{"id":621,"date":"2025-03-31T05:01:00","date_gmt":"2025-03-31T05:01:00","guid":{"rendered":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/chapter\/example-6-13-calculation-of-the-ultimate-geotechnical-strength-and-of-the-pile-head-deflection-of-a-laterally-loaded-pile-driven-in-drained-soil\/"},"modified":"2026-03-16T14:14:58","modified_gmt":"2026-03-16T14:14:58","slug":"example-6-13-calculation-of-the-ultimate-geotechnical-strength-and-of-the-pile-head-deflection-of-a-laterally-loaded-pile-driven-in-drained-soil","status":"publish","type":"chapter","link":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/chapter\/example-6-13-calculation-of-the-ultimate-geotechnical-strength-and-of-the-pile-head-deflection-of-a-laterally-loaded-pile-driven-in-drained-soil\/","title":{"raw":"Example 6.13","rendered":"Example 6.13"},"content":{"raw":"A steel pipe pile of external diameter Dext<\/sub><\/em> = 0.4 m and internal diameter Dint<\/sub><\/em> = 0.39 m is driven through a layer of uniform, dry loose-to-medium sand with friction angle \u03c6<\/em>\u2032 <\/em>= 33\u00b0 and unit weight \u03b3<\/em> =18 kN\/m3<\/sup>. The design yield strength of the pile\u2019s steel material is \u03c3<\/em>y <\/sub>= 450 MPa, and the Young\u2019s modulus of the steel material is Ep<\/sub><\/em> = 210 GPa. The necessary pile length to safely carry the vertical load is L <\/em>= 10 m. Assuming the pile is connected with a rigid cap (fixed-head),<\/em> determine the collapse lateral load Hf<\/sub><\/em>, and the deflection of the pile\u2019s head at a serviceability load equal to half the collapse lateral load Hw<\/sub><\/em> = Hf<\/sub><\/em>\/2.\n

Answer:<\/h2>\nBoth long<\/em> and short<\/em> pile failure must be considered to determine the collapse lateral load, according to AS2159 (Example 6.12<\/a>). In the previous Example 6.12<\/a>, we have found that the yield moment of the pile\u2019s section to be My <\/sub><\/em>= 351 kNm.\n\nAs 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):\n\n[latex]{H_f} = \\dfrac{3}{2}\\gamma {K_P}{L^2}{D_{ext}}[\/latex]\n\nwith the Rankine passive pressure coefficient being:\n\n[latex]{K_P} = {\\tan ^2}\\left( {{{45}^ \\circ } + \\dfrac{{\\varphi '}}{2}} \\right) = 3.39[\/latex]\n\nsubstituting:\n\n[latex]{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}}[\/latex]\n\nand the maximum bending moment corresponding that collapse load (Eq. 6.122) is:\n\n[latex]{M_{\\max }} = \\gamma {K_P}{L^3}{D_{ext}} = 18 \\times 3.39 \\times {10^3} \\times 0.4 = 24408{\\rm{ \\:kNm}} > {M_y}[\/latex]\n\nThe 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:\n\n[latex]{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}}} }}[\/latex]\n\nwhich yields Hf<\/sub><\/em> = 343.8 kN.\n\nHowever, 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:\n\n[latex]{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}}[\/latex]\n\nThus the long pile failure mode is critical, as it results in the lower collapse lateral load.\n\nThe pile head displacement for the working lateral load Hw <\/sub><\/em>= Hult<\/sub><\/em>\/2 = 171.9 kN will be estimated while assuming the nh<\/sub><\/em> factor to be nh <\/sub><\/em>= 5500 kN\/m3<\/sup> from Table 6.16 i.e., the lower bound for medium-dense dry sand.\n\nIn order to use Figure 6.108<\/a>, we must first estimate the factor \u03b7<\/em> from Eq. 6.127:\n\n[latex]\\eta = {\\left( {\\dfrac{{{n_h}}}{{{E_p}{I_p}}}} \\right)^{0.20}}[\/latex]\n\nwhere:\n\n[latex]{I_p} = \\pi \\left( {\\dfrac{{D_{ext}^4 - D_{{\\mathop{\\rm int}} }^4}}{{64}}} \\right) = 0.000121{\\rm{ }}{{\\rm{\\:m}}^4}[\/latex]\n\nis the moment of inertia of the hollow pile\u2019s cross-section, and\n\n[latex]\\eta = {\\left( {\\dfrac{{5500}}{{210 \\times {{10}^6} \\times 1.21 \\times {{10}^{ - 4}}}}} \\right)^{0.20}} = 0.736[\/latex]\n\nConsidering the curve for fixed-head piles of Figure 6.108<\/a>:\n\nwe conclude that:\n\n[latex]y(0)\\dfrac{{{{\\left( {{E_p}{I_p}} \\right)}^{0.6}}{n_h}^{\\tfrac{2}{3}}}}{{HL}} = 0.01[\/latex]\u00a0or y<\/em>(0) \u2248 0","rendered":"

