smLecture7 Calculation of Settlement.pptx

Calculation of settlement

1. Calculation of Settlement

2. H z x Settlement of a single layer

3. H z x (1) Settlement of a single layer e e zz v S H e e = = = - + D D 1

4. H z x (1) Settlement of a single layer e e zz v S H e e thus S e H e = = = - + = - + D D D D 1 1

5. sub-layer 1 sub-layer 2 sub-layer n Soil profile divided into a number of sub-layers =

6. Thus For sub layer i S e H e i i i i - = - + D D 1 Soil profile divided into a number of sub-layers

7. (2) Thus For sub layer i S e H e so that Total Settlemen t S S e H e i i i i i n i i i n - = - + = å = - + å D D D D 1 1 1 1 [ ] Soil profile divided into a number of sub-layers

8. 5m 4m 4m Gravel Clay A B Layered soil deposit 2m Example of Settlement Calculation W.T.

9. 5m 4m 4m Gravel Clay A B Layered soil deposit 2m Stress increase at A=100 KPa Stress increase at B= 60 KPa Example of Settlement Calculation W.T.

10. Properties: Gravel Relatively incompressible Example of Settlement Calculation

11. Properties: Gravel Relatively incompressible Clay e = 0.8 G s = 2.7 C c = 0.20; C r = 0.05 ; Example of Settlement Calculation s pc =120 kPa at A

12. Voids V s = 1 m 3 Distribution of Volume Example of Settlement Calculation V v = e V s = 0.8 m 3 Skeletal material

13. Voids V s = 1 m 3 Distribution of Volume Distribution of Weight Example of Settlement Calculation V v = e V s = 0.8 m 3 Skeletal material

14. Voids V s = 1 m 3 Distribution of Volume Distribution of Weight Example of Settlement Calculation g sat w s v s W W V V kN m kN m = + + = + + = 7 84 26 46 8 1 19 06 3 3 . . . / . / V v = e V s = 0.8 m 3 Skeletal material

15. Voids V s = 1 m 3 Distribution of Volume Distribution of Weight Example of Settlement Calculation g g g sat w s v s sat s w W W V V kN m kN m or G e e kN m = + + = + + = = + + = 7 84 26 46 8 1 19 06 1 19 06 3 3 3 . . . / . / ( ) . / V v = e V s = 0.8 m 3 Skeletal material

16. The next step is to calculate the initial and final effective stress at the centre of each sub-layer Example of Settlement Calculation

17. The next step is to calculate the initial and final effective stress at the centre of each sub-layer Example of Settlement Calculation · Initial State at A Total stress s zz = 2 ´ 18 + 3 ´ 22 + 2 ´ 19.06 = 140.12 kPa

18. The next step is to calculate the initial and final effective stress at the centre of each sub-layer Example of Settlement Calculation · Initial State at A Total stress s zz = 2 ´ 18 + 3 ´ 22 + 2 ´ 19.06 = 140.12 kPa Pore water pressure u w = 5 ´ 9.8 kPa = 49 kPa (3a)

19. The next step is to calculate the initial and final effective stress at the centre of each sub-layer Example of Settlement Calculation · Initial State at A Total stress s zz = 2 ´ 18 + 3 ´ 22 + 2 ´ 19.06 = 140.12 kPa Pore water pressure u w = 5 ´ 9.8 kPa = 49 kPa (3a) Effective stress s ¢ zz = s zz - u w = 140.12 - 49 = 91.12 kPa

20. The next step is to calculate the initial and final effective stress at the centre of each sub-layer Example of Settlement Calculation · Initial State at A Total stress s zz = 2 ´ 18 + 3 ´ 22 + 2 ´ 19.06 = 140.12 kPa Pore water pressure u w = 5 ´ 9.8 kPa = 49 kPa (3a) Effective stress s ¢ zz = s zz - u w = 140.12 - 49 = 91.12 kPa Notice the initial effective stress is less than s ¢ pc =120 kPa thus the clay is initially over-consolidated.

21. Example of Settlement Calculation · Final State at A Total stress s zz = 100 + 2 ´ 22 + 3 ´ 22 + 2 ´ 19.06 = 248.12 kPa

22. Example of Settlement Calculation · Final State at A Total stress s zz = 100 + 2 ´ 22 + 3 ´ 22 + 2 ´ 19.06 = 248.12 kPa Pore water pressure u w = 7 ´ 9.8 kPa = 68.6 kPa (3b)

23. Example of Settlement Calculation · Final State at A Total stress s zz = 100 + 2 ´ 22 + 3 ´ 22 + 2 ´ 19.06 = 248.12 kPa Pore water pressure u w = 7 ´ 9.8 kPa = 68.6 kPa (3b) Effective stress s ¢ zz = s zz - u w = 248.12 - 68.6 = 179.52 kPa

24. Example of Settlement Calculation · Final State at A Total stress s zz = 100 + 2 ´ 22 + 3 ´ 22 + 2 ´ 19.06 = 248.12 kPa Pore water pressure u w = 7 ´ 9.8 kPa = 68.6 kPa (3b) Effective stress s ¢ zz = s zz - u w = 248.12 - 68.6 = 179.52 kPa Notice that the final effective stress exceeds the initial preconsolidation stress and thus the clay moves from being initially over-consolidated to finally normally consolidated.

