Piles - 2

1CE5107 Pile foundation Lecture 2 : Pile capacityCE5107 Pile foundation Lecture 2 : Pile capacity

Lecture 2Lecture 2

Pile capacity

Associate Professor C F LeungDepartment of Civil Engineering, NUSE-mail : [email protected]

CE5107 Pile foundation Lecture 2 : Pile capacityCE5107 Pile foundation Lecture 2 : Pile capacity

Load Transfer Mechanism

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33 1. Structural load (e.g. column)2. Pile cap

3. Pile (material)

4. Soil (geotechnical)

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2.1 Structural Capacity(allowable material stress)

Bored Cast In-Situ PilesQa = 0.2 fcu A or 0.25 fcu A

(depending on specifications and limited to Grade 35 concrete for design)

Precast Reinforced Concrete PilesQa = 0.25 fcu A

Prestressed Concrete Spun Piles:Qa = 0.25 (fcu - prestress after losses) A


Steel PilesQa = 0.3 fy A (driven piles)Qa = 0.5 fy A (jacked-in piles)

Timber PilesQa = 0.25 fc A

where Qa = allowable working loadA = cross-sectional area of pilefcu = concrete strength fy = yield stress of steel

(Grade 43A: 275 N/mm2; Grade 50A: 355 N/mm2)fc = compressive strength of timber


Partially embedded piles (e.g. piles for jetties)

For the portion of pile above ground, it should be designed as column in accordance with BS8110 which recommends the following effective length and fixity:

In firm ground, the lower point of contraflexure is taken at 1 m below the ground surface. In soft soils, the point may be taken as one-half of the depth of penetration into the stratum or 3 m, whichever is lower.


2.2 Reinforcement

Pile subject to axial loading only

0.5% nominal reinforcement for the first 6 m of pile (first 12 m is preferred in some cases).

[Appropriate reinforcement needs to be provided to resist the bending moment for piles subject to lateral loads, eccentric loads, uplift or tension loads.]


Pile Cap Appropriate depth of pile cap over piles

- transmission of vertical load- to resist high shear forces and to avoid punching shear

Adequately reinforced against - bending due to eccentric loading- high stress induced by differential settlement between piles

Adequate attachment of pile head to pile cap - transmission of lateral loads


Pile Cap (cont.)

Method of attachment

- Square top, treated with preservative and embedded in concrete pile cap.

Timber pile

- Adequate length of pile embedment, otherwise fitted with cleats

Steel Pile- Project into cap for bondingRC pile

MethodType of pile

Refer to BS8110 and see also Tomlinson


2.3 Geotechnical capacity

Ultimate bearing capacity of pile (Qu):Qu = Qs + Qb

where Qs = shaft capacity = (fsi Asi)Qb = end bearing capacity = qb Abfsi = unit shaft resistance for soil layer iAsi = total shaft area of pile in soil layer iqb = unit base resistanceAb = base area of pile

Load

Pile base, Qb

Pile shaft Qs


Geotechnical capacity (cont.)Allowable working load

All pile types: Qa = Qu/Fwhere F = overall factor of safety (usually 2 to 3)

Alternative for bored piles: Qa = Qs/Fs + Qb/Fbwhere Fs = factor of safety for shaft (usually 1.5)

Fb = factor of safety for pile toe (usually 3)


2.4 Cohesive soils (clay) Unit base resistance

qb = Nc cbwhere Nc = bearing capacity factor (= 9) ;

may be conservative for local soilscb = undrained cohesion of clay at pile base

Unit shaft friction (total stress approach)fs = cu

where cu = undrained shear strength = adhesion factor (generally correlated with cu)


Limiting fs value for clayfs < 100kPa (or 200kPa to be confirmed by load test)

Driven piles: from the work of Tomlinson (1986) [refer to Fig. 2.1]

Bored piles: = 0.3 to 0.6 (overconsolidated clays) = 0.8 to 1.0 (normally consolidated clays) = 0.45 (if no previous data available)


Fig. 2.1 Values of adhesion factor (after Tomlinson)

Piles driven through overlying or sandy gravels


Fig. 2.1 (cont.)

