{"id":574,"date":"2025-03-31T04:46:00","date_gmt":"2025-03-31T04:46:00","guid":{"rendered":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/chapter\/example-6-10-calculation-of-settlement-of-a-group-of-piles-connected-with-rigid-and-flexible-pile-cap\/"},"modified":"2026-03-16T14:13:31","modified_gmt":"2026-03-16T14:13:31","slug":"example-6-10-calculation-of-settlement-of-a-group-of-piles-connected-with-rigid-and-flexible-pile-cap","status":"publish","type":"chapter","link":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/chapter\/example-6-10-calculation-of-settlement-of-a-group-of-piles-connected-with-rigid-and-flexible-pile-cap\/","title":{"raw":"Example 6.10","rendered":"Example 6.10"},"content":{"raw":"Consider the floating pile group depicted in the figure below, consisting of six concrete piles of diameter D<\/em> = 300 mm driven into a deep clay layer. The group is subjected to a vertical compressive load Qw,group<\/sub><\/em> = 3000 kN. A load test was performed on a single pile, and resulted in settlement of 15 mm under load 500 kN. The ratio of the Young\u2019s modulus of the pile over the Young\u2019s modulus of the clay is Ep<\/sub><\/em>\/Es <\/sub><\/em>= 2000. Determine the immediate settlement of the pile group if i) the pile cap is rigid, and ii) the pile cap is flexible.\n\n[caption id=\"attachment_573\" align=\"aligncenter\" width=\"450\"]\"Schematic Example 6.10. Problem description and input parameters.[\/caption]\n\nWe need first to determine the pile stiffness factor Kp <\/sub><\/em>from Eq. 6.93. Since the piles are solid it is Kp <\/sub><\/em>= Ep<\/sub><\/em>\/Es <\/sub><\/em>= 2000.\n\nCorner piles 1, 3, 4 and 6 will behave identically, and we will call them Pile Type A. Mid piles 2 and 5 will also behave identically, and we will call them Pile Type B. We will also denote the serviceability load on piles A is B as QA<\/sub><\/em> and QB<\/sub><\/em>, respectively.\n\nWe can find the interaction factors \u03b1<\/em>i,ij<\/sub> from the chart in Figure 6.77<\/a> for L<\/em>\/D <\/em>= 25, interpolating between Kp <\/sub><\/em>= 1000 and Kp <\/sub><\/em>=\u221e. We need interaction factors for two piles types: Pile Type A (e.g. for Pile 1) and Pile Type B (e.g. for Pile 2).\n\n\n\n\n\n\n\n\n\n
Example 6.10: Interaction factors \u03b1ij<\/sub><\/em><\/caption>\n
PIle\u00a0j<\/em><\/th>\nPile 1 (Pile type A)<\/th>\nPile 2 (Pile type B)<\/th>\n<\/tr>\n
s\/D<\/em><\/td>\n\u03b1i,ij<\/sub><\/em><\/td>\ns\/D<\/em><\/td>\n\u03b1i,ij<\/sub><\/em><\/td>\n<\/tr>\n
1<\/td>\n0<\/td>\n-<\/td>\n5<\/td>\n0.42<\/td>\n<\/tr>\n
2<\/td>\n5 (i.e., 1500 mm\/300 mm)<\/td>\n0.42<\/td>\n0<\/td>\n-<\/td>\n<\/tr>\n
3<\/td>\n10<\/td>\n0.27<\/td>\n5<\/td>\n0.42<\/td>\n<\/tr>\n
4<\/td>\n5<\/td>\n0.42<\/td>\n7.07<\/td>\n0.35<\/td>\n<\/tr>\n
5<\/td>\n7.07<\/td>\n0.35<\/td>\n5<\/td>\n0.42<\/td>\n<\/tr>\n
