Geotechnical Engineering 1 – Soil Properties & Problem-Set #1 Notes

Fundamental Concepts and Key Equations

• 3-phase soil system

  • Soil is considered as a combination of
  • Soil solids (mass = $W<em>s$, volume = $V</em>s$)
  • Water (mass = $W<em>w$, volume = $V</em>w$)
  • Air (mass ≈ $0$, volume = $V_a$)
  • Total (moist) quantities
  • Weight $W<em>m = W</em>s + W_w$
  • Volume $V = V<em>s + V</em>w + V_a$

• Index parameters

  • Moisture (water-content)
    $w = \frac{W<em>w}{W</em>s}$
    (Expressed as decimal or %.)
  • Unit weights and densities (any consistent unit)
  • Moist unit weight $\gamma<em>m = \frac{W</em>m}{V}$
  • Dry unit weight  $\gamma<em>d = \frac{W</em>s}{V}=\frac{\gamma_m}{1+w}$
  • Saturated unit weight (no air voids)
    $\gamma<em>{sat}=\frac{(G</em>s+e)\,\gamma_w}{1+e}$
  • Specific gravity of soil solids
    $G<em>s = \frac{\rho</em>s}{\rho<em>w}=\frac{\gamma</em>s}{\gamma_w}$
  • Void ratio and porosity (measure the volume of voids)
    $e = \frac{V<em>v}{V</em>s} = \frac{V<em>a+V</em>w}{V<em>s}$$n = \frac{V</em>v}{V} = \frac{e}{1+e}$
  • Degree of saturation
    $S = \frac{V<em>w}{V</em>v}=\frac{w\,G_s}{e}$
    $S = 0\;(\text{dry}) \qquad S = 1\;(\text{fully saturated})$
  • Relationship combining $\gamma<em>m,w,G</em>s,e,S$
    $\gamma<em>m = \frac{(G</em>s+S e)\,\gamma_w}{1+e}$

• Practical interpretations

  • $\gamma_d$ expresses how closely soil particles are packed; used in compaction control.
  • $e,n$ govern hydraulic conductivity, compressibility and strength.
  • $S$ indicates how much of the void space is filled with water; important for seepage and effective stress.
  • Adding/removing water changes $w$ which in turn shifts $\gamma_m$ and $S$.

Problem 1 (0.10 ft³ sample, $W<em>m = 12.5\;\text{lb},\;w = 14\%,\;G</em>s = 2.71$)

• Moist unit weight
  • $\gamma_m = \frac{12.5}{0.1}=125\;\text{lb/ft}^3$
• Dry unit weight
  • $\gamma_d = \frac{125}{1+0.14}=109.65\;\text{lb/ft}^3$
• Void ratio
  • $e = \frac{G<em>s\,\gamma</em>w}{\gamma_d}-1 = \frac{2.71\,(62.4)}{109.65}-1 = 0.54222$
• Porosity
  • $n = \frac{e}{1+e}=0.35158$
• Degree of saturation
  • $S = \frac{w\,G_s}{e}=\frac{0.14\times2.71}{0.54222}=0.69972\;(69.97\%)$
• Water quantities
  • Weight of water $W<em>w = w\,W</em>s = 0.14\,(\gamma_d V)=1.5351\;\text{lb}$
  • Volume of water $V<em>w = \frac{W</em>w}{\gamma_w}=0.02460\;\text{ft}^3$

Problem 2 ($\gamma<em>m = 19.2\;\text{kN/m}^3,\;G</em>s = 2.69,\;w = 9.8\%$)

• Dry unit weight
  • $\gamma_d = \frac{19.2}{1+0.098}=17.49\;\text{kN/m}^3$
• Void ratio
  • $e = \frac{G<em>s\,\gamma</em>w}{\gamma_d}-1 = \frac{2.69\,(9.81)}{17.49}-1 = 0.50880$
• Degree of saturation
  • $S = \frac{0.098\times2.69}{0.50880}=0.5181\;(51.81\%)$

Problem 3 (Water to be added per $\text{m}^3$ to reach higher $S$; base values from Problem 2)

• Dry unit weight (carry-over) $\gamma<em>d = 17.49\;\text{kN/m}^3$, (w{old}=0.098)
• Target 90 % saturation
  • New water-content $w<em>{90}=\frac{0.90\,e}{G</em>s}=0.17023$
  • Added water weight $W<em>w = (w</em>{90}-w<em>{old})\,\gamma</em>d =1.26\;\text{kN/m}^3$
• Target 100 % saturation
  • $w_{100}=0.1891$
  • Added water $1.69\;\text{kN/m}^3$
• Field meaning: roughly 1.3–1.7 kN of water must be sprinkled per cubic metre to bring the soil close to or fully saturated.

Problem 4 ($w = 26\%,\;S = 72\%,\;\gamma_m = 108\;\text{lb/ft}^3$)

• Simultaneous solution of
  $\gamma<em>m = \frac{(G</em>s+S e)\,\gamma<em>w}{1+e},\qquad S=\frac{w G</em>s}{e}$
• Results
  • Specific gravity $G_s = 2.73$
  • Void ratio $e = 0.9858$
• Saturated unit weight
  • $\gamma_{sat}=\frac{(2.73+0.9858)\,62.4}{1+0.9858}=116.76\;\text{lb/ft}^3$
• Insight: even with a high moisture content (26 %), the soil is still < 100 % saturated because of the large void ratio.

