{"id":198,"date":"2025-03-28T03:42:56","date_gmt":"2025-03-28T03:42:56","guid":{"rendered":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/chapter\/example-3-1-stresses-in-soil-due-to-a-water-tank-founded-on-a-ring-type-foundation\/"},"modified":"2026-03-16T13:57:56","modified_gmt":"2026-03-16T13:57:56","slug":"example-3-1-stresses-in-soil-due-to-a-water-tank-founded-on-a-ring-type-foundation","status":"publish","type":"chapter","link":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/chapter\/example-3-1-stresses-in-soil-due-to-a-water-tank-founded-on-a-ring-type-foundation\/","title":{"raw":"Example 3.1","rendered":"Example 3.1"},"content":{"raw":"An H <\/em>= 4 m tall water tank with radius rexternal <\/sub><\/em>= 5 m is going to be founded on a ring foundation, with the dimensions depicted in the following plan view. Assuming the tank is filled with water, determine the vertical stress due to the weight of the tank below the centre of the foundation, at a depth z <\/em>= 5 m.\n\n[caption id=\"attachment_197\" align=\"aligncenter\" width=\"350\"]\"The Example 3.1. Problem description and input parameters.[\/caption]\n

Answer:<\/h2>\nFirst we have to determine the maximum load, Qext<\/sub><\/em> on the foundation, assuming that the tank is filled with water:\n\nAtank<\/sub> = <\/em>\u03c0rexternal<\/sub>2 <\/sup><\/em>= 78.54 m2<\/sup>\n\nQext<\/sub> = A \u00d7 H \u00d7 y<\/em>w <\/sub>= 78.54 m2<\/sup> x 4 m x 10 kN\/m3<\/sup> = 3141 kN where 10 kN\/m3<\/sup> is the unit weight of water.\n\nThe area of the ring footing on which the load is applied is:\n\nAfooting<\/sub>= <\/em>\u03c0rexternal<\/sub>2 <\/sup>- <\/em>\u03c0rinternal<\/sub>2 <\/em>=78.54 m2<\/sup>-50.26 m2<\/sup> = 28.28 m2<\/sup>\n\nthus the pressure applied on the ring foundation of the tank will be:\n\nqext<\/sub> = Qext<\/sub>\/Afooting <\/sub><\/em>= 111 kPa\n\nEmploying the principle of superposition discussed in Chapter 3.5<\/a>, this ring pressure is equivalent, in terms of stresses applied to the soil, to a circular pressure qext<\/sub><\/em>,1<\/sub> = 111 kPa on a radius rexternal <\/sub><\/em>= 5 m, plus a circular pressure qext<\/sub><\/em>,2<\/sub> = -111 kPa on a radius rinternal <\/sub><\/em>= 4 m (see also Figure 3.14<\/a>). Negative pressure values correspond to tension. The additional vertical stress in the soil below the axis of symmetry of a circular pressure is provided by Eq. 3.13. At depth z <\/em>= 5 m, the positive pressure of radius rexternal<\/sub> <\/em>= 5 m will result in a compressive stress:\n\n[latex]\\Delta {\\sigma _{z,ext}} = {q_{ext,1}}\\left[ {1 - {{\\left( {\\dfrac{1}{{1 + {{\\left( {\\dfrac{{{r_{external}}}}{z}} \\right)}^2}}}} \\right)}^{\\frac{3}{2}}}} \\right] = 111\\left[ {1 - {{\\left( {\\dfrac{1}{{1 + {{\\left( {\\tfrac{5}{5}} \\right)}^2}}}} \\right)}^{\\frac{3}{2}}}} \\right] = 71.75{\\rm{ \\:kPa}}[\/latex]\n\nwhile the negative pressure of radius rinternal<\/sub> <\/em>= 4 m in a tensile stress:\n\n[latex]\\Delta {\\sigma _{z,{\\mathop{\\rm int}} }} = {q_{ext,2}}\\left[ {1 - {{\\left( {\\dfrac{1}{{1 + {{\\left( {\\dfrac{{{r_{{\\mathop{\\rm internal}}}}}}{z}} \\right)}^2}}}} \\right)}^{\\frac{3}{2}}}} \\right] = - 111\\left[ {1 - {{\\left( {\\tfrac{1}{{1 + {{\\left( {\\tfrac{4}{5}} \\right)}^2}}}} \\right)}^{\\frac{3}{2}}}} \\right] = - 58.15{\\rm{ \\: kPa}}[\/latex]\n\nThe total stress increment due to the loading applied on the ring foundation is found from Eq. 3.21 as:\n\n[latex]\\Delta {\\sigma _z} = \\Delta {\\sigma _{z,ext}} + \\Delta {\\sigma _{z,{\\mathop{\\rm int}} }} = 13.6{\\rm{\\:kPa }}[\/latex]","rendered":"

