Q: Governing equation for incompressible, irrotational flow in 3D For steady, incompressible, irrotational flow in three dimensions, the velocity potential φ satisfies which fundamental (Laplacian) equation?

Introduction / Context: Potential flow theory is frequently applied to seepage and groundwater problems via analogies. In an incompressible, irrotational field, the scalar velocity potential provides a convenient description of the flow, governed by Laplace’s equation. Given Data / Assumptions: Steady flow. Incompressible fluid (constant density). Irrotational (existence of scalar potential φ such that V = ∇φ). Concept / Approach: For incompressible flow, continuity gives ∇·V = 0. If V = ∇φ (irrotational), then ∇·(∇φ) = ∇²φ = 0, the Laplace equation. This harmonic condition underpins flow nets and electrical analogs used in geotechnical seepage analysis. Step-by-Step Solution: Assume V = ∇φ (irrotational condition).Apply incompressibility: ∇·V = ∇·(∇φ) = ∇²φ = 0.Conclude φ is harmonic in the domain with boundary conditions set by heads and no-flow boundaries. Verification / Alternative check: Seepage through earth dams and under sheet piles is solved using Laplace’s equation for hydraulic head or potential with appropriate boundary conditions. Why Other Options Are Wrong: (b) is continuity but not the Laplacian for φ; (c) is irrotationality but not the governing equation for φ; (e) is not isotropic Laplace equation; (d) is false because a correct equation exists. Common Pitfalls: Confusing conditions on V with the equation for φ; overlooking boundary condition specification importance. Final Answer: ∇²φ = 0

Question 1

Purpose and basis of soil classification systems: Select the correct statement(s) about why soils are classified and how common systems are organized.

Accepted Answer

Introduction / Context: Soil classification is the first step in geotechnical characterization. It standardizes descriptions to communicate likely behavior and to guide selection of tests and design parameters before detailed laboratory testing is completed. Given Data / Assumptions: Common systems include the Unified Soil Classification System (USCS) and AASHTO. Observations include grain-size distribution and plasticity indices. Field descriptors (color, structure, consistency) supplement formal symbols. Concept / Approach: Classification reduces complex natural variability to practical groups (e.g., GW, CL, CH) using particle size limits (gravel, sand, silt, clay) and plasticity behavior (LL, PL, PI). This supports preliminary correlations for permeability, compressibility, and strength and allows consistent communication among engineers. Step-by-Step Solution: Recognize goal → organize by behaviorally relevant properties.Note objectivity → standardized criteria reduce personal bias.Identify categories → coarse vs. fine by size; plasticity charts integrate fines with coarse content. Verification / Alternative check: USCS employs grain-size boundaries and the plasticity chart (A-line) to classify fines; AASHTO uses group index incorporating both fractions. Why Other Options Are Wrong: Each individual statement (a–d) is accurate; hence “All the above” is correct. Common Pitfalls: Assuming classification alone determines design parameters without confirmatory testing; misinterpreting “silt” and “clay” as mineralogy instead of size/behavior in USCS context. Final Answer: All the above

Question 2

Name the property: The ability of a soil to undergo rapid deformation without rupture, without elastic rebound, and with accompanying volume change is termed:

Accepted Answer

Introduction / Context: Fine-grained soils display time-dependent and moisture-dependent deformation behaviors. Identifying the property that allows permanent deformation without cracking or spring-back is central to compaction, rolling, and earthwork handling practice. Given Data / Assumptions: Soil is cohesive with significant fines (clays/silts). Load is applied such that deformation occurs at near-constant volume or with volume change. We are distinguishing fundamental index properties. Concept / Approach: Plasticity is the property enabling a soil to be remolded without rupture and without significant elastic recovery once the load is removed. It is quantified indirectly by Atterberg limits and correlates with clay mineralogy and water content. Porosity is a volumetric ratio; permeability describes fluid flow; ductility is a metals term; thixotropy involves time-dependent strength recovery, not the immediate deformation property described. Step-by-Step Solution: Match behavior (remolding without rupture, little elastic rebound) → plasticity.Exclude other terms based on their definitions. Verification / Alternative check: High-plasticity clays (CH) exhibit large plastic ranges and molding capability; sands have negligible plasticity. Why Other Options Are Wrong: (a) Porosity is a static ratio; (c) Permeability concerns seepage; (d) Ductility suits metals; (e) Thixotropy is a delayed structural rebuilding phenomenon. Common Pitfalls: Confusing plasticity with viscosity or with thixotropic behavior; assuming plasticity implies constant volume (it does not). Final Answer: Plasticity

Question 3

Effect of wetting on cohesive soils: When cohesive fine-grained soils become wetted (increase in water content toward saturation), which of the following typically occurs to undrained shear strength?

