{"id":610,"date":"2025-03-31T04:56:59","date_gmt":"2025-03-31T04:56:59","guid":{"rendered":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/chapter\/example-6-12-calculation-of-the-ultimate-geotechnical-strength-and-of-the-pile-head-deflection-of-a-laterally-loaded-pile-driven-in-undrained-soil\/"},"modified":"2026-03-16T14:14:39","modified_gmt":"2026-03-16T14:14:39","slug":"example-6-12-calculation-of-the-ultimate-geotechnical-strength-and-of-the-pile-head-deflection-of-a-laterally-loaded-pile-driven-in-undrained-soil","status":"publish","type":"chapter","link":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/chapter\/example-6-12-calculation-of-the-ultimate-geotechnical-strength-and-of-the-pile-head-deflection-of-a-laterally-loaded-pile-driven-in-undrained-soil\/","title":{"raw":"Example 6.12","rendered":"Example 6.12"},"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, stiff overconsolidated clay with undrained shear strength Su <\/sub><\/em>= 100 kPa. 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 also to AS2159. To do that, we must first estimate the yield moment My<\/sub><\/em> of the pile\u2019s section, which is the maximum moment that the pile\u2019s section can carry before a plastic hinge is formed. For a hollow pipe pile and homogeneous material, the plastic moment is estimated from Table 6.14 as:\n\n[latex]{M_y} = \\left( {\\dfrac{{D_{ext}^3 - D_{{\\mathop{\\rm int}} }^3}}{6}} \\right){\\sigma _y} = \\left( {\\dfrac{{{{0.4}^3} - {{0.39}^3}}}{6}} \\right)450000 = 351{\\rm{ \\:kNm}}[\/latex]\n\nWe will start by applying the short-pile equations. If the resulting maximum pile moment Mmax<\/sub><\/em> is higher than the yield moment (351 kNm), the long-pile equations should be used instead, as they will result in a lower collapse load. For the case herein, the collapse lateral load considering short-pile failure is (Eq. 6.106):\n\n[latex]{H_f} = 9{S_u}{D_{ext}}\\left( {L - 1.5{D_{ext}}} \\right) = 9 \\times 100 \\times 0.4 \\times \\left( {10 - 1.5 \\times 0.4} \\right) = 3384{\\rm{ \\:kN}}[\/latex]\n\nand the maximum moment corresponding the ultimate lateral load (Eq. 6.107):\n\n[latex]{M_{\\max }} = 4.5{S_u}{D_{ext}}\\left( {{L^2} - 2.25D_{ext}^2} \\right) = 4.5 \\times 100 \\times 0.4 \\times \\left( {{{10}^2} - 2.25 \\times {{0.4}^2}} \\right) = 17935{\\rm{ \\:kNm}} > {M_y}[\/latex]\n\nIt is thus clear that the pile will develop a plastic hinge before a short-pile failure mode fully develops, and the long-pile equations must be used instead. The collapse later load considering long-pile failure is given by substituting Eq. 6.104 into Eq. 6.109 as:\n\n[latex]{H_f} = \\dfrac{{2{M_y}}}{{\\left( {1.5{D_{ext}} + 0.5\\dfrac{{{H_f}}}{{9{S_u}{D_{ext}}}}} \\right)}} = \\dfrac{{2 \\times 351}}{{1.5 \\times 0.4 + 0.5\\left( {\\dfrac{{{H_f}}}{{9 \\times 100 \\times 0.4}}} \\right)}}\\[\/latex]\n\n[latex]{H_f} = 527.03{\\rm{ \\:kN}}[\/latex]\n\nHowever, we have to further check whether the collapse lateral load that will result while considering intermediate pile failure is not lower than 527.03 kN. Employing Eq. 6.110 and substituting f<\/em> from Eq. 6.104 yields:\n\n[latex]{H_f} = \\dfrac{{{M_y} + 2.25{S_u}{D_{ext}}{{\\left( {L - 1.5D - \\dfrac{{{H_f}}}{{9{S_u}{D_{ext}}}}} \\right)}^2}}}{{\\left( {1.5D + 0.5\\dfrac{{{H_f}}}{{9{S_u}{D_{ext}}}}} \\right)}}[\/latex]\n\nThe above equation can be solved for Hf<\/sub><\/em>, and results in \u0397<\/em>f <\/em><\/sub>= 1333.6 kN > 527.03 kN, which means that long pile failure is critical. Indeed, if we substitute Hf<\/sub><\/em> in Eq. 6.111 for the bending moment that will develop at a depth 1.5D<\/em>+f<\/em>:\n\n[latex]{M_{\\max }} = 2.25{S_u}{D_{ext}}{\\left( {L - 1.5{D_{ext}} - \\dfrac{{{H_f}}}{{9{S_u}{D_{ext}}}}} \\right)^2} = 2.25 \\times 100 \\times 0.4 \\times {\\left( {10 - 1.5 \\times 0.4 - \\dfrac{{527.03}}{{9 \\times 100 \\times 0.4}}} \\right)^2} = 5668{\\rm{ \\:kNm}} > {M_y}[\/latex]\n\nwhich implies that two plastic hinges will be formed along the pile.\n\nThe pile head displacement for the working lateral load Hw <\/sub><\/em>= Hf<\/sub><\/em>\/2 = 263.5 kN will be estimated while calculating the modulus of subgrade reaction according to Davidson (1972) as:\n\n[latex]{K_h} = 67\\dfrac{{{S_u}}}{{{D_{ext}}}} = 16750{\\rm{ \\:kN\/}}{{\\rm{m}}^3}[\/latex]\n\nIn order to use Figure 6.101<\/a>, we must first estimate the factor \u03b2<\/em> from Eq. 6.115:\n\n[latex]\\beta = {\\left( {\\dfrac{{{K_h}{D_{ext}}}}{{4{E_p}{I_p}}}} \\right)^{0.25}}[\/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]\\beta = {\\left( {\\dfrac{{16750\\times 0.4}}{{4 \\times 210 \\times {{10}^6} \\times 1.21 \\times {{10}^{ - 4}}}}} \\right)^{0.25}} = 0.50[\/latex]\n\nConsidering the curve for fixed-head piles of Figure 6.101<\/a> and \u03b2<\/em>L<\/em> = 5, we obtain:\n\n[latex]y(0) = 5\\left( {\\dfrac{{{H_w}}}{{{K_h}{D_{ext}}L}}} \\right) = 5\\left( {\\dfrac{{263.5}}{{16750 \\times 0.4 \\times 10}}} \\right) = 0.019{\\rm{ m}}[\/latex]","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, stiff overconsolidated clay with undrained shear strength Su <\/sub><\/em>= 100 kPa. 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 also to AS2159. To do that, we must first estimate the yield moment My<\/sub><\/em> of the pile\u2019s section, which is the maximum moment that the pile\u2019s section can carry before a plastic hinge is formed. For a hollow pipe pile and homogeneous material, the plastic moment is estimated from Table 6.14 as:<\/p>\n

