A 4 m clay layer drains to sand on both faces. If the underlying sand is replaced by impermeable rock, the time to reach 50% consolidation

quadruples, because time varies as the square of the drainage path

doubles, because the drainage path doubles

quadruples, because time varies as the square of the drainage path

is unchanged, because the layer thickness is unchanged

halves, because water now flows in one direction only

For a saturated clay, which construction is most critical immediately after it is built, becoming safer with time?

Neither, since clay strength does not change with time

A dense sand and a loose sand of the same mineral are sheared at the same effective confining pressure. Compared with the loose sand, the dense sand

shows a peak strength and then falls to a similar ultimate strength, and dilates while shearing

shows no peak and contracts throughout

has a higher ultimate strength as well as a higher peak

behaves identically, since both are the same mineral

Which of the following will increase the effective stress at a point in the ground? (One or more)

Lowering the water table by pumping

Placing a surcharge fill at the surface

Applying a vacuum beneath a sealed membrane at the surface

Lowering the water table by pumping

Placing a surcharge fill at the surface

A rise in the water table after heavy rain

Applying a vacuum beneath a sealed membrane at the surface

Soil Mechanics II: Effective Stress, Consolidation, Shear Strength and Stress Paths

One idea carries this entire chapter, and it is the idea that makes soil mechanics a subject of its own rather than a branch of solid mechanics. Soil grains do not feel the total stress; they feel σ′ = σ − u. Water in the pores carries part of the load and, being a fluid, carries it without shear — so it neither strengthens the soil nor stiffens it. Everything that matters mechanically — strength, stiffness, volume change — is a function of the effective stress alone. Two consequences follow immediately, and they look contradictory until the principle is properly held. Raise the water table without adding any load and the soil gets weaker, because u rose and σ′ fell. Pump the water table down without removing any load and the soil settles, because u fell and σ′ rose. Neither is caused by a change in loading, which is why dewatering for a basement can crack buildings two streets away. From there the chapter divides by time. Consolidation is what happens in a clay when σ′ increases: water has to leave before the skeleton can compress, so the settlement is not immediate but spread over months or years, and the two questions — how much and how long — have separate answers with separate parameters, C_c and c_v. Shear strength obeys τ = c′ + σ′ tan φ′, and the whole drained-versus-undrained distinction is about whether the pore pressures generated by the loading have had time to dissipate. And stress paths plot where a sample travels rather than only where it ends, which is the only way to see that two soils reaching the same failure point got there by completely different routes.

Effective Stress, and the Two Opposite Things It Explains

Total vertical stress at a depth is just the weight of everything above: sum γ h layer by layer, using the saturated unit weight below the water table and the moist one above it. Pore pressure at the same point is u = γ_w h_w, the height of water standing above it. Subtract, and you have the effective stress, which is the part carried by grain-to-grain contact — and it is the only part that produces friction, resists shear, or squeezes voids out of a clay. The rule for computing it is mechanical; the rule for interpreting it is where the subject lives.

🎯 Pumping the water table down adds a third of the load, without adding any load

Take a profile: 3 m of moist sand at 17 kN/m³ over a water table, then 5 m of saturated sand at 20 kN/m³. At 8 m depth the total stress is 3(17) + 5(20) = 151 kPa, the pore pressure is 5(9.81) = 49.05 kPa, and the effective stress is 151 − 49.05 = 101.95 kPa. Now pump the water table down to 8 m for a basement excavation. Nothing has been built, nothing has been loaded, and no soil has been added. But the sand that was submerged is now moist, so the total stress falls to 8(17) = 136 kPa — and the pore pressure falls to zero, so the effective stress rises to 136 kPa. The grains are now carrying 34 kPa more than before, a 33 per cent increase, and every compressible layer in that profile will consolidate under it. This is not a curiosity: it is the mechanism by which dewatering for one basement cracks buildings two streets away, and why groundwater abstraction has sunk parts of cities by metres. Read the principle the other way and it explains the opposite hazard. Raise the water table — a wet season, a leaking main, a reservoir filled behind a slope — and u rises while σ barely changes, so σ′ falls and with it every frictional strength in the ground. That is why slopes fail after rain rather than during the dry season, and why the strength of a sand is reported against effective stress and never against total. One principle, two hazards that look unrelated until it is written down.

