.756-6.3395) =2.'11..Ifthe tests were conducted at 27C, Gmor 0.99654 G = W"x 0.99654 = 0.66 x 0.99654. W"..Wa-W") 0.66-(6.756-6.3395)=2.7011 ...2.'10.

. The dry unit weight of a sand sample in the loosest state is 13.34 kN/m3 and in thedensest state, it is 21.19 kN/m3 Determine the density index of this sand when it has aporosity of 33%. Assume the grain specific gravity as 2.68.

'Ymin(loosesttate) = 13.34 kN/m3'Ymu(denseststate) = 21.19kN/m3Porosity, n = 33%Void ratio, neo = (1- n) = 33167 = 0.49

G'Yw'Y o = (1-eo) 2.68x9.81(1+0,49) kN/m3 = 17.64 kN/m3

Density Index,ID(by Eq. 3.10)= ('Y IJW[) ( Y o -Ym in )( Y o ) Ymax-Ymin21.19 {17.64-13.34} 21.19 4.30= - - x . -- x--= 0.658 = 65.8%17.64 (21.19-13.34) 17.64 7.85

Alternatively:G yw' Y m i n = (1- e I D a J [ ) or 2.68x 9.8113.34 = (1+mu:)

elJW[= 0.971a.v;1, = ormax (1-emin)emin = 0.241

(emu - eo)ID = ( ) > (byEq. 3.8)emax-emin

(0.971- 0.49) 0.48= (0.971- 0.241)= 0.73 = 56.8%.

2.68x 9.8121.19 = ( )l+emin

37


25/43

SECTION CHORT ANSWER TYPE QUESTIONS.Whatarethe various parameters that effect the permeability of soil in the field?ns. The following soil characteristics haveinfluence on permeability:

1.Grain-size2.Void ratio3. Composition4. Fabric or structural arrangementof particles5.Degreeof saturation6. Presenceof entrapped air andother foreign matter.

; Estimate the coefficient of permeability for a uniform sand where a sieve analysisindicates that the 010 size is 0.12mm.

ns. 010 = 0.12.mm= 0.012cm.According to Allen Hazen's relationship,k = 1000102where k is permeability in cm/s and 010 is effective size in cm.k = 100)( (0.012)2 100)(0.000144= 0.0144cm/sPermeability coefficient = 1.44)( 10-1mm/s.

3. What are the conditions necessary for Darcy's law to be applicable for flow of waterthrough soil?

Ans. The grains grouping around void spaces larger than the grain-size are flocs andflocs grouping around void spaces larger than eventhe flocs result in the formation of a'flocculent' structure. When inter-particle repulsive forces are brought back into playeither by remoulding or by the transportation process, a more parallel arrangementorreorientation of the particles occurs. This means more face-to-face contacts occur forthe flaky particles when these are in a dispersed state and results low permeability ascompare to Flocculated structure.

4. Why is the permeability of a clay soil with flocculated structure greater than that for it inthe remoulded state?

Ans. Darcy demonstrated experimentally that "For laminar flow conditions in saturatedsoils, the rate of percolation is directly proportional to the hydraulic gradient" .q = k[( h1 - hz ) ! L] x A = k.i.Awhereq = the rate of flow or dischargek = a constant, now known asDarcy's coefficient of permeabilityh1 = the height above datum which the water rose in a standpipe inserted at the

entrance of the sand bed,hz = the height above datumwhich the water rose in a stand pipe insertedatthe exitend of the sand bed.

43


26/43

L = the length of the sample.A =the areaof cross-section of the sand bed normal to the general direction of flow.i = (hi - n2)IL, the hydraulic gradient

tatethe principle of Darcy's law for laminar flow of water through saturated soil.s. The principle of capillary says that water is "pulled up" in the capillary tube to aheight, dependent upon the diameter of the tube, the magnitude of surface tension, andthe unit weight of water. Smaller the diameter of tube the morewill be the capillary rise.So in fine grain soil, due to availability of small capillary tube or pores the capillary risebecomes more. While for coarse-grained soils, the voids are larger and resulting lowcapillary rise.A 1.25m layer of the soil (G = 2.65and porosity = 35%)is subject to an upward seepageheadof 1.85m.What depth of coarse sand would be required above the soil to provide afactor of safety of 2.0 against piping assuming that the coarse sand has the sameporosity and specific gravity as the soil and that there is negligible head loss in thesand.

