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SkyCiv Foundation

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  4. NET ZO 2159 & 3600 Beton stapel ontwerp

NET ZO 2159 & 3600 Beton stapel ontwerp

Single pile design in accordance with AS 2159 (2009) & 3600 (2018)

In case of high lateral load or unfavorable soil conditions, pile foundation is more preferred over shallow foundations. Attempts such as soil modification methods can be made to avoid piles, echter, these methods may involve expensive processes, wherein this case, piles maybe even cheaper.

SkyCiv Foundation Design module includes the design of piles conforming to American Concrete Institute (ACI 318) en Australische normen (NET ZO 2159 & 3600).

Wil je de Foundation Design-software van SkyCiv proberen?? Met onze gratis tool kunnen gebruikers lastdragende berekeningen uitvoeren zonder download of installatie!

Design geotechnical strength of a pile

Vertical loads applied on piles are carried by the end-bearing of the pile and the skin or shaft-friction along its length. The design geotechnical strength (Rd,g) is equal to the ultimate geotechnical strength (Rd,ug) multiplied by a geotechnical reduction factor (Og) as specified on NET ZO 2159 Sectie 4.3.1.

\({R}_{d,g} = {O}_{g} × {R}_{d,ug}\) (1)

Rd,g = Design geotechnical strength

Rd,ug = Ultimate geotechnical strength

Og = Geotechnical reduction factor

Ultimate Geotechnical Strength (Rd,ug)

The ultimate geotechnical strength is equal to the sum of the factored skin friction of the pile (fm,s ) multiplied by the lateral surface area and base resistance multiplied by the cross-sectional area at the tip of the pile.

\( {R}_{d,ug} = [{R}_{s} × ({f}_{m,s} × {EEN}_{s} )] + ({f}_{b} × {EEN}_{b} )\) (2)

Rs = Reduction factor for shaft resistance

fm,s = Shaft-frictional resistance

EENs = Lateral surface area

fb = Base resistance term

EENb = Cross-sectional area at the tip of the pile

Voor een meer gedetailleerde gids, bekijk ons ​​artikel over rekenen de huidwrijvingsweerstand en het einddraagvermogen.

Geotechnical Reduction Factor (Og)

The geotechnical reduction factor is a risk-based calculation for the ultimate design which takes into account different factors, zoals locatievoorwaarden, pile design, and installation factors. Its value ranges commonly from 0.40 naar 0.90. NET ZO 2159 4.3.1 also states how to estimate its value as shown in equation (3).

\( {O}_{g} = {O}_{gb} + [K × ({O}_{tf} – {O}_{gb})] ​ {O}_{gb} \) (3)

Ogb = Basic geotechnical strength reduction factor

Otf = Intrinsic test factor

K= Testing benefit factor

Intrinsic test and testing benefit factors both rely on which type of load testing used on the piles. Their values are specified in Table 1 and on equations (4) en (5). Pile load testing is discussed briefly in Section 8 van AS 2159.

Intrinsic Test Factor (Otf)
Static load testing 0.90
Rapid load testing 0.75
Dynamic load testing of preformed piles 0.80
Dynamic load testing of other than preformed piles 0.75
Bi-directional load testing 0.85
No testing 0.80

Tafel 1: Intrinsic Test Factor Values

Testing benefit factor for static load testing:

\( K = \frac{1.33 × p}{p + 3.3} ≤ 1\) (4)

Testvoordeelfactor voor dynamische belastingtests:

\( K = \frac{1.13 × p}{p + 3.3} ≤ 1\) (5)

p = Percentage of the total piles that are tested and meet the acceptance criteria

The basic geotechnical strength reduction factor is evaluated using a risk assessment procedure discussed in Section 4.3. van AS 2159. The outcome of the said procedure is Individual Risk Rating (IRR) en een algemeen ontwerp Gemiddelde risicoscore (ARR) which shall be used to determine the value of øgb zoals weergegeven in de tabel 2.

