Problem Solution Manual for Unsaturated Soil Mechanics by Ning Lu and William J. Likos, Wiley, 2004 This manual is prepared by the following individuals: Phillip J. Wolfram, Alexandra Wayllace, William J. Likos 1 , and Ning Lu 1
2007 1 All correspondences should be addressed to
[email protected] or
[email protected]
Contents 1
State of Usaturated Soil
1
2
Material Variables
6
3
Interfacial Equilibrium
11
4
Capillarity
20
5
State of Stress
28
6
Shear Strength
36
7
Suction and Earth Pressure Profiles
52
8
Steady Flows
66
9
Transient Flows
71
10 Suction Measurement
80
11 Hydraulic Conductivity Measurement
87
12 Suction and Hydraulic Conductivity Mo dels
90
i
Acknowledgments This Solution Manual was not possible without the generous help of Silvia Simoni, Adam Prochaska, and Tom Bonnie. Bonnie. These individual individualss were participan participants ts in the Unsaturated Soil Mechanics course at the Colorado School of Mines and their fine homeworks served as an initial starting points for chapters 5, 6, 7, 10, and 11.
ii
Chapter 1
State of Usaturated Soil 1.1 Where are the regions in the U.S. where unsaturated soils are likely encountered to significant depth below the ground surface? Unsaturated soils are likely encountered to significant depth below the ground surface in large portions of the United States in the arid or semiarid regions (Figure 1.8) 1.2 What kind of climatic conditions tent to lead to the formation of a thick unsaturated zone? Precipitation, evaporation, and evapotranspiration are factors that contribute to the depth and extent of the unsaturated zone. A thick unsaturated zone generally occurs in regions where potential evaporation outweighs annual precipitation by a factor ranging between 2 and 20 within 40 degree north of equator. 1.3 What is the fundamental difference between unsaturated soils and saturated soils in terms of pore water pressure? Pore water pressure in a saturated soil is generally compressive and isotropic, in contrast to unsaturated soils where pore water pressure is generally, but not necessarily, tensile (p. 20-21). 1.4 Describe and illustrate the Mohr-Coulomb failure criterion. The Mohr Coulomb criterion delineates a failure envelope for a material defining critical states of stresses. It is described in terms of (c) and internal friction angle (f ). It is described in terms of states of cohesion ( c ) and internal friction angle(φ ) to dictate a failure shear stress at a given normal effective stress (σn ) in τ f = c + σn tan φ . Graphically, the Mohr-Coulomb failure criterion plots as a straight line on the effective normal and shear stress graph as shown in S1.1. It is also important to note that the Mohr-Coulomb failure criterion is a linear approximation and therefore valid only near the range of values from which it was derived.
1
Figure S1.1: Mohr-Coulomb Failure Criterion
2
1.5 When the state of stress (i.e., Mohr circle) in a soil reaches the Mohr-Coulomb criterion, what is the state of stress called? Failure state. 1.6 Give three examples of unsaturated soil mechanics problems in geotechnical engineering. Transient and steady seepage in unsaturated embankment dams, consolidation and settlement of unsaturated soils, bearing capacity for shallow foundations under moisture loading, slope stability, and land sliding. 1.7 For a given unsaturated soil under either a dry or wet condition, which one has a higher suction? The dry soil has a higher suction (p. 39, 42-43). 1.8 What are state variables, material variables, and constitutive laws? State variables are those variables that completely describe the state of the system for the given phenomenon. Material variables generally vary with state variables and describe the physical characteristics of the material. These variables are intrinsic material properties. Constitutive laws describe the governing physical principles which demonstrate interrelationships between or among state variables and material variables. Constitutive laws are used as the mathematical connection between state and material variables for the purpose of prediction and explanation of phenomena (p. 26-28). 1.9 What are the principal differences between saturated and unsaturated soil profiles of pore water pressure, total stress, and effective stress? Pore water Pressure Profiles- Generally vary linearly with depth, increasing hydrostatically below the water table (saturated soils), and decreasing hydrostatically above the water table (unsaturated soils). Total Stress Profiles- For unsaturated soils, total stress decreases due to change in the selfweight when the material is dewatered. Therefore, the total stress profile for a saturated soil extends to a greater magnitude than the profile for the same soil under unsaturated conditions. Effective Stress Profiles- The effective stress for a saturated soil at ground surface is 0. It is also important to note that effective stresses for an unsaturated soil are greater than for a saturated soil due to the tensile pore water pressure in the unsaturated soil (p. 22-23). Figures 1.12 and 1.13 graphically demonstrate these concepts. For both saturated and unsaturated conditions, horizontal stresses are dependent upon vertical stresses according to formula 1.5a (p. 23). 1.10 According to Bishop’s effective stress concept, which state, saturated or unsaturated, has a higher effective stress? Why? According to Bishop’s effective stress concept, unsaturated conditions have higher effective stress since pore water pressures are negative, leading to a greater effective stress. This is demonstrated by examining the equation: σ = (σ ua ) + χ(ua uw ) where matric suction (ua uw ) is positive.
