Worked Examples Design of Pile Foundations

Eurocodes: Background & Applications GEOTEC GEO TECHNI HNICAL CAL DESIGN DESIGN wi th wor ked ex amples

13-14 1314 Jun e 2013, 2013, Dubli n

Worked examples – es gn o p e f n i n Dr. Trevor Orr Trinity College Dublin Convenor SC7/EG3


Worked example 1 – desi de si n of il ile e fou founda ndati tion ons s fr from om lo load ad te test sts s

DESIGN SITUATION

13-14 1314 Jun e 2013, 2013, Dubli n


Worked example 1 – desi de si n of il ile e fou founda ndati tion ons s fr from om lo load ad te test sts s

DESIGN SITUATION

13-14 1314 Jun e 2013, 2013, Dubli n

Eurocodes: Background & Applications w

w o r e ex am p es

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Design situation for a pile designed from static •

Pile foundations are required to support the following loads from a buildin : •

Characteristic permanent vertical load

Gk = 6.0 MN =

•

.

•

It has been decided to use bored piles 1.2m in diameter and 15m

•

The pile foundation design is to determine how many piles are requ re

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©2013 Trevor Orr. All rights reserved.


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Pile load test results

•

e resu ts o oa -sett ements curves are plotted in Figure 1

• Ado t settlement of the ile top equal to10% of the pile base diameter as the "failure" . . .

0,5

1

1,5

2

n

oa ,

2,5

3

0

20

40

m 60 m , t n e 80 m e l t t e S100

120

140

160

4

u

Pile settlements 0

• Load tests have been erformed on site on four iles of the same diameter and same length

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Pile 2 Pile 3 Pile 4 ean

Eurocodes: Background & Applications GEOTECHNICAL DESIGN wi th wor ked ex amples

Worked example 1 – desi n of ile foundations from load tests

SOLUTION

13-14 Jun e 2013, Dubli n


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Measured pile resistances •

Adopting the pile load at a settlement of the top of the piles equal to 10% of the ile diameter as the ultimate resistance means using the measured resistances at a settlement of: 12.0 x (10/100) x 103 = 120mm

•

From the load-settlement graphs for each pile this gives: Pile 1 Pile 3 Pile 4

•

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Rm = 2.14 MN . m Rm = 1.73 MN Rm = 2.33 MN

Hence the mean and minimum measured pile resistances are : Rm, mean = 2.04 MN = . m, min ©2013 Trevor Orr. All rights reserved.


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Characteristic resistance •

The characteristic pile resistance is obtained by dividing the mean and minimum measured pile resistances by the correlation factors ξ1 and ξ2 and choosing the minimum value: R c; k

(R ) ⎫ ⎧ (R ) = Min ⎨ c;m mean ; c; m min ⎬ 1

2

•

For four load tests, recommended ξ1 and ξ2 values are: = 1.1 ξ2 = 1.0

•

Hence the characteristic pile resistance: R c;k

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⎧ 2 .04 1 .73 ⎫ = Min ⎨ ; ⎬ = Min {1 .85;1 .73 } = 1 .73 ⎩ 1 .1 1 . 0 ⎭



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Design Approach 1 partial factors •

•

•

Combinations of sets of partial factors DA1.C1

A1 “+” M1 “+” R1

DA1.C2

A2 “+” M1 or M2 “+” R4

Partial actions factors γG = . A2 γG = 1.0

γQ = . γQ = 1.3

Partial material factors M1 and M2 not relevant

•

Partial resistance factors R1 γt = 1.15 (Total/combined compression) R4

8

(γφ’ = 1.0 and not used)

γt = 1.5


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Design Approach 1 - pile design Design equation DA1.C1 DA1.C1

Fc,d

≤ R c,d

Fc,d = 1.35 Gk+1.5 Qk = 1.35x6.0+1.5x3.2 Fc,d = 1.0 Gk+1.3 Qk = 1.0x6.0+1.3x3.2

For a single pile DA1.C1 DA1.C2

= 12.9 MN = 10.2 MN

Rc,d = Rc,k /γt = 1.73 / 1.15 = 1.50 MN Rc,d = Rc,k /γt = 1.73 / 1.5 = 1.15 MN

Assuming no pile group effect, for n piles, resistance = n x R c,d Hence

DA1.C1

n

≥ Fc,d /Rc,d

= 12.9/1.5 = 8.6

DA1.C2

n

≥ Fc,d /Rc,d

= 10.2/1.15 = 8.9

Therefore DA1.C2 controls and no. piles required: n = 9 9



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Combination of sets of partial factors DA2

