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1. Casa
2. Fundação SkyCiv
3. Pilhas
4. COMO 2159 & 3600 Projeto de estacas de concreto

# COMO 2159 & 3600 Projeto de estacas de concreto

## 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, Contudo, 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) e padrões australianos (COMO 2159 & 3600).

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## 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 factorg) as specified on COMO 2159 Seção 4.3.1.

$${R}_{d,g} = {ø}_{g} × {R}_{d,ug}$$ (1)

Rd,g = Design geotechnical strength

Rd,ug = Ultimate geotechnical strength

øg = 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.

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

Rs = Reduction factor for shaft resistance

fm,s = Shaft-frictional resistance

UMAs = Lateral surface area

fb = Base resistance term

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

Para um guia mais detalhado, confira nosso artigo sobre cálculo a resistência à fricção da pele e a capacidade de suporte da extremidade.

### Geotechnical Reduction Factor (øg)

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

$${ø}_{g} = {ø}_{gb} + [K × ({ø}_{tf} – {ø}_{gb})] ≥ {ø}_{gb}$$ (3)

øgb = Basic geotechnical strength reduction factor

øtf = 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) e (5). Pile load testing is discussed briefly in Section 8 of AS 2159.

Intrinsic Test Factortf)
Dynamic load testing of preformed piles 0.80
Dynamic load testing of other than preformed piles 0.75
No testing 0.80

Mesa 1: Intrinsic Test Factor Values

Testing benefit factor for static load testing:

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

Testing benefit factor for dynamic load testing:

$$K = \frac{1.13 × p}{p + 3.3} ≤ 1$$ (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. of AS 2159. The outcome of the said procedure is Individual Risk Rating (IRR) and an overall design Average Risk Rating (ARR) which shall be used to determine the value of øgb como mostrado na Tabela 2.

Basic Geotechnical Strength Reduction Factorgb)
Average Risk Rating (ARR) Categoria de Risco øgb for low redundancy systems øgb 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 Baixo 0.56 0.64
2.5 < ARR ≤ 3.0 Baixo a moderado 0.52 0.60
3.0 < ARR ≤ 3.5 Moderado 0.48 0.56
3.5 < ARR ≤ 4.0 Moderado a alto 0.45 0.53
4.0 < ARR ≤ 4.5 Alto 0.42 0.50
ARR > 4.5 Muito alto 0.40 0.47

Mesa 2: Valores para fator de redução geotécnica básica, (COMO 2159 Mesa 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 pilhas.

## Design Structural Strength

Piles are structurally designed almost the same as a column. Design structural strength (Rd,s) requires ultimate capacities, such as axial and shear forces, e momento de flexão. The design structural strength of a concrete pile is equivalent to the ultimate design strength (Rnós) reduced by a strength reduction factors) and a concrete placement factor (k), as stated by Section 5.2.1 of AS 2159.

$${R}_{d,s} = {ø}_{s} × k × {R}_{nós}$$ (6)

øs = Strength reduction factor

k = Concrete placement factor

Rnós = Ultimate design strength

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

Fatores de redução de força (ø)
Axial force without bending 0.65
Bending without axial forcepb) 0.65 ≤ 1.24 – [(13 × kuo)/12] ≤ 0.85
Bending with axial compression:
(eu) Nvocê ≥ Nub 0.60
(ii) Nvocê < Nub 0.60 + {(øpb – 0.66) × [1 – (Nvocê/Nub)]}
Cisalhamento 0.70

Mesa 3: Strength reduction factors (Mesa 2.2.2, COMO 3600-18)

### Capacidades axial e flexural de uma única estaca

Similar to columns, piles may also be subjected to combined compression and bending load. As capacidades axial e flexural são verificadas usando um diagrama de interação. 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.

Figura 1: Diagrama de interação de coluna

The squash load point is a point on the diagram where the pile will fail in pure compression. Neste ponto, 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.

$${ϕN}_{uo} = ø × [({UMA}_{g} – {UMA}_{s}) × ({uma}_{1} × f’c) + ({UMA}_{s} × {f}_{seu})]$$ (7)

UMAg = Gross cross-sectional area

UMAs = Total area of steel

uma1 = 1.0 – (0.003 × f’c) [0.72 ≤α1 ≤0.85]

f’c = Concrete strength

fseu = Yield Strength of steel

$${d}_{q} = frac{[(b × D) – {UMA}_{s}] × ({uma}_{1} × f’c) × \sum_{i=1}^{n} ({UMA}_{bi} × {f}_{seu} × {d}_{fazer})}{{N}_{uo}}$$ (8)

b = Pile cross-sectional width

D = Pile cross-sectional depth or diameter

UMAbi = Area of reinforcing bar being considered

dfazer = 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 e a deformação na fibra de tração extrema é zero. Strength of the pile between the squash load and the decompression points can be calculated by linear interpolation with strength reduction factors) de 0.6.

### Decompression point through to pure bending

O ponto de flexão puro é onde a capacidade de carga axial é zero. The transition from decompression point to pure bending uses a strength reduction factor of 0.6 para 0.8 and an input parameter (kvocê) is introduced. The value of kvocê starts at 1 at decompression point and decreases until pure bending is reached. Between the transition of the two points, uma condição equilibrada é alcançada. Neste ponto, a deformação do concreto está no limite (ec= 0,003) e a deformação externa do aço atinge o rendimento (es= 0,0025), The value of kvocê at this point is approximately 0.54 with a strength reduction factor of 0.6.

