M.Tech(Geo-Tech Engg), UNIT-07 CONSOLIDATION OF SOILS€¦ · Compiled by: Prof.B.S.Chawhan...

Compiled by: Prof.B.S.Chawhan M.Tech(Geo-Tech Engg), Asst.Professor,CED,Government.Engineering College,Haveri-581110(12/4/2011-Till date) UNIT-07

CONSOLIDATION OF SOILS Definition: Consolidation may be defined as the gradual expulsion of pore water from voids is defined as consolidation. or 1. It is a line-dependent phenomenon. 2. Soil is completely saturated. 3. Volume reduction is due to expulsion of pore water from voids. 4. It occurs on account of load placed on the soil is called consolidation..

Δh W W V1 S S V2 (a) (b)

Fig.1. Before consolidation After consolidation Terzaghi’s One dimensional consolidation theory: Assumptions:

The following are the assumptions of one dimensional consolidation theory; a.The soil is homogeneous and fully saturated. b.Soil particles and water are incompressible. c.Darcy’s law for the velocity of flow of water through soil is perfectly valid. d.Coefficient of permeability,k is constant during the process. e.Soil is laterally confined so that the compression is one dimensional. f.Excess pore water drains out only in a vertical direction. g.Linear relationship between effective pressure and void ratio exist are constant for every stage of consolidation. h.The time log of consolidation is due entirely to the low permeability of soil, and thus the secondary consolidation is disregarded.

The following figure shows one dimensional consolidation theory; Sand Drainage face Δσ total vertical stress A Ui=ΔσI

z d dx dy H/2 t=0t4 t4 t3 t2 t1

Clay σIz t=00

Uz t=00 H

Consolidated layer H/2 Sand b c Fig.a. Section of clay layer Fig.b. Excess pore pressure distribution.

σIz=Vertical effective pressure at depth Z.

Uz=Pore pressure at depth Z. Fig.2.

A layer of clay is shown sandwiched between two sand strata. The clay layer is subjected to an increase in total vertical stress Δσ distributed uniformly a semi-infinite area. At the instant of loading, that is,t=0, the excess pore pressure will increase uniformly by ui (ui= Δσ) over the thickness H of the layer as shown by abcd in figure(b). After a time t has elapsed, because of the drainage into the sand layers above and below, the distribution of excess pore water pressure will have changed to the shape shown by the unshaded portion in figure(b). H is the maximum distance that water has to travel to reach a drainage face; that is, the length of the longest drainage path. If there are two drainage surfaces, one of the top and another at bottom of the consolidation layer, it is called double drainage layer and

Compiled by: Prof.B.S.Chawhan M.Tech(Geo-Tech Engg), Asst.Professor,CED,Government.Engineering College,Haveri-581110(12/4/2011-Till date) H will then be equal to half of the thickness of the clay layer(H=H/2). If there is only one drainage surface, it is called single drainage layer and H will be equal to the thickness of clay layer (i.e. H=H). In the Terzaghi’s solution, he consider the three factors i.e., a. Drainage path(H) b. Time factor(Tv) c. Coefficient of Consolidation(Cv). i.e.,

Time factor 2HtC

T vv = (1)

wv

v mkCγ

= (2)

( )

wvv a

ekCγ*

1+= (3)

Where, Tv= time factor. t= time in seconds. H=Thickness of permeability in cm/sec or m/sec. k=Co-efficient of volume change in cm2/sec or m2/sec. mv= Co-efficient of volume change in cm2/gm or m2/KN. Cv= Co-efficient of consolidation in cm2/sec or m2.sec γw= Unit weight of water in gm/cm3 or KN/m3. av= Co-efficient of compressibility. e=void ratio.

)0

0

(*)1( oef

fv ppe

eem

−+

−= (4).

Normally Consolidated, Under Consolidated and Over Consolidated Soils: a. Normally Consolidated Soils: A soil is said to be normally loaded if the present effective overburden pressure Po is the maximum pressure to which the layer has ever been subjected at any time in its history is called normally consolidated soils. i.e. Pc=Po where, Po is called Overburden pressure Po=γz in KN/m2 Pc=Preconsolidated pressure=Pc=load/area in KN/m2.

Fig.3. Normally consolidated soils.

(Total load of building)

Pc=structural load

G L G γ= density of soil Z γ b

Compiled by: Prof.B.S.Chawhan M.Tech(Geo-Tech Engg), Asst.Professor,CED,Government.Engineering College,Haveri-581110(12/4/2011-Till date) b. Over Consolidated soils: A soil layer is said to be precompressed (over consolidated) if the layer was subjected at one time in its history to a greater effective over burden pressure is called over consolidated pressure.

