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Example 6.13

Calculation of the ultimate geotechnical strength and of the pile head deflection of a laterally loaded pile driven in drained soil

A steel pipe pile of external diameter Dext = 0.4 m and internal diameter Dint = 0.39 m is driven through a layer of uniform, dry loose-to-medium sand with friction angle φ′ = 33° and unit weight γ =18 kN/m3. The design yield strength of the pile’s steel material is σy = 450 MPa, and the Young’s modulus of the steel material is Ep = 210 GPa. The necessary pile length to safely carry the vertical load is L = 10 m. Assuming the pile is connected with a rigid cap (fixed-head), determine the collapse lateral load Hf, and the deflection of the pile’s head at a serviceability load equal to half the collapse lateral load Hw = Hf/2.

Answer:

Both long and short pile failure must be considered to determine the collapse lateral load, according to AS2159 (Example 6.12). In the previous Example 6.12, we have found that the yield moment of the pile’s section to be My = 351 kNm.

As in the case of the pile driven in clay under undrained conditions, the solution starts by applying the short-pile equations. If the resulting maximum pile moment is higher than the yield moment, the long-pile equations are used instead. For the case herein, the collapse lateral load considering short-pile failure is (Eq. 6.121):

{H_f} = \dfrac{3}{2}\gamma {K_P}{L^2}{D_{ext}}

with the Rankine passive pressure coefficient being:

{K_P} = {\tan ^2}\left( {{{45}^ \circ } + \dfrac{{\varphi '}}{2}} \right) = 3.39

substituting:

{H_f} = \dfrac{3}{2}\gamma {K_P}{L^2}{D_{ext}} = \frac{3}{2} \times 18 \times 3.39 \times {10^2} \times 0.4 = 3661.2{\rm{ \:kN}}

and the maximum bending moment corresponding that collapse load (Eq. 6.122) is:

{M_{\max }} = \gamma {K_P}{L^3}{D_{ext}} = 18 \times 3.39 \times {10^3} \times 0.4 = 24408{\rm{ \:kNm}} > {M_y}

The pile cannot sustain the maximum bending moment that is calculated while considering short-pile failure. It will develop a plastic hinge before the lateral load reaches 3661.2 kN, and the long-pile equations must be used instead. The collapse later load considering long-pile failure is given by solving Eq. 6.124:

{H_f} = \dfrac{{2{M_y}}}{{0.544\sqrt {\dfrac{{{H_f}}}{{\gamma {K_P}{D_{ext}}}}} }} = \dfrac{{2 \times 351}}{{0.544\sqrt {\dfrac{{{H_f}}}{{18 \times 3.39 \times 0.4}}} }}

which yields Hf = 343.8 kN.

However, we have to further check whether the collapse lateral load that will result while considering intermediate pile failure from Eq. 6.125 is not lower than 343.8 kN:

{H_f} = \dfrac{{{M_y}}}{L} + \dfrac{1}{2}\gamma {D_{ext}}{K_P}{L^2} = \dfrac{{351}}{{10}} + \dfrac{1}{2} \times 18 \times 0.4 \times 3.39 \times {10^2} = 1255.5{\rm{ \:kN}}

Thus the long pile failure mode is critical, as it results in the lower collapse lateral load.

The pile head displacement for the working lateral load Hw = Hult/2 = 171.9 kN will be estimated while assuming the nh factor to be nh = 5500 kN/m3 from Table 6.16 i.e., the lower bound for medium-dense dry sand.

In order to use Figure 6.108, we must first estimate the factor η from Eq. 6.127:

\eta = {\left( {\dfrac{{{n_h}}}{{{E_p}{I_p}}}} \right)^{0.20}}

where:

{I_p} = \pi \left( {\dfrac{{D_{ext}^4 - D_{{\mathop{\rm int}} }^4}}{{64}}} \right) = 0.000121{\rm{ }}{{\rm{\:m}}^4}

is the moment of inertia of the hollow pile’s cross-section, and

\eta = {\left( {\dfrac{{5500}}{{210 \times {{10}^6} \times 1.21 \times {{10}^{ - 4}}}}} \right)^{0.20}} = 0.736

Considering the curve for fixed-head piles of Figure 6.108:

we conclude that:

y(0)\dfrac{{{{\left( {{E_p}{I_p}} \right)}^{0.6}}{n_h}^{\tfrac{2}{3}}}}{{HL}} = 0.01 or y(0) ≈ 0

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Fundamentals of foundation engineering and their applications Copyright © 2025 by University of Newcastle & G. Kouretzis is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.

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