Fundamentele geotechnische sterktereductiefactor (Ogb)
Gemiddelde risicoscore (ARR) Risicocategorie Ogb for low redundancy systems Ogb for high redundancy systems
ARR ≤ 1.5 Very low 0.67 0.76
1.5 < ARR ≤ 2.0 Very low to low 0.61 0.70
2.0 < ARR ≤ 2.5 Laag 0.56 0.64
2.5 < ARR ≤ 3.0 Low to moderate 0.52 0.60
3.0 < ARR ≤ 3.5 Matig 0.48 0.56
3.5 < ARR ≤ 4.0 Moderate to high 0.45 0.53
4.0 < ARR ≤ 4.5 hoog 0.42 0.50
ARR > 4.5 Very high 0.40 0.47

Tafel 2: Values for Basic Geotechnical Reduction Factor, (NET ZO 2159 Tafel 4.3.2)

Low redundancy systems are heavily loaded single piles while high redundancy systems include large pile groups under large pile caps or pile groups with more than 4 stapels.

Design Structural Strength

Piles are structurally designed almost the same as a column. Design structural strength (Rd,s) vereist ultieme capaciteiten, zoals axiale en afschuifkrachten, en buigend moment. The design structural strength of a concrete pile is equivalent to the ultimate design strength (Rus) verminderd met een sterktereductiefactor (Os) en een concrete plaatsingsfactor (k), zoals vermeld door Sectie 5.2.1 van AS 2159.

\( {R}_{d,s} = {O}_{s} × k × {R}_{us} \) (6)

Os = Strength reduction factor

k = Concrete placement factor

Rus = Ultimate design strength

The values for the strength reduction factor are shown in Table 3. The concrete placement factor ranges from 0.75 naar 1.0, depending on the pile construction method. Echter, for piles other than concrete and grout, k shall be taken as 1.0.

Krachtverminderingsfactoren (O)
Axial force without bending 0.65
Bending without axial force (Opb) 0.65 ≤ 1.24 – [(13 × kuo)/12] ≤ 0.85
Bending with axial compression:
(ik) Nu ≥ Nub 0.60
(ii) Nu < Nub 0.60 + {(Opb – 0.66) × [1 – (Nu/Nub)]}
Schuintrekken 0.70

Tafel 3: Strength reduction factors (Tafel 2.2.2, NET ZO 3600-18)

Axiale en buigcapaciteiten van een enkele paal

Similar to columns, piles may also be subjected to combined compression and bending load. Axiale en buigcapaciteiten worden gecontroleerd met behulp van een interactiediagram. This diagram is a visual representation of the behavior of the bending and axial capacities caused by an increase in load from pure bending point until a balanced point is reached.

 

Figuur 1: Kolom interactie diagram

Squash Load (Nuo)

The squash load point is a point on the diagram where the pile will fail in pure compression. Op dit punt, the axial load is applied on the plastic centroid of the section to remain in compression without bending. Squash load (Nuo) and the location of the plastic centroid (dq) are computed as shown in equations (7) & (8). Although location of the plastic centroid can be taken as 1/2 of the total depth of the cross-section for symmetrical sections with symmetrical reinforcement layout.

\( {N}_{uo} = ø × [({EEN}_{g} – {EEN}_{s}) × ({een}_{1} × f’c) + ({EEN}_{s} × {f}_{de opwaartse bodemdruk veroorzaakt bidirectionele buiging met trekspanningen aan het bodemoppervlak})] \) (7)

EENg = Gross cross-sectional area

EENs = Total area of steel

een1 = 1.0 – (0.003 × f’c) [0.72 ≤α1 ≤0.85]

f’c = Concrete strength

fde opwaartse bodemdruk veroorzaakt bidirectionele buiging met trekspanningen aan het bodemoppervlak = Yield Strength of steel

\( {d}_{q} = frac{[(b × D) – {EEN}_{s}] × ({een}_{1} × f’c) × \sum_{i=1}^{n} ({EEN}_{bi} × {f}_{de opwaartse bodemdruk veroorzaakt bidirectionele buiging met trekspanningen aan het bodemoppervlak} × {d}_{Doen})}{{N}_{uo}} \) (8)

b = Pile cross-sectional width

D = Pile cross-sectional depth or diameter

EENbi = Area of reinforcing bar being considered

dDoen = Depth of reinforcing bar being considered

Squash load point through to decompression point

Decompression point is where the concrete strain at the extreme compressive fibre is equal to 0.003 en de spanning in de vezel met extreme treksterkte is nul. Strength of the pile between the squash load and the decompression points can be calculated by linear interpolation with strength reduction factor (Os) van 0.6.

