Deep Foundations

Deep Foundations

Axial Load Capacity based on Analytical Methods (Chapter 14) Downdrag Loads (Chapter 18)

Deep Foundation Load Transfer P

P

Ps

Ps

Hard Stratum P t’

P t’ CV3301 - LEC (2008)

Lecture 6

2

Toe Bearing Resistance

q ult = cN c + qN q + 0.5γBN γ

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Toe Bearing Resistance – Sands (Driven Piles) * * q t ' = BγN γ + σ'zD N q

When D/B > 5, first term is negligible.

N *γ and N*q depends on both shear strength and compressibility as any three modes of shear failure governs. Need to define a rigidity index, I r : E Ir = 2(1 + ν )(σ'zD tan φ') CV3301 - LEC (2008)

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Toe Bearing Resistance – Sands (Driven Piles) (cont’d) Typically 10 < I r < 400. High I r ⇒ general shear failure Low I r ⇒ soil compressibility is important, local or punching shear E can be estimated as E s of Schmertmann analysis or from SPT and CPT.

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Toe Bearing Resistance – Sands (Driven Piles) (cont’d)

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Example 14.1 – A 400-mm square prestressed concrete pile is to be driven 19 m into the soil profile shown in Figure 14.6. Compute the net ultimate toe-bearing capacity. The water table is at a depth of 3m below the ground surface. σ'zD = ∑ γH − u = (17.8)(3) + (18.2 )(16 ) − (9.8)(16 ) = 187.8 kPa E s = β0 OCR + β1 N 60

(See Eq. 7.17)

= (5000 ) 1 + (1200 )(25) = 35000 kPa E Ir = 2(1 + ν )σ'zD tan φ' 16 m

=

35000 = 99 o 2(1 + 0.3)(187.8) tan 36

From figures, N *γ = 13.8 and N*q = 75 q 't = BγN *γ + σ'zD N*q

= (0.4 )(18.2 − 9.8)(13.8) + (187.8)(75) = 14131 kPa

(

)

P't = (14131) 0.4 2 = 2261 kN

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Toe Bearing Resistance – Sands (Drilled Shaft) For N 60 < 50, q't = 57.5 N 60 ≤ 2900 kPa where N 60 is mean SPT N between toe and depth 2Bb below toe Bb is base diameter q't is assumed to be mobilized within 25 mm settlement 1200 For B b > 1200 mm, reduce q't to q'tr = q 't Bb

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Toe Bearing Resistance – Clays For s u < 250 kPa , q t ' = s u N *c

(O' Neill and Reese 1999)

where : N*c = 6.5 for s u = 25 kPa N*c = 8.0 for s u = 50 kPa N*c = 9.0 for s u ≥ 100 kPa s u is undrained shear strength between toe and 2Bb below toe

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Toe Bearing Resistance – Clays (cont’d) For Bb > 1900 mm, Reduce q't to q'tr = Fr q't 2.5 Fr = ≤ 1.0 ψ1Bb + 2.5ψ 2

( Bb in m )

⎛D⎞ ψ1 = 0.28Bb + 0.083⎜⎜ ⎟⎟ ⎝ Bb ⎠ ψ 2 = 0.065 s u

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(s u in kPa )

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A driven pile in clay is shown in the figure below. The pile has a diameter of 406mm. Determine the net toe bearing capacity.

π 2 π 2 D = (0.406 ) = 0.1295m 2 4 4 Pt ' = A t q't = A t s u N*c

At =

= (0.1295)(100 )(9 ) = 116.6kN

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Toe Bearing Resistance – Intermediate Geomaterials O’Neill and Reese (1999): Hard soils, soft rocks: N60 > 50 250 kPa < su < 2500 kPa Rocks: su ≥ 2500 kPa (qu ≥ 5000 kPa)

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Toe Bearing Resistance – Cohesive Intermediate Geomaterial and Rock O' Neill and Reese (1999) : For RQD = 100% and D/B ≥ 1.5, q't = 2.5q u For 70% < RQD < 100% (all joints closed and nearly horizontal) and q u > 500 kPa, q't = 4830(q u )

0.51

If material is jointed with joints having random orientation and where joints' condition can be evaluated :

[

(

q't = t 0.5 + mt 0.5 + t

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)

0.5

]q

u

Lecture 6

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Toe Bearing Resistance – Cohesive Intermediate Geomaterial and Rock

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Toe Bearing Resistance – Cohesive Intermediate Geomaterial and Rock

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Toe Bearing Resistance – Noncohesive Intermediate Geomaterial O' Neill and Reese (1999) : q't = 0.59[(N1 )60 ] σ'zD 0.8

where N1 = N 60

100 kPa σ'zD

1200 mm For B b > 1200 mm, reduce q't to q'tr = q't Bb CV3301 - LEC (2008)

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A steel H-pile (At = 0.52m2) is driven to a layer of sandstone. The unconfined compressive strength of the sandstone is qu = 17 MPa. Estimate the net toe bearing capacity.

