Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay

Size: px

Start display at page:

Download "Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay"

Stewart Anderson
6 years ago
Views:

1 13

2 Permeability and Seepage -2

3 Conditions favourable for the formation quick sand Quick sand is not a type of sand but a flow condition occurring within a cohesion-less soil when its effective stress is reduced to zero due to upward flow of water. Quick sand occurs in nature when water is being forced upward under pressurized conditions. In this case, the pressure of the escaping water exceeds the weight of the soil and the sand grains are forced apart. The result is that the soil has no capability to support a load. Why does quick condition or boiling occurs mostly in fine sands or silts?

4 Some practical examples of quick conditions Excavations in granular materials behind cofferdams alongside rivers Any place where artesian pressures exist (i.e. where head of water is greater than the usual static water pressure). -- When a pervious underground structure is continuous and connected to a place where head is higher. Behind river embankments to protect floods

5 Some practical examples of quick conditions Contrary to popular belief, it is not possible to drown in quick sand, because the density of quick sand is much greater than that of water. ρ ρ quicksand >> water Consequently, it is literally impossible for a person to be sucked into quicksand and disappear. So, a person walking into quicksand would sink to about waist depth and then float.

6 Boiling condition Example The sand layer of the soil profile shown in figure is under artesian pressure. A trench is to be excavated in the clay up to a depth of 4m. Determine the depth of water h to avoid boiling. 6m Water level 4m h stiff clay γ = 18kN/m 3 3m P Sand 5m Rock

7 Boiling condition Solution: At P: σ = hγ w + (6-4)γ clay = hγ w + 2γ clay At P: u = 5γ w At P: σ = hγ w + 2γ clay - 5γ w σ = 0 (boiling condition) hγ w + 2γ clay - 5γ w = 0 h = (5γ w - 2γ clay )/γ w = (5 x x18)/9.81 h = 1.33m 6m Water level 4m h = 1.33 m stiff clay γ = 18kN/m 3 3m P Sand Rock 5m

8 Seepage forces B Water flowing past a soil particle exerts a drag force on the particle in the direction of flow. h W d (F s ) c The drag force is caused by pressure gradient and by viscous drag. L

9 Seepage forces B Direction of flow W W d (F s ) c FBD of grain FBD of soil At critical condition; h = h c Gs + e 1+ e γ wal = + ( h ) c L Aγ w i c = hc L = Gs 1 1+ e

10 Considering FBD of grain in the direction of flow at i = i c ( ) ( ) ( ) ( ) ( ) ( ) w w s c s w V w s c s w s w s c s d c s c s d AL n AL e G F V V V V AL e G F V AL e G F B W F B F W γ γ γ γ γ γ ) ( = + = + = = + = V i F V e G F w c c s w s c s γ γ = + = ) ( 1 1 ) ( Seepage pressure p s = i c γ w If h < h c then seepage force J s is iγ w V

11 Seepage force Area per unit width of section = l x 1 Volume affected by seepage force = l 2 x 1 h 1 > h 2 h 1 h 2 h Direction of flow Force applied to sand particles = γ w h 1 -γ w h 2 ( l x 1) l l

12 Seepage force Force applied to sand particles = γ w ( h) ( l x 1) = γ w ( h/ l) ( l 2 x 1) J = i γ w V Seepage pressure p s = Seepage force per unit volume = i γ w

13 Critical hydraulic gradient Quick condition The quick condition occurs at a critical upward hydraulic gradient i c, when the seepage force just balances the buoyant weight of an element of soil. (Shear stresses on the sides of the element are i c neglected.) = Gs 1 υ The critical hydraulic gradient is typically around 1.0 for many soils. Fluidized beds in chemical engineering systems rely on deliberate generation of quick conditions to ensure that the chemical process can occur most efficiently.

14 Example problem Determine and plot total stress, PWP and effective stress diagrams if: (i) h = 1 m; (ii) h = 4 m; and (iii) h = 2 m. i = h 2 Case I; i = 0.5; Case II; i = 2; Case III; i = 1 a b 1 m c 1 m d Datum h 2 m γ sat =20 kn/m 3

15 Case - I i = 0.5 Point PH [m] EH [m] TH [m] a b d c

16 a h = 1 m 2 m Case - I i = 0.5 σ u σ b 1 m c 1 m d Datum γ sat =20 kn/m Stress units are in kn/m 2

17 Case - II i = 2 Point PH [m] EH [m] TH [m] a b d c 5 1 6

18 a Case - II h = 4 m 2 m Stress units are in kn/m 2 σ u σ b 1 m c 1 m d Datum γ sat =20 kn/m Quick sand condition would have already occurred

19 Case - III i = 1 Point PH [m] EH [m] TH [m] a b d c 4 1 5

20 Case - III Stress units are in kn/m 2 a h = 2 m σ u σ 0 2 m b 1 m c 1 m d Datum Just subjected to Quick sand condition 0 0 0

21 Example problem A large open excavation was made in a stratum of clay with a saturated unit weight of 17.6 kn/m 3. When the depth of the excavation reached 7.5 m, the bottom rose, gradually cracked and was flooded from below by a mixture of sand and water. Subsequent borings showed that the clay was underlain by a bed of sand with its surface at a depth of 11 m. Compute the elevation for which the water would have risen from the sand surface into a drill hole before the excavation was started.