Pile driven through overlying weak clay

Piles without different overlying strata


Cohesive soils (cont.) Unit shaft friction (effective stress approach)

Burland proposed that fs = v

where = dimensionless parameter depending on soil type= OCR (1 sin ) tan

= drained friction angle of clay,OCR = overconsolidation ratio of clay

(= 1 for normally consolidated clay)(>> 1 for heavily overconsolidated clay)

and v = average effective vertical stress for given pile segment


Cohesive soils (cont.) Preliminary design --

take = 0.3 for soft clay and = 0.8 for stiff clay

For long piles with pile length / width ratio ( L/B) >15, a reduction factor of ( 1 log L/15B) may be applied to .

v at a given elevation is the effective overburden pressure at thegiven elevation

= (Hi i)where Hi = thickness of soil layer i and

i = effective unit weight of soil layer i = i - w(w = unit weight of water; use full for soils above water table)


2.5 Cohesionless soils (Sand)

Unit base resistanceqb = Nq b

where Nq = bearing capacity factor for sand, see Fig. 2.2and subjected to limiting value of(a) Driven piles, see Fig. 2.3(b) Bored piles : 2 to 11 MPa

b = Effective overburden pressure at pile base


Fig. 2.2

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Fig. 2.3 Limiting static cone resistance versus for sand (Meyerhof 1976)


Unit shaft friction (sand)

fs = Ks v tan (limiting value 100 kPa; higher value to be verified by load tests)

Values of Ks tan Sand Driven piles Bored pilesLoose ( < 33o) 0.8 0.3Medium (33o < < 38o) 1.0 0.5Dense ( > 38o) 1.5 0.8

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2.6 Design based on in-situ tests

(a)Standard penetration resistance N value (blow count) Clay -- Use cu = 5N (for conservative design) Sand Use Fig. 1.3 to determine or

Bored piles:fs = 2N (local data)fs = 13 N (Db/d) kPa (Meyerhof)Driven pilesfs = 3N (local data)qb = 40 N (Db/d) kPa (Meyerhof)where Db is the pile penetration in bearing stratum (stiff soil) & d is pile diameter.[Need to verify limit values for qb as before]


SPTFor clay,Undrained apparent

cohesion (kPa)cu = 5N to 6N

Friction angle,

Fig. 1.3 Relationship between N value and (for sand)

Sand

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Design based on in-situ tests (cont.)(b) Cone penetration test

Use Fig. 1.1 to determine for sand orQ = 2Cs As + qcb Ab

where qcb = cone resistance at base of pile,Cs = average shaft friction along pile

as measured on friction jacket= 0.005 qc for driven concrete and timber piles= 0.0025 qc for driven steel H-piles,

and qc = cone resistance at given elevation.

(c) Pressuremeter TestRefer to Lecture 1 for the determination of soil

parameters from pressuremeter test results.


Friction angle, Fig. 1.1 Correlation between static cone resistance qcand friction angle of sand

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Example 2.1 : Piles in clay

Given: Working column load = 6000 kN

Soil condition

Determine the number of piles and penetration required to support the column.

500100Hard silty clay25 - 3510020Stiff silty clay5 255010Firm silty clay0 -5

Av. cu (kPa)Av. N valueSoilDepth (m)


Solution

(A) Try precast RC pilefcu = 40 N/mm2 , X-section 320 mm x 320 mm, Desired factor of safety, F = 2.5

(1) Structural material strengthQa = allowable load = 0.25 fcu Ap

= 0.25 x 40 x 0.322 x 103 kN= 1024 kN

6 number of piles required for working load of 6000kNWorking load on each pile = 1000 kN

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(2) Geotechnical considerationsTry penetration depth of 25m (Consider Why?!)(a) Total stress () approachBase resistance at 25 m (cu = 500 kPa & Nc = 9), Qb = 9 x 0.32 x 0.32 x 500 = 460 kN (conservative)

Qs = 320 + 2176 = 2496 kN see next slideThis is a friction pile with most resistance from the shaft!Qa = Q/F = (2496 + 460)/ 2.5 = 1182 kN > 1000 kN Okay!!


