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# Mathcad - Ex 13.4 CFA pile design from cone tests.xmcd by panniuniu

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```									                             Example 13.4
Design of continuous flight auger piles from cone tests
Verification of strength (limit state GEO)
Design situation
Consider the design of continuous flight auger (CFA) piles for a site in
Twickenham, London. Ground conditions at the site comprise dense, becoming
loose gravelly, SAND. Cone penetration tests have been perfomed at the site
to a depth of 8m. (Data courtesy CL Associates.) The limiting average unit
shaft resistance p s and limiting unit base resistance pb at each cone location
are estimated to be:
⎛120kPa ⎞            ⎛2800kPa ⎞
⎜       ⎟          ⎜        ⎟
⎜120kPa ⎟          ⎜3000kPa ⎟
ps =   ⎜       ⎟   pb =   ⎜        ⎟
⎜100kPa ⎟          ⎜2000kPa ⎟
⎜120kPa ⎟          ⎜3000kPa ⎟
⎝       ⎠          ⎝        ⎠
A group of N = 6 piles with diameter D = 400mm and length L = 6m are
required to carry between them a permanent action FGk = 2100kN together

with a variable action FQk = 750kN. The weight density of reinforced

kN
concrete is γck = 25           (as per EN 1991-1-1 Table A.1).
3
m
Design Approach 1

Actions and effects
⎛ π × D2 ⎞
⎜        ⎟
The self-weight of pile is WGk =     ⎜ 4 ⎟ × L × γck = 18.8 kN
⎝        ⎠
⎛A1 ⎞      ⎛1.35 ⎞          ⎛1.5 ⎞
Partial factors from Sets      ⎜ ⎟: γG = ⎜      ⎟ and γQ = ⎜ ⎟
⎝A2 ⎠      ⎝ 1 ⎠            ⎝1.3 ⎠
Design total action per pile is:

Fcd =
(               )
γG × FGk + WGk + γQ × FQk ⎛664 ⎞
=⎜    ⎟ kN
N          ⎝516 ⎠

Calculated shaft resistance
Number of cone penetration tests n = 4
⎛905 ⎞
⎜    ⎟
⎜905 ⎟
Calculated shaft resistance R s =    π × D × L × ps = ⎜    ⎟   kN
⎜754 ⎟
⎜905 ⎟
⎝    ⎠

∑ Rs
Mean calculated shaft resistance R s,mean =                  = 867 kN
n

( )
Minimum calculated shaft resistance R s,min = min R s = 754 kN

Calculated base resistance
⎛352 ⎞
⎜    ⎟
⎛ π × D2 ⎞
⎜        ⎟   ⎜377 ⎟
Calculated base resistance R b =    ⎜ 4 ⎟ × pb = ⎜251 ⎟        kN
⎝        ⎠   ⎜    ⎟
⎜377 ⎟
⎝    ⎠

∑ Rb
Mean calculated base resistance R b,mean =               = 339 kN
n

( )
Minimum calculated base resistance R b,min = min R b = 251 kN

Calculated total resistance
Mean calculated total resistance R t,mean = R s,mean + R b,mean = 1206 kN

Minimum calculated total resistance R t,min = R s,min + R b,min = 1005 kN

Characteristic resistance
Correlation factor on mean measured resistance ξ3 = 1.31

Correlation factor on minimum measured resistance ξ4 = 1.20
For a pile group that can transfer load from weak to strong piles
(§7.6.2.2.(9)), ξ may be divided by 1.1 (but ξ 3 cannot fall beneath 1.0).
⎛ ξ3
⎜          ⎞
⎟                 ξ4
Thus ξ3 = max ⎜     , 1.0⎟ = 1.19 and ξ4 =     = 1.09
⎝ 1.1      ⎠                 1.1
R t,mean                   R t,min
Calculated resistances              = 1013 kN and              = 922 kN
ξ3                         ξ4
Characteristic resistance should therefore be based on the minimum value.
R s,min
Characteristic shaft resistance is R sk =         = 691 kN
ξ4

