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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: (ft3, m3)

Weights: (lbs, kg, kN)

 

Phase Relationships:

Volume-volume relationship:

 

Weight-weight relationship:

Note! Moisture content at fully saturation is not 100%!
100% moisture content means the weight is equally divided into water and solid or in other words the weight of soil particles is equal to the weight of water.
See: Solution of The Value of The Moisture When Fully Saturated

 

Weight-Volume relationship: 

(Unit weight or density, lbs/ft3, g/cm3, kN/m3)

 

Note! Following relations are very handy in solving problems:

gamma_{d} = frac{gamma_{t}}{1+omega}=frac{gamma_{s}}{1+e}=frac{gamma_{sat}}{1+frac{e}{Gs}}

 

gamma_{s} > gamma_{sat} > gamma_{t} > gamma_{d} > gamma_{b}

 

gamma_{d}=gamma_{sat}-ngamma_{w}=gamma_{sat}-(frac{e}{1+e})gamma_{w}

 

gamma_{d}=frac{G_{s}gamma_{w}(1-A)}{1+omega G_{s}}

 

A=n(1-Sr)

 

eS_r=omega G_s

 

e=frac{gamma_s}{gamma_{d}}-1

 

Unit weight to unit weight relationship

 


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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)

Requirements:

Determine moist unit weight of soil, dry unit weight of soil, and water content.

Problem solving technique:

  1. Moist unit weight gt= Wt / Vt  (Wt = 100 lbs, Vt=1 ft3,  are given)
  2. Dry unit weight, gd  = Ws / Vt  (Weight of solid is weight of soil after dried in oven ,Ws = 80 lbs, Vt=1 ft3,  are given)
  3. Water content, w (%) = Ww/Ws (Ws = 80 lbs , weight of water, Ww not known)
  4. Find weight of water, from phase relationship diagram, Ww = Wt – Ws.

Solution:

  1. Moist (total) unit weight, gt = Wt / Vt = 100/1 = 100 pcf (lbs/ft3)

  2. Dry unit weight, gd  = Ws / Vt = 80/1= 80 pcf (lbs/ft3).

  3. Weight of water = 100-80=20 lbs

  4. Water (Moisture) content: w (%) = Ww/Ws ´ 100 (%) = 20/80x100% = 25%

 

Example 2: Determine unit weights, water content, based on known volume and weight (SI units)

Given: (SI units)

Problem solving technique:

  1. Moist unit weight gt = Wt / Vt  (both value are given)
  2. Dry unit weight, gd  = Ws / Vt  (both value are given)
  3. Water content, w (%) = Ww/Ws (Weight of solid is weight of soil after dried in oven is given, weight of water not known)
  4. Find weight of water, from phase relationship diagram, Ww = Wt – Ws.

Requirements:

Determine moist unit weight of soil, dry unit weight of soil, and water content.

Solution:

  1. Moisture (total) unit weight, gt = Wt / Vt = 45.5/0.0283 = 1608 kg/m3 = 1.608 g/cm3

  2. Dry unit weight, gd  = Ws / Vt = 36.4/0.0283= 1286 kg/m3=1.286 g/cm3

  3. Weight of water = 45.5-36.4=9.1 lbs

  4. Water (Moisture) content: w (%) = Ww/Ws ´ 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)

Requirements:

Determine void ratio, porosity, and degree of saturation

Problem solving technique:

  1. Void ratio, e = Vv/Vs   (Vv, Vs, not given)
  2. Find Vs = Ws/gs          (Ws = 100 lbs, gs is not given)
  3. Find gs = Gsgw            (Gs is given, gw =62.4 lbs/ft3 is a know value)
  4. Find Vv = 1-Vs            (e can be calculated)
  5. Porosity, n = Vv/Vt      (Vv from step 4, Vs from step 2)
  6. Degree of saturation, S = Vw/Vv          (Vv from step 4, need to find Vw)
  7. Vw =Ww/gw               (Ww, not given, gw=62.4 lbs/ft3)
  8. Find Ww = Wt – Ws    (Both Wt, Ww are given)

Solution:

  1. Solid unit weight, gs = Gsgw=2.65*62.4=165.4 lbs/ft3

  2. Volume of solid, Vs = Ws/gs = 100/165.4=0.6 ft3

  3. Volume of void = Vt – Vs = 1 –0.6=0.4 ft3

  4. Void ratio, e = Vv/Vs = 0.4/0.6=0.66

  5. Porosity, n = Vv/Vt = 0.4/1 = 0.4

  6. Weight of water = 125-100=25 lbs

  7. Volume of water, Vw = Ww/gw = 25/62.4=0.4 ft3

  8. Degree of saturation, S = Vw/Vv = 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)

Requirements:

Determine void ratio, porosity, and degree of saturation

Problem solving technique:

  1. Void ratio, e = Vv/Vs   (Vv, Vs, not given)
  2. Find Vs = Ws/gs          (Ws = 45.5 kg, gs is not given)
  3. Find gs = Gsgw            (Gs is given, gw =1 g/cm3 is a know value)
  4. Find Vv = 1-Vs            (e can be calculated)
  5. Porosity, n = Vv/Vt      (Vv from step 4, Vs from step 2)
  6. Degree of saturation, S = Vw/Vv          (Vv from step 4, need to find Vw)
  7. Vw =Ww/gw               (Ww, not given, gw=62.4 lbs/ft3)
  8. Find Ww = Wt – Ws    (Wt = 56.6 kg, Ws = 45.5 kg are given)

Solution:

  1. Solid unit weight, gs = Gsgw=2.65*1=2.65 g/cm3 = 2650 kg/m3

  2. Volume of solid, Vs = Ws/gs = 45.5/2650=0.0171 m3

  3. Volume of void = Vt – Vs = 0.0283 –0.0171=0.0112 m3

  4. Void ratio, e = Vv/Vs = 0.0112/0.0171=0.65

  5. Porosity, n = Vv/Vt = 0.0111/0.0283 = 0.39

  6. Weight of water = 56.6-45.5=11.1 kg

  7. Volume of water, Vw = Ww/gw = 11.1 kg/1 g/cm3= 11100 cm3= 0.0111m3

  8. Degree of saturation, S = Vw/Vv = 0.0111/0.0111x100% = 100%.

 


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