General Analytical Solution for Laterally-Loaded Pile-Based Miche Model

Abstract

Piles subjected to lateral loading can create problems in soil-structure interaction. Several differing methods of analysis have been proposed to solve the problem of laterally loaded piles, resulting in the determination of pile bending and the bending moment as a function of depth below soil surface. These piles are widely used to support laterally loaded piles, such as bridge pillars, offshore platforms, communication towers and others. This study presents an analytical solution to Miche’s problem as a continuous function of depth: deflection and moment, as well as a dimensional plots to be used in projects involving piles subjected to laterally loading only including data concerning laterally loading test and pile geometry. A new formula is presented to calculate the pile head displacement as well as an equation to determine maximum moment for a generalized Miche model and further analysis. In addition, this paper proposes an equation for the determination of constant horizontal subgrade reaction $(n_{h})$ based on the CPT in-situ test and the geometric characteristics of the pile. Calibration of the analytical model showed good fit and conservative results regarding inclinometer data from an bored pile and good agreement with the literature results.

Appendix

By substituting Eq. (6) for Eq. (1), and developing the sum to $n=4$ in the first part:

$$\begin{aligned}&\sum _{n=4}^{\infty }n(n-1)(n-2)(n-3) a_{n} z^{n-4}\nonumber \\&\quad +\beta ^5 \sum _{n=0}^{\infty } a_{n} z^{n+1}=0 \end{aligned}$$

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Turning $n \rightarrow n+4$ in the first installment of the equation and $n \rightarrow n-1$ for the second installment, it has:

$$\begin{aligned}&\sum _{n=0}^{\infty }(n+1)(n+2)(n+3)(n+4) a_{n+4} z^{n}\nonumber \\&\quad +\beta ^5 \sum _{n=1}^{\infty } a_{n-1} z^{n}=0 \end{aligned}$$

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Rearranging this, we have

$$\begin{aligned} \begin{aligned}&1\cdot 2\cdot 3\cdot 4 a_{4}+\sum _{n=1}^{\infty }[(n+1)(n+2)(n+3)(n+4)\\&\quad a_{n+4}+\beta ^5 a_{n-1}] z^{n} \end{aligned} \end{aligned}$$

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So that $a_{4}=a_{9}=\cdots =a_{5n+4}=0$, and $a_{0}$, $a_{1}$, $a_{2 }$ and $a_{3}$ are arbitrary, so the recurrence relationship is provided by

$$\begin{aligned} a_{n+4}=-\frac{\beta ^5 a_{n-1}}{(n+1)(n+2)(n+3)(n+4)}. \end{aligned}$$

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Considering $n=1, 2, 3, 4$, in Eq. (37), it has

$$\begin{aligned} a_{5}&=\frac{-\beta ^5 a_{0}}{2\cdot 3\cdot 4\cdot 5}, \\ a_{6}&=\frac{-\beta ^5 a_{1}}{3\cdot 4\cdot 5\cdot 6},\\ a_{7}&=\frac{-\beta ^5 a_{2}}{4\cdot 5\cdot 6\cdot 7},\\ a_{8}&=\frac{-\beta ^5 a_{3}}{5\cdot 6\cdot 7\cdot 8}.\\ \end{aligned}$$

Furthermore, for $n=6,7,8,9$, in Eq. (37), and using the previous equations:

$$\begin{aligned} a_{10}&=\frac{\beta ^{10} a_{0}}{2\cdot 3\cdot 4\cdot 5\cdot 7\cdot 8\cdot 9\cdot 10} \\ a_{11}&=\frac{\beta ^{10} a_{1}}{3\cdot 4\cdot 5\cdot 6\cdot 8\cdot 9\cdot 10\cdot 11}\\ a_{12}&=\frac{\beta ^{10} a_{2}}{4\cdot 5\cdot 6\cdot 7\cdot 9\cdot 10\cdot 11\cdot 12}\\ a_{13}&=\frac{\beta ^{10} a_{3}}{5\cdot 6\cdot 7\cdot 8\cdot 10\cdot 11\cdot 12\cdot 13}\\ \end{aligned}$$

By applying the same process to infinite values of n, it has the behavior of a generalized hypergeometric function, as defined in Eq. (3). The general solution of Eq. (2) is achieved by substituting the terms $a_{n}$ found in the Taylor series given by Eq. (3), therefore,

$$\begin{aligned} y[z]&= a_{0}\left( 1-\frac{(\beta ^5z^5)^1}{2\cdot 3\cdot 4\cdot 5}+\frac{(\beta ^5z^5)^2}{2\cdot 3\cdot 4\cdot 5\cdot 7\cdot 8\cdot 9\cdot 10}\right. \\ &\quad\left. -\,\frac{(\beta ^5 z^5)^3}{2\cdot 3\cdot 4\cdot 5\cdot 7\cdot 8\cdot 9\cdot 10\cdot 12\cdot 13\cdot 14\cdot 15}+\cdots \right) \\ &\quad + a_{1}z\beta \left( 1-\frac{(\beta ^5 z^5)^1}{3\cdot 4\cdot 5\cdot 6}+\frac{(\beta ^5 z^5)^2}{3\cdot 5\cdot 6\cdot 8\cdot 9\cdot 10\cdot 11}\right. \\ &\quad\left. -\,\frac{(\beta ^5 z^5)^3}{3\cdot 4\cdot 5\cdot 6\cdot 8\cdot 9\cdot 10\cdot 11\cdot 13\cdot 14\cdot 15\cdot 16}+\cdots \right) \\ &\quad a_{2} z^2 \beta ^2 \left( 1-\frac{(\beta ^5 z^5)^1}{4\cdot 5\cdot 6\cdot 7}+\frac{(\beta ^{5}z^5)^2 }{4\cdot 5\cdot 6\cdot 7\cdot 9\cdot 10\cdot 11\cdot 12}\right. \\ &\quad\left. -\,\frac{(\beta ^{5}z^5)^3}{4\cdot 5\cdot 6\cdot 7\cdot 9\cdot 10\cdot 11\cdot 12\cdot 14\cdot 15\cdot 16\cdot 17}+\cdots \right) \\ &\quad a_{3}z^3\beta ^3 \left( 1-\frac{(\beta ^5 z^5)^1}{5\cdot 6\cdot 7\cdot 8} +\frac{(\beta ^5z^5)^2}{5\cdot 6\cdot 7\cdot 8\cdot 10\cdot 11\cdot 12\cdot 13}\right. \\ &\quad\left. -\,\frac{(\beta ^{5}z^5)^3}{5\cdot 6\cdot 7\cdot 8\cdot 10\cdot 11\cdot 12\cdot 13\cdot 15\cdot 16\cdot 17\cdot 18}+\cdots \right) \end{aligned}$$

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