Soil Mechanics 1: Soil Physical Properties PDF

Chapter II: SOIL PHYSICAL PROPERTIES Intended Learning Outcome:  Understand the fundamental concepts of soil physical properties.  Explain the importance of soil phase relationships in geotechnical engineering.  Calculate mass‐volume and weight‐volume relationships for differe...

Chapter II: SOIL PHYSICAL PROPERTIES Intended Learning Outcome:  Understand the fundamental concepts of soil physical properties.  Explain the importance of soil phase relationships in geotechnical engineering.  Calculate mass‐volume and weight‐volume relationships for different soil samples. Chapter Overview: The Significance of Soil: Nature's Foundation Soil is often overlooked but is, in fact, the very foundation upon which our natural world and human civilization are built. It's not just the stuff beneath our feet; it's a dynamic and intricate substance that supports life, ecosystems, and human infrastructure. Without a deep understanding of soil, it's impossible to fully appreciate the intricate web of interactions it has with our environment and the critical role it plays in various fields, particularly geotechnical engineering. Soil in Geotechnical Engineering: Building on a Solid Ground Geotechnical engineering is the branch of civil engineering that focuses on the behavior of soil and rock in the context of construction and infrastructure projects. Soil is the canvas on which engineers sketch the blueprints of our roads, bridges, buildings, and more. Its properties and behaviors directly impact the safety, stability, and longevity of these structures. Therefore, to be effective geotechnical engineers, we must become soil detectives, deciphering the hidden clues within its physical properties. Discussion: II.1 Soil Phase Relationships: Unveiling the Secrets of Composition Soil isn't just a homogenous material; it's composed of solid particles, water, and air. Understanding how these phases interact and influence each other is like peering into the soul of the soil. It allows engineers to predict how soil will behave under different conditions – whether it will expand or shrink, become stable or erode. This knowledge is fundamental for building safe and resilient structures. Think of soil like a three‐part system. Imagine it's like a sandwich with layers. First, you have the solid stuff, like tiny pieces of rock and dirt. Then, there's the liquid, which is usually water, but sometimes it might have a bit of salt mixed in. And finally, there's the gas, which is just the air around the soil. We can talk about these parts using measurements like how much stuff there is (mass) or how much space they take up (volume). It's kind of like describing how thick the layers of your sandwich are. Understanding how these parts work together is important because it helps us figure out what the soil is like and what we can do with it. Mass‐Volume Relationships: Quantifying Soil Density Knowing how much soil weighs and how densely it's packed is vital. Mass‐volume relationships enable us to calculate parameters like porosity and void ratio, which, in turn, determine how much water soil can hold or how easily it can be compacted. In geotechnical engineering, these calculations are the building blocks of designing foundations, assessing slope stability, and mitigating the risk of landslides. From the where: figure shown: VT  total volume VT  Vv  Vs Vs  volume of solids Vv  Vw  Va Vw  volume of water m T  m w  ms Va  volume of air m T  total mass m w  mass of water m a  mass of airሺzeroሻ Weight of Water, Ww It is the unit weight of water multiplied by the volume of water. Ww   wVw ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ eq. 1 Where;  w = unit weight of water, 9.81 kN/m3 or 9810 N/m3 or 62.4 lb/ft3 Vw = volume of water Mass of Water, mw It is the mass density of water multiplied by the volume of water. mw   wVw ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ eq.2 Where;  w = unit weight of water, 1000 kg/m3 or 1.0 g/cm3 or 1.938 slugs/ft3 Vw = volume of water Unit weight of solids,  s It is the specific gravity of solids multiplied by the unit weight of water.  