{"id":301,"date":"2025-03-28T05:12:45","date_gmt":"2025-03-28T05:12:45","guid":{"rendered":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/chapter\/example-4-6-numerical-simulation-of-1-d-consolidation\/"},"modified":"2026-03-16T14:01:38","modified_gmt":"2026-03-16T14:01:38","slug":"example-4-6-numerical-simulation-of-1-d-consolidation","status":"publish","type":"chapter","link":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/chapter\/example-4-6-numerical-simulation-of-1-d-consolidation\/","title":{"raw":"Example 4.6","rendered":"Example 4.6"},"content":{"raw":"Calculate the degree of consolidation of the compressible clay layer depicted in the figure below a) 100 days, and b) 5 years after the application of an infinite strip pressure. Consider the bottom layer of dense sand with gravels to be incompressible and permeable. Compare the settlement below the center of the pressure at <em>t <\/em>= 5 years with the settlement estimated analytically in <a href=\"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/chapter\/example-4-3-estimation-of-1-d-primary-consolidation-settlement-assuming-linear-elastic-soil-response\/\">Example 4.3<\/a>.\n\n&nbsp;\n\n[caption id=\"attachment_300\" align=\"aligncenter\" width=\"600\"]<img class=\"wp-image-294 size-full\" src=\"https:\/\/libraryresources.nse.org.ng\/wp-content\/uploads\/sites\/9\/2025\/02\/example-4.6-brief-HR-e1743559862717.png\" alt=\"Graph showing a pressure q_ext = 90 kPa of width B = 16 m applied on the surface of a soft saturated clay layer of thickness H = 4 m. The properties of the clay layer are E' = 1 MPa, v' = 0.333, k=10^(-4) m\/day and \u03b3 = 15 kN\/m^3. The clay layer is underlaid by dense sand, with gravels. The water table is at the ground surface.\" width=\"600\" height=\"269\"> Example 4.6. Problem description and input parameters.[\/caption]\n<h2>Answer:<\/h2>\nAs discussed in <a href=\"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/chapter\/example-4-3-estimation-of-1-d-primary-consolidation-settlement-assuming-linear-elastic-soil-response\/\">Example 4.3<\/a>, we can assume 1-D consolidation conditions, given that the width of the loaded area is large, compared to the thickness of the compressible clay layer. This suggests that we do not need to simulate the full two-dimensional problem: a 1m-wide soil column (unit cell) would suffice, provided that the correct boundary conditions are applied (<a href=\"https:\/\/oercollective.caul.edu.au\/app\/uploads\/sites\/143\/2025\/03\/4.43-HR.png\">Figure 4.43<\/a>). Note that the width 1 m is arbitrary, results are independent of the selected width of the unit cell, provided that pressure is applied across the entire width of the unit cell. As the strip pressure extents \u201cinfinitely\u201d along the out-of-plane dimension, the problem will be simulated as plane-strain (<a href=\"https:\/\/oercollective.caul.edu.au\/app\/uploads\/sites\/143\/2025\/01\/3.4-replace-e1747991384822.png\">Figure 3.4a<\/a>).\n\n[caption id=\"attachment_300\" align=\"aligncenter\" width=\"200\"]<img class=\"wp-image-295 size-full\" src=\"https:\/\/libraryresources.nse.org.ng\/wp-content\/uploads\/sites\/9\/2026\/03\/4.43-HR-e1743559878450.png\" alt=\"Graph showing a soil column of width 1m and thickness 4m. A pressure q_ext=90 kPa is applied across the entire width of the soil column.\" width=\"200\" height=\"422\"> Figure 4.43. Equivalent 1-D problem.[\/caption]\n\nAs also discussed in <a href=\"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/chapter\/4-8-evolution-of-primary-consolidation-settlement-with-time\/\">Chapter 4.8<\/a>, for the simulation of consolidation we also have to account for permeability boundary conditions, on top of displacement boundary conditions (<a href=\"https:\/\/oercollective.caul.edu.au\/app\/uploads\/sites\/143\/2025\/03\/4.44-HR.png\">Figure 4.44<\/a>). <em>\u201cStandard fixities\u201d <\/em>in PLAXIS will apply the necessary displacement boundary conditions for the plane-strain problem<em>.<\/em> As the bottom layer of dense sand with gravels is incompressible, we can choose not to explicitly introduce it in the model, and fix the displacement at the bottom of the clay layer in both directions (<a href=\"https:\/\/oercollective.caul.edu.au\/app\/uploads\/sites\/143\/2025\/03\/4.44-HR.png\">Figure 4.44<\/a>). Thus the dimensions of the model will be 1 m x 4 m.