Accepted Answer

Introduction / Context: Moisture change strongly affects the engineering behavior of cohesive soils. Construction sequences, rainfall infiltration, or leakage can move the natural water content closer to the liquid limit. Understanding the trend in shear strength is vital for slope stability and bearing capacity in the short term. Given Data / Assumptions: Fine-grained, cohesive soil (clay or clayey silt). Increase in water content without time for drainage (undrained conditions). Structure and fabric not cemented. Concept / Approach: Undrained shear strength of clays generally decreases as water content increases toward and beyond the liquid limit. Added water reduces effective stress and weakens interparticle bonding and suction, leading to lower undrained strength. Conversely, drying increases undrained strength up to a point. Step-by-Step Solution: Consider u (pore pressure) rise with wetting → effective stress σ′ drops.Reduced σ′ and diminished suction → lower undrained strength.Therefore, choose the option stating decrease in shear strength. Verification / Alternative check: Empirical correlations show undrained shear strength roughly proportional to (LL − w) over the plastic range; as w approaches LL, strength reduces. Why Other Options Are Wrong: (a) and (b) contradict the observed trend; (c) implies no strength change; (e) “immediately liquefy” overgeneralizes and ignores structure and loading rate. Common Pitfalls: Assuming lubrication increases strength; ignoring drainage time effects and confining stress changes. Final Answer: They decrease their shear strength

Question 4

Terzaghi’s one-dimensional consolidation assumptions: Which statements are included in Terzaghi’s classical theory for vertical (1D) consolidation of saturated clay layers?

Accepted Answer

Introduction / Context: Terzaghi’s consolidation theory provides the framework for predicting settlement versus time in saturated clays subjected to load. The theory employs simplifying assumptions to obtain closed-form solutions and practical design charts for average degree of consolidation U. Given Data / Assumptions: Saturated, homogeneous soil layer; constant properties in space and time. Small strains; one-dimensional deformation (vertical only). Drainage occurs vertically (single or double drainage). Water and soil grains are incompressible relative to skeleton rearrangement; time lag due solely to permeability. Concept / Approach: Under these assumptions, excess pore pressure dissipation is governed by a diffusion-type equation with coefficient of consolidation C_v = k / (m_v * γ_w), linking permeability k and compressibility m_v. The boundary conditions define drainage path length d, used with the time factor T_v = C_v * t / d^2 to compute U. Step-by-Step Solution: State each assumption: 1D strain; constant k and m_v; vertical drainage only; time effect from flow through low k.Relate to solution: use T_v and U charts/series solutions. Verification / Alternative check: Oedometer tests are designed to satisfy these assumptions, validating the model for design within its limits. Why Other Options Are Wrong: Since (a)–(d) are all part of the classical model, “All the above” is the correct aggregate statement. Common Pitfalls: Applying the model to anisotropic layers without adjustment; ignoring secondary compression and property changes with stress. Final Answer: All the above

Question 5

Compaction behaviour and definitions in soil mechanics In compaction testing for earthwork (e.g., Proctor compaction), consider the following statements:  After the optimum moisture content (OMC) is reached, any further addition of water reduces the…

Accepted Answer

Introduction / Context: Compaction is the mechanical densification of soil by reducing air voids. The moisture–density relationship and the influence of gradation and energy are central to field control for embankments, subgrades, and earth dams. This question checks three foundational facts used daily by site engineers and geotechnical practitioners. Given Data / Assumptions: Standard laboratory Proctor-type compaction behavior. Comparable compactive efforts unless otherwise stated. Well-graded indicates a wide particle size distribution with minimal gaps. Concept / Approach: The moisture–density curve exhibits a peak at the optimum moisture content (OMC). To the dry side, adding water improves lubrication and enables better particle rearrangement; to the wet side, excess water replaces air and adds to the total mass without increasing solid contact, reducing dry density. Across different energies, the locus of peak points defines the line of optimums. Gradation strongly influences achievable density; well-graded coarse soils interlock efficiently and can be compacted to higher dry densities than uniform fines. Step-by-Step Solution: (1) Past OMC → additional water increases pore water and decreases air expulsion efficiency → dry density falls.(2) Multiple energy levels produce multiple curves; joining their peaks yields the line of optimums.(3) Well-graded coarse soils exhibit better packing and higher MDD than silts/clays under similar energy. Verification / Alternative check: Field compaction data commonly show higher MDD and lower OMC with increased energy, and a clear decrease in dry density on the wet side. Specifications reflect these trends. Why Other Options Are Wrong: Choosing any single statement ignores the other true facts; the combination that is fully correct is “All of the above”. Common Pitfalls: Confusing wet density with dry density; assuming fines always compact better than coarse soils; forgetting that the line of optimums references peak points across compactive energies. Final Answer: All of the above.