•

•

A1 “+” M1 “+” R2

Partial actions factors γG = 1.35 A1 Partial material factors M1 not relevant

•

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γQ = 1.5 (γφ’ = 1.0 and not used)

Partial resistance factor R2 γt = 1.1 (Total/combined compression)


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Design Approach 2 - pile design Design equation

Fc,d

≤ R c,d

Fc,d = 1.35 Gk + 1.5 Qk = 1.35·x 6.0 + 1.5·x 3.2 = 12.9 MN For a single pile Rc,d = Rc,k /γt = 1.73 / 1.1 = 1.57 MN Assuming no pile group effect, for n piles, resistance = n x R c,d Hence

Fc,d

≤ n

T ere ore no. pi es require : n = Fc,d

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R c,d

Rc,d = 12.9

1.57 = 9 p es


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Combination of sets of partial factors: DA3

A1 “+” M1 “+” R3

•

Partial resistance factor: γt = 1.0 (Total/combined compression) R3

•

Since the R3 recommended partial resistance factor is equal to . , to calculated the design pile resistance from pile load test results

•

Design Approach 3 and the recommended partial resistance factor

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Conclusions from pile worked example 1 •

The same design number of piles, 9 is obtained for both DA1 and DA2

•

Since the recommended partial resistance factors are 1.0 for DA3, this Desi n A roach should not be used for the desi n of iles from pile load test results unless the partial resistance factors are increased

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Worked example 2 – desi n of ile foundations from test rofiles

DESIGN SITUATION



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Design situation for a pile designed from a •

The piles for a building are each required to support the following loads: •


Gk = 300 kN

•

Characteristic variable vertical load

Qk = 150 kN

•

The ground consists of dense sand beneath loose sand with soft clay and peat to 16.5m as shown in figure on next slide

•

It has been decided to use 0.45m diameter bored piles

•

The pile foundation design involves determining the length of the pi es

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B’hole log/CPT profile • 1 CPT was carried out

Loose sand, soft clay and some

Cla with peat seams

Cautious average qc = 2.5 MPa •

ronger ayer o me um o dense sand starts at depth of 16.5m Cautious average qc = 12.5 MPa

16.5m

Medium to dense sand

• Assume the soil above 16.5m prov es no s a res s ance 16

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qc

•

loose sand, soft clay and some peat over 5.5m of clay with peat seams

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Unit pile resistances • The base and shaft pile resistances

are calculated using Tables D.3 . single cautious average qc value in stronger soil to the unit base and , b s • Assume the ULS settlement of the

pile head, sg so that the normalised settlement is 0.1 • Interpret linearly between relevant

values to obtain these tables: pb = 2.5 MPa . s 17

and

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Table D.3 Unit base resistance p b of cast in-situ piles in coarse soil with little or no fines

Unit base resist ance p b , in MPa, at average cone penetration resistance q c (CPT) in MPa qc = 10 qc = 15 qc = 20 qc = 25 , , , , , 0,03 0,90 1,35 1,80 2,25 0,10 (= sg) 2,00 3,00 3,50 4,00 NOTE Intermediate values may be interpolated linearly. In the case of cast in-situ piles with pile base enlargement, the values , . s is the normalised pile head settlement Ds is the diameter of the pile shaft Db is the diameter of the pile base sg is the ultimate settlement of pile head Normalised settlement s /Ds ; s /Db

Table D.4 Unit shaft resistance p s of cast in-situ piles in coarse soil with little or no fines

Av erage co ne penetrat io n resistance q c (CPT) MPa 0 5 10 NOTE

Unit shaft resist ance p s

MPa 0 0,040 0,080 , Intermediate values may be interpolated linearly



Worked example 2 – desi n of ile foundations from test rofiles

SOLUTION



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Characteristic pile resistance Pile diameter Pile base cross sectional area

D = 0.45m Ab = π x 0.452 / 4 = 0.159 m2 2 . = . s = Length of pile in stronger layer providing shaft resistance = L s Calculated compressive pile resistance for the one profile of test results: Rc;cal = Rb;cal + Rs;cal

= A b x p b + As x L s x ps = (0.159 x 2.5 + 1.414 x L s x 0.10) x 103 kN =

+

Hence, applying the recommended correlation factors ξ3 and ξ4, which are both the same and equal to 1.4 for one profile of test results because the mean and m n mum ca cu a e res s ances are e same so a 3 an 4 = = . an the characteristic base and shaft compressive pile resistances are:

19

R b;k = Rb;cal /ξ = 398/1.4

= 284 kN

R s;k = Rs;cal /ξ = 141 x Ls /1.4

= 101 Ls



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•

•

Combinations of sets of partial factors: DA1.C1

A1 “+” M1 “+” R1

DA1.C2

A2 “+” M1 or M2 “+” R4

Partial actions factors: γG = . A2 γG = 1.0

γQ = . γQ = 1.3

Partial material factors: M1 and M2 not relevant

•

Partial resistance factors: R1 γb = 1.25 R4

20

γb = 1.6

(γφ’ = 1.0)

