Home Courses Articles Watch! Tools Downloads Request ----------------------------------------------------- Soil Mechanics > Physical Properties of Soil > Soil Phase Relationships > Friction Angle of Soils + Typical Values > Cohesion Intercept of Soils + Typical Values > Lateral Earth Pressure > At-Rest State > Rankine's Lateral Earth Pressure > Coulomb's Lateral Earth Pressure > Slope Stability > Effects of Water on Slope Stability > Infinite Slope Analysis > Circular Arc Failure of Slope Analysis > Critical Failure Surface > Geotechnical Laboratory and In-Situ Testing Methods > 101 How To Plan A Construction Site Soil Investigations Program > 101 How to Write a Geotechnical Report > Determination of Water Content > Specific Gravity of Soils > Advanced Soil Mechanics Topics > Soil Liquefaction: Factor of Safety, Calculations, and Simplified Procedures Foundation Analysis and Design > Settlement of Shallow Foundations > Settlement types, definitions, and general equation > Immediate settlement computations > Primary Consolidation > Secondary Compression > Bearing Capacity of Shallow Foundations > Terzaghi's Method > Meyerhof's Method > Brinch Hansen's Method > Bearing from SPT Number > Effect of Ground Water Table > Bearing Capacity of Piles (Deep Foundations) > Ultimate Tip Resistance of Piles > Shaft Resistance of Piles > Sheet-pile Walls: Cantilevered and Anchored > Cantilever Sheet Pile Walls Penetrating Sandy Soils > Anchored Sheet Pile Walls Penetrating Sandy Soils > Factors of Safety for Cantilevered Sheet Pile Walls Load Calculation > Structural Loading > Dead Load vs Live Load > Load Combinations Reinforced Concrete Design > General Topics of Concrete Material and Design > What Is Concrete? > Concrete Properties > Section Properties of Reinforcing Bars & Cement Types > Load Combinations of Concrete Design > Design of Concrete Members > Reinforced Concrete Beam Design > Flexural Design of Reinforced Concrete Beams > Serviceability of Reinforced Concrete Beams > Shear Design of Reinforced Concrete Beams Structural Steel Design > General Topics of Steel Material and Design > Steel Structures > Connections in Steel Structures > Cold Formed Steel Sections Construction > Elements of construction > Construction Site Layout Planning Elements > All Types of Roofs And Their Details > Types of Foundations From Construction Point of View > All Types of Foundation Materials Timber Design > Design of Timber Members > Design of Sawn Timber Beams or Joists > Design of Sawn Timber Columns and Compressive Members Masonry Design Finite Elements Method > Basics of Finite Elements > Introduction to Finite Elements And the Big Picture > One-dimensional Bars/Springs > Plane Trusses Transportation Surveying > Instruments and Distance Measurements > Introduction to Surveying > Distance Measurement > Leveling, Bench marks, Turning points > Leveling > Horizontal and Vertical Angles, Azimuths, Traverses, Closure Error > Angle Measurement > Topographic Data, Base Mapping > Topographic Surveys > Creating a Plan and Profile Introduction Concepts & Formulas Videos Solved problems Download Files Soil Phase Relationships Courses > Soil Mechanics > Physical Properties of Soil > Soil Phase Relationships
Introduction on Soil Phase Relationships :
Soil mass is generally a three-phase system. It consists of solid particles, liquid, and gas. For all practical purposes, the liquid may be considered to be water (although in some cases, the water may contain some dissolved salts) and the gas as air.The phase system may be expressed in SI units either in terms of mass-volume or weight-volume relationships. The interrelationships of the different phases are important since they help to define the condition or the physical make-up of the soil.
