* ! " " ) % )!( & ' % # $. 0 1 %./ +, - 7 : %8% 9 ) 7 / ( * 7 : %8% 9 < ;14. " > /' ;-,=. / ١

Size: px

Start display at page:

Download "*** ***! " " ) * % )!( & ' % # $. 0 1 %./ +, - 7 : %8% 9 ) 7 / ( * 7 : %8% 9 < ;14. " > /' ;-,=. / ١"

James Harper
6 years ago
Views:

1 ١ ******!" #$ % & '!( ) % * ") +,-./ % ( /4 ) 8%9 % : 7 ;14 < 8%9 % : *7./ = ;-, >/'."

2 Soil Permeability & Seepage ٢

3 Soil Permeability- Definition ٣

4 What is Permeability? Permeability is the measure of the soil s ability to permit water to flow through its pores or voids water Loose soil - easy to flow - high permeability Dense soil - difficult to flow - low permeability

5 Importance of Permeability The following applications illustrate the importance of permeability in geotechnical design: The design of earth dams is very much based upon the permeability of the soils used. The stability of slopes and retaining structures can be greatly affected by the permeability of the soils involved. Filters made of soils are designed based upon their permeability. ٥

6 Importance of Permeability ٦

7 Importance of Permeability Estimating the quantity of underground seepage (Ch. 8) Solving problems involving pumping seepage water from construction excavation ٧

8 Flow rate through soil What is the flow rate through a soil? Concrete dam SOIL Flow rate = Q [m 3 /sec]

9 ٩

10 ١٠

11 tank of water A B

12 Why does water flow? If flow is from A to B, the energy is higher at A than at B. Energy is dissipated in overcoming the soil resistance and hence is the head loss. water A B ١٢

13 Soil Permeability -Definition Soils consists of solid particles with interconnected voids where water can flow from a point of high energy to a point of low energy water ١٣

14 Bernoulli s Equation The energy of fluid comprise of: 1. Potential energy - due to elevation (z) with respect to a datum fluid particle z - due to pressure 2. Kinetic energy - due to velocity datum ١٤

15 ١٥

16 ١٦

17 Soil piezometer h p A u = γ w h p Pressure head at A. The pore water pressure at A is

18 Hydrostatic pore water pressure u h = γ w h p z w1 u h = γ w z w1 z w z w2 u h = γ w z w 2 Depth, z

19 Bernoulli s Equation Then at any point in the fluid, the total energy is equal to Total Energy = Potential energy + Pressure energy + Kinetic energy = mgh + P + ½mv 2 Expressing the total energy as head (units of length) Total Head = Elevation Head + Pressure Head + Velocity Head h = Z + u γ + w 2 2 v g 0 For flow through soils, velocity (and thus velocity head) is very small. Therefore, v 2 /2g = zero.

20 ٢٠

21 ٢١

22 Bernoulli s Equation At any point w u Z h γ + = The head loss between A and B + + = = w B B w A A B A u Z u Z h h h γ γ Head loss in non-dimensional form L h i = Hydraulic gradient Distance between points A and B Difference in total head

23 W.T. Impervious Soil )h = h A - h B Water In h A Impervious Soil W.T. Datum h B Head Loss or Head Difference or Energy Loss i = Hydraulic Gradient h A (q) Water out h B Z A Z B Datum

24 Seepage Through Porous Media Water In L = Drainage Path i = Hydraulic Gradient Head Loss or Head Difference or Energy Loss h A Water out h B A Soil B Datum Porous Stone L Porous Stone

25 Seepage Through Porous Media Water In L = Drainage Path i = Hydraulic Gradient Head Loss or Head Difference or Energy Loss )h =h A - h B h A Water out h B A Soil B Z A Porous Stone L Porous Stone Z B Datum

26 Tricky case!! Remember always to look at total head P B /γ w P A /γ w

27 Hydraulic Gradient In the field, the gradient of the head is the head difference over the distance separating the 2 wells. i H H X 1 2 = ٢٧