γs = 1.0 γs = 1.3


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Fc,d

≤ R c,d

es gn ac ons DA1.C1 Fc,d = 1.35 Gk+1.5 Qk = 1.35x300+1.5x150 DA1.C1 Fc,d = 1.0 Gk+1.3 Qk = 1.0x300+1.3x150

= 630 kN = 495 MN

Design resistances DA1.C1 Rc,d = Rb,k /γb + Rs,k /γs = 284/1.25 + 101 x L / s 1.0 . . . c,d b,k b s,k s s Equating actions and resistances DA1.C1 630 = 284/1.25 + 101 x L / 1.0

→

DA1.C2

495 = 284/1.6 + 101 x L / s 1.3

Hence DA1.C2 controls and DA1 desi n 21

→

L = 3.99 m Ls = 4.08 m

ile len th L = 16.5 + L = 21m



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Design Approach 2 - pile design Partial action factors same as for Example 1 Partial resistance factors: γb = 1.1 R2 Design equation

Fc,d

γs = 1.1 ≤ R c,d

Design action Fc;d = 1.35 Gk+1.5 Qk = 1.35x300+1.5x150

= 630 kN

Design resistance Rc;d = Rb;k /γb + Rs;k /γs = 284/1.1 + 101 x Ls/ 1.1 quat ng act ons an res stances 630 = 284/1.1 + 101 x L / s 1.1

Ls = 4.05 m

→

. 22


s


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Design Approach 3 - pile design •

Combination of sets of partial factors: DA3 A1 “+” M1 “+” R3

•

Partial resistance factors: γb = γs = 1.0 R3

•

Since the R3 recommended partial resistance factors are both equal to 1.0, no safety margin is provided if these factors are test profile

partial resistance factors applied to the characteristic pile resistance obtained from a CPT test profile 23



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The same design pile length, 21m is required for both DA1 and DA2

•

Since the recommended partial resistance factors are 1.0 for DA3, this Desi n A roach should not be used for the desi n of iles from profiles of ground test results unless the partial resistance factors are increased

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Worked example 3 – design of pile foundations from soil parameters

DESIGN SITUATION

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Design situation for a pile designed from soil •

The piles for a proposed building in Dublin are each required to support the following loads: •


Gk = 600 kN

•

Characteristic variable vertical load

Qk = 300 kN

•

The ground consists of about 3m Brown Dublin Boulder Clay over Black Dublin Clay to great depth

•

It has been decided to use 0.45m diameter driven piles

•

The pile foundation design involves determining the length of the piles

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Characteristic undrained •

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Brown DBC

Figure shows plot of SPT N values Black DBC

•

Shaft resistance in Brown Dublin Boulder Clay is ignored

•

Average N va ue in B ac Du in Bou er Clay = 57

•

A cautious average N value = 45

•

PI of the Dublin Boulder Clay = 14%

•

From Stroud and Butler plot of f 1 vs. N op

•

27

1

=

Hence characteristic undrained shear strength cu;k = f 1 x N = 6 x 45 = 270 kPa ©2013 Trevor Orr. All rights reserved.

Average N = 57 =

cu = f 1 x N


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Pile resistances Pile diameter

D = 0.45m

Pile base cross sectional area

Ab = π x 0.452 / 4 = 0.159 m2

Pile shaft area per metre length

As = π x 0.45 = 1.414 m2 /m

Length of pile in Black Dublin Clay providing shaft resistance = L s The characteristic unit pile base and shaft resistances, q b;k and qs;k are obtained as follows: qb;k = Nq x cu

= 9 cu

qs;k = α x cu

= 0.4 x cu

Characteristic base resistance b;k

b

b;k

.

Characteristic shaft resistance Rs k = As x Ls x qs k = 1.414 x Ls x 0.4 x 270 28


= 153 Ls


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Design Approach 1 partial and model factors •

•

•

Combinations of sets of partial factors: DA1.C1

A1 “+” M1 “+” R1

DA1.C2

A2 “+” M1 or M2 “+” R4

Partial actions factors: γG = . A2 γG = 1.0

γQ = . γQ = 1.3

Partial material factors: M1 and M2 not relevant (resistances factored not soil parameters)

•

•

Partial resistance factors: γb = 1.0 R1 γb = 1.3 R4

γs = 1.0 γs = 1.3

Model factors:

