Department of Civil & Geological Engineering GEOE 218.3 Engineering Geology Assignment #3, Head, Pore Pressure & Effective Stress Due 08 Oct, 2010 NOTE: Numbered subscripts indicate depth, in metres, below liquid surface 1. (5 marks) Consider two tanks Tank#1 has dimensions 1m x 1m (LxW) and 4 m tall. Tank #2 is 2m x 2m (LxW) and 1 m tall. Each tank is completely full of water. Assume the mass of each tank is very small compared to the mass of water in the tank. a) what is the mass (in T) and weight (in kn) of water in each tank? Tank 1: M = 4 T W = 39.2 kn Tank 2: M = 4 T W = 39.2 kn b) what is the total force (kn) exerted on the bottom of the tank? Tank 1: F = W = 39.2 kn Tank 2: F = W = 39.2 kn c) Remembering that pressure (or stress) is Force per unit area, what is the pressure on the bottom of each tank in kpa (1 Pa = 1 N/m2) Tank 1: P 4 = 39.2 kpa Tank 2: P 1 = 9.8 kpa d) A simple definition of a fluid (i.e. a liquid or a gas) is that it cannot support any shearing stress. What that means, as you will be discovering over the next short while, is that the pressure (or stress) is the same in all directions. Knowing this, what is the pressure on the side wall of each tank at exactly 1 m below the surface of the water?
Tank 1: Tank 2: P 1 = 9.8 kpa P 1 = 9.8 kpa e) For Tank #1, what is the pressure on the sidewall of the tank at depths 2, 3 and 4 m below the surface of the water? i) P 2 = 19.6 kpa ii) P 3 = 29.4 kpa iii) P 4 = 39.2 kpa 2. (5 marks) Redo Q#1 assuming both tanks are full of oil with a specific gravity of 0.8. a) Tank 1: M = 3.2 T W = 31.4 kn Tank 2: M = 3.2 T W = 31.4 kn b) Tank 1: F = W = 31.4 kn Tank 2: F = W = 31.4 kn c) Tank 1: P 4 = 31.4 kpa Tank 2: P 1 = 7.8 kpa d) Tank 1: P 1 = 7.8 kpa Tank 2: P 1 = 7.8 kpa e) i) P 2 = 15.7 kpa ii) P 3 = 23.5 kpa iii) P 4 = 31.4 kpa 3. (2 marks) For the tanks in Q#1 (full of water), what would be the water pressure at 0 and 1 m below the top of each tank if a pressure gauge connected on the top of each tank read 50 kpa? Consider the points mid-height on the vertical walls (Point A1 and A2). What is the horizontal water pressure exerted on the sidewall under this condition? i) Tank 1: P 0 = 50 kpa P 1 = 59.8 kpa Tank 2: P 0 = 50 kpa P 1 = 59.8 kpa ii) Tank 1: P 2 = 69.6 kpa Tank 2: P 0.5 = 54.9 kpa
4. (4 marks) Consider two cylindrical pipes, each 0.4 m in diameter and 6 m long and completely full of water. Pipe A is supported vertically and pipe B is lying on its side. What is the pressure at the centre of the circular ends of each pipe? What is the pressure at the midpoint (3 m from each end) of Pipe A? For Pipe B, describe how the pressure at this midpoint (3 m from each end) would change from the top (up) direction to the bottom (down) direction. i) Pipe A: P 0 = 0 kpa P 6 = 58.8 kpa Pipe B: P 0.2 = 2.0 kpa ii) P 3 = 29.4 kpa iii) Would increase linearly from 0 kpa at the top to 3.9 kpa at the bottom. 5. (3 marks) A cylindrical pipe of diameter 0.4 m is supported as shown and is full of water. What is the pressure of water in the horizontal direction at the centre of the sidewalls of the pipe at points X (A), Y (B) & Z (C)? i) P A = 58.8 kpa ii) P B = 58.8 kpa iii) P C = 96.0 kpa
