Basic design of stilts based on the CPT

Julio R. ValdesGeo-Innovations Research Group

Civil EngineeringSDSU

CPT

Parameters

Design – method

Example

CPT

‘coin’

Hydraulic jack

Steel tubes

Continuous penetration at 2 cm/sec

Cone

Parameters

By means of electric sensors,

the CPT provides

Four parameters

Which can be used to calculate

The resitance, stiffness, and In some cases, permeability

Of the soil.

Parameters

fs

qt

Vs

u2

Tip resistance = qt

Sleve friction= fs

Pore Pressure = u2

Shear wave velocity= Vs

Tip resitance

There needs to be a “correction” of qt because the actual values of u are different at the top and at the bottom of the tip.

qT = qt + (1-an)u2

an = 0.8 = fn (cone)

Force sensor

Sleve friction

The “sleve” is forced upward during the test because of the soil-sleve friction

fs

Pore pressureThe pore presure changes during the test because the soil is subjected to forces during penetration..

If k is low, u2 ≠ uh

If k is high, u2 = uh

uh = ahp

hp

During penetration (beneath the ground)

Velocity

Hitting the beam with the hammer creates a shear wave propagates through the soil towards the cone .

A geophone inside of the cone captures the arrival of the wave.

Initial Arrival

(t = tp) Time t

Hammer hit (t = 0)

Velocity

Vs = D / tp

Measurements in real time(computer)

Sand

Clay

Crust

Results (Example)

Design of stilts

Design by analysis

Drilling stilts

Compacting stilts

Brown 2008 Agra foundations 08

layer #1

layer #2

Capacity

L = G1 + G2

Q = Qf + Qp

G1

G2

diameter = d

FS = Load/ Q

Tip capacity

Friction capacity

Calculate for z = L

Tip capacity Qp

Qp = (qp)(Ap)

qp = qT – uh

d2/4 layer #1

layer #2

diameter = d

qp = (qT – ’vo)/k2

Compactor and Drill (Esllami & Fellenius 2006)

Compators in clay (Powell et al. 2001)

L =

G1 +

G2

k2 = 1.7

Friction

T = N

Coefficient of friction (block-floor)

Given the magnitude of N, how large does T have to be for the block to move?

blockT

Weight of block = N

floor

In other words, when T = N , the system fails; FS = 1.

T is the resistance of the system if T = N

layer #1

layer #2

Friction capacity Qf

L = G1 + G2

Qf = (As1)(f1) + (As2)(f2)

f1 = (’ho1) tan(’) CM CK

f1 = Ko1(’vo1) tan(’) CM CK

dG1G1

G2

diameter = d

Needed: G , Ko , vo , ’ , CM , CK

N...for soil?

CM y CK

materialInstalation

f1 = Ko1(’vo1) tan(’) CM CK

Needed: Ko , vo , ’ , CM , CK

CM CK

concrete

steel

drill

compactor

1.0

0.8

0.9

1.1

Ko

Ko = [1 – sin(’)] OCR sin ’ Mayne & Kulhawy (1982)

Friction angle

´ = 17.6 + 11 log Kulhawy & Mayne (1990)

pa = 100 kPa ’vo = Effective stress at the center of the particle

)(avo

0.5T

pσ

q

Needed: Ko , vo , ’ , CM , CK

?

OCR (fine soils)

OCR = ’p / ’vo

’p = 0.60 (qT – u2) Mayne (2005)

Needed: Ko , vo , ’ , CM , CK

OCR (coarse soils)

Mayne (2005)

OCR

0.192q T

p a

0.22

1 sin ( )( ) vo

p a

0.31

1

sin ( ) 0.27

.

OCR (Fine soils)

OCR = ’p / ’vo

’p = 0.60 (qT – u2) Mayne (2005)

Needed: Ko , vo , ’ , CM , CK

OCR (Coarse soils)

Mayne (2005)

OCR

0.192q T

p a

0.22

1 sin ( )( ) vo

p a

0.31

1

sin ( ) 0.27

.

