soil mechanics (eciv 3 second term chapter 10 ...site.iugaza.edu.ps/nhindi/files/chapter10.pdf ·...

1

Soil Mechanics (ECIV 3351)

Chapter 10: Compressibility of Soil

Second Term

2011

1- Types of settlement:

1-1 Elastic (immediate) settlement (Se).

1-2 Primary consolidation settlement (Sc).

1-3 Secondary consolidation settlement (Sp).

Calculation of elastic settlement:

p

s

se I

EBS

21

: Applied pressure = Load / area.

B: Width of the foundation

s :Poison's ratio.

sE :Modulus of elasticity of soil.

pI :Influence factor given from table 10.1 P.262

Depend on the value (m1 = Length of foundation "L" / Width of foundation "B").

21 so

EFGe

eE

IIIBS

BLBe

4

oE :Modulus of elasticity of soil just below the foundation.

GI :Influence factor for variation of E (Fig 10.4 P.264)

FI :Influence factor for rigidity (Fig 10.5 P.265).

EI :Influence factor for foundation embedment (Fig 10.6 P.266)

ثابحة للحزبة بدون Eالعالقة السابقة جىاولث حساب الهبىط المزن باعحماد بعض الفزضُات ، مثل وجىد قُمة

جغُز مع العمق ، وهذا غُز مىجىد فٍ الىاقع، وبالحالٍ هىاك عالقات أخزي أخذت بعُه االعحبار هذا الحغُز و

بعض األمىر األخزي كما َلٍ:

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Second Term

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Second Term

2011

Problem1:( 10.1 Text book)

Find the immediate settlement of rigid column footing 4.5 ft in diameter and carries 20 tons. Es =

1500psi and =0.25

Solution:

p

s

se I

EBS

21

psfA

P04.2515

5.425.0

2000202

From table 10.1 P.262 pI =0.79

inchftSe 465.0038.079.012/1500

25.015.404.2515

2

2

____________________________________________________________________________________

Calculation of primary consolidation settlement:

In general,

o

Ce

eHS

1

- For normal consolidated clay:

'

''

log1

o

o

o

c

Ce

HCS

cC : Compressibility index.

'

o :Present effective pressure at the middle of the layer under consideration.

' :Increase in stress at the mid of the layer.

إذا أعطانا في السؤال أن هناك إجهاد متىشع على كامل التستح فنعتثسه مثاشسج هى'.

إذا أعطانا في السؤال وجىد حمل على قاعدج محددج المساحح ، فيجة حساب' المتىسطح خالل الطثقح المساد حساب

:الهثىط لها كما يلي

6

4 '''' bottommidTop

- For over consolidated clay:

If ''' oc

'

''

log1

o

o

o

S

Ce

HCS

H .سماكة الطبقة المزاد حساب الهبىط لها :

eo .وسبة الفزاغات األولُة :

e( حم الحصىل علُها مه المىحىً الذٌ َزبط بُه وسبة الفزاغات و اإلجهاد (.e – : الحغُز فٍ وسبة الفزاغات.َ

1Metric tons = 2204.6 Ib

1 ton or US ton = 2000 Ib

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Second Term

2011

If '''' oco

'

''

'

'

log1

log1

c

o

o

C

o

c

o

SC

e

HC

e

HCS

SC : Swell index or called recompression index (Cr).

'

c : Pre-consolidation pressure.

'

'

cOCR

The most famous relation for Cc is:

Cc = 0.009 (LL - 10)

cS CC

10

1

5

1

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Second Term

2011

Problem2: ( 10.8 text book)

Solution:Find the unit weights of the soil layers:

- For dry sand:

pcfe

G wsdry 35.103

6.01

4.6265.2

1

Solution:

- For saturated sand:

pcf

e

eG wssat 75.126

6.01

4.626.065.2

1

- For saturated clay:

e = 2.7 x 0.3 = 0.81

pcf

e

eG wssat 121

81.01

4.6281.065.2

1

Find the present effective pressure:

psfo 19704.621212

104.6275.1261035.10310'

psfo 417022001970''

psfc 3400'

)4170()3400()1970( '''' oco

Find Cc and Cs:

Cc = 0.009 ( 35 – 10 ) = 0.225

Cs = Cc / 5 = 0.225 / 5 = 0.045

inftS

e

HC

e

HCS

C

c

o

o

C

o

c

o

SC

03.2169.03400

4170log

81.01

10225.0

1970

3400log

81.01

10045.0

log1

log1 '

''

'

'

A soil profile as shown, the pre-consolidation

of the clay is 3400 psf. Estimate the primary

consolidation settlement that will take place as

the result of surcharge equal 2200 psf.

Assume Cs = Cs5

1.

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Second Term

2011

Problem 2: Calculate the final voids ratio of a thin clay layer when the vertical stress increases by 65 kPa

if the initial void ratio is 1.11, the initial vertical effective stress is 50 kPa, OCR = 2, and the compression

and recompression indexes are Cc = 0.4 and Cr = 0.08 respectively.

Solution:

11.211.111

eH

eH

e

eHS

o

C

KpaOCR ccc 100

502 '

'

'

'

)115()100()50(

50

1156550

''''

'

''

oco

o

o

kpa

kpa

0616.104836.011.1

04836.011.104836.00229.011.211.2

0229.0

0229.0100

115log

11.11

4.0

50

100log

11.11

08.0

log1

log1

22

2

'

''

'

'

ee

eee

HH

HHH

S

e

HC

e

HCS

C

c

o

o

C

o

c

o

SC

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Second Term

2011

Sand

=15kN/m3

Sand

sat=18kN/m3

Clay (NC)

Gs=2.7,w=35%

LL = 38

Problem 3 :( 10.21 text book)

For the shown rectangular footing, calculate the

primary consolidation pressure given that:

B = 1.5 m

L = 2.5 m

Q = 120kN.