28 Also..

29 WaterMovement in Soil Two Principles to Remember: 1. Darcy s Law 2. Continuity Equation: mass in = mass out + change in storage

30 Darcy s Law Assumptions: flow is laminar soil properties do not with time

31 Darcy s Law Since velocity in soil is small, flow can be considered laminar v i v = discharge velocity = i = hydraulic gradient v = k i k = coefficient of permeability Q = kia Cross-sectional area to flow Hydraulic conductivity permeability [cm/s] Hydraulic gradient

32 k Units are in cm/sec but k = velocity

33 Flow in Soil W.T. A Impervious Soil h = h A - h B L W.T. h A = total head B Datum Impervious Soil h B = total head i = ( h A L h B ) = h L Q = k i A = k h L A

34 Solution Q = kia k = 4x10-2 cm/sec i = h/l = (167.3m 165m) / 256m = A = (3.2 m) (1000 m) = 3200 m 2 Q = kia = m 3 /sec = 41.5 m 3 /hr

36 ٣٦

37 Hydraulic Conductivity The hydraulic conductivity k is a measure of how easy the water can flow through the soil. The hydraulic conductivity is expressed in the units of velocity (such as cm/sec and m/sec). ٣٧

38 k Measure of a soil-fluid system s resistance to flow depends on soil Void size Fabric (structure) Void continuity Specific surface (drag) fluid Viscosity Mass density

39 Hydraulic Conductivity Hydraulic conductivity of soils depends on several factors: Fluid viscosity (η): as the viscosity increases, the hydraulic conductivity decreases Pore size distribution Temperature Grain size distribution Degree of soil saturation It is conventional to express the value of k at a temperature of 20 o C. ٣٩

40 ٤٠

41 To determine the quantity of flow, two parameters are needed * k = hydraulic conductivity * i = hydraulic gradient k can be determined using 1- Laboratory Testing [constant head test & falling head test] 2- Field Testing [pumping from wells] 3- Empirical Equations i can be determined 1- from the head loss and geometry 2- flow net (chapter 8)

42 Laboratory Testing of Hydraulic Conductivity Two standard laboratory tests are used to determine the hydraulic conductivity of soil The constant-head test The falling-head test. ٤٢

43 Constant Head Test ٤٣

44 Constant Head Test From Darcy s Law Then compute: ٤٤

45 Constant Head Test ٤٥

47 Falling Head Test The falling head test is mainly for fine-grained soils. Simplified Procedure: Record initial head difference, h 1 at t 1 = 0 Allow water to flow through the soil specimen Record the final head difference, h 2 at time t = t 2 Then compute: ٤٧

48 Falling Head Test ٤٨

49 Limitations of Laboratory tests for Hydraulic Conductivity i. It is generally hard to duplicate in-situ soil conditions (such as stratification). ii. The structure of in-situ soils may be disturbed because of sampling and test preparation. iii. Small size of laboratory samples lead to effects of boundary conditions.

50 ٥٠

51 ٥١

52 ٥٢

53 ٥٣

54 ٥٤

55 Equivalent Hydraulic Conductivity on Stratified Soils Horizontal flow Constant hydraulic gradient conditions Analogous to resistors in series

56 Equivalent Coefficient of Vertical Permeability (kv ) q = A* v = H * kh'* i average H. kh'. i = k. H. i + k. H. i k. H. i n n kh k. H + k. H ' = 2 H k n. H n

57 Equivalent Hydraulic Conductivity on Stratified Soils Vertical flow Constant velocity Analogous to resistors in parallel

58 Equivalent Coefficient of Vertical Permeability (kv ) Equivalent Coefficient of Vertical Permeability (kv ) Equivalent Coefficient of Vertical Permeability (kv ) Equivalent Coefficient of Vertical Permeability (kv ) Basic Concept q in = q out v constant n n n H h k H h k H h k i kv v ' = = = = = v h k H... v h k H ; v h k H ; v h k H n n n = = = = n n n k H... k H k H k H v h... v h v h v h = H H H H H n = n H n k k H k H k H H kv =... '