Fig.4. Over Consolidated soils: i.e. Pc>Po where, Po=γZ in KN/m2. Pc=Total design load of structure in KN/m2. c. Under Consolidated Soils: A soil layer is said to be compressed (under consolidated) if the layer was subjected at one time in its history to a lesser effective over burden pressure. i.e. Pc<Po Pc loose material placed or water logged area

i.e. Pc=Total structural load Po=Overburden Pressure=γZ

Fig.5. Under Consolidated Soils. d. Over Consolidated ratio (OCR): Over consolidation ratio may be defined as it is the ratio of preconsolidation pressure to the over burden pressure is called Over Consolidation ratio. i.e. OCR=Pc/Po Therefore for a normally consolidated soil, OCR=1 and for a preconsolidated or over consolidated soil OCR>1. If the OCR<1 then it is said to be under consolidated soils. Preconsolidation pressure and its determination by Casagrande’s method and Log-Log method: Preconsolidation Pressure, Pc: Preconsolidation pressure may be defined as, it is the total structural load per unit area due to void ratio and effective over burden pressure is called preconsolidation pressure ,Pc. The following graph shows pressure-void ratio curve.

(Total load of building) Pc=structural load G L G L Z=depth of foundation Clay layer or depth of foundation. γ= density of soil

G L G L Z γ

Compiled by: Prof.B.S.Chawhan M.Tech(Geo-Tech Engg), Asst.Professor,CED,Government.Engineering College,Haveri-581110(12/4/2011-Till date)

Y

Virgin Compression Curve

Void ratio(e)

Recompression or shrinkage Curve c B D Clay Expansion or Virgin Curve swelling Curve E X

0 1 2 3 4 5 6 7 8 σ' (Applied effective pressure) KN/m2

Fig.6. Pressure-void ratio Curve. Procedure: 1. Now increased the load from A to B then void ratio decreased, then the portion AB called Vergin Compression Curve. 2. Then now decreased or remove the load after certain time then the pressure decreased from point B to C and increased the void ratio from point B to C then the point BC is called Expansion Curve. 3. Once again increased the pressure C to D point at a certain load and decreased the void ratio from point C to D is called Recompression Curve. 4.Now then goon increased the vertical load continuously upto point E then the void ratio will decreased continuously upto till end, point E is called Vergin Curve. Determination of preconsolidation Pressure,Pc: by a. Casagrande’s method. b. Log-Log method. a. Casagrande’s method: Y α

Tangent line AC B Horizontal line AB

A E α/2 α/2 D Bisection line AD This AD line is required for c locate the pc' by backward extension

Void

ratio

(e)

pc'

X

logp' Fig.7. Pressure-Void ratio Curve.

Compiled by: Prof.B.S.Chawhan M.Tech(Geo-Tech Engg), Asst.Professor,CED,Government.Engineering College,Haveri-581110(12/4/2011-Till date) Procedure: 1. The point A of maximum curvature is selected by inspection on the curved portion.

2. Draw a horizontal line AB and the line AC tangent to the curve are drawn through the point A. 3. The angle BAC is then bisected by the line AD. 4. The point of intersection E of the bisector AD. 5. Now draw the backward extension of the lower straight portion of the curve represents the probable preconsolidation pressure Pc.

b. Log-Log method or Burmister method: Y

A

5 F H

1 1 C 3 B 2 O 1 D

Voi

d ra

tio(e

)

6

4

k E logp' X

Fig.8. e-logP curve by Burmister method. Procedure:

1. As soon as the straight line portion of the e-logP curve is approached, the sample is rebounded and reloaded. 2. Now increased the pressure from A to B. 3. Then unloaded from B to C referred as rebound curve. 4. Now reloaded from C to D referred as reloaded curve. 5. Once again continue the load increased then the void ratio decreased. 6. Now draw the straight line from ED through F then BOD becomes a triangle hatched portion. 7. Then shifted the hatched portion to the above line BF. 8. Now draw the vertical line, point H to point downward then the point k is called Pc (Preconsolidation Pressure).

Laboratory One Dimensional Consolidation Test: The test results are used for estimating the rate of settlement and amount of settlement of structure. In addition, the effect of saturation on consolidation, the permeability of the loaded specimen and the amount of swelling when unloaded, may also be determine in this test. A thinner specimen requires a short time for consolidation and its side friction is also less.

While the pressure-voids ratio curve is essentially independent of specimen size, the coefficient of consolidation is found to increase with size. Therefore, some uniformity in the size of the specimens is desirable.

The test specimen is allowed to consolidate under a number of successive increments of vertical pressure, each pressure increment being maintained constant until the compression vertically ceases, generally 24hours.

