Contoh Soal Tiang Pancang

CE 366 – PILE FOUNDATIONS ULTIMATE BEARING CAPACITY OF SINGLE PILES Determine the ultimate bearing capacity of the 800mm diameter concrete, bored pile given in the figure below. Assume Dc r=15*diameter ;

and the pile-friction angle =0.75φ’

Dc r=15*0.8=12m Clay Cu=60 kPa γ=18 kN/m3

4m

18x4=72kPa

Sand c’=0 ,φ’=300 γsat=20 kN/m3

6m

72+6(20-9.8)=133.2kPa Clay Cu=100 kPa γ=20 kN/m3

5m

Dcr=12m

133.2+2(20-9.8)=153.6kPa

153.6

Ultimate capacity of pile: Qul t=Qp+Qs •

Qp= Nc.cu.Ap= 9cuAp =9 * 100 * [π * (0.8)2/4] = 452 kN Nc : Fig 4.6 - Lec. Notes

•

Qs = Qs1 + Qs2 + Qs3

Qs1= α.cu.As = 0.8 * 60 * (π * 0.8 * 4) = 483kN α : Fig 1 – on sheet distributed (or Table 7.3 – Lecture Notes)

Qs2= Ks.σvo' .tanδ.As ; where Ks=0.5(Table 1 - Lec. Notes) , δ=0.75 φ’ =22.50 , As= p.L = (π * 0.8) * 6 =15.08 m2 Qs2=0.5 * [(72+133.2)/2] * tan22.5 * 15.08 =320 kN Qs3= α.cu. As =0.58 * 100 * (π * 0.8 * 5) = 729kN

Qult=452+(483+320+729)=1984 kN

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PILE GROUP CAPACITY Find the allowable bearing capacity of a single pile in the group of piles given below, by using: a) Terzaghi-Peck Method b) Converse-Labarre Formula 4.0m

1.2m Clay L=12m

2.8m 1.2m

1.2m

1.2m

1.2m

Diameter : 400mm Cu=50kPa, γ=18kN/m3 F.S.=2.5

Qult=Qp+Qs •

Qp=9cuAp =9 * 50 * π * (0.4)2/4 = 56.5 kN

•

Qs=α.cu.As =0.84 * 50 * π * 0.4 * 12 = 633kN

Qult=56.5+633=689.5 kN Qall=689.5 / 2.5 ≅ 275.8 kN

Converse-Labarre(Group action reduction)

⎡ ( n − 1 )m + ( m − 1 )n ⎤ E = 1 − θ⎢ ⎥⎦ 90mn ⎣ m= # of rows=3 n= # of piles in a row=4 θ = arctan(D/s)=arctan(0.4/1.2)=18.40 D=diameter s=spacing(center to center)

2

⎡ ( 4 − 1 )3 + ( 3 − 1 )4 ⎤ E = 1 − 18.4 ⎢ ⎥⎦ = 0.71 90 x3 x 4 ⎣ (Qall)single pile in the group=275.8x0.71 = 195.8 kN

Terzaghi_Peck Group Reduction Method: Skin fric.

Tip resist.

(Qg)ult =pDfcu+A(qult)net

where (qult)net = qnf = cu Nc

=pDf cu+Aqult-ADf γ

p=2(2.8+4.0)=13.6m A=2.8 * 4= 11.2m2

2.8m

4.0m

Qg=13.6 * 12*50+11.2(50*8.6)=12976kN (Qult)for single pile=12976/12=1081kN (Qall)for single pile=1081/2.5=432.5kN>275.8 kN >195.8 kN B.C. is not controlled by Qgroup,therefore: (qall)single pile=195.8 kN

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SETTELEMENT OF PILE GROUPS Calculate the consolidation settlement of the 12 pile group consisting of 10 m long end bearing piles. Piles are 80 cm in diameter and spaced at 2m center to center in either direction. Pile group carries a vertical load of 5000 kN including the weight of the pile cap. Use 2:1(V:H) pressure distribution. 6.8m

5000kN

2m

3m

2m

2m

sand

3m 3m

4.8m

Soft clay cu=25kPa

Df=10.0m

.∆σ

1

.∆σ

Clay mv=0.0001

1 2

2

∆σ = 5000/(6.8*4.8) =153.2 kN/m2 ∆σ1 = 5000/((6.8+4.5)*(4.8+4.5)) =47.6 kN/m2 ∆σ2 = 5000/((6.8+7.5)*(4.8+7.5)) =28.4 kN/m2

s=3*0.0001*(47.6+28.4)=0.0228m=2.28cm

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2m

2m

LOAD ON PILES Find the maximum and minimum vertical load in the pile group given below Q2=2000kN

Q1=1500kN

M2=1750kN.