2176850.851005 -25 m

320501.0500 5 m

Shaft friction, Qs (kN)

Unit shaft friction,fs (kPa)

Fig. 2.1

cu(kPa)

Shaft friction

Calculation of shaft friction

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(b) Effective stress () approachAssume water table is at ground level for the worst scenario (one can use the actual water table elevation for a less conservative design)

Firm silty clay (0 to 5 m)Assume effective unit weight, = w = 9 kN/m3

Overconsolidation ratio OCR = 3Friction angle = 27 = OCR ( 1 sin ) tan = 0.482

At 0 m , v (effective overburden) = 0At 5 m, v = 9 x 5 = 45 kPa

(v)av = 22.5 kPa

Tests should be carried out to determine these values


fs = (v )av = 0.482 x 22.5 = 10.84 kPawhich is lower than 50 kPa for the approach. This is due to the small v values.

Qs = 10.84 x 0.32 x 4 x 5 = 69.4 kN

Stiff silty clay (5 to 25 m)Assume = 10.5 kN/m3 , OCR = 5, = 29

= OCR ( 1 sin ) tan = 0.639A reduction factor may need to be applied for long pile, see lecture note.

At 25 m , v = 45 + 10.5 x (25-5) = 255 kPa(v)av = (45 + 255)/2 = 150 kPa

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fs = 0.639 x 150 = 96 kPaQs = 96 x 0.32 x 4 x (25-5) = 2454 kN

This value may be lower if one has applied a reduction factor to for long piles.

Base resistance is the same as before. Qa can hence be determined using the same procedure described earlier.

Notes: Qa will increase (or verse versa) considerably for method if

(i) water table is lower (ii) is higher

Both imply higher v


(B) Try bored pilefcu = 25 N/mm2 , Diameter = 0.8 m,

(1) Structural material strengthQa = 0.25 fcu Ap

= 0.25 x 25 x x 0.42 x 103 kN= 3142 kN

2 number of 800-mm diameter bored piles are required for working load of 6000 kN

Working load on each pile = 3000 kN

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(2) Geotechnical considerationsFor a desired factor of safety of 2.5,Ultimate load = 2.5 x 3000 kN = 7500 kN

Base resistance = Nc cb Ab = 9 x 500 x x 0.42 kN= 2262 kN

Shaft resistance from 0 to 5 m = cu As = 0.45 x 50 x x 0.8 x 5= 283 kN

Shaft resistance from 5 to 25 m = 0.45 x 100 x x 0.8 x 20= 2262 kN

For bored piles, one can try to achieve an optimum design by penetrating deeper to arrive at the same geotechnical and structural capacities.


Shaft resistance required from below 25 m= 7500 2262 283 2262 = 2693 kNUnit shaft friction fs for hard silty clay = cu = 0.45 x 500

= 225 kPa

A maximum fs = 200 kPa may be used but such value needs to be confirmed by pile load test. Say use fs = 200 kPa

Contact length required = 2693 kN / (200 kPa x x 0.8 m)= 5.357 5.4 m

Total penetration required = 25 + 5.4 m = 30.4 m (say 31 m)

Poser: Why bored piles is usually lower than driven pile?

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Example 2.2 : Piles in sand (from Tomlinson)

An isolated 350 mm x 350 mm reinforced concrete pile (fcu = 35) in a jetty structure is required to carry a maximum compression load of 400 kN and a net uplift load of 320 kN. The soil consists of a loose to medium-dense saturated fine sand (average N = 12) extending to a depth of 9 m below sea bed followed by dense sand and gravel (average N = 40). Determine the required depth of penetration of the pile to resist the compression load. No erosion is expected.


Solution

Seabed to 9 m (N = 12) = 31 from Fig. 1.3; = 7.36 kN/m3 (estimated)Ks tan = 0.8 for driven piles(v)av = [ 0 + 7.36 x 9] / 2 = 33.1 kPaShaft friction = 0.8 x 33.1 x 4 x 0.35 x 9 = 334 kN

Try penetration of 4 m into dense sand and gravel N = 40 = 38.5o Ks tan = 1.5 (Fig. 1.3 & table) = 11.77 kN/m3 (estimated)(v)av = [ 7.36 x 9] + [ 11.77 x 4] /2 = 89.74 kPaUnit shaft friction = Ks tan (v)av > 100 kPa (use limiting value 100 kPa)

Shaft friction = 100 x 4 x 0.35 x 4 = 560 kN

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Base resistance For = 38.5o, Fig. 2.2 gives Nq = 120b = 7.36 x 9 + 11.77 x 4 = 113.2 kPa

Unit base resistance = b Nq = 113.2 x 120 = 13.6 MPaFig. 2.3 gives limiting qc = 15 MPa and thus use 13.6 MPaBase resistance = 13.6 x 0.352 = 1.666 MN = 1666 kNTotal geotechnical capacity = 334 + 560 + 1666 = 2560 kN

F = 2560/400 = 6.4 >> 2.5

This indicates that the pile penetration may be reduced from geotechnical capacity viewpoint. You are encouraged to tryshorter pile penetrations until you reach F = 2.5.