Consolidation: How Much, and How Long — Two Questions, Two Parameters

When effective stress increases on a saturated clay, the skeleton cannot compress until water leaves the voids — and water leaves slowly, because a clay’s permeability is minute. So the settlement is not an elastic response but a process, and it needs two answers. How much comes from the compression index: ΔH = H (C_c/(1+e₀)) log₁₀[(σ₀′ + Δσ)/σ₀′], the logarithm being what makes a clay’s e–log σ′ curve straight over the normally consolidated range. How long — the time rate of consolidation — comes from the coefficient of consolidation through the time factor, T_v = c_v t/d², with T_v = 0.197 at 50 per cent consolidation and 0.848 at 90 per cent. The two parameters are independent: a clay can be very compressible and drain quickly, or barely compressible and take a decade.

⚠️ Tripling the load doubles the settlement, and halving the drainage path quarters the time

Take a clay layer 4 m thick with e₀ = 0.9, C_c = 0.3, under an existing σ₀′ = 100 kPa. Add Δσ = 100 kPa and the settlement is 4 × (0.3/1.9) × log₁₀(200/100) = 190.1 mm. Now triple the increment to 300 kPa: the settlement becomes 4 × (0.3/1.9) × log₁₀(400/100) = 380.3 mm, which is exactly double, not triple. The reason is in the logarithm — log₁₀ 4 = 2 log₁₀ 2 — and it is the single most useful intuition about consolidation: settlement responds to the ratio of stresses, not their difference, so the first increment of load always costs more settlement than the second. Doubling σ′ from 100 to 200 and doubling it again from 200 to 400 produce equal settlements. Now the timing, which is a completely separate calculation. With c_v = 3 m²/year and drainage on both faces, the drainage path is half the layer, d = 2 m, so t₅₀ = 0.197(2²)/3 = 0.263 years = 96 days and t₉₀ = 0.848(2²)/3 = 1.13 years = 413 days. Put an impermeable stratum under the clay so it can only drain upward and d becomes the full 4 m: the same 50 per cent now takes 383 days and 90 per cent takes 4.5 years. Four times longer, from one boundary condition — because t ∝ d². Two exam traps live here. The drainage path is not the layer thickness, it is the longest distance water must travel to a free-draining boundary, so it is H/2 for double drainage and H for single. And a change in the boundary conditions alters the time and leaves the magnitude untouched: the same 190 mm arrives either way, just four times later.

One 4 m clay layer, c_v = 3 m²/year: the drainage boundary decides the programme
Condition	Drainage path d	50% consolidation	90% consolidation
Sand above and below	2 m (`H/2`)	96 days	413 days
Rock below, sand above	4 m (`H`)	383 days	4.5 years
Ratio	× 2	× 4	× 4
Final settlement	—	190 mm either way	190 mm either way

Shear Strength: Drained, Undrained, and Which One the Site Is In

The Mohr-Coulomb criterion is τ_f = c′ + σ′ tan φ′, and both parameters are effective-stress quantities because that is what the grains feel. In a triaxial test the same criterion becomes a relation between the cell pressure and the failure stress: σ₁ = σ₃ N_φ + 2c√N_φ with N_φ = tan²(45 + φ/2). At φ = 30° that flow factor is exactly 3, so a cohesionless sample at σ₃ = 100 kPa fails at σ₁ = 300 kPa, and adding c = 20 kPa lifts it to 300 + 2(20)√3 = 369.3 kPa. What separates the test types is not the criterion, which never changes, but whether the pore pressures generated by the loading are allowed to dissipate.