Critical hydraulic gradient,

n 0.35G = 2.65; n = 35% = 0.35: e = --= -=/13, (1-n) 0.65_ (G-1). (2.65-1) 1.65x 13l = = = = 1.0725c (1+e) (1+ 7113) 20

Determinethe coefficient of permeability from the following data:Length of sand sample = 25 emArea of cross section of the sample = 30 C1tl2Head of water = 40 cmDischarge = 200 ml in 110 s.Ans. L = 25 emA = 30 cm2h = 40 cm (assumed constant)Q = 200 mi. t = 110 sq = Q lt = 200/110 mils = 20111 = 1.82 em3/si = h lL = 40125 = 815 = 1.60

20q = k . i . A k = qliA = 11X 16X 30 = 0.03788 cmls= 3.788 x 10-1 mm/s.44


27/43

glass cylinder 5 cm internal diameter and with a ecreen at the bottom was used asafalling head permeameter. The thickness of the samplewas 10 cm. With the water levelin the tube at the start of the test as 50 cm above the tail water, it dropped by 10cm inone minute, the tail water level remaining unchanged. Calculate the value of ~ for thesampleof the soil. Comment on the natureof the soil.;

Falling head permeability test:hI =50 em; h2=40emtl = 0; t2= 60s ... t = t2- tl = 60sA = (1fI4)x 5 2 = 6.251tem2 ;L = 10 emSincea isnot gi.yen. let us assume a =A.a.

k = 2.803 At ;loglD (ht1h,_)= 2.303 x (10/60) loglD (50/40) cmJs= 0.0372 cmls= 3.'12 )( 10-1mmlsThe soil may be coarse sand or fine Kravel

What is Quick sand phenomenon?ns. If the seepage pressure equals to the pressure due to submerged weight of soil, theeffective pressure reduces to zero. At this stage a cohesionless soil particles tend to belifted up along with the flowing water. This phenomenon is called quick condition orquick sand..Definepermeability and coefficient of permeability of soil?ns. The property of soil masswhich permits the flow of water through its interconnectedvoids, is called permeability.The coefficient of permeability is defined as the average velocity of flow that occursthrough the total cross sectional areaof the soil sample under unit hydraulic gradient. Itis expressed in cmisec or m/day

11.What is the difference between Discharge velocity and Seepage velocity of soil waterflow?

Ans.i) Discharge velocity : The rate of discharge of soil water through unit area of crosssection of the soil sample (both the area of soilds and voids), is known as dischargevelocity.ii) Seepagevelocity: The rate of discharge of percolatingwater per,unit cross sectionalareaof voidsonly perpendicular to the direction of flow,is known a~!,~eepagevelocity.

. . , ; _ - : : - : ; / i . . " - - " - "

45


28/43

2. How does the flow net help In the determination of seepage pressure? r .ns. The hydraulic potential h at any point located after n drops, each of value Ah Isglve~z\_ 6

1 ; 1l \The seepage pressure at any point equals the hydraulic potential or the balancehydraulic head multiplied' ~y the unit weight of water and hence, Is given by

p~.1,

h= H-nAh

P,= h z, = (H - nAh)Twhe pressure acta In the direction of flow.


29/43

SECTION 0NG TYPE QUESTIONS1. State and prove the validity of Darcy's law.

s. Darcy's LawH. Darcy of France performed a classical experiment in 1856. In orderto study theproperties of the flow of water through a sand filter bed; By measuring the value ofthe rate of flow or discharge, q for various values of the length of the sample, L, andpressure of water at top and bottom the sample, h1 and h 2 ' Darcy found that q wasproportional to (h1 - h2)ILor the hydraulic gradient, ; :q = k[(h 1 - h2)/L ] ) ( A = k.i.Awhereq = the rate of flow or dischargek = a constant, now known as Darcy's coefficient of permeabilityh1 = the height above datum which the water rose in a standpipe inserted at the

entrance of the sand bed,h2 = the height above datum which the water rose in a stand pipe inserted at the exit:

end of the sand bed.L = the length of the sample.A = the area of cross-section of the sand bed normal to the general direction of flow.; = (h1 - h2)IL, the hydraulic gradient.The Equation is known as Darcy's law and is valid for laminar flow. It is of utmostimportance in geotechnical engineering in view of the its w)de range of applicability.Validity of Darcy's LawReynolds found a lower limit of critical velocity for transition of the flow fromlaminar to a turbulent one. The values of R for which the flow in porous mediabecome turbulent have been measured as low as 0.1 and as high as 75. According toScheidegger, the probable reason that porous media do not exhibit a definite criticalReynold's number is because soil can be no means be accurately represented as abundle of straight tubes. He further discussed several reasons why flow throughvery small openingsmay not follow Darcy's law. There is overwhelming evidence which shows that

