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Soil Phase Relationships

A soil mass is commonly treated as a three-phase material made of soil solids, water, and air. For most board problems, the weight of air is neglected, so the total weight is the sum of the water weight and solid weight.

$$V = V_s + V_v = V_s + V_w + V_a$$
$$V_v = V_w + V_a$$
$$W = W_s + W_w$$

The usual symbols are $V_s$ for volume of solids, $V_w$ for volume of water, $V_a$ for volume of air, $V_v$ for volume of voids, $W_s$ for weight of solids, and $W_w$ for weight of water.

Indices of Soil

These ratios describe how much void, water, and solid material exists in the soil mass.

$$e = \frac{V_v}{V_s}$$
$$n = \frac{V_v}{V} = \frac{e}{1 + e}$$
$$S = \frac{V_w}{V_v}$$
$$w = \frac{W_w}{W_s}$$

Where $e$ is void ratio, $n$ is porosity, $S$ is degree of saturation, and $w$ is water content or moisture content.

Unit Weight Formulas

Use consistent units. These formulas assume $G_s$ is the specific gravity of soil solids and $\gamma_w = 9.81 \text{ kN/m}^3$ unless another value is specified.

$$\gamma = \frac{W}{V} = \frac{(G_s + Se)\gamma_w}{1 + e}$$
$$\gamma_d = \frac{W_s}{V} = \frac{G_s\gamma_w}{1 + e}$$
$$\gamma_{sat} = \frac{(G_s + e)\gamma_w}{1 + e}$$
$$\gamma' = \gamma_{sat} - \gamma_w = \frac{(G_s - 1)\gamma_w}{1 + e}$$
$$\gamma = \gamma_d(1 + w)$$
$$w = \frac{Se}{G_s}$$
$$S = \frac{wG_s}{e}$$
$$e = \frac{G_s\gamma_w}{\gamma_d} - 1$$

Quicksand Condition

Quicksand condition occurs when upward seepage reduces the effective stress to zero. The critical hydraulic gradient is obtained from the buoyant unit weight divided by the unit weight of water.

$$i_c = \frac{\gamma'}{\gamma_w} = \frac{G_s - 1}{1 + e}$$

At the critical gradient, the net effective unit weight of the soil mass becomes zero because the upward seepage force balances the buoyant weight.

Problem: Unit Weight from Void Ratio and Saturation

A soil sample has a specific gravity of 2.6, a void ratio of 0.40, and a degree of saturation of 0.60. What is the unit weight of the soil in kN/m3?

$$\gamma = \frac{(G_s + Se)\gamma_w}{1 + e}$$
$$\gamma = \frac{(2.6 + 0.60(0.40))(9.81)}{1 + 0.40}$$
$$\gamma = 19.90 \text{ kN/m}^3$$

Answer: 19.90 kN/m3.

Problem: Void Ratio from Dry Unit Weight

A soil sample has a dry unit weight of 16 kN/m3. If the specific gravity of the solid grains is 2.67, obtain the void ratio.

$$e = \frac{G_s\gamma_w}{\gamma_d} - 1$$
$$e = \frac{2.67(9.81)}{16} - 1$$
$$e = 0.637$$

Answer: 0.637.

Problem: Dry Unit Weight, Porosity, and Saturation

A soil sample has $G_s = 2.74$, moist unit weight of 20.6 kN/m3, and moisture content of 16.6 percent. Compute the dry unit weight, porosity, and degree of saturation.

$$\gamma_d = \frac{\gamma}{1+w} = \frac{20.6}{1.166} = 17.67 \text{ kN/m}^3$$
$$e = \frac{2.74(9.81)}{17.67} - 1 = 0.521$$
$$n = \frac{e}{1+e} = \frac{0.521}{1.521} = 0.343$$
$$S = \frac{wG_s}{e} = \frac{0.166(2.74)}{0.521} = 0.872$$

Answer: $\gamma_d = 17.67$ kN/m3, $n = 34.3\%$, and $S = 87.2\%$.