Concepts and Formulas of Soil Phase Relationships:
Phase relationship diagram:
In a mass of soil, there are three physical components: solid, water, and air. A phase relationship diagram is normally used to represent the relationship as follows:
Definitions:
Volume: (ft^{3} , m^{3} )
V_{t} : Total volume
V_{s} : Volume of solid
V_{v} : Volume of void
V_{w} : volume of water
V_{a} : Volume of air
Weights: (lbs, kg, kN)
W_{t} : total weight
W_{s} : weight of solid
W_{w} : weight of water
Weight of air = 0
Phase Relationships:
Volume-volume relationship:
Weight-weight relationship:
Water (Moisture) content (%):
Weight-Volume relationship:
(Unit weight or density, lbs/ft^{3} , g/cm^{3} , kN/m^{3} )
Moisture (total) unit weight:
Dry unit weight:
Solid unit weight:
Saturated unit weight (when soil is completely saturated, S = 100%,V_{a} =0):
Submerged (buoyant) unit weight (when soil is below ground water table, S = 100%):
Following relations are very handy in solving problems:
Unit weight to unit weight relationship
Specific gravity:
(Unit weight of water = 62.4 lbs/ft^{3} = 1 g/cm^{3} = 9.8 kN/m^{3} )
Average value of Gs for granular soils is 2.65, while the average value of Gs for cohesive soils is 2.80.
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Solved sample problems of Soil Phase Relationships:
Example 1: Determine unit weights, water content, based on known volume and weight (English units)
Given: (English units)
Volume of soil mass: 1 ft^{3} .
Weight of soil mass at moist condition: 100 lbs
Weight of soil after dried in oven: 80 lbs
Requirements:
Determine moist unit weight of soil, dry unit weight of soil, and water content.
Problem solving technique:
Moist unit weight g_{t} = W_{t} / V_{t} (W_{t} = 100 lbs, V_{t} =1 ft^{3} , are given)
Dry unit weight, g_{d} = W_{s} / V_{t} (Weight of solid is weight of soil after dried in oven ,Ws = 80 lbs, V_{t} =1 ft^{3} , are given)
Water content, w (%) = W_{w} /W_{s} (W_{s} = 80 lbs , weight of water, W_{w} not known)
Find weight of water, from phase relationship diagram, W_{w} = W_{t} – W_{s} .
Solution:
Moist (total) unit weight, g_{t} = W_{t} / V_{t} = 100/1 = 100 pcf (lbs/ft^{3} )
Dry unit weight, g_{d} = W_{s} / V_{t} = 80/1= 80 pcf (lbs/ft^{3} ).
Weight of water = 100-80=20 lbs
Water (Moisture) content: w (%) = W_{w} /W_{s} ´ 100 (%) = 20/80x100% = 25%
Example 2: Determine unit weights, water content, based on known volume and weight (SI units)
Given: (SI units)
Volume of soil mass: 0.0283 m^{3} .
Weight of soil mass at moist condition: 45.5 kg
Weight of soil after dry in oven: 36.4 kg
Problem solving technique:
Moist unit weight g_{t} = W_{t} / V_{t} (both value are given)
Dry unit weight, g_{d} = Ws / V_{t} (both value are given)
Water content, w (%) = W_{w} /W_{s} (Weight of solid is weight of soil after dried in oven is given, weight of water not known)
Find weight of water, from phase relationship diagram, W_{w} = W_{t} – W_{s} .
Requirements:
Determine moist unit weight of soil, dry unit weight of soil, and water content.
Solution:
Moisture (total) unit weight, g_{t} = W_{t} / V_{t} = 45.5/0.0283 = 1608 kg/m3 = 1.608 g/cm3
Dry unit weight, g_{d} = Ws / V_{t} = 36.4/0.0283= 1286 kg/m^{3} =1.286 g/cm^{3}
Weight of water = 45.5-36.4=9.1 lbs
Water (Moisture) content: w (%) = W_{w} /W_{s} ´ 100 (%) = 9.1/36.4x100% = 25%
Example 3: Determine void ratio, porosity, and degree of saturation based on known volume, weight, and specific gravity (English units)
Given: (English units)
Volume of soil mass: 1 ft^{3} .