"'( .Darcy's law holds in silts as well as medium sands and also fQr a steady state flowthrough clays. For soils more pervious than medium sand, the actual relationship

47


30/43

between the hydraulic gradient and velocity should be obtained onlyexperiments for the particular soil and void ratio under study.

v -DR = c = 20001)

where R =Reynold's numberv = Velocity of flowD = Diameter ofpipelpore\ ) = Kinematic viscosityofwater

' Y w = Unit weight ofwater11= Viscosity of water and

g = Acceleration due to gravity.

f7 '~_?-- : :~~~~~~~_: - :~-= : :~~ ~~~~~~~~~~~~- - - - - - - - - - - - - - - - - - - - -- - - - - - - - - - - - - - - - - - - - -- - - - - - - - - - - - - - - - - - - - -- - - - - - - - - - - - - - - - - - - - -- - - - - - - - - - - - - - - - - - - - -- - - - - - - - - - - - - - - - - - - -

FigureWhat are the two laboratory methods for determining the coefficient of permeability?Explain the Falling head permeability test.s. (a) constant head & Falling head permeability method)Falling or Variable Head Permeameter .simple set-up of the falling, or variable head permeameter is shown in Fig.


31/43

Standp i peorluette(Cros s - sec t i ona l a r e a . a)

~~5~~~:~~~~ou s a .Q"oss-sedionalarea A)Figure

better set-up in. which the top of the standpipe is closed, with manometers anduum supply, may also be used to enhance the accuracy of the observations. Theling head permeameter is used for relatively less permeable soils where the dischargesmall. The water level in the stand-pipe falls continuously as water flows through theil specimen. Observations should be taken after a steady state of flow has reached. Ifhead or height of water level in the standpipe above that in the constant head

amber falls from ho to hi, corresponding to elapsed times foand ti, the coefficient ofk, can be shown to be:


32/43

= area of crcss-sectlcn of standpipeand A = length and area of cross-sectlon of the soll sample and the other quantities as

defined.This can be derived as follows:Let - dh be the change In head In a small interval of time dt. (NegatiyeJdgn Indicatesthe head decreases with Increase In elapsed time).

F rom Darc y's law ,

" ..Q = (-tJdhVdt = iia

- adh ldt = K-A-h/Ldh( k h I L ) - A = - a d t

(kA la L) . t i t = - dhlhrIntegrating both s id es and app ly ing the lim its to and tl fo r t, and k o 8lld A l fot~.',~ t d t = - t ~ = t~

(WaL ) ( t1 - t~ = 1 , (hJA 1) = 2_3logm (hJh1)..Transplsing the terms,

A horizontal stratified soil deposit consists of three layers each uniform In itself. Thepermeabilities of these layers are 8 )( 10-4 em/s, 52 )( 10-4 em/s, and 6 )( 10-' em/s,and their thicknesses are 7, 3 and 10 m respectively. Find the effective avera'!epermeability of the deposit in the horizontal and vertical directions.

k l = 8)(1 0 " " cmls 1 1 . 1 = 7 mk 2 = 52 )( 10 "" emfs 1 1 . 2 = 3 mk a = 6 X 1 0 " "~m1s ha: ' 1 0m

k (or k );: (k1n ! +~ ~+ k a h a )1 1 % (~ + ~ + h a )50


33/43

= (8x7+52x3+6x 10) )(1~'20= 13.6)( 10-4 e m l s

, ', E ffec tiv e average permeab ility inthe ho rizon ta l d ire ction= 13 .6 )('10 -3mmll1 1

20= - - - - - - - -1 -4 ( 7 /8 + 3 1 5 2 + 1 0 1 8 ]10= 7.7)(lit' emlsr. E f fec tiv e av erag e p erme ab ility in the vert ica l d irec t ion

= 1 .1 )( 1 0 -3mmll.A large excavation was made In a stratum of stiff clay with a saturated unit weight of18.64 kN/m3 When the depth of excavation reached 8 m, the excavation falled as amixture of sand and water rushed In. Subsequent borings Indicated that the clay wasunderlain by a bed of sand with Its top surface at a depth of 12.5 m . To what heightwould the water have risen above the stratum of sand Into a drill hole before theexcavation was started?

