{"id":627,"date":"2025-03-31T05:04:15","date_gmt":"2025-03-31T05:04:15","guid":{"rendered":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/chapter\/example-6-14-calculation-of-pile-head-deflection-and-rotation-with-the-zheng-et-al-2024-method\/"},"modified":"2026-03-16T14:15:24","modified_gmt":"2026-03-16T14:15:24","slug":"example-6-14-calculation-of-pile-head-deflection-and-rotation-with-the-zheng-et-al-2024-method","status":"publish","type":"chapter","link":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/chapter\/example-6-14-calculation-of-pile-head-deflection-and-rotation-with-the-zheng-et-al-2024-method\/","title":{"raw":"Example 6.14","rendered":"Example 6.14"},"content":{"raw":"A solid concrete pile (Ep<\/sub><\/em> = 20 GPa) of length L<\/em> = 10 m and diameter D<\/em> = 1 m is embedded in a deep H<\/em> = 40 m soil layer. The Young\u2019s modulus of the soil is Es<\/sub><\/em> = 20 MPa. 1) Calculate the deflection of the pile head y<\/em>(0) and the bending moment that will develop at the pile head Mw<\/sub><\/em> if the pile head is fixed against rotation (\u03c9<\/em>(0)=0) and is subjected to lateral working load Hw<\/sub><\/em> = 200 kN. 2)\u00a0 Calculate the deflection of the pile head y<\/em>(0) and the rotation of the pile head \u03c9<\/em>(0) if the pile head is free to rotate and is subjected to lateral working load Hw<\/sub><\/em> = 200 kN.\n

Answer:<\/h2>\nFrom Figure 6.110<\/a> we obtain for Ep<\/sub><\/em>\/Es<\/sub><\/em> = 1000, L<\/em>\/D<\/em> = 10 and H<\/em>\/D<\/em> = 40 the components of the stiffness matrix:\n\n[latex]{K_{hh}} = 4.5{E_s}D = 90000{\\rm{\\: kN\/m}}[\/latex]\n\n[latex]{K_{hr}} = 7.1{E_s}{D^2} = {\\rm{ 142000 \\:kN}}[\/latex]\n\n[latex]{K_{rr}} = 28{E_s}{D^3} = 560000{\\rm{ \\:kNm}}[\/latex]\n\nWe can now form the stiffness and compliance matrixes of the pile (Eq. 6.128):\n\n[latex]\\left[ {\\begin{array}{*{20}{c}}{{H_w}}\\\\{{M_w}}\\end{array}} \\right] = \\left[ {\\begin{array}{*{20}{c}}{90000{\\rm{ \\:kN\/m}}}&{142000{\\rm{ \\:kN}}}\\\\{142000{\\rm{ \\:kN}}}&{560000{\\rm{ \\:kNm}}}\\end{array}} \\right]\\left[ {\\begin{array}{*{20}{c}}{y(0)}\\\\{\\omega (0)}\\end{array}} \\right][\/latex]\n\n[latex]\\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] =[\/latex]\n\n[latex]= \\dfrac{1}{{30.236 \\times {{10}^9}{\\rm{\\:k}}{{\\rm{N}}^{\\rm{2}}}}}\\left[ {\\begin{array}{*{20}{c}}{560000{\\rm{ \\:kNm}}}&{ - 142000{\\rm{ \\:kN}}}\\\\{ - 142000{\\rm{ \\:kN}}}&{90000{\\rm{ \\:kN\/m}}}\\end{array}} \\right]\\left[ {\\begin{array}{*{20}{c}}{{H_w}}\\\\{{M_w}}\\end{array}} \\right][\/latex]\n\n1. Pile head fixed against rotation (\u03c9<\/em>(0) = 0), and pile subjected to lateral working load\u00a0Hw<\/sub><\/em> = 200 kN\n\nHere we have two unknowns: The deflection of the pile head y<\/em>(0) and the bending moment that will develop on the pile head Mw<\/sub><\/em>. These are calculated as:\n\n[latex]\\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][\/latex]\n\n[latex]0 = - 939.28 \\times {10^{ - 6}} + 2.976 \\times {10^{ - 6}}{M_w} \\Rightarrow {M_w} = - 315.5{\\rm{ \\:kNm}}[\/latex]\n\n[latex]y(0) = 18.521 \\times {10^{ - 6}} \\times 200 - 4.696 \\times {10^{ - 6}} \\times \\left( { - 315.5} \\right) = 5.18{\\rm{ \\:mm}}[\/latex]\n\n2. Pile head free to rotate, and pile subjected to lateral working load\u00a0Hw<\/sub><\/em> = 200 kN\n\nAgain we have two unknowns, the deflection y<\/em>(0) and the rotation \u03c9<\/em>(0) of the pile head. These are calculated as:\n\n[latex]\\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][\/latex]\n\n[latex]y(0) = 18.521 \\times {10^{ - 6}} \\times 200 = 3.7{\\rm{\\: mm}}[\/latex]\n\n[latex]\\omega (0) = - 4.696 \\times {10^{ - 6}} \\times 200 = - 9.39 \\times {10^{ - 4}}{\\rm{ rad}} = - 0.0538\\deg[\/latex]","rendered":"