−
−
−
1.11 What is the shape of the pore water pressure profile under hydrostatic conditions in saturated and unsaturated states, respectively? 3
The shape of the pore pressure profile under the hydrostatic condition for saturated and unsaturated states is linear, as shown in Figure S1.2 where z = 0 is the ground surface:
Figure S1.2: Comparison of saturated and unsaturated hydrostatic soil profiles 1.12 If an unsaturated soil has a water potential of 1000 J/kg , what is the equivalent soil suction value? If the soil at the air dry condition has a matric suction of 100 MP a, what is the soil water potential in joules per kilogram?
−
For 1000 J/kg potential the equivalent soil suction is 1000 kP a, since 100 J/kg = 100 kP a. Soil with a matric suction of 100 M P a has a soil water potential of 100 , 000 J/kg since 0.1 M P a = 100 J/kg (p. 40).
−
−
1.13 Three soils- clay, silt, and sand- are all equilibrated at the same matric suction, which soil has the highest water content and why? If clay, silt, and sand are all equilibrated at the same matric suction, then clay has the highest water content as shown on Figure 1.20 (p. 42). Clay has the highest water content at a given matric suction due its charged surfaces and very high specific surface area as (p. 42) Sand and silt have lower specific surface areas than clay. 1.14 Describe the major physical and physicochemical mechanisms responsible for soil suction. Soil suction is caused by the physical and physicochemical mechanisms that decrease the potential of the pore water relative to a reference potential of free water. These mechanisms include capillary effects, short-range adsorption effects composed of particle-pore water interaction, and osmotic effects. Capillary effects are caused by curvature of the air-water interface. Short-range adsorption effects are composed of electrical double layer and van der Waals force field interactions at the solid-liquid interface. Osmotic effects are the result of dissolved solute in the pore water. With the same chemical concentration, osmotic pressure of pure solution 4
could be different with that of pore water as interaction between solute and solid surface of soil particles could occur. Matric suction is generally used to group the aggregate of capillary and short-range adsorption effects. Osmotic suction refers to the aggregate of osmotic effects (p. 34-35).
5
Chapter 2
Material Variables 2.1 What are the state variables that control the density of air? What is the average air density at your location? Temperature, pressure, and relative humidity are state variables that control the density of air as evidenced by the equation below. The average air density for Golden, CO during a typical winter, assuming a 10% relative humidity, an average temperature of 0 and an average pressure of 85 kPa is as follows: Moist air density is estimated using Tables 2.8 and 2.9. ud ωd ρa,moist = RT
− 0.611
ωd ωv
−
T 273.2 1 exp 17.27 T 36
− −
ωv RH RT
The first part of the expression is estimated as 1.084 kg m (from Table 2.8), and the second part of the expression can be estimated as 0.000 (from Table 2.9), resulting in 3
ρa,moist = 1.084
kg m3
2.2 What is the physical meaning of relative humidity? Relative humidity (RH) is the ratio of absolute humidity in equilibrium with any solution to the absolute humidity in equilibrium with free water at the same temperature. RH is also equivalent to the ratio of vapor pressure in equilibrium with a given solution and the saturated vapor pressure in equilibrium with free water. 2.3 At 25 and 101.3 kPa (1 atm), what is the ratio of the viscosity of water to the viscosity of air? The viscosity of which phase, air or water, is more sensitive to temperature changes between 0 and 100 ? −4
8.77×10 kg/m·s ν w /ν a = 1.845 = 47.53 50. Water is more sensitive to viscocity changes than ×10− kg/m·s air, changing by a magnitude of about 10, between 0 and 100 . 5
≈
in the night and 30 in the afternoon at a certain location. 2.4 Temperature varies between 15 If the ambient vapor pressure remains constant at 1.6 kPa, what is the range of the relative
6
humidity variation? If the vapor pressure remains unchanged, at what temperature will dew formation occur? RH =
uv uv,sat .