51


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12.5m

he effective stress at the top of sand stratum goes on getting reduced as thexcavation proceeds due to relief of stress, the neutral pressure in sand remainingonstant. The excavation would fall when the effective stress reached zero value at the ..op of sand.ffective stress at the top of sand stratum,

o r

ij=t l & 1 t - h y wh T I I I =t .T a t

k = ~ . 1 5 1 t = ( 1 2 .5 - 8 ) X 1 S .6 4 = 8 . 5 5 m1 m 9 . 8 1

Therefore, the water would have risen to a height of 8.55 m above the stratum of sand :into the drill hole before excavation under the influence of neutral pressure.

he earth dam of homogenous section with a horizontal filter is shown in the figure. If the-t,jtcoefficient of permeability of the soil is 3x103mm/s find the quantity of seepage per unit ..

.~length of the dam.' . ~ I> 1


35/43

IIIII II 1 3 2 mI II II II IIN IB'

9 0 m ~

30m

III: D ir e c t r ix o fr b a s e p a r a b o l aIII

.-------------~Om------------~~

S = (J92 2 ~3n2 -92 ) m = 4.77mThe, quanti ty o f seepage per met re unit length of the dam

q=k .8= 3 )( 1(t2 x lQ-S x 4.77 mIl s= 14.81 x lQ -5 m S / S= 143.1 m 1 I s .

6; A deposit of cohesion less soil with a permeability of 3 )( 1O ~2 cm/s has a depth of 10mwith an impervious ledge below. A sheet pile wall is driven into this deposit to a depth of7.5 m. The wall extends above the surface of the soil and a 2.5 m depth of water acts onone side. Sketch the flow net and determine the seepage quantity per metre length of thewall.Ans. The flow net is shown.

Number o f flow channels,Number o f equipotential drops,Quantity of s.eepagepermeter}length of wall ..

n,=4nd=14

53


36/43

4=3 x 10-' x 2.5)( - mS/sec/metre run14= 2.148' x 1()-4 mS/sec/ meter run= 214.3 m1JsecJmetre run.

h=2..5m

ImperviousWater flows at the rate of 0.09 mils in an upward direction through a sand sample with acoefficient of permeability of 2.7 )C 10-2 mm/s. The thickness of the sample Is 120 mmand the area of cross-section is 5400 mm2 Taking the saturated unit weight of the sandas 18.9 kN/m\ determine the effective pressure at the middle and bottom of the sample.

Here , q :::0.09mlIS? ~Drnm l/s .k = 2.7 x 10-2mml sA= 5400mm2. 90t= qlkA ::.. Z = 0.6173. 2.7xlO- x5400'f =1u. t - " fw = (18 .9'O -9.81 )kN/ml = 9.09 kN/ml= 9.. 0 9 ) ( 1()-6 N f m . n i 3

F or the bo ttom of the s am p le , i::. 1 20 m m'ij = " I t - i z ' r 1 / J '

or dow nw ard flow ,con s idering the effec t of s eep ag e p res su re ... ij= (9 .0 9 x 1r )(120-.0.6173)( 1 20 )( 9.81 x 1 0 -6) N/mm2

::. 1 20 x 1 0 -' (9.0 9 - 0 .61 73 )( 9.81 ) = 0 .364 )( 1~ N /mm !l: : .3G4N/m.2

For the m idd le o f th e s am ple .1 -= 60 romij= 1 '2 - iz 'Y w::. (9 ..0 9 x 1 0 -')( 6 .0 - .o.(173)t 60 x . 9.81 x 1 0-').N /mm 2= 0 .1 82. x 1 n-J N lmm2::. 182 N /m :l. .-


37/43

( a v % = dV .1= D am v = b U . h ' ) .d x . a y ~ "Z'dt'flow occurs on account ofhydrostatic excess pressure (h = ulyw)'

.he discharge of water collected from a constant head permeameter In a period of 15minutes is 500 mi. The internal diameter of the permeameter is 5 cm and the measureddifference in head betWeen two gauging points 15cm vertically apart Is 40 em, Calculatethe coefficient of permeability .If the dry weight of the 15cm long sample Is 4.86 Nand tl1e ~pecific gravity of the solidsis 2.65, calculate the seepage velocity.