A solid concrete pile (Ep<\/sub><\/em> = 20 GPa) of length L<\/em> = 10 m and diameter D<\/em> = 1 m is embedded in a deep H<\/em> = 40 m soil layer. The Young\u2019s modulus of the soil is Es<\/sub><\/em> = 20 MPa. 1) Calculate the deflection of the pile head y<\/em>(0) and the bending moment that will develop at the pile head Mw<\/sub><\/em> if the pile head is fixed against rotation (\u03c9<\/em>(0)=0) and is subjected to lateral working load Hw<\/sub><\/em> = 200 kN. 2)\u00a0 Calculate the deflection of the pile head y<\/em>(0) and the rotation of the pile head \u03c9<\/em>(0) if the pile head is free to rotate and is subjected to lateral working load Hw<\/sub><\/em> = 200 kN.<\/p>\n

Answer:<\/h2>\n

From Figure 6.110<\/a> we obtain for Ep<\/sub><\/em>\/Es<\/sub><\/em> = 1000, L<\/em>\/D<\/em> = 10 and H<\/em>\/D<\/em> = 40 the components of the stiffness matrix:<\/p>\n

[latex]{K_{hh}} = 4.5{E_s}D = 90000{\\rm{\\: kN\/m}}[\/latex]<\/p>\n

[latex]{K_{hr}} = 7.1{E_s}{D^2} = {\\rm{ 142000 \\:kN}}[\/latex]<\/p>\n

[latex]{K_{rr}} = 28{E_s}{D^3} = 560000{\\rm{ \\:kNm}}[\/latex]<\/p>\n

We can now form the stiffness and compliance matrixes of the pile (Eq. 6.128):<\/p>\n

[latex]\\left[ {\\begin{array}{*{20}{c}}{{H_w}}\\\\{{M_w}}\\end{array}} \\right] = \\left[ {\\begin{array}{*{20}{c}}{90000{\\rm{ \\:kN\/m}}}&{142000{\\rm{ \\:kN}}}\\\\{142000{\\rm{ \\:kN}}}&{560000{\\rm{ \\:kNm}}}\\end{array}} \\right]\\left[ {\\begin{array}{*{20}{c}}{y(0)}\\\\{\\omega (0)}\\end{array}} \\right][\/latex]<\/p>\n

[latex]\\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] =[\/latex]<\/p>\n

[latex]= \\dfrac{1}{{30.236 \\times {{10}^9}{\\rm{\\:k}}{{\\rm{N}}^{\\rm{2}}}}}\\left[ {\\begin{array}{*{20}{c}}{560000{\\rm{ \\:kNm}}}&{ - 142000{\\rm{ \\:kN}}}\\\\{ - 142000{\\rm{ \\:kN}}}&{90000{\\rm{ \\:kN\/m}}}\\end{array}} \\right]\\left[ {\\begin{array}{*{20}{c}}{{H_w}}\\\\{{M_w}}\\end{array}} \\right][\/latex]<\/p>\n

1. Pile head fixed against rotation (\u03c9<\/em>(0) = 0), and pile subjected to lateral working load\u00a0Hw<\/sub><\/em> = 200 kN<\/p>\n

Here we have two unknowns: The deflection of the pile head y<\/em>(0) and the bending moment that will develop on the pile head Mw<\/sub><\/em>. These are calculated as:<\/p>\n

[latex]\\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][\/latex]<\/p>\n

[latex]0 = - 939.28 \\times {10^{ - 6}} + 2.976 \\times {10^{ - 6}}{M_w} \\Rightarrow {M_w} = - 315.5{\\rm{ \\:kNm}}[\/latex]<\/p>\n

[latex]y(0) = 18.521 \\times {10^{ - 6}} \\times 200 - 4.696 \\times {10^{ - 6}} \\times \\left( { - 315.5} \\right) = 5.18{\\rm{ \\:mm}}[\/latex]<\/p>\n

2. Pile head free to rotate, and pile subjected to lateral working load\u00a0Hw<\/sub><\/em> = 200 kN<\/p>\n

Again we have two unknowns, the deflection y<\/em>(0) and the rotation \u03c9<\/em>(0) of the pile head. These are calculated as:<\/p>\n

[latex]\\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][\/latex]<\/p>\n

[latex]y(0) = 18.521 \\times {10^{ - 6}} \\times 200 = 3.7{\\rm{\\: mm}}[\/latex]<\/p>\n

[latex]\\omega (0) = - 4.696 \\times {10^{ - 6}} \\times 200 = - 9.39 \\times {10^{ - 4}}{\\rm{ rad}} = - 0.0538\\deg[\/latex]<\/p>\n","protected":false},"author":1,"menu_order":40,"template":"","meta":{"pb_show_title":"","pb_short_title":"Example 6.14","pb_subtitle":"Calculation of pile head deflection and rotation with the Zheng et al. (2024) method","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-627","chapter","type-chapter","status-publish","hentry"],"part":421,"_links":{"self":[{"href":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-json\/pressbooks\/v2\/chapters\/627","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-json\/wp\/v2\/users\/1"}],"version-history":[{"count":1,"href":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-json\/pressbooks\/v2\/chapters\/627\/revisions"}],"predecessor-version":[{"id":628,"href":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-json\/pressbooks\/v2\/chapters\/627\/revisions\/628"}],"part":[{"href":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-json\/pressbooks\/v2\/parts\/421"}],"metadata":[{"href":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-json\/pressbooks\/v2\/chapters\/627\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-json\/wp\/v2\/media?parent=627"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-json\/pressbooks\/v2\/chapter-type?post=627"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-json\/wp\/v2\/contributor?post=627"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-json\/wp\/v2\/license?post=627"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}