\n\n[caption id=\"attachment_300\" align=\"aligncenter\" width=\"410\"]<img class=\"wp-image-296 size-full\" src=\"https:\/\/libraryresources.nse.org.ng\/wp-content\/uploads\/sites\/9\/2026\/03\/4.44-HR-e1743550320769.png\" alt=\"The figure on the left shows a soil column of width B = 1 m and thickness H = 4 m. A pressure q_ext = 90 kPa is applied across the entire width of the column. The left boundary of the column is denoted as axis of symmetry. The figure on the top right shows the displacement boundary conditions applied to that soil column: Rollers are applied to the sides of the column prohibiting horizontal displacement, while is bottom boundary is fixed along both horizontal and vertical directions. The figure on the bottom right shows the flow boundary conditions applied to that soil column: Impermeable boundaries are applied to the sides of the column prohibiting horizontal flow, while the top and bottom boundary are modelled as permeable.\" width=\"410\" height=\"500\"> Figure 4.44. Displacement and drainage boundary conditions of the equivalent 1-D problem.[\/caption]\n\n<em>\u201cPermeable\u201d <\/em>boundary conditions are applied at the top and bottom of the clay layer to specify that water is free to flow from these directions. <em>\u201cImpermeable\u201d <\/em>boundary conditions at the lateral boundaries account for symmetry and for vertical-only water flow. Note that the axis of symmetry always corresponds to impermeable boundary conditions. The considered boundary conditions ensure that lateral strains will be zero <em>\u03b5<\/em><sub>x<\/sub><em> = <\/em><em>\u03b5<\/em><sub>h<\/sub> = 0 and that excess pore water pressure dissipates only vertically.\n\nThe undrained (A) linear elastic model is used to simulate clay behaviour under the applied loading, which allows for the calculation of (excess) pore water pressures in plastic stages. Compressibility is defined in terms of effective parameters i.e., the drained Young\u2019s modulus <em>E\u2032 <\/em>and Poisson\u2019s ratio <em>v\u2032.<\/em>\u00a0Note however that the undrained behavior stiffness parameters (<em>E<sub>u<\/sub><\/em>, <em>v<sub>u <\/sub><\/em>= 0.495) are calculated internally by PLAXIS, and the coefficient of earth pressure at-rest <em>K<\/em><sub>0<\/sub> is automatically set equal to <em>K<\/em><sub>0<\/sub> = 1. Note that in PLAXIS we do not input <em>v<sub>u <\/sub><\/em>= 0.50, but a value very close to 0.5. That it because the (undrained) bulk modulus of soil <em>K<sub>u<\/sub><\/em> = <em>E<sub>u<\/sub><\/em>\/3(1-2<em>v<sub>u<\/sub><\/em>) becomes infinite if <em>v<sub>u <\/sub><\/em>= 0.50 (soil is incompressible), resulting in a division-by-zero. The permeability of the clay, <em>k<\/em> is also entered as a material parameter.\n\nThe analysis consists of three stages:\n<ul>\n \t<li><em>Step 1: Initial phase.<\/em> The geostatic stresses and hydrostatic pore water pressures before the application of the external pressure are initialised with the \u201cK<sub>0<\/sub>-procedure\u201d. Drainage boundary conditions and groundwater table level are defined, as in <a href=\"https:\/\/oercollective.caul.edu.au\/app\/uploads\/sites\/143\/2025\/03\/4.44-HR.png\">Figure 4.44<\/a>.<\/li>\n \t<li><em>Step 2: Sudden application of the loading.<\/em> A \u201cPlastic\u201d step is introduced, where the external pressure is instantaneously applied on the ground surface i.e., the duration of this step (time interval) is 0.0 days.<\/li>\n \t<li><em>Step 3: Consolidation.<\/em> A \u201cConsolidation\u201d step follows the previous plastic step, where the external pressure will remain on the ground surface for a prescribed time interval, and excess pore pressures developed during Step 2 will be allowed to dissipate through the free-draining boundaries. <em>\u201cReset displacements to zero\u201d<\/em> should not be active, so that any immediate settlement will be co-estimated.<\/li>\n<\/ul>\nResults at the end of Step 2 are presented in <a href=\"https:\/\/oercollective.caul.edu.au\/app\/uploads\/sites\/143\/2025\/03\/4.45-HR.png\">Figure 4.45<\/a> in the form of contours of total vertical stress along the thickness of the clay layer, excess pore pressure distribution, and settlement. Notice that excess pore water pressures are not equal to the external applied pressure as you might expect, but somewhat lower, and some immediate settlement develops. This is because PLAXIS considers a large, yet realistic bulk modulus for water, unlike the analytical solution where we assumed that water is incompressible.