R b,min
Characteristic base resistance is R bk =             = 230 kN
ξ4

Design resistance
⎛R1 ⎞        ⎛1 ⎞          ⎛ 1.1 ⎞
Partial factors from Sets   ⎜ ⎟:    γs = ⎜ ⎟ and γb = ⎜      ⎟
⎝R4 ⎠        ⎝1.3 ⎠        ⎝1.45 ⎠
R sk     R bk ⎛901 ⎞
Design resistance is R cd =      +         =⎜     ⎟ kN
γs        γb     ⎝691 ⎠

Verification of compression resistance
Fcd ⎛74 ⎞
Degree of utilization ΛGEO,1 =     =⎜ ⎟ %
R cd ⎝75 ⎠

Design is unacceptable if degree of utilization is > 100%

Design Approach 2
Actions and effects
Partial factors from set A1: γG = 1.35 and γQ = 1.5

Design total action per pile is Fcd =
(            )
γG × FGk + WGk + γQ × FQk
= 664 kN
N

Design resistance
Characteristic shaft and base resistances are unchanged from DA1
Partial factors from set R2: γs = 1.1 and γb = 1.1
R sk R bk
Design resistance is R cd =     +      = 838 kN
γs     γb
Verification of compression resistance
Fcd
Degree of utilization ΛGEO,2 =          = 79 %
R cd

Design is unacceptable if degree of utilization is > 100%

Design Approach 3

Actions and effects
Partial factors from set A1: γG = 1.35 and γQ = 1.5

Design total action per pile is Fcd =
(           )
γG × FGk + WGk + γQ × FQk
= 664 kN
N

Characteristic resistance
Partial factors from set M2 should be applied to material properties... but
since there are no material properties to factor, we will factor the
resistances instead using γφ = 1.25 . Since resistances are governed by the

minimum calculated resistance (as per DAs 1 and 2)...
R s,min
Characteristic shaft resistance is R sk =         = 553 kN
ξ4 × γφ

R b,min
Characteristic base resistance is R bk =         = 184 kN
ξ4 × γφ

Design resistance
Partial factors from set R3: γs = 1 and γb = 1

R sk R bk
Design resistance is R cd =     +     = 737 kN
γs    γb

Verification of compression resistance
Fcd
Degree of utilization ΛGEO,3 =      = 90 %
R cd

Design is unacceptable if degree of utilization is > 100%
Design to UK National Annex to BS EN 1997-1

Characteristic resistance
Correlation factor on mean measured resistance ξ3 = 1.38

Correlation factor on minimum measured resistance ξ4 = 1.29
For a pile group that can transfer load from weak to strong piles
(§7.6.2.2.(9)), ξ may be divided by 1.1 (but ξ 3 cannot fall beneath 1.0).
⎛ ξ3
⎜          ⎞
⎟                 ξ4
Thus ξ3 = max ⎜     , 1.0⎟ = 1.25 and ξ4 =     = 1.17
⎝ 1.1      ⎠                 1.1
R t,mean                           R t,min
Calculated resistances              = 961.6 kN and                    = 857 kN
ξ3                                ξ4

Characteristic resistance should therefore be based on the minimum value,
so...
R s,min
Characteristic shaft resistance is R sk =         = 643 kN
ξ4

R b,min
Characteristic base resistance is R bk =                    = 214 kN
ξ4

Design resistance
⎛R1 ⎞        ⎛1 ⎞          ⎛1 ⎞
Partial factors from Sets   ⎜ ⎟:    γs = ⎜ ⎟ and γb = ⎜ ⎟
⎝R4 ⎠        ⎝1.6 ⎠        ⎝2 ⎠
R sk     R bk ⎛857 ⎞
Design resistance is R cd =      +         =⎜     ⎟ kN
γs        γb     ⎝509 ⎠
Verification of compression resistance
Fcd        ⎛77 ⎞
Degree of utilization ΛGEO,1 =             =   ⎜ ⎟      %
R cd       ⎝101 ⎠
Design is unacceptable if degree of utilization is > 100%

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