Soil Mech Ques1 Nw

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Reg. No.

B.E / B.Tech. DEGREE EXAMINATIONS, NOV/DEC 2011

III SEMSTER

CIVIL ENGINEERING

U4CEA12 / Soil Mechanics

Execution PlanSl.No Activities Time in Minutes1. To study the Question Paper and to Choose those to attempt 52. Part A 3 Minutes x 15 Questions 453. Part-B 25 Minutes x 5 Questions 1254. Quick revision & Winding up 5

Total 180

Time: Three Hours Maximum: 100 marksPART - A (15 X 2 Marks = 30 Marks)

Answer all the Questions. Each question carries 2 marks

1. Sketch the soil phase system

2. Define density index

3. Define void ratio and porosity.

4. Differentiate between compaction and consolidation.

5. What is the use of Casagrande apparatus?

6. List out the factors affecting permeability.

7. What is mean by seepage pressure

8. Write the Laplace equation for two dimensional flow

9. Define influence chart.

10. What is mean by shear strength of soil?


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11. What are the methods involved in shearing resistance test?

12. Write the advantages of Triaxial test.

13. Is the drained test called as slow test? Why?

14. Define finite slope

15.Sketch the Culmanns slip plane and force triangle.

Part B (5 x 14 marks = 70 marks)

(Answer all the questions. Each question carries 14 marks)

16.a i)A soil sample has a porosity of 40%. The specific gravity of soilds is 2.70. Calculate a) void ratio

b) dry density c) unit weight if the soil is 50% saturated and d) unit weight the soil is completely

saturated. (10)

a ii) Define plasticity index and shrinkage limit (4)

(or)

16. b i) an undisturbed statured specimen of clay has a volume of 18.9 cm3 .And mass of 30.2 g. on the

oven drying he mass reduces to 18.0g. the volume of dry specimen has determined by displacement

of mercury is 9.9 cm3 . Determine shrinkage limit, specific gravity, shrinkage ratio, and volumetric

shrinkage. (10)

b ii) List out the factors affecting compaction. (4)

17. a) The water table in a deposit of sand 8 m thick is at a depth of 3m below the surface. Above the

water table, the sand is saturated with capillary water. The bulk density of sand is 19.62 kN/m3.


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Calculate the effective pressure at 1m, 3m and 8m below the surface. Hence plot the variation of total

pressure, neutral pressure and effective pressure over the depth of 8m. (14)

(OR)

17. b) In a falling head permeameter test, the initial head (t=0) is 40 cm. The head drops by 5 cm in 10

minutes. Calculate the time required to run the test for the final head to be at 20 cm. If the sample is 6

cm is height and 50 cm2 in cross-sectional area, calculate the coefficient of permeability, taking area

of stand pipe = 0.5 cm2. (14)

18. a. A water tank is supported by a ring foundation having outer diameter of 10m and inner diameter

of 7.5 m. the ring foundation transmits uniform load intensity of 160 KN/m2. Compute the vertical

stress induced at a depth of 4 m , below the centre of ring foundation , using (a) Boussinesq analysis

and ii) Westergaard`s analysis taking =0. (12)

aii) Define contact pressure (2)

(Or)

18.b. Briefly Explain laboratory consolidation test

(14)

19. a. Explain Mohr Coulomb failure theory. (14)

(OR)

19. b. Explain Triaxial compression Test. (14)

20. a. Briefly explain the friction circle method. (14)

(OR)


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20. b. Write short notes on Bishop`s method. (14)

KEY FORU4CEA12 / Soil Mechanics

PART - A (15 X 2 Marks = 30 Marks)

Answer all the Questions. Each question carries 2 marks

1.

2. It is defined as the ratio of difference between the voids ratio of soil in its looest state emax and itsnatural void ratio e to the difference between the voids ratios in the loosest and densest state

ID=emax-e/(emax-emin)3. Void ratio :e=Vv/Vs, porosity: n=Vv/V

4. Compaction : artificially rearranged and packed together into closer state of contact by mechanicalmeans in order to decrese the porosity of the soil and thus increase the dry density.

Consolidation: consequent escape of pore water is termed as consolidation.