The vertical compression of the specimen is measured by means of a dial gauge.i.e.,

Applied Pressure,kg/cm2 or KN/m2

Time in minutes Final dial gauge in mm 0.25 or 25 0.25 0.50 or 50 0.50 1. or 100 1.00 2.0 or 200 2.00 4.0 or 400 4.00 8.0 or 800 8.00

16.0 or 1600 15.00 32.0 or 3200 30.00

BC-Rebound Curve CD-Reloaded Curve

Original triangle hatched

Compiled by: Prof.B.S.Chawhan M.Tech(Geo-Tech Engg), Asst.Professor,CED,Government.Engineering College,Haveri-581110(12/4/2011-Till date) 64.0 or 6400 60.00

128.0 or 12800 120.00 256.0 or 25600 240.00 512.0 or 51200 480.00

1024.0 or 102400 1440.00

After the completion of consolidation under the desired maximum vertical pressure, the specimen is unloaded and allowed to swell. After the completion of swelling, the final dial readings are taken and the specimen is taken out the consolidometer and dried to determine its water content. Pressure Void ratio Curves for determination of void ratio: The pressure-void ratio curve can be obtained if the void ratio of the sample at the end of each increment of load is determined. The following parameters are required to determine void ratio (e); a. Cross-sectional area of sample A. b. The specific gravity, G of the solids. c. The dry weight, Ws of the soil sample. d. The sample thickness h, at any stage of the test. Let Vs= Volume of the solids in the sample. where, Vs= Ws/Gγw or Vs=hsa or hs=Vs/a where, hs= thickness of solid matter then e is the void ratio of the sample, then;

h

hhe

hhh

AhAhAh

e s

s

s

s

s −=∴

−=

−= (1)

where, h=thickness of sample Determination of Consolidation Characteristics of Soils: a. Compression Index, Cc: Compression index Cc is a dimensionless quantity which represents the slope of the linear portion of the pressure-void ratio curve on a semi-log plot.

By referring the e-logP curve; Y eo Cc=compression index ef Peo Pef KN/sqm

Fig.9. e-logP curve

Compression Index of

foc pp

eeC

loglog −

−= (1)

And according to Terzaghi’s and Peck formula; Compression Index )10(009.0 −= Lc WC (2) b. Co-efficient of Consolidation ,Cv;

Co-efficient of Consolidation is applied to indicate the combined effects of permeability and compressibility of a soil on the rate of volume change is referred as co-efficient of consolidation, Cv,

i.e., ( )

wv

o

wvv a

ekor

mkC

γγ+

=1

in m2/sec. (3)

c. Co-efficient of Volume Change, mv: The change in volume of a soil per unit of initial volume due to a given increase in pressure is called co-efficient of volume change.


i.e., ( )( )ofo

fov ppe

eem

−+

−=

1 in m2/KN. (4)

d. Co-efficient of Compressibility, av: It is the decrease in void ratio per unit increase in pressure is called coefficient of compressibility av and this is given by the following relation;

of

fov pp

eea

−

−= (5)

e. Degree of Consolidation, U: The degree of consolidation is depends on the following factors i.e. a. Number of drainage faces

b. Thickness of clay layer (d) c. Coefficient of permeability (k) d. Coefficient of compressibility (av) e. Magnitude of the consolidating pressure.

Therefore the degree of consolidation may be expressed by the following empirical expressions; when, U<60%;

2

1004⎟⎠⎞

⎜⎝⎛=

UTvπ

(6)

when, U>60% Tv=1.781-0.933log10(100-U%) (7) Determination of Co-efficient of Consolidation by a. Square root of time fitting method. b. Logarithmic time fitting method and . c. Rectangular hyperbola method. a. By square root of time fitting method:

X √t (minutes) Fig. 10. Time Consolidation Curve (Square root of Time fitting method). Procedure:

1. Now plot the t V/s Dial gauge readings in mm and locate the points i.e, x1, x2,x3, x4,……+etc due to pressure increment.

Y (Ro to Rc called initial consolidation) 10 9 Ro 8 Rc

(R0=Initial dial gauge reading at pressure increment, i.e. t=0 or U=0%)

7 6 Rc called corrected zero reading 5 X1 4 X1' X1'=1.15x1 along X-axis 3 Dial gauge2 X2 Reading 1 X2'=1.15X2 along x-axis (10-2 mm) X3 X3'=1.15X3 along x-axis X4 X4'=1.15X4 along X-axis R90, U=90% C 0 Rf

Compiled by: Prof.B.S.Chawhan M.Tech(Geo-Tech Engg), Asst.Professor,CED,Government.Engineering College,Haveri-581110(12/4/2011-Till date) 2. Now locate the Ro point on Y-axis @ initial dial reading. 3. Then draw a line passing through all these points. 4. Now then locate x1'= 1.15x1, x2'= 1.15x2, x3', x4'……….+ xn' along with respect to X-axis. 5. Then now draw a straight line along these points and coincide on a curve and locate the point Re at Y- axis. 6. Now from this joined point C draw a horizontal line parallel to x-axis and draw a vertical down ward line at perpendicular to the x-axis. 7. This joined point d is called time t taken at 90% consolidation R90 on the Y-axis. 8. Rf is called approximate 100% consolidation.