[

Pt ' = A t q 't = A t 4830(q u )

0.51

= (0.052 )(4830 )(17000 )

]

0.51

= 36098kN

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Contact Areas At and As In open-ended piles, the soil plug may be considered rigidly embedded if D/B > 10 to 20 for clays D/B > 25 to 35 for sands

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Review „

„

What is the difference in bearing capacity of shallow foundations and toe bearing resistance of pile? How do you compute the toe bearing resistance for a pile in different soil types?

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Side Friction „ „

Effective Stress Analysis – β Method Total Stress Analysis – α Method

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Side Friction – Effective Stress Analysis P σ’x fs P+ΔP

σ' x f s = σ'x tan φf , K = σ' z ⎛ K ⎞ ⎡⎛ φf ⎞ ⎤ ⎟⎟ tan ⎢⎜⎜ ⎟⎟φ'⎥ f s = K o σ'z ⎜⎜ ⎝ K o ⎠ ⎣⎝ φ' ⎠ ⎦ - β Method = βσ 'z ⎛ K ⎞ ⎡⎛ φf ⎞ ⎤ ⎟⎟ tan ⎢⎜⎜ ⎟⎟φ'⎥ i.e. β = K o ⎜⎜ ⎝ K o ⎠ ⎣⎝ φ' ⎠ ⎦

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Side Friction – Effective Stress Analysis (cont’d)

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Side Friction – Effective Stress Analysis (cont’d)

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Displacement piles „

„

Large-displacement piles include all solid piles such as precast concrete piles, and steel or concrete tubes closed at the lower end by a driving shoe or a plug, i.e. cast-inplace piles. Small-displacement piles include rolled steel sections such as H-piles and open-ended tubular piles. However, these piles will effectively become large displacement piles if a soil plug forms.

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Construction of Piles – Jetted piles Jetting is the process of forcing water under pressure around and under the pile to lubricate and/or to displace the surrounding soil

From www.fhwa.dot.gov

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Construction of Piles – Driven piles

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Construction of Piles – Driven piles

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Construction of Piles – Drilled Shaft

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Construction of Piles – Drilled Shaft

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Side Friction – Effective Stress Analysis (cont’d) Typical β values: For sands: 0.25 ≤ β ≤ 1.20 For gravels (> 50% gravel size): 0.25 ≤ β ≤ 3.00 For gravelly sands ( 25 to 50% gravel size): 0.25 ≤ β ≤ 1.80 For silts: 0.27 ≤ β ≤ 0.50 For clays: 0.25 ≤ β ≤ 0.35

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Example: A concrete pile is 25 m long and 305 mm x 305 mm in cross section. The pile is fully embedded in sand, for which γ = 17.5 kN/m3 and φ’ = 35o. The groundwater table is at the surface. Calculate the total side resistance of the pile for K = 1.3 and φf = 0.8φ’. ⎛ K ⎞ ⎡⎛ φf ⎞ ⎤ ⎟⎟ tan ⎢⎜⎜ ⎟⎟φ'⎥ = K tan φf = 1.3 tan (0.8)(35o ) = 0.691 β = K o ⎜⎜ ⎝ K o ⎠ ⎣⎝ φ' ⎠ ⎦ ⎛ 25m ⎞ Average σ z ' = γ ' z centre = 17.5kN / m 3 − 9.8 ⎜ ⎟ = 96.25kPa ⎝ 2 ⎠ f s = βσ z ' = (0.691)(96.25kPa ) = 66.5kPa

[

(

]

)

Total side resistance = A s f s = 4BLfs

= 4(0.305m )(25m )(66.5kPa ) = 2028kN

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Side Friction – Total Stress Analysis (α Method for insensitive clays, St < 4)

f s = αs u

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Side Friction – Total Stress Analysis (α Method) (cont’d) Most commonly used α is from API (1974) : For s u < 25 kPa : α = 1.0 ⎛ s u − 25 kPa ⎞ ⎟⎟ For 25 kPa < s u < 75 kPa : α = 1.0 - 0.5⎜⎜ ⎝ 50 kPa ⎠ For s u > 75 kPa : α = 0.5

