18-Geol-A6 Soil Mechanics · May 2016
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
National Exams — May 2016 — 04-Geol-A6 Soil Mechanics. Three-hour, closed-book exam; a Casio or Sharp approved calculator, compass and ruler are permitted. Six questions constitute the complete 100-mark paper: Questions 1–5 are compulsory; Question 6 offers five 5-mark optional items of which the source asks for three.
Reference texts: Das, Principles of Geotechnical Engineering, 9th ed. — USCS classification, phase relations, permeability/seepage, consolidation; Craig's Soil Mechanics (Craig & Knappett), 8th ed. — effective stress, seepage/flow nets, consolidation theory, shear strength.
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 discharge velocity through a saturated porous medium is proportional to the hydraulic gradient: $v=ki$, or as a total flow rate $q=vA=kiA$, 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 full cross-section $A$, not the true (faster) velocity within the pores. $k$ is the hydraulic conductivity (units of velocity, e.g. m/s), a property of both the soil (pore size/structure) and the permeant fluid.
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.
Consider a horizontal plane through a saturated soil mass, cut so that it passes through the grain-to-grain contact points rather than always through the pore water. Total normal force on a gross area $A$ is carried partly by the water pressure acting on the (much larger) non-contact area, and partly by the intergranular contact forces acting on the (very small) true contact area $A_c$: $\sigma A=\sigma'_{contact}A_c+u(A-A_c)$, where $\sigma'_{contact}A_c$ is defined as the effective stress force. Because $A_c\ll A$ for typical granular contacts, this reduces to Terzaghi's principle: $\boxed{\sigma=\sigma'+u}$, i.e. total stress equals effective (intergranular) stress plus pore water pressure.
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.
Surface tension $T_s$ at the air–water meniscus pulls water up a narrow tube against gravity until the vertical component of the tension force balances the weight of the raised column: $h_c=\dfrac{4T_s\cos\alpha}{\gamma_w d}$, where $d$ is the tube diameter and $\alpha$ the contact angle. Smaller tubes (smaller $d$) produce greater rise — the same principle explains capillary rise in soil pores, where the "tube diameter" is set by the pore size. In an unsaturated soil, this pore-size dependence is exactly what a water-retention (soil-water characteristic) curve captures: fine-grained soils (small pores) hold water at much higher suction (matric potential) for a given saturation than coarse-grained soils (large pores), and the "air-entry value" — the suction at which air first invades the largest pores — is the retention-curve analogue of $h_c$ for the coarsest connected pore throat.
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.
Given. $V=1165\text{ cm}^3$, $M_{wet}=2600$ g, $M_{dry}=1645$ g.
Find. (a) Field dry density; (b) field water content.
Approach. Bulk density from wet mass/volume, dry density from dry mass/volume, water content from the mass of water lost on drying.
| Quantity | Value |
|---|---|
| Field dry density, $\rho_d$ | 1.412 g/cm³ (13.85 kN/m³) |
| Field water content, $w$ | 58.1% |
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.
The groundwater table (water table) is the surface within the ground at which pore water pressure equals atmospheric (zero gauge) pressure — below it the soil is saturated and pore pressure is positive (hydrostatic); above it (the vadose/unsaturated zone) pressure is at or below atmospheric. Taking the datum at the base of the 5 m sand layer: above the water table (depth 0–1.5 m) pressure head is taken as zero, so total head equals elevation head and decreases linearly toward the water table. Below the water table (depth 1.5–5 m), hydrostatic equilibrium (no vertical flow) means total head is constant with depth, equal to the water table's own elevation above the datum (3.5 m); pressure head increases linearly with depth below the water table exactly enough to offset the falling elevation head.