5. Apparatus used to find liquid limit.

6. A)Effect of size and shape of particle b) effect of properties of pore fluid c)effect of structuralarrangement of particles and stratification d) effect of degree of saturation and other foregin matter e)effect of adsorbed water.

7. By virtue of the viscous friction exerted on water flowing through soil pores, an energy transfer effectbetween the water and the soil. The force corresponding to the this energy is called seepage pressure.

8. 2/x2 + 2/y2 =0

AIR

WATER

SOLIDS


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9. Vertical stress at any point under a uniformly loaded area of any shape can be determine with thehelp of influence chart or influence diagram.

10. shear strength means resistance to the deformation by continuous shear displacement of soilparticles.

11. Direct shear test , triaxial test , unconfined compression test, vane shear test.12. Stress distribution on the failure plane is uniform, b) precise measurements of the pore

pressure and volume change during the test are possible c) the shear test under all threedrainage conditions can be performed complete control.13. Yes. Drained test consolidated under normal load and then sheared load sufficiently slowly.

so that complete dissipation of pore pressure takes place. so it is called slow test.

14. .If the slope of limited extent is called as finite slope. Example are the inclined faces of arth

dam, embankments an cuts etc


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15. .


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Part B ( 5 x 14 marks = 70 marks)(8)

16. ai). GIVEN:

a) e=n/n-1= (0.40/91-0.4)= 0.667 Bed width b= 6 m

b) d = GW/(1+e)= 2.7 x 9.81)/1+ 0.667)= 15.89 KN/m3C) e = Wg/S= (OR)= W=Es/G=0.667 X0.5)/(2.70)= 0.124

=d(1+W)=15.89 X 1.123= 17.85 KN/m3(or)

=

W/(G+eS)/(1+e)=(9.812 x (2.70+0.667 x 0.5)/(1+0.667)= 17.85 KN/m3

D) when the soil is fully saturated e=Wsat.G

Wsat = e/G=0.667/2.70=0.247

sat= d(1+Wsat)= 15.89 x 1.247=19.81 KN/m3

aii) Plastic index is defined as the numerical diffeernece between the liquid limit and plastic

limit of a soil Ip=WL-WP

shrinkage limit is defined the max .water content at which a reduction n water cotent at whichsoil can still be completely saturated.

(OR)

16.b) i)Given: M1=30.2 G Md=18g w=1 g/ cm (10)V1=18.9 cm3, V2=9.9 cm3

i) Ws=(M1-Md)/Md-(V1-V2)w)/Md= X100=17.8 %

ii)G=Md/9V1-(M1-Md)= 2.69

iii)Shrinkage ratio SR= d/w=d/w= 1.82


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iv) volumetric Shrinkage ratio = VS=( W1-W2) X SR= 91%

bii) factors affect compaction (4)

1.Water content 2. Amount of compaction 3. Method of compaction 4. Type of soil 5.adition of

amixtres

17.a) a) stress at D 8 M BELOW GROUND (14)

If we inserted piezometric tube at D water will rise through a heiht hw=5m in it

=(3+ 5) sat = 8 x 19.62 =156.96 KN/m2

u= hww= 5 x9.81 = 49.05 KN/m2

`= u = 107.91 KN/m2

b) stress at C 3m below ground level

=sat= 3 x 19.62= 58.86 u=0;`=58.86

c) stress at A GROUND LEVEL=Ou=-hcw=-3(9.81)= -29.43 KN/m2

`=29.43 KN/m2

d) stress at B, 1 m below ground level=1sat= 19.62 KN/m2

u=-2w =-2 x 9.81 =-19.62 KN/m2

`= (-u)=39.24 KN/m2

(OR)

17.b) Given: (14)...

