Question 6 of 6: Optional Questions

Question 6: Optional Questions (5 marks each; choose 3 of 8 — all 8 solved below as a complete study set)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

6.1 — Darcy's Law

Darcy's Law states that the flow rate of water through a saturated porous medium is directly proportional to the hydraulic gradient driving it: $$q=kiA=k\frac{\Delta h}{L}A, \qquad v=\frac{q}{A}=ki$$ where $q$ is the volumetric flow rate, $k$ is the hydraulic conductivity of the soil (a material property combining the pore geometry and the fluid's viscosity/density), $i=\Delta h/L$ is the hydraulic gradient (head loss $\Delta h$ over flow-path length $L$), and $A$ is the total cross-sectional area (soil plus voids) perpendicular to flow. $v$ is the discharge velocity (a bulk, apparent velocity across the full area $A$) — it is smaller than the true average pore (seepage) velocity $v_s=v/n$, since water only actually flows through the void fraction $n$ of that area. Darcy's Law is empirical but is now understood as the low-Reynolds-number (laminar) limit of the Navier-Stokes equations averaged over a representative elementary volume of soil, and it holds for essentially all groundwater flow situations except very coarse, high-velocity gravels where turbulence violates the linear proportionality.

6.2 — Effective stress between two sand grains

Consider a horizontal plane cutting through a saturated soil mass at some depth, passing partly through mineral-to-mineral grain contacts and partly through pore water. The total (applied) stress $\sigma$ acting on that plane is carried by two parallel mechanisms: the inter-granular contact forces summed over the plane's area (the effective stress, $\sigma'$) and the pressure in the pore water ($u$), which acts equally in all directions and pushes the grains apart without itself transmitting shear resistance. Because the true contact area between grains is a tiny fraction of the total plane area, Terzaghi's simplification treats the pore-water contribution as acting over the FULL plane area: $$\sigma=\sigma'+u \quad\Rightarrow\quad \boxed{\sigma'=\sigma-u}$$ This is the single most important equation in soil mechanics because shear strength, compressibility, and volume-change behaviour are all governed by the effective (inter-granular) stress alone — the pore water carries load but, being a fluid, contributes essentially no shear resistance of its own.

6.3 — Capillary rise and water retention curves

When a narrow tube of diameter $d$ is placed in water, surface tension $T$ acting around the meniscus's circumference pulls water up the tube until the weight of the suspended water column balances the vertical component of the surface-tension force. Equating the two gives the capillary rise: $$h_c=\frac{4T\cos\alpha}{\gamma_w d}$$ where $\alpha$ is the contact angle (near $0^{\circ}$ for water on clean glass or most mineral surfaces, so $\cos\alpha\approx1$). Smaller tube diameters give greater rise — exactly analogous to a soil's pore network, where the "tube diameter" is set by the pore/particle size. A soil's water retention curve (degree of saturation $S$ vs. matric suction) is the macroscopic expression of the same capillary physics: below the air-entry value (AEV) the soil stays essentially saturated (suction is too low to overcome the smallest meniscus radii in the largest pores), and once suction exceeds the AEV, $S$ falls off steeply as progressively smaller pores are desaturated, flattening out toward a residual saturation where remaining water is held in isolated, disconnected films that further suction increases cannot easily remove. Fine-grained soils (small effective pore "tube diameter") have both a higher AEV and a higher capillary rise than coarse soils, by the same $1/d$ relationship.

6.4 — Sand cone test

Given. Excavated hole volume $V=1165\text{ cm}^3$; moist mass $M_{\text{wet}}=2600\text{ g}$; dry mass $M_{\text{dry}}=1645\text{ g}$.

Find. Field dry density; field water content.

1. Field dry density. $\rho_{d}=\dfrac{M_{\text{dry}}}{V}=\dfrac{1645}{1165}=\boxed{1.412\text{ g/cm}^3 = 1412\text{ kg/m}^3}$
2. Field water content. $w=\dfrac{M_{\text{wet}}-M_{\text{dry}}}{M_{\text{dry}}}=\dfrac{2600-1645}{1645}=\boxed{58.1\%}$

6.5 — Groundwater table and total head

Groundwater table (GWT) is the depth (or elevation) at which pore-water pressure in a soil equals atmospheric pressure ($u=0$) — the top surface of the fully saturated zone below which pore pressures are positive (hydrostatic, increasing with depth) and above which the soil is unsaturated with pore pressures effectively negative (capillary/matric suction).

Given. 5 m thick sand layer, GWT at 1.5 m depth; datum taken at the ground surface.

For point $P$ at the base of the layer ($z=-5\text{ m}$ from the datum): elevation head $h_e=z=-5\text{ m}$; pressure head is the height of the water column above $P$ up to the water table, $h_p=5-1.5=3.5\text{ m}$; total head $h_t=h_e+h_p=-5+3.5=\boxed{-1.5\text{ m}}$ — which equals the water table's own elevation ($z=-1.5\text{ m}$), confirming the expected result that under static (no-flow) hydrostatic conditions, total head is uniform everywhere in the saturated zone and equal to the free water surface elevation.

