{"id":519,"date":"2025-03-31T04:07:02","date_gmt":"2025-03-31T04:07:02","guid":{"rendered":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/chapter\/example-6-5-estimation-of-the-ultimate-geotechnical-strength-of-a-pile-subjected-to-axial-compressive-load-based-on-spt-test-results\/"},"modified":"2026-03-16T14:11:02","modified_gmt":"2026-03-16T14:11:02","slug":"example-6-5-estimation-of-the-ultimate-geotechnical-strength-of-a-pile-subjected-to-axial-compressive-load-based-on-spt-test-results","status":"publish","type":"chapter","link":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/chapter\/example-6-5-estimation-of-the-ultimate-geotechnical-strength-of-a-pile-subjected-to-axial-compressive-load-based-on-spt-test-results\/","title":{"raw":"Example 6.5","rendered":"Example 6.5"},"content":{"raw":"Determine the ultimate geotechnical strength <em>Q<sub>f<\/sub><\/em> of the pile shown below. The pile is subjected to axial compressive load and is embedded in homogeneous sand. Use the SPT-based method (<a href=\"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/chapter\/6-11-ultimate-geotechnical-strength-of-piles-subjected-to-axial-compressive-load-from-spt-test-results\/\">Chapter 6.11<\/a>) and the SPT results provided below. Ignore the weight of the pile in the calculations.\n\n[caption id=\"attachment_518\" align=\"aligncenter\" width=\"600\"]<img class=\"wp-image-518 size-full\" src=\"https:\/\/libraryresources.nse.org.ng\/wp-content\/uploads\/sites\/9\/2025\/03\/example-6.5brief-hr-e1743553575403.png\" alt=\"The figure on the left presents a pile in sand, loaded with axial compressive force Qf on its head. The length of the pile is 10 m and its diameter is 1 m. The groundwater table is found below the ground surface. The figure on the right presents the variation of N60 values with depth, up to a depth of 15 m. N60 values range between 5 and 28. The average N60 value along the length of the pile is N60=9.5. The average N60 value from 10D above to 4D below the pile toe is N60=13.8.\" width=\"600\" height=\"473\"> Example 6.5. Problem description and input parameters.[\/caption]\n<h2>Answer:<\/h2>\n1. Calculation of skin friction resistance, <em>Q<sub>sf<\/sub><\/em>:\n\nWe will use the upper bound of Eq. 6.21. for driven piles. The average SPT value along the length of the pile <em>N<sub>60,shaft<\/sub><\/em> \u00a0is equal to <em>N<sub>60,shaft<\/sub><\/em> = 9.5, thus:\n\n[latex]{Q_{sf}} = 0.02{p_a}{N_{60,shaft}}\\pi DL = 0.02 \\times 100 \\times 9.5 \\times \\pi \\times 1 \\times 10 = 597{\\rm{ \\:kN}}[\/latex]\n\n2. End-bearing resistance, <em>Q<sub>b<\/sub><\/em>:\n\nFor the estimation of the end-bearing resistance, we have to find the average SPT value at the vicinity of the pile toe, considering blow counts about 10<em>D <\/em>= 10m above and 4<em>D <\/em>= 4m below the pile toe (<a href=\"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/chapter\/6-11-ultimate-geotechnical-strength-of-piles-subjected-to-axial-compressive-load-from-spt-test-results\/\">Chapter 6.11<\/a>). As depicted from the figure above, the average SPT value to be considered for the estimation of the end-bearing resistance is <em>\u039d<sub>60,toe <\/sub><\/em>= 13.8. Substituting in Eq. 6.22:\n\n[latex]{Q_b} = 0.4{p_a}{N_{60,toe}}\\dfrac{L}{D}\\left( {\\pi {{\\left( {\\dfrac{D}{2}} \\right)}^2}} \\right) = 0.4 \\times 100 \\times 13.8 \\times \\dfrac{{10}}{1} \\times \\left( {\\pi {{0.5}^2}} \\right) = 4335{\\rm{ \\:kN}} \\le 4{p_a}{N_{60,toe}}\\left( {\\pi {{\\left( {\\dfrac{D}{2}} \\right)}^2}} \\right) = 4335{\\rm{ \\:kN}}[\/latex]\n\n3. Ultimate geotechnical strength, <em>Q<sub>f<\/sub><\/em>:\n\nThe ultimate geotechnical strength of the pile is again the sum of its skin friction and end-bearing resistance:\n\n[latex]{Q_f} = {Q_b} + {Q_{sf}} = 4932{\\rm{ \\:kN}}[\/latex]\n\nNotice that considering the full extent of the possible failure surface from <a href=\"https:\/\/oercollective.caul.edu.au\/app\/uploads\/sites\/143\/2025\/03\/6.22-extent-of-possible-failure-surface.png\">Figure 6.22<\/a> to get the average <em>N<sub>60<\/sub><\/em> value implies that toe failure for this short pile with <em>L\/D<\/em>=10 will reach the soil surface. This is probably unrealistic, yet it is a conservative assumption, as generally <em>N<sub>60<\/sub><\/em> values increase with depth.\n\nWhile using similar methodologies that are based on non-site specific correlations, care must be taken so that appropriate, conservative risk ratings are introduced during the estimation of the average risk rating after AS2159.","rendered":"<p>Determine the ultimate geotechnical strength <em>Q<sub>f<\/sub><\/em> of the pile shown below. The pile is subjected to axial compressive load and is embedded in homogeneous sand. Use the SPT-based method (<a href=\"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/chapter\/6-11-ultimate-geotechnical-strength-of-piles-subjected-to-axial-compressive-load-from-spt-test-results\/\">Chapter 6.11<\/a>) and the SPT results provided below. Ignore the weight of the pile in the calculations.