For 90% consolidation, the coefficient of consolidation is given by ;

( )

90

290

tdT

C vv = (1)

where, Tv=Time factor for 90% consolidation (U=90%) from table (10.1) as per Alamsingh and Chawdhary page 26. Tv=0.848 d=Average drainage path or drainage path

90

2848.0t

dCv =∴ (2)

For 50% consolidation then Tv=0.197

50

2197.0t

dCv =∴ (3)

Or

vCdt

2197.0=∴ (4)

Where, 24

HHiorHH

d fi Δ−+= for double drainage face for t90

24HHior

HHd fi Δ−+= for double drainage face for t50 and ΔH=compression of sample at t50

Note:-1. Always you locate the dial gauge readings from top to bottom on Y-axis. 2. Always locate the pressure increment from origin O. 3. These square root time fitting method and logarithmic time fitting methods are called Curve fitting methods. b. By Logarithmic of Time fitting method: DGR in mm

Logarithmic of time fitting (logt-min)

Fig.11, Logarithmic of time fitting method Curve Procedure:

1. Two straight portions of the curve intersect at 100% U; the corresponding dial gauge reading designated as R100. 2. A time t1, say 1min, is located on X-axis and locate the point (x) on the early origin curve.

A t50, U=50%

t1/4 t1Rf

BR=100,U=100%

R=50,U=50%

Z

Z Rc, U(%)

Compiled by: Prof.B.S.Chawhan M.Tech(Geo-Tech Engg), Asst.Professor,CED,Government.Engineering College,Haveri-581110(12/4/2011-Till date) 3. A second point correspond to t1/4 is select and locate the point(x) on the curve. 4. A horizontal line at a height equal to the vertical distance between the second point (z=t1-t1/4).

Now the ordinate of this horizontal line is the corrected zero reading Rc. 5. The consolidation from Rc to R100 is called the primary consolidation. 6. From R100 to Rf= is called secondary consolidation 7. Now locate 50% consolidation, U=50% in between Rc to R100 i.e. R50 is on the curve and t50 on the X- axis.

Therefore, the co-efficient of consolidation is calculated by the equation;

( )

50

250

tdT

C vv = or

10

2197.0t

d= (1)

C. By Rectangular hyperbola Method: Y C m b d D X O Time, t Fig.12. t v/s (t/ΔH) Curve. In this rectangular hyperbola method, the following procedure is recommended for the determination of Cv.

1. Obtain the curve t and the specimen deformation (ΔH) from the laboratory consolidation test. 2. Plot the graph of t/ ΔH against t as shown in figure above. 3. Identify the straight-lime portion bc and project it back to point d. Determine the intercept D. 4. Determine the slope m of the line bc. 5. Calculate, Cv as

Cv=0.3(mh2dr)/D----------------- (1)

Note that because the unit of D is time/length and the unit of m is (time/length)/time=1/lemgth, the unit of Cv is (1/length)(length)2=(length)2/time (time/length) wher,hdr is drainage path in cm or m, m is slope obtained from graph. If the drainage on bothe side then hdr

becomes 0.5times thickness of drainage path and if the drainage on one side then hdr become thickness of drainage path.

The rectangular hyperbola method is fairly simple to use, and it gives good results for U=60% to 90%. Determination of settlement: The probable settlement is given by;

o

o

o

ct p

ppe

HCS

Δ++

= 10log*1

(2)

where, St= Total settlement in cm or mm H=Thickness of layer in cm or mm eo=Initial void ratio. po=initial pressure in kg/cm2 or KN/m2 Δp=Excess pressure =pf-po in kg/cm2

Cc=compression Index This above formula is given on the basis of e-logp curve. Jan/2005 and Dec.08/Jan.09 1. In a consolidation test the void ratio of soil sample decreases from 1.20 to 1.10 when the pressure is increased from 160 to 320KN/m2. Calculate the co-efficient of consolidation if the co-efficient of permeability is 8X10-7mm/sec.

t/ΔH

Compiled by: Prof.B.S.Chawhan M.Tech(Geo-Tech Engg), Asst.Professor,CED,Government.Engineering College,Haveri-581110(12/4/2011-Till date) Solution: Given data; a.eo=1.20, b.ef=1.10,c.Peo=16kN/m2, d. Pef=320kN/m2, e.k=8*10-7mm/sec=8*10-10m/sec.

Y eo Cc ef Peo Pef

Fig: 13. e-logP curve To find: i. Cv=?. Procedure: a.We have, Coefficient of volume compressibility, mv;

( ) ( ) ( ) ( ) kNmppe

eem

eoefo

fv /10*84.2

160320*2.111.12.1

*1240 −=

−+−

=−+

−= .

b. We have, Co-efficient of consolidation, Cv;

sec/10*87.281.9*10*84.2

10*8*

274

10

mm

kCwv

v−

−

−

===γ

.