6. (6 marks) Consider Tank #1 of Question #1. a) If all the water was removed and the tank was instead filled with dry sand with a (dry) density of 1.8 T/m3 what would be the mass of material in the tank? What would be the downward vertical pressure applied by the sand on the bottom of the tank? i) M = 7200 kg ii) P S4 = 70.6 kpa b) If water were slowly and carefully added to the tank full of sand (which initially had a dry density = 1.8 T/m3) until all the pores or voids in the sand which had initially been filled with air become filled with water, what would be the water pressure at depths 1, 2, 3 and 4 metres below the top of the tank? i) P W1 = 9.8 kpa ii) P W2 = 19.6 kpa iii) P W3 = 29.4 kpa iv) P W4 = 39.2 kpa 7. (6 marks) A cube-shaped box with dimensions exactly 1 m is filled with uniform sand. The specific gravity of the sand mineral grains is 2.66. What mass of perfectly dry sand would have been added to the box if the porosity of the sand in the box ended up being exactly i) 30% ii) 25% iii) 38%. What would be the dry density and bulk density of the dry sand in each case? What would be the total downward vertical pressure exerted on the bottom of the box? i) M = 1862 kg ρ D = ρ B = 1862 kg/m 3 P 1 = 18.2 kpa ii) M = 1995 kg ρ D = ρ B = 1995 kg/m 3 P 1 = 19.6 kpa iii) M = 1649 kg ρ D = ρ B = 1649 kg/m 3 P 1 = 16.2 kpa
8. (5 marks) For the box of initially-dry sand of problem 6 7; a) what mass of water would be required to exactly saturate the sand in each of cases i), ii) and iii)? i) M W = 300 kg ii) M W = 250 kg iii) M W = 380 kg b) What would be the resulting dry density and bulk density in each case? i) ρ D = 1862 kg/m 3 ρ B = 2162 kg/m 3 ii) ρ D = 1995 kg/m 3 ρ B = 2245 kg/m 3 iii) ρ D = 1649 kg/m 3 ρ B = 2029 kg/m 3 c) What would be the total mass and weight of the sand and water in the box fir each case? i) M T = 2162 kg W = 21.2 kn ii) M T = 2245 kg iii) M T = 2029 kg W = 22.0 kn W = 19.9 kn d) What would be the resulting total pressure on the bottom of the box? i) P = 21.2 kpa ii) P = 22.0 kpa iii) P = 19.9 kpa e) What would be the water pressure applied on the bottom of the box? P W = 9.8 kpa for all cases
9. (3 marks) Consider a perfectly spherical rock with no voids or fractures, a volume of 0.1 m 3 and a specific gravity of 2.65. a) What is the weight of this rock i.e. what force is required to suspend this rock in air? W A = 2.60 kn b) What force would be required to suspend this rock in water (without it touching thebottom)? Does the answer depend on the depth of water or the depth you have submerged the rock? i) W W = 1.62 kn ii) No. c) what force would be required to suspend the rock in a very large bowl of porridge with a density of 1.2 T/m3? W P = 1.42 kn 10. (4 marks) Return to the box of sand of questions 6 7, but now let s make the box 2 metres high while remaining 1m x 1m in the horizontal plane. Let s assume we filled it with dry sand with a porosity of 30%. Consider a free body diagram of the upper metre of dry sand and ignore friction on the walls of the box. What is the vertical stress applied upward at 1 metre depth? If the sand becomes saturated with water, what is the vertical total stress applied across this imaginary horizontal plane at depth 1m? What is the water pressure acting across this same surface? What is the difference between the total stress and the water pressure? What do you think is the mechanism that supports that portion of the stress? Assume: G S = 2.66 i) Dry: σ 1 = 18.2 kpa ii) Saturated: σ 1 = 21.1 kpa iii) u = 9.8 kpa iv) σ = 11.3 kpa v) Buoyancy.
