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1 Civil Engineering Department College of Engineering Course: Soil Mechanics (CE 359) Lecturer: Dr. Frederick Owusu-Nimo
2 FREQUENCY CE 260 Results (2013) Civil Geological Below MARKS
3 CE 260 Results Civil Engineering Geological Engineering
4 CE 359 Course Outline and Schedule
5 Fluid flow through soils (Seepage)
6 Dam failures due to Seepage Idaho, US(June 5, 1976) For irrigation and hydroelectric power Water eroded subsoil beneath dam 14 people died $1 billion in property damage TETON DAM FAILURE
7 Dam failures due to Seepage Georgia, US(Nov 6, 1977) For hydroelectric power Piping and progressive erosion 39 people died KELLY BARNES DAM $2.5 million in property damage
8 Dam Failures
9 Fluid flow through soils Objective - Introduction to the flow of water (seepage) through soils Significance - Important to understand the principles of fluid flow to be able to: o Determine the rate at which water flows through soil (e.g. seepage through earth dam) It is estimated that more people have lost their lives as a result of failures of dams and levees due to seepage and piping than to all the other failures of civil engineering works combined o Determine rate of foundation settlement o Evaluating the factor of safety of embankments
10 Introductory Concepts Review of CE 260 and CE 256 o Composition of soil? o Properties of fluid or fluid parameters? o Flow through channels? For 1 D flow o Fluid parameters such as pressure, velocity, temperature are constant in any cross section perpendicular to the direction of flow o Fluid parameters can vary from section to section along direction of flow Flow path in a soil medium
11 Darcy s Law Study of flow properties of water through clean sands
12 Darcy s law Q h L A Q = k h L i = L A k = permeability = ydraulic gradient Hydraulic gradient is the total head difference between two points divided by the distance between the points Limitations of Darcy s Law: Generally not valid for o Liquid flow at very high velocities o Gas flow at very low or very high velocities
13 Darcy s Law In terms of soil or particle size, Darcy's law is valid for silt through medium sand ( mm)
14 Flow Velocity From Darcy Q = kia v = ki Q = va For same flow rate, v will change if A changes Velocity of flow from 1-2 is Approach velocity Velocity from 2-3 is Seepage velocity Which of these velocities is the actual velocity of flow through a soil medium?
15 Flow Velocity Seepage velocity is the true or actual velocity of water flowing through the pores of a soil Approach/ Discharge velocity is the fictitious velocity of water flowing through a unit cross sectional area of a soil medium (Area of solids + voids) Approach velocity is however a statistically convenient engineering velocity and is often used Which velocity is used in the Darcy s law? Seepage and Approach velocity which is greater? Prove
16 Hydraulic Head How do we determine total energy at any point in a system with water flowing through it? Total Energy = v2 + P + gz per unit mass at a point 2 ρ w where v = velocity of fluid flow at the point P = pressure at the point ρ w = density of fluid g = acceleration due to gravity Z = elevation of the point above a reference point (Datum) What if total energy between two points are the same?
17 Hydraulic Head Normalizing energy equation by gravity P Total Energy = v2 + + z per unit weigt at a point 2g ρ w g Each term has unit of length so we refer to each of them in terms of head Total Head = v2 2g + P ρ w g + z Velocity Head, h v Pressure Head, h p Elevation Head, h e For flow through soils velocity is so small it is neglected Water will flow from one point to another when there is difference in total heads between the points
18 Determining Elevation Head Elevation head at any point is the vertical distance above or below a reference point (Datum) Determine the elevation heads at points A, B, C, D, E and F.