Problem: Specific Gravity, Saturated Unit Weight, and Quicksand

A soil sample has a dry unit weight of 17 kN/m3 and a void ratio of 0.60. Evaluate the specific gravity, the saturated unit weight, and the hydraulic gradient at quicksand condition.

$$G_s = \frac{\gamma_d(1+e)}{\gamma_w} = \frac{17(1.60)}{9.81} = 2.77$$
$$\gamma_{sat} = \frac{(G_s + e)\gamma_w}{1+e} = \frac{(2.77 + 0.60)(9.81)}{1.60} = 20.68 \text{ kN/m}^3$$
$$i_c = \frac{G_s - 1}{1+e} = \frac{2.77 - 1}{1.60} = 1.11$$

Answer: $G_s = 2.77$, $\gamma_{sat} = 20.68$ kN/m3, and $i_c = 1.11$.

Problem: Void Ratio from Moist and Saturated Unit Weight

A soil sample has a unit weight of 21.1 kN/m3 at a moisture content of 9.8 percent. When completely saturated, the unit weight becomes 22.58 kN/m3. Evaluate the void ratio.

$$\gamma_d = \frac{21.1}{1.098} = 19.22 \text{ kN/m}^3$$
$$\gamma_{sat} - \gamma_d = \frac{e\gamma_w}{1+e}$$
$$\frac{e}{1+e} = \frac{22.58 - 19.22}{9.81} = 0.343$$
$$e = \frac{0.343}{1-0.343} = 0.522$$

Answer: 0.522.

Problem: Dry Unit Weight 15 kN/m3 and Void Ratio 0.50

A soil sample has a dry unit weight of 15 kN/m3 and a void ratio of 0.50. Evaluate the specific gravity, saturated unit weight, and hydraulic gradient at quicksand condition.

$$G_s = \frac{15(1.50)}{9.81} = 2.29$$
$$\gamma_{sat} = \frac{(2.29+0.50)(9.81)}{1.50} = 18.27 \text{ kN/m}^3$$
$$i_c = \frac{2.29 - 1}{1.50} = 0.862$$

Answer: $G_s = 2.29$, $\gamma_{sat} = 18.27$ kN/m3, and $i_c = 0.862$.

Problem: Field Density and Particle Density

The field unit mass of a soil sample is 1800 kg/m3, the unit mass of the soil particles is 2000 kg/m3, and the moisture content is 12 percent. Evaluate the void ratio, dry unit weight in kN/m3, and degree of saturation.

$$\rho_d = \frac{1800}{1.12} = 1607.14 \text{ kg/m}^3$$
$$\gamma_d = \frac{1607.14(9.81)}{1000} = 15.77 \text{ kN/m}^3$$
$$e = \frac{G_s\rho_w}{\rho_d} - 1 = \frac{2000}{1607.14} - 1 = 0.244$$
$$S = \frac{wG_s}{e} = \frac{0.12(2.00)}{0.244} = 0.982$$

Answer: $e = 0.244$, $\gamma_d = 15.77$ kN/m3, and $S = 98.2\%$.

Problem: Unit Weights from Gs, e, and S

A soil sample has $G_s = 2.67$, void ratio $e = 0.45$, and degree of saturation $S = 40\%$. Evaluate the moist unit weight, dry unit weight, and saturated unit weight.

$$\gamma = \frac{(2.67 + 0.40(0.45))(9.81)}{1.45} = 19.28 \text{ kN/m}^3$$
$$\gamma_d = \frac{2.67(9.81)}{1.45} = 18.06 \text{ kN/m}^3$$
$$\gamma_{sat} = \frac{(2.67 + 0.45)(9.81)}{1.45} = 21.11 \text{ kN/m}^3$$

Answer: 19.28 kN/m3, 18.06 kN/m3, and 21.11 kN/m3.

