6.44: Example 6.14

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Page ID: 123657

George Kouretzis
The University of Newcastle, Australia via Council of Australian University Librarians Initiative

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Calculation of pile head deflection and rotation with the Zheng et al. (2024) method

A solid concrete pile (E_p = 20 GPa) of length L = 10 m and diameter D = 1 m is embedded in a deep H = 40 m soil layer. The Young’s modulus of the soil is E_s = 20 MPa. 1) Calculate the deflection of the pile head y(0) and the bending moment that will develop at the pile head M_w if the pile head is fixed against rotation (ω(0)=0) and is subjected to lateral working load H_w = 200 kN. 2) Calculate the deflection of the pile head y(0) and the rotation of the pile head ω(0) if the pile head is free to rotate and is subjected to lateral working load H_w = 200 kN.

Answer:

From Figure 6.110 we obtain for E_p/E_s = 1000, L/D = 10 and H/D = 40 the components of the stiffness matrix:

${K_{hr}} = 7.1{E_s}{D^2} = {\rm{ 142000 \:kN}}$

${K_{rr}} = 28{E_s}{D^3} = 560000{\rm{ \:kNm}}$

We can now form the stiffness and compliance matrixes of the pile (Eq. 6.128):

$\left[ {\begin{array}{*{20}{c}}{y(0})\\{\omega (0})\end{array}} \right] = {\left[ {\bf{K}} \right]^{ - 1}}\left[ {\begin{array}{*{20}{c}}{{H_w}}\\{{M_w}}\end{array}} \right] = \dfrac{1}{{\det {\bf{K}}}}\left[ {\begin{array}{*{20}{c}}{{K_{rr}}}&{ - {K_{hr}}}\\{ - {K_{rh}}}&{{K_{hh}}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{H_w}}\\{{M_w}}\end{array}} \right] = \dfrac{1}{{{K_{rr}}{K_{hh}} - {K_{hr}}{K_{rh}}}}\left[ {\begin{array}{*{20}{c}}{{K_{rr}}}&{ - {K_{hr}}}\\{ - {K_{rh}}}&{{K_{hh}}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{H_w}}\\{{M_w}}\end{array}} \right] =$

1. Pile head fixed against rotation (ω(0) = 0), and pile subjected to lateral working load H_w = 200 kN

Here we have two unknowns: The deflection of the pile head y(0) and the bending moment that will develop on the pile head M_w. These are calculated as:

$\left[ {\begin{array}{*{20}{c}}{y(0})\\0\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{18.521 \times {{10}^{ - 6}}\dfrac{{\rm{m}}}{{{\rm{kN}}}}}&{ - 4.696 \times {{10}^{ - 6}}\dfrac{{\rm{1}}}{{{\rm{kN}}}}}\\{ - 4.696 \times {{10}^{ - 6}}\dfrac{{\rm{1}}}{{{\rm{kN}}}}}&{2.976 \times {{10}^{ - 6}}\dfrac{{\rm{1}}}{{{\rm{kNm}}}}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{200}\\{{M_w}}\end{array}} \right]$

$0 = - 939.28 \times {10^{ - 6}} + 2.976 \times {10^{ - 6}}{M_w} \Rightarrow {M_w} = - 315.5{\rm{ \:kNm}}$

$y(0) = 18.521 \times {10^{ - 6}} \times 200 - 4.696 \times {10^{ - 6}} \times \left( { - 315.5} \right) = 5.18{\rm{ \:mm}}$

2. Pile head free to rotate, and pile subjected to lateral working load H_w = 200 kN

Again we have two unknowns, the deflection y(0) and the rotation ω(0) of the pile head. These are calculated as:

$\left[ {\begin{array}{*{20}{c}}{y(0})\\{\omega (0})\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{18.521 \times {{10}^{ - 6}}\dfrac{{\rm{m}}}{{{\rm{kN}}}}}&{ - 4.696 \times {{10}^{ - 6}}\dfrac{{\rm{1}}}{{{\rm{kN}}}}}\\{ - 4.696 \times {{10}^{ - 6}}\dfrac{{\rm{1}}}{{{\rm{kN}}}}}&{2.976 \times {{10}^{ - 6}}\dfrac{{\rm{1}}}{{{\rm{kNm}}}}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{200}\\0\end{array}} \right]$

$y(0) = 18.521 \times {10^{ - 6}} \times 200 = 3.7{\rm{\: mm}}$

$\omega (0) = - 4.696 \times {10^{ - 6}} \times 200 = - 9.39 \times {10^{ - 4}}{\rm{ rad}} = - 0.0538\deg$

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