That is why a clay is quoted with two sets of parameters that describe the same soil at different times. In the short term, immediately after a rapid load, the water has not had time to move, the loading is undrained, and the strength is a single number c_u with φ_u = 0 — not because the clay is frictionless, but because with no volume change permitted the effective stress does not rise no matter what total stress is applied, so all Mohr circles at failure have the same diameter. In the long term the excess pore pressures dissipate, effective stress rises, and the drained parameters c′ and φ′ govern. The design consequence is the part worth carrying: for a clay, an embankment is most critical immediately after construction, when the undrained strength is all there is, and it becomes safer with time; a cut is most critical in the long term, because unloading generates negative pore pressures that gradually equalise and reduce the effective stress. The same soil, two failure times, opposite trends — and for a sand, which drains as fast as it is loaded, the distinction barely arises.

Stress Paths, and Why the K_f Line Is Flatter Than the Envelope

A Mohr circle records a state; a stress path records a journey. Plot p = (σ₁ + σ₃)/2 against q = (σ₁ − σ₃)/2 — the centre and radius of the Mohr circle — and each loading stage becomes a single point, so the whole test is a line rather than a family of overlapping circles. The value is that two samples reaching the same failure state by different routes are now visibly different: a sample loaded by increasing σ₁ at constant σ₃ travels up and to the right at 45°, while one unloaded by reducing σ₃ travels up and to the left, and the second is the path an excavation actually follows.

ℹ️ tan α = sin φ, and why that is not a typo

Join the failure points of several stress paths and you get the K_f line, which is the failure envelope expressed in p–q space. It is not the same line as the Mohr-Coulomb envelope, and the relation between them is worth deriving once because it looks like a misprint. At failure the Mohr circle touches the envelope, so the radius q and the centre p satisfy q = (p + c cot φ) sin φ. Expanding, q = p sin φ + c cos φ — a straight line in p–q with slope tan α = sin φ and intercept a = c cos φ. So a soil with φ = 30° has a K_f line at α = arctan(0.5) = 26.57°, and one with φ = 35° has α = 29.84°. The K_f line is always flatter than the failure envelope it represents, and the gap widens as φ grows. With c = 20 kPa and φ = 30° the intercept is a = 20 cos 30° = 17.32 kPa, not 20. Getting this backwards — reading α off a p–q plot and reporting it as φ — understates the friction angle, and by nearly 3.5 degrees at φ = 30°, which is more than enough to change a bearing capacity. Two habits make the topic safe. Always state which pair you are plotting, since the alternative p′ = (σ₁′ + 2σ₃′)/3 with q = σ₁′ − σ₃′ is equally common and has a different slope relation, M = 6 sin φ′/(3 − sin φ′). And read the direction of the path, not only where it ends: a path moving left is an unloading problem, and for a clay that means negative excess pore pressures and a long-term hazard rather than a short-term one.

Key takeaways

Grains feel σ′ = σ − u, and nothing else. Raising the water table weakens soil without unloading it; pumping it down settles soil without loading it — the 8 m profile gains 33% of effective stress from dewatering alone.
Settlement follows the ratio of stresses, not the difference: tripling Δσ from 100 to 300 kPa only doubles the settlement, because log 4 = 2 log 2. The first increment always costs the most.
How much and how long are separate questions with separate parameters. t ∝ d², and d is the longest path to a draining face — so blocking one face quadruples the time and leaves the settlement unchanged.
For a clay, φ_u = 0 is not frictionlessness — it is what undrained loading looks like when no volume change is allowed. An embankment is critical at once and improves; a cut is critical in the long term and worsens.
The K_f line has tan α = sin φ and a = c cos φ, so it is always flatter than the failure envelope — 26.57° for a soil with φ = 30°. Reading α as φ understates the friction angle by 3.4 degrees.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