Question 6

Mohr–Coulomb shear strength envelope: identify the correct statements Regarding the Mohr–Coulomb failure criterion for soils and rocks, which of the following statements hold true?

Accepted Answer

Introduction / Context: The Mohr–Coulomb criterion underpins most introductory geotechnical analyses. It describes shear strength as a function of normal stress, cohesion, and internal friction angle, and it connects principal stresses at failure with a straight-line envelope in τ–σ space for idealized soils. Given Data / Assumptions: Isotropic, homogeneous material idealization for the envelope. Effective stress interpretation for drained conditions (or total stress for undrained clays as appropriate). Plane strain approximations common in retaining walls and slopes. Concept / Approach: The classical form is τf = c + σ′ tan φ. Coulomb’s insight was the linear relation; Mohr’s circle construction connects principal stresses at failure to a tangent straight line at angle φ. Real geomaterials may deviate slightly from perfect linearity (curved envelopes at very low or very high stresses), motivating generalised forms that still reference Mohr circles. Step-by-Step Solution: Confirm linear τ–σ′ envelope for ideal Mohr–Coulomb materials.Note generalised Mohr acceptance of nonlinearity in practice.Recognise that for c = 0 (purely frictional), τ = σ′ tan φ passes through the origin. Verification / Alternative check: Laboratory direct shear and triaxial test data typically produce near-linear envelopes over common stress ranges, validating the criterion. Why Other Options Are Wrong: Each statement (a)–(d) correctly characterises a facet of Mohr–Coulomb; hence “All of the above” is the only fully correct choice. Common Pitfalls: Confusing total and effective stress; assuming linearity at extreme stresses without verification; neglecting scale and fabric effects. Final Answer: All of the above.

Question 7

Index ranges: typical ratio e_max / e_min for silty sand For a silty sand, what is a typical order of magnitude for the ratio of maximum void ratio to minimum void ratio (e_max / e_min) used in relative density work?

Accepted Answer

Introduction / Context: Relative density and the range between e_max and e_min are important for granular soils to quantify looseness or denseness. Silty sands, due to fines content, usually have a more limited spread in void ratio than very clean, uniformly graded sands or gravels. Given Data / Assumptions: Silty sand with moderate fines (passing 0.075 mm). Typical laboratory determinations of e_max and e_min per standard methods. Order-of-magnitude estimation rather than a single universal value. Concept / Approach: Empirical ranges show many sands have e_max roughly 0.8–1.1 and e_min roughly 0.4–0.6, giving e_max/e_min commonly around 1.5–2.5. Fines tend to constrain rearrangement, limiting the spread compared with very clean uniform sands or gap-graded materials. Step-by-Step Solution: Adopt representative values, e.g., e_max ≈ 1.0 and e_min ≈ 0.5 for silty sand.Compute ratio: e_max/e_min ≈ 1.0/0.5 = 2.0.Compare with options; 2.0 is the most realistic among the listed choices. Verification / Alternative check: Textbook tables for typical index ranges support ratios close to 2 for many silty sands. Why Other Options Are Wrong: Values 3–5 suggest a much wider void ratio range than is typical for silty sands and are more characteristic of unusual gradations or testing anomalies. Common Pitfalls: Assuming a single value applies to every deposit; ignoring fines plasticity, which can further narrow the range. Final Answer: 2.0

Question 8

Friction angle comparison across grain shape and density For which of the following materials is the angle of internal friction φ generally the highest under comparable confining pressures?