Q = 500 ml ; t = 15 x 60 = 900 s ..J \ = (1t I4) x ~2 = 6.251t cm2 ; L = 15 em ; h = 40 CIn;_QL = 500)( 15 w .V s _k _ At h 6.25 xx x 900 x 40 _ 0.106 mmls

500Superficial velocity v = QIAt = 900 x 6.25x.cmls:: 0.0283 cmfs=0.283 mm/s

Dryweight of sample, = 4.86 NVolume of sample. = A .L = 6.25 x n x 15 em3 ::i: 294.52 em3Drydensity~ 'Yd = 4.86 N/cm3 = 16.5 kN/ma294.52

1 . - G rwd- (1+e)

(1 ) - 2.65X 10. _ 1606' 10kN J 3. + e - . -. , smce 'Y .w". m16.5 .e = 0.606n = (1+e) =0.8778 = 37.73%

r: Seepage velocity, v = vln = 0.283 = 0.750 mmls.:I 0.3773e

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Explain the effect of surface tension and Capillarity on soil.s. Effects of Surface Tension and CapillarityAt the level of the menlscua the surface tension imposes a compressive force onto thesoil grains in contact with the meniscus of magnitude equal to the weight of water in thecapillary column. This effect applies to both a meniscus resulting from capillary rise andfor pore water suspended above a capillary zone. The compressive force imposed onthe soil in contact with the held column of water causes compression or shrinkage ofthe soil.When the ground water drops subsequent to the time of formation of a clay deposit,internal compressive stresses in the clay mass due to the surface tension and capillaryforces make it firm and strong. This is referred to as drying by desiccation and the claysare therefore called "desiccated clays" . Sometimes, such desiccated clays may overliesoft and weak deeper clays. However, the strength and thickness of the desiccated zonemay be such that roads and light buildings could be satisfactorily supported by it. Sincethe intergranular pressure in the capillary zone is increased by capillary pressures, theprocedure for determination of the effective stress when such a zone overlies asaturated soil mass, gets modified as Illustrated below Let a saturated soil mass ofdepth hs be overlain by a capillary zone of height he assumed saturated by capillarity.

Rr--------------------.~CapiBary zone . . jL(saturated) ' , " 1 :

. . . . . . .. . . . . .. . . .. . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . .p ....::: ..: ..:::::: ..: ..:::::: ..::

. . . . . . . . . . . .. . . . . . . . . . . .. . . . . . . . . . . . .. . . . . . . . . . . . .. .. . . . .. . . ,:::: : ..: :SaturatedSOir::: : ... :. . . . . . . . . . . .

capillary zo n e -c om puta tio n o f e ffe c tive s tre s s

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., E ffec tive s tres s a = (1 - u= h Y u t + h , : ' Y 1 J I - h yw=h y' +h ~Y IJI

When h = h~,this b ec ome s:a = h , : " ( + h c Y w = h ~ Y H t ' as earlier.

The su rface ten s ion phenom enon also con tribu tes to th e strength of the so il m as s ins atu ra te d c oa rs e-g ra in ed soi l s . The mois ture w ill b e in the shape ofw edges a t g ra inn ta cts w h ile the c en tra l p ortio n of the void is f il le d With air. T hus , an air-w ater interface ism ed. T he s urfa ce te ns io n in th is men i s cus im po ses a com pres s iv e fo rc e on the s o il g ra inS .the fric tio n b etw e en the grains an d consequen t ly th e shear strength.This strength gain in partially s atu ra ted gran ular soi ls due totension is te rm ed 'A p paren t Cohesion ' (Terzaghi) . This gain can be significant in s ome

tuations . This apparen t cohes ion d is appears on full saturation an d hence canno t a lw ays beie d u p on .

: . E f f e c t i v e s t r e s s ij. =0- U= h Y a t + h ( J w - h y w= h y ' + h e y w

W h e n h = h , t h i s b e c o m e s :e' 0 = h 'V' + h 'II = h 'V t' as earlier.el erw elsa

A t l e v e l P P :T o t a l s t r e s s (J ;; ( h I : + h , ) l a t

u ; : h , " I 1 DG : : (J - U :: h t Y i l t + h . Y llt - h a l l : : h t l M t + h a Y '

N e u t r a l s t r e s sE f f e c t i v e s t r e s sA t l e v e l Q Q :T o t a l s t r e s s (J :: h t .l"tN e u t r a l s t r e s s . u = z e r o .E f f e c t i v e s t r e s s G : : (J - U :: h t . Y i l t .T h i s is b e c a u s e th e c a p i l l a r y p h e n o m e n o n i n t t e a s e S th e e ff e c t iV e o r in t e r g ra n u l a r s t t e s s

b y a m a g n i t u d e e q u a l t o t h e n e g a t i v e pmp r e s s u r e h e . l U I a t t h e t o p o f t h e c a p i l l a r y f r i n g e , th ep o r e p r e s s u r e b e i n g z e r o a t t h e b o t t o m o f th e c a p i l l a r y f r i n g e .

