Download presentation

Presentation is loading. Please wait.

Published byHilary Hoover Modified over 2 years ago

1
Basic design of stilts based on the CPT Julio R. Valdes Geo-Innovations Research Group Civil Engineering SDSU

2
CPT Parameters Design – method Example

3
CPT

4
‘coin’

5
Hydraulic jack Steel tubes Continuous penetration at 2 cm/sec Cone

8
Parameters

9
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.

10
Parameters fsfs qtqt VsVs u2u2 Tip resistance = q t Sleve friction= f s Pore Pressure = u 2 Shear wave velocity= V s

11
Tip resitance There needs to be a “correction” of q t because the actual values of u are different at the top and at the bottom of the tip. q T = q t + (1-a n )u 2 a n = 0.8 = fn (cone) Force sensor

12
Sleve friction The “sleve” is forced upward during the test because of the soil-sleve friction f s

13
Pore pressure The pore presure changes during the test because the soil is subjected to forces during penetration.. If k is low, u 2 ≠ u h If k is high, u 2 = u h u h = a h p hphp During penetration (beneath the ground)

14
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.

15
Initial Arrival (t = t p ) Time t Hammer hit (t = 0) Velocity V s = D / t p

16
Measurements in real time(computer) Sand Clay Crust Results (Example)

17
Design of stilts

18
Design by analysis

19
Drilling stilts

20
Compacting stilts Brown 2008 Agra foundations 08

21
layer #1 layer #2 Capacity L = G 1 + G 2 Q = Q f + Q p G1G1 G2G2 diameter = d FS = Load/ Q Tip capacity Friction capacity

22
Calculate for z = L Tip capacity Q p Q p = (q p )(A p ) q p = q T – u h d 2 /4 layer #1 layer #2 diameter = d q p = (q T – ’ vo )/k 2 Compactor and Drill (Esllami & Fellenius 2006) Compators in clay (Powell et al. 2001) L = G 1 + G 2 k 2 = 1.7

23
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? block T 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

24
layer #1 layer #2 Friction capacity Q f L = G 1 + G 2 Q f = (As 1 )(f 1 ) + (As 2 )(f 2 ) f 1 = ( ’ ho1 ) tan( ’) C M C K f 1 = K o1 ( ’ vo1 ) tan( ’) C M C K dG 1 G1G1 G2G2 diameter = d Needed: G, K o, vo, ’, C M, C K N ...for soil?

25
C M y C K material Instalation f 1 = K o1 ( ’ vo1 ) tan( ’) C M C K Needed: K o, vo, ’, C M, C K CMCM CKCK concrete steel drill compactor 1.0 0.8 0.9 1.1

26
K o K o = [1 – sin( ’)] OCR sin ’ Mayne & Kulhawy (1982) Friction angle ´ = 17.6 + 11 log Kulhawy & Mayne (1990) p a = 100 kPa ’ vo = Effective stress at the center of the particle Needed: K o, vo, ’, C M, C K ?

27
OCR (fine soils) OCR = ’ p / ’ vo ’ p = 0.60 (q T – u 2 ) Mayne (2005) Needed: K o, vo, ’, C M, C K OCR (coarse soils) Mayne (2005)

28
OCR (Fine soils) OCR = ’ p / ’ vo ’ p = 0.60 (q T – u 2 ) Mayne (2005) Needed: K o, vo, ’, C M, C K OCR (Coarse soils) Mayne (2005)

29
Example 5 12 34 6 sand clay 2300 20 ~0 0 d = 0.7m L = 15m P = 4000 kN Compacting stilts of concrete c = 28kN/m 3

30
Example G 1 =4m, =18kN/m 3, q t = 5MPa, u 2 = 0kPa =15m sand clay G 3 =3m, =20kN/m 3, q t = 12MPa, u 2 = 0kPa G 2 =2m, =18kN/m 3, q t = 12MPa, u 2 = 0kPa G 4 =2.5m, =20kN/m 3, q t = 34MPa, u 2 = 20kPa G 5 =3.5m, =20kN/m 3, q t = 6MPa, u 2 = 2300kPa ’ vo =(2)(18)=36kPa q T = q t + (1-a n )u 2 = 5000 + (1-0.8)(0) = 5000kPa Needed: K o, vo, ’, C M, C K OCR = (long equation; Thick soils) = 4.29 K o = [1 – sin( ’)] OCR sin ’ = [1-sin(38.7)] (4.29) sin(38.7) = 0.93

31
G 1 =4m, =18kN/m 3, q t = 5MPa, u 2 = 0kPa sand clay =15m G 3 =3m, =20kN/m 3, q t = 12MPa, u 2 = 0kPa G 2 =2m, =18kN/m 3, q t = 12MPa, u 2 = 0kPa G 4 =2.5m, =20kN/m 3, q t = 34MPa, u 2 = 20kPa G 5 =3.5m, =20kN/m 3, q t = 6MPa, u 2 = 2300kPa ’ vo =(4)(18)+(2)(18)+(3)(20)+(2.5)(20)+(1.75)(20) – (9.81)(3+2.5+1.75) =182kPa q T = q t + (1-a n )u 2 = 6000 + (1-0.8)(2300) = 6460 kPa Needed: K o, vo, ’, C M, C K ’ p = 0.60 (q T – u 2 ) = 0.60 (6460-2300) = 2496 kPa OCR = ’ p / ’ vo = 2496 / 182 = 13.7 K o = [1 – sin( ’)] OCR sin ’ = [1-sin(36)] (13.7) sin(36) = 1.91

32
G 1 =4m, =18kN/m 3, q t = 5MPa, u 2 = 0kPa sand clay =15m G 3 =3m, =20kN/m 3, q t = 12MPa, u 2 = 0kPa G 2 =2m, =18kN/m 3, q t = 12MPa, u 2 = 0kPa G 4 =2.5m, =20kN/m 3, q t = 34MPa, u 2 = 20kPa G 5 =3.5m, =20kN/m 3, q t = 6MPa, u 2 = 2300kPa Soil Layer

33
f 1 = K o1 ( ’ vo1 ) tan( ’) C M C K A s1 = dG 1 A p = d 2 /4 q p = (q T – ’ vo )/k 2 FS = Q / P = 8145 / 4000 = 2.04 COMPACTING STILT OF CONCRETE FRICTIONAL CAP. TIP CAP.

35
OCR & K o OCR K o = 0.192 ( q T / p a ) 0.22 ( ´ vo / p a ) -0.31 OCR 0.27 (1) K o = [1 – sin( ’)] OCR sin ’ (2) p a = 100 kPa (1)Mayne (1995) (2)Mayne & Kulhawy (1982) a) Calculate ´ b) Vary OCR until the two values of K o (eq. 1 y 2) are similar.

36
CPT-parameters D r = relative density (sands) D r = 100 if unknown, use OCR = 1 e = void ratio e = 1.152 – 0.233·log(q C1 ) + 0.043 log(OCR)

Similar presentations

OK

SOFT SOIL (PROBLEMS & STABILISATION METHOD) Session 2 - 7

SOFT SOIL (PROBLEMS & STABILISATION METHOD) Session 2 - 7

© 2018 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on history of english literature Download ppt on cybercrime and security Ppt on water softening techniques of integration Ppt on dispersal of seeds by animals Ppt on acid-base titration curve Ppt on landing gear system for motorcycles Ppt on statistics and probability examples Ppt on pi in maths games Ppt on history of world wide web Ppt on panel discussion format