An H <\/em>= 4 m tall water tank with radius rexternal <\/sub><\/em>= 5 m is going to be founded on a ring foundation, with the dimensions depicted in the following plan view. Assuming the tank is filled with water, determine the vertical stress due to the weight of the tank below the centre of the foundation, at a depth z <\/em>= 5 m.<\/p>\n

\"The
Example 3.1. Problem description and input parameters.<\/figcaption><\/figure>\n

Answer:<\/h2>\n

First we have to determine the maximum load, Qext<\/sub><\/em> on the foundation, assuming that the tank is filled with water:<\/p>\n

Atank<\/sub> = <\/em>\u03c0rexternal<\/sub>2 <\/sup><\/em>= 78.54 m2<\/sup><\/p>\n

Qext<\/sub> = A \u00d7 H \u00d7 y<\/em>w <\/sub>= 78.54 m2<\/sup> x 4 m x 10 kN\/m3<\/sup> = 3141 kN where 10 kN\/m3<\/sup> is the unit weight of water.<\/p>\n

The area of the ring footing on which the load is applied is:<\/p>\n

Afooting<\/sub>= <\/em>\u03c0rexternal<\/sub>2 <\/sup>– <\/em>\u03c0rinternal<\/sub>2 <\/em>=78.54 m2<\/sup>-50.26 m2<\/sup> = 28.28 m2<\/sup><\/p>\n

thus the pressure applied on the ring foundation of the tank will be:<\/p>\n

qext<\/sub> = Qext<\/sub>\/Afooting <\/sub><\/em>= 111 kPa<\/p>\n

Employing the principle of superposition discussed in Chapter 3.5<\/a>, this ring pressure is equivalent, in terms of stresses applied to the soil, to a circular pressure qext<\/sub><\/em>,1<\/sub> = 111 kPa on a radius rexternal <\/sub><\/em>= 5 m, plus a circular pressure qext<\/sub><\/em>,2<\/sub> = -111 kPa on a radius rinternal <\/sub><\/em>= 4 m (see also Figure 3.14<\/a>). Negative pressure values correspond to tension. The additional vertical stress in the soil below the axis of symmetry of a circular pressure is provided by Eq. 3.13. At depth z <\/em>= 5 m, the positive pressure of radius rexternal<\/sub> <\/em>= 5 m will result in a compressive stress:<\/p>\n

[latex]\\Delta {\\sigma _{z,ext}} = {q_{ext,1}}\\left[ {1 - {{\\left( {\\dfrac{1}{{1 + {{\\left( {\\dfrac{{{r_{external}}}}{z}} \\right)}^2}}}} \\right)}^{\\frac{3}{2}}}} \\right] = 111\\left[ {1 - {{\\left( {\\dfrac{1}{{1 + {{\\left( {\\tfrac{5}{5}} \\right)}^2}}}} \\right)}^{\\frac{3}{2}}}} \\right] = 71.75{\\rm{ \\:kPa}}[\/latex]<\/p>\n

while the negative pressure of radius rinternal<\/sub> <\/em>= 4 m in a tensile stress:<\/p>\n

[latex]\\Delta {\\sigma _{z,{\\mathop{\\rm int}} }} = {q_{ext,2}}\\left[ {1 - {{\\left( {\\dfrac{1}{{1 + {{\\left( {\\dfrac{{{r_{{\\mathop{\\rm internal}}}}}}{z}} \\right)}^2}}}} \\right)}^{\\frac{3}{2}}}} \\right] = - 111\\left[ {1 - {{\\left( {\\tfrac{1}{{1 + {{\\left( {\\tfrac{4}{5}} \\right)}^2}}}} \\right)}^{\\frac{3}{2}}}} \\right] = - 58.15{\\rm{ \\: kPa}}[\/latex]<\/p>\n

The total stress increment due to the loading applied on the ring foundation is found from Eq. 3.21 as:<\/p>\n

[latex]\\Delta {\\sigma _z} = \\Delta {\\sigma _{z,ext}} + \\Delta {\\sigma _{z,{\\mathop{\\rm int}} }} = 13.6{\\rm{\\:kPa }}[\/latex]<\/p>\n","protected":false},"author":1,"menu_order":8,"template":"","meta":{"pb_show_title":"","pb_short_title":"Example 3.1","pb_subtitle":"Stresses in soil due to a water tank founded on a ring-type foundation","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-198","chapter","type-chapter","status-publish","hentry"],"part":165,"_links":{"self":[{"href":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-json\/pressbooks\/v2\/chapters\/198","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\/198\/revisions"}],"predecessor-version":[{"id":199,"href":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-json\/pressbooks\/v2\/chapters\/198\/revisions\/199"}],"part":[{"href":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-json\/pressbooks\/v2\/parts\/165"}],"metadata":[{"href":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-json\/pressbooks\/v2\/chapters\/198\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-json\/wp\/v2\/media?parent=198"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-json\/pressbooks\/v2\/chapter-type?post=198"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-json\/wp\/v2\/contributor?post=198"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-json\/wp\/v2\/license?post=198"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}