\n\n[caption id=\"attachment_300\" align=\"aligncenter\" width=\"1024\"]<img class=\"wp-image-297 size-large\" src=\"https:\/\/libraryresources.nse.org.ng\/wp-content\/uploads\/sites\/9\/2026\/03\/4.45-HR-1024x532.png\" alt=\"The figure on the left shows cartesian total vertical stress contours. The maximum value at the surface is -90 kPa and the minimum value at the bottom is - 150 kPa. The figure on the middle shows excess pore pressure contours. A uniform value of -89.28 kPa is predicted. The figure on the right shows total displacement contours. The maximum value of zero is predicted at the bottom, and the minimum value of -1.916mm is predicted at the top.\" width=\"1024\" height=\"532\"> Figure 4.45. (Left) Total vertical stress, (middle) Excess pore pressure, and (right) Settlement contours immediately after the application of the external pressure, at the end of Step 2.[\/caption]\n\n[caption id=\"attachment_300\" align=\"aligncenter\" width=\"1024\"]<img class=\"wp-image-298 size-large\" src=\"https:\/\/libraryresources.nse.org.ng\/wp-content\/uploads\/sites\/9\/2026\/03\/4.46-HR-1024x516.png\" alt=\"The figure on the left shows cartesian total vertical stress contours. The maximum value at the surface is -90 kPa and the minimum value at the bottom is - 150 kPa. The figure on the middle shows excess pore pressure contours. The maximum value of zero is predicted at the free-draining top and bottom boundaries of the model, while the minimum value of -49.39 kPa is predicted at the mid of the model. The figure on the right shows total displacement contours. The maximum value of zero is predicted at the bottom, and the minimum value of -155.6mm is predicted at the top.\" width=\"1024\" height=\"516\"> Figure 4.46. (Left) Total vertical stress, (middle) Excess pore pressure, and (right) Settlement contours at the end of Step 3, after t = 100 days.[\/caption]\n\nResults at the end of Step 3, after <em>t <\/em>= 100 days, are presented in <a href=\"https:\/\/oercollective.caul.edu.au\/app\/uploads\/sites\/143\/2025\/03\/4.46-HR.png\">Figure 4.46<\/a>, again in terms of total vertical stress, excess pore pressure distribution and settlement contours, for comparison. As expected, total stresses do not change. Excess pore water pressures have partly dissipated, but the maximum value at the middle of the sample is <em>\u0394<\/em><em>u <\/em>= 49.4 kPa (<a href=\"https:\/\/oercollective.caul.edu.au\/app\/uploads\/sites\/143\/2025\/02\/4.47-HR.png\">Figure 4.47<\/a>) which suggests that consolidation has not been completed yet (<em>U <\/em>&lt; 100 %), and settlement has not attained its maximum value. Notice also in <a href=\"https:\/\/oercollective.caul.edu.au\/app\/uploads\/sites\/143\/2025\/02\/4.47-HR.png\">Figure 4.47<\/a> that excess pore pressure at the top and bottom boundary, where drainage is free, is zero.\n\nIf Step 3 is run for <em>t <\/em>= 1825 days (5 years), excess pore pressure will become practically zero along the full thickness of the clay (<a href=\"https:\/\/oercollective.caul.edu.au\/app\/uploads\/sites\/143\/2025\/03\/4.48-HR.png\">Figure 4.48a<\/a>), which implies that consolidation has been completed, and the full primary consolidation settlement has developed (<a href=\"https:\/\/oercollective.caul.edu.au\/app\/uploads\/sites\/143\/2025\/03\/4.48-HR.png\">Figure 4.48b<\/a>).\n\n[caption id=\"attachment_300\" align=\"aligncenter\" width=\"800\"]<img class=\"wp-image-299 size-full\" src=\"https:\/\/libraryresources.nse.org.ng\/wp-content\/uploads\/sites\/9\/2026\/03\/4.47-HR-e1743559913857.png\" alt=\"The figure on the left shows variation of excess pore pressure with depth measure from the bottom, at t = 0 and at t = 100 days. At t = 0 the excess pore pressure is uniform, and equal to about 89 kPa. At t = 100 days the excess pore pressure is zero at the top and bottom, and attains its maximum value at the mid. The figure on the right shows that variation of the consolidation degree with depth measured from the bottom of the sample. Results for t = 100 days are shown. The consolidation degree is equal to 1 at the top and bottom, and attains its minimum value at the mid.\" width=\"800\" height=\"635\"> Figure 4.47. (Left) Excess pore pressure, and (right) Degree of consolidation profiles along the thickness of the clay layer at the end of Step 3, after t = 100 days.[\/caption]\n\nTotal stresses will of course remain unchanged. The degree of consolidation will be equal to <em>U <\/em>\u2248 100% along the thickness of the soil layer. The value of settlement is, as expected, practically equal to the value computed analytically in <a href=\"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/chapter\/example-4-3-estimation-of-1-d-primary-consolidation-settlement-assuming-linear-elastic-soil-response\/\">Example 4.3<\/a> (<em>\u03c1<\/em><sub>pc<\/sub><em>\u00a0<\/em>= 240 mm).