Problem 5 ($G<em>s = 2.74,\;\gamma</em>m = 20.6\;\text{kN/m}^3,\;w = 16.6\%$)

• Dry unit weight $\gamma_d = \frac{20.6}{1+0.166}=17.67\;\text{kN/m}^3$
• Void ratio $e = \frac{2.74\,9.81}{17.67}-1 = 0.52119$
• Porosity $n = 0.34262$
• Degree of saturation $S = \frac{0.166\times2.74}{0.52119}=0.8727\;(87.27\%)$
• Practical note: soil is nearly saturated; only ~13 % of void space remains air.

Problem 6 ($\rho<em>m = 1750\;\text{kg/m}^3,\;w = 23\%,\;G</em>s = 2.73$)

• Dry density $\rho_d = \frac{1750}{1+0.23}=1422.76\;\text{kg/m}^3$
• Void ratio $e = \frac{2.73\,1000}{1422.76}-1 = 0.9188$
• Porosity $n = 0.4788$
• Degree of saturation $S = \frac{0.23\times2.73}{0.9188}=0.6834\;(68.34\%)$
• Water needed for full saturation
  • Target $w<em>{sat}=\frac{e}{G</em>s}=0.3366$
  • Added water mass $(0.3366-0.23)\,\rho_d = 151.67\;\text{kg/m}^3$

Problem 7 ($\rho_d = 2180\;\text{kg/m}^3,\;n = 0.30$)

• Void ratio $e = \frac{n}{1-n}=0.4286$
• Specific gravity $G<em>s = \rho</em>d\,(1+e)/\rho_w = 2180\,(1+0.4286)/1000 = 3.11$
• Significance: high $G_s$ (> 2.7) suggests mineralogy richer in heavy constituents (e.g., iron-bearing sands).

Problem 8 ($V = 0.25\;\text{ft}^3,\;W<em>m = 30.75\;\text{lb},\;w = 9.8\%,\;G</em>s = 2.66$)

• Moist unit weight $\gamma_m = \frac{30.75}{0.25}=123\;\text{lb/ft}^3$
• Dry unit weight $\gamma_d = \frac{123}{1+0.098}=112.02\;\text{lb/ft}^3$
• Void ratio $e = \frac{2.66\,62.4}{112.02}-1 = 0.4817$
• Volume of water present
  • Water weight $W<em>w = w\,W</em>s = 0.098\,(\gamma_d V)=2.7445\;\text{lb}$
  • Volume $V<em>w = \frac{W</em>w}{\gamma_w}=0.04398\;\text{ft}^3$

Problem 9 ($w = 17\%,\;\gamma<em>d = 105\;\text{lb/ft}^3,\;G</em>s = 2.69$)

• Void ratio $e = \frac{2.69\,62.4}{105}-1 = 0.5986$
• Degree of saturation $S = \frac{0.17\times2.69}{0.5986}=0.7639\;(76.39\%)$

Problem 10 (Partial statement)

• Initial condition $S<em>1 = 55\%,\;\gamma</em>{m1}=106\;\text{lb/ft}^3$
• After wetting $S<em>2 = 82.2\%,\;\gamma</em>{m2}=114\;\text{lb/ft}^3$
• Using $\gamma<em>m = \frac{(G</em>s+S e)\,\gamma<em>w}{1+e}$ and equal $G</em>s,e$ before & after, one can back-solve for unknown $e$ and $G_s$ or compute change in water content. (Transcript cuts off before full numerical solution.)

Cross-Problem Observations & Connections

• Formula symmetry helps: knowing any three of $\gamma<em>m,\gamma</em>d,e,w,S,G_s$ usually fixes the rest.
• Adding water at constant total volume V raises $w$, $S$, and $\gamma_m$; the dry unit weight stays unchanged.
• Typical ranges (coarse guidelines)
  • $G_s$: 2.60–2.80 for quartz-dominated sands, >3 for heavy minerals, ≈2.1 for organic soils.
  • $e$: 0.3–0.8 for sands, up to 2–3 for soft clays and peat.
  • $\gamma_d$: 14–20 kN/m³ for compacted fills, 9–13 kN/m³ for loose silts.
• Ethical & practical implications
  • Accurate index testing prevents settlement failures (e.g., over-estimation of compaction in roadbeds).
  • Water management governs both slope stability and construction scheduling; knowing how many kN/m³ (or kg/m³) of water to add can save resources and avoid over-saturation.

Quick Reference (All Key Equations Together)

• $w = \frac{W<em>w}{W</em>s}$
• $\gamma<em>m = \frac{W</em>m}{V}$ $\gamma<em>d = \frac{\gamma</em>m}{1+w}$
• $e = \frac{G<em>s\,\gamma</em>w}{\gamma_d}-1$
• $n = \frac{e}{1+e}$
• $S = \frac{w\,G_s}{e}$
• $\gamma<em>m = \frac{(G</em>s + S e)\,\gamma_w}{1+e}$
• $\gamma<em>{sat}=\frac{(G</em>s+e)\,\gamma_w}{1+e}$