[latex]{M_y} = \\left( {\\dfrac{{D_{ext}^3 - D_{{\\mathop{\\rm int}} }^3}}{6}} \\right){\\sigma _y} = \\left( {\\dfrac{{{{0.4}^3} - {{0.39}^3}}}{6}} \\right)450000 = 351{\\rm{ \\:kNm}}[\/latex]<\/p>\n

We will start by applying the short-pile equations. If the resulting maximum pile moment Mmax<\/sub><\/em> is higher than the yield moment (351 kNm), the long-pile equations should be used instead, as they will result in a lower collapse load. For the case herein, the collapse lateral load considering short-pile failure is (Eq. 6.106):<\/p>\n

[latex]{H_f} = 9{S_u}{D_{ext}}\\left( {L - 1.5{D_{ext}}} \\right) = 9 \\times 100 \\times 0.4 \\times \\left( {10 - 1.5 \\times 0.4} \\right) = 3384{\\rm{ \\:kN}}[\/latex]<\/p>\n

and the maximum moment corresponding the ultimate lateral load (Eq. 6.107):<\/p>\n

[latex]{M_{\\max }} = 4.5{S_u}{D_{ext}}\\left( {{L^2} - 2.25D_{ext}^2} \\right) = 4.5 \\times 100 \\times 0.4 \\times \\left( {{{10}^2} - 2.25 \\times {{0.4}^2}} \\right) = 17935{\\rm{ \\:kNm}} > {M_y}[\/latex]<\/p>\n

It is thus clear that the pile will develop a plastic hinge before a short-pile failure mode fully develops, and the long-pile equations must be used instead. The collapse later load considering long-pile failure is given by substituting Eq. 6.104 into Eq. 6.109 as:<\/p>\n

