Question 6 of 6: Optional Questions

Question 6: Optional Questions (5 marks each; source asks for 3 of 8, all 8 answered)

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 discharge velocity through a saturated porous medium is proportional to the hydraulic gradient: $$v=ki \qquad\text{or, as a total flow rate,}\qquad q=vA=kiA$$ where $i=\Delta h/L$ is the hydraulic gradient (head loss per unit length of flow path), $v$ is the discharge (superficial) velocity — a bulk quantity averaged over the whole cross-section $A$, not the faster true velocity within the pores — and $k$ is the hydraulic conductivity (units of velocity), a property of both the soil's pore structure and the permeant fluid.

(6.2) Effective stress between two grains

At any horizontal plane through a saturated soil, the total normal stress $\sigma$ is carried partly by inter-particle contact forces and partly by the pressure of the water filling the voids. Consider a wavy plane threading between grain contacts, of gross area $A$: the total force is $\sigma A$. The pore water carries $u(A-a_c)\approx uA$ (contact area $a_c$ is a negligible fraction of $A$ for real soils), leaving the grain-to-grain contact forces to carry the rest: $$\sigma A = N' + u(A-a_c) \approx N'+uA \quad\Rightarrow\quad \sigma'=\frac{N'}{A}=\sigma-u$$ $\sigma'$ (the effective stress) is what actually controls soil strength and compressibility, since only the inter-granular contact forces resist shearing or squeeze the skeleton together — the pore pressure $u$ acts equally in all directions on the (nearly incompressible) grains and does no work compressing the skeleton.

(6.3) Capillary rise and unsaturated retention curves

In a thin capillary tube inserted into free water, surface tension $T_s$ at the curved air-water meniscus pulls water up until the vertical component of the tension force balances the weight of the risen column: $$h_c=\frac{4T_s\cos\alpha}{\gamma_w d}$$ where $d$ is the tube diameter and $\alpha$ the contact angle (near $0^\circ$ for a clean, water-wet tube). Smaller pores (smaller effective $d$) support a taller capillary rise and a larger negative (suction) pore pressure. Soil pores act as an irregular bundle of such capillaries: the soil-water characteristic (retention) curve plots degree of saturation against matric suction, and is essentially a distribution of the soil's effective pore sizes — fine-grained soils (small pores, high capillary suction) retain water at much higher suction than coarse sands, giving a retention curve that stays near full saturation over a much wider suction range before draining.

(6.4) Sand cone field compaction test

Given. Volume excavated $V=1165\text{ cm}^3$; wet mass $=2600$ g; dry mass $=1645$ g.

1. (a) Field dry density. $$\rho_d=\frac{M_{dry}}{V}=\frac{1645}{1165}=1.412\text{ g/cm}^3=\boxed{1412\text{ kg/m}^3}\ (\gamma_d=\rho_d g=13.85\text{ kN/m}^3)$$
2. (b) Field water content. $$w=\frac{M_{wet}-M_{dry}}{M_{dry}}=\frac{2600-1645}{1645}=\boxed{58.1\%}$$

(6.5) Groundwater table and total head components

The groundwater table (GWT, or water table) is the surface within a soil mass at which pore pressure equals atmospheric ($u=0$); below it the soil is (in the simplest static case) fully saturated with a hydrostatic pore-pressure distribution, and above it the soil is unsaturated (or, immediately above, can carry negative pore pressure by capillarity). For a 5 m sand layer with the GWT 1.5 m below the ground surface, total head $h_t$ is constant with depth below the GWT under static (no-flow) conditions — there is no hydraulic gradient to drive flow — while elevation head $z$ decreases and pressure head $h_p=u/\gamma_w$ increases linearly with depth below the GWT, so that $h_t=z+h_p$ stays fixed.

(6.6) Consistency, water content, and Atterberg limits

A clay's strength and stiffness fall steeply as its water content rises through the Atterberg limits. Below the plastic limit ($PL$), the soil is in the semi-solid/solid state: stiff, brittle, and relatively strong, with particle-to-particle contacts largely intact. Between $PL$ and the liquid limit ($LL$) the soil is plastic: it can be remoulded without cracking, but strength and stiffness both decrease steadily as $w$ increases toward $LL$, because the added water thickens the double layers separating clay platelets and reduces inter-particle attraction. Above $LL$, the soil behaves as a viscous liquid with negligible shear strength. The liquidity index $I_L=(w-PL)/(LL-PL)$ quantifies where a soil's current water content sits within this range, and correlates directly with undrained shear strength: $I_L$ near 0 (near $PL$) indicates a stiff, strong clay, while $I_L$ near 1 (near $LL$) indicates a soft, weak one.

(6.7) Falling head permeability test

Given. Specimen $D=5$ cm, $L=10$ cm; burette $d=0.5$ cm; head falls $h_1=125\text{ cm}\to h_2=115\text{ cm}$ in $t=35\text{ min}=2100$ s.

1. (a) Hydraulic conductivity. $$k=\frac{aL}{At}\ln\frac{h_1}{h_2},\qquad a=\frac{\pi}{4}(0.5)^2=0.1963\text{ cm}^2,\ A=\frac{\pi}{4}(5)^2=19.63\text{ cm}^2$$ $$k=\frac{0.1963\times10}{19.63\times2100}\ln\left(\frac{125}{115}\right)=\boxed{3.97\times10^{-6}\text{ cm/s}}$$
2. (b) Soil type. $k\approx4\times10^{-6}$ cm/s falls in the $10^{-5}$–$10^{-7}$ cm/s band typical of a low-permeability fine-grained soil — $\boxed{\text{a silt (or clayey silt), not a sand}}$; a falling-head (rather than constant-head) apparatus is itself standard practice specifically because sands drain too fast for a falling-head test to resolve.

(6.8) Phase relations from $w$, $e$, $G_s$

Given. $w=15\%$, $e=0.54$, $G_s=2.6$.

1. Dry density. $$\gamma_d=\frac{G_s\gamma_w}{1+e}=\frac{2.6\times9.81}{1.54}=\boxed{16.56\text{ kN/m}^3}\ (\rho_d=1688\text{ kg/m}^3)$$
2. Degree of saturation. $$S=\frac{wG_s}{e}=\frac{0.15\times2.6}{0.54}=\boxed{72.2\%}$$
3. Volumetric water content. $n=e/(1+e)=0.54/1.54=0.351$; $$\theta=Sn=0.722\times0.351=\boxed{25.3\%}$$