A profile has 3 m of moist sand (γ = 17 kN/m³) over a water table, then saturated sand (γ_sat = 20 kN/m³) to a depth of 8 m. Taking γ_w = 9.81 kN/m³, the effective vertical stress at 8 m depth in kPa is ______ (round to two decimals)
Numerical answer — type the value.
Show answer
Answer: 101.95
101.95 kPa. Total stress is the weight of everything above, layer by layer with the right unit weight for each: σ = 3(17) + 5(20) = 51 + 100 = 151 kPa. Pore pressure is the height of water standing above the point: the water table is at 3 m and the point at 8 m, so u = 5(9.81) = 49.05 kPa. Then σ′ = 151 − 49.05 = 101.95 kPa. The most common error is using the submerged unit weight and subtracting u — that double-counts the buoyancy and gives 51 + 5(10.19) = 101.95 by a different route only if done consistently; done half-way it gives nonsense. Both routes are valid if kept separate: either total minus pore pressure, or sum γ h above the water table plus γ_sub h below, which here is 3(17) + 5(20 − 9.81) = 51 + 50.95 = 101.95 ✓. Pick one and stay in it. The interesting part is what happens next. Pump the water table down to 8 m for an excavation and the sand that was submerged becomes moist, so σ = 8(17) = 136 kPa and u = 0, giving σ′ = 136 kPa — the effective stress rises by 34 kPa, a third, with nothing built and nothing loaded. That is the mechanism behind dewatering-induced settlement of neighbouring buildings, and it is the clearest demonstration that it is σ′ and not σ that the ground responds to.
A 4 m clay layer has e₀ = 0.9 and C_c = 0.3. The existing effective stress at mid-depth is 100 kPa and a load increases it by 100 kPa. The consolidation settlement in mm is ______ (round to one decimal)
Numerical answer — type the value.
Show answer
Answer: 190.1
190.1 mm. ΔH = H (C_c/(1+e₀)) log₁₀[(σ₀′ + Δσ)/σ₀′] = 4 × (0.3/1.9) × log₁₀(200/100) = 4 × 0.15789 × 0.30103 = 0.19012 m. Three points about the formula are worth more than the arithmetic. It uses log₁₀, not the natural logarithm — the compression index is defined against the base-10 plot, and using ln inflates the answer by a factor of 2.303, which is the single commonest error in this calculation. It uses 1 + e₀ in the denominator because the settlement is a change in void volume expressed over the total original volume. And it uses the ratio of final to initial effective stress, which has a consequence worth testing yourself on: triple the load increment to 300 kPa and the settlement becomes 4 × 0.15789 × log₁₀ 4 = 380.3 mm, exactly double, not triple, since log 4 = 2 log 2. Each doubling of effective stress costs the same settlement, so the first increment of load is always the expensive one and a heavily pre-loaded ground settles far less under the same addition. Note also what this calculation does not tell you — how long it takes. That is a separate question with a separate parameter, c_v, and the same 190 mm can arrive in three months or four years depending only on where the clay can drain to.
A 4 m clay layer drains to sand on both faces. If the underlying sand is replaced by impermeable rock, the time to reach 50% consolidation
1. doubles, because the drainage path doubles
2. quadruples, because time varies as the square of the drainage path
3. is unchanged, because the layer thickness is unchanged
4. halves, because water now flows in one direction only
Show answer
Answer: B — quadruples, because time varies as the square of the drainage path
It quadruples. From T_v = c_v t/d², reaching a given degree of consolidation means reaching a fixed T_v, so t = T_v d²/c_v and time varies as the square of the drainage path. With sand on both faces water escapes to the nearer boundary, so the longest journey is half the layer, d = H/2 = 2 m. Replace the lower sand with rock and every particle of water must reach the top, so d = H = 4 m. Doubling d multiplies the time by four. Running the numbers with c_v = 3 m²/year: t₅₀ goes from 0.197(4)/3 = 96 days to 0.197(16)/3 = 383 days, and t₉₀ from 413 days to 4.5 years. Option A is the trap and it is the answer you get by remembering that the path doubles but not that the relation is quadratic. Option C confuses the two lengths that matter — the layer thickness is indeed unchanged, but the drainage path is not the layer thickness, it is the longest distance water must travel to a free-draining face. Option D has the direction of the effect backwards; fewer escape routes can only make drainage slower. Two things this question is really testing. First, that d is a boundary condition, not a soil property, so an unnoticed impermeable stratum can turn a nine-month settlement into a four-year one. Second, that this changes only the timing: the final settlement is fixed by C_c, e₀ and the stress ratio, and arrives at the same 190 mm either way — just four times later.
