►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