Decompression point through to pure bending

Zuiver buigpunt is waar de axiale belastingscapaciteit nul is. The transition from decompression point to pure bending uses a strength reduction factor of 0.6 naar 0.8 and an input parameter (ku) is introduced. The value of ku starts at 1 at decompression point and decreases until pure bending is reached. Between the transition of the two points, een evenwichtige toestand wordt bereikt. Op dit punt, de betonspanning is aan zijn limiet (ec=0.003) en de buitenste staalstam bereikt opbrengst (es=0,0025), The value of ku at this point is approximately 0.54 with a strength reduction factor of 0.6.

Once a value of ku is selected, tensile and compressive forces of the section can be calculated. The axial load on the section is equivalent to the sum of tensile and compressive forces, while the bending moment is calculated by resolving these forces about the neutral axis. Calculation for the compressive and tensile forces are enumerated below

Force due to concrete (Fcc):

\( {F}_{cc} = {een}_{2} × f’c × {EEN}_{c} \) (9)

een2 = 0.85 – (0.0015 × f’c) [een2 ​0.67]

EENc = Compression block area (zie figuur 2)

= geb × γ × ku × d (rectangular cross-section)

=(1/2) × (θ – sinθ) × (D/2)2 (circular cross-section)

= 0.97 – (0.0025 × f’c) [c0.67]

figure-compression-block-piles

Figuur 2: Concrete Compression Block Area

Kracht (FEn) en moment (Mik) contributed by each individual bar:

Each reinforcing bar of the section exerts a force that could either be compressive or tensile, depending on the value bar strain (eEn) shown in equation (10).

\( {e}_{En} = frac{{e}_{c}}{({k}_{u} × d)} × [({k}_{u} × d) – {d}_{Doen}] \) (10)

dDoen = Depth to the bar being considered

ec= Concrete strain = 0.003

If εEn < 0 (bar is in tension)

If εEn > 0 (bar is in compression)

Bar in compression:

\( {F}_{En} = {σ}_{En} × {EEN}_{bi} \) (11)

σEn = Stress in bar = Minimum [(eEn × Es ), fde opwaartse bodemdruk veroorzaakt bidirectionele buiging met trekspanningen aan het bodemoppervlak]

Es = Modulus elasticity of steel

EENbi = Bar area

Bar in tension:

\( {F}_{En} = [{σ}_{En} – ({een}_{2} × f’c)] × {EEN}_{bi} ​ 0\) (12)

σEn = Stress in bar = Minimum [(eEn × Es ), –fde opwaartse bodemdruk veroorzaakt bidirectionele buiging met trekspanningen aan het bodemoppervlak]

Es = Modulus elasticity of steel

EENbi = Bar area

Moment by each bar:

\( {M}_{ik} = {F}_{En} × {d}_{Doen} \) (13)

Axial capacity of the pile:

\( {øN}_{u} = ø × [ {F}_{cc} + {S}_{i=1}^{n} {F}_{En}]\) (14)

Flexural capacity of the pile:

\( {ontstoken}_{u} = ø × [ ({N}_{u} × {d}_{q}) – ({F}_{cc} × {en}_{c}) – {S}_{i=1}^{n} {M}_{ik}] \) (15)

Design bending moment:

Sectie 7.2 specifies that piles are required to have a out-of-position tolerance of 75mm for the horizontal positioning of the piles. This requirement may induce a bending moment equal to axial load multiplied by the eccentricity of 75mm. Bovendien, a minimum design moment shall also be considered which is equivalent to the axial force multiplied by 5% of the overall minimum width of the pile. Daarom, the design bending moment should be the greater value between equations 16a and 16b.