Problem: Water Added to Reach Full Saturation

The moist unit weight of a soil is 16.5 kN/m3. If it has a moisture content of 15 percent, determine the mass of water in kg/m3 to be added to reach full saturation. The specific gravity of soil is 2.7.

$$\gamma_d = \frac{16.5}{1.15} = 14.35 \text{ kN/m}^3$$
$$e = \frac{2.7(9.81)}{14.35} - 1 = 0.846$$
$$S = \frac{0.15(2.7)}{0.846} = 0.479$$
$$\Delta W_w = \frac{(1-S)e}{1+e}\gamma_w = 2.34 \text{ kN/m}^3$$
$$\Delta m_w = \frac{2.34(1000)}{9.81} = 239 \text{ kg/m}^3$$

Answer: Approximately 239 kg of water per m3 of soil.

Problem: Void Ratio from Water Content and Moist Unit Weight

A soil sample has a water content of 20 percent and a moist unit weight of 18 kN/m3. The specific gravity of the solid is 2.65. Obtain the void ratio.

$$\gamma_d = \frac{18}{1.20} = 15.00 \text{ kN/m}^3$$
$$e = \frac{2.65(9.81)}{15.00} - 1 = 0.733$$

Answer: 0.733.

Problem: Specific Gravity from Critical Gradient

If the hydraulic gradient for quicksand condition is 1.13 with a void ratio of 0.50, compute the specific gravity of the soil sample.

$$i_c = \frac{G_s - 1}{1+e}$$
$$G_s = i_c(1+e) + 1$$
$$G_s = 1.13(1.50) + 1 = 2.70$$

Answer: 2.70.

Problem: Effective Unit Weight at Quicksand Condition

A soil having a void ratio of 0.50 has a specific gravity of 2.70. Compute the effective unit weight of the soil at quicksand condition.

$$\gamma' = \frac{(G_s-1)\gamma_w}{1+e} = \frac{(2.70-1)(9.81)}{1.50} = 11.12 \text{ kN/m}^3$$
$$i_c\gamma_w = \gamma'$$

At quicksand condition, the upward seepage force equals the buoyant unit weight, so the net effective unit weight is zero.

Answer: $\gamma' = 11.12$ kN/m3 before seepage balance; net effective unit weight at quick condition is 0.

Concept: Degree of Saturation

The ratio between the volume of water and the volume of voids in a soil mass is called the degree of saturation.

$$S = \frac{V_w}{V_v}$$

Answer: Degree of saturation.

Concept: Porosity

The ratio between the volume of voids and the total volume of the soil mass is called porosity.

$$n = \frac{V_v}{V}$$

Answer: Porosity.

Problem: Porosity, Buoyant Unit Weight, and Added Water

A soil sample has $G_s = 2.74$, moisture content of 16.6 percent, and a moist unit weight of 20.6 kN/m3. Compute the porosity, buoyant unit weight, and weight of water to be added per cubic meter of soil for 90 percent degree of saturation.

$$\gamma_d = \frac{20.6}{1.166} = 17.67 \text{ kN/m}^3$$
$$e = \frac{2.74(9.81)}{17.67} - 1 = 0.521$$
$$n = \frac{0.521}{1.521} = 0.343 = 34.3\%$$
$$\gamma' = \frac{(2.74-1)(9.81)}{1.521} = 11.22 \text{ kN/m}^3$$
$$S = \frac{0.166(2.74)}{0.521} = 0.872$$
$$\Delta W_w = (0.90 - 0.872)n\gamma_w = 0.093 \text{ kN/m}^3$$

Answer: $n = 34.3\%$, $\gamma' = 11.22$ kN/m3, and water to add is 0.093 kN/m3.

