How to calculate the ultimate load-carrying capacity of a single pile

Load-Carrying Capacity

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Evaluating the ultimate load-carrying capacity of a single pile is one of the most important aspects of pile design, and can sometimes be complicated. This article will walk through the governing equations for single pile design as well as an example.

To easily understand the load transfer mechanism of a single pile, imagine a concrete pile of length L with diameter D, wie in Abbildung gezeigt 1.

Zahl 1: Load transfer Mechanism for piles

The load Q applied on the pile shall be transferred directly to the soil at the bottom of the pile. Part of this load will be resisted by the sides of the pile using something called “skin friction” developed along the shaft (Q._s), and the rest will be resisted by the soil that the pile is bearing on (Q._p). Deshalb, the ultimate load-carrying capacity (Qu) of a pile shall be given by the equation (1). There are multiple methods available to estimate the values of Q_p and Q_s.

\( {Q.}_{u} = {Q.}_{p} + {Q.}_{s} \) (1)

Q._u = maximale Tragfähigkeit

Q._p = Endlagertragfähigkeit

Q._s = Hautreibungswiderstand

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End-bearing Capacity, Q._p

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Ultimate end-bearing capacity is theoretically the maximum load per unit area that can be supported by the soil in bearing, without failure. The following equation of Karl Von Terzaghi, the father of soil mechanics, is one of the first and most commonly used theory when evaluating the ultimate bearing capacity of foundations. Terzaghi’s equation for ultimate bearing capacity can be expressed as:

\( {q}_{u} = (c × {N.}_{c}) + (q × {N.}_{q}) + (\frac{1}{2} × γ × B × {N.}_{γ }) \) (2)

q_u = Ultimate end-bearing capacity

c = Cohesion of soil

q = Effective soil pressure

γ = Soil unit weight

B = Cross-sectional depth or diameter

N._c, N._q, N._γ = Bearing factors

Since q_u is in terms of load per unit area or pressure, multiplying it by the cross-sectional area of the pile will result in the end-bearing load capacity (Q._p) of the pile. The resulting value of the last term of Equation 2 is negligible due to a relatively small pile width, daher, it may be dropped from the equation. So, the ultimate end-bearing load capacity of the pile can be expressed as shown in equation (3). This modified version of Terzaghi’s equation is used in the SkyCiv Foundation module when designing piles.

\( {Q.}_{p} = {EIN}_{p} × [(c × {N.}_{c}) + (q × {N.}_{q}) ] \) (3)

EIN_p = Cross-sectional area of pile

Bearing factors N_c and N_q are non-dimensional, empirically derived, and are functions of the soil friction angle (Phi). Researchers have already completed the calculations required to find bearing factors. Tabelle 1 summarizes the values of N_q according to Naval Facilities Engineering Command (NAVFAC DM 7.2, 1984). The value of N_c is approximately equal to 9 for piles under clayey soils.

Bearing Factor (N.q)
Reibungswinkel (Ø)	26	28	30	31	32	33	34	35	36	37	38	39	40
Driven Piles	10	15	21	24	29	35	42	50	62	77	86	120	145
Bored Piles	5	8	10	12	14	17	21	25	30	38	43	60	72

Tabelle 1: N._q values from NAVFAC DM 7.2

Skin-frictional Resistance Capacity, Q._s

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Skin-frictional resistance of piles is developed along the length of the pile. Allgemein, the frictional resistance of a pile is expressed as:

\( {Q.}_{s} = ∑ (p × ΔL × f) \) (4)

p = Perimeter of the pile

ΔL = Incremental pile length over which p and f are taken

f = Unit frictional resistance at any depth

Estimating the value of the unit frictional resistance (f) requires several important factors to consider, such as the nature of pile installation and soil classification. Gleichungen (5) und (6) shows the computational method to find the unit frictional resistance of piles in sandy and clayey soils, beziehungsweise. Tabellen 2 und 3 present the recommended effective earth pressure coefficient (K.) and the soil-pile frictional angle (δ’), according to NAVFAC DM7.2.