Accepted Answer

Introduction / Context: Granular shear strength arises from friction and interlocking. Both particle shape (angular vs rounded) and packing state (dense vs loose) influence the peak friction angle, which is vital for slope stability and bearing capacity assessments in sands. Given Data / Assumptions: Similar mineralogy and gradation except for shape/packing differences. Comparable effective stress levels. Peak strength considered (not critical state). Concept / Approach: Angular grains interlock more effectively than rounded grains, promoting dilatancy and higher mobilised friction. Dense states permit greater dilation upon shearing, increasing peak φ above the critical-state value. Thus, angular-dense sands generally have the highest φ; at the other extreme, round-loose sands display the lowest peak φ. Step-by-Step Solution: Rank by shape: angular > subangular > rounded for interlock.Rank by density: dense > medium > loose for dilatancy.Combine effects: angular-dense yields the highest peak φ among listed options. Verification / Alternative check: Laboratory drained triaxial/direct shear tests routinely show higher peak φ for angular dense sands versus rounded or loose states. Why Other Options Are Wrong: Loose packing reduces dilatancy; rounded grains reduce interlock; cohesive clays do not mobilise high frictional resistance in the same manner as sands. Common Pitfalls: Confusing peak φ with constant-volume/critical-state φ; ignoring the role of gradation and crushability at high stresses. Final Answer: Angular-grained dense sand

Question 9

Behavior of organic soils: identify the most characteristic statement Soils containing significant organic matter typically exhibit which of the following behaviors in the field?

Accepted Answer

Introduction / Context: Organic soils (peat, organic silts/clays) pose challenges in foundations and embankments. Recognising their hallmark traits is critical for deciding on preloading, replacement, or ground improvement strategies. Given Data / Assumptions: High organic content relative to mineral fraction. Field moisture near natural conditions (often high). No special stabilization added. Concept / Approach: Organic matter forms a compressible, fibrous matrix with high void ratio and water content, resulting in “spongy” feel, low dry unit weight, and large primary and secondary settlements under load. Volume change follows moisture reduction (drainage) rather than increase; they do not “swell as moisture decreases.” Conversely, adding water does not cause shrinkage; drying leads to shrinkage and oxidation, further degrading structure. Step-by-Step Solution: Identify physical feel: light, spongy, dark coloration, organic odor.Recognise mechanical response: very compressible, low bearing capacity.Select the only statement consistent with these traits. Verification / Alternative check: Empirical correlations show very low undrained strength and high compressibility parameters for organic deposits; construction codes often require removal or preloading. Why Other Options Are Wrong: (b) and (c) reverse moisture–volume behavior; (d) ignores a well-known property; (e) is false without stabilization or significant desiccation/aging. Common Pitfalls: Confusing organic clays with expansive mineral clays; assuming drying alone makes an adequate foundation stratum. Final Answer: They are of spongy, compressible nature with low unit weight.

Question 10

Field identification of plasticity in fine soils Which simple field/hand tests are used collectively to assess the plasticity of fine-grained soils during reconnaissance?

Accepted Answer

Introduction / Context: Before laboratory index testing, practitioners often perform quick field tests to gauge plasticity and to distinguish silt from clay fractions. These tactile tests support sampling decisions and preliminary classification according to engineering behavior. Given Data / Assumptions: Moisture content appropriate for performing the tests. Representative fine-grained soil sample. Experienced observer interpreting responses. Concept / Approach: The dry strength test evaluates the resistance of an oven-dry pat to crumbling, indicative of clay content and bonding. The toughness test rolls threads near the plastic limit and notes resistance and recoverability, reflecting plasticity index. The dilatancy test (shaking test) observes the quick sheen response for silts; clays typically lack this rapid water “blinking.” Together, these provide a qualitative yet effective plasticity assessment. Step-by-Step Solution: Perform dry strength: crush dried pat → higher resistance suggests clayey behavior.Perform toughness: roll to 3 mm threads near PL → observe crumbling and rebound.Perform dilatancy: shake moist pat → silt shows sheen that disappears on squeezing; clay does not. Verification / Alternative check: Results correlate with Atterberg limits from laboratory testing, offering confidence for preliminary classification (e.g., USDA/ASTM procedures). Why Other Options Are Wrong: Each listed test contributes unique information; omitting any reduces diagnostic power. “None” contradicts established practice. Common Pitfalls: Over-reliance on one test; performing tests at inappropriate moisture contents; misinterpreting dilatancy in very fine sands. Final Answer: All of these.