s  Gs  w ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ eq. 3 Where; Gs = specific gravity of solids  w = unit weight of water, 9.81 kN/m3 or 9810 N/m3 or 62.4 lb/ft3 Weight of solids, Ws It is the specific gravity of solids multiplied by the unit weight of water. Ws   sVs ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ eq. 4 Combine equation 3 and 4. Ws  Gs wVs ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ eq. 5 Void Ratio, e It is the ratio of the volume of voids to the volume of solids. Vv e Vs ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ eq. 6 Porosity, e It is the ratio of the volume of voids to the total volume. Vv n VT ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ eq. 7 Water or Moisture Content, ω It is the ratio of the weight of the water to the weight of solids. Ww  Ws ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ eq. 8 Degree of Saturation, S It is the ratio of the volume of water to the volume of voids. Vw S Vv ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ eq. 9 Relationship between e and n Vv Vv e n Vs VT and Let’s combine the two formulas so we left e and n in our equation. Derivation; Vv  eVs VT  Vv  Vs eVs n Vv  Vs eVs n eVs  Vs  Cancel Vs eVs n Vs 1  e  e n 1 e ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ eq. 10 Relationship between Gs ,  , e,S Let’s combine the equation 1, 5, 6, 8 and 9. Derivation; Vv Ww Vw Ww   wVw Ws  Gs wVs e  S Vs Ws Vv Combine eq. 6 and 9. Vv  eVs + Vw  SVv = Vw  SeVs Then substitute equation 1. Ww   wVw + Vw  SeVs = Ww   w SeVs Then substitute equation 8. Ww   w SeVs + Ww  Ws = Ws   w SeVs Then substitute equation 5. Ws   w SeVs + Ws  Gs wVs = Gs wVs   w SeVs Then simplify, cancel  w and Vs ; Gs wVs   w SeVs  Gs  Se Se  Gs ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ eq. 11 Wet Unit Weight (Bulk Unit Weight) of Soil,  wet It is the ratio of the total weight of the soil to the total volume of the soil. Derivation; Vv Ww WT  Ws  Ww VT  Vs  Vv e  Vs Ws Ws  W w  wet  Vs  Vv Then substitute eq. 6 and 8. Ws  Ws  wet  Vs  eVs W 1     wet  s Vs 1  e  Then substitute equation 5. Ws 1     wet  Vs 1  e  Gs  wVs 1    + Ws  Gs  wVs =  wet  Vs 1  e  Then simplify, cancel Vs ; G s  w 1     wet  1 e ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ eq. 12 Dry Unit Weight of Soil,  dry It is the ratio of the dry weight of the soil to the total volume of the soil. Where  = 0 and S = 0 Derivation; VT  Vs  Vv V e v Vs Ws  dry  V s  Vv Then substitute eq. 6. Then substitute equation 5. Ws Ws  dry   wet  Vs  eVs Vs 1  e  + Ws  Gs  wVs Ws  dry  Vs 1  e   wet  G s  wVs Vs 1  e  Then simplify, cancel Vs ; Gs  w  wet  1 e ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ eq. 13 Saturated Unit Weight,  sat Saturated unit weight occurs when the volume of voids is fully occupied by water. Where S =100% Derivation; Gs  w 1     wet  1 e  w G s  G s    wet  1 e  w Gs  Se  wet  1 e  Where G s   Se For saturated soil, S = 1  w G s  e   sat  1 e ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ eq. 14  Submerge or Buoyant Unit Weight,  sub If the soil is submerged, the density of the submerged soil  sub , is equal to the density of the saturated soil reduced by the density of water.  sub   sat   w  w G s  1  sat  1 e ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ eq. 15 Percent air voids, na Percent air voids is defined as the ratio of the volume of air voids to the total volume of the soil mass and is expressed as percentage. Va na   100 VT ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ eq. 16 Air Content, ac The air content is defined as the ratio of volume of air voids to the volume of voids. Va ac  1S Vv ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ eq. 17 Examples: Example 2.1 A 0.0047 m3 of soil sample taken from the site weighs 112.8 N. After oven‐drying, it weighed 101.9 N. If the specific gravity of soil solids is 2.96, determine the following: a) Moist unit weight b) Dry unit weight c) Water content d) Void ratio e) Porosity f) Degree of saturation g) Volume occupied by water Given: WT  112.80 N W s  101.90 N G s  2.96 Solution: a) Moist unit weight W  wet  T VT 112.80 N  wet  0.0047 m 3  wet  24,000 N / m 3 or 24.00kN / m 3  b) Dry unit weight e) Porosity W e  dry  s n VT 1 e 101.90 N 0.339  dry  n 0.0047 m 3 1  0.339  dry  21,680.85 N / m 3 or 21.68kN / m 3 n  0.253 c) Water Content G  (1   ) f) Degree of Saturation  