25. Example of Settlement Calculation · Settlement of the first sub-layer The soil in the first sub layer moves from being over-consolidated to normally consolidated and so the calculation of the change in voids ratio must be made in two stages.

26. ESTIMATION OF CONSOLIDATION SETTLEMENT BY USING OEDOMETER TEST DATA Settlement calculation from e- logp curves A general equation for computing oedometer consolidation settlement may be written as follows. Normally consolidated clays Over consolidated clays for P o + Δ p < p c

27. If the thickness of the clay stratum is more than 3 m the stratum has to be divided into layers of thickness less than 3 m. Further, e is the initial void ratio and p Q , the effective overburden pressure corresponding to the particular layer; Δ p is the increase in the effective stress at the middle of the layer due to foundation loading which is calculated by elastic theory. The compression index, and the swell index may be the same for the entire depth or may vary from layer to layer. Settlement calculation from e-p curve Eq. (13.35) can be expressed in a different form as follows: where m v = coefficient of volume compressibility

28. SKEMPTON-BJERRUM METHOD OF CALCULATING CONSOLIDATION SETTLEMENT (1957) In a oedometer test the lateral yield remains constant. In view of the lateral yield, the ratios of the minor and major principal stresses due to a given loading condition at a given point in a clay layer do not maintain a constant Ko. By the one-dimensional method, consolidation settlement S oc is expressed as By the Skempton-Bejerrum method, consolidation settlement is expressed as

29. Example of Settlement Calculation · Settlement of the first sub-layer The soil in the first sub layer moves from being over-consolidated to normally consolidated and so the calculation of the change in voids ratio must be made in two stages. e log s ’ Slope C r Slope C c

30. Example of Settlement Calculation Stage 1 Soil over-consolidated ( s ¢ < s ¢ pc (initial) ) D e 1 = - C r ´ log 10 ( s ¢ pc (initial) / s ¢ I )

31. Example of Settlement Calculation Stage 1 Soil over-consolidated ( s ¢ < s ¢ pc (initial) ) D e 1 = - C r ´ log 10 ( s ¢ pc (initial) / s ¢ I ) Stage 2 Soil normally consolidated ( s ¢ = s ¢ pc ) (3c) D e 2 = - C c ´ log 10 ( s ¢ F / s ¢ pc (initial) )

32. Example of Settlement Calculation Stage 1 Soil over-consolidated ( s ¢ < s ¢ pc (initial) ) D e 1 = - C r ´ log 10 ( s ¢ pc (initial) / s ¢ I ) Stage 2 Soil normally consolidated ( s ¢ = s ¢ pc ) (3c) D e 2 = - C c ´ log 10 ( s ¢ F / s ¢ pc (initial) ) e log s ’ Slope C r Slope C c D e 1 D e 2 s ¢ pc

33. e e Example of Settlement Calculation Now: D D S H = - + 1

34. e e Example of Settlement Calculation Now: D D S H = - + 1

35. 120 91 4 1 . e e Example of Settlement Calculation Now: D D S H m = - + = ´ = + ´ 1 8 120 05 00 12 0911 2 179 52 00 10 10 [ . log ( . . ) . . log ( . . ) ] (3d)

36. Settlement of the second sub-layer is calculated in similar fashion to the settlement of the first sub-layer Example of Settlement Calculation e log s ’ Slope C c

37. Settlement of the second sub-layer is calculated in similar fashion to the settlement of the first sub-layer The settlement is then the sum of the settlements of each of the sub-layers Example of Settlement Calculation e log s ’ Slope C c

38. Calculation of Settlement To calculate the settlement it is necessary to find the initial and final effective stress. The initial stress can be calculated from a knowledge of the overburden (the weight of the overlying soil). The initial and final pore pressures can be determined from the positions of the water table. In order to find the final total stress it is necessary to find the increase in total stress due to the applied loads. Under many circumstances this can be adequately approximated using the theory of elasticity.