60 EXAMPLE 2 Questions : - determine h - determine q in cc/sec q Section 1 Section 2

61 EXAMPLE 2 Determination of h Section 1 Section 2 q = k 1 1.i1. A1 q 1 q 2 = k 2.i 2. A2 50 h h 5 = q2 = q = q (50 h) = ( h 5) Bina Nusantara h = cm

62 EXAMPLE 2 Determination of water flow rate q = k 1 1.i1. A1 or q = k 2 2.i 2. A2 q = q = 0.15 cc/s Bina Nusantara

63 Determination of Hydraulic conductivity in the Field 1. Pumping Wells with observation holes 2. Borehole test. 3. Packer Test.

Pumping Well with Observation holes Pumping Well in an Unconfined Aquifer q k = 2.303 q π ( h 2 2.

64 Pumping Well with Observation holes Pumping Well in an Unconfined Aquifer q k = q π ( h 2 2.log h ) r r 2 1 k OR r2 q.ln r = π ( h h ) If q, h 1, h 2, r 1, r 2 are known, k can be calculated

65 Stress Concept

66 Stresses in Soils 1. Geostatic Stresses 2. Induced Stresses Due to soil s self weight Due to added loads (structures) 3. Dynamic Stresses e.g., earthquakes AS i g t i

67 ٦٧

68 ٦٨

69 ٦٩

70 Effective Stress Spring Analogy σ = σ u P σ σ = effective stress σ= total stress * u = pore pressure X σ u

71 Geostatic Stresses SHEAR STRESSES If ground surface is flat, all geostatic shear stresses = zero

72 ٧٢

73 ٧٣

74 Geostatic Stresses TOTAL VERTICAL STRESS AT A POINT Ground surface z = depth = 5 m Soil, γ = 18 kn/m 3 A σ = γ z A A total vertical stress at A

75 ٧٥

76 ٧٦

77 Pore water pressures u hydrostatic = u h = due to hydrostatic condition only u excess = u e = due to additional processes u = u + u e h

78 Geostatic Stresses PORE WATER PRESSURE AT A POINT Ground surface z = 5 m Soil, γ = 18 kn/m 3 h pa A u A = γ w h p A pore water pressure at A

79 ٧٩

80 ٨٠

81 ٨١

82 ٨٢

83 ٨٣

84 ٨٤

85 EXAMPLE Plot the variation of total and effective vertical stresses, and pore water pressure with depth for the soil profile shown below in Fig. ٨٥

86 Solution: Within a soil layer, the unit weight is constant, and therefore the stresses vary linearly. Therefore, it is adequate if we compute the values at the layer interfaces and water table location, and join them by straight lines. ٨٦

87 Pore Water Pressure ٨٧

88 Effective Stress General Expression ٨٨

89 Methods of Computations Effective Stress ٨٩

90 In Flow Piezometer 3 ft No Seepage D C Out Flow 2 ft 14 ft u = 14 x 62.4 u = 6 x 62.4 B 4 ft 12 ft W s 8 ft W s W s W s W s A Buoyancy 3 ft 3 ft Datum

91 W s W s W s W s W s Buoyancy ٩١

92 No Seepage 1 W.T. γ 1 =110 pcf 3 ft 4 ft 2 - = 6 ft 3 12 ft 4 Total Stress Pore Water Pressure Effective Stress W W s W s s W s W s Buoyancy

93 No Seepage 1 1 γ 1 =110 pcf 2 2 W.T. 3 ft 4 ft = 6 ft ft 5 5 Total Stress Pore Water Effective Stress Pressure Total Stress Pore Water Pressure Effective Stress σ1 = σ2 = σ3 = u1 = u2 = u3 = σ1 = σ2 = σ3 = σ4 = u4 = σ4 = σ5 = u5 = σ5 = W s W s W s W s Buoyancy W s