<\/p>\n<figure id=\"attachment_518\" aria-describedby=\"caption-attachment-518\" style=\"width: 600px\" class=\"wp-caption aligncenter\"><img decoding=\"async\" class=\"wp-image-518 size-full\" src=\"https:\/\/libraryresources.nse.org.ng\/wp-content\/uploads\/sites\/9\/2025\/03\/example-6.5brief-hr-e1743553575403.png\" alt=\"The figure on the left presents a pile in sand, loaded with axial compressive force Qf on its head. The length of the pile is 10 m and its diameter is 1 m. The groundwater table is found below the ground surface. The figure on the right presents the variation of N60 values with depth, up to a depth of 15 m. N60 values range between 5 and 28. The average N60 value along the length of the pile is N60=9.5. The average N60 value from 10D above to 4D below the pile toe is N60=13.8.\" width=\"600\" height=\"473\" srcset=\"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-content\/uploads\/sites\/9\/2025\/03\/example-6.5brief-hr-e1743553575403.png 600w, https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-content\/uploads\/sites\/9\/2025\/03\/example-6.5brief-hr-e1743553575403-300x237.png 300w, https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-content\/uploads\/sites\/9\/2025\/03\/example-6.5brief-hr-e1743553575403-65x51.png 65w, https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-content\/uploads\/sites\/9\/2025\/03\/example-6.5brief-hr-e1743553575403-225x177.png 225w, https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-content\/uploads\/sites\/9\/2025\/03\/example-6.5brief-hr-e1743553575403-350x276.png 350w\" sizes=\"(max-width: 600px) 100vw, 600px\" \/><figcaption id=\"caption-attachment-518\" class=\"wp-caption-text\">Example 6.5. Problem description and input parameters.<\/figcaption><\/figure>\n<h2>Answer:<\/h2>\n<p>1. Calculation of skin friction resistance, <em>Q<sub>sf<\/sub><\/em>:<\/p>\n<p>We will use the upper bound of Eq. 6.21. for driven piles. The average SPT value along the length of the pile <em>N<sub>60,shaft<\/sub><\/em> \u00a0is equal to <em>N<sub>60,shaft<\/sub><\/em> = 9.5, thus:<\/p>\n<p>[latex]{Q_{sf}} = 0.02{p_a}{N_{60,shaft}}\\pi DL = 0.02 \\times 100 \\times 9.5 \\times \\pi \\times 1 \\times 10 = 597{\\rm{ \\:kN}}[\/latex]<\/p>\n<p>2. End-bearing resistance, <em>Q<sub>b<\/sub><\/em>:<\/p>\n<p>For the estimation of the end-bearing resistance, we have to find the average SPT value at the vicinity of the pile toe, considering blow counts about 10<em>D <\/em>= 10m above and 4<em>D <\/em>= 4m below the pile toe (<a href=\"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/chapter\/6-11-ultimate-geotechnical-strength-of-piles-subjected-to-axial-compressive-load-from-spt-test-results\/\">Chapter 6.11<\/a>). As depicted from the figure above, the average SPT value to be considered for the estimation of the end-bearing resistance is <em>\u039d<sub>60,toe <\/sub><\/em>= 13.8. Substituting in Eq. 6.22:<\/p>\n<p>[latex]{Q_b} = 0.4{p_a}{N_{60,toe}}\\dfrac{L}{D}\\left( {\\pi {{\\left( {\\dfrac{D}{2}} \\right)}^2}} \\right) = 0.4 \\times 100 \\times 13.8 \\times \\dfrac{{10}}{1} \\times \\left( {\\pi {{0.5}^2}} \\right) = 4335{\\rm{ \\:kN}} \\le 4{p_a}{N_{60,toe}}\\left( {\\pi {{\\left( {\\dfrac{D}{2}} \\right)}^2}} \\right) = 4335{\\rm{ \\:kN}}[\/latex]<\/p>\n<p>3. Ultimate geotechnical strength, <em>Q<sub>f<\/sub><\/em>:<\/p>\n<p>The ultimate geotechnical strength of the pile is again the sum of its skin friction and end-bearing resistance:<\/p>\n<p>[latex]{Q_f} = {Q_b} + {Q_{sf}} = 4932{\\rm{ \\:kN}}[\/latex]<\/p>\n<p>Notice that considering the full extent of the possible failure surface from <a href=\"https:\/\/oercollective.caul.edu.au\/app\/uploads\/sites\/143\/2025\/03\/6.22-extent-of-possible-failure-surface.png\">Figure 6.22<\/a> to get the average <em>N<sub>60<\/sub><\/em> value implies that toe failure for this short pile with <em>L\/D<\/em>=10 will reach the soil surface. This is probably unrealistic, yet it is a conservative assumption, as generally <em>N<sub>60<\/sub><\/em> values increase with depth.<\/p>\n<p>While using similar methodologies that are based on non-site specific correlations, care must be taken so that appropriate, conservative risk ratings are introduced during the estimation of the average risk rating after AS2159.<\/p>\n","protected":false},"author":1,"menu_order":20,"template":"","meta":{"pb_show_title":"","pb_short_title":"Example 6.5","pb_subtitle":"Estimation of the ultimate geotechnical strength of a pile subjected to axial compressive load based on SPT test results","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-519","chapter","type-chapter","status-publish","hentry"],"part":421,"_links":{"self":[{"href":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-json\/pressbooks\/v2\/chapters\/519","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\/519\/revisions"}],"predecessor-version":[{"id":520,"href":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-json\/pressbooks\/v2\/chapters\/519\/revisions\/520"}],"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\/519\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-json\/wp\/v2\/media?parent=519"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-json\/pressbooks\/v2\/chapter-type?post=519"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-json\/wp\/v2\/contributor?post=519"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-json\/wp\/v2\/license?post=519"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}