July/2005 and Dec.08/Jan.09 2. In a consolidation test voids ration decreases from 0.7o to 0.65, when the load was changed from 50KN/m2 to 100KN/m2. Compute compression index and coefficient of volume change. Solution: Given data; a.eo=0.70, b. ef=0.65,c. Peo=50KN/m2, d.Pef=100KN/m2.

Fig.14. e-logP curve To find: i) Cv=?. Procedure: a. We have, Coefficient of volume compressibility,mv;

( ) ( ) ( ) ( ) KNmppe

eem

eoefo

fv /10*8824.5

50100*7.0165.070.0

*1240 −=

−+−

=−+

−=

b.We have, Co-efficient of consolidation(Cv):

Y eo Cc ef X Peo pef


116.050log10100log

65.07.01log10

0 =−

−=

−

−=

efef

fc plodp

eeC

Jan/2004 3. Saturated soil of 5m thick lies above an impervious stratum below a pervious stratum. It has compression index 0.25 and k is 3.2*10-10m/sec. Its void ratio at a stress of 147KN/m2 is 1.9, Calculate; i. The change in void ratio due to increase of stress to 190KN/m2. ii. Coefficient of volume compressibility,iii. Co-efficient of consolidation, iv. Time required for 50% consolidation. Solution: Given data a. H=5m thick=d, b. Cc= 0.25, c. k=3.2*10-10m/sec, d..eo=1.9, e. po=147KN/m2 and f. pf = 196KN/m2. Impervious Cc Saturated soil eo 5m ef Peo Pef Pervious

Fig.15. e-logP curve To find: a.Δe=?, b.ef=?, c.mv=?. d.Cv=?, e.t50=?, i.e.U=50%. Procedure:

1. of

foc pp

eeC

loglog −

−=

868.1147log196log

9.125.0 =∴

−

−= f

f ee

2. Change in void ratio )( eΛ ;

031.0868.19.1 =−=−=Δ feeoe 3. Co-efficient of volume compressibility ,mv;

( )( )ofo

fov ppe

eem

−+

−=

1= ( ) ( ) KNm /10*25.2

147196*9.11868.19.1 24−=−+

−

4. Coefficient of consolidation Cv.

sec/10*448.181.9*10*25.2

10*2.3 274

10

mm

kCwv

v−

−

−

===γ

4. Time taken t;

v

v

CdT

t2

=∴ = sec93.3400017510*448.1

5*197.0197.07

22

50 ===∴ −vCdt or 393.52days.

July/2004 4. A saturated soil stratum of 5m thick lies above an impervious stratum. It has a compression index of 0.25 and a Coefficient of permeability of 3.2*10-3mm/sec. It has a void ratio of 1.9 at normal stress of 0.15N/mm2. Compute; i. The void ratio due to increase of stress to 0.2N/mm2. ii. Settlement of soil stratum due to the above increase in stress. Solution: Given data

a. H=5m=5000mm=d, b. Cc= 0.25, c. k=3.2*10-3mm/sec, d.eo=1.9, e. po=0.15N/mm2 and f. pf = 0.2N/mm2.

b. Draw the diagram of e-logp curve

Compiled by: Prof.B.S.Chawhan M.Tech(Geo-Tech Engg), Asst.Professor,CED,Government.Engineering College,Haveri-581110(12/4/2011-Till date) c.

Impervious eo Cc Saturated soil 5m ef Peo Pef Pervious

Fig.16. e-logP curve To find: a.Δe=?, b.ef=?, c.mv=?. d.Cv=?, e.t50=?, i.e.U=50%. Procedure:

1. of

foc pp

eeC

loglog −

−=

868.115.0log2.0log

9.125.0 =∴

−

−= f

f ee

2. Change in void ratio )( eΛ ;

0312.0868.19.1 =−=−=Δ feeoe 3. Co-efficient of volume compressibility (mv);

( )( )ofo

fov ppe

eem

−+

−=

1= ( ) ( ) Nmm /22.0

15.02.0*9.11868.19.1 2=−+

−

4. Settlement of soil stratum due to the above increase in stress is given by; ( ) mmpHmS vt 172.5515.02.0*5000*22.0** =−=Δ=

July/2005 5.In a consolidation test voids ratio decreased from 0.7 to 0.65 hen the load was changed from 50KN/m2 to 100KN/m2.