γR;d = 1.75 (In Irish NA and applied to γb, γs and γt) 29



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Design Approach 1 - pile design to Irish NA Design equation

Fc,d

≤ R c,d

Design actions DA1.C1 Fc;d = 1.35 Gk+1.5 Qk = 1.35 x 600+1.5 x 300 DA1.C1 Fc;d = 1.0 Gk+1.3 Qk = 1.0 x 600+1.3 x 300

= 1260 kN = 990 kN

Desi n resistances DA1.C1 Rc;d = Rb;k /(γb x γR;d) + Rs;k /(γs x γR;d) DA1.C2

R

= 386/(1.0x1.75)+ 153 x Ls /(1.0x1.75) = 221 + 87.4 Ls kN =R +R , = 386/(1.3x1.75) + 153 x Ls /(1.3x1.75) = 170 + 67.3 Ls kN

Equating actions and resistances . = + . DA1.C2

s

990 = 170 + 67.3 Ls

→

→

s

. m

Ls = 12.2 m

. 30

=

. ©2013 Trevor Orr. All rights reserved.

s

.


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Combination of sets of partial factors DA2 A1 “+” M1 “+” R2

•

Partial actions factors A1

•

•

•

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γG = 1.35

γQ = 1.5

Partial material factors M1 not relevant

(γφ’ = 1.0)

Partial resistance factor γb = 1.1 R2

γs = 1.1

o e actors: γR;d = 1.75 (In Irish NA and applied to γb, γs and γt) γR;d = 1.27 (In German NA giving γb = 1.1x1.27 = 1.4, γs = 1.1x1.27 = 1.4



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Design Approach 2 - pile design to Irish NA Design equation

Fc;d

≤ R c;d

Design action Fc;d = 1.35 Gk+1.5 Qk = 1.35 x 600+1.5 x 300

=

1260 kN

Design resistance Rc;d = Rb;k /(γb x γR;d) + Rs;k /(γs x γR;d) = 386 1.1 x 1.75 + 153 x L

1.1 x 1.75 = 201 + 79.5 L kN

Equating design action and resistance 1260 = 201 + 79.5 Ls

→

Ls = 13.3 m

Hence the DA2 design pile length L = 3.0 + L s = 16.5m

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Design Approach 2 - pile design to German NA Design equation

Fc;d

≤ R c;d

es gn ac on Fc;d = 1.35 Gk+1.5 Qk = 1.35 x 600+1.5 x 300

= 1260 kN

Rc;d = Rb,k /γb + Rs,k /γs Note: R1 partial factors of 1.1 have been increased to 1.4, i.e. a model . . = . = 386 /1.4 + 153 x Ls /1.4 = 276+109Ls kN qua ng es gn ac on an res s ance 1260 = 276 + 109 Ls

→

Ls = 9.0 m

. 33


s

.


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Combination of sets of partial factors: DA3

•

A1* or A2† “+” M2 “+” R3

Partial actions factors: γ = 1.35 A1 A2 γG = 1.0

•

Partial material factor: γcu = .

•

Partial resistance factors: R3

•

= 1.0

γ = 1.5 γQ = 1.3

= 1.0

Model factor:

γR;d = 1.75 (applied to γb, γs and γt)

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*

on structural actions

†

on geotechnical actions


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Fc,d

≤ R c,d

Desi n action Fc,d = 1.35 Gk+1.5 Qk = 1.35 x 600+1.5 x 300

= 1260 kN

Design resistance c,d

=

b,d

+

s,d

= Ab x qb;d + As x Ls x qs;d = (AbxNqxcu;k /γcu)/(γR;dxγb) + (AsxLsxαxcu;k /γcu)/(γR;dxγs) = (0.159x9x270/1.4)/(1.75x1.0) + (1.414xLsx0.4x270/1.4)/(1.75x1.0) = 158 + 62.3 Ls 1260 = 158 + 62.3 Ls

→

Ls = 17.7 m

Hence the DA3 design pile length L = 3.0 + L s = 21 m 35



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The design pile lengths obtained from ground strength parameters using the alternative procedure and the model factor in the Irish and German National Annexes are: −

DA1 (Irish NA)

L = 15.5 m

−

DA2 (Irish NA)

L = 16.5 m .

− −

DA2 (German NA)

L = 12.0 m

•

Application of the model factor of 1.75 as well as the material factor of 1.4 to obtain the design resistance when using DA3, results in DA3 providing a longer design pile length and hence the least economical Design Approach in Ireland

•

The longer design pile length of 16.5 m when using the Irish NA with DA2 compared to 12.0 m when using the German NA is because of the model factor of 1.75 in the Irish NA and 1.27 in the German NA

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Worked Examples Design of Pile Foundations

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