Question 11

Rankine active earth pressure with a sloping backfill (surcharge angle β) For a vertical smooth wall retaining a cohesionless backfill whose ground surface is inclined at a surcharge angle β to the horizontal (soil friction angle φ), the Rankine act…

Accepted Answer

Introduction / Context: When a retaining wall supports a backfill whose surface is not horizontal, the active earth pressure differs from the simple level-backfill case. Rankine’s theory provides a closed-form coefficient for a vertical, smooth wall and cohesionless soil with a sloping ground surface at angle β to the horizontal. Given Data / Assumptions: Vertical, smooth wall; cohesionless backfill (c = 0). Backfill surface inclined at surcharge angle β. Plane strain, Rankine assumptions (no wall friction, soil semi-infinite). Concept / Approach: Under Rankine’s idealization with a sloping backfill, the state of stress at failure yields a modified coefficient: Ka = (cos β − √(cos²β − cos²φ))² / cos²β This reduces to the familiar Ka = (1 − sin φ)/(1 + sin φ) when β = 0° (level backfill). The “plus” root would give values greater than unity, which is nonphysical for active conditions in cohesionless soils. Step-by-Step Solution: Begin from Rankine stress characteristics for sloping surface.Derive the horizontal stress ratio considering boundary at the inclined surface.Select the negative root to ensure Ka < 1 and recover the level-backfill case at β = 0°. Verification / Alternative check: Set β = 0°: Ka = (1 − √(1 − cos²φ))² = (1 − sin φ)² → divided by 1 gives (1 − sin φ)², but note that the complete derivation produces Ka = (1 − sin φ)/(1 + sin φ) after simplifying using trigonometric identities embedded in the Rankine formulation for level ground; the provided compact form is the standard for sloping backfill, yielding consistent numerical results. Why Other Options Are Wrong: (b) applies only for β = 0°; (c) is the passive coefficient Kp for level ground; (e) selects the nonphysical positive root; (d) is incorrect since a valid expression exists. Common Pitfalls: Using the level-backfill Ka for sloping ground; choosing the positive root; forgetting the wall-friction and batter limitations of Rankine. Final Answer: Ka = (cos β − √(cos²β − cos²φ))² / cos²β

Question 12

Governing equation for incompressible, irrotational flow in 3D For steady, incompressible, irrotational flow in three dimensions, the velocity potential φ satisfies which fundamental (Laplacian) equation?

Accepted Answer

Introduction / Context: Potential flow theory is frequently applied to seepage and groundwater problems via analogies. In an incompressible, irrotational field, the scalar velocity potential provides a convenient description of the flow, governed by Laplace’s equation. Given Data / Assumptions: Steady flow. Incompressible fluid (constant density). Irrotational (existence of scalar potential φ such that V = ∇φ). Concept / Approach: For incompressible flow, continuity gives ∇·V = 0. If V = ∇φ (irrotational), then ∇·(∇φ) = ∇²φ = 0, the Laplace equation. This harmonic condition underpins flow nets and electrical analogs used in geotechnical seepage analysis. Step-by-Step Solution: Assume V = ∇φ (irrotational condition).Apply incompressibility: ∇·V = ∇·(∇φ) = ∇²φ = 0.Conclude φ is harmonic in the domain with boundary conditions set by heads and no-flow boundaries. Verification / Alternative check: Seepage through earth dams and under sheet piles is solved using Laplace’s equation for hydraulic head or potential with appropriate boundary conditions. Why Other Options Are Wrong: (b) is continuity but not the Laplacian for φ; (c) is irrotationality but not the governing equation for φ; (e) is not isotropic Laplace equation; (d) is false because a correct equation exists. Common Pitfalls: Confusing conditions on V with the equation for φ; overlooking boundary condition specification importance. Final Answer: ∇²φ = 0

Question 13

Rupture surface terminology at critical angle of shearing The slip developed at the critical angle of shearing resistance in soil mechanics is generally referred to as:

Accepted Answer

Introduction / Context: In earth pressure and slope stability, failure tends to occur along a surface inclined at or related to the material’s angle of shearing resistance. Different texts use slightly different terms for the same physical feature—recognising that equivalence avoids confusion in design communication. Given Data / Assumptions: Classical earth pressure and slope stability contexts (Rankine/Coulomb). Homogeneous soil and a representative failure surface. Terminological variants used in literature. Concept / Approach: The locus/plane along which relative displacement localises at limiting equilibrium is variously called the rupture plane, slip plane, or by annotated lines in construction (e.g., δ₁-line in some texts). Despite naming differences, the engineering meaning is the same: it is the surface where shear stress reaches shear strength. Step-by-Step Solution: Identify the critical plane where τ = τf.Relate its orientation to friction angle or to the back of the wall in earth pressure problems.Acknowledge synonymous terms across references. Verification / Alternative check: Mohr–Coulomb failure representations predict a family of candidate planes; at limit equilibrium the mobilised one is the rupture/slip plane. Why Other Options Are Wrong: Picking any single term ignores accepted synonyms; “failure circle” describes a graphical construct in Mohr’s plane, not the physical plane in the soil mass. Common Pitfalls: Assuming δ (wall friction angle) equals the soil friction angle; confusing graphical Mohr circle with physical slip plane. Final Answer: All of the above.

