Question 6 of 6: General Questions

Question 6: General Questions (5 marks each)

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.

(a) Darcy's law

Darcy's law states that the seepage discharge velocity through a saturated soil is directly proportional to the hydraulic gradient: $$v=ki,\qquad q=vA=kiA$$ where $v$ is the discharge (Darcy) velocity, $k$ is the coefficient of permeability (hydraulic conductivity) of the soil, $i=\Delta h/L$ is the hydraulic gradient (head loss per unit length of flow path), $q$ is the total volumetric flow rate, and $A$ is the total cross-sectional area of soil (solids + voids) perpendicular to flow. Because $v$ is defined over the total area rather than just the pore area, the true average velocity of water through the pores (the seepage velocity) is higher: $v_s=v/n$, where $n$ is porosity.

(b) Water content, strength/stiffness, and Atterberg limits

A clay soil's mechanical behaviour changes fundamentally as its water content crosses the Atterberg limits, which mark transitions between four physical states. Below the shrinkage limit ($w_S$), the soil is a brittle solid: it is stiff and strong, but fails suddenly with little deformation, and further drying causes no further volume change. Between $w_S$ and the plastic limit ($w_P$), the soil is a semi-solid: still relatively stiff and strong but capable of some plastic deformation before cracking. Between $w_P$ and the liquid limit ($w_L$), the soil is plastic: it can be remoulded without cracking, and both strength and stiffness decrease steadily as water content rises toward $w_L$, because the added water increasingly separates and lubricates the clay particles, weakening interparticle (van der Waals and diffuse double-layer) bonds. Above $w_L$, the soil behaves as a viscous liquid with negligible shear strength.

The liquidity index $I_L=(w-w_P)/(w_L-w_P)$ quantifies where a soil's current water content sits within this plastic range: $I_L\approx0$ (near $w_P$) indicates a stiff, high-strength consistency, while $I_L\approx1$ (near $w_L$) indicates a soft, low-strength consistency close to the point of losing all shear strength. A soil with $I_L>1$ (water content above $w_L$) is a sensitive or "quick" clay that can lose most of its strength if disturbed, since its natural water content already exceeds the liquid limit measured on a remoulded sample.

(c) Falling head permeability test

Given. Specimen diameter $d=5\text{ cm}$, height $L=10\text{ cm}$; burette diameter $d_a=5\text{ mm}$; $h_1=1.25\text{ m}$, $h_2=1.15\text{ m}$; $t=35\text{ min}$.

Find. (i) Hydraulic conductivity $k$. (ii) Soil type consistent with that $k$.

Approach. Apply the falling-head formula directly, using the burette (standpipe) area $a$ and the specimen area $A$.

1. Areas and time. $$a=\frac{\pi}{4}(0.5\text{ cm})^2=0.1963\text{ cm}^2,\qquad A=\frac{\pi}{4}(5\text{ cm})^2=19.63\text{ cm}^2,\qquad t=35(60)=2100\text{ s}$$
2. Falling-head conductivity. $$k=\frac{aL}{At}\ln\frac{h_1}{h_2}=\frac{(0.1963)(10)}{(19.63)(2100)}\ln\frac{125}{115}=\boxed{3.97\times10^{-6}\text{ cm/s}}$$
3. Part (ii) — soil type. A conductivity of order $10^{-6}\text{ cm/s}$ falls within the typical range for silt / silty clay (roughly $10^{-3}$ to $10^{-7}\text{ cm/s}$), well below the range for clean sand ($10^{-1}$ to $10^{-3}\text{ cm/s}$) and above the range for a well-sealed, low-permeability intact clay ($<10^{-7}$ to $10^{-9}\text{ cm/s}$) — the falling-head method itself (rather than a constant-head test) confirms this is a fine-grained, low-permeability soil, since falling-head is the appropriate test for $k$ below about $10^{-4}\text{ cm/s}$.

(d) Phase relations from water content, void ratio and $G_s$

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

Find. Dry density $\rho_d$, volumetric water content $\theta$, degree of saturation $S$.

1. Dry density. $$\rho_d=\frac{G_s\rho_w}{1+e}=\frac{2.6(1.0)}{1.54}=\boxed{1.69\text{ g/cm}^3}\ (\gamma_d=16.6\text{ kN/m}^3)$$
2. Porosity and degree of saturation. From $Se=wG_s$: $$n=\frac{e}{1+e}=\frac{0.54}{1.54}=\boxed{0.351}\ (35.1\%),\qquad S=\frac{wG_s}{e}=\frac{0.15(2.6)}{0.54}=\boxed{0.722}\ (72.2\%)$$
3. Volumetric water content. $$\theta=nS=0.351(0.722)=\boxed{0.253}\ (25.3\%)$$