\n\n[caption id=\"attachment_300\" align=\"aligncenter\" width=\"750\"]<img class=\"wp-image-300 size-full\" src=\"https:\/\/libraryresources.nse.org.ng\/wp-content\/uploads\/sites\/9\/2026\/03\/4.48-HR-e1743559939370.png\" alt=\"The figure on the left shows excess pore pressure contours. The maximum value of zero is predicted at the free-draining top and bottom boundaries of the model, while the minimum value of -0.22 kPa is predicted at the mid of the model. The figure on the right shows total displacement contours. The maximum value of zero is predicted at the bottom, and the minimum value of -239.9mm is predicted at the top.\" width=\"750\" height=\"597\"> Figure 4.48. (Left) Excess pore pressure, and (right) Settlement contours at the end of Step 3, after t = 1825 days.[\/caption]","rendered":"<p>Calculate the degree of consolidation of the compressible clay layer depicted in the figure below a) 100 days, and b) 5 years after the application of an infinite strip pressure. Consider the bottom layer of dense sand with gravels to be incompressible and permeable. Compare the settlement below the center of the pressure at <em>t <\/em>= 5 years with the settlement estimated analytically in <a href=\"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/chapter\/example-4-3-estimation-of-1-d-primary-consolidation-settlement-assuming-linear-elastic-soil-response\/\">Example 4.3<\/a>.<\/p>\n<p>&nbsp;<\/p>\n<figure id=\"attachment_300\" aria-describedby=\"caption-attachment-300\" style=\"width: 600px\" class=\"wp-caption aligncenter\"><img decoding=\"async\" class=\"wp-image-294 size-full\" src=\"https:\/\/libraryresources.nse.org.ng\/wp-content\/uploads\/sites\/9\/2025\/02\/example-4.6-brief-HR-e1743559862717.png\" alt=\"Graph showing a pressure q_ext = 90 kPa of width B = 16 m applied on the surface of a soft saturated clay layer of thickness H = 4 m. The properties of the clay layer are E' = 1 MPa, v' = 0.333, k=10^(-4) m\/day and \u03b3 = 15 kN\/m^3. The clay layer is underlaid by dense sand, with gravels. The water table is at the ground surface.\" width=\"600\" height=\"269\" srcset=\"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-content\/uploads\/sites\/9\/2025\/02\/example-4.6-brief-HR-e1743559862717.png 600w, https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-content\/uploads\/sites\/9\/2025\/02\/example-4.6-brief-HR-e1743559862717-300x135.png 300w, https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-content\/uploads\/sites\/9\/2025\/02\/example-4.6-brief-HR-e1743559862717-65x29.png 65w, https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-content\/uploads\/sites\/9\/2025\/02\/example-4.6-brief-HR-e1743559862717-225x101.png 225w, https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-content\/uploads\/sites\/9\/2025\/02\/example-4.6-brief-HR-e1743559862717-350x157.png 350w\" sizes=\"(max-width: 600px) 100vw, 600px\" \/><figcaption id=\"caption-attachment-300\" class=\"wp-caption-text\">Example 4.6. Problem description and input parameters.<\/figcaption><\/figure>\n<h2>Answer:<\/h2>\n<p>As discussed in <a href=\"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/chapter\/example-4-3-estimation-of-1-d-primary-consolidation-settlement-assuming-linear-elastic-soil-response\/\">Example 4.3<\/a>, we can assume 1-D consolidation conditions, given that the width of the loaded area is large, compared to the thickness of the compressible clay layer. This suggests that we do not need to simulate the full two-dimensional problem: a 1m-wide soil column (unit cell) would suffice, provided that the correct boundary conditions are applied (<a href=\"https:\/\/oercollective.caul.edu.au\/app\/uploads\/sites\/143\/2025\/03\/4.43-HR.png\">Figure 4.43<\/a>). Note that the width 1 m is arbitrary, results are independent of the selected width of the unit cell, provided that pressure is applied across the entire width of the unit cell. As the strip pressure extents \u201cinfinitely\u201d along the out-of-plane dimension, the problem will be simulated as plane-strain (<a href=\"https:\/\/oercollective.caul.edu.au\/app\/uploads\/sites\/143\/2025\/01\/3.4-replace-e1747991384822.png\">Figure 3.4a<\/a>).<\/p>\n<figure id=\"attachment_300\" aria-describedby=\"caption-attachment-300\" style=\"width: 200px\" class=\"wp-caption aligncenter\"><img decoding=\"async\" class=\"wp-image-295 size-full\" src=\"https:\/\/libraryresources.nse.org.ng\/wp-content\/uploads\/sites\/9\/2026\/03\/4.43-HR-e1743559878450.png\" alt=\"Graph showing a soil column of width 1m and thickness 4m. A pressure q_ext=90 kPa is applied across the entire width of the soil column.