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# Chapter 4 Settlement and consolidation.ppt

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# Chapter 4 Settlement and consolidation.ppt

Settlement and Consolidation

Settlement and Consolidation

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### Chapter 4 Settlement and consolidation.ppt

1. 1. Settlement and Consolidation CHAPTER 4
2. 2. §4 Settlement and Consolidation §4.1 General §4.2 Oedometer test §4.3 Preconsolidation pressure §4.4 Consolidation settlement § 4.5 Terzaghi’s theory § 4.6 Degree of consolidation
3. 3. §4.1 General §4 Settlement and Consolidation  compressibility –volume changes in a soil when subjected to pressure –giving AMOUNTS of settlement  consolidation -rate of volume change with time – giving TIME to produce an amount of settlement required
4. 4. §4.2 Oedometer test §4 Settlement and Consolidation P1 s1 e1 e0 P t e s t P2 s2 e2 P3 s3 e3
5. 5. e H e H    1 1 1 0 0 ) 1 ( 0 0 0 e H s e e    curve curve curve
6. 6. Compression coefficient a p1 p2 e1 e2 M1 M2 e0 e p curve e-p △p △e p e a d d   1 2 2 1 p p e e p e a       ＝ Evaluation of compression with a1-2  a1-2＜0.1MPa-1 , Low compressibility  0.1MPa-1≤a1-2＜0.5MPa-1, Middle compressibility  a1-2≥0.5MPa-1, High compressibility 1 2 2 1 ＝ p p e e p e a slope      
7. 7. e - lgσ′Curve compression index Cc p C dp de a c v    1 2 2 1 1 2 2 1 lg ) ( lg lg p p e e p p e e Cc       p dp de p p e e C c c lg lg 2 1    
8. 8. Preconsolidation pressure-the maximum effective vertical stress that has acted on the clay in the past p s OCR    OCR=1： lack consolidation OCR>1： normal consolidation OCR<1： over consolidation How to obtain the preconsolidation pressure: 1 Produce back the straight-line part (BC) . 2 Determine the point (D) of maximum curvature on the recompression part (AB) of the curve. 3 Draw the tangent to the curve at D and bisect the angle between the tangent and the horizontal through D. 4 The vertical through intersection point of the bisector and CB produced gives the approximate value of the preconsolidation pressure. §4.3 Preconsolidation pressure §4 Settlement and Consolidation
9. 9. s c d S S S S     coefficient of volume compressibility or the compression index  Consider a layer of saturated clay of thickness H.  an elemental layer of thickness dz at depth z. 1 2 2 1 p p e e p e a       ＝ 1 1 1 2 1 ) ( 1 H E p H p p e a s s      1 1 2 1 2 1 1 H e e e H H s      §4.4 Consolidation settlement §4 Settlement and Consolidation
10. 10. Curve of gravity stress and additional stress at the central of base Determine the calculation depth Determine the layer The settlement of each layer The whole settlement i i i i i h e e e s 1 2 1 1        n i i s s 1 d depth σc σz z i-1 1 2 3 4 5 6 1 2 aip0 ai-1p0 z i Ai Ai-1
11. 11. Example 1
12. 12. The assumptions made in the theory are: 1 The soil is homogeneous and fully saturated. 2 There is a unique relationship, independent of time, between void ratio and effective stress. 3 The solid particles and water are incompressible. 4 Compression and flow are one-dimensional (vertical). 5 Strains are small. 6 Darcy’s law is valid at all hydraulic gradients. 7 The coefficient of permeability and volume compressibility remain constant. §4.5 Terzaghis theory of one-dimensional consolidation §4 Settlement and Consolidation
13. 13.   2 1 2 w k 1 e u u t a z        The total stress increment soil skeleton increasing effective stress the excess pore water pressure decreases   a e k cv ) 1 ( 1   ) 4 / exp( 2 sin 1 4 2 2 1 2 , v m z t z T m H m m u          t H c T v v 2 
14. 14. c ct s s U  the consolidation settlement at time t being given  by the product of U and  the final settlement.            S S H e a dz e a dz dz U t z t z z t z t 1 1 , , 1 1     §4.6 Degree of consolidation §4 Settlement and Consolidation
15. 15. Determine the degree of consolidation Time factor permeable impermeable degree of consolidation Curve 1 Curve 2 Curve 3 P
16. 16.  H=10m ; e1=0.8; a=0.00025kPa-1; k=0.02m/year ？ ① settlement after one year St ② time(t) when Uz=0.75 ③ if the bottom layer is permeable，time(t) when Uz=0.75 Example 2 157kPa 235kPa H p clay impermeable permeable
17. 17. Solution  1. When t=1 year mm H e a S z 273 1 1     year m a e k c w v / 4 . 14 ) 1 ( 2 1     144 . 0 2   t H c T v v 5 . 1 157 235   a From figure Ut=0.45 mm S U S z t 123    2. If Uz=0.75 From Uz=0.75，a＝1.5 then Tv＝0.47 year 26 . 3 2   v v c H T t  3. If open layer, Uz=0.75 From Uz=0.75,a＝1,H5m then Tv＝0.49 year 85 . 0 2   v v c H T t