With uv constant, RH depends on the saturated vapor pressure which is related
−273.2 to change in temperature according to the following equation: uv,sat = 0.611 exp 17.27 T T . −36 At 15 , u v,sat = 1.70 kPa. At 30 , u v,sat = 4.24 kPa. With u v = 1.6 kPa, RH ranges from 93.8% to 37.7 %. If the vapor pressure remains unchanged, dew formation will occur when uv,sat = u v = 1.6 kPa. This condition is met when T = 287.2 K, or 14.0 . Dew formation will therefore occur at 14.0 .
2.5 If a saturated swelling soil has a specific gravity of 2.7 and gravimetric water content of 300 %, what is the volumetric water content? θ =
1 1+
1 Gs w
=
1 1+
1 2.7(3)
= 0.89 = 89%
2.6 A closed room is filled with humid air. If the temperature rises significantly, does the relative humidity increase or decrease? In a closed room filled with humid air, a significant temperature rise will result in a decrease of relative humidity (RH) since RH is inversely proportional to temperature (by Equations 2.11 and 2.12). 2.7 Can the vapor pressure of soil gas be greater than the saturation pressure at the same temperature and pressure? Why or why not? No, the vapor pressure of soil gas cannot be greater than saturation pressure at the same temperature and pressure because the saturation pressure is the maximum pressure corresponding to a given equilibrium state (state of same temperature and pressure). 2.8 Can volumetric water content be greater than 100% in unsaturated soil? Volumetric water content cannot be greater than 100 % in unsaturated soil as mathematically evidenced by the following definition: θ =
V w V w = V t V w + V s + V a
where θ = volumetric water content, V w = volume of water, V s = volume of solids, and V a = volume of air. 2.9 Is degree of saturation a mass-based or volume-based quantity? Degree of saturation (S ) is a volume-based quantity since it is defined as: S =
V w V v
where V w = volume of water, and V v = volume of voids.
7
2.10 When the temperature of unsaturated soil increases, does the surface tension at the air-water interface increase or decrease? When the temperature of unsaturated soil increases, the surface tension at the air-water interface decreases, as shown in Figure 2.12 and Table 2.10 (Lu, 2004).
and 95 2.11 What is the density of dry air if the prevailing temperature and pressure are 25 kPa, respectively? What is the relative change in dry-air density if the tempreature rises to 40 and the air pressure remains unchanged? If the temperature is kept at a constant value of 25 , how much pressure change is required to cause the dry-air density to decrease by 15% compared to 95 kPa? The density of dry air at T = 25 ρd =
and P = 95 kPa is
ud ωd (95 kPa)(28.966 10−3 kg/mol) = = 1.110 kg/m3 RT (8.314 N m/mol K)(298.2 K)
·
If the temperature rises to 40 ρd =
× ·
and the air pressure remains unchanged, the air density is
(95 kPa)(28.966 10−3 kg/mol) = 1.057 kg/m3 (8.314 N m/mol K)(313.2 K)
× ·
·
Relative change of 0.053 kg/m3 or a 4.8 % decrease. If temperature is kept constant at 25 , the pressure change required to cause the dry-air density to decrease by 15 % compared to 95 kPa (ρd = 0.85 1.110 kg/m3 = 0.944 kg/m3 ) is computed as follows:
·
(ρd )(RT ) = ωd
ud =
(0.944 kg/m3 )((8.314 N m/mol K)(298.2 K)) = 80.76 kPa 28.966 10−3 kg/mol
×
·
·
This value corresponds to a decrease in pressure of 14.24 kPa. 2.12 Estimate the viscosity of air and water at a temperature of 50 . Given a mean pore size for a sandy soil as 10 −3 m, and a specific discharge for both air and water as 10 −2 m/s, identify the flow regimes for the air and water, respectively. Estimates for the viscosity of air and water at a temperature of 50 , are as follows: µa = 1.96 10−5 kg/(m s) µw = 5.32 10−4 kg/(m s) Assume that density of water is 1000 kg /m3 and the density of air is 1 kg/m3 . Given a mean pore size for a sandy soil as 10 −3 m, and a specific discharge, q, for both air and water as 10−2 m/s, the Reyonds number for air and water are:
×
·
Rea =
×
·
ρvd (1 kg/m3 )(10−2 m/s)(10−3 m) = = 0.51