19 Determine Pressure Head Pressure head, p = P ρ w g Pressure of the water surface opened to the atmosphere is zero If surface is not opened to atmosphere? Pressure head increases with depth in water and vice versa Rate of change in pressure head changes in different medium Determine the pressure heads at Points A -F
20 Determine Pressure Head Piezometers give the pressure head at a point. Piezometer level is the height of water in the piezometer as measured from the point where the piezometer inlet is located
21 Determine Pressure Head Determine pressure heads at points B, C, D Piezometer level is the height of water in the piezometer as measured from the point where the piezometer inlet is located
22 Total Head Total head = Elevation head + Pressure head Point H e H p H t H l A B C D E 0??? F The heads vrs their corresponding points (elevation) can be plotted on a graph
23 Hydraulic head It is possible for the elevation head, h e as well as pressure head, h p to be negative depending on geometry of problem The important thing is that the total head must equal the sum of h p and h e at all times The direction of flow is determined by the head difference. Any head loss occurs in soils
24 Worked Examples Which of these set ups will have upward, downward of no flow
25 Worked Example Determine the elevation head, pressure head, total head and head loss at points A, B and C. Determine the direction flow.
26 Worked Example Two soil types with different permeability, k 1 = 4k 2. L 1 = 4cm, L 2 = 6cm Determine h e, h p, h t and h l at A, B, C, D, E Plot variation of heads for points A-E with horizontal distance
27 Effective Stress Effective stress principle o Effective stress, σ V = total stress, σ v pore water pressure, u Importance of effective stress - Controls the engineering properties of soils o Increasing effective stress increases soil compression leading to settlement o Changes in effective stress changes shear strength ( slope stability, bearing capacity, depends on shear strength of soils) o Changes in effective stress changes void ratio and thus to some extent permeability
28 Effective stresshydrostatic conditions Determine effective stress at soil surface A-A, B-B and C-C
29 Effective stresshydrostatic conditions If the water table is in soil 1 ( at D-D) as shown, determine effective stress at section C-C
30 Effective Stress Hydraulic conditions As water flows through a soil it exerts a force on the soil grains This force, known as seepage force or drag force, acts in the direction of flow in isotropic soils The pressure induced in the soil as a result of the force is termed seepage pressure Seepage pressures can be determined by considering the boundary water pressures on the top and bottom of soil
31 Calculating seepage pressure Hydrostatic pressures are the components which would occur if there were no flow Seepage pressures are the components responsible for flow of water Seepage pressures can be converted to forces and vice versa Upward flow
32 Effective Stress- fluid flow Determine effective stress at points A, B and C
33 Stresses due to Flow Static Situation (No flow) h w L z X At X, v = w h w + sat z soil u = w (h w + z) v ' = ' z
34 Downward Flow Stresses due to Flow At X, flow v = w h w + sat z as for static case h L h w u = w h w u = w h w + w (L-h L )(z/l) = w h w + w (z-iz) = w (h w +z) - w iz L z X soil Reduction due to flow v ' = ' z + w iz u = w (h w +L-h L ) Increase due to flow
35 Stresses due to Flow Upward Flow At X, v = w h w + sat z as for static case u = w h w + w (L+h L )(z/l) = w h w + w (z+iz) = w (h w +z) + w iz Increase due to flow v ' = ' z - w iz flow h L h w L z X soil u = w h w u = w (h w +L+h L ) Reduction due to flow
36 Quick Condition Seepage force acts in the direction of flow and can combine with soil weight to either improve stability or worsen it Effective stress represent interparticle forces. The higher the effective stress the higher the interparticle forces and hence strength For downward flow, effective stress increases as seepage force increases interparticle forces For upward flow, effective stress decreases as seepage force decreases interparticle forces Quick condition occurs when the shear strength of soil become zero due to the absence of effective stress Quick condition occurs in cohesionless soils
37 Quick Condition Prove that for the unloaded condition (i.e. a=0 and q s =0) with water level at the ground level, the quick condition is given by Where i c = γ γ w i c is te critical ydraulic gradient γ is te submerged unit weigt of soil i. e(γ sat γ w ) γ w is te unit weigt of water
38 Quick Condition The critical hydraulic gradient, i c is the hydraulic gradient at which effective stress becomes zero and quick conditions occur The quick condition is also known as boiling condition In practice, boiling conditions are restricted to soils in the silt and fine sand range In general, for quick conditions γ γ w and thus i c 1 Quick refers to a condition and not a material
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