A steel pipe pile of external diameter Dext<\/sub><\/em> = 0.4 m and internal diameter Dint<\/sub><\/em> = 0.39 m is driven through a layer of uniform, dry loose-to-medium sand with friction angle \u03c6<\/em>\u2032 <\/em>= 33\u00b0 and unit weight \u03b3<\/em> =18 kN\/m3<\/sup>. The design yield strength of the pile\u2019s steel material is \u03c3<\/em>y <\/sub>= 450 MPa, and the Young\u2019s modulus of the steel material is Ep<\/sub><\/em> = 210 GPa. The necessary pile length to safely carry the vertical load is L <\/em>= 10 m. Assuming the pile is connected with a rigid cap (fixed-head),<\/em> determine the collapse lateral load Hf<\/sub><\/em>, and the deflection of the pile\u2019s head at a serviceability load equal to half the collapse lateral load Hw<\/sub><\/em> = Hf<\/sub><\/em>\/2.<\/p>\n

Answer:<\/h2>\n

Both long<\/em> and short<\/em> pile failure must be considered to determine the collapse lateral load, according to AS2159 (Example 6.12<\/a>). In the previous Example 6.12<\/a>, we have found that the yield moment of the pile\u2019s section to be My <\/sub><\/em>= 351 kNm.<\/p>\n

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):<\/p>\n

[latex]{H_f} = \\dfrac{3}{2}\\gamma {K_P}{L^2}{D_{ext}}[\/latex]<\/p>\n

with the Rankine passive pressure coefficient being:<\/p>\n

[latex]{K_P} = {\\tan ^2}\\left( {{{45}^ \\circ } + \\dfrac{{\\varphi '}}{2}} \\right) = 3.39[\/latex]<\/p>\n

substituting:<\/p>\n

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

and the maximum bending moment corresponding that collapse load (Eq. 6.122) is:<\/p>\n

[latex]{M_{\\max }} = \\gamma {K_P}{L^3}{D_{ext}} = 18 \\times 3.39 \\times {10^3} \\times 0.4 = 24408{\\rm{ \\:kNm}} > {M_y}[\/latex]<\/p>\n

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:<\/p>\n

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

which yields Hf<\/sub><\/em> = 343.8 kN.<\/p>\n

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:<\/p>\n

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

Thus the long pile failure mode is critical, as it results in the lower collapse lateral load.<\/p>\n

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

In order to use Figure 6.108<\/a>, we must first estimate the factor \u03b7<\/em> from Eq. 6.127:<\/p>\n

[latex]\\eta = {\\left( {\\dfrac{{{n_h}}}{{{E_p}{I_p}}}} \\right)^{0.20}}[\/latex]<\/p>\n

where:<\/p>\n

[latex]{I_p} = \\pi \\left( {\\dfrac{{D_{ext}^4 - D_{{\\mathop{\\rm int}} }^4}}{{64}}} \\right) = 0.000121{\\rm{ }}{{\\rm{\\:m}}^4}[\/latex]<\/p>\n

is the moment of inertia of the hollow pile\u2019s cross-section, and<\/p>\n

[latex]\\eta = {\\left( {\\dfrac{{5500}}{{210 \\times {{10}^6} \\times 1.21 \\times {{10}^{ - 4}}}}} \\right)^{0.20}} = 0.736[\/latex]<\/p>\n

Considering the curve for fixed-head piles of Figure 6.108<\/a>:<\/p>\n

we conclude that:<\/p>\n

[latex]y(0)\\dfrac{{{{\\left( {{E_p}{I_p}} \\right)}^{0.6}}{n_h}^{\\tfrac{2}{3}}}}{{HL}} = 0.01[\/latex]\u00a0or y<\/em>(0) \u2248 0<\/p>\n","protected":false},"author":1,"menu_order":38,"template":"","meta":{"pb_show_title":"","pb_short_title":"Example 6.13","pb_subtitle":"Calculation of the ultimate geotechnical strength and of the pile head deflection of a laterally loaded pile driven in drained soil","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-621","chapter","type-chapter","status-publish","hentry"],"part":421,"_links":{"self":[{"href":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-json\/pressbooks\/v2\/chapters\/621","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\/621\/revisions"}],"predecessor-version":[{"id":622,"href":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-json\/pressbooks\/v2\/chapters\/621\/revisions\/622"}],"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\/621\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-json\/wp\/v2\/media?parent=621"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-json\/pressbooks\/v2\/chapter-type?post=621"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-json\/wp\/v2\/contributor?post=621"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-json\/wp\/v2\/license?post=621"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}