In a time interval t= 10 mintues the head drops from intial value of h1=40 to h2= 40-5=35cm

K=2.3 aL/At log10 h1/h2= 2.3 aL/Aklog10 h1/h2 = m log10 h1/h2M=2.3 aL/At constant

10= m log10 40/35m = 172.5 units

t= m log10 h1/h2=172.5log10 h1/h2h1=40 and h2 = 20 cmt= 51.9 minm = 2.3 aL/Ak

k= 1.33 x 10-5 cm/sec

18. ai) Bussinesq analysis (12)Inner radius Ri= 7.5/2= 3.75 m

Outer radius Ro= 10/2= 5mRi/z= 3.75/4= 0.9375


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Ro/z= 5/4=1.25z = 160( 1-(1/(1+(1.25)2) 3/2 )-160 (1-(1/( 1+ (0.9375)2 )3/2= 23.11 KN/m2

B) WESTGUAARD`S ANALYSISa = RO (OR ) Ri for outer and inner radii respectively

R/Z= 3.75/ 4 = 0.9375 Ro=/z= 1.25z =17.57 KN/m2

aii) contact pressure: vertical pressure act at the surface of contact between the base ofa footing and underlying soil mass.

(2)

18.a) Consolidation is the process of gradual transfer ofan applied load from the pore water to the soil structure aspore water is squeezed out of the voids. The amount of waterthat escapes depends on the size of the load andcompressibility of the soil, The rate at which it escapesdepends on the coefficient of permeability, thickness, andcompressibility of the soil. The rate and amount of

consolidation with load are usually determined in thelaboratory by the one-dimensional consol -idation test. In thistest, a laterally confined soil is subjected to successivelyincreased vertical pressure, allowing free drainage from thetop and bottom surfaces.


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T h e various -___ _-_- L __.- _corrosive material. All-elastic or combination plastic and metal consoli-

3 metal narts of the consolidometer shall be of the same nondometersmay also be used to reduce electrochemical effects. Theconsolidometershall conform to the following requirements,(1) Fixed-ring consolidometer shall have a rigid base with arecess for supporting the bottom porous stone and for seating andattachingthe consolidation ring. The upper surface of the recess shall begrooved to permit drainage, The base shall also have (a) an inundationVIII - 2

ring to permit submergence of the specimen in water toprevent evaporationof water from the specimen during thetest, and (b) suitable connectionsand a standpipe for makingpermeability tests,(2) Floating-ring consolidometerjy shall have a rigid basefor supporting the bottom porous stone, The base shall belarge enoughto permit free vertical movement of theconsolidation ring and shall havea chamber surrounding thering for submergence of the specimen.l?* Consolidation ring shall completely and rigidly confine and

support the specimen laterally, The inside diameter of the ringshouldnot be less than 2-3/4 in. and preferably not less than 4in.; use of largerrings for specimens of larger diameter,particularly with the fixed-ringconsolidometer, will reduce thepercentage of applied load carried byside friction andconsequently will provide .more accurate results.NNoqrrmmaallllyy, , ththee rraatitoio oof f tthhee hheeiigghhttooff rriinngg ttoo iinnssiiddee diiameetteerr of ring sshoulldbe between 1/4 and l/6. The consolidation ring may be linedwith a ma

terial ssuucchh aass TTeefflloonn ttoo rreedduuccee tthheeffrricicttioionn bbeettwweeeenn, tthhee rriinngg aanndd aassppeecci-imeaof fine-grained soil. A stainless steel ring ispreferable for specimenscontaining abrasive particles.


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c. Porous stones more pervious than the specimen of soilshouldbe used to permit effec ive drainage. For routine testing,stones ofmedium porosity are satisfactory. The diarrie the porous

stonest In the floating-ring consolidometer the friction between theinside of thering and the specimen is less than that in thefixed-ring type. oweverwhen very soft soils are te with thefloating-ring conthe side friction will not su ort the weight ofthe ring, aoccurs towar the middle o he specimen from top andbefloating-ring device is suit y for comparatively stihas thedisadvantage that it c be used for permshall be such as to prevent the squeezing out of soil throughthe clearancespaces between the ring and stone and to permit

free compressionof the specimen without binding; tominimizethepossibility of binding,the sides of the upper porous stoneofthe fixed-ring consolidometershould be slightly tapered away_-- - - -

from the specimen, while both porousstones of the floating-ring-2O-CAGE BRASS

consolidometer should be tapered.A clearance of about 0.010to 0.015PLAN in. around the stone generally willbe adequate; however, ifvery soft

soils are tested, a smaller clear-SECTION A-Aance may be desirable or retainer-

Figure 3, rataiqer rings may be used as shown inFigure 1. Details of a typical reerring are shown in Figure 3,The porous stones should 1)e cleaned r every test, preferablyin anultrasonic cleaner or by boiling and flushing.d. Loading devicesof various types may be used to apply load toI The most

commonly used is the beam-and-weight mechanism.Theloading device should be capable of transmitting axial loadtothe specimen quickly and gently, Also, the equipment should