A cohesionless soil with φ = 30° is tested in a triaxial cell at a confining pressure of 100 kPa. The major principal stress at failure, in kPa, is ______
Numerical answer — type the value.
Show answer
Answer: 300
300 kPa. At failure σ₁ = σ₃ N_φ + 2c√N_φ with the flow factor N_φ = tan²(45 + φ/2). With c = 0 the second term vanishes, and at φ = 30° the flow factor is tan²(60°) = (√3)² = 3 exactly, so σ₁ = 100 × 3 = 300 kPa. That exact 3 is why φ = 30° is a favourite in question papers, and it is worth recognising so the arithmetic can be checked by eye. Add cohesion and the same soil at the same confining pressure would fail higher: with c = 20 kPa, σ₁ = 300 + 2(20)(√3) = 300 + 69.28 = 369.3 kPa. Two things to keep straight. The parameters in the Mohr-Coulomb criterion are effective-stress quantities, so a question quoting φ for a sand is quoting φ′ whether it says so or not, since a sand drains as fast as it is loaded and the distinction does not arise. For a clay it very much does: the same soil has an undrained strength c_u with φ_u = 0 in the short term, and drained c′, φ′ in the long term, and these are not two soils but one soil at two times. The φ_u = 0 result deserves a moment — it does not mean the clay has no friction. It means that with no volume change permitted, raising the total confining pressure raises the pore pressure by the same amount, leaving the effective stress unchanged, so every Mohr circle at failure has the same diameter and the envelope is horizontal.
For a saturated clay, which construction is most critical immediately after it is built, becoming safer with time?
1. A deep cut or excavation
2. An embankment or a footing
3. Both, equally
4. Neither, since clay strength does not change with time
Show answer
Answer: B — An embankment or a footing
An embankment or a footing — a loading problem. Placing load on a saturated clay generates positive excess pore pressure, because the water carries the increment first. Effective stress is therefore at its lowest immediately after construction, and the available strength is only the undrained c_u. As the excess pore pressure dissipates, σ′ rises, strength rises, and the structure gets safer — which is why a critical embankment is sometimes built in stages, pausing to let each lift consolidate before adding the next. A cut does the opposite. Removing load generates negative excess pore pressure, which temporarily holds the effective stress high and the clay strong. As those negative pressures equalise with the surrounding groundwater, σ′ falls and the slope becomes progressively less stable, so a cutting can stand for years and then fail — the long-term drained condition governs, and delayed failures in clay cuttings are a documented and recurring pattern. Option C ignores the sign of the pore pressure change, which is the whole distinction. Option D is contradicted by everything above: the strength of a clay is very much a function of time, because the strength depends on effective stress and effective stress depends on pore pressure, which is still moving. The practical consequence is that the two cases are analysed differently — an embankment on clay is checked with a total stress analysis using c_u for the short term, and a cutting with an effective stress analysis using c′ and φ′ for the long term.
For a soil with φ = 30°, the inclination of the K_f line in p–q space, in degrees, is ______ (round to two decimals)
Numerical answer — type the value.
Show answer
Answer: 26.57
26.57°. The K_f line is the failure envelope expressed in terms of the Mohr circle’s centre p = (σ₁+σ₃)/2 and radius q = (σ₁−σ₃)/2. At failure the circle touches the Mohr-Coulomb envelope, which gives q = (p + c cot φ) sin φ, and expanding that is q = p sin φ + c cos φ — a straight line of slope tan α = sin φ and intercept a = c cos φ. So α = arctan(sin 30°) = arctan(0.5) = 26.5651°. The relation tan α = sin φ looks like a misprint and is not, and the consequence is that the K_f line is always flatter than the failure envelope it represents: 26.57° against 30° here, and 29.84° against 35° for a denser sand. Reading a p–q plot and reporting its slope angle as φ therefore understates the friction angle — by 3.4 degrees at φ = 30°, which is easily enough to change a bearing capacity factor by tens of per cent. The intercept behaves the same way: with c = 20 kPa and φ = 30°, a = 20 cos 30° = 17.32 kPa, not 20. One caution to carry into the exam. There is a second convention in wide use, p′ = (σ₁′ + 2σ₃′)/3 with q = σ₁′ − σ₃′, in which the failure line has slope M = 6 sin φ′/(3 − sin φ′) instead — a completely different number, 1.2 at φ′ = 30°. Always state which pair is being plotted, because the two look alike on a page and share none of their algebra.