\" width=\"200\" height=\"422\" srcset=\"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-content\/uploads\/sites\/9\/2026\/03\/4.43-HR-e1743559878450.png 200w, https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-content\/uploads\/sites\/9\/2026\/03\/4.43-HR-e1743559878450-142x300.png 142w, https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-content\/uploads\/sites\/9\/2026\/03\/4.43-HR-e1743559878450-65x137.png 65w\" sizes=\"(max-width: 200px) 100vw, 200px\" \/><figcaption id=\"caption-attachment-300\" class=\"wp-caption-text\">Figure 4.43. Equivalent 1-D problem.<\/figcaption><\/figure>\n<p>As also discussed in <a href=\"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/chapter\/4-8-evolution-of-primary-consolidation-settlement-with-time\/\">Chapter 4.8<\/a>, for the simulation of consolidation we also have to account for permeability boundary conditions, on top of displacement boundary conditions (<a href=\"https:\/\/oercollective.caul.edu.au\/app\/uploads\/sites\/143\/2025\/03\/4.44-HR.png\">Figure 4.44<\/a>). <em>\u201cStandard fixities\u201d <\/em>in PLAXIS will apply the necessary displacement boundary conditions for the plane-strain problem<em>.<\/em> As the bottom layer of dense sand with gravels is incompressible, we can choose not to explicitly introduce it in the model, and fix the displacement at the bottom of the clay layer in both directions (<a href=\"https:\/\/oercollective.caul.edu.au\/app\/uploads\/sites\/143\/2025\/03\/4.44-HR.png\">Figure 4.44<\/a>). Thus the dimensions of the model will be 1 m x 4 m.<\/p>\n<figure id=\"attachment_300\" aria-describedby=\"caption-attachment-300\" style=\"width: 410px\" class=\"wp-caption aligncenter\"><img decoding=\"async\" class=\"wp-image-296 size-full\" src=\"https:\/\/libraryresources.nse.org.ng\/wp-content\/uploads\/sites\/9\/2026\/03\/4.44-HR-e1743550320769.png\" alt=\"The figure on the left shows a soil column of width B = 1 m and thickness H = 4 m. A pressure q_ext = 90 kPa is applied across the entire width of the column. The left boundary of the column is denoted as axis of symmetry. The figure on the top right shows the displacement boundary conditions applied to that soil column: Rollers are applied to the sides of the column prohibiting horizontal displacement, while is bottom boundary is fixed along both horizontal and vertical directions. The figure on the bottom right shows the flow boundary conditions applied to that soil column: Impermeable boundaries are applied to the sides of the column prohibiting horizontal flow, while the top and bottom boundary are modelled as permeable.\" width=\"410\" height=\"500\" srcset=\"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-content\/uploads\/sites\/9\/2026\/03\/4.44-HR-e1743550320769.png 410w, https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-content\/uploads\/sites\/9\/2026\/03\/4.44-HR-e1743550320769-246x300.png 246w, https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-content\/uploads\/sites\/9\/2026\/03\/4.44-HR-e1743550320769-65x79.png 65w, https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-content\/uploads\/sites\/9\/2026\/03\/4.44-HR-e1743550320769-225x274.png 225w, https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-content\/uploads\/sites\/9\/2026\/03\/4.44-HR-e1743550320769-350x427.png 350w\" sizes=\"(max-width: 410px) 100vw, 410px\" \/><figcaption id=\"caption-attachment-300\" class=\"wp-caption-text\">Figure 4.44. Displacement and drainage boundary conditions of the equivalent 1-D problem.<\/figcaption><\/figure>\n<p><em>\u201cPermeable\u201d <\/em>boundary conditions are applied at the top and bottom of the clay layer to specify that water is free to flow from these directions. <em>\u201cImpermeable\u201d <\/em>boundary conditions at the lateral boundaries account for symmetry and for vertical-only water flow. Note that the axis of symmetry always corresponds to impermeable boundary conditions. The considered boundary conditions ensure that lateral strains will be zero <em>\u03b5<\/em><sub>x<\/sub><em> = <\/em><em>\u03b5<\/em><sub>h<\/sub> = 0 and that excess pore water pressure dissipates only vertically.<\/p>\n<p>The undrained (A) linear elastic model is used to simulate clay behaviour under the applied loading, which allows for the calculation of (excess) pore water pressures in plastic stages. Compressibility is defined in terms of effective parameters i.e., the drained Young\u2019s modulus <em>E\u2032 <\/em>and Poisson\u2019s ratio <em>v\u2032.