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Side Friction – Total Stress Analysis (α Method) (cont’d) For drilled shafts

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Side Friction – Total Stress Analysis (α Method) (cont’d) For compression load For drilled shafts (O’Neill and Reese 1999): maximum fs = 260 kPa See Example 14.4

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Underreamed Pile

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Example: A driven pile in clay is shown in the figure below. The pile has a diameter of 406mm. Determine the total side resistance by the α method using API formulas. ⎛ s − 25kPa ⎞ α1 = 1 − 0.5⎜ u ⎟ = 0.95 ⎝ 50kPa ⎠ Ps1 = A s1f s1 = πDL1α1s u

= π(0.406 )(5m )(0.95)(30kPa ) = 181.8kN

⎛ s − 25kPa ⎞ α 2 = 1 − 0.5⎜ u ⎟ = 0.95 ⎝ 50kPa ⎠ Ps 2 = A s 2 f s 2 = πDL 2 α 2s u

= π(0.406)(5m )(0.95)(30kPa ) = 181.8kN

α 3 = 0.5 Ps 3 = A s 3f s 3 = πDL3α 3s u

= π(0.406 )(20m )(0.5)(100kPa ) = 1275.5kN

Ps = Ps1 + Ps 2 + Ps 3 = 181.8 + 181.8 + 1275.5 = 1639.1kN CV3301 - LEC (2008)

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Upward Load Capacity When a deep foundation is subjected to downward loads, it experience some elastic compression and, due to Poisson effect, a small increase in diameter. The opposite occurs when the pile is subjected to upward loading. For this reason, fs may be conservatively reduced by 25% for design i.e. f = 0.75f

( s )upward

s

Alternatively, a higher factor of safety is usually used when computing the upward loading capacity of the pile. CV3301 - LEC (2008)

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Upward Load Capacity (cont’d) Additional capacity due to enlarged base may be estimated for clays as (O' Neill and Reese 1999) :

(P

)

upward a

=

(s u N u + σ zD )⎛⎜ π ⎞⎟(B2b − Bs2 )

α=0

1.5m

⎝4⎠ F

For unfissured clays, N u = 3.5

Db ≤9 Bb α=0

D For fissured clays, N u = 0.7 b ≤ 9 Bb

2Bb

Important: Neglect fs from bottom to 2Bb from bottom See Tutorial Q11.1 CV3301 - LEC (2008)

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Pile Groups „ „

„

„

A single pile usually does not have enough capacity Piles are located with low degree of precision and can easily be 150 mm or more from the desired location. The eccentricity would generate unwanted moments and deflections in both pile and column. Multiple piles provide redundancy, and thus can support the structure even if one pile is defective. The lateral compression during pile driving is greater and therefore side friction capacity can be greater than that of a single pile.

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Typical configurations of pile caps

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Pile supported or pile-enhanced mats

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Pile Group

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Group Effects Piles are usually installed in groups of three or more. If the piles are too close (< 2 to 2.5 B or 600 mm), there may not be enough room for positioning and alignment errors. If the pile spacing, s, is too wide, pile cap will be large and expensive. Typically, 2.5 < s/B < 3.0 CV3301 - LEC (2008)

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Group Effects (cont’d) In a pile group, there are interactions between the piles and the adjacent soil and Pultg ≠ NPult Therefore,

Pag = ηNPa where η = group efficiency factor

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Group Effects (cont’d) η depends on several factors, including: „ The number, length, diameter, arrangement, and spacing of piles „ The load transfer mode (side friction vs end bearing) „ The construction procedures used to install the piles „ The sequence of pile installation „ The soil type „ The elapsed time since the piles were driven „ The interaction, if any, between the pile cap and the soil „ The direction of applied load

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Group Effects (cont’d) Considering pile group geometry, Converse - Labarre Formula (Bolin 1941) :

( n - 1)m + (m − 1)n η = 1- θ

90 m n where m = number of rows of piles n = number of piles per row ⎛B⎞ θ = tan -1 ⎜ ⎟ (in degrees) ⎝s⎠ B = pile diameter s = center - to - center spacing of piles

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Group Effects (cont’d) Considering individual pile failure and group failure : 2s(m + n − 2 ) + 4B η= ≤1 πmn B (n − 1)s + B (m − 1)s + B

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Pile Group Failure

Qs

Block failure : Q t + Qs η= ≤1 NPsin gle

Qt CV3301 - LEC (2008)

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Group Effects (cont’d) Tests in sands suggest (O’Neill 1973): „

„

„

In loose sands, η ≥ 1 and reaches a peak at s/B ≈ 2. It also seems to increase with no. of piles in the group. In dense sands with 2 ≤ s/B ≤ 4, η is usually slightly greater than 1 so long as pile was installed without predrilling or jetting. Piles installed by predrilling or jetting, and drilled shafts have lower η, as low as 0.7.