A

B

C

D

1 CL 2.3m 0.5m 1.2m

2.0m

1.2m

0.5m

0.5m

2.0m

2.0m

ΣQ=3500kN

2.0m

0.5m

1

e=1/3500(2000(2.3)+1750-1500*2.3)=0.83m M=ΣQ*e=3500*0.83=2900kN.m I1-1=2*(2*12+2*32)=40pile-m4 due to piles

Qi=ΣQ /n ± M*di /I

QD=3500/8+2900*3.0 /40 = 655 kN(max) QA=3500/8-2900*3.0 /40 = 220 kN(min)

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BATTER PILES One of the legs of a steel structure rests on a concrete pedestal footing which is supported by a group of 18 piles. Outside rows are formed from batter piles (1 to 4) as shown below. Each pile is permitted to resist an horizontal force of 25 kN. a) Calculate axial loads on all piles. b) Could the unbalanced horizontal force be overcome? c) If the allowable bearing value of single pile is equal to 600 kN, state whether or not this pile foundation is safe.

Q = 6000 kN H = 600 kN

6.0 m

1

1

4

4

y 18 pile group

x

1.0 m

1.5 m

1.5 m

6

1.0 m

SOLUTION: • •

•

M = 6 ∗ 600 = 3600kN − m I = 2 ∗ 3 ∗ 1.5 2 + 4 ∗ 2.5 2 = 63.5 pile − m 2

(

)

Unbalanced horizontal force

Vertical Component of pile loads: Q M Qi = m ∗ xi n I

4QE= 1900 kN

6000 3600 − ∗ 2.5 = 192kN 18 63.5 6000 3600 QB = − ∗1.5 = 248kN 18 63.5 6000 QC = = 333kN 18 6000 3600 QD = + ∗1.5 = 418kN 18 63.5 6000 3600 QE = + ∗ 2.5 = 475kN 18 63.5

QA =

•

Q=6000 kN

4QD= 1254 kN

4QC= 1332 kN

3QB= 744 kN

Axial loads: 1 tan α = → cos α = 0.970 4 QA 192 = = 198 PA = cos α 0.970 PB = QB = 248kN PB = QC = 333kN PB = QD = 418kN Q 475 = 490kN PE = E = cos α 0.970

4PA= 792 kN

H=600 kN NOT SCALED

•

Unbalanced horizontal force: 1 1 H u = 600 + 4 ∗ 192 ∗ − 4 ∗ 475 ∗ = 317kN 4 4

•

Available horizontal resistance: H r = 18 ∗ 25 = 450kN . Since H u < H r ∴ Unbalanced force is overcome.

•

Maximum axial load:

Pmax = 490kN < Pall = 600kN ∴ Foundation is safe.

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INTERPRETATION OF PILE LOAD TEST The load-settlement data shown in the figure were obtained from a full-scale load test on a 400 mm square, 17 m long concrete pile (28-day compressive strength of concrete, fc' = 40 MPa). Use Davisson’s method to compute the ultimate downward load capacity.

Davisson’s method defines the ultimate capacity as that which occurs at a settlement of 0.012Br + 0.1B/Br + PL/(AE). The last term in this formula is the elastic compression of a pile that has no skin friction.

B= Diameter of the pile Br= Reference width = 300 mm P= Applied load L= Length of the pile A= Cross-section area of the pile E= Modulus of elasticity of the pile

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200,000 MPa for stell 11,000 MPa for pine 15,200σr (fc'/σr)0.5 for concrete where σr = 0.1 MPa E = 15,200σr (fc'/σr)0.5 =(15,200)(100 kPa)(40 MPa / 0.10 MPa) 0.5 = 30.4 * 106 kPa

0.012 Br +

0.10 B PL + Br AE

= 0.012 * 300 +

0.10(400) P(17,000) + 2 300 400 (30.4 *10 6 )(1 *10 −6 m 2 / mm 2 )

= 3.7 mm + 0.0035 P

Plotting this line on the load-displacement curve produces Pult =1380 kN

PILE DRIVING ENERGY CORRECTION 400 mm square concrete piles are to be driven to depth of 12.5. m. What should be the stroke of a 1.5 t rammer, due to the Danish Formula, if the amount of penetration should be about 5 mm/blow for the safe working load of 2000 kN on the pile? (Econc = 30 · 106 kN/m2)

Solution A pile = 0.4 2 = 0.16 m 2 c2 =

2 ⋅ (1.5 ⋅ 9.8) ⋅ Η ⋅ 12.5

0.16 ⋅ 30 ⋅ 10 14.7 ⋅ H Q = 2000 = 6

=

Η 13100

Η 0.005 + 0.5 ⋅ 13100

⇒ H = 1.4 m

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