<\/em>\u00a0Note however that the undrained behavior stiffness parameters (<em>E<sub>u<\/sub><\/em>, <em>v<sub>u <\/sub><\/em>= 0.495) are calculated internally by PLAXIS, and the coefficient of earth pressure at-rest <em>K<\/em><sub>0<\/sub> is automatically set equal to <em>K<\/em><sub>0<\/sub> = 1. Note that in PLAXIS we do not input <em>v<sub>u <\/sub><\/em>= 0.50, but a value very close to 0.5. That it because the (undrained) bulk modulus of soil <em>K<sub>u<\/sub><\/em> = <em>E<sub>u<\/sub><\/em>\/3(1-2<em>v<sub>u<\/sub><\/em>) becomes infinite if <em>v<sub>u <\/sub><\/em>= 0.50 (soil is incompressible), resulting in a division-by-zero. The permeability of the clay, <em>k<\/em> is also entered as a material parameter.<\/p>\n<p>The analysis consists of three stages:<\/p>\n<ul>\n<li><em>Step 1: Initial phase.<\/em> The geostatic stresses and hydrostatic pore water pressures before the application of the external pressure are initialised with the \u201cK<sub>0<\/sub>-procedure\u201d. Drainage boundary conditions and groundwater table level are defined, as in <a href=\"https:\/\/oercollective.caul.edu.au\/app\/uploads\/sites\/143\/2025\/03\/4.44-HR.png\">Figure 4.44<\/a>.<\/li>\n<li><em>Step 2: Sudden application of the loading.<\/em> A \u201cPlastic\u201d step is introduced, where the external pressure is instantaneously applied on the ground surface i.e., the duration of this step (time interval) is 0.0 days.<\/li>\n<li><em>Step 3: Consolidation.<\/em> A \u201cConsolidation\u201d step follows the previous plastic step, where the external pressure will remain on the ground surface for a prescribed time interval, and excess pore pressures developed during Step 2 will be allowed to dissipate through the free-draining boundaries. <em>\u201cReset displacements to zero\u201d<\/em> should not be active, so that any immediate settlement will be co-estimated.<\/li>\n<\/ul>\n<p>Results at the end of Step 2 are presented in <a href=\"https:\/\/oercollective.caul.edu.au\/app\/uploads\/sites\/143\/2025\/03\/4.45-HR.png\">Figure 4.45<\/a> in the form of contours of total vertical stress along the thickness of the clay layer, excess pore pressure distribution, and settlement. Notice that excess pore water pressures are not equal to the external applied pressure as you might expect, but somewhat lower, and some immediate settlement develops. This is because PLAXIS considers a large, yet realistic bulk modulus for water, unlike the analytical solution where we assumed that water is incompressible.<\/p>\n<figure id=\"attachment_300\" aria-describedby=\"caption-attachment-300\" style=\"width: 1024px\" class=\"wp-caption aligncenter\"><img decoding=\"async\" class=\"wp-image-297 size-large\" src=\"https:\/\/libraryresources.nse.org.ng\/wp-content\/uploads\/sites\/9\/2026\/03\/4.45-HR-1024x532.png\" alt=\"The figure on the left shows cartesian total vertical stress contours. The maximum value at the surface is -90 kPa and the minimum value at the bottom is - 150 kPa. The figure on the middle shows excess pore pressure contours. A uniform value of -89.28 kPa is predicted. The figure on the right shows total displacement contours. The maximum value of zero is predicted at the bottom, and the minimum value of -1.916mm is predicted at the top.\" width=\"1024\" height=\"532\" srcset=\"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-content\/uploads\/sites\/9\/2026\/03\/4.45-HR-1024x532.png 1024w, https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-content\/uploads\/sites\/9\/2026\/03\/4.45-HR-300x156.png 300w, https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-content\/uploads\/sites\/9\/2026\/03\/4.45-HR-768x399.png 768w, https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-content\/uploads\/sites\/9\/2026\/03\/4.45-HR-1536x798.png 1536w, https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-content\/uploads\/sites\/9\/2026\/03\/4.45-HR-65x34.png 65w, https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-content\/uploads\/sites\/9\/2026\/03\/4.45-HR-225x117.png 225w, https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-content\/uploads\/sites\/9\/2026\/03\/4.45-HR-350x182.png 350w, https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-content\/uploads\/sites\/9\/2026\/03\/4.45-HR.png 1883w\" sizes=\"(max-width: 1024px) 100vw, 1024px\" \/><figcaption id=\"caption-attachment-300\" class=\"wp-caption-text\">Figure 4.45. (Left) Total vertical stress, (middle) Excess pore pressure, and (right) Settlement contours immediately after the application of the external pressure, at the end of Step 2.