94 1 No Seepage W.T. γ 1 =110 pcf 2 3 ft 3 ft 4 ft 3 - = 6 ft 4 12 ft 5 Total Stress Pore Water Pressure Effective Stress W W s W s s W s W s Buoyancy

95 Downward Seepage Piezometer D In Flow 3 ft Out Flow u = 6 x u C 4 ft 2 ft 10 ft u = 17 x 62.4 B 12 ft Seepage Force 8 ft W s W s W s W s W s A Buoyancy - Seepage Force 3 ft 3 ft Datum

96 Seepage Force W s W s W s W s W s Buoyancy - Seepage Force ٩٦

97 Downward Seepage 1 γ 1 =110 pcf 1 W.T. 3 ft 3 ft 4 ft = 6 ft ft Total Stress 4 Pore Water Pressure 4 Effective Stress Seepage Force Total Stress Pore Water Pressure Effective Stress W s W s W s W s W s Buoyancy - Seepage Force

98 In Flow Piezometer Upward Seepage u D 3 ft u = 6 x u C Out Flow 4 ft 2 ft 17 ft u = 17 x 62.4 B 8 ft 12 ft W s W s W s W s W s A Buoyancy + Seepage Force 3 ft 3 ft Datum

99 W s W s W s W s W s Buoyancy + Seepage Force ٩٩

100 Upward Seepage 5 ft 2 1 γ 1 =110 pcf W.T. 3 ft 4 ft 3 - = 6 ft 4 12 ft Total Stress 54 Total Stress Pore Water Pressure Pore Water Pressure 4 Effective Stress Effective Stress W s W s W s W s W s Buoyancy + Seepage Force

101 γ 1 =110 pcf W.T. 3 ft W.T. 3 ft 4 ft 4 ft 6 ft 6 ft 12 ft 12 ft

102

103

104

105

106 SEEPAGE FORCE γ w. h 2. A H L Soil weight = γ t.l.a h 1 h 2 γ w. h 1. A L TOTAL FORCE F = γ. L. A γ.( h h A t w 1 2 ). BODY FORCE Body Force( F) = Total Force volume

107 SEEPAGE FORCE F F F γ L A h = L. A H + L = γ t γ w L = γ i. γ t.. γ w.( 1 2 bouyancy w h ). A = γ γ t w (1 + i) γ bouyancy = γ t - γ w SEEPAGE BODY FORCE (j)= CRITICAL CONDITION i. γ γ i w c bouyant = γ i. γ bouyant γ w w = = G 0 s 1+ 1 e

108 EXAMPLE : k = 1x10-3 cm/s n = 0.67 Questions : 1. Water Flow Rate 2. Flow Velocity 3. Seepage Velocity 4. Seepage Force at point A

109 Water Flow Rate H 4 q = k.i.a i = = = 1 L 4 Flow Velocity v = k.i v = 1x10.1 = 1x m / s q = 1x10.1.A = 1x A Seepage Velocity k.i v ' = = n v n 1*10 5 v ' = = 1.5x10 5 m / s 0.67 Seepage Force F s = i. γ w F = 1*1000 = 1000 kg / m s 2

Laplace equation of Continuity In reality, the flow of water through soil is not in one direction only, nor is it uniform over the entire area perpendicular to the

110 Laplace equation of Continuity In reality, the flow of water through soil is not in one direction only, nor is it uniform over the entire area perpendicular to the flow. The flow of water in two dimensional is described using Laplace equation. Laplace equation is the combination of the equation of continuity and Darcy s law. ١١٠

111 Laplace equation of Continuity Flow in: v x dydz v z dxdy Flow out: v x vx + dx dydz x v Flow in = Flow out (Continuity equation) z + v z z dz dxdy By simplification, we get v v + x dx dydz + v v + z dz dxdy v dydz x z x z = x + v z dxdy v x x dxdydz + v z z dxdydz = 0 v x x + v z z = 0 ١١١

