Every chapter that follows — earth pressure, retaining walls, slopes, settlement, bearing capacity, piles — is built on five ideas from CEE 240. If any of them is shaky, the rest will feel like memorising formulas. This chapter makes sure they aren't.
Phase RelationsEffective StressMohr–CoulombStress Distribution
01
What This Course Is Actually About
Structural engineering gets to specify its material. You order 30 MPa concrete and 500 MPa steel, and that is what arrives. Geotechnical engineering gets whatever is already there — variable, layered, full of water, and different at every borehole.
🏗️
A structural engineer designs the ship; a geotechnical engineer surveys the sea. You cannot order calmer water. You can only measure what you have, understand how it behaves, and design something that survives it. That is why this course spends so much effort on investigating and characterising before it ever designs anything.
Core Idea
Every design in this course reduces to comparing a demand against a capacity, both computed from the same handful of soil parameters: unit weight γ, shear strength (c', φ'), compressibility (Cc, cv), and the position of the water table. Learn where those five come from and everything else follows.
02
Soil as Three Phases
Soil is not a solid. It is a skeleton of grains with voids between them, and those voids hold water, air, or both. Almost every soil property is really a statement about the proportions of those three phases.
Volume ratios
\[ e = \frac{V_v}{V_s}, \qquad n = \frac{V_v}{V} = \frac{e}{1+e} \]
Void ratio e compares voids to solids; porosity n compares voids to the total. Consolidation is written in terms of e because the solids never change.
Water content & saturation
\[ w = \frac{W_w}{W_s}, \qquad S = \frac{V_w}{V_v} = \frac{G_s w}{e} \]
S = 1 means saturated — the condition assumed throughout the consolidation and undrained-strength chapters.
γ' is the one to watch. Submerged unit weight is roughly half the saturated value, and it is what drives the collapse of the friction term in the slope stability chapter and the loss of bearing capacity under a high water table. Every dramatic result later traces back to this one subtraction.
03
Phase Relations Calculator
Give it Gs, w and e, and it returns every other phase quantity — including the γ' you will need constantly.
04
Classification & Why It Matters
Coarse-grained (sand, gravel)
Behaviour set by grain size and packing
Drains fast, so loading is effectively drained — pore pressures dissipate as quickly as you apply load. Strength is almost pure friction (c' ≈ 0). Settlement is immediate and essentially over by the end of construction.
Fine-grained (silt, clay)
Behaviour set by mineralogy and water
Drains very slowly, so short-term loading is undrained. Has real cohesion. Settlement continues for years. Characterised by Atterberg limits — liquid limit LL, plastic limit PL, plasticity index PI = LL − PL.
Core Idea
Classification is not paperwork — it decides which analysis you are allowed to run. Sand → drained, effective stress, c' = 0. Clay short-term → undrained, φ = 0, use cu. Clay long-term → drained, use c' and φ'. Choosing the wrong one is the most common conceptual error in geotechnical design.
05
The Effective Stress Principle
If you remember one equation from CEE 240, this is it. Terzaghi's effective stress principle is the single most important idea in all of soil mechanics.
Terzaghi, 1925
\[ \sigma = \sigma' + u \qquad\Longrightarrow\qquad \sigma' = \sigma - u \]
🏊
Stand in a swimming pool. Your weight has not changed, but you press on the floor far more gently — the water is carrying part of you. Soil grains behave identically: the pore water holds part of the load, and only the remainder, pressed grain against grain, generates friction. Water has no shear strength, so every kilopascal carried by water is a kilopascal that produces no strength.
Total stress σ
Weight of everything above, soil and water together. What a pressure cell would read.
Pore pressure u
Pressure in the water in the voids. Hydrostatic below a static water table: u = γwzw.
Effective stress σ'
The grain-to-grain contact stress. Controls strength, stiffness and volume change — nothing else does.
⚡
This one principle explains the whole course. Rain raises u, drops σ', and slopes fail. Loading clay raises u first and σ' only slowly — that is consolidation. A high water table raises u under a footing and halves its bearing capacity. Every dramatic number in the following chapters is this subtraction doing its work.
06
Effective Stress Profile
Build a two-layer profile and see σ, u and σ' plotted against depth. Drag the water table up and watch σ' fall while σ stays put — the mechanism behind almost every failure in this course.
07
Permeability & Seepage
Darcy's law
\[ v = ki, \qquad i = \frac{\Delta h}{L}, \qquad q = kiA \]
k spans an extraordinary range: gravel ~10⁻¹ m/s down to clay ~10⁻¹⁰ m/s — nine orders of magnitude.
Why that range matters
It is the entire reason "drained" and "undrained" are different analyses. Load sand and the water leaves before you finish; load clay and the water is still leaving a decade later. Permeability, not strength, decides which case you are in — and it also sets cv, the rate of consolidation.