wet  s w Se  G s  1 e 2.960.1070 24.0kN / m 3    2.96 9.81kN / m 3 (1   ) S 0.339 1  0.339 S  93.43%   10.70% d) Void ratio g) Volume occupied by water. Gs w WT  W s  W w  dry  112.80 N  101.90 N  W w 1 e 21.68kN / m 3   2.96 9.81kN / m 3  W w  10.90 N 1 e W w   wV w e  0.339 10.90 N  9810 N / m 3V w V w  0.0011 m 3 Example 2.2 The moist unit weight of a soil is 16.5 kN/m3. If the water content is 15% and sp.gr. of 2.7, determine the following: a) Dry unit weight b) Saturated unit weight c) Submerge unit weight Given;  wet  16.50kN / m 3   15.0% G s  2.70 Solution: a) Dry unit weight  wet   dry (1   ) 16.50kN / m 3   dry (1  0.15)  dry  14.35kN / m 3 b) Saturated unit weight Se  G s  1(e)  2.70(0.15) e  0.41  G  e   sat  w s 1 e 9.81kN / m 3 2.70  0.41  sat  1  0.41  sat  21.64kN / m 3  c) Submerge Unit Weight.  sub   sat   wet  sub  21.64kN / m 3  9.81kN / m 3  sub  11.83kN / m 3 Example 2.3 Dry sand is placed in a container having a volume of 0.34 ft3. The dry weight of the sample is 42 lb. A volume of 0.056 ft3 of water is carefully added to the container so as not to disturb the condition of the sand. If the specific gravity of the sand is 2.89, determine the: a) Void ratio b) Degree of saturation c) Unit weight at zero air voids Given; VT  0.34 ft 3 W s  42 lb V w  0.056 ft 3 G s  2.89 Solution: a) Void ratio Vv W s  G s  wV s e Vs 0.107 ft 3 42.0 lb  2.89(62.4 lb/ft 3 )V s e 0.233 ft 3 V s  0.233 ft 3 e  0.46 VT  Vs  Vv 0.34 ft 3  0.233 ft 3  V v V v  0.107 ft 3 b) Degree of Saturation V S w Vv 0.056 ft 3 S 0.107 ft 3 S  52.34% c) Unit weight at zero air voids Volume of void is fully occupied by the volume of water (fully saturated). S 1 W  w Ws 62.4lb / ft 3 (0.056 ft 3 )  42lb   8.32%  Gs  w  dry  1 Gs 2.89(62.4lb / ft 3 )  dry  1  2.89(0.08)  dry  146.47lb / ft 3 Example 2.4 A soil sample in its natural state has, when fully saturated, a water content of 32.5%. Determine the void ratio, dry and total unit weights. Calculate the total weight of water required to saturate a soil mass of volume 10 m3. Assume Gs = 2.69. a) Void ratio b) Total unit weight of the soil sample c) Dry unit weight of the soil sample d) Weight of water required to saturate a soil mass of 10 m3. Given; VT  0.34 ft 3 W s  42 lb V w  0.056 ft 3 G s  2.89 Solution: c) Dry unit weight a) Void ratio  wet   dry (1   ) Fully saturated state (S=1) 18.66kN / m 3   dry (1  0.325) G s   eS  dry  14.08kN / m 3 2.69(0.325)  e(1) d) Weight of water required to saturate the sample e  0.874 WT   wet VT b) Total unit weight WT  18.66kN / m 3 (10m 3 )  G 1     wet  w s W s   dry VT 1 e 3 9.81kN / m ( 2.69)(1  0.325)  wet  WT  18.66kN / m 3 (10m 3 ) 1  0.874  wet  18.66kN / m 3 W w  WT  W s   W w  18.66kN / m 3  14.08kN / m 3 (10m 3 ) W w  45.80kN Practice Problems: Problem 2.1 The in situ‐density of an embankment, compacted at a water content of 12% was determined with the help of core cutter. The empty mass of the cutter was 1286 grams and the cutter full of soil has a mass of 3195 grams. The volume of the cutter being 1000 cm3. Determine the following: a) bulk density b) dry density c) degree of saturation d) Void ratio If the embankment becomes fully saturated during rains, what would be its water content and saturated unit weight? Assume no volume changes in soil on saturation. Take specific gravity of soil to be 2.70. Ans: (a)18.73 kN/m3 (b) 2.72 kN/m3 (c)55.5 % (d) 20.34 kN/m3 Problem 2.2 The mass specific gravity (apparent specific gravity) of a soil equals 1.64. The specific gravity of solids is 2.70. Determine the void ratio under the assumption that the soil is perfectly dry. What would be the void ratio if the sample is assumed to have a water content of 8%. Ans: 0.646, 0.78 Problem 2.3 A natural soil deposit has a bulk unit weight of 18.44 kN/m3 and water content of 5%. Calculate the amount of water required to be added to 1 cubic meter of soil to raise the water content to 15%. Assume the void ratio to remain constant. What will then be the degree of saturation? Take specific gravity of soil to be 2.67. Ans: 179 liters, 81.7% Problem 2.4 A soil has a bulk unit weight of 20.11 kN/m3 and water content of 15%. Calculate the water content if the soil partially dries to a unit weight of 19.42 kN/m3 and the void ratio remains unchanged. Ans: 