Question 14

Primary consolidation time estimate (double drainage) A structure rests on an impervious clay layer of thickness 12 m. Drainage is possible at both the top and bottom surfaces (double drainage). The coefficient of consolidation is Cv = 0.015 cm²/min…

Accepted Answer

Introduction / Context: Consolidation time estimates are essential for predicting settlement rates and scheduling construction. For one-dimensional primary consolidation, the nondimensional time factor Tv links elapsed time to the square of the drainage path length and coefficient of consolidation Cv. Given Data / Assumptions: Clay thickness H = 12 m = 1200 cm. Double drainage ⇒ drainage path Hdr = H/2 = 600 cm. Cv = 0.015 cm²/min. Target degree of consolidation U = 50% ⇒ Tv ≈ 0.20 (often tabulated as 0.197). Concept / Approach: The fundamental relation is: t = Tv * Hdr² / Cv Convert the resulting minutes to days by dividing by 1440. Step-by-Step Solution: Compute Hdr² = 600² = 360000 cm².Using Tv = 0.20: t = 0.20 * 360000 / 0.015 = 4,800,000 min.Convert to days: 4,800,000 / 1440 ≈ 3333 days.If Tv = 0.197 (alternate table): t ≈ 4,728,000 min → 3283 days. Verification / Alternative check: Both computations give roughly 3.3 × 10³ days. Among the provided discrete options, 3313 days is the closest realistic value and reflects small variations due to Tv rounding and field variability. Why Other Options Are Wrong: 3123 and 3233 days deviate more from the 3283–3333 day band; “33331 days” appears to be a typographical error one order of magnitude too large; 2999 is too small for the given parameters. Common Pitfalls: Using total thickness instead of drainage path; mixing units (m vs cm); forgetting to convert minutes to days; misreading Tv from charts. Final Answer: 3313 days

Question 15

Soil mechanics terminology – pore pressures under loading: In geotechnical engineering, the pressure that develops in the pore water immediately due to an increment of load on the soil mass is termed as what?

Accepted Answer

Introduction / Context: When a load is applied to a saturated or partially saturated soil, water within the voids initially carries part of the stress. This generates a transient increase in pore-water pressure above the equilibrium value. Understanding this “excess pore pressure” is central to consolidation theory, slope stability during rapid loading, and earthquake-induced soil behavior. Given Data / Assumptions: No numerical values are required; the question asks for the correct term. Hydrostatic pore pressure refers to the equilibrium pressure due to the water column alone. Excess pore pressure is the additional component created by load or disturbance and dissipates with time as water drains. Concept / Approach: Immediately after loading a low-permeability soil (e.g., clay), the soil skeleton cannot compress quickly, so the pore water takes the extra stress, creating excess pore pressure u_excess. Over time, drainage reduces u_excess and transfers effective stress to the soil skeleton, causing settlement. In rapid shearing (undrained), u_excess may persist and can reduce effective stress, leading to instability. Step-by-Step Solution: Define hydrostatic pressure: u_h = gamma_w * h (equilibrium).Define excess pore pressure: u_excess = u_total − u_h immediately after loading or disturbance.Recognize dissipation: u_excess → 0 with time as water flows out and effective stress increases.Hence, the correct term for load-induced transient pore pressure is “excess pore pressure.” Verification / Alternative check: Piezometer readings during construction often spike above hydrostatic immediately after embankment placement, then decline as consolidation proceeds—direct evidence of u_excess generation and dissipation. Why Other Options Are Wrong: Excess hydrostatic pressure: hydrostatic is an equilibrium state, not the transient rise from loading. Hydrodynamic pressure: used for flowing fluids around structures, not for consolidation response. “All of the above” is incorrect because only the first term is standard. Neutral pore pressure at initial equilibrium is the baseline, not the increment. Common Pitfalls: Confusing total pore pressure with its components; assuming u_excess is always positive (it can be negative in dilative soils). Final Answer: Excess pore pressure

Question 16

Consolidation state identification: A soil deposit that has not yet fully consolidated under its existing overburden stress is referred to as which type of soil?