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Group Effects (cont’d) Tests in clays suggest : „

„ „

Generally, η < 1 and decreases with no. of piles in the group. η can be as low as 0.5. η increases with time.

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Settlement „

„

Most piles designed as covered so far will not have settlement greater than 25 mm – acceptable for nearly all structures. Therefore settlement computations are usually not needed.

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Settlement (cont’d) However certain conditions can produce excessive settlement and warrants evaluation: „ „

„

„

The structure is especially sensitive to settlements. The foundation has a large B and a large portion of the allowable capacity is due to toe bearing. One or more compressible strata are present, especially if these strata are below the toe. Downdrag loads might develop during the life of the structure.

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Load-Settlement Response The load settlement response may be approximated as follows (adapted from Fellenius 1999) :

(q't )m q't

(f s )m fs

⎛ δ ⎞ = ⎜⎜ ⎟⎟ ⎝ δu ⎠

g

⎛ δ = ⎜⎜ ⎝ δu

h

⎞ ⎟⎟ ≤ 1 ⎠

where :

(q't )m = mobilized net unit toe bearing resistance (f s )m = unit side - friction resistance δu

= settlement required to mobilize ultimate resistance = B/10 for toe bearing = 10 mm for side friction

g h

= 0.5 (clay) - 1.0 (sand) = 0.02 - 0.5

CV3301 - LEC (2008)

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Load-Settlement Response (cont’d) Deep foundations also experience elastic compression, which is another source of " apparent" settlement : δe =

Pz c AE

where : δ e = settlement due to elastic compression of foundation z c = depth to centroid of soil resistance (typically 0.75D) A = cross - sectional area of a single foundation E = modulus of elasticity of foundation = 200 GPa for steel = 4700 f'c MPa for concrete See Example 14.7 in book

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Load-Settlement Response for Drilled Shafts – O’Neill and Reese (1999) Method

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Load-Settlement Response – O’Neill and Reese (1999) Method (cont’d) See Example 14.8

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Load-Settlement Response – Pile Group Imaginary footing method (Equivalent raft method)

(See Tutorial Q11.3)

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Downdrag Loads (Negative Skin Friction)

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Downdrag Loads (cont’d) Both β and α methods may be used to compute negative skin friction Negative friction load, Psn = ∑ f sn A sn

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Downdrag Loads – Design by CP4 (2003) For design purposes, Depth to neutral plane = 0.6 L s for friction pile = 1.0 L s for end - bearing pile Allowable structural capacity, f 'c A (Pa )st = ≥ P + ζPsn F(= 4) where ζ is the degree of mobilization, typically 0.67, although 1.0 may be used for special cases involving low capacity piles in highly compressible clay stratum. CV3301 - LEC (2008)

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Downdrag Loads – Design by CP4 (2003) (cont’d) Allowable geotechnical capacity, Pt '+ Psp ≥ P + ζPsn Pa = F where F may be taken as between 2.0 to 2.5.

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Downdrag Reduction Techniques „

„

„ „ „

„

Coat the pile with bitumen. This method is very effective so long as the pile is not driven through an abrasive soil, such as sand, that might scrape off the bitumen coating. Drive the piles before placing the fill, wrap the exposed portions with lubricated polyethylene sheets or some other low-friction material and place the fill around the piles. Use a large diameter predrill hole, possibly filled with bentonite, thus reducing K. Use a pile tip larger in diameter than the pile, thus making a larger hole as the pile is driven. Drive an open-end steel pipe pile through the consolidating soils, remove the soil plug, then drive a smaller diameter load-bearing pile through the pipe and into the lower bearing strata. This isolates the inner pile from the downdrag loads. Accelerate the settlement using surcharge fills or other techniques, and then install the foundations after the settlement is complete.

CV3301 - LEC (2008)

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Review „

„

„ „ „ „

How do you compute side resistance using the α and β methods? How do you compute the load settlement behaviour of pile? How do you compute the uplift resistance of a pile? What is negative skin friction? How do you compute negative skin resistance? What methods can be used for reducing negative skin resistance?

CV3301 - LEC (2008)

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