6<\/td>\n11.2<\/td>\n0.25<\/td>\n7.07<\/td>\n0.35<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\nWe can now cast Eq. 6.100 for all piles. It is:\n\nSettlement of pile 1 (and all Type A piles), \u03c1\u0391<\/sub><\/em>:\n\n[latex]{\\rho _A} = {\\rho _1}\\left[ {{Q_A}\\left( {0.27 + 0.42 + 0.25} \\right) + {Q_B}\\left( {0.42 + 0.35} \\right) + {Q_A}} \\right] \\to \\dfrac{{{\\rho _A}\\,}}{{{\\rho _1}}} = 1.94{P_A} + 0.77{P_B}[\/latex]\n\nSettlement of pile 2 (and all Type B piles), \u03c1B<\/sub><\/em>:\n\n[latex]{\\rho _B} = {\\rho _1}\\left[ {{Q_A}\\left( {0.42 + 0.42 + 0.35 + 0.35} \\right) + {Q_B}\\left( {0.42} \\right) + {Q_B}} \\right] \\to \\dfrac{{{\\rho _B}\\,}}{{{\\rho _1}}} = 1.54{P_A} + 1.42{P_B}[\/latex]\n\nIn addition, the force equilibrium Eq. 6.101 is:\n\n[latex]{Q_{w,group}} = 3000 = 4{Q_A} + 2{Q_B}[\/latex]\n\nWe have three equations, with four unknowns, the two settlements and the two loads acting on each pile type. So we need one more equation.\n\nCase 1: Piles connected with a rigid pile cap\n\nIn this case the extra equation results from considering that the settlement of all piles will be the same, therefore \u03c1\u0391<\/sub><\/em>\/\u03c1<\/em>1<\/sub> = \u03c1\u0392<\/sub><\/em>\/\u03c1<\/em>1<\/sub>. In addition, settlement of a pile under unit load \u03c1<\/em>1<\/sub> is \u03c1<\/em>1<\/em> <\/sub>= 15\/500 = 0.03 mm\/k\u039d. As such:\n\n[latex]1.54{Q_A} + 1.42{Q_B} = 1.94{Q_A} + 0.77{Q_B} \\to \\dfrac{{{Q_A}}}{{{Q_B}}} = 1.625[\/latex]\n\nand\n\n[latex]3000 = 4{Q_A} + 2{Q_B}[\/latex]\n\n[latex]\\to {Q_B} = 352.9{\\rm{\\: kN}}[\/latex]\n\n[latex]\\to {Q_A} = 573.5{ }{\\rm{\\:kN}}[\/latex]\n\nAs expected, when the pile cap is rigid the corner piles are carrying larger axial loads, relatively to the mid piles.\n\n[latex]\\dfrac{{{\\rho _A}\\,}}{{{\\rho _1}}} = \\dfrac{{{\\rho _B}\\,}}{{{\\rho _1}}} = 1.94{Q_A} + 0.77{Q_B} = 1384.32{\\rm{\\:kN}}[\/latex]\n\n[latex]\\to {\\rho _A} = {\\rho _B} = 41.5{\\rm{ \\:mm}}[\/latex]\n\nNote that for a six-pile group (n<\/em> = 6) and \u03a6<\/em> = 0.5 Eq. 6.96 yields Rs<\/sub><\/em> = 2.44. In other words, the simplified method of Flemming et al.<\/em> predicts that settlement of each pile, carrying load 3000\/6 = 500 kN, will be 500 x 0.03 x 2.44 = 36.6 mm. The advantage of this, more rigorous, method, apart of course of considering the distribution of the vertical load on piles, is that it does not require assuming a value for the factor \u03a6<\/em>.\n\nCase 2: Piles connected with a flexible pile cap\n\nIf the piles are connected with a flexible pile cap, we can assume that they will all be carrying the same vertical load. Therefore our fourth equation is QA<\/sub><\/em> = QB<\/sub><\/em> = 500 kN. Accordingly:\n\n[latex]\\dfrac{{{\\rho _A}\\,}}{{{\\rho _1}}} = 1.94{Q_A} + 0.77{Q_B} = 1355{\\rm{ \\:kN}}[\/latex]\n\n[latex]\\to {\\rho _A} = 40.65{\\rm{\\:mm}}[\/latex]\n\n[latex]\\dfrac{{{\\rho _B}\\,}}{{{\\rho _1}}} = 1.54{P_A} + 1.42{P_B} = 1480{\\rm{\\:kN}}[\/latex]\n\n[latex]\\to {\\rho _B}\\, = 44.4{\\rm{\\:mm}}[\/latex]\n\nNotice that there is some differential settlement between piles connected with a flexible cap, with the mid piles 2 and 5 (Piles B) settling, as expected, more. This differential settlement may be important, and should be considered in the design of the pile group.","rendered":"