Once a value of kvocê 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 (Finclui cálculos detalhados passo a passo):

$${F}_{inclui cálculos detalhados passo a passo} = {uma}_{2} × f’c × {UMA}_{c}$$ (9)

uma2 = 0.85 – (0.0015 × f’c) [uma2 ≥0.67]

UMAc = Compression block area (consulte a Figura 2)

= b × γ × kvocê × d (rectangular cross-section)

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

γ = 0.97 – (0.0025 × f’c) [c0.67]

Figura 2: Concrete Compression Block Area

### Força (Fe) Os três tipos de forças internas que se espera que as conexões de aço transmitam incluem força axial (Meu) 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 (ee) shown in equation (10).

$${e}_{e} = frac{{e}_{c}}{({k}_{você} × d)} × [({k}_{você} × d) – {d}_{fazer}]$$ (10)

dfazer = Depth to the bar being considered

ec= Concrete strain = 0.003

If εe < 0 (bar is in tension)

If εe > 0 (bar is in compression)

Bar in compression:

$${F}_{e} = {σ}_{e} × {UMA}_{bi}$$ (11)

σe = Stress in bar = Mínimo [(ee × Es ), fseu]

Es = Modulus elasticity of steel

UMAbi = Bar area

Bar in tension:

$${F}_{e} = [{σ}_{e} – ({uma}_{2} × f’c)] × {UMA}_{bi} ≥ 0$$ (12)

σe = Stress in bar = Mínimo [(ee × Es ), –fseu]

Es = Modulus elasticity of steel

UMAbi = Bar area

Moment by each bar:

$${M}_{eu} = {F}_{e} × {d}_{fazer}$$ (13)

Axial capacity of the pile:

$${øN}_{você} = ø × [ {F}_{inclui cálculos detalhados passo a passo} + {Σ}_{i=1}^{n} {F}_{e}]$$ (14)

Flexural capacity of the pile:

$${dolorido}_{você} = ø × [ ({N}_{você} × {d}_{q}) – ({F}_{inclui cálculos detalhados passo a passo} × {Y}_{c}) – {Σ}_{i=1}^{n} {M}_{eu}]$$ (15)

Design bending moment:

Seção 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. Além disso, 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. Portanto, the design bending moment should be the greater value between equations 16a and 16b.

$${M}_{d} = {{M}^{*}}_{Deflexão de viga simplesmente apoiada Outro exemplo de deflexão é a deflexão de uma viga simplesmente apoiada} + ({N}^{*} × 0.075 m)$$ (16uma)

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

Md = Design bending moment

M*Deflexão de viga simplesmente apoiada Outro exemplo de deflexão é a deflexão de uma viga simplesmente apoiada = Applied moment

D = Pile width

## Capacidade de cisalhamento de uma única pilha

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

$${øV}_{você} = ø × ({V}_{uc} + {V}_{nós}) ≤ {øV}_{você,max}$$ (17)

### Resistência ao cisalhamento do concreto (Vuc)

A contribuição do concreto para a capacidade de cisalhamento é calculada conforme mostrado na equação (18) que é definido na Seção 8.2.4.1 of AS 3600. This section also requires the value of √f’c shall not exceed 9.0 MPa. The values for the parameter kv e θv are determined by using a simplified method suggested by Section 8.2.4.3 of AS 3600.

$${V}_{uc} = {k}_{v} × b × {d}_{v} [object Window]{f’c}$$ (18)

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

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

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

$$\fratura{{UMA}_{sv.min}}{s} = frac{0.08 [object Window]{f’c} × b}{{f}_{sy.f}}$$

fsy.f = Yield strength of shear reinforcing bars

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

Pra (UMAsv/s) < (UMAsv.min/s):

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

Pra (UMAsv/s) ≥ (UMAsv.min/s):

$${k}_{v} = 0.15$$

### Resistência ao cisalhamento de barras de aço (Vnós)

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 of AS 3600.

$${V}_{nós} = frac{{UMA}_{sv} × {f}_{sy.f} × {d}_{v}}{s} × cot{θ}_{v}$$ (19)

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

### Maximum shear strength (Vu.max)

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

$${V}_{u.max} = 0.55 × [ (f’c × b × {d}_{v}) [object Window]{cot{θ}_{v} + cot{uma}_{v}}{1 + cot^{2}{θ}_{v} }]$$ (20)

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

### Ultimate shear strength (Vvocê)

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

$${V}_{você} = ({V}_{uc} + {V}_{nós} ) ≤ {V}_{u.max}$$ (21)

### Resistência ao cisalhamento do projeto (øVvocê)

O fator de redução de capacidade que deve ser aplicado para a resistência ao cisalhamento final é ø = 0.7. Portanto, a resistência ao cisalhamento de projeto da estaca é dada por:

$${øV}_{você} = ø × ({V}_{uc} + {V}_{nós} )$$ (22)

## Referências

• Pack, Lonnie (2018). Guia australiano para engenheiros estruturais. CRC Press.
• Projeto e instalação de estacas (2009). COMO 2159. Padrão Australiano
• Estruturas de concreto (2018). COMO 3600. Padrão Australiano
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