Weight of soil mass at moist condition: 125 lbs
Weight of soil after dry in oven: 100 lbs
Specific gravity of solid = 2.65
Requirements:
Determine void ratio, porosity, and degree of saturation
Problem solving technique:
Void ratio, e = V_{v} /V_{s} (Vv, Vs, not given)
Find V_{s} = W_{s} /g_{s} (Ws = 100 lbs, gs is not given)
Find g_{s} = G_{s} g_{w} (Gs is given, g_{w} =62.4 lbs/ft^{3} is a know value)
Find V_{v} = 1-V_{s} (e can be calculated)
Porosity, n = V_{v} /V_{t} (V_{v} from step 4, Vs from step 2)
Degree of saturation, S = V_{w} /V_{v} (V_{v} from step 4, need to find V_{w} )
V_{w} =W_{w} /g_{w} (W_{w} , not given, g_{w} =62.4 lbs/ft3)
Find W_{w} = W_{t} – W_{s} (Both W_{t} , W_{w} are given)
Solution:
Solid unit weight, g_{s} = G_{s} g_{w} =2.65*62.4=165.4 lbs/ft^{3}
Volume of solid, V_{s} = W_{s} /g_{s} = 100/165.4=0.6 ft^{3}
Volume of void = V_{t} – V_{s} = 1 –0.6=0.4 ft^{3}
Void ratio, e = V_{v} /V_{s} = 0.4/0.6=0.66
Porosity, n = V_{v} /V_{t} = 0.4/1 = 0.4
Weight of water = 125-100=25 lbs
Volume of water, V_{w} = W_{w} /g_{w} = 25/62.4=0.4 ft^{3}
Degree of saturation, S = V_{w} /V_{v} = 0.4/0.4x100% = 100%.
Example 4: Determine void ratio, porosity, and degree of saturation based on known volume, weight, and specific gravity (English units)
Given: (metric units)
Volume of soil mass: 0.0283 m^{3} .
Weight of soil mass at moist condition: 56.6 kg
Weight of soil after dry in oven: 45.5 kg
Specific gravity of solid = 2.65
Requirements:
Determine void ratio, porosity, and degree of saturation
Problem solving technique:
Void ratio, e = V_{v} /V_{s} (V_{v} , V_{s} , not given)
Find V_{s} = W_{s} /g_{s} (W_{s} = 45.5 kg, g_{s} is not given)
Find g_{s} = G_{s} g_{w} (G_{s} is given, g_{w} =1 g/cm^{3} is a know value)
Find V_{v} = 1-V_{s} (e can be calculated)
Porosity, n = V_{v} /V_{t} (V_{v} from step 4, V_{s} from step 2)
Degree of saturation, S = V_{w} /V_{v} (V_{v} from step 4, need to find V_{w} )
V_{w} =W_{w} /g_{w} (W_{w} , not given, g_{w} =62.4 lbs/ft^{3} )
Find W_{w} = W_{t} – W_{s} (W_{t} = 56.6 kg, W_{s} = 45.5 kg are given)
Solution:
Solid unit weight, g_{s} = G_{s} g_{w} =2.65*1=2.65 g/cm^{3} = 2650 kg/m^{3}
Volume of solid, V_{s} = W_{s} /g_{s} = 45.5/2650=0.0171 m^{3}
Volume of void = Vt – Vs = 0.0283 –0.0171=0.0112 m3
Void ratio, e = V_{v} /V_{s} = 0.0112/0.0171=0.65
Porosity, n = V_{v} /V_{t} = 0.0111/0.0283 = 0.39
Weight of water = 56.6-45.5=11.1 kg
Volume of water, V_{w} = W_{w} /g_{w} = 11.1 kg/1 g/cm^{3} = 11100 cm^{3} = 0.0111m^{3}
Degree of saturation, S = V_{w} /V_{v} = 0.0111/0.0111x100% = 100%.
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