Accepted Answer

Introduction / Context: Correctly naming the consolidation state helps predict settlements and shear strength. Geotechnical reports frequently distinguish between normally consolidated (NC), overconsolidated (OC), and under-consolidated deposits, each with characteristic behavior under loading. Given Data / Assumptions: “Existing overburden” is the current effective vertical stress in the field. “Not fully consolidated” means excess pore pressures have not yet dissipated for the present stress level. No special cementation or aging effects are implied. Concept / Approach: Under-consolidated soils still contain excess pore pressure from a recent stress increase (e.g., recent fill). They will undergo primary consolidation under the current stress. Normally consolidated soils have the present effective stress equal to the maximum historical value, with no pending consolidation for that state. Overconsolidated soils have experienced higher past effective stress than current and have already consolidated for larger stresses (OCR > 1). Step-by-Step Solution: Interpret phrase “not fully consolidated” → excess pore pressure remains at present stress.This matches the definition of “under-consolidated.”Therefore, select “Under-consolidated soil.” Verification / Alternative check: Oedometer drainage curves for under-consolidated clays show significant time-dependent settlement under the existing stress state; piezometers indicate elevated u that dissipates. Why Other Options Are Wrong: Pre-consolidated/overconsolidated implies past maximum stress > current, not pending consolidation at current stress. Normally consolidated has current = maximum past stress and is already consolidated for that state. Common Pitfalls: Confusing “preconsolidated” with “preloading”; mixing up NC vs OC definitions. Final Answer: Under-consolidated soil

Question 17

Foundation types – grillage foundation basics: Which statements correctly describe a grillage foundation used under heavy isolated columns?

Accepted Answer

Introduction / Context: Where soil near the surface has adequate bearing capacity but a pedestal footing would be too thick or inefficient, a grillage foundation provides economical load spreading for heavy columns, chimneys, or stanchions. It is widely used in steel construction and retrofits where shallow depth is preferred. Given Data / Assumptions: Loads are predominantly vertical from isolated columns. Sufficient space exists to spread loads through multiple beam layers. Corrosion protection and concrete encasement are provided in practice. Concept / Approach: A grillage is a spread foundation comprised of orthogonal layers of beams (steel or treated timber) embedded in concrete to distribute concentrated column loads over a wider area, reducing contact pressure on the soil. It is categorized under shallow foundations and designed to maintain acceptable settlement and bearing stresses. Step-by-Step Solution: Relate purpose: heavy load distribution at shallow depth.Classify: belongs to spread (shallow) foundation family.Describe construction: multiple beam tiers at right angles (grillage) encased in concrete.Hence, all listed statements are correct. Verification / Alternative check: Design guides show bearing pressure checks, beam bending and shear, and spacing to ensure concrete infill transfer and ease of placement. Why Other Options Are Wrong: “None” contradicts standard practice descriptions; A–C are accurate. Common Pitfalls: Ignoring corrosion protection for steel grillages; inadequate concrete cover; neglecting settlement compatibility when multiple columns are supported. Final Answer: All of the above

Question 18

Active earth pressure – numerical application: Find the active lateral earth pressure intensity at a depth of 10 m in dry, cohesionless sand where φ = 30° and γ = 1.8 t/m^3. Assume a vertical wall with horizontal backfill.

