Presentation is loading. Please wait.

Presentation is loading. Please wait.

9. Axial Capacity of Pile Groups CIV4249: Foundation Engineering Monash University CIV4249: Foundation Engineering Monash University.

Similar presentations


Presentation on theme: "9. Axial Capacity of Pile Groups CIV4249: Foundation Engineering Monash University CIV4249: Foundation Engineering Monash University."— Presentation transcript:

1

2 9. Axial Capacity of Pile Groups CIV4249: Foundation Engineering Monash University CIV4249: Foundation Engineering Monash University

3 Axial Capacity W P base Bearing failure at the pile base P shaft Shear failure at pile shaft F u F u + W = P base + P shaft

4 P shaft,t Shear failure at pile shaft T u - W = P shaft,t < P shaft,c Tension Capacity

5 Applications LowWeight Soft to Firm Clay Large Distributed Weight Very Large Concentrated Weight Dense Sand Strong Rock

6 Group Capacity Overlapping stress fields Overlapping stress fields Progressive densification Progressive densification Progressive loosening Progressive loosening Case-by-case basis Case-by-case basis Overlapping stress fields Overlapping stress fields Progressive densification Progressive densification Progressive loosening Progressive loosening Case-by-case basis Case-by-case basis Pile Cap P ug P ug n.P up P ug = n.P up P ug n.P up P ug = n.P up

7 Efficiency, Soil Type Soil Type Clay Sand Rock Number of Piles, n Number of Piles, n n = 5 x 5 = 25 Spacing/Diameter Spacing/Diameter sd s/d typically > 2 to 3 Pile Cap

8 Types of Groups Rigid Cap Capped Groups Flexible Cap Free-standing Groups

9 Feld Rule for free-standing piles in clay A B A B C B B C B B C B A B A reduce capacity of each pile by 1/16 for each adjoing pile 13/1611/16 8/16 e = 1/15 * (4 * 13/16 + 8 * 11/16 + 3 * 8/16) = 0.683

10 Converse-Labarre Formula for free-standing piles in clay = 1 - (n-1)m + (m-1)n = 1 - (n-1)m + (m-1)n 90 mn 90 mn m = # rows = 3 n = # cols = 5 s = 0.75 d=0.3 = tan -1 (d/s) = tan -1 (d/s) = 0.645 = 0.645

11 Flexible Cap P ug = min (nP up,P BL ) D L,B P BL = BLc b N c + 2(B+L)Dc s cscscscs cbcbcbcb Block Failure N c incl shape & depth factors

12 Empirical Modification n nP up P BL = BLc b N c + 2(B+L)Dc s P ug = min (nP up,P BL ) 1 1 1 1 1 1 P 2 ug = n 2 P 2 up + P 2 BL 1 1 1 1 1 1 P 2 ug = n 2 P 2 up + P 2 BL 1 = 1 + n 2 P 2 up 1 = 1 + n 2 P 2 up 2 2 1 = 1 + n 2 P 2 up 1 = 1 + n 2 P 2 up 2 2 P 2 BL

13 Flexible Cap D = 20m L = B = 5m c s = c b = 50 kPa Block Failure d = 0.3m

14 Rigid Cap P total = P group + P cap for group block failure, P cap = c cap N c [B c L c - BL] for single pile failure, P cap = c cap N c [B c L c - nA p ] Capped Groups B x L B c x L c

15 Efficiency increases s/d 1234 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 7 2 free-standing 7 2 capped

16 Piles in Granular Soils End bearing - little interaction, = 1 Shaft - driven –For loose to medium sands, > 1 –Vesic driven : 1.3 to 2 for s/d = 3 to 2 –Dense/V dense - loosening? Shaft - bored –Generally minor component, = 1

17 Pile Settlement

18 Elastic Analysis Methods based on Mindlins equations for shear loading within an elastic halfspacebased on Mindlins equations for shear loading within an elastic halfspace Poulos and Davis (1980)Poulos and Davis (1980) assumes elasticity - i.e. immediate and reversibleassumes elasticity - i.e. immediate and reversible OK for settlement at working loads if reasonable FOSOK for settlement at working loads if reasonable FOS use small strain modulususe small strain modulus

19 Definitions Area Ratio, A p R A = A p / A s ApApApAp AsAsAsAs Pile Stiffness Factor, K K = R A.E p /E s EpEpEpEp EsEsEsEs

20 Floating Pile % load at the base Pile top settlement = o C K C = o C K C = P.I o R K R L R / E s d = P.I o R K R L R / E s d d EpEpEpEp E s, E s, Rigid Stratum h L Solutions are independent of soil strength and pile capacity. Why?

21 Floating pile example 0.5 E p = 35,000 MPa E p = 35,000 MPa E s = 35 MPa Rigid Stratum 32 25 = 0.3 = 0.3 = o C K C = o C K C = P.I o R K R L R / E s d = P.I o R K R L R / E s d P = 1800 kN o = 0.038 o = 0.038 C K = 0.74 C = 0.79 =.022 =.022 P b = 40 kN I o = 0.043 R K = 1.4 R L = 0.78 R = 0.93 = 4.5mm = 4.5mm Effect of : L = 15m d b /d = 2 h = 100m

22 Pile on a stiffer stratum % load at the base Pile top settlement = o C K C b C = o C K C b C = P.I o R K R b R / E s d = P.I o R K R b R / E s d d E s, E s, Stiffer Stratum E b > E s L EpEpEpEp

23 Layered Soils E 1, 1 Stiffer Stratum E b > E s Ld EpEpEpEp E 2, 2 E s = 1 E i h i L L

24 Stiffer base layer example 0.5 E p = 35,000 MPa E p = 35,000 MPa E b = 70 MPa 25 = o C K C b C = o C K C b C = P.I o R K R b R / E s d = P.I o R K R b R / E s d P = 1800 kN o = 0.038 o = 0.038 C K = 0.74 C = 0.79 C b = 2.1 =.0467 =.0467 P b = 84 kN I o = 0.043 R K = 1.4 R b = 0.99 R = 0.93 = 4.5 mm = 4.5 mm E s = 35 MPa = 0.3 = 0.3 Effect of: E s = 15 MPa to 15m

25 Movement Ratios M R is ratio of settlement to PL/AE Focht (1967) - suggested in general : 0.5 < M R < 2 See Poulos and Davis Figs 5.23 and 5.24

26 Pile group settlment Floating Piles End bearing piles Single pile settlement is computed for average working load per pile


Download ppt "9. Axial Capacity of Pile Groups CIV4249: Foundation Engineering Monash University CIV4249: Foundation Engineering Monash University."

Similar presentations


Ads by Google