6.6 — Soil behaviour and consistency limits

A clay's mechanical behaviour changes dramatically with water content, and the Atterberg (consistency) limits mark the boundaries between distinct behavioural states. At very low water content (below the shrinkage limit, $SL$), the clay is a brittle solid: it is stiff, strong, and further drying causes no further volume change (all pore water has already been expelled from between particles). Between $SL$ and the plastic limit ($PL$), the clay is a semi-solid: still stiff and relatively strong, but it will crumble rather than deform smoothly if worked. Between $PL$ and the liquid limit ($LL$), the clay is in its plastic range — it can be remoulded into new shapes without cracking or losing strength catastrophically, because thin water films between particles let them slide past one another while van der Waals/electrochemical attraction still holds the fabric together; within this range, both undrained shear strength and stiffness decrease steadily and roughly log-linearly as water content rises toward $LL$. Above $LL$, the clay behaves as a viscous liquid: particle-to-particle contact forces are essentially lost, and the material has negligible shear strength, flowing under its own weight. The wider the plastic range ($I_p=LL-PL$), the more water content the clay can absorb before losing usable strength, which is why plasticity index is such a strong practical predictor of a fine soil's engineering behaviour and its sensitivity to moisture changes in the field.

6.7 — Falling head permeability test

Given. Specimen diameter $5\text{ cm}$, height $L=10\text{ cm}$; burette (standpipe) diameter $5\text{ mm}$; head falls from $h_1=1.25\text{ m}$ to $h_2=1.15\text{ m}$ in $t=35\text{ min}$.

Find. Hydraulic conductivity $k$ (cm/s); the likely soil type.

1. Areas. Specimen: $A=\tfrac{\pi}{4}(5)^2=19.63\text{ cm}^2$. Burette: $a=\tfrac{\pi}{4}(0.5)^2=0.1963\text{ cm}^2$ (5 mm $=0.5$ cm). $t=35\times60=2100\text{ s}$.
2. Falling-head formula. $$k=\frac{aL}{At}\ln\!\left(\frac{h_1}{h_2}\right)=\frac{(0.1963)(10)}{(19.63)(2100)}\ln\!\left(\frac{125}{115}\right)$$ $$\boxed{k = 3.97\times10^{-6}\text{ cm/s} \approx 4.0\times10^{-8}\text{ m/s}}$$
3. Soil type. A conductivity of order $10^{-6}\text{ cm/s}$ is far too low for a sand or silty sand (typically $10^{-3}$–$10^{-5}\text{ cm/s}$) and sits in the range typical of a low-permeability silt (or silty clay); it is also consistent with the choice of test method itself — a falling-head test (rather than constant-head) is specifically used for fine-grained soils whose flow is too slow to measure accurately with a constant-head setup. $$\boxed{\text{Soil type} \approx \text{silt (or silty clay)}}$$

6.8 — Shear strength: test types and strength definitions

(a) CD vs. CU triaxial. Both tests begin identically: the specimen is saturated and then consolidated under the cell pressure $\sigma_3$, with drainage allowed until excess pore pressure is zero. They differ entirely in the shearing stage. In a consolidated-drained (CD) test, drainage valves stay OPEN during shear and loading is applied slowly enough that no excess pore pressure ever builds up — total stress equals effective stress throughout shearing, and the test directly yields effective strength parameters $c'$, $\phi'$. In a consolidated-undrained (CU) test, drainage valves are CLOSED during shear, so excess pore pressure develops as the specimen is loaded (positive for contractive/normally-consolidated soils, negative for dilative/heavily overconsolidated soils); if pore pressure is measured during the test, CU results can still back out effective-stress parameters, but the test also directly gives the soil's total-stress (undrained) strength response, and it runs much faster than a CD test because no drainage-controlled loading rate is required.

(b) Undrained vs. drained strength. Drained strength is the shear resistance available once all excess pore pressure has fully dissipated — it is governed by the effective-stress Mohr-Coulomb parameters, $\tau_f=c'+\sigma'\tan\phi'$, and is the relevant strength for long-term stability problems (e.g., a slope or foundation years after construction, once consolidation is complete). Undrained strength ($s_u$ or $c_u$) is the shear resistance available immediately after loading, before any pore pressure has had time to dissipate — for saturated clays under rapid (undrained) loading this is often expressed as a single total-stress parameter ($\phi_u=0$ under fully undrained conditions), and it governs short-term, "end of construction" stability, which for a soft, low-permeability clay is frequently the MORE critical (lower-strength) case, since the soil has not yet gained the extra effective-stress strength that consolidation will eventually provide.