112 Laplace equation of Continuity From Darcy s Law: dh vx = kx vz = dx k z dh dz Replace in the continuity equation k x 2 h + k 2 x 2 h = 0 2 z If soil is isotropic (i.e. k x = k z = k) z 2 h 2 x + 2 h 2 z = 0 Laplace equation This equation governs the steady flow condition for a given point in the soil mass ١١٢

113 q = A k i = A k h L Equipotential Lines Flow Lines

114 Principles of the Flow Net Equipotential Lines Flow Element

115 Principles of the Flow Net )h = head loss = one drop Piezometer Flow Element Datum Equipotential Lines Total heads along this line are the same

116 Flow nets Flow nets are a graphical solution method of Laplace equation for 2D flow in a homogeneous, isotropic aquifer. In an isotropic medium, the continuity equation represents two orthogonal families of curves: 1. Flow lines: the line along which a water particle will travel from upstream to the downstream side in the permeable soil medium 2. Equipotential lines: the line along which the potential (pressure) head at all points is equal. Datum ١١٦

equipotential lines in such away that: 1.

117 Flow nets Flow nets are the combination of flow lines and equipotential lines. To complete the graphic construction of a flow net, one must draw the flow and equipotential lines in such away that: 1. The equipotential lines intersect the flow lines at right angles. 2. The flow elements formed are approximate squares. Equipotential line Flow channel Flow line ١١٧

118 Boundary Conditions ١١٨

119 H 0 H ١١٩ H-3 h

120 ١٢٠

121 Seepage Calculation from Flow Net In a flow net, the strip between any two adjacent flow lines is called a flow channel. The drop in the total head between any two adjacent equipotential lines is called the potential drop. Flow element If the ratio of the sides of the flow element are the same along the flow channel, then: 1. Rate of flow through the flow channel per unit width perpendicular to the flow direction is the same. q 1 = q 2 = q 3 = q 2. The potential drop is the same and equal to: h h = h h = h h = Where H: head difference between the upstream and downstream sides. N ١٢١ d : number of potential drops H N d

122 Seepage Calculation from Flow Net ١٢٢ From Darcy s Equation, the rate of flow is equal to: If the number of flow channels in a flow net is equal to N f, the total rate of flow through all the channels per unit length can be given by: f = N d N H k q = = = = = N d H k q l l h h k l l h h k l l h h k q

123 Example on estimating the total flow under dam Example:if k = 10-7 m/sec, what would be the flow per day over a 100 m length of wall? 50 m of water Dam cutoff 5 m of water Low permeability rock ١٢٣

124 Calculations N f = 5 N d = 14 h= 45 m k= 10-7 m/sec Answer: = 10-7 (5/14) 45 x 100 m length = m 3 /sec = 13.9 m 3 /day ١٢٤

125 Pressure head at any point Total head = h L - # of drops from upstream x h Elevation head = -z Pressure head = Total head Elevation head = +ve h N L d datum H T = h L concrete dam H T = 0 h L z -ve h X impervious strata

126 In Flow )h )h )h )h )h )h )h )h 3 in Out Flow 2 in 8 14 in F eff = *( soil + * ( water - ( - )h) * ( water u = [14 - (3. )h)].( water 4 W s 3 W s W s 2 W s W s 1 2 Buoyancy + Seepage Force

127 Uplift Pressure under Hydraulic structures At point a: u a /γ w = (7-1x1) - (-2) = 8 kn/m 2 At point b: Datum u a /γ w = (7-2x1) - (-2) = 7 kn/m 2 At point f: u a /γ w = (7-6x1) - (-2) = 3 kn/m 2 h = h L /N d = 7/7=1

128 Example of Dam Failure Around 7:00 am on June 5, 1976 a leak about 30 m from the top of Teton dam was observed. ١٢٨

129 Example of Dam Failure The Dam Broke at 11:59 AM ١٢٩

130 ١٣٠

131 ١٣١

132 ١٣٢

133 ١٣٣

134 Flow Nets: an example Posit ion: A B C D E F G H I J Dist ance f rom f ront t oe (ft) n The flow net is drawn with: m = 5 n =