Problem: Saturated Soil with Water Content

A saturated soil has a water content of 23 percent and $G_s = 2.67$. Determine the saturated unit weight, dry unit weight, and moist unit weight when the degree of saturation becomes 70 percent.

$$e = wG_s = 0.23(2.67) = 0.614$$
$$\gamma_{sat} = \frac{(2.67+0.614)(9.81)}{1.614} = 19.96 \text{ kN/m}^3$$
$$\gamma_d = \frac{2.67(9.81)}{1.614} = 16.23 \text{ kN/m}^3$$
$$\gamma = \frac{(2.67+0.70(0.614))(9.81)}{1.614} = 18.84 \text{ kN/m}^3$$

Answer: $\gamma_{sat} = 19.96$ kN/m3, $\gamma_d = 16.23$ kN/m3, and $\gamma = 18.84$ kN/m3.

Problem: Unit Weight with Gs 2.5, e 0.40, and S 40 Percent

A soil sample has a specific gravity of 2.5 for its solid grains. It has a void ratio of 0.40 and a degree of saturation of 40 percent. Evaluate the unit weight of the soil in kN/m3.

$$\gamma = \frac{(2.5 + 0.40(0.40))(9.81)}{1.40} = 18.64 \text{ kN/m}^3$$

Answer: 18.64 kN/m3.

Problem: Porosity from Void Ratio

A sample of soil has a void ratio of 0.40. Evaluate the porosity of the soil.

$$n = \frac{e}{1+e} = \frac{0.40}{1.40} = 0.286$$

Answer: 28.6 percent.

Problem: Field Weight and Soil Particle Unit Mass

The field mass of a soil sample is 1900 kg/m3 and the unit mass of the soil particles is 2660 kg/m3. Compute the dry unit weight if the moisture content is 11.5 percent, then compute the void ratio and degree of saturation.

$$\rho_d = \frac{1900}{1.115} = 1704.0 \text{ kg/m}^3$$
$$\gamma_d = \frac{1704.0(9.81)}{1000} = 16.72 \text{ kN/m}^3$$
$$e = \frac{2660}{1704.0} - 1 = 0.561$$
$$S = \frac{0.115(2.66)}{0.561} = 0.546$$

Answer: $\gamma_d = 16.72$ kN/m3, $e = 0.561$, and $S = 54.6\%$.

Concept: Relative Density

Relative density $D_r$ describes how densely a granular soil is packed relative to its loosest and densest possible states. It applies mainly to sands and gravels, where denseness controls settlement behavior and liquefaction potential.

$$D_r = \frac{e_{max}-e}{e_{max}-e_{min}}$$
$$D_r = \frac{\frac{1}{\gamma_{d,min}}-\frac{1}{\gamma_d}}{\frac{1}{\gamma_{d,min}}-\frac{1}{\gamma_{d,max}}}$$

A relative density of 0 percent means the soil is in its loosest state. A value of 100 percent means it is at its densest. Loose sands have $D_r < 35\%$, medium-dense sands fall between 35 and 65 percent, and dense sands exceed 65 percent. High relative density is associated with lower compressibility, higher friction angle, and greater resistance to liquefaction during earthquakes.

Problem: Phase Diagram from Specimen Measurements

A soil specimen has a total volume of 0.006 m3, a moist mass of 11.4 kg, and a dry mass of 10.2 kg. The specific gravity of the solids is 2.68. Compute the water content, void ratio, degree of saturation, and moist unit weight.

$$w = \frac{11.4-10.2}{10.2}=0.1176=11.76\%$$
$$V_s = \frac{m_s}{G_s\rho_w}=\frac{10.2}{2.68(1000)}=3.806\times10^{-3}\text{ m}^3$$
$$V_v = 0.006-3.806\times10^{-3}=2.194\times10^{-3}\text{ m}^3$$
$$e = \frac{V_v}{V_s}=\frac{2.194\times10^{-3}}{3.806\times10^{-3}}=0.577$$
$$V_w = \frac{m_w}{\rho_w}=\frac{1.2}{1000}=1.2\times10^{-3}\text{ m}^3$$
$$S = \frac{V_w}{V_v}=\frac{1.2\times10^{-3}}{2.194\times10^{-3}}=0.547=54.7\%$$
$$\gamma = \frac{11.4(9.81)/1000}{0.006}=18.63\text{ kN/m}^3$$

Answer: $w=11.76\%$, $e=0.577$, $S=54.7\%$, and $\gamma=18.63$ kN/m3.