Compute compression index and co-efficient of volume change. Solution: Given data a. eo=0.7, b. ef=0.65, c. po=50KN/m2, d.pf=100KN/m2. Cc eo ef Peo Pef

Fig.17. e-logP curve To find: a.Cc=?, b.mv=? Procedure:


1. of

foc pp

eeC

loglog −

−=

166.050log100log

65.070.0=

−−

=cC

2. Co-efficient of volume compressibility (mv);

( )( )ofo

fov ppe

eem

−+

−=

1= ( ) ( ) KNm /10*882.5

50100*7.0165.070.0 24−=−+

−

July/2004 6. A stratum of clay 8 m deep, has WL=45%. The surface of clay is at 10m below the present ground level, w=40% and Gs=2.78 for clay. Between ground surface and clay, the subsoil consists of fine grained sand. The ground water level is 4.5m below ground level. The average submerged unit weight of sand is 10.4KN/m3 and the unit weight of sand above the water table is 17KN/m3. The clay is normally consolidated. The weight of structure coming on top of the sand above the clay increases the overburden pressure on clay by 40KN/m2. Calculate the settlement of the building. Solution; Given data a. WL=45%, b. H=8m, c.w=40%, d. Gs=2.78, e. depth of W.T=4.5m from GL, f.γ' sand=10.4KN/m3,g).γd sand=17KN/m3, h. The clay is normally consolidated, Sr=1, i. Δp=40KN/m2. GL G

4.5m γd Sand W.T 10m 5.5m γ' Sand 8m Clay layer

Fig.18. Soil Starata. To find: a. St=? Procedure: 1. We know that the total settlement is given by the following equation;

⎥⎦

⎤⎢⎣

⎡ Δ++

=o

o

o

ct p

ppe

HCS 10log

1 (1)

2. According to Terzaghi’s; ( ) ( ) 315.01045009.010009.0 =−=−= Lc WC

3. Initial void ratio is given by the equation below;

112.11

78.2*4.0===

ro S

wGe …….(Sr=1 for normally consolidated).

4. Overburden pressure Po is given by; po=4.5*17+5.5*10.4+8/2*(γsat-γw)

where, γsat= 81.9*112.11112.178.21 ⎥

⎦

⎤⎢⎣

⎡+

+=⎥

⎦

⎤⎢⎣

⎡++

wo

o

eeG

γ =18.077KN/m3.

po=133.7+4*(18.077-9.81)=166.768KN/m2.

∴ ⎥⎦

⎤⎢⎣

⎡ Δ++

=o

o

o

ct p

ppe

HCS 10log

1= m114.0

768.16640768.166log

112.11315.0*8

10 =⎥⎦⎤

⎢⎣⎡ +

+ or 11.14cm

Jan/2007 7. 20mm thick undisturbed sample of saturated clay is tested in laboratory with drainage allowed through top and bottom. Sample reaches 50% consolidation in 35minutes. If clay layer from which sample was obtained is 3.0m thick and is free to drain through top and bottom surfaces, calculate the time required for degree of consolidation in the field. What is the time required if the drainage in the field is only through the top? Procedure: Given data 1st Case: a. 20mm thick sample, b. U=50%, c. t=35minutes, d. H=3m or 300cm thick. To find: a. Cv=? 2nd Case:

Compiled by: Prof.B.S.Chawhan M.Tech(Geo-Tech Engg), Asst.Professor,CED,Government.Engineering College,Haveri-581110(12/4/2011-Till date) a. Tv=0.197 (U=50%), b. H=3m or 300cm, c. Cv=2.11cm2/sec To find: a. t=? Procedure: 1st case; We know that the following equation;

a. tdT

C vv

2

=

b. 197.05050

41004

22

=⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛=

ππ UTv

c. sec/11.260*35

2300*197.0

2

2

cmCv =⎟⎠⎞

⎜⎝⎛

=

2nd Case: We know that the following equation;

a. sec84.840211.2

2300*197.0

2

2

=⎟⎠⎞

⎜⎝⎛

==v

v

CdT

t or 140.047minutes.

July/2007. 8. A one dimensional consolidation test was conducted on a clay sample, with double drainage condition. The dial gauge readings recorded for a pressure increment of 100kpa to 200kpa are shown in the table below. One division of the dial gauge corresponds to 1*10-3mm. The thickness of the clay sample at 100kpa overburden pressure was 16mm. Determine the value of coefficient of consolidation of the clay by rectangular hyperbola method.

Elapsed time, minutes 0 0.25 1 2 4 6 9 12 16 25 36 50 60 Dial readings, divisions 340 360 370 378 386 394 402 410 416 426 434 440 443

Solution: Procedure:a. Calculate ΔH and t/ ΔH

Elapsed time, minutes

0 0.25 1 2 4 6 9 12 16 25 36 50 60

ΔH*10-3mm 0 20 30 38 46 54 62 70 76 86 94 100 103 t/ ΔH 0 12.5 33.34 52.6 86.9 11.11 14.51 17.14 21.05 29.06 38.29 50 58.25b.

b. Plot the curve t/ ΔH(Y-axis) v/s t(X-axis) on natural graph sheet; Y C 60 50 m m=dy/dx=(55-30)810-2/(56-26.25)=8.4034*10-3 D=8*10-2 40 30 b 20 d 10 D 0 X O 0 10 20 30 40 50 60 70 Time, t

d. Cv=0.3(mh2

dr)/D, where, h2dr=16+(16-0.105)/2*2=7.9737mm

e. Cv=0.30(8.4034*10-3*7.9737*7.9737)/8*10-2=1.985mm2/minute.