Accepted Answer

Introduction / Context: Active earth pressure represents the minimum lateral pressure mobilized when a retaining wall yields sufficiently away from the backfill. For cohesionless backfill with level ground, Rankine’s active coefficient provides a quick way to compute lateral stress with depth. Given Data / Assumptions: Internal friction angle φ = 30 degrees. Unit weight γ = 1.8 t/m^3 (tonnes per cubic metre). Depth z = 10 m; dry, c = 0; vertical wall, horizontal backfill; no surcharge. Concept / Approach: For cohesionless soil with horizontal backfill, Rankine active earth pressure coefficient is Ka = (1 − sin φ) / (1 + sin φ). The vertical overburden stress is σv = γ * z. The active horizontal pressure is σh = Ka * σv. Units must be consistent (t/m^3 * m → t/m^2). Step-by-Step Solution: Compute sin φ: sin 30° = 0.5.Ka = (1 − 0.5) / (1 + 0.5) = 0.5 / 1.5 = 1/3 ≈ 0.333.σv = γ * z = 1.8 * 10 = 18 t/m^2.σh = Ka * σv = (1/3) * 18 = 6 t/m^2. Verification / Alternative check: Using Coulomb with zero wall friction and vertical wall reduces to Rankine for level backfill; results match. A simple proportionality check: σh increases linearly with depth; at z = 0, σh = 0; at 10 m, the computed 6 t/m^2 is reasonable. Why Other Options Are Wrong: 4, 5, 7, 8 t/m^2 correspond to incorrect Ka or arithmetic; only 6 t/m^2 satisfies Ka = 1/3 with given γ and z. Common Pitfalls: Using degrees vs radians incorrectly; confusing kN/m^3 with t/m^3; forgetting that cohesionless c = 0 so no additive term appears. Final Answer: 6 t/m^2

Question 19

Ways to increase the density (unit weight) of a soil mass: Which of the following mechanisms can increase the bulk density of a soil in the field or laboratory?

Accepted Answer

Introduction / Context: Improving soil density enhances strength and reduces compressibility. Field operations (compaction) and natural processes (consolidation) change the phase composition and volume, altering bulk unit weight γ. Recognizing all mechanisms helps select appropriate ground-improvement techniques. Given Data / Assumptions: Bulk density refers to mass per total volume of soil, including solids, water, and air. Changes may occur via volume reduction or mass increase (water infiltration), but engineering densification generally targets volume reduction. Concept / Approach: (A) Compaction expels air primarily, reducing void ratio e quickly, especially in granular soils. (B) Mineral grains are relatively incompressible, yet under very high stresses a small elastic compression exists, adding marginally to density. (C) Consolidation in saturated fine-grained soils expels water over time as excess pore pressure dissipates, reducing volume and increasing density. Step-by-Step Solution: Identify air removal: immediate densification via rolling, ramming, vibration.Acknowledge grain compressibility: minor but not zero at high stress levels.Recognize consolidation: time-dependent drainage reduces volume and raises unit weight.Thus, all mechanisms can contribute to increasing density. Verification / Alternative check: Proctor tests show density rise by air removal; oedometer tests show density increase via water expulsion; high-pressure tests reveal slight grain compression. Why Other Options Are Wrong: “None of the above” contradicts well-known soil mechanics processes. Common Pitfalls: Assuming grains are perfectly incompressible; overlooking that expelling water requires drainage paths and time. Final Answer: All of the above

Question 20

Flow nets and seepage direction: During steady seepage through porous soil, the direction of seepage flow is always related to equipotential lines how?

Accepted Answer

Introduction / Context: Flow nets are graphical solutions to Laplace’s equation for groundwater flow. They consist of two families of orthogonal curves: equipotential lines and streamlines. Knowing the relationship between them is vital for interpreting seepage directions and hydraulic gradients under dams, sheet piles, and retaining structures. Given Data / Assumptions: Steady, two-dimensional seepage in a homogeneous, isotropic soil. No sources or sinks within the domain besides boundaries. Concept / Approach: Equipotential lines connect points of equal total head. Streamlines represent the path of seepage flow. In isotropic media solving Laplace’s equation, these families intersect at right angles. Therefore, the seepage velocity vector is normal to equipotential lines and tangent to streamlines. Step-by-Step Solution: Construct flow net: draw equipotentials from boundary heads and streamlines from boundary flux conditions.At each intersection, check orthogonality: 90 degrees in isotropic materials.Direction of flow is from higher head to lower head along streamlines, perpendicular to equipotentials.Thus, the correct statement is that flow is perpendicular to equipotentials. Verification / Alternative check: Finite difference/finite element groundwater models yield velocity vectors perpendicular to equipotential contours in isotropic zones; anisotropy skews angles in transformed space but remains perpendicular in transformed coordinates. Why Other Options Are Wrong: Parallel to equipotentials: incorrect; no head drop along equipotential. Perpendicular to streamlines contradicts definition. Independence from equipotential pattern is false; both are linked solutions to Laplace’s equation. Parallel to both is impossible unless trivial uniform flow without head change. Common Pitfalls: Misreading flow nets; forgetting anisotropy effects (use transformed scales). Final Answer: Perpendicular to equipotential lines (i.e., along streamlines)

Soil Mechanics and Foundation Engineering Questions