17.49 kN/m3, 11% Problem 2.5 It is required to prepare a compacted cylindrical specimen of 40 mm diameter and 80 mm length from oven dry soil. The specimen is required to have water content of 10% and percent air voids of 18%. Taking Gs = 2.70, determine the mass of soil and mass of water, required for the preparation of the above specimen. Ans: 155.4 grams, 24.9 grams  PROBLEM SET 1) The moist weight of 0.0068 m3 of a soil is 112.4 N. The moisture content and the specific gravity of soil solids are determined in the laboratory to be 12% and 2.5 respectively. Calculate the following: a) Moist unit weight b) Dry unit weight c) Void ratio d) Porosity e) Degree of saturation Ans. a) 16,529.41 N/m3 , b) 14,758.40 N/m3 , c) 0.66 , d) 45% 2) A sample of saturated clay was placed in a container and weighed. The weight was 6 N. The clay in its container was placed in an oven for 24 hours at 105°C. The weight was reduced to a constant value of 5 N. The weight of the container is 1 N. G = 2.7. Determine the following: a) Water content of the soil specimen b) Void ratio in percent c) Effective unit weight of the soil in kN/m3 Ans. a) 25% , b) 67.5 , c) 9.96 kN/m3 3) A soil has bulk density of 20.1 kN/m3 and water content of 15%. Calculate the water content if the soil partially dries to a density of 19.4 kN/m3 and the void ratio remains unchanged. Ans. 10.86% 4) The specimen of sand has a porosity of 46%, and the specific gravity of its solids is 2.73. Compute the specific weight of this soil in the submerge state, in lb/ft3. Ans. 58.35 5) A hand carved sample of soft saturated clay weighs 420 grams and has a volume of 250 cc. After oven‐drying, it weighs 315 grams. Calculate the following: a) Moisture content in percent. b) Specific gravity of solids. c) Porosity in percent. Ans. a) 33.33% , b) 2.17 , c) 42% 6) The moist weight of 0.20 ft3 of a soil is 24 lb. The moisture content and the specific gravity of solids are determined in the laboratory to be 11% and 2.7, respectively. a) What is the moist unit weight in lb/ft3? b) What is the dry unit weight in lb/ft3? c) What is the degree of saturation in percent? Ans. a) 120 , b) 108 , c) 53% 7) The mass of a sample of saturated soil is 540 grams. The dry mass, after oven‐drying is 405 grams. Assuming Gs = 2.7, calculate the effective unit weight of the soil mass, in kN/m3? Ans. 8.78 8) A fully saturated clay sample has a mass of 1.526 grams. After oven drying, its mass was reduced to 1.053 grams. The specific gravity of soil particles is 2.7. a) Calculate the natural water content of the sample in percent. b) Calculate the void ratio in percent. c) Calculate the porosity in percent. Ans. a) 44.92% , b) 121.3 , c) 55% 9) The mass of a moist soil sample collected from the field is 465 g. and its oven dry mass is 405.76 g. The specific gravity of the soil solids was determined in the laboratory to be 2.68. lf the void ratio of the soil in the natural state is 0.83, find the following: a) The moist density of the soil in the field (kg/m3) b) The dry density of the soil in the field (kg/m') c) The mass of water, in kilograms to be added per cubic meter of soil in the field for saturation. Ans. a) 1,678.30 kg/m3 , b) 1,459.02 kg/m3 , c) 66.6 g 10) For a saturated soil, show that  w 1     sat  n  11) For a saturated soil, the following are given: ω = 2I% and G =2.67. Determine the following: a) Saturated unit weight b) Dry unit weight c) Moist unit weight when the degree of saturation becomes 70%. Ans. a) 20.31 kN/m3 , b) 16.79 kN/m3 , c) 19.25 kN/m3 12) For a silty clay soil, the moist unit weight for 50% degree of saturation = 20.03 kN/m3 and moist unit weight for 80% degree of saturation = 20.93 kN/m3. Determine a) The void ratio of the soil b) The moisture content if the degree of saturation is 40% Ans. a) 0.43, b) 19.7 kN/m3 13) For a moist soil, given volume = 0.25 m3, mass = 472 kg, moisture content = 16%, G = 2.7. Determine the following; a) Moist density (kg/m3) b) Dry density c) Void ratio d) Porosity e) Degree of saturation Ans. a) 1,888 kg/m3 , b) 1,627.59 kg/m3 , c) 0.66, d) 0.398, e) 65% 14) The volume of the soil sample measured before drying is 102 cm3. The wet weight of soil is 189 grams and the dried weight of the soil sample is 156 g. If the sp.gr. of the soil is 