<\/figcaption><\/figure>\n<figure id=\"attachment_300\" aria-describedby=\"caption-attachment-300\" style=\"width: 1024px\" class=\"wp-caption aligncenter\"><img decoding=\"async\" class=\"wp-image-298 size-large\" src=\"https:\/\/libraryresources.nse.org.ng\/wp-content\/uploads\/sites\/9\/2026\/03\/4.46-HR-1024x516.png\" alt=\"The figure on the left shows cartesian total vertical stress contours. The maximum value at the surface is -90 kPa and the minimum value at the bottom is - 150 kPa. The figure on the middle shows excess pore pressure contours. The maximum value of zero is predicted at the free-draining top and bottom boundaries of the model, while the minimum value of -49.39 kPa is predicted at the mid of the model. The figure on the right shows total displacement contours. The maximum value of zero is predicted at the bottom, and the minimum value of -155.6mm is predicted at the top.\" width=\"1024\" height=\"516\" srcset=\"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-content\/uploads\/sites\/9\/2026\/03\/4.46-HR-1024x516.png 1024w, https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-content\/uploads\/sites\/9\/2026\/03\/4.46-HR-300x151.png 300w, https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-content\/uploads\/sites\/9\/2026\/03\/4.46-HR-768x387.png 768w, https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-content\/uploads\/sites\/9\/2026\/03\/4.46-HR-1536x774.png 1536w, https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-content\/uploads\/sites\/9\/2026\/03\/4.46-HR-65x33.png 65w, https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-content\/uploads\/sites\/9\/2026\/03\/4.46-HR-225x113.png 225w, https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-content\/uploads\/sites\/9\/2026\/03\/4.46-HR-350x176.png 350w, https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-content\/uploads\/sites\/9\/2026\/03\/4.46-HR.png 1760w\" sizes=\"(max-width: 1024px) 100vw, 1024px\" \/><figcaption id=\"caption-attachment-300\" class=\"wp-caption-text\">Figure 4.46. (Left) Total vertical stress, (middle) Excess pore pressure, and (right) Settlement contours at the end of Step 3, after t = 100 days.<\/figcaption><\/figure>\n<p>Results at the end of Step 3, after <em>t <\/em>= 100 days, are presented in <a href=\"https:\/\/oercollective.caul.edu.au\/app\/uploads\/sites\/143\/2025\/03\/4.46-HR.png\">Figure 4.46<\/a>, again in terms of total vertical stress, excess pore pressure distribution and settlement contours, for comparison. As expected, total stresses do not change. Excess pore water pressures have partly dissipated, but the maximum value at the middle of the sample is <em>\u0394<\/em><em>u <\/em>= 49.4 kPa (<a href=\"https:\/\/oercollective.caul.edu.au\/app\/uploads\/sites\/143\/2025\/02\/4.47-HR.png\">Figure 4.47<\/a>) which suggests that consolidation has not been completed yet (<em>U <\/em>&lt; 100 %), and settlement has not attained its maximum value. Notice also in <a href=\"https:\/\/oercollective.caul.edu.au\/app\/uploads\/sites\/143\/2025\/02\/4.47-HR.png\">Figure 4.47<\/a> that excess pore pressure at the top and bottom boundary, where drainage is free, is zero.<\/p>\n<p>If Step 3 is run for <em>t <\/em>= 1825 days (5 years), excess pore pressure will become practically zero along the full thickness of the clay (<a href=\"https:\/\/oercollective.caul.edu.au\/app\/uploads\/sites\/143\/2025\/03\/4.48-HR.png\">Figure 4.48a<\/a>), which implies that consolidation has been completed, and the full primary consolidation settlement has developed (<a href=\"https:\/\/oercollective.caul.edu.au\/app\/uploads\/sites\/143\/2025\/03\/4.48-HR.png\">Figure 4.48b<\/a>).<\/p>\n<figure id=\"attachment_300\" aria-describedby=\"caption-attachment-300\" style=\"width: 800px\" class=\"wp-caption aligncenter\"><img decoding=\"async\" class=\"wp-image-299 size-full\" src=\"https:\/\/libraryresources.nse.org.ng\/wp-content\/uploads\/sites\/9\/2026\/03\/4.47-HR-e1743559913857.png\" alt=\"The figure on the left shows variation of excess pore pressure with depth measure from the bottom, at t = 0 and at t = 100 days. At t = 0 the excess pore pressure is uniform, and equal to about 89 kPa. At t = 100 days the excess pore pressure is zero at the top and bottom, and attains its maximum value at the mid. The figure on the right shows that variation of the consolidation degree with depth measured from the bottom of the sample. Results for t = 100 days are shown. The consolidation degree is equal to 1 at the top and bottom, and attains its minimum value at the mid.