Geology and Soil Mechanics, UW-Stout

Example 1

Consider an extreme example of a sandy soil.   I just recently was working on a project that required knowledge about the engineering properties of play sand purchased at Fleet/Farm here in Menomonie, WI.   I made the following laboratory measurements:

M_cup = 151.0 g
M_{dry sand+cup} = 558.6 g
M_{water+sand+cup} = 633.6 g  (100% saturated)
V_{dry sand}=V_water+sand=250 cm³

Determine the void ratio,  porosity, dry unit weight, moist (100% saturated) unit weight, specific gravity of solids, and the 100% saturated water content.

Solution:

One should first convert to some standard units in which to work.  Converting all the weights into Newtons and volumes into cubic meters:

M_{dry sand} = M_{dry sand+cup} - M_cup = 558.6 - 151.0 = 408 g

W_{dry sand} = 0.408 kg(9.8 m/s²) = 4.0 N

and the weight of the sample when 100% saturated is

M_water+sand =  M_{water+sand+cup} - M_cup = 633.6 - 151.0 = 483 g  or  W_water+sand = 4.73 N.

The weight and volume of the water can now be determined:

W_water = W_water+sand   - W_sand = 4.73 N - 4.0 N = 0.73 N.

and by using the unit weight of water, g_water=9.8 kN/m³, we can determine the volume of this water by

V_w =W_w/g_w = 7.3x10^-4 kN/(9.8 kN/m³) = 7.4x10^-5 m³.

We can now determine many other parameters: 250 cm³(1 m/100cm)³ = 2.5x10^-4 m³

V_s = V - V_w = 2.5x10^-4 m³ - 7.4x10^-5 m^-3 = 1.76x10^-4 m³.

void ratio,  e = V_v/V_s = (7.4x10^-5 m³)/(1.76x10^-4 m³) = 0.42

water content (100% saturated), w = (W_w/W_s)(100%) = (0.73 N/4.0N)(100%) = 18.3 %

porosity, n = (V_v/V)(100%) = (7.4x10^-5 m³/2.5x10^-4 m³)(100%) = 29.6%

100% saturated unit weight, g = W/V = (4.73 N)/(2.5x10^-4 m³) = 18920 N/m³ = 18.9 kN/m³.

dry unit weight, g_d = W_s/V = (4.0 N)/(2.5x10^-4 m³) = 16000 N/m³ = 16.0 kN/m³.

specific gravity of the solids, G_s = (W_s/V_s )/g_w = (4.0 N/1.76x10^-4 m³)/(9810 N/m³) = 2.32

Example 2

You take a sample of soil at a construction site to determine soil properties.   The moist weight is 210 N, the dry weight is 170 N, the total volume is 0.015 m³, and the void ratio equals 0.45.   Determine the degree of saturation, porosity, moist unit weight, and specific gravity of solids of the sample.

Solution:

The moist unit weight is probably the easiest property to determine:

g = W/V = (210 N)/(0.015 m³) = 14000 N/m³ = 14.0 kN/m³.

The next step is a bit more difficult.  The volumes of the air, water, and solids are not given to you directly.  So you need to find an algebraic relationship between these variables that enable you to solve for them.

Lets first solve for the volume of the water:

W_w = W - W_d = 210 N - 170 N = 40 N, then  V_w =W_w/g_w = (40 N)/(9810 N/m³) = 0.0041 m³.

By definition

e = V_v/V_s = (V_w+V_a)/V_s

0.45 = (0.0041 + V_a)/V_s          (equation A)

V = V_s+V_w+V_a

0.015 = V_s+0.0041+V_a                (equation B)

Equations A and B represents two equations and two unknowns which is solvable by substitution.  Changing equation A around

V_s = (0.0041+ V_a)/0.45   putting this into equation B for V_s:

0.015 = ((0.0041+ V_a)/0.45) + 0.0041 + V_a ,

solving this for V_a:

0.015 = 9.1x10^-3 + 2.22V_a + 0.0041 + V_a

1.8x10^-3 = 3.22V_a

V_a = 5.6x10^-4 m³.  Now we can solve for some other variables:

V_s = (0.0041+ 5.6x10^-4)/0.45 = 0.0104 m³.

degree of saturation, S = (V_w/V_v)(100%) = ((0.0041)/(0.0041+5.6x10^-4))(100%) = 88.0%

porosity, n = (V_v/V)(100%) = ((0.0041+5.6x10^-4)/0.015)(100%) = 65%

specific gravity of the solids, G_s = (W_s/V_s )/g_w = (170 N/0.0104 m³)/(9810 N/m³) = 1.67