Compressibility of Soil-Soil Mechanics and Foundations-Lecture 29 Slides-Civil and Environmental Engineering, Slides of Soil Mechanics and Foundations

Download Compressibility of Soil-Soil Mechanics and Foundations-Lecture 29 Slides-Civil and Environmental Engineering and more Slides Soil Mechanics and Foundations in PDF only on Docsity! CE 240 Soil Mechanics & Foundations Lecture 10.2 Compressibility of Soil IV (Das, Ch. 10) Class Outline • Secondary consolidation settlement • Time rate of consolidation • Calculation of consolidation settlement under a foundation • Consolidation settlement acceleration • Summary Chs. 7-10 Secondary consolidation settlement is more important than primary consolidation of organic and highly compressible inorganic soils. In overconsolidated inorganic clays, the secondary compression index C is very small and of less practical significance. Time rate of consolidation For calculating the primary consolidation settlement (10.24, 10.26, and 10.27, see below), no information regarding the time rate is provided. (10.24, normally consolidated) (10.26, over-consolidated) (10.27), for Time rate of consolidation Terzaghi (1925) derived the time rate of consolidation based on the following assumptions: 1, homogeneous clay-water system; 2, complete saturation; 3, zero compressibility for water; 4, zero compressibility for solid grains; 5, 1D flow; 6, Darcy’s law is valid. Eq. (10.50) is the basic differential equation of Terzaghi’s consolidation theory and can be solved with the following boundary conditions: z=0, u=0 z= 2H, u=0 t=0, u=Uo The solution yields (10.53) where m = an integer M = (a/2)(2m + 1) Ug = initial excess pore water pressure (10.54) The time factor is a nondimensional number. Because consolidation progresses by the dissipation of excess pore water pres- sure, the degree of consolidation at a distance z at any time ris Ug — uu, u £3 Se loietiye es phir: ie (10.55) where u, = excess pore water pressure at time 7. Equations (10.53) and (10.55) can be combined to obtain the degree of con- solidation at any depth z. This is shown in Figure 10.29. The average degree of consolidation for the entire depth of the clay layer at any time t can be written from Eq. (10.55) as (10.56) where U = average degree of consolidation Si) = settlement of the layer at time f S. = ultimate settlement of the layer from primary consolidation Substitution of the expression for excess pore water pressure u, given in Eg. (10.53) into Eq. (10.56) gives u-1-"5 ane (10.57) Consolidation starts from top and bottom and gradually reaches the center of the clay layer. Calculation of Consolidation Settlement under a Foundation Chapter 9 showed that the increase in the vertical stress in soil caused by a load ap- plied over a limited area decreases with depth z measured from the ground surface downward. Hence to estimate the one-dimensional settlement of a foundation, we can use Eq. (10.24), (10.26), or (10.27). However, the increase of effective stress, Ao’, in these equations should be the average increase in the pressure below the center of the foundation. The values can be determined by using the procedure described in Chapter 9. Assuming that the pressure increase varies parabolically, using Simpson’s rule, we can estimate the value of Ac, as (10.66) where Ag;, Aoj,,, and Ao; represent the increase in the effective pressure at the top, middle, and bottom of the layer, respectively.  Calculate the settlement of the 10-ft-thick clay layer (Figure 10.37) that will result - e from the load carried by a 5-ft-square footing, The clay is normally consolidated. E Use the weighted average method [Eq. (10.66)] to calculate the average increase of effective pressure in the clay layer. where 200 kips C. = 0.009(LL — 10) = 0.009(40 — 10) = 0.27 Sistah H = 10 X 12 = 120in. i Footing size 1 Sit x5 ft 2p = 1.0 Fete Lies i eee ena lent heat ai : 10 | eae mie ee ee = LOL X Yanisung) + 10 ft] Yssusens) ~ 62-4] + S-[ Yenc ~ 62-4] “OEE Yin = 100 pef = 10 x 100 + 10(120 — 62.4) + 5(110 — 62.4) = 1814 lb/ft From Eq. (10.66), ~ Groundwater ta _ da, + 4ha,, + Aci, at = 120 pet howe ; Aq, Ac},,and Ao}, below the center of the footing can be obtained from Eq. (9.39). Now we can prepare the following table (Note: L/B = 5/5 = 1): Zz b= B/2 q Ao' = ql, m, (ft) (ft) ny =2/b (kip/ft?) I, (kip/ft’) 1 15 25 6 es 0.051 0.408 = Ac! 5x5 i 20 a8 8 8 0.029 0.232 = Ao’, 1 25 25 10 8 0.019 0.152 = Ao}, E] pry sand [2] sand f&) Clay So, i 0.408 + (4)(0.232) + 0.152 BEIGE bat, = eo sha Sane fi 6 Solution For normally consolidated clay, from Eq. (10.24), Hence, GH 6a, ke _ (0.27)(20) 1814 1248 S. = —=— log 2 ———* ens pene aera Oa Me : —1te To Consolidation settlement acceleration (By draining the groundwater out from the clays) Reviews : Das, Ch. 7-10 Ch. 7: Seepage flow CH. 8: in situ stress (effective stress) CH. 9: (vertical) stress in soil mass Ch. 10: compressibility/consolidation Reviews : Ch. 7: Flow rate calculation from flow net (isotropic case) Figure 7.6 Seepage through a flow channel with square elements the flow rate in an element: AG > AG; = ha = = Aq oo (* - =) 2 (2 Jp = (2), — ... Darcy’s law ae 1 2 i 3 = hy — hy = hy — h3 = higee Te —> a Reviews : Ch. 8: Effective stress, effective stress in the case of downward flow (over pressure) When pore pressure is built up at depth, the effective stress is lessened. When the pore pressure is reduced at depth, the effective stress is increased. Vertical stress caused by a rectangular areal load EXAMPLE II A building 20 m x 20 m results in a uniform surface contact pressure of 150 kPa. Using the Newmark Influence Chart obtain the vertical pressure depth of 10 m below (a) the centre of the building (b) a corner of the building. Estimate the additional pressure at both locations of a tower 5 m x 5 m placed at the centre of the building imposing 300 kPa uniform additional pressure. Increase in stress below centre of building: = 4 x IV x M x q = 4 x 0.001 x 177 x 150 = 106 kPa Increase in stress below corner of building: = IV x M x q = 0.001 x 232 x 150 = 35 kPa Increase in stress below centre due to tower: = 0.001 x 4 x 27x300 = 32 kPa Increase in stress below corner due to tower: = 0.001 x 7 x 300 = 2 kPa Reviews : Ch. 10: Consolidation settlement under a foundation