Based on loading, select pile type and material.

Work out number of piles and cross-sectional area of pile required based on allowable material stress (structural capacity).

Determine pile penetration required based on geotechnical considerations. If geotechnical capacity is a problem, (a) increase the number of piles for driven piles, or (b) increase the penetration depth for bored piles such that structural capacity = geotechnical capacity.

Pile selection and design

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2.7 Pile Driving Formula Monitoring of pile penetration for every 10 hammer blows.

Each hammer blow causes (i) permanent pile penetration (set) and (ii) temporary pile penetration during pile driving, see Fig. 2.4 (a). Typical record of blow counts for pile penetration with depth is shown in Fig. 2.4 (b). Pile is deemed to be driven to set (refusal) if the set is < 2 mm per blow or 20 mm per 10 blows.

Estimated required set can be determined using Hiley formula (1925).


Fig. 2.4(a)

10 hammer blowsPermanent set10 mm for 10 blows

TemporaryCompression (T/C)

HydraulicHammer1.10 m drop

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Fig. 2.4(b)

Soft soilEasy pile penetration

Stiff soilDifficult pile penetration


Hiley formula (1925)Assume Energy in = Work + Impact loss + losses in (driving assemble + pile + soil) The ultimate pile capacity Qu is given as

where ef = hammer efficiency (Table 2.1)W = hammer weightH = hammer drop heightn = coefficient of restitution (Table 2.2)Wp= weight of pileS = set (pile penetration in mm/hammer blow)

C1, C2, C3 = temporary compression due to pile head and cap, pile and soil compression, respectively (Table 2.3)

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Pile driving assembly

Hammer Energy = weight x drop height

Cushion and helmet


Table 2.1 Values of Hammer Efficiency, ef, (from Chellis 1969)

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Table 2.2 Values of Coefficient of Restitution, n, (after Whitaker 1970)


Table 2.3 Values of C1, C2, C3 for Hiley Formula (after Chellis 1961)

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(1) Limitations of Hiley and other pile driving formula {Important!!}

(2) Although Qu can be determined by trial and error using Hiley formula. Pile capacity Qu so obtained needs to be re-checked based on geotechnical capacity for the given pile penetration.

(3) The use of actual measured temporary compression at a given site may be used to recalculate the required set.


Example 2.3 Pile Driving Formula

Precast RC pile 320 mm x 320 mmQa = Working Load = 1000kNTry Kobe K 35 diesel hammer W = 3500 kg 35 kNDrop height H = 3 mF >> 3 preferred Pile capacity Q = 3000 kN

where ef = 1 (Table 2.1) n = 0.4 (Table 2.2)Wp = conc x Ap x length

= 25 kN/m3 x 0.322 m2 x 27 m = 69.1 kN

As example 2.1

Depending on helmet

L = 25 m in Example 2.1. Add 2 m for driving

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C1 = 0.05 in = 1.27 mm (Table 2.3 for medium driving)

C2 = QL / AEp = (3000 x 27) / (0.322 x 32 x 106 )= 0.0247 m = 24.7 mm [This is the major component!]

C3 = 0.1 in = 2.54 mm

Temporary compression C = C1 + C2 + C3 = 28.51 mm

After calculation, set s = 1.23 mm for last blow.or 12.3 mm for last 10 blows


2.8 Pile groupsFailure mechanism Capacity of pile group may not be equal to the sum of capacities of individual piles. The ratio of the two capacity is termed the efficiency of a pile group. Pressure bulbs of neighbouring piles tend to overlap, creating greater stress concentration on the surrounding soil. This leads to greater settlement of the pile group and is termed as group interaction. Soil will fail in shear (local failure) or the pile group will settle excessively (block failure)

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BS 8004 recommends that Friction piles -- pile centre to centre spacing > perimeter of square piles; or > 3 times pile diameter for circular piles End bearing piles -- distance between surface of shafts of adjacent piles > least width of the pile.