135 Flow Nets: the solution Solve for the flow per unit width: q = (m/n) K h = (5/17)(150)(35) = 1544 ft 3 /day per ft 135

136 Flow Nets: An Example There is an earthen dam 13 meters across and 7.5 meters high.the Impounded water is 6.2 meters deep, while the tail water is 2.2 meters deep. The dam is 72 meters long. If the hydraulic conductivity is 6.1 x 10-4 centimeter per second, what is the seepage through the dam if n = 21 K = 6.1 x 10 4 cm/sec = m/day 136

137 Flow Nets: the solution From the flow net, the total head loss, H, is = 4.0 meters. There are 6 flow channels (m) and 21 head drops along each flow path (n): Q = (KmH/n) x dam length = (0.527 m/day x 6 x 4m / 21) x (dam length) = 0.60 m 3 /day per m of dam = 43.4 m 3 /day for the entire 72-meter length of the dam 137

$through rock openings Hydraulic fracture Differential settlement and cracking Settlement in$

138 Example of Dam Failure Post Failure Investigation Seepage piping and internal erosion Seepage through rock openings Hydraulic fracture Differential settlement and cracking Settlement in bedrock ١٣٨

139 Piping in Granular Soils At the downstream, near the dam, the exit hydraulic gradient i exit = h l H concrete dam l h = total head drop impervious strata

140 Piping in Granular Soils If i exit exceeds the critical hydraulic gradient (i c ), first the soil grains at exit get washed away. This phenomenon progresses towards the upstream, forming a free passage of water ( pipe ). concrete dam H no soil; all water impervious strata

141 Critical hydraulic gradient, i c The critical hydraulic gradient (i c ), γ = γ sat γ w Consequences: no stresses to hold granular soils together soil may flow boiling or piping = EROSION ١٤١

142 Piping in Granular Soils Piping is a very serious problem. It leads to downstream flooding which can result in loss of lives. Therefore, provide adequate safety factor against piping. F piping = i i c exit > 3 concrete dam impervious strata

143 Filters Used for: facilitating drainage preventing fines from being washed away Used in: earth dams retaining walls Filter Materials: granular soils geotextiless ١٤٣

144 Granular Filter Design The proper design of filters should satisfy two conditions: granular filter Condition 1: The size of the voids in the filter material should be small enough to hold the larger particles of the protected material in place Condition 2: The filter material should have a high hydraulic conductivity to prevent buildup of large seepage forces and hydrostatic pressures in the filters.

145 Granular Filter Design Condition 1: Condition 2: D D 15( filter) 85( soil ) 4 to 5 D D 15( filter) 15( soil ) 4 to 5 after Terzaghi & Peck (1967) D D 15( filter) 85( soil ) < 5 D D 15( filter ) 15( soil ) > 4 D D 50( filter) 50( soil ) < 25 after US Navy (1971) D15( filter) < 20 GSD D15( Curves soil ) for the soil and filter must be parallel

146 Granular Filter Design D D 15( filter) 85( soil ) 4 to 5 D D 15( filter) 15( soil ) 4 to 5 GSD of Soil GSD Curves for the soil and filter must be parallel

147 EXAMPLE A stiff clay layer underlies a 12 m thick silty sand deposit. A sheet pile is driven into the sand to a depth of 7 m, and the upstream and downstream water levels are as shown in the figure. Permeability of the silty sand is cm/s. The stiff clay can be assumed to be impervious. The void ratio of the silty sand is 0.72 and the specific gravity of the grains is (a) Estimate the seepage beneath the sheet pile in m 3 /day per meter. (b) What is the pore water pressure at the tip of the sheet pile? (c) Is the arrangement safe against piping? ١٤٧

148 EXAMPLE ١٤٨

149 Solution ١٤٩

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