Solution:

4

.325.25.1

120

qI

kpaq

دج مستطيلح و النقطح المدزوسح سنقىم تحساب الصيادج المتىسطح في اإلجهاداخ في الطثقح الطينيح أسفل منتصف القاعدج، وحيث أن القاع

Influence factorفي منتصف المستطيل فلمعسفح قيمح نرهة إلى جدول 9.8 .942صفحح

6

4 '''' bottommidTop

Layer Z(m) m1(L/B) n1(2Z/B) I =32xI(kpa)

Top 1.5 1.67 2 0.4058 13

Middle 2.75 1.67 3.67 0.1925 6.16

Bottom 4 1.67 5.33 0.09 2.88

Kpa38.186

88.216.6413'

1.5

m

1.5

m

2.5

m

Top

Middle

Bottom

Z

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Second Term

2011

For clay layer, e = Gs w = 2.7*0.35=0.945

./38.18

945.01

81.9945.07.2

1

3mkNe

eG wssat

Kpao 5.4581.938.182

5.281.9185.1155.1'

Cc = 0.009 ( 38 – 10) = 0.252

It is normal consolidated clay:

mmmS

e

HCS

C

o

o

o

cC

2002.05.45

75.64.45log

945.01

5.2252.0

log1 '

''

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Second Term

2011

Time rate of consolidation:

t

HTC drvv

2

t: The required time to reach a specific degree of consolidation.

Hdr: Height of drainage.

Cv: Coefficient of consolidation.

To find Cv:

vw

Vm

kC

K: Permeability coefficient.

mv: Coefficient of volume compressibility.

2,

1

1001

10

eee

e

ee

m avgavg

v

To find Tv: Look Table 10.5 P293

Or from equations:

%60%100log933.0781.1

%600100

%

4

2

UU

UU

TV

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Second Term

2011

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Second Term

2011

Problem 4 :( 10.10+10.11+10.12 Text book)

The coordinates of two points in the virgin compression curve are as follow:

2

12

2

11

/40054.1

/20082.1

mkNe

mkNe

1- Determine the coefficient of volume compressibility for the pressure range stated above.

2- Given that Cv = 0.003 cm2/sec, Determine (k) in cm/sec corresponding to average void ratio.

3- What would be the effective pressure ρ' corresponding to e=1.7

4- What would be the void ratio corresponding to effective pressure ρ' = 500kN/m2.

Solution:

1- Find mv:

2- Find K:

vw

Vm

kC

sec/10537.1sec/10537.1

102238.581.910003.0

79

4

4

cmmK

k

3- Find ρ' corresponding to e=1.7:

Slope of the virgin line = 93.0200log400log

54.182.1

loglog 12

21

ee

At e = 1.7:

.19.269200loglog

7.182.193.0 3

3

kpa

4- Find e corresponds to ρ' = 500kN/m2:

45.1200log500log

82.193.0 3

3

e

e

e

log scale

1.82

45.4

922 422

kNme

ee

mavg

v /102238.5

2

54.182.11

200400

54.182.1

1

2412

21

12



Second Term

2011

Problem 5 :( 10.17 Text book)

For normally consolidated clay, the following results were obtained:

dayftk

etsf

eetsf

oo

oo

/108.1

96.04

21.12

4

''

'

1- How long will it take in days for 9 ft thick clay (Drained in one side) to reach 60% consolidation.

2- What is the settlement at that time?

Solution:

1- Find t60:

.8.2409286.0

096.0

286.0%60

/096.01034.62

108.1

/1032000/06.0/06.0

2

96.021.11

24

96.021.1

1

60

60

2

60

2

60

2

5

4

25212

21

daysttt

HTc

TU

dayftm

kc

Ibfttonfte

ee

m

drvv

vw

v

avg

v

2- Sc =? At 60% rate of consolidation:

8304.02log4log

96.021.1

CC

inftSU

ftS

e

HCS

c

C

o

o

o

c

C

33.761.0018.16.0%60

018.12

4log

21.11

98304.0

log1 '

''

13



Second Term

2011

Problem 6 :( 10.19 text book)

For laboratory sample:

H = 25mm (One way drainage).

t50 = 11 min.

For field:

H = 4m (Two way drainage):

In the field, find t50 and t90.

Solution:

t

HTC drvv

2

At the lab:

U = 50% → T50 = 0.197

min/10119.111

025.0197.0 252

mCv

At the field:

daystt

daystt

100min1440572403.0

10119.1

49min704202197.0

10119.1

70

70

25

50

50

25

14



Second Term

2011

Problem 7:

A 3m thick clay layer is able to drain from both its upper and lower surface and after 3 years a settlement

of 32 mm is observed. After many years a settlement of 40 mm is measured. If a specimen of this soil, 20

mm thick, was to be tested in an oedometer apparatus with two-way drainage what would be the time

for 50% consolidation.

Solution:

At 100 % consolidation → Sc = 40mm

At U %? Consolidation → Sc = 32 mm

%8040

10032

U

daysmC

daysyearst

mH

TU

t

HTC

V

dr

drvv

/10165.11095

5.1567.0

10953

5.12/3

567.0%80

232

80

2

In the lab.:

min34.240169.001.0197.0

10165.150

23

2

dayst

t

HTC drvv

soil mechanics (eciv 3 second term chapter 10 ...site.iugaza.edu.ps/nhindi/files/chapter10.pdf ·...

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