For sandy soils:

\( f = K × σ’× tan(δ’) \) (5)

K = Effective earth pressure coefficient

σ’ = Effective vertical stress at the depth under consideration

D’ = Soil-pile frictional angle

For clayey soils:

\( f = α × c \) (6)

α = Empirical adhesion factor

Soil-Pile Frictional Angle (δ’)
Pile Type	D’
Steel Pile	20Um es zu berechnen
Timber Pile	3/4 × Φ
Concrete Pile	3/4 × Φ

Tabelle 2: Soil-Pile Frictional Angle Values (NAVFAC DM7.2, 1984)

Lateral Earth Pressure Coefficient (K.)
Pile Type	Compression Pile	Tension Pile
Driven H-piles	0.5-1.0	0.3-0.5
Driven displacement piles (round, rechteckig)	1.0-1.5	0.6-1.0
Driven displacement piles (tapered)	1.5-2.0	1.0-1.3
Driven jetted piles	0.4-0.9	0.3-0.6
Bored piles (<24″ Durchmesser)	0.7	0.4

Tabelle 3: Lateral Earth Pressure Coefficient (K.) Values (NAVFAC DM7.2, 1984)

Adhesion Factor (ein)
c/pein	ein
≤ 0.1	1.00
0.2	0.92
0.3	0.82
0.4	0.74
0.6	0.62
0.8	0.54
1.0	0.48
1.2	0.42
1.4	0.40
1.6	0.38
1.8	0.36
2.0	0.35
2.4	0.34
2.8	0.34

Hinweis: p_ein = atmospheric pressure ≈ 100 kN / m²

Tabelle 4: Adhesion Factor Values (p-Delta-Effekte, Peck, and Mesri, 1996)

Beispiel: Calculating the capacity of piles in sand

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A 12-meter long concrete pile with a diameter of 500 mm is driven into multiple sand layers with no groundwater present. Find the ultimate load-carrying capacity (Q._u) of the pile.

Einzelheiten
Sektion
Durchmesser	500 mm
Länge	12 m
Layer 1-Bodeneigenschaften
Dicke	5 m
Gewichtseinheit	17.3 kN / m3
Reibungswinkel	30 Abschlüsse
Zusammenhalt	0 kPa
Grundwasserspiegel	Nicht anwesend
Layer 2-Bodeneigenschaften
Dicke	7 m
Gewichtseinheit	16.9 kN / m3
Reibungswinkel	32 Abschlüsse
Zusammenhalt	0 kPa
Grundwasserspiegel	Nicht anwesend

Schritt 1: Compute the end-bearing load capacity (Q._p).

At the tip of the pile:

EIN_p = (π/4) Axiale Tragfähigkeit des Pfahls² = (π/4) × 0.5²

EIN_p = 0.196 m²

c = 0 kPa

θ = 32º

N._q = 29 (Aus der Tabelle 1)

Effective soil pressure (q):

q = (γ ₁ × t₁) + (γ ₂ × t₂) = (5 m × 17.3 kN / m³) + (7 m × 16.9 kN / m³)

q = 204.8 kPa

Then use equation (3) for the end-bearing load capacity:

Q._p = EIN_p × [(c × N_c) + (q × N_q)]

Q._p = 0.196 m² × ( 204.8 KPa × 29)

Q._p = 1,164.083 kN

Schritt 2: Compute the skin-frictional resistance (Q._s).

Using equations (4) und (5), calculate the skin-frictional per soil layer.

Q._s = ∑ (p × ΔL × f)

p = π × D = π × 0.5 m

p = 1.571 m

Ebene 1:

ΔL = 5 m

f₁ = K × σ’₁× tan(δ’)

K = 1.25 (Tabelle 3)

D’ = 3/4 × 30º

D’ = 22.50º

σ’₁ = γ₁ × (0.5 × t₁) = 17.3 kN / m³ × (0.5 × 5 m)

σ’₁ = 43.25 kN / m²

f₁ = 1.25 × 43.25 kN / m² × tan(22.50Um es zu berechnen)

f₁ = 22.393 kN / m²

Q._s1 = p × ΔL × f₁ = 1.571 m ×5 m × 22.393 kN / m²

Q._s1 = 175.897 kN

Ebene 2:

ΔL = 7 m

f₂ = K × σ’₂× tan(δ’)

K = 1.25 (Tabelle 3)

D’ = 3/4× 32º

D’ = 24º

σ’₂ = (γ ₁ × t₁) + [γ ₂ × (0.5 × t₂)] = (17.3 kN / m³ × 5 m) + [16.9 kN / m³ ×(0.5 × 7 m)]

σ’₂ = 145.65 kN / m²

f₂ = 1.25 × 145.65 kN / m² × tan(24Um es zu berechnen)

f₂ = 81.059 kN / m²

Q._s2 = p × ΔL × f₂ = 1.571 m ×7 m × 81.059 kN / m²

Q._s2 = 891.406 kN

Total skin-frictional resistance:

Q._s = Q_s1+ Q._s2 = 175.897 kN + 891.406 kN

Q._s = 1,067.303 kN

Schritt 3: Compute for the ultimate load-carrying capacity (Q._u).

Q._u = Q_p+ Q._s = 1,164.083 kN + 1,067.303 kN

Q._u = 2,231.386 kN

Beispiel 2: Calculating the capacity of piles in clay

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Consider a 406 mm diameter concrete pile with a length of 30m embedded in layered, saturated clay. Find the ultimate load-carrying capacity (Q._u) of the pile.

Einzelheiten
Sektion
Durchmesser	406 mm
Länge	30 m
Layer 1-Bodeneigenschaften
Dicke	10 m
Gewichtseinheit	8 kN / m3
Reibungswinkel	0Um es zu berechnen
Zusammenhalt	30 kPa
Grundwasserspiegel	5 m
Layer 2-Bodeneigenschaften
Dicke	10 m
Gewichtseinheit	19.6 kN / m3
Reibungswinkel	0Um es zu berechnen
Zusammenhalt	0 kPa
Grundwasserspiegel	Fully submerged

Schritt 1: Compute the end-bearing load capacity (Q._p).

At the tip of the pile:

EIN_p = (π/4) Axiale Tragfähigkeit des Pfahls²= (π/4) × 0.406²

EIN_p = 0.129 m²

c = 100 kPa

N._c = 9 (Typical value for clay)

Q._p = (c × N_c) × A_p = (100 kPa × 9) × 0.129 m²

Q._p = 116.1 kN

Schritt 2: Compute the skin-frictional resistance (Q._s).

Using equations (4) und (6), calculate the skin-frictional per soil layer.

Q._s = ∑ (p × ΔL × f)

p = π × D = π × 0.406 m

p = 1.275 m

Ebene 1:

ΔL = 10 m

ein₁ = 0.82 (Tabelle 4)

c₁ = 30 kPa

f₁= α₁ × c₁ = 0.82 × 30 kPa

f₁ = 24.6 kN / m²

Q._s1 = p × ΔL × f₁ = 1.275 m × 10 m × 24.6 kN / m²

Q._s1 = 313.65 kN / m²

Ebene 2:

ΔL = 20 m

ein₂= 0.48 (Tabelle 4)

c₂ = 100 kPa

f₂ = α₂ × c₂ = 0.48 × 100 kPa

f₂ = 48 kN / m²

Q._s2 = p × ΔL × f₂ = 1.275 m × 20 m × 48 kN / m²

Q._s2 = 1,224 kN / m²

Total skin-frictional resistance:

Q._s = Q_s1+ Q._s2 = 313.65 kN + 1224 kN

Q._s = 1,537.65 kN

Schritt 3: Compute for the ultimate load-carrying capacity (Q._u).

Q._u = Q_p+ Q._s = 116.1 kN + 1537.65 kN

Q._u = 1,653.75 kN

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Verweise:

• Das, B.M. (2007). Grundlagen des Grundbaus (7th Ausgabe). Global Engineering
• Rajapakse, R.. (2016). Pile Design and Construction Rule of Thumb (2nd Ausgabe). Elsevier Inc.
• Tomlinson, M.J. (2004). Pile Design and Construction Practice (4th Ausgabe). E. & FN Spon.