2.65, determine the following: a) void ratio b) degree of saturation c) dry unit weight at zero air voids Ans. a) 0.73, b) 76.8%, c) 1.698 g/cm3 15) The effective unit weight of soil is 10.68 kN/m3 and the void ratio is 0.58. a) Compute the specific gravity of soil. b) If the water content of the soil is 14%, compute for the degree of saturation. c) Compute the bulk unit weight of soil. Ans. a) 2.72, b) 0.66, c) 19.25 kN/m3 16) The volume of the soil sample before drying is 106 cm3 and the volume of the dried soil is 68 cm3. Specific gravity of soil is 2.6. Compute the following: a) Void ratio b) Porosity c) Degree of saturation if it has a moisture content of 17%. Ans. a) 0.56, b) 0.36, c) 78.9% 17) A sample of wet soil was extruded from a sampling tube of diameter 100mm in a soil testing laboratory. The length of extruded sample was 200mm. The mass of the wet soil was 3.15kg. Following a water content determination, the mass of the dry soil was found to be 2.82kg. Determine the following; a) Bulk density b) Water content c) Dry density Ans. a) 2.0 Mg/m3, b) 011.70%, c) 1.80 Mg/m3 18) A natural soil deposit has a bulk unit weight of 18.44 kN/m3 and water content of 5%. Calculate the amount of water in liters required to be added to 1 cubic meter of soil to raise the water content to 15%. Assume the voids to remain constant. Determine also the degree of saturation after the water is added. Assume specific gravity of solid is 2.67. Ans. 179 liters, 81.70 % 19) Calculate the saturated unit weight of a soil composed of 60% quartz, 25% mica and 15% iron oxide. Assume that both soils are saturated and have void ratio of 0.63. Take average specific gravity for quartz 2.66, for mica 3.0 and for iron oxide 3.8. Ans. 21.36 kN/m3 20) A soil has a bulk unit weight of 20.11 kN/m3 and water content of 15%. Calculate the water content if the soil partially dries to a unit weight of 19.42 kN/m3 and the void ratio remains unchanged. Ans. 11% 21) A soil is composed of 60% quartz, 25% mica and 15% iron oxide. Assuming that both compositions are saturated and have a void ratio of 0.63. Take the specific gravity for quartz = 2.66, mica = 3.0 and iron oxide = 3.8. Determine the saturated unit weight of the given composition of the soil in kN/m3. Ans. 21.34 22) An embankment for a highway of 30 m wide and 1.5 m in compacted thickness is to be constructed from a borrow pit. The water content of the sandy soil in the borrow pit is 15% and its void ratio is 0.69. The specification requires in the embankment is to be compacted to a dry unit weight of 18 kN/m3, specific gravity of soil is 2.70. For a 1 km length of embankment, compute the following. a) Volume of borrow pit soil required to construct the embankment b) The weight of water per truckload of sandy soil, if the capacity of 1 truckload is 10 m3 c) Degree of saturation Ans. 51,682, 23.51, 58.70% 23) The dry unit weight of a silty sand is 16.43 kN/m3. It has a specific gravity of 2.65. The silty sand has a void ratio in the loosest condition equal to 0.88 and a void ratio of 0.36 in densest condition. Determine the relative density of the sand, the porosity and the degree of saturation. Ans. 57.26%, 0.37, 72.82% 24) The sample of soil weighing 30.6 kg had a volume of 0.0183 m³. The sample is allowed to dry in an oven and the weight was reduced to 27.2 kg. The specific gravity of the solids was found to be 2.65. Determine the moist unit weight of the soil, the dry unit weight and moisture content of the soil. Ans. 16.4 kN/m3, 14.6 kN/m3, 0.125 25) A fully saturated clay sample has a mass of 1526 grams. After oven drying, its mass was reduced to 1053 grams. The specific gravity soil particles is 2.70. Calculate the natural water content, void ratio, and porosity. Ans. 44.92%, 121.3%, 54.81% References:  1. Geotechnical Engineering Principles and Practices of Soil Mechanics and Foundation Engineering V.N.S. Murthy, CRC Press, 2002  2. Basic Fundamentals of Geotechnical Engineering 5th Edition, Besavilla  3. Fundamentals of Geotechnical Engineering, DIT Gillesania  4. Principles of Geotechnical Engineering, 9th Edition, Braja M. Das, Khaled Sobhan, CENGAGE Learning, 2018  5. Introduction to Geotechnical Engineering, 2nd Edition, Braja M. Das, Nagaratnam Sivakugan, CENGAGE Learning, 2016