![\"Graph](\"https:\/\/libraryresources.nse.org.ng\/wp-content\/uploads\/sites\/9\/2025\/03\/example-4.4-brief-HR-e1743559695624.png\")
Example 4.4. Problem description and input parameters.[\/caption]\n![Figure 4.33<\/a>) separately.\n\n[caption id=\"attachment_272\" align=\"aligncenter\" width=\"400\"]](\"https:\/\/oercollective.caul.edu.au\/app\/uploads\/sites\/143\/2025\/03\/4.33was4.32-replace-e1747993293472.png\"){kind=link}
![\"Graph](\"https:\/\/libraryresources.nse.org.ng\/wp-content\/uploads\/sites\/9\/2026\/03\/4.33was4.32-replace-e1747993293472.png\")
<\/a> Figure 4.33. Dividing the compressible layer into two sublayers.[\/caption]\n\nAccordingly, we sum individual settlements to find the settlement of the entire compressible layer. In the corresponding equations, H<\/em> will be the thickness of each sublayer (Hi<\/sub><\/em>), and the effective stresses will be calculated at the middle of each sublayer.\n\nAs the clay is normally consolidated, we will apply Eq. 4.34 for the estimation of the settlement of each sublayer. The necessary calculations are presented in the table below:\n| <\/td>\n | Geostatic total stress \u03c3z0<\/sub><\/em> = \u03b3z<\/em><\/th>\n| Hydrostatic pore pressure\u00a0u0<\/sub> = \u03b3w<\/sub>z<\/em><\/th>\n | Geostatic effective stress \u03c3<\/em>\u2032z0<\/sub> = \u03c3z0<\/sub><\/em> - u0<\/sub><\/em><\/th>\n | Final effective stress \u03c3\u2032fin<\/sub><\/em> = \u03c3\u2032z0<\/sub><\/em> + qext<\/sub><\/em><\/th>\n | log(\u03c3\u2032fin<\/sub><\/em>\/\u03c3\u2032z0<\/sub><\/em>)<\/th>\n | Settlement of each sublayer, \u03c1pc<\/sub><\/em><\/th>\n<\/tr>\n | Sublayer 1 (Hi<\/sub><\/em> = 2m)<\/th>\n | 15 kPa<\/td>\n | 10 kPa<\/td>\n | 5 kPa<\/td>\n | 95 kPa<\/td>\n | 1.278<\/td>\n | 0.14 m<\/td>\n<\/tr>\n | Sublayer 2 (Hi<\/sub><\/em> = 2m)<\/th>\n | 45 kPa<\/td>\n | 30 kPa<\/td>\n | 15 kPa<\/td>\n | 105 kPa<\/td>\n | 0.845<\/td>\n | 0.09 m<\/td>\n<\/tr>\n | Total settlement:<\/strong><\/td>\n | 0.23 m<\/strong><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\nAs expected, settlement of sublayers 1 and 2 is not equal, as the preconsolidation stress (equal in this case to the initial, geostatic effective stress as the clay is normally consolidated) increases with depth, while the applied pressure on both is 90 kPa. This is illustrated in the figure below, where the path followed by a soil element at the middle of each sublayer is presented, and is a consequence of introducing soil non-linear response in consolidation calculations.\n\n[caption id=\"attachment_272\" align=\"aligncenter\" width=\"1024\"] | ![\"The](\"https:\/\/libraryresources.nse.org.ng\/wp-content\/uploads\/sites\/9\/2026\/03\/example-4.4-voidraio-logsigma-plot-HR-1024x494.png\") Example 4.4. Stress paths followed by soil elements at the middle of layer 1 and layer 2.[\/caption]","rendered":"Calculate again the primary consolidation settlement due to the strip pressure, using this time the oedometer test results depicted in the figure.<\/p>\n As discussed in Example 4.3<\/a>, the width of the loaded area is large, compared to the thickness of the compressible layer so we can well assume 1-D consolidation conditions. The additional effective stress \u0394\u03c3\u2032<\/em>z <\/sub>due to the application of the pressure will be equal to the applied pressure \u0394\u03c3\u2032<\/em>z <\/sub>= qext<\/sub> <\/em>= 90 kPa.<\/p>\n To improve accuracy when the problem involves rather thick compressible layers (H <\/em>> 2 m) and we are using non-linear consolidation theory<\/em>, we can divide the layer into sublayers, and find the settlement of each sublayer (Figure 4.33<\/a>) separately.<\/p>\n |
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