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be capable ofmaintaining the load constant for at least 24 hr.The equipment should becalibrated to ensure that the loads inr4inal.p.dg~rethnse actuallv- -

(or) 18.b) It seems reasonable to assume that a sliding failure of a soil will occur if on acertain plane the shear stress is too large, compared to the normal..... stress. On other planesthe shear stress is sufficiently small compared to the normal stress to prevent sliding failure.It may be illustrative to compare the analogous situation of a rigid block on a slope, seeFigure 20.2. Equilibrium of forces shows that the shear force in the plane of the slope is T =W sin _ and that the normal force acting on the slopeis N = W cos _, where W is the weight of the block. The ratio of shear force to normal force is

T/N = tan _. As long as this is smaller than a certain critical value, the friction coefficient f,the block will remain in equilibrium. However, if the slope angle _ becomes so large that tan

_ = f, the block will slide down the slope. On steeper slopes the block can never be inequilibrium.In 1776 Charles-Augustin de Coulomb, a French scientist who also made importantcontributions to the theory of electricity, used the analogy with a sliding block load to

propose that the maximum possible shear stress _fin a soil body is _f= c + _0 tan _. Here _0is the normal (effective) stress on the plane considered. The quantity c is the cohesion, and_ is the angle of internal friction or the friction angle. An elementary interpretation is that ifthe shear stress on a certain plane is smaller than the critical value _f, then thedeformations will be limited, but if the shear stresses on any single plane reaches the criticalvalue, then the shear deformations are unlimited,indicating shear failure. The cohesion cindicates that even when the normal stress is zero, a certain shear stress is necessary toproduce shear failure. In the case of two rough surfaces sliding over each other (e.g. twoblocks of wood), this may be due to small irregularities in the surface. In the case of twovery smooth surfaces molecular attractions may play a role.For soils the formula (20.1) should be expressed in terms of effective stresses, as thestresses acting from one soil particle on another determine the eventual sliding. For thisreason the soil properties are often denoted as c0 and _0, in order to stress that these

quantities refer to effective stresses.Coulomb failure criterion this failure criterion has also been indicated, in the form of twostraight lines, making an angle _ with the horizontal axis. Their intersections with the verticalaxis is at distances c. In order to underline that failure of a soil is determined by theeffective stresses, the stresses in this figure have been indicated as _0. There are twoplanes, defined by the points C and D in Fig. in which the stress state is critical. On all otherplanes the shear stress remains below the critical value. Thus it can be conjectured thatfailure will start to occur whenever Mohrs circle just touches the Coulomb envelope.

This is called the Mohr-Coulomb failure criterion. If the stress circle is completely within theenvelope no failure will occur, because on all planes the shear stress remains well below thecritical value, as given by equation (20.1). Circles partly outside the envelope areimpossible, as the shear stress on some planes would be larger than the critical value.When the circle just touches the envelope there are two planes, making angles _/4 _/2

with the direction of the major principal stress, on which the stresses arecritical. Sliding failure may occur on these planes. It can be expected that the soil may slidein the directions of these two critical planes. In the case represented by the figures in thischapter, in which it is assumed that the vertical direction is the direction of the majorprincipal stress, see Figure 20.3, the planes on which the stresses are most critical make anangle _/4 _/2 with the vertical direction. Thus it can be expected that sliding failure will


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occur in planes that are somewhat steeper than 45_. If for instance _ = 30_, which is anormal value for sands, failure will occur by sliding along planes that make an angle of 30_with the vertical direction.. . . . . . .. . . . . . .. . . . . . . . .. . . . . . .

19.a) Boussinesqs Method:

When a point load Q acting on the surface of a semi infinite solid, a vertical

stress z produces at any point in addition to lateral and shear stress.