Local Failure Block Failure


(1) Pile group in clay(a) Local failure

Capacity of pile group = QG = m n Qwhere = pile group efficiency (Fig. 2.5, after Whitaker),

=

(after Converse Labarre)m = number of rows of piles,n = number of piles in a row,d = diameter of piles,s = centre-to-centre spacing between piles = tan-1 (d/s), in degrees,

and Q = ultimate capacity of single pile.

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Fig. 2.5 Bearing capacity of pile group in cohesive soils


(b) Block failureThe ultimate bearing capacity of the whole block

QG = cbNcBgLg + cu [ 2 D (Bg + Lg)]where D = depth of pile in bearing stratum

Bg = width of pile groupand Lg = length of pile group

Choose the lower QG from (a) local or (b) block failure for design.

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(2) Pile group in sand Pile driving compacts the sand to a radius of at least three times the pile width.

For loose sands, efficiency > 1 due to sand densification between piles.

For dense sand, pile driving causes loosening and efficiency less than 1 may result.

Block failure to be considered if centre-to-centre pile spacing < 7 pile diameters.

Capacity of pile group is the lesser of (a) sum of capacity of individual piles, and (b) load capacity of pile group block.


Example 2.4 Pile Group

Using the same data as Example 2.1. That is:Column load 6000 kN6 Precast RC piles 320 mm by 320 mm each of allowable working load of 1000 kN and ultimate geotechnical capacity Qu of 2956 kN are chosen and arranged in a pattern of n = 2 by m = 3 as shown above with pile centre-to-centre arbitrarily taken to be 0.72 m. The piles are assumed to rest on the very stiff soils at 25 m below ground surface.

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(a) Check against local failureEquivalent pile diameter d for the precast pile

= [(0.32 x 0.32)/] x 2 = 0.361 mPile centre-to-centre spacing is hence 2 pile diameters.

Refer to Fig. 2.5 -- If pile number is less than 9, use 3 x 3pile group lines as this would be conservative. If pile length isshorter than 24 pile diameters, use 24 diameter lines as this would also be conservative. If pile number is greater than 81, use 9 x 9 pile group lines as the pile group efficiency for larger pile groups does not decrease much. If pile length is longer than 48 diameters, use 48 diameter lines for similar reason.


In the present case, pile number is 6 (48). Therefore use the 48 pile diameter 3x3 pile group line and

pile group efficiency = 0.7 for pile spacing of 2 diameters from Fig. 2.5

Pile group geotechnical capacity = m n Qu= 0.7 x 3 x 2 x 2956 kN = 12415 kN

Factor of safety against local failure = 12415/6000 > 2 OKAY

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(b) Check against block failureAssume failure as a single block comprising all 6 piles.

Length of block Lg (consisting of 3 piles) = 0.32/2 + 2 x 0.72 + 0.32/2 = 1.76 m

Width of block Bg (consisting of 2 piles)= 0.32/2 + 0.72 + 0.32/2 = 1.04 m

Geotechnical capacity of single block= block shaft resistance + block base resistance= cu[2D(Bg + Lg)] + cb Nc Bg Lg= 1 x 50 x 2 x 5 (1.04 + 1.76) + 0.85 x 100 x 2 x

(25 5) (1.04 + 1.76) + 500 x 9 x 1.04 x 1.76 = 1400 + 9520 + 8237 = 19157 kN (1st component may be ignored for crude estimation)


Note: For , cu, pile contact length for each soil layer, refer to Example 2.1

Factor of safety against block failure = 19157/6000 >> 2 OKAY

Posers: (1) Under what pile group configuration and soil conditions

that block failure would be a concern in pile design.(2) If the factor of safety of either local or block failure falls

below 2, suggest means to improve the situation!

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2.9 Uplift Capacity

Piles may be required to resist uplift forces structure foundation subjected to large overturning moments.

Examples : tall chimneys, transmission towers, or jetty structure.

If uplift load is temporary (e.g. wind load) engineers may use a lower safety factor.

If uplift load is permanent higher safety factor is desired.


Fig 2.6 Lateral Load on Raked Piles

.

