18-Geol-A6 Soil Mechanics · May 2015
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
National Exams — May 2015 — 04-Geol-06, Soil Mechanics. Three-hour, closed-book exam; one of two approved calculators permitted, plus a compass and ruler. Candidates must answer Questions 1 to 5 and choose three of the five sub-questions in Question 6 (5 marks each); all five Question-6 parts are answered here as a complete study resource.
Reference texts: Das, Principles of Geotechnical Engineering — USCS classification (Q1), phase relations (Q2), consolidation theory (Q4), Boussinesq/Newmark stress-influence factors (Q4), flow nets and seepage (Q5), sand-cone compaction (Q6d); Craig, Craig's Soil Mechanics — Mohr-circle shear strength and limit-equilibrium slope stability (Q3), effective stress and capillarity (Q6b, Q6c); EGBC Geoscience Professional Practice Guidelines for assumption-disclosure conventions.
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.
Darcy's law states that the seepage discharge through a saturated soil is directly proportional to the hydraulic gradient driving it:
$$v=ki,\qquad q=vA=kiA$$Here $v$ is the discharge velocity (a superficial, cross-section-averaged velocity, not the true pore velocity, since flow only occurs through the void space), $k$ is the coefficient of permeability (hydraulic conductivity) of the soil, and $i=\Delta h/L$ is the hydraulic gradient — the head loss $\Delta h=h_1-h_2$ between two points divided by the flow-path length $L$ between them, as shown in the figure. $A$ is the total (gross) cross-sectional area of soil perpendicular to flow, and $q$ is the total volumetric discharge. Darcy's law is empirical but is now understood as the low-Reynolds-number (laminar) limit of flow through a porous medium, valid for essentially all groundwater and seepage problems in soils, and it underlies every seepage, consolidation and flow-net calculation in this paper.
Consider a horizontal plane cutting through a saturated soil mass at some points of grain-to-grain contact and elsewhere through the pore water. The total normal stress $\sigma$ on this plane is the total weight of everything above it (soil solids + water) divided by the gross area $A$. This stress is carried by two mechanisms in parallel: the pore water pressure $u$, acting equally in all directions on the (much larger) area of water at the plane, and the sum of the inter-granular contact forces $N'$, acting over the (much smaller) actual solid-to-solid contact area $A_c$. Summing vertical force equilibrium over the gross area $A$:
$$\sigma A = \sum N' + u(A-A_c)$$Dividing through by $A$ and defining the effective stress as the average inter-granular force per unit gross area, $\sigma'=\sum N'/A$:
$$\sigma = \sigma' + u\left(1-\frac{A_c}{A}\right)$$Since the actual contact area $A_c$ is experimentally found to be a negligibly small fraction of the gross area $A$ for real soils (contacts are near-point contacts), $A_c/A\approx0$, giving Terzaghi's effective stress principle:
$$\boxed{\sigma' = \sigma - u}$$This is the single most important relation in soil mechanics: strength and volume-change behaviour of soil depend only on $\sigma'$, the stress actually carried grain-to-grain, not on the total stress $\sigma$ alone — which is why raising the pore pressure $u$ (holding $\sigma$ fixed) can trigger shear failure or excess settlement even with no change in the external load.
When a fine glass tube is inserted into a free water surface, surface tension $T_s$ acting around the tube's wetted perimeter pulls water up the tube against gravity, forming a curved meniscus. Vertical equilibrium of the suspended water column (weight balanced by the vertical component of the surface-tension force around the circumference) gives
$$h_c=\frac{2T_s\cos\alpha}{r\gamma_w}$$where $\alpha$ is the contact angle (near $0^{\circ}$ for a clean glass–water interface) and $r$ is the tube radius: the narrower the tube, the higher the rise. In an unsaturated soil, the pore network behaves like a bundle of capillary tubes of varying effective radius (controlled by the pore/grain size), so the same physics produces matric suction — a negative pore pressure that holds water in the soil above the free water table. Because a real soil has a continuous distribution of pore sizes rather than one uniform radius, the relationship between suction and the resulting degree of saturation is not a single rise height but a continuous curve: the soil–water characteristic curve (SWCC). At low suction, nearly all pores (large and small) remain saturated; as suction increases, progressively smaller pores are able to resist drainage, so the largest pores desaturate first and saturation falls gradually, giving the SWCC's characteristic S-shape (steep desaturation once the air-entry suction of the coarsest pores is exceeded, flattening again as only the finest pores remain water-filled).
Given. Excavated hole volume $V=1165\text{ cm}^3$; wet soil mass $=2230$ g; dry soil mass $=1852$ g.
Find. (a) Field dry density. (b) Field water content.
| Quantity | Value |
|---|---|
| Field dry density | 1.590 g/cm³ (15.6 kN/m³) |
| Field water content | 20.4% |
The groundwater table (water table, phreatic surface) is the level in a soil deposit at which pore water pressure equals atmospheric pressure ($u=0$); below it the soil is saturated and pore pressure increases hydrostatically with depth, while above it (in the absence of significant capillary rise) the soil is unsaturated and pore pressure is taken as zero. For a 10 m thick sand layer with the water table 2 m below the ground surface, taking the elevation datum $z=0$ at the base of the layer: the ground surface is at $z=10$ m and the water table at $z=8$ m.
Below the water table, in a static (no-flow) groundwater condition, total head is constant with depth and equal to the elevation of the water table itself, $h_{total}=z_{WT}=8$ m; elevation head $z$ decreases linearly with depth while pressure head $\psi=h_{total}-z$ grows correspondingly, reaching $\psi=8$ m at the base of the layer ($z=0$). Above the water table, pressure head is taken as zero (ignoring capillary rise for this plot), so total head simply equals elevation head and varies from $8$ m (at the water table) up to $10$ m (at the ground surface) — this region is not part of the same hydrostatically-connected system, which is exactly why total head is not continuous across the water table in this simplified picture.