39. Point load of magnitude P z x Point load on an elastic half-space Useful Elastic solutions for Soil Mechanics

40. (6) Boussinesq’s solution

41. 5m p=100 kPa z r A 2m B 5m Circular loaded area on a deep elastic layer

42. (7a) (7b) p = 100 kPa a = 2.5m z = 2m Formula for Stress increase on the centre-line of the loaded area Data Stress increase at A (Note stress increase is independent of elastic parameters) Circular loaded area on a deep elastic layer

43. Circular loaded area on a deep elastic layer Note that this result can also be obtained from the chart giving the influence factors for a uniformly loaded circular area (Figure 7). In this case r/a = 0, z/a = 0.8, and hence I s = 0.8

44. 10 -3 10 -2 10 -1 1 2 4 6 8 10 10 9 8 7 6 5 4 3 2.5 2.0 1.5 1.25 1.00 0.0 z/a Values on curves are values of r/a z/a=2/2.5=0.8 r/a=0/2.5=0.0 p=100 kPa

45. Circular loaded area on a deep elastic layer Note that this result can also be obtained from the chart giving the influence factors for a uniformly loaded circular area (Figure 7). In this case r/a = 0, z/a = 0.8, and hence I s = 0.8 For the stress increase at B analytical solutions are not avaliable. The chart must be used. In this case r/a = 2, z/a = 0.8, and hence I s = 0.03

46. 10 -3 10 -2 10 -1 1 2 4 6 8 10 10 9 8 7 6 5 4 3 2.5 2.0 1.5 1.25 1.00 0.0 z/a Values on curves are values of r/a z/a=2/2.5=0.8 r/a=5/2.5=2.0 p=100 kPa

47. A B D C Stress increase below a rectangular loaded area Elevation Plan q

48. Point of interest A B D C z Stress increase below a rectangular loaded area Chart gives the stress at a distance z, beneath a corner of a rectangular loaded area Elevation Plan X q L B

49. 0.2 0.4 0.6 0.8 1.0 2.0 3.0 8 10 1 0.1 0.01 (n=L/z) 0.00 0.05 0.10 0.15 0.20 0.25 Note m & n are interchangeable (m = B/z)

50. Point of interest O A B D C X Y Z T z Stress increase below an interior point Chart gives the stress at a distance z, beneath a corner of a rectangular loaded area Must use Superposition O Elevation Plan

51. O A B D C X Y Z T 2m 3m 3m 2m Example For rectangular loading OZCT m = L/z =1 n = B/z =1

52. 0.2 0.4 0.6 0.8 1.0 2.0 3.0 8 10 1 0.1 0.01 (n=L/z) 0.00 0.05 0.10 0.15 0.20 0.25 Note m & n are interchangeable m

53. O A B D C X Y Z T 2m 3m 3m 2m Example For rectangular loading OZCT m = L/z =1 n = B/z =1 thus I s = 0.175 and so D s zz = p I s = 100 ´ 0.175 = 17.5 kPa (9a)

54. Determination of stress below an exterior point Area ABCD has uniform stress q Stress required at depth z below point O, exterior to loaded region. A B D C O (q)

55. Stage 1 +q on OXAY A B D C X Y Z T O (q) (q) (q) (q) Determination of stress below an exterior point Area ABCD has uniform stress q Stress required at depth z below point O, exterior to loaded region. A B D C O (q)

56. Stage 2 +q on OZCT A B D C X Y Z T O (q) (q) ( q) (2q) Determination of stress below an exterior point

57. Stage 2 +q on OZCT A B D C X Y Z T O (q) (q) ( q) (2q) Stage 3 -q on OZBY A B D C X Y Z T O (q) (q) (0) (q) Determination of stress below an exterior point

58. Stage 2 +q on OZCT A B D C X Y Z T O (q) (q) ( q) (2q) Stage 3 -q on OZBY A B D C X Y Z T O (q) (q) (0) (q) Stage 4 -q on OXDT A B D C X Y Z T O (0) (q) (0) (0) Determination of stress below an exterior point

59. Determination of stress below an arbitrary shape Newmark’s Chart 1. The scale for this procedure is determined by the depth z at which the stress is to be evaluated, thus z is equal to the distance OQ shown on the chart.

60. O Q z z

61. Determination of stress below an arbitrary shape Newmark’s Chart 1. The scale for this procedure is determined by the depth z at which the stress is to be evaluated, thus z is equal to the distance OQ shown on the chart. 2. Draw the loaded area to scale so that the point of interest (more correctly its vertical projection on the surface) is at the origin of the chart, the orientation of the drawing does not matter

62. O Q z z

63. Determination of stress below an arbitrary shape Newmark’s Chart 1. The scale for this procedure is determined by the depth z at which the stress is to be evaluated, thus z is equal to the distance OQ shown on the chart. 2. Draw the loaded area to scale so that the point of interest (more correctly its vertical projection on the surface) is at the origin of the chart, the orientation of the drawing does not matter 3. Count the number of squares (N) within the loaded area, if more than half the square is in count the square otherwise neglect it.

64. Determination of stress below an arbitrary shape Newmark’s Chart 1. The scale for this procedure is determined by the depth z at which the stress is to be evaluated, thus z is equal to the distance OQ shown on the chart. 2. Draw the loaded area to scale so that the point of interest (more correctly its vertical projection on the surface) is at the origin of the chart, the orientation of the drawing does not matter 3. Count the number of squares (N) within the loaded area, if more than half the square is in count the square otherwise neglect it. 4. The vertical stress increase D s zz = N ´ [scale factor(0.001)] ´ [surface stress (p)]

65. O Q 4m Loaded Area