\( {M}_{d} = {{M}^{*}}_{toegepast} + ({N}^{*} × 0.075 m) \) (16een)

\( {M}_{d} = {N}^{*} × (0.05 × D) \) (16b)

Md = Design bending moment

M*toegepast = Applied moment

N* = Axial load

D = Pile width

Afschuifcapaciteit van een enkele paal

Calculation for the strength in shear shall be in accordance with Section 8.2 van AS 3600. Shear strength is equivalent to a combined shear capacities of the concrete and the steel reinforcement (vergelijking 17).

\( {øV}_{u} = ø × ({V }_{uc} + {V }_{us}) ≤ {øV}_{u,max} \) (17)

Afschuifsterkte van beton (V uc)

De bijdrage van beton aan de afschuifcapaciteit wordt berekend zoals weergegeven in de vergelijking (18) die is gedefinieerd op Sectie 8.2.4.1 van AS 3600. This section also requires the value of √f’c shall not exceed 9.0 MPa. The values for the parameter kv en θv are determined by using a simplified method suggested by Section 8.2.4.3 van AS 3600.

\( {V }_{uc} = {k}_{v} × b × {d}_{v} × sqrt{f'c} \) (18)

dv = Effective shear depth = Maximaal [(0.72 × D ), (0.90 × d )]

Determination of the minimum area of shear reinforcement (EENsv.min) & kv:

The area of the shear reinforcement (EENsv) is the total bar area of all the provided steel bars tied in the same direction of the applied load. Sectie 8.2.1.7 van AS 3600 provided the equation for the minimum transverse shear reinforcements, which shall be:

\( \frac{{EEN}_{sv.min}}{s} = frac{0.08 × sqrt{f'c} × b}{{f}_{sy.f}} \)

fsy.f = Yield strength of shear reinforcing bars

s= Center-to-center spacing of shear reinforcing bars

Voor (EENsv/s) < (EENsv.min/s):

\( {k}_{v} = frac{200}{[1000 + (1.3 × {d}_{v} )]} ≤ 0.10\)

Voor (EENsv/s) ​ (EENsv.min/s):

\( {k}_{v} = 0.15 \)

Afschuifsterkte van stalen staven (V us)

The contribution of the transverse shear reinforcements to the shear capacity calculated is shown in equation (19), which is defined in Section 8.2.5 van AS 3600.

\( {V }_{us} = frac{{EEN}_{sv} × {f}_{sy.f} × {d}_{v}}{s} × cot{θ}_{v} \) (19)

θv= angle of inclination of the compression strut = 36º

Maximum shear strength (V u.max)

Shear capacity is limited and in no case shall exceed the maximum value specified on Section 8.2.6 van AS 3600 (vergelijking 20).

\( {V }_{u.max} = 0.55 × [ (f’c × b × {d}_{v}) × frac{cot{θ}_{v} + cot{een}_{v}}{1 + cot^{2}{θ}_{v} }] \) (20)

eenv= angle between the inclined shear reinforcement and the longitudinal tensile reinforcement≈ 90º

Ultimate shear strength (V u)

The total shear strength contributed by the concrete and shear reinforcements shall be less than or equal to the limiting value of Vu.max

\( {V }_{u} = ({V }_{uc} + {V }_{us} ) ≤ {V }_{u.max} \) (21)

Ontwerp schuifsterkte (øVu)

Capacity reduction factor that shall be applied for the ultimate shear strength is ø = 0.7. Daarom, the design shear strength of the pile is given by:

\( {øV}_{u} = ø × ({V }_{uc} + {V }_{us} ) \) (22)

Referenties

  • Pack, Lonnie (2018). Australian Guidebook For Structural Engineers. CRC Press.
  • Piling Design and Installation (2009). NET ZO 2159. Australische standaard
  • Betonnen constructies (2018). NET ZO 3600. Australische standaard
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