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

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CPT Parameters Design – method Example

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CPT

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‘coin’

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

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Parameters

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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.

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

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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

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

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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)

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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.

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

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

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Design of stilts

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Design by analysis

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Drilling stilts

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Compacting stilts Brown 2008 Agra foundations 08

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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

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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

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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

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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?

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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

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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 ?

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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)

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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)

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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

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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

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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

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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

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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.

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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.

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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)

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