11. (4 marks) The definition of the water table is that it is an imaginary horizontal plane below the water table the ground is saturated, and above it, the soil or rock is perfectly dry (that s a simplification because of capillary effects but let s go with that for now). a) What is the pore water pressure at the exact elevation of the water table? P 0 = 0 kpa b) What is the pore water pressure 1, 2 or 3 metres below the elevation of the water table? Does the answer depend on the depth that the water table is below the ground surface? i) P 1 = 9.8 kpa ii) P 2 = 19.6 kpa iii) P 3 = 29.4 kpa iv) No. 12. (3 marks) What is the total stress, pore water pressure and effective stress 5 metres below the ground: a) if the soil is saturated all the way to surface and the saturated bulk density is 2 T/m3? i) σ 5 = 98.0 kpa ii) u 5 = 49.0 kpa iii) σ 5 = 49.0 kpa b) if the water table is located 1 m below the ground surface? and If ρ B = 2000 kg/m 3, ρ D = 1600 kg/m 3 Assume: - Soil is perfectly dry above the water table i) σ 5 = 94.1 kpa ii) u 5 = 39.2 kpa iii) σ 5 = 54.9 kpa
c) if the water table is located 3 m below the ground surface? i) σ 5 = 86.2 kpa ii) u 5 = 19.6 kpa iii) σ 5 = 66.6 kpa 13. (4 marks) Consider the sketch below. The lake is 10 metres deep, followed by a clayey formation 20 metres thick with a bulk density of 1.9 T/m3, which is in turn underlain by a sand aquifer 5 m thick. (a permeable water- bearing formation into which one might drill a well is called an aquifer). Assume the water level in the sand aquifer is the same as the water level in the lake. In other words, if you stuck a pipe into the aquifer (but it was sealed and watertight above that) the water level inside the pipe (we call it a standpipe or a standpipe piezometer or sometimes just a well would rise up to the same elevation as the water level in the lake. That s what the little triangle symbol represents. a) What would be the total stress at the top, midpoint and bottom of the clayey layer (a low permeability formation is what we call an aquitard ) which overlies the aquifer. i) σ top = 98.0 kpa ii) σ mid = 284 kpa iii) σ bot = 470 kpa
b) If the water level in the aquifer (i.e. the elevation of the triangle symbol on the standpipe) were lower by 5 metres of elevation, what would happen to the pore pressure at the midpoint of the clay layer, would it increase, decrease or stay the same? Can you estimate how much it would change? What would be the effective stress at the midpoint of the clay layer under this condition? i) Decrease ii) Δu = -24.5 kpa iii) σ = 113 kpa c) If the water level in the standpipe is 6 metres higher than the elevation of the water level of the lake, what would be to the total stress, pore pressure and effective stress at the midpoint of the clay layer? i) σ = 284 kpa ii) u = 225 kpa iii) σ = 59 kpa d) Calculate the height of water in the standpipe (we call that the hydraulic head in the aquifer) that would be required in order for the effective stress at the bottom of the clay layer to be equal to 0. What would be the resulting pore pressure at the top of the aquifer? Can you express this in terms of the following definition? Hydraulic head = elevation + pore pressure/unit weight of water h =z+u/γ W Where: γ W = ρ W *g = 1.0 T/m 3 * 9.8 kn/t = 9.8 kn/m 3 i) h = 48.0 m ii) u = 470 kpa iii) Datum at top of aquifer: z = 0 m u/γ W = 48.0m h = 0 m + 48.0 m
14. (2 marks) Consider Stop D Artesian Borrow Pit in the Guide Book you were given for the GeoE218 Field Trip. Evaluate the information presented (in particular the head measurements and simplified schematic of the 2nd last page of this section) and determine approximately what the head in the Fielding Aquifer would have been in order to cause blowout. According to the water levels on Page 18, what was the Factor of Safety against blowout in 1997/98? Remember, F.S. = Forces restoring / Forces causing failure Assuming saturated unit weight of till is 17.5 kn/m 3 Given: h in aquifer = 531.9 m Elevation of top of aquifer = 517 m Therefore: u = 146.02 kpa Before Excavation: Ground elevation = 531 m Therefore: Total stress overlying aquifer, σ = 245 kpa F.S. = 1.68 After Excavation: Ground Elevation = 526.8 m Therefore: Total stress overlying aquifer, σ = 171.5 kpa F.S. = 1.17 1.17 > 1.00 but failure STILL occurred. This is because of uncertainty, and that is why we use a factor of safety. To calculate what the head in the aquifer really was we set the factor of safety, after excavation, to F.S. = 1.00: Therefore: u = σ = 171.5 kpa h = 517 m + 171.5 kpa/9.8 kn/m 3 h = 534.5 m