Compiled by: Prof.B.S.Chawhan M.Tech(Geo-Tech Engg), Asst.Professor,CED,Government.Engineering College,Haveri-581110(12/4/2011-Till date) Dec.2010 9. A soil sample 20mm thick takes 20min to reach 20% consolidation. Find the time taken for a clay layer of 6m thickness in the field to reach 40% consolidation. Sample in laboratory and clay in field are same. Assume double drainage in both cases. Solution: a. Tv=π/4(U/100)2= π/4(20/100)=0.0314 b. Cv=Tvd2/t=0.0314*3000*3000/(20*60)=2.617*10-3002/sec. c. Field u=40%, Tv=0.125 d. t=Tv*d2/Cv=5003days. May/June.2010 10. In a consolidation test, a soil sample 20mm in thickness took 28minutes to reach 90% consolidation under two-way drainage condition. For the same soil in the field what would be the time taken in days for 50% and 90% consolidation, if the thickness of soil layer is 4m and if there is i) One way drainage and ii) Two-way drainage?. Solution: a. Cv=0.848d2/t90=0.848(400*400)/(28*60)=5.05*10-4cm2/sec b. t90=3110days for one way drainage and 777.5days for two way drainage. c. Cv=0.197*d2/t50, then t50=722.4 days for one ways drainage and 180.6days for two ways drainage. Dec-2011 11. A 20mm thick clay specimen under double drainage undergoes 50% consolidation in 10minutes in the laboratory. Under the similar drainage conditions, what time is required for 90% consolidation for a 2m thick clay layer in the field? Ans: 1. Tv=Cvt/H2=0.196. 2. Cv=0.196*12/10=0.0196cm2/min.. 3. T90=Cvt/H2=0.848. 4. t=0.848*1002/0.0196=432653min or 72010Hr or 300.45days. 2002 scheme.June-July.2009 12. In laboratory consolidation test a clay specimen 20mm thick under double drainage , the time required for 50% consolidation of the clay is 30minutes. Calculate the coefficient consolidation. Calculate time required for 90% consolidation of the same clay in field it is 2m thick and drains on one face only. Take T50=0.196; T90=0.848. Solution: a. Cv=0.196*10*10/(30*60)=0.0109mm/sec b.Cv=Tv1*d1

2/t1= Tv2*d22/t2 then, t2=0.848(2000*2000)/0.0109=311510204.4sec=9.88years.

13. Soil investigation at a site gave the following information. The top soil upto a depth of 10.6m is fine sand, and below this lies soft clay layer of 7.6m thick. The water table is at 4.6m below the ground surface. The submerged unit weight of sand γb, is 10.4KN/m3

, and et unit weight above water table is 17.6KN/m3. the water content of the normally consolidated clay Wn=40% its liquid limit, WL=45%, and specific gravity of the solid particles is 2.78. The proposed construction will transmit a net stress of 120KN/m2. Find the average settlement of the clay layer. Solution: Given data a. WL=45%, b. H=7.6m, c.w=40%, d. Gs=2.78, e. Depth of W.T=4.6m from GL, f.γ' sand=10.4KN/m3, g. γd sand=17.6KN/m3, h. The clay is normally consolidated, Sr=1, i. Δp=120KN/m2. GL G

4.6m γd Sand W.T 10.6m 6m γ' Sand 7.6m Clay layer

Fig.19. Soil Starata. To find: a. St=? Procedure: 1. We know that the total settlement is given by the following equation;

⎥⎦

⎤⎢⎣

⎡ Δ++

=o

o

o

ct p

ppe

HCS 10log

1 (1)

2. According to Terzaghi’s; ( ) ( ) 32.01045009.010009.0 =−=−= Lc WC

3. Initial void ratio is given by the equation below;

112.11

78.2*4.0===

ro S

wGe …….(Sr=1 for normally consolidated).

4. Overburden pressure Po is given by; po=4.6*17.6+6*10.4+7.6/2*(γsat-γw)


where, γsat= 81.9*112.11112.178.21 ⎥

⎦

⎤⎢⎣

⎡+

+=⎥

⎦

⎤⎢⎣

⎡++

wo

o

eeG

γ =18.077KN/m3.

po=143.36+3.8*(18.077-9.81)=174.90KN/m2.