T h i s i s i n t e r e s t i n g b e c a u s e t h e e f f e c t i v e s t r e s s i n c re a s e s f r o m h e." ( to he' r a t a t t h eb o t t om o f t h e e a p i l l a r y z o n e , w h e n t h e s a t u r a t i o n is by c a p i l l a r i t y a n d n o t b y s u b m e r g e n c e .


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AtlevelRR:E f f e c t i v e S t re s s a = c a p il l a r y p r e s s u r e = h eY ID

T h e e f f e c t o f a c a p il l a r y frinp ofheight h e 1 s a n a l o g o u s to that o f a surcharp h e . 1 .p l a c e d o n th e s a t u r a t e d s o i l m a s s .A t depth h b e l o w t h e s u r f a te (h < h ) :E t r e c t i v e stress G = = h y ' + h e ' Y . . ,T h i s m a y b e s h o w I i a s f o l l o w s :T o t a l s U e s s (J = h . ' Y II,N ~ t r a l stress, u = -(Ptessure d u e towe ight o f w a t e r hanging ' b e l o w t h a t l e v e l )

=-(h -h~1 1 1 1 1. Explain the difference between Constant head Permeameter and Falling headPermeameteron experimental basis.s. Constant-HeadPermeameter

A simple set-up of the constant-head permeameteris shownW a te r s u p p l y

L

Rubber Stepper h


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e principle in this set-up is that the hydraulic head causing flow is maintainednstant; the quantity of water flowing through a soil specimen of known cross-ctional area and length in a given time is measured. In highly impervious solis theuantity of water that can be collected will be small and, accurate measurements areifficult to. make. Therefore, the constant head permeameter is mainly application cablerelatively pervious soils, although, theoretically speaking, It can be used for any typesoli.f the length of the speclmen is large, the head lost over a chosen convenient length ofe specimen may be obtained by inserting piezometers at the end of the specifiedength. If Q is the total quantity of water collected in the measuring jar after flowinghrough the soil in an elapsed time t, from Darcy's law,

= Qlt = k.i.A.k= (Qlt).(1IiA) = (Qlt).(UAh) = QLlthAhere

k = Darcy's coefficient of permeabilityL and A = length and area of cross-section of soil specimenh = hydraulic head causing flow.The water should be collected only after a steady state of flow has been established.The constant head permeameter is widely used owing to its simplicity in principle.However, certain modifications will be required in the set-up in order to getreasonabte precision In the case of soils of low permeability.

Falling or Variable Head PermeameterA simple set-up of the falling, or variable head permeameter is shown below.A better set-up in which the top of the standpipe is closed, with manometers andvacuum supply, may also be used to enhance the accuracy of the observations (Lambeand Whitman, 1969). The falling head permeameter is used for relatively less permeablesoils where the discharge is small.

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Stint pipeorburetl:e( C l 'O S !H S e C t io n a l a r e a a )

Po ro us s to n e SemenSoi l sa"",e(Cross-sec t i ona l

,_-_....._-_'...... area A ) \

-------

OverfttM

a ~ a r e a o f c ro s s - s e c t i o n /o f s t a n d p ip eL a n d A ~ l e n g t h and a r e a ~ C r o s s - s e c t i o n o f the s o il s a m p le and t h e o t h e r q u a n t i t i e s a sined. .


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T h i s c a n be d e r i v e d a s f o I lu w a ;L e t - dk be the m a n s e inb e a d in a s m a l l inteMl o f t ime dt. ( N e g a t i v e s ig n i n d i c a t e s

the " h e e d d e c r e a s e s wi th i n c r e a s e in e l a p s e d t i m e ) ,F romDa r~ s law t

Q ; ; ; ( - a.dh~dt ;;; ta.- ad/ddt = = KAh l tt . t . . / ' " dh{ I U ' ~)A 2!-4-dt(kA laL ) .d t ~ -dh lh

l n t e g r a . t i n g both s i d e s and a p p l y i n g the limits t o and t1 f o r t , and h o and hI f o r h ,kA ~.l}a t . . _ ~ d h ; ; ; J I t o d ka L~ J l I o h In hr. ( k A l a L X tl - t o > ~ log,(hlh1) ~ . 2 . 8 1 o g lO ( h o ' h 1)Transpnsingthe t e rm s ,

.

soil mechanics question and answers

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