consolidation part 2

8/11/2019 Consolidation Part 2

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Part 2

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Assumption :1. The soil is homogeneous ( kz is independent of z).2. The soil is completely saturated ( S = 100%).3. The soil grains and water are virtually incompressible ( w is constant and

volume change of soil is only due to change in void ratio).4. The behaviour of infinitesimal masses in regard to expulsion of pore water

and consequent consolidation is no different from that of largerrepresentative masses.

5. The compression is one-dimensional ( u varies with z only).

6. The flow of water in the soil voids is one-dimensional, Darcys law beingvalid.7. Certain soil properties such as permeability and modulus of volume change

are constant;these actually vary somewhat with pressure. ( k and m v areindependent of pressure).

8. The pressure versus void ratio relationship is taken to be the idealised one ( a v is constant).

9. Hydrodynamic lag alone is considered and plastic lag is ignored, although itis known to exist. (The effect of k alone is considered on the rate ofexpulsion of pore water).

Karl Terzaghi1883-1963

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A mechanistic model for consolidation (adapted from Taylor, 1948)

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Consolidation of a clay sample with double drainage

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For one-dimensional flow situation, this reduces to:

S = 100% or unity, and

The flow is only due to the hydrostatic excess pressure,h =u/ w , where w = unit weight of water.

The change in volume per unit volume, V, may be written, as per the definition ofthe modulus of volume change, mv;

V = mv. = mv. u , since an increase represents a decrease u.

a)

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From equation a) and b)

The coefficient of consolidation may also be written in terms of the coefficientof compressibility:

That is the basic differential equation of consolidation according toTerzaghis theory of one-dimensional consolidation. Its units can be shownto be mm2/s or L2T1.

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COEFfISIENT OF CONSOLIDATION (c v)

cv = kmvw

- use to determine rate of settlement

- Base on one dimensional flow and fully saturated soil- Unit : m 2/year

- Constant during consolidation

k = coefisient of permeability

mv = modulus of volume change

Time Factor (T v)

Tv = cvt

d2

9

d= length of the longest drainage path


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ISOCHRONThe progress of consolidation can be shown by plotting a

series of curves of u e against z for different values of t. Suchcurves are called isochrones

2d

Tv = 0.8Tv = 0.3

Tv = 0.05

Tv = 0u1

(b)

d

Tv = 0.7Tv = 0.3 Tv = 0.05

Tv = 0

(c)

2d

Open layer

dslope = i

Half closed layer

(a)

u1 = constant

Open layer

Half closed layer10

t/d2cvTv =


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2d

kurva 1 kurva 3kurva 2

LAPISAN TERBUKA

d


LAPISAN SETENGAH TERBUKA

Variasi tekanan air pori berlebih11


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Derajat konsolidasi U)

Pada permukaan bebas u=0dan konsolidasi telah selesaiUz = 100%

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U


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perkembangan proses konsolidasi pada penambahan

tegangan total tertentuPada permukaan bebas, u = 0, U z = 100%

e0

e

e1

0

Asumsi hubungan linear e-

ui u

Uz = e0 - ee0 - e 1

=

-

0 1- 0 = ui - u = 1 - u

ui ui

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2d


LAPISAN TERBUKA

2d


LAPISAN TERBUKA


LAPISAN TERBUKA

d



d



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INITIAL VARIATIONS OF EXCESS PORE WATER PRESSURES

0.001 0.01 0.1 1 21.00

0.90

0.80

0.70

0.60

0.50

0.40

0.30

0.20

0.10

0

U

(3)

(1)(2)

Tv = c v t/d 2

OPEN LAYER HALF CLOSED LAYERCURVE 1 CURVE 2 CURVE 3

CURVE 1 CURVE 2 CURVE 3


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LOG Time method ( CASAGRANDE)

cv = 0.196d2 t 50

Primary consolidation

Initialcompression

Theoritical curve

Secondaryconsolidation

log T v

U

10

0

A

B

a 100

a 50

a f

a s a 0

t50 0.1 1 10 100

10

1000 10000log t (minute)