[latex]{H_f} = \\dfrac{{2{M_y}}}{{\\left( {1.5{D_{ext}} + 0.5\\dfrac{{{H_f}}}{{9{S_u}{D_{ext}}}}} \\right)}} = \\dfrac{{2 \\times 351}}{{1.5 \\times 0.4 + 0.5\\left( {\\dfrac{{{H_f}}}{{9 \\times 100 \\times 0.4}}} \\right)}}\\[\/latex]<\/p>\n

[latex]{H_f} = 527.03{\\rm{ \\:kN}}[\/latex]<\/p>\n

However, we have to further check whether the collapse lateral load that will result while considering intermediate pile failure is not lower than 527.03 kN. Employing Eq. 6.110 and substituting f<\/em> from Eq. 6.104 yields:<\/p>\n

[latex]{H_f} = \\dfrac{{{M_y} + 2.25{S_u}{D_{ext}}{{\\left( {L - 1.5D - \\dfrac{{{H_f}}}{{9{S_u}{D_{ext}}}}} \\right)}^2}}}{{\\left( {1.5D + 0.5\\dfrac{{{H_f}}}{{9{S_u}{D_{ext}}}}} \\right)}}[\/latex]<\/p>\n

The above equation can be solved for Hf<\/sub><\/em>, and results in \u0397<\/em>f <\/em><\/sub>= 1333.6 kN > 527.03 kN, which means that long pile failure is critical. Indeed, if we substitute Hf<\/sub><\/em> in Eq. 6.111 for the bending moment that will develop at a depth 1.5D<\/em>+f<\/em>:<\/p>\n

[latex]{M_{\\max }} = 2.25{S_u}{D_{ext}}{\\left( {L - 1.5{D_{ext}} - \\dfrac{{{H_f}}}{{9{S_u}{D_{ext}}}}} \\right)^2} = 2.25 \\times 100 \\times 0.4 \\times {\\left( {10 - 1.5 \\times 0.4 - \\dfrac{{527.03}}{{9 \\times 100 \\times 0.4}}} \\right)^2} = 5668{\\rm{ \\:kNm}} > {M_y}[\/latex]<\/p>\n

which implies that two plastic hinges will be formed along the pile.<\/p>\n

The pile head displacement for the working lateral load Hw <\/sub><\/em>= Hf<\/sub><\/em>\/2 = 263.5 kN will be estimated while calculating the modulus of subgrade reaction according to Davidson (1972) as:<\/p>\n

[latex]{K_h} = 67\\dfrac{{{S_u}}}{{{D_{ext}}}} = 16750{\\rm{ \\:kN\/}}{{\\rm{m}}^3}[\/latex]<\/p>\n

In order to use Figure 6.101<\/a>, we must first estimate the factor \u03b2<\/em> from Eq. 6.115:<\/p>\n

[latex]\\beta = {\\left( {\\dfrac{{{K_h}{D_{ext}}}}{{4{E_p}{I_p}}}} \\right)^{0.25}}[\/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]\\beta = {\\left( {\\dfrac{{16750\\times 0.4}}{{4 \\times 210 \\times {{10}^6} \\times 1.21 \\times {{10}^{ - 4}}}}} \\right)^{0.25}} = 0.50[\/latex]<\/p>\n

Considering the curve for fixed-head piles of Figure 6.101<\/a> and \u03b2<\/em>L<\/em> = 5, we obtain:<\/p>\n

[latex]y(0) = 5\\left( {\\dfrac{{{H_w}}}{{{K_h}{D_{ext}}L}}} \\right) = 5\\left( {\\dfrac{{263.5}}{{16750 \\times 0.4 \\times 10}}} \\right) = 0.019{\\rm{ m}}[\/latex]<\/p>\n","protected":false},"author":1,"menu_order":36,"template":"","meta":{"pb_show_title":"","pb_short_title":"Example 6.12","pb_subtitle":"Calculation of the ultimate geotechnical strength and of the pile head deflection of a laterally loaded pile driven in undrained soil","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-610","chapter","type-chapter","status-publish","hentry"],"part":421,"_links":{"self":[{"href":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-json\/pressbooks\/v2\/chapters\/610","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\/610\/revisions"}],"predecessor-version":[{"id":611,"href":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-json\/pressbooks\/v2\/chapters\/610\/revisions\/611"}],"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\/610\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-json\/wp\/v2\/media?parent=610"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-json\/pressbooks\/v2\/chapter-type?post=610"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-json\/wp\/v2\/contributor?post=610"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-json\/wp\/v2\/license?post=610"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}