08
Shear Strength — Mohr–Coulomb
Soil does not fail in tension or compression. It fails in shear — one block sliding past another. Everything in this course that "fails" is a shear failure on some surface.
The failure criterion
\[ \tau_f = c' + \sigma'\tan\phi' \]
Note it is σ', not σ. Strength depends on effective stress — this is where Section 5 meets Section 8.
Cohesion c'
Strength at zero confinement — grains genuinely sticking together. Real in clay, essentially zero in clean sand.
Friction angle φ'
How much extra strength each unit of confinement buys. Interlocking and sliding between grains. Dominant in sand.
Undrained strength cu
For clay loaded fast: use φ = 0 and τf = cu. Not a different material — a different drainage condition.
Core Idea
Two soils can have identical τf today for completely different reasons — one from cohesion, one from friction — and behave nothing alike when conditions change. The Chattogram case study in the slope chapter is exactly this: the site with the highest cohesion failed worst, because it had almost no friction to fall back on when rain softened it.
09
Stress Increase with Depth
A footing applies pressure at the surface, but the layer that matters may be 5 m down. How much of that pressure reaches it?
The 2:1 approximation
\[ \Delta\sigma = \frac{qBL}{(B+z)(L+z)} \]
Assume the load spreads outward at 2 vertical to 1 horizontal. Crude, quick, and good enough for most settlement estimates.
Boussinesq
The rigorous elastic solution for a point or distributed load on a semi-infinite elastic half-space, usually applied through influence factors and charts. More accurate, more work.
⚡
This is what sets borehole depth. The rule from the soil investigation chapter — bore until Δσ ≤ 0.1q — is just this equation solved for z. It is also how you get the Δσ' that goes into the settlement calculation.
10
Where Each Idea Reappears
γ' = γsat − γwslope stability, bearing capacity
σ' = σ − uevery single chapter
τf = c' + σ'tanφ'earth pressure, slopes, bearing capacity
Void ratio esettlement (Cc, Cs)
Permeability ktime rate of consolidation cv
Drained vs. undrainedwhich strength parameters are legal
Δσ with depthsettlement, borehole depth
Atterberg limitsclassification, α for piles
Core Idea
If a later chapter stops making sense, the problem is almost always one of these eight rows — most often the difference between σ and σ', or between drained and undrained. Come back here first.
11
Worked Examples
Example 1
Phase Relations from Three Numbers
Problem: A soil has Gs = 2.68, water content w = 18%, void ratio e = 0.72. Find porosity, degree of saturation, dry unit weight, moist unit weight, saturated unit weight and submerged unit weight.
1
Porosity: \( n = \dfrac{e}{1+e} = \dfrac{0.72}{1.72} = 0.419 \)
Answer: n = 0.419, S = 67%, γd = 15.29, γ = 18.04, γsat = 19.39, γ' = 9.58 kN/m³. Note γ' is half of γsat — remember that ratio; it drives most of this course.
Example 2
Effective Stress Profile
Problem: 3 m of moist sand (γ = 17 kN/m³) sits over 5 m of saturated sand (γsat = 20 kN/m³). The water table is at 3 m depth. Find σ, u and σ' at 3 m and at 8 m.
1
At 3 m (just at the water table): \( \sigma = 17(3) = 51.0\text{kPa} \), \( u = 0 \), so \( \sigma' = 51.0\text{kPa} \)
2
At 8 m, total stress: \( \sigma = 51.0 + 20(5) = 151.0\text{kPa} \)
Sanity check using γ' directly: \( \sigma' = 51.0 + (20-9.81)(5) = 51.0 + 51.0 = 102.0\text{kPa} \) ✓ — the two routes must agree.
Answer: At 8 m, σ = 151.0 kPa but σ' = 101.9 kPa. Water is carrying a third of the load. Raise the water table to the surface and σ' drops further still — try it in the calculator above.
12
Quick Reference & Quick Check
Effective stressσ' = σ − u
Porosityn = e/(1+e)
SaturationS = Gsw/e
Dry unit weightγd = Gsγw/(1+e)
Saturated unit weightγsat = (Gs+e)γw/(1+e)
Submerged unit weightγ' = γsat − γw ≈ ½γsat
Shear strengthτf = c' + σ'tanφ'
Darcyv = ki, q = kiA
2:1 stress spreadΔσ = qBL/[(B+z)(L+z)]
γw9.81 kN/m³
1. The water table rises to the ground surface. What happens to total stress σ and effective stress σ' at 5 m depth?
2. Why does a clay need a different analysis in the short term than in the long term?
3. Two soils both have τf = 40 kPa at the current stress. Soil A is all cohesion, soil B is all friction. Rain saturates both. Which is in more trouble?
4. Why is submerged unit weight roughly half the saturated value?