∴ ⎥⎦

⎤⎢⎣

⎡ Δ++

=o

o

o

ct p

ppe

HCS 10log

1= m26.0

9.1741209.174log

112.11315.0*6.7

10 =⎥⎦⎤

⎢⎣⎡ +

+ or 26cm

14. An Odometer test is performed on a 2cm thick clay sample. After 5 minutes, 50% consolidation is reached. After ho long a time would the same degree of consolidation be achieved in the field where the clay layer is 3.7m thick?. Assume the sample and the clay layer has the same drainage boundary conditions (double drainage). Solution: Given data 1st case: a. d1=H/2=2/2=1cm, b. t1=5minutes, c. U=50% i.e. Tv=0.197 To find; a. Cv=? 2nd case: a. d2=H/2=3.7/2=1.85cm, b. U=50%. To find: a. t2=? Procedure: 1st case; a. We know that the following equation;

tdT

C vv

2

= (1)

22

22

1

1 2**1d

tCd

tC vv =

utestddt min1125.175*

185.1*

2

1

2

1

22 =⎥⎦

⎤⎢⎣⎡=⎥

⎦

⎤⎢⎣

⎡=

15. A normally consolidated clay of thickness 4m is sandwiched between to sand layers with outlets. The stress at the mid-height of the clay layer was 1.9t/m2 before any loading was placed. However, due to placement of a fill on the ground surface the stress at the midle-height of the clay layer increases by 1.1t/m2. The initial void ratio and compression index of clay were 1.2 and 0.3 respectively. Determine the total compression of the clay a long time after placement of the fill and time required for 20% and 80% of ultimate compression if co-efficient of consolidation of clay is 4*10-3cm2/sec. Solution: Given data a. H=4m, b. po=1.9t/m2, c.Δp=1.1t/m2, d. eo= 1.2, e. Cc=0.3, f. Cv=4*10-3cm2/sec.

GL G

Sand 4m 2m Clay layer 2m Sand

Fig.20. Soil strata location. To find: a. St=?. b. t=? for U=20%, c.t=? for U=80%. Procedure: 1. We know that the total settlement is given by the following equation;

⎥⎦

⎤⎢⎣

⎡ Δ++

=o

o

o

ct p

ppe

HCS 10log

1 (1)

cmSt 82.109.1

1.19.1log2.113.0*4

10 =⎥⎦⎤

⎢⎣⎡ +

+= or 0.1082m

2. For 20% consolidation;

0314.05020

41004

22

=⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛=

ππ UTv

3. We have;


sec25.31415910*4

2400*0314.0

3

2

2

=⎟⎠⎞

⎜⎝⎛

== −v

v

CdT

t or *1/60*60*60=3.636days.

4. For 80% consolidation; ( ) ( ) 567.080100log933.0781.1%100log933.0781.1 =−−=−−= UTv

5. sec14.567139010*4

2400*567.0

3

2

2

=⎟⎠⎞

⎜⎝⎛

== −v

v

CdT

t or 65.641days.

16. During a consolidation test, a sample of fully saturated clay 3m thick is consolidated under a pressure increment of 200KN/m2. When equilibrium is reached, the sample thickness is reduced to 2.6cm, the pressure is then removed and the sample is allowed to expand and adsorb water . The final thickness is observed as 2.8cm and the final moisture content is determined as 24%. If the specific gravity of the soil solids is 2.7, find the void ratio of the sample before and after consolidation. 17. A recently completed fill was 10m thick and its initial average void ratio was 1.0. the fill was loaded on the surface by constructing an embankment covering a large area of the fill. Some months after the embankment was constructed, measurements of the fill indicated an average void ratio of 0.8. Estimate the compression of the fill. 18.Soil investigation at a site gave the following information. Fine sand exists to a depth of 10.6m and below this lie a soft clay layer 7.6m thick. The water table is at 4.6m below the ground surface. The submerged unit weight of sand γb is 10.4KN/m3, and the wet unit weight above the water table is 17.6KN/m3.The water content of the normally consolidated clay Wn-40%, its liquid limit WL=45%, and the specific gravity of the solid particles is 2.78. The proposed construction will transmit a net stress of 120KN/m2 at the centre of the clay layer. Find the average settlement of the clay layer. 19. A strata of normally consolidated clay of thickness 3m is drained on one side only. It has a hydraulic conductivity of k=5*10-8 cm/sec and a coefficient of volume compressibility mv=125*10-2cm2/sec. Determine the ultimate value of the compression of the stratum by assuming a uniformity distributed load of 250KN/m2 and determine the time required for 20percent and 80percent consolidation.

GOOD-LUCK

M.Tech(Geo-Tech Engg), UNIT-07 CONSOLIDATION OF SOILS€¦ · Compiled by: Prof.B.S.Chawhan...

Documents

Transcript of M.Tech(Geo-Tech Engg), UNIT-07 CONSOLIDATION OF SOILS€¦ · Compiled by: Prof.B.S.Chawhan...