Problem: Relative Density of Sand

A sand deposit has a maximum void ratio of 0.85, a minimum void ratio of 0.42, and a current void ratio of 0.60. Find the relative density and classify the state of packing of the sand.

$$D_r = \frac{e_{max}-e}{e_{max}-e_{min}}=\frac{0.85-0.60}{0.85-0.42}$$
$$D_r = \frac{0.25}{0.43}=0.581=58.1\%$$

Since $35\% < D_r < 65\%$, the sand is in a medium-dense state.

Answer: $D_r = 58.1\%$, medium-dense.

Problem: Zero Air Voids Unit Weight

For a soil with $G_s = 2.70$, compute the zero air voids unit weight at water contents of 10, 15, and 20 percent. The zero air voids line represents the theoretical maximum dry unit weight achievable at each moisture level if no air remains in the voids.

At zero air voids, $S = 1$ so $e = wG_s$. The dry unit weight becomes:

$$\gamma_{zav} = \frac{G_s\gamma_w}{1+wG_s}$$
$$w=0.10:\quad \gamma_{zav}=\frac{2.70(9.81)}{1+0.10(2.70)}=21.27\text{ kN/m}^3$$
$$w=0.15:\quad \gamma_{zav}=\frac{2.70(9.81)}{1+0.15(2.70)}=19.98\text{ kN/m}^3$$
$$w=0.20:\quad \gamma_{zav}=\frac{2.70(9.81)}{1+0.20(2.70)}=18.83\text{ kN/m}^3$$

The zero air voids line always plots above the compaction curve. No compacted point can lie to the right of this line.

Answer: 21.27, 19.98, and 18.83 kN/m3 at 10, 15, and 20 percent water content, respectively.

Problem: Settlement from Void Ratio Change

A 3 m thick clay layer has an initial void ratio of 1.20. After a structural load is applied, laboratory tests show the final void ratio will decrease to 1.05. Compute the expected settlement of the clay layer.

$$S = H\frac{e_0-e_1}{1+e_0}$$
$$S = 3\left(\frac{1.20-1.05}{1+1.20}\right)=3\left(\frac{0.15}{2.20}\right)$$
$$S = 0.2045\text{ m}$$

Answer: 0.205 m or about 205 mm of settlement.

Problem: Full Phase Analysis of Borrow Pit Sample

In a borrow pit, a 1 m3 sample has a moist unit weight of 17.5 kN/m3, a moisture content of 14 percent, and $G_s = 2.65$. Determine the weight of soil solids, weight of water, volume of water, volume of solids, volume of voids, and volume of air in this 1 m3 sample.

$$W = 17.5\text{ kN},\quad W_s = \frac{W}{1+w}=\frac{17.5}{1.14}=15.35\text{ kN}$$
$$W_w = 17.5-15.35=2.15\text{ kN}$$
$$V_w = \frac{W_w}{\gamma_w}=\frac{2.15}{9.81}=0.219\text{ m}^3$$
$$V_s = \frac{W_s}{G_s\gamma_w}=\frac{15.35}{2.65(9.81)}=0.590\text{ m}^3$$
$$V_v = 1.0-0.590=0.410\text{ m}^3$$
$$V_a = V_v-V_w=0.410-0.219=0.191\text{ m}^3$$

Answer: $W_s=15.35$ kN, $W_w=2.15$ kN, $V_w=0.219$ m3, $V_s=0.590$ m3, $V_v=0.410$ m3, and $V_a=0.191$ m3.

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