Pile Foundation Design

►KDA - FACTORY II (GEN.TRIAS, CAVITE)

DESIGN OF MICROPILES Reference: Geotechnical Geotechnical Engineering by Braja M. Das Foundation Analysis and Design 5th Ed by Joseph E. Bowles

A SITE SITE CONDI CONDITIO TIONS NS From Geotechnical Report, supplied by client. Silt General Soil Type:

Pile Capacity Parameters: Skin Friction, f  Micro Pile End Bearing

30 5000

kPa kPa

B PILE PILE DESI DESIGN GN Location:

As shown in drawings SERVIC RVICE E ULTI ULTIMA MATE TE 300 470 Support reaction, P = kN 60 90 Support reaction, M = kNm 16 36 Support Shear, V = kN L=1,475 775

   0    0    1  ,    1   =    W

P

2

Qu = ultimate pile load Qt = load-carrying capacity at the pile point Qs = frictional resistance fz = frictional resistance per unit area p = perimeter of the pile cross section

Q u =Q t +Q s ∆Q z f z = p∆z Q s =

P

1

 p∆Lf 

Determine max and min reaction of pile due to service load: + : Compression P Mx - : Tension σ= N I

±

Pile Reaction kN 1 73 2 227 227 max = 73 min = Pile

Reference: F-1 Compute for Required Pile Capacity: 9.5 Length of Piles = 250 Diameter of Piles = Area of Piles = Width of Columns = Frictional capacity of Soil, Qs = Bearing capacity of Soil, Qt = Total Capacity of Soil,Qu = Factor of Safety = Allow. Soil capacity, Q req'd = Number of Piles = AASHTO §10.7.1.2 minimum o.c. spacing = AASHTO §10.7.1.2 edge clearance = Minimum edge of columnedge of pile cap =

2

49087 500 214.41 245.44 459.85 2 229.00 2

kN pcs

775

mm

225

mm

300

mm

mm mm kN kN kN

Determine Moment Moment of Inertia of Group of Piles Pile x 1 -387.5 2 387.5 Ix=0.30m2

Determine max and min reaction of pile due to ultimate load: + : Compression P Mx - : Tension σ= N I

±

Pile Reaction kN 1 119 2 351 351 max = 119 min =

Pile

kN kN

m mm

kN kN

Design Pile against Required Capacity: 50 Clear Cover = 49087 Pile Gross Area, Ag = 24 f'c = 414 fy = 16 Longit. Bar f = 12 Tie Bars f = 6 No. of bars = Area of reinforcement, As = 1206.37 2.46% Steel Reinforcement Ratio, r =

Check for slenderness effect: 1.20 K= 9.50 Lu = 0.0003 Moment of Inertia = 0.0814 r= Young's modulus of concrete, Ec = 23172173 1.00 Cm = 0.51 Reduction factor, f = EI = (Ec I/2.5)/(1+B) 3017.21 229.14 Pc = Pi ^2 EI /(kLu)² 235.00 Factored Axial Load, Pu = f Pc = 117.76

mm mm2 Mpa Mpa mm mm pcs mm2 Ok!

Design Pile against Required Capacity: 4.80 0.20f'c = Mpa 4.63 Mpa Max Axial Stress, smax = IF : smax >= 0.20fc', f = 0.50 smax < 0.20fc', f = 0.90- smax(0.40/0.20fc') THEN, f = 0.5139 DESIGN LOADS

1

29

Moment Mod. Factor => 1

 AXIAL

MY

MZ

V

 AXIAL/f

MY/f

MZ/f

V

351.13

0.00

0.00

36.00

683.24

0.00

0.00

36.00 C/D

Middle

Mgraph

C

D

40.00

20.56

0.00

Left

Middle

Right

Total

6

0

6

12

Axial Force (kN) dP

5

0

4

8

Right

351

0

351

75

276

0

426

57

294

0

408

25

57

294

0

408

0

9

6

22

Middle

21

5 7

Percent Differenc

Left

29

3

22

Ok in Axial!

D = demand

Column Shear Forces (kN)

Right

2 4

139.99 1.00

(kLu) / r =

PCA LOADS

1.3 x Mn (kN-m) Left

kN kN

If (kLu)/r > 22 consider slenderness effect

C = capacity

Step

m m4 m kPa

21 5

0

4

9

9

OK! Design Shear Reinforcement of Pile 67 Shear Ultimate 34 Concrete Shear Capacity 12 try f = 226 Av s = 100 try s = Av prov'd =

kN kN

mm mm2 mm mm2

bws Avmin = f y

3

s min =

; smin=

Avmin3fy bw

749.16 mm2

C PILE CAP DESIGN

Pu Mu Wsoily

Wsoily x' t=

900 x

   N     k    9    1    1

   0    5  .    9

   N     k    1    5    3

Wsoilx

Wsoilx y'

Wsoilx

   N     k    1    5    3

Wx

Check reinforcement parameters: Solve for rmax: ρmax=

′ 0.85β1 f  c 600 fy 600+fy

0.85 b1 = rmax = 0.02169

500 500 5600 1100.00 1475 900 613 21.00 414.00 20 20 0.00 0.0 0.0

mm mm mm mm mm mm mm Mpa Mpa mm mm m kN kN

Check Punching Shear: (Assume no piles inside the punching area) 351.13 kN Ultimate Shear, Vu = 17.24 Max Pile Reaction = kN Ok! Concrete shear cap., fVc = 4457.14 kN Design Reinforcements Longitudinal bars: x= x' = Mu = Rn = r= As reqd = s required = s supplied = no. of bars =

137.50 243.75 48.28 0.130 0.0028 1865.96 185.20 175.00 6

mm mm kNm

mm2 mm mm pcs

Transverse bars:

Solve for rmin: ′ f  c ρmin=

4fy 0.00277 rmin =

rtemp = 0.002bt=

Footing Parameters: Column Width, w = Column Length, b = Punching perimeter, Bo = Footing Width, W = Footing Length, L = Footing thickness, t = Ftg eff. Depth, d = f'c = fy = Longit. Bar f = Trans. Bar f = Ht. of soil = Wt. of soil y = Wt. of soil x =

1808.4 mm2

y= y' = Mu = Rn = r= As reqd = s required = s supplied = no. of bars =

150.00 0.00 0.000 0.0028 2502.08 185.20 175.00 8

mm mm kNm

mm2 mm mm pcs

C LATERAL SUPPORT REACTION

V=36kN El.-1900

kh=120Su

Nvalu

Su, kPa

El.-3900

kh =

5760

kN/m

10

48

El.-4900

kh =

5760

kN/m

10

48

El.-5900

kh =

23040 kN/m

50

192

El.-6900

kh =

23040 kN/m

50

192

El.-7900

kh =

23040 kN/m

50

192

El.-8900

kh =

23040 kN/m

50

192

El.-9900

kh =

23040 kN/m

50

192

El.-10900

kh =

23040 kN/m

50

192

El.-11900

kh =

23040 kN/m

50

192

Column Interaction Diagram from SP Column

P= M=

P ( kN) 1600

351 kN 24 kNm

Analysis Result: Section adequate.

(Pmax) fs=0

fs=0 fs=0.5fy

fs=0.5fy 1

0

35 Mx ( kNm)

(Pmin) -600

get from STAAD model