Example5

12

34

6

sand

sand

sand

clay2300

20

~0

0

d = 0.7m

L = 15m

P = 4000 kN

Compacting stilts of concrete

c = 28kN/m3

Example

G1=4m , =18kN/m3 , qt = 5MPa , u2 = 0kPa

=15m

sand

sand

sand

clay

G3=3m , =20kN/m3 , qt = 12MPa , u2 = 0kPa

G2=2m , =18kN/m3 , qt = 12MPa , u2 = 0kPa

G4=2.5m , =20kN/m3 , qt = 34MPa , u2 = 20kPa

G5=3.5m , =20kN/m3 , qt = 6MPa , u2 = 2300kPa

’vo=(2)(18)=36kPa

qT = qt + (1-an)u2 = 5000 + (1-0.8)(0) = 5000kPa

Needed: Ko , vo , ’ , CM , CK

o0.5

avo

0.5T .

100)(36log.

pσ

qlog.'

)(738

50001161711617

OCR = (long equation; Thick soils) = 4.29

Ko = [1 – sin(’)] OCR sin ’ = [1-sin(38.7)] (4.29)sin(38.7) = 0.93

G1=4m , =18kN/m3 , qt = 5MPa , u2 = 0kPa

sand

sand

sand

clay

=15m

G3=3m , =20kN/m3 , qt = 12MPa , u2 = 0kPa

G2=2m , =18kN/m3 , qt = 12MPa , u2 = 0kPa

G4=2.5m , =20kN/m3 , qt = 34MPa , u2 = 20kPa

G5=3.5m , =20kN/m3 , qt = 6MPa , u2 = 2300kPa

’vo=(4)(18)+(2)(18)+(3)(20)+(2.5)(20)+(1.75)(20) – (9.81)(3+2.5+1.75) =182kPa

qT = qt + (1-an)u2 = 6000 + (1-0.8)(2300) = 6460 kPa

Needed: Ko , vo , ’ , CM , CK

o

.

avo

0.5T

100182log.

pσ

qlog.'

)(36

64601161711617

50

’p = 0.60 (qT – u2) = 0.60 (6460-2300) = 2496 kPa

OCR = ’p / ’vo = 2496 / 182 = 13.7

Ko = [1 – sin(’)] OCR sin ’ = [1-sin(36)] (13.7)sin(36) = 1.91

G1=4m , =18kN/m3 , qt = 5MPa , u2 = 0kPa

sand

sand

sand

clay

=15m

G3=3m , =20kN/m3 , qt = 12MPa , u2 = 0kPa

G2=2m , =18kN/m3 , qt = 12MPa , u2 = 0kPa

G4=2.5m , =20kN/m3 , qt = 34MPa , u2 = 20kPa

G5=3.5m , =20kN/m3 , qt = 6MPa , u2 = 2300kPa

SUELOS

G qt u2 qT 'voEstrato Suelo (m) (Mpa) (kPa) (kPa) (kPa) OCR Ko

1 arena 4 5 0 5000 36 38.7 4.29 0.932 arena 2 12 0 12000 90 40.7 17.9 1.243 arena 3 12 0 12000 123 40 4.9 0.794 arena 2.5 34 20 34004 151 44.4 1.85 0.725 arcilla 3.5 6 2300 6460 182 36 13.7 1.91

L = 15

SoilLayer

PILOTE HINCADO DE CONCRETO W = 51.45 kNd(m) = 0.7 CM = 1 CK = 1.1

CAP. FRICCIONAL CAP. DE PUNTAf As f As 'vo qp Ap qp Ap

(kPa) (m2) (kN) (kPa) (kPa) (m2) (kN)29.5 8.796 259.539

105.6 4.398 464.41389.9 6.597 593.155

117.3 5.498 645.146277.8 7.697 2138.34 199.7 10643 0.38485 4096

Qf = 4100.59 kN Qp = 4096 kN

Q = Qf + Qp - WQ = 8144.87 kN

f1 = Ko1(’vo1) tan(’) CM CK

As1 = dG1

Ap = d2/4

qp = (qT – ’vo)/k2

FS = Q / P = 8145 / 4000 = 2.04

COMPACTING STILT OF CONCRETE

FRICTIONAL CAP. TIP CAP.

OCR & Ko

OCR

Ko = 0.192 ( qT / pa)0.22 (´vo / pa)-0.31 OCR 0.27 (1)

Ko = [1 – sin(’)] OCR sin ’ (2)

pa = 100 kPa

(1) Mayne (1995)

(2) Mayne & Kulhawy (1982)

a) Calculate ´

b) Vary OCR until the two values of Ko (eq. 1 y 2) are similar.

CPT-parameters

Dr = relative density (sands)

Dr = 100

if unknown, use OCR = 1 e = void ratio e = 1.152 – 0.233·log(qC1) + 0.043 log(OCR)

OCR305

q2.0

1C

5.0