2.00

2.50

3.00

3.50

4.00

4.50

5.00

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2

50

H

t cv

Determine a 100 which is intersectionbetween primary consolidation tangent andsecondary consolidation tangent. Determine point B at consolidation curvethat has horizontal axixs t 1=1 minute, thenpoint out C which is found from line verticalthrough horizontal axis t 2 = 4 t 1 hence thevertical distance between B and C is x draw horzontal line such as so as to thedistance from B to s horisontal line DE equalto x. Intersection of horizontal line withvertical axis is d s that is 0% consolidation. Determine d 50 which is equal to 50% of iprimery consolidation (1/2 of d 10 0 -d 0), fromline through d

50intersect line of i primery

consolidation at C and obtained point F thatis t 50 Hence for 50% degree of consolidation :

T50 = c v t50 /H 2

Untuk mendapatkan t 50 :

cv = 0.197 H 2 /t50 17


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Root square method (due to TAYLOR)

cv = 0.848 d 2

t 90

Mendapatkan t 90 : gambar garis pada bagian

linier kurva danmemotomg titik D. U = 0

gambar garis lurus DEdengan absis 1.15 kaligaris pertama

perpotongan garis DEdengan kurva (a 90) samadengan U = 90%

didapat t 90

Konsolidasiprimer

kompresiawal

Konsolidasisekunder

P e m b a c a a n

j a r u m

p e n g u k u r

( m m

)

a 0

a s

a 90

a 100

a f

D

E

5.00

4.50

4.00

3.50

3.00

2.50 0 42 6 8 10 12 14t90 t90 (min)

0

1.00 1

0.9

U

Tv

A B

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Contoh soal

cv = 0.848 d 2

t90

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Contoh soalDari lapisan lempung setebal 12m dengan double drainage c v=8.0 x 10-8 m2/s.Tentukanlah derajat konsolidasi setelah 5 tahun pembebanan pada kedalaman 3, 6, 9

dan 12 mPenyelesaian:1. Hitung faktor waktu Tv = Tv=0.352. Dari grafik degree of consolidation 2H=12m Hr=6m dan T = 0.35

Z z/H U3 0.5 61%6 1 46%9 1.5 61%

12 2 100%

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memperpendek alur pengaliran pada lempung mempercepat laju konsolidasi pada lempung jenuhdengan permeabilitas rendah

hanya mempengaruhi laju penurunan, tidakmempengaruhi nilai akhirnya

timbunan lapisan drainasihorizontal

drainasi vertikal

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Jarak antar pengaliran < ketebalan lapisan lempung

Parameter yang diperlukan :- ch (koefisien konsolidasi arah horizontal)- cv (koefisien konsolidasi arah vertikal)Secara normal, rasio c h/ cv berkisar antara 1 - 2Semakin basar rasionya, maka penggunaan drainasi akansemakin berguna

PROBLEM YANG TERJADI :smear effect (pengurangan nilai koefisien akibatremoulding selama palaksanaan)berkurangnya penambahan tegangan vertikal dan nilaitekanan air pori berlebih akibat diameter drainasi pasir

yang besar

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FAKTOR WAKTU UNTUK KONSOLIDASI

S

SR

r d

R = 0.564SPola bujur sangkar

S

S

Rr

d

R = 0.525SPola segitiga

d

2R2r d

Akibat drainasi vertikal :T v = cvt

d2

Akibat drainasi radial :T v = cht

4R2

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DERAJAT KONSOLIDASI RATA-RATA (U) AKIBATKOMBINASI DRAINASI VERTIKAL DAN RADIAL

Uv = (T

v)

Ur = (T r)

(1 - U) = (1 - Uv)(1 -Ur)

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Example


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ExampleThe final consolidation settlement of a layer of clay 5.0 m thick iscalculated to be 280mm. The coefficient of consolidation for theloading range is 0.955 mm/min. There is two-way drainage, upwardand downward.Calculate the time required for (a) 90% consolidation settlement, (b)a settlement of 100 mm(a) Drainage path length, d = 5.0/2 = 2.50 m = 2500 mm

For U90, T90 = 0.848. Then

(b) For 100 mm settlement,

Ut = 100/280 = 0.357and since Ut < 0.6, Tv = 0.357 x /4 = 0.100Then time for 100mm settlement

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Bangkok Clay :

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consolidation part 2

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