\" width=\"800\" height=\"635\" srcset=\"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-content\/uploads\/sites\/9\/2026\/03\/4.47-HR-e1743559913857.png 800w, https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-content\/uploads\/sites\/9\/2026\/03\/4.47-HR-e1743559913857-300x238.png 300w, https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-content\/uploads\/sites\/9\/2026\/03\/4.47-HR-e1743559913857-768x610.png 768w, https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-content\/uploads\/sites\/9\/2026\/03\/4.47-HR-e1743559913857-540x430.png 540w, https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-content\/uploads\/sites\/9\/2026\/03\/4.47-HR-e1743559913857-65x52.png 65w, https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-content\/uploads\/sites\/9\/2026\/03\/4.47-HR-e1743559913857-225x179.png 225w, https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-content\/uploads\/sites\/9\/2026\/03\/4.47-HR-e1743559913857-350x278.png 350w\" sizes=\"(max-width: 800px) 100vw, 800px\" \/><figcaption id=\"caption-attachment-300\" class=\"wp-caption-text\">Figure 4.47. (Left) Excess pore pressure, and (right) Degree of consolidation profiles along the thickness of the clay layer at the end of Step 3, after t = 100 days.<\/figcaption><\/figure>\n<p>Total stresses will of course remain unchanged. The degree of consolidation will be equal to <em>U <\/em>\u2248 100% along the thickness of the soil layer. The value of settlement is, as expected, practically equal to the value computed analytically in <a href=\"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/chapter\/example-4-3-estimation-of-1-d-primary-consolidation-settlement-assuming-linear-elastic-soil-response\/\">Example 4.3<\/a> (<em>\u03c1<\/em><sub>pc<\/sub><em>\u00a0<\/em>= 240 mm).<\/p>\n<figure id=\"attachment_300\" aria-describedby=\"caption-attachment-300\" style=\"width: 750px\" class=\"wp-caption aligncenter\"><img decoding=\"async\" class=\"wp-image-300 size-full\" src=\"https:\/\/libraryresources.nse.org.ng\/wp-content\/uploads\/sites\/9\/2026\/03\/4.48-HR-e1743559939370.png\" alt=\"The figure on the left shows excess pore pressure contours. The maximum value of zero is predicted at the free-draining top and bottom boundaries of the model, while the minimum value of -0.22 kPa is predicted at the mid of the model. The figure on the right shows total displacement contours. The maximum value of zero is predicted at the bottom, and the minimum value of -239.9mm is predicted at the top.\" width=\"750\" height=\"597\" srcset=\"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-content\/uploads\/sites\/9\/2026\/03\/4.48-HR-e1743559939370.png 750w, https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-content\/uploads\/sites\/9\/2026\/03\/4.48-HR-e1743559939370-300x239.png 300w, https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-content\/uploads\/sites\/9\/2026\/03\/4.48-HR-e1743559939370-540x430.png 540w, https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-content\/uploads\/sites\/9\/2026\/03\/4.48-HR-e1743559939370-65x52.png 65w, https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-content\/uploads\/sites\/9\/2026\/03\/4.48-HR-e1743559939370-225x179.png 225w, https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-content\/uploads\/sites\/9\/2026\/03\/4.48-HR-e1743559939370-350x279.png 350w\" sizes=\"(max-width: 750px) 100vw, 750px\" \/><figcaption id=\"caption-attachment-300\" class=\"wp-caption-text\">Figure 4.48. (Left) Excess pore pressure, and (right) Settlement contours at the end of Step 3, after t = 1825 days.<\/figcaption><\/figure>\n","protected":false},"author":1,"menu_order":15,"template":"","meta":{"pb_show_title":"","pb_short_title":"Example 4.6","pb_subtitle":"Numerical simulation of 1-D consolidation","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-301","chapter","type-chapter","status-publish","hentry"],"part":215,"_links":{"self":[{"href":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-json\/pressbooks\/v2\/chapters\/301","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\/301\/revisions"}],"predecessor-version":[{"id":302,"href":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-json\/pressbooks\/v2\/chapters\/301\/revisions\/302"}],"part":[{"href":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-json\/pressbooks\/v2\/parts\/215"}],"metadata":[{"href":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-json\/pressbooks\/v2\/chapters\/301\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-json\/wp\/v2\/media?parent=301"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-json\/pressbooks\/v2\/chapter-type?post=301"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-json\/wp\/v2\/contributor?post=301"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/libraryresources.nse.org.ng\/fundamentalsoffoundationengineering\/wp-json\/wp\/v2\/license?post=301"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}