Lateral Load

Tension Compression

Lateral Load

Pile in Tension (Uplift)

Pile in Compression

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Single Pile Ultimate uplift resistance = pile weight + mobilized negative shaft resistance. Actual shaft resistance in uplift may be < resistance in compression uplift resistance = x calculated shaft resistance for downward loading (this can be a conservative estimation)

Best to design piles subjected to uplift loading by uplift testing Structural capacity should be checked to ensure sufficient reinforcement in concrete piles. Additional resistance through underreaming or enlarging pile base. In such case, uplift capacity produced by pile shaft can be ignored.


Approximate approach to determine the uplift capacity of pile with underream (Meyerhof & Adams)

(1)Short term stability (soil parameter cu)

Uplift capacity of pile in clay is given by the lesser of(a) S x k + W

where S = Shear resistance of a vertical cylinder above the base,k = 1 to 1.25 for soft clays

0.7 for medium stiff clays0.5 for stiff clays0.25 for stiff fissured clays (low value due to

premature tension failure of clay) W = weight of soil and pile above the base.

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(b) The uplift capacity of the base plus W, i.e. Qu = (/ 4) (db2 d2) cu Nc + W

where db = diameter of underream,d = diameter of pile shaft,

and Nc = bearing capacity factor (as for downward load)

Underream diameter db

Pile shaft diameter d


(2) Long term stability (Soil parameters c and )The ultimate uplift capacity is given by the lesser of (a) The sum of net base bearing capacity, side adhesion of shaft and pile weight. i.e.

Qu = (/4) (db2 d2) (c Nc + vb Nq) + As qs + Wwhere Nq = bearing capacity factor (as in downward loading), and

vb = effective vertical stress at level of pile base.

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(b) By assuming underream behave as an anchor, for shallow underream (pile length, L < db )

Qu = c db L + s (/2) db L2 Ku tan + Wfor L > limiting height of failure surface above base, H

Qu = c db H + s (/2) db (2L H) H Ku tan + Wwhere s = shape factor = 1 + mL/db

with a maximum value of 1+ mH/db , = soil unit weight (use for soil below water table),m = coefficient depending on (Table 2.4), andKu = earth pressure coefficient (approximately 0.9 to 0.95 for value between 25 and 40)


Table 2.4 Factors for uplift analysis (from Meyerhof and Adams, 1968)

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Pile GroupUltimate uplift capacity of pile group is the lesser of(Meyerhof & Adams)

(1) Sum of uplift capacity of individual piles.(2) Uplift capacity of an equivalent pier foundation

consisting of piles and enclosed soil mass


Example 2.2 (continued)

An isolated 350 mm x 350 mm reinforced concrete pile (fcu = 35) in a jetty structure is required to carry a net uplift load of 320 kN. Using the same soil properties and pile penetration depth, evaluate the factor of safety against uplift failure of the pile.

Taking the uplift resistance as half of the compressive shaft resistance (may be conservative) -- Qu = 0.5 (334 + 560) = 447 kNFactor of safety against uplift failure = 447/320 = 1.40This is not acceptable. (a) Either increase the number of piles to enhance uplift resistance, or (b) Further evaluate the uplift shaft resistance which may be more than half of the compressive resistance.

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.

Slender Pile

Very soft soil

Stiff soil

Need to check buckling of Pile

Slender Column

Beam

Beam

2.10 Buckling of Piles

May be experienced by long and slender piles driven in soft soils


(a) Single piles (Whitaker)Minimum buckling load, Qcr = 2 (kEI)where k = coefficient of soil lateral reaction

= 67 cu (Davisson, 1970)E = modulus of elasticity of pile materialI = second moment of area of pile section

(b) Pile GroupIf piles are closely spaced, critical buckling load is reduced. The value of k can be arbitrarily reduced (Poulos & Davis) by 2 times for 2-pile group,

by 3 times for 3pile & 4pile groups, andby 4 times for 5 or more piles in a group.

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Example 2.5 Buckling of pile

Using the same data as Example 1, evaluate the buckling resistance of the pile.

Minimum buckling load, Qcr = 2 (kEI)Where k = 67 cu = 67 x 50 = 3350 kPa

Econc = 30 x 106 kPaI = 0.32 x 0.323/12 = 8.74 x 10-4 m4

Thus Qcr = 18742 kN which >> 1000 kN working load OKAY!

Take note that even if Qcr is smaller than the working load, the pile may not buckle but further checks are warranted, refer to Poulosand Davis for more details.

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Piles - 2