Consider the floating pile group depicted in the figure below, consisting of six concrete piles of diameter D<\/em> = 300 mm driven into a deep clay layer. The group is subjected to a vertical compressive load Qw,group<\/sub><\/em> = 3000 kN. A load test was performed on a single pile, and resulted in settlement of 15 mm under load 500 kN. The ratio of the Young\u2019s modulus of the pile over the Young\u2019s modulus of the clay is Ep<\/sub><\/em>\/Es <\/sub><\/em>= 2000. Determine the immediate settlement of the pile group if i) the pile cap is rigid, and ii) the pile cap is flexible.<\/p>\n

\"Schematic
Example 6.10. Problem description and input parameters.<\/figcaption><\/figure>\n

We need first to determine the pile stiffness factor Kp <\/sub><\/em>from Eq. 6.93. Since the piles are solid it is Kp <\/sub><\/em>= Ep<\/sub><\/em>\/Es <\/sub><\/em>= 2000.<\/p>\n

Corner piles 1, 3, 4 and 6 will behave identically, and we will call them Pile Type A. Mid piles 2 and 5 will also behave identically, and we will call them Pile Type B. We will also denote the serviceability load on piles A is B as QA<\/sub><\/em> and QB<\/sub><\/em>, respectively.<\/p>\n

We can find the interaction factors \u03b1<\/em>i,ij<\/sub> from the chart in Figure 6.77<\/a> for L<\/em>\/D <\/em>= 25, interpolating between Kp <\/sub><\/em>= 1000 and Kp <\/sub><\/em>=\u221e. We need interaction factors for two piles types: Pile Type A (e.g. for Pile 1) and Pile Type B (e.g. for Pile 2).<\/p>\n\n\n\n\n\n\n\n\n\n\n
Example 6.10: Interaction factors \u03b1ij<\/sub><\/em><\/caption>\n
PIle\u00a0j<\/em><\/th>\nPile 1 (Pile type A)<\/th>\nPile 2 (Pile type B)<\/th>\n<\/tr>\n
s\/D<\/em><\/td>\n\u03b1i,ij<\/sub><\/em><\/td>\ns\/D<\/em><\/td>\n\u03b1i,ij<\/sub><\/em><\/td>\n<\/tr>\n
1<\/td>\n0<\/td>\n–<\/td>\n5<\/td>\n0.42<\/td>\n<\/tr>\n
2<\/td>\n5 (i.e., 1500 mm\/300 mm)<\/td>\n0.42<\/td>\n0<\/td>\n–<\/td>\n<\/tr>\n
3<\/td>\n10<\/td>\n0.27<\/td>\n5<\/td>\n0.42<\/td>\n<\/tr>\n
4<\/td>\n5<\/td>\n0.42<\/td>\n7.07<\/td>\n0.35<\/td>\n<\/tr>\n
5<\/td>\n7.07<\/td>\n0.35<\/td>\n5<\/td>\n0.42<\/td>\n<\/tr>\n
6<\/td>\n11.2<\/td>\n0.25<\/td>\n7.07<\/td>\n0.35<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

We can now cast Eq. 6.100 for all piles. It is:<\/p>\n

Settlement of pile 1 (and all Type A piles), \u03c1\u0391<\/sub><\/em>:<\/p>\n

[latex]{\\rho _A} = {\\rho _1}\\left[ {{Q_A}\\left( {0.27 + 0.42 + 0.25} \\right) + {Q_B}\\left( {0.42 + 0.35} \\right) + {Q_A}} \\right] \\to \\dfrac{{{\\rho _A}\\,}}{{{\\rho _1}}} = 1.94{P_A} + 0.77{P_B}[\/latex]<\/p>\n

Settlement of pile 2 (and all Type B piles), \u03c1B<\/sub><\/em>:<\/p>\n

[latex]{\\rho _B} = {\\rho _1}\\left[ {{Q_A}\\left( {0.42 + 0.42 + 0.35 + 0.35} \\right) + {Q_B}\\left( {0.42} \\right) + {Q_B}} \\right] \\to \\dfrac{{{\\rho _B}\\,}}{{{\\rho _1}}} = 1.54{P_A} + 1.42{P_B}[\/latex]<\/p>\n

In addition, the force equilibrium Eq. 6.101 is:<\/p>\n

[latex]{Q_{w,group}} = 3000 = 4{Q_A} + 2{Q_B}[\/latex]<\/p>\n

We have three equations, with four unknowns, the two settlements and the two loads acting on each pile type. So we need one more equation.<\/p>\n