22 Solution 11m γ sat = 17.6 kn/m 3

23 Solution Elevation for which the water would have risen from the sand surface into a drill hole before the excavation was started. h = 6.16 m A 7.5 m h For σ = 0 at A: σ = 17.6 x (11 7.5) - 10 x h = 0

24 Example problem Determine and plot the total vertical stress, pore water and effective vertical stress distribution at levels A, B and C. 4 m A 6 m B 3 m C 3 m Silty clay; G s = 2.70; w = 45.2 % Coarse sand; G s = 2.65; e = 0.833

25 Example problem An artesian pressure exists in the lower sand layer. 1 m 1 m γ d = 18.5 kn/m 3 Determine σ, u, and σ at A, B, and C. 2 m A γ sat = 19 kn/m 3 (TH) A = = 5m 1.33 m 2 m (TH) B = = 7m 2 m 1 m C B γ clay = 17 kn/m 3 DATUM

26 Solution At point B: σ = 18.5 x x x 3 = kn/m 2 u = 10 x 7 = 70 kn/m 2 σ = = 37.5 kn/m 2 At point A: σ = 18.5 x x 2 = 56.5 kn/m 2 u = 10 x 2 = 20 kn/m 2 σ = = 36.5 kn/m 2

27 Solution At point c: σ = 18.5 x x x 2 = 90.5 kn/m 2 u = 10 x (PH) c = 10 x 5.33 = 53.3 kn/m 2 σ = = 38.2 kn/m 2 (TH) C = (TH) B - i Z = 7 (2/3) x 1 = 6.33 m (PH) C = (TH) C Z c = = 5.33 m

28 1 m 1 m 2 m 2 m A m C B DATUM

29 Measurement of soil permeabilities The rate of flow of water q (volume/time) through cross-sectional area A is found to be proportional to hydraulic gradient i according to Darcy s law: q v = = A ki i = h L where v is flow velocity and k is coefficient of permeability with dimensions of velocity (length/time). The coefficient of permeability of a soil is a measure of the conductance (i.e. the reciprocal of the resistance) that it provides to the flow of water through its pores.

30 Measurement of soil permeabilities The value of the coefficient of permeability k depends on: (i) Average size of the pores and is related to the particle sizes and their packing, d = ed10 (ii) Particle shape, and d10 d = (iii) Soil structure. 5 The ratio of permeabilities of typical sands/gravels to those of typical clays is of the order of A small proportion of fine material in a coarse-grained soil can lead to a significant reduction in permeability.

31 Measurement of soil permeabilities A number of tests can be used to measure or estimate the permeability of soils Laboratory methods: The constant head test (used for highly permeable soils) The falling head test (used for relatively impermeable soils) Indirect methods: computation from grain size distribution and During Oedometer test Field methods: 1. Pumping tests and 2. Borehole tests

32 Laboratory measurement of permeability In determining permeability of coarse grained material, large quantities of flow occur in short periods of time and small quantities of flow occur over long periods of time for fine grained soils. Two aspects that need careful attention for all types of soils are: (i) To ensure that flow occurs only through the soil and not at the interface between soil and the mould in which the soil is contained. (ii) The soil sample is fully saturated before recording observations

33 Determination of coefficient of permeability in the laboratory Water Constant supply water level Constant Head test Q = Avt h i = L Q = Ak QL k = Aht h L t A = area of C/S of the specimen L h

34 Main features of constant head test It is suitable for soils having a coefficient of permeability in the range of 10-2 m/s to 10-5 m/s, which applies to clean sand and sand-gravel mixtures with less than 10 % fines. It can be suitable for soils when used in their completely disturbed or remolded states such as for drainage materials and filters to confirm that their performance will be adequate.

35 dh Falling head test or Variable head test h h 2 h 1 Cross-sectional area of stand pipe = a Cross-sectional area of specimen = A At time t o head h 1 L At time t 1 head h 2 Let h be the head of water at any time t. Let in time dt the head drop by amount dh

36 Falling head test Quantity of water flowing through the sample in time dt from Darcy s law: dq = kia( dt) Quantity of discharge = k h L A( dt) k a = can also be expressed as: h2 ka t = 1 dh dt h1 L t t0 al log 10 At = dq = a(dh) h h 1 2 As the time increases head decreases!!

37 Main features of constant head and falling head tests In the constant head test, permeability is computed on the basis of fluid that passes through the soil sample. While in the falling head test, k is computed on the basis of fluid flowing into the sample. With theconstant head test, time is required to accumulate the fluid volume necessary to perform computation. Extreme care would be required to prevent leaks in the apparatus and evaporation of discharge water. With the falling head test, the duration of the test is shortened and care is required to prevent evaporation of water in the inlet tube.

38 Problem Determine and plot the total vertical stress, pore water and effective vertical stress distribution at levels A, B and C. 4 m A 6 m B 3 m C 3 m Silty clay; G s = 2.70; w = 45.2 % Coarse sand; G s = 2.65; e = 0.833

Instructor : Dr. Jehad Hamad. Chapter (7)

Instructor : Dr. Jehad Hamad. Chapter (7) Instructor : Dr. Jehad Hamad Chapter (7) 2017-2016 Soil Properties Physical Properties Mechanical Properties Gradation and Structure Compressibility Soil-Water Relationships Shear Strength Bearing Capacity