Assumptions of Boussinesq theory:

a. For soil, the soil mass is elastic, isotropic, homogeneous and semi-

infinite.

b. The soil is weightless.

c. For load, the load is vertical, concentrated acting on the surface.


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d. Hooks Low Applied, it is mean that the constant ratio between stress

and strain.

Boussinesqs Formula:

Boussinesqs solved the problem of stresses

produced at any point (A) due to point load Q.

At point (A)

z=

z =

= R5 = (r2 + Z2)5/2

=

=

=


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=

z = I

p= Ip

Where:

z : Vertical stress at point A as shown in fig.(5)

Z : Vertical dimension for point A at load

Ip : Influence factor depend on ( z

r

) = F( z

r

)

Q : Point load

The variation of Ip for various value of z

r

is gi

(Or)

19.b) TRIAXIAL TEST:The failure of a soil sample under shear could perhaps best beinvestigated in a laboratory test in which the sample is subjected to pure distorsion, atconstant volume. The volume could be kept constant by taking care that the isotropic stress

_0 = 1 3 (_1 + _2 + _3) remains constant during the test, or, better still, by using a test setup inwhich the volume change can be measured and controlled very accurately, so that thevolume change can be zero. In principle such a test is possible, but it is much simpler to

perform a test in which the lateral stress is kept constant, the triaxial test, see Figure. Inorder to avoid the complications caused by pore pressure generation, it will first be assumedthat the soil is dry sand. The influence of pore water pressures will be considered later. Inthe triaxial test a cylindrical soil sample is placed in a glass or plastic cell, with the samplebeing enclosed in a rubber membrane. The membrane is connected to circular plates at thetop and the bottom of the sample,with two o-rings ensuring a water tight connection. The


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cell is filled with water, with the pressure in the water (the cell pressure) being controlled bya pressure unit, usually by a connection to a tank in which the pressure can be controlled.Because the sample is completely surrounded by water, at its cylindrical surface and at thetop, a pressure equal to the cell pressure is generated in the sample. The usual, andsimplest, test procedure is to keep the cell pressure constant during the test. In addition tothe lateral (and vertical) loading by the cell pressure, the sample can also be loaded by a

vertical force, by means of a steel rod that passes through the top cap of the cell. The usualprocedure is that in the second stage of the test the rod is being pushed down, at a constantrate, by an electric motor. This means that the vertical deformation rate is constant, andthat the force on the sample gradually increases. The force can be measured using a straingauge or a compression ring, and the vertical movement During the test the verticaldisplacement of the top of the sample increases gradually as a function of time, because themotor drives the steel rod at a very small constant velocity downwards. The vertical force onthe sample will also gradually increase, so that the difference of the vertical stress and thehorizontal stressgradually increases, but after some time this reaches a maximum, andremains constant afterwards, or shows some small additional increase, or decreasessomewhat. The maximum of the vertical force indicates thatthe sample starts to fail. Usuallythe test is continued up to a level where it is quite clear that the sample has failed, byrecording large deformations, up to 5 % or 10 %. This can often be observed in the shape ofthe sample too, with the occurrence of some distinct sliding planes. It may also be, however,that the deformation of the sample remains practically uniform, with a considerableshortening and at the same time a lateral extension of the sample. In the interior of thesample many sliding planes may have formed, but these may not be observed at itssurface.The test is called the triaxial test because stresses are imposed in three directions.

This can be accomplished in many different ways, however, and there even exist tests inwhich the stresses applied in three orthogonal directions onto a cubical soil sample enclosedin a rubber membrane) can all be different, the true triaxial test. This gives many morepossibilities, but it is a much more complex apparatus, and the testing procedures are morecomplex as well.In the normal triaxial test the sample is of cylindrical shape, and the twohorizontal stresses are identical. The usua diameter of the sample is 3.8 cm (or 1.5 inch, asthe test was developed in England), but there also exist triaxial cells in which larger sizesamples can be tested. For tests on gravel a diameter of 3.8 cm seems to be insufficient toachieve a uniform state of stress. For clay and sand it is sufficient to guarantee that in every


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cross section there is a sufficient number of particles for the ress to be well defined.

(14)


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(OR) 20. b) i)


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Soil Mech Ques1 Nw