A dense sand and a loose sand of the same mineral are sheared at the same effective confining pressure. Compared with the loose sand, the dense sand
1. shows a peak strength and then falls to a similar ultimate strength, and dilates while shearing
2. shows no peak and contracts throughout
3. has a higher ultimate strength as well as a higher peak
4. behaves identically, since both are the same mineral
Show answer
Answer: A — shows a peak strength and then falls to a similar ultimate strength, and dilates while shearing
A peak, then a fall to a similar ultimate strength, with dilation on the way. A dense sand has its grains interlocked, so shearing them requires lifting one layer over another — the sample increases in volume as it shears, which is dilatancy, and the work done against the confining pressure to expand shows up as extra apparent strength. That gives the peak. Past the peak the structure has been disturbed, the interlock is lost, and the sample loosens towards a critical state at which it shears at constant volume. A loose sand starts on the other side: it contracts as it shears, has no interlock to break, shows no peak, and arrives at the same critical state from below. The important result is that both end at nearly the same ultimate strength, because the critical-state friction angle is a property of the mineral and grain shape rather than of the packing — so option C is wrong in its second half even though it is right in the first. Option B describes the loose sand, not the dense one. Option D ignores the density, which is the one thing that differs and the thing that matters. Two engineering consequences. Design on the peak strength of a dense sand is unconservative wherever strains can concentrate, because the material past a shear band has already softened to critical state while its neighbours are still at peak — this progressive failure is why residual and critical-state parameters are used for slopes. And the same dilatancy explains why damp beach sand goes pale and firm under your foot: shearing it tries to increase the void volume, and the water cannot arrive fast enough to fill it.
Which of the following will increase the effective stress at a point in the ground? (One or more)
1. Lowering the water table by pumping
2. Placing a surcharge fill at the surface
3. A rise in the water table after heavy rain
4. Applying a vacuum beneath a sealed membrane at the surface
Show answer
Answer: A — Lowering the water table by pumping; B — Placing a surcharge fill at the surface; D — Applying a vacuum beneath a sealed membrane at the surface
A, B and D. Work each through σ′ = σ − u. A lowers u a great deal while lowering σ only slightly, since the soil above the new water table changes from saturated to moist — on the profile in this chapter, dewatering from 3 m to 8 m raises σ′ from 101.95 to 136 kPa, a third, with nothing built. B raises σ and leaves u unchanged in the long term, which is the ordinary way of loading ground; note that in a clay the increase in σ′ is not immediate, since the pore water carries the increment first and hands it over as consolidation proceeds. D is the one that looks exotic and is the purest illustration of the principle: a vacuum beneath a sealed membrane reduces the pore pressure without adding any weight at all, so σ is unchanged and σ′ rises by the applied suction. That is exactly how vacuum consolidation works on soft ground, and it has a practical advantage over a surcharge fill — it cannot cause a bearing failure, because no load is being added to a soil that may be too weak to carry one. C is the odd one out: a rising water table increases u faster than it increases σ, so σ′ falls and the ground gets weaker. That is the mechanism behind slope failures after prolonged rain, and it is the same equation read in the opposite direction. Seeing all four together is the point of the question — three quite different site operations and one weather event, all governed by a single subtraction.