Case 1: Piles connected with a rigid pile cap<\/p>\n

In this case the extra equation results from considering that the settlement of all piles will be the same, therefore \u03c1\u0391<\/sub><\/em>\/\u03c1<\/em>1<\/sub> = \u03c1\u0392<\/sub><\/em>\/\u03c1<\/em>1<\/sub>. In addition, settlement of a pile under unit load \u03c1<\/em>1<\/sub> is \u03c1<\/em>1<\/em> <\/sub>= 15\/500 = 0.03 mm\/k\u039d. As such:<\/p>\n

[latex]1.54{Q_A} + 1.42{Q_B} = 1.94{Q_A} + 0.77{Q_B} \\to \\dfrac{{{Q_A}}}{{{Q_B}}} = 1.625[\/latex]<\/p>\n

and<\/p>\n

[latex]3000 = 4{Q_A} + 2{Q_B}[\/latex]<\/p>\n

[latex]\\to {Q_B} = 352.9{\\rm{\\: kN}}[\/latex]<\/p>\n

[latex]\\to {Q_A} = 573.5{ }{\\rm{\\:kN}}[\/latex]<\/p>\n

As expected, when the pile cap is rigid the corner piles are carrying larger axial loads, relatively to the mid piles.<\/p>\n

[latex]\\dfrac{{{\\rho _A}\\,}}{{{\\rho _1}}} = \\dfrac{{{\\rho _B}\\,}}{{{\\rho _1}}} = 1.94{Q_A} + 0.77{Q_B} = 1384.32{\\rm{\\:kN}}[\/latex]<\/p>\n

[latex]\\to {\\rho _A} = {\\rho _B} = 41.5{\\rm{ \\:mm}}[\/latex]<\/p>\n

Note that for a six-pile group (n<\/em> = 6) and \u03a6<\/em> = 0.5 Eq. 6.96 yields Rs<\/sub><\/em> = 2.44. In other words, the simplified method of Flemming et al.<\/em> predicts that settlement of each pile, carrying load 3000\/6 = 500 kN, will be 500 x 0.03 x 2.44 = 36.6 mm. The advantage of this, more rigorous, method, apart of course of considering the distribution of the vertical load on piles, is that it does not require assuming a value for the factor \u03a6<\/em>.<\/p>\n

Case 2: Piles connected with a flexible pile cap<\/p>\n

If the piles are connected with a flexible pile cap, we can assume that they will all be carrying the same vertical load. Therefore our fourth equation is QA<\/sub><\/em> = QB<\/sub><\/em> = 500 kN. Accordingly:<\/p>\n

[latex]\\dfrac{{{\\rho _A}\\,}}{{{\\rho _1}}} = 1.94{Q_A} + 0.77{Q_B} = 1355{\\rm{ \\:kN}}[\/latex]<\/p>\n

[latex]\\to {\\rho _A} = 40.65{\\rm{\\:mm}}[\/latex]<\/p>\n

[latex]\\dfrac{{{\\rho _B}\\,}}{{{\\rho _1}}} = 1.54{P_A} + 1.42{P_B} = 1480{\\rm{\\:kN}}[\/latex]<\/p>\n

[latex]\\to {\\rho _B}\\, = 44.4{\\rm{\\:mm}}[\/latex]<\/p>\n

Notice that there is some differential settlement between piles connected with a flexible cap, with the mid piles 2 and 5 (Piles B) settling, as expected, more. This differential settlement may be important, and should be considered in the design of the pile group.<\/p>\n","protected":false},"author":1,"menu_order":30,"template":"","meta":{"pb_show_title":"","pb_short_title":"Example 6.10","pb_subtitle":"Calculation of settlement of a group of piles connected with rigid and flexible pile cap","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-574","chapter","type-chapter","status-publish","hentry"],"part":421,"_links":{"self":[{"href":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-json\/pressbooks\/v2\/chapters\/574","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\/574\/revisions"}],"predecessor-version":[{"id":575,"href":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-json\/pressbooks\/v2\/chapters\/574\/revisions\/575"}],"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\/574\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-json\/wp\/v2\/media?parent=574"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-json\/pressbooks\/v2\/chapter-type?post=574"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-json\/wp\/v2\/contributor?post=574"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-json\/wp\/v2\/license?post=574"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}