Q: Active earth pressure in cohesionless backfill (horizontal ground): In the active state of plastic equilibrium for a cohesionless soil with a horizontal ground surface, which principal stress is vertical?

Introduction / Context: Active earth pressure corresponds to the state where lateral confinement is reduced until the soil mass reaches limiting equilibrium and lateral stress becomes the minor principal stress. Understanding the orientation of principal stresses clarifies wall design and interpretation of K-active relationships. Given Data / Assumptions: Backfill is cohesionless with a level surface. Retaining wall yields sufficiently to mobilize active conditions. Gravity loading yields vertical overburden stress sigma_v = gamma * z. Concept / Approach: Under active conditions, the horizontal earth pressure equals sigma_h = Ka * sigma_v, where Ka < 1. Thus sigma_h is the minor principal stress, and sigma_v is the major principal stress. Conversely, in passive conditions, sigma_h becomes the major principal stress. For horizontal ground and no surcharge, principal stresses align with vertical and horizontal directions. Step-by-Step Solution: Establish overburden: vertical stress increases with depth due to self-weight.At active limit, sigma_h reduces to Ka * sigma_v and is the minimum principal stress.Therefore, sigma_v > sigma_h and is the major principal stress.So the major principal stress is vertical and the minor is horizontal. Verification / Alternative check: From Mohr–Coulomb, active condition occurs when the Mohr circle is tangent to the failure envelope on the “active” side, giving sigma_1 = sigma_v and sigma_3 = sigma_h. Why Other Options Are Wrong: Option A reverses roles of sigma_v and sigma_h. Option B mislabels the minor principal stress as vertical. Option D is incorrect for horizontal backfill; principal directions align with vertical and horizontal axes. Common Pitfalls: Confusing active and passive states; forgetting that sufficient wall movement is required to mobilize active pressure; mixing sign conventions. Final Answer: Major principal stress is vertical

Question 1

Soil classification terminology – particle size only: What is the name of the soil classification system that categorizes composite soils exclusively on the basis of particle-size distribution (no plasticity indices)?

Accepted Answer

Introduction / Context: Different classification systems serve different purposes. Some rely entirely on grain-size distribution, while others incorporate plasticity characteristics. Knowing which system uses only particle-size data helps in preliminary identification and communication among engineers and geologists. Given Data / Assumptions: Composite natural soils may include gravel, sand, silt, and clay fractions. The question restricts the basis strictly to particle-size distribution (PSD). Concept / Approach: Textural classification (e.g., triangular charts of sand–silt–clay) assigns names based solely on relative proportions of size fractions. It does not use Atterberg limits. By contrast, the Unified Soil Classification System (USCS) combines PSD with plasticity data (plasticity chart). The HRB/AASHTO system also uses both gradation and plasticity for roadwork suitability, not PSD alone. “Particle classification” is a generic phrase rather than a standard system name in practice. Step-by-Step Solution: Identify systems that rely only on particle size: textural classification fits this criterion.Eliminate systems that also use plasticity (USCS, HRB/AASHTO).Discard nonstandard terminology (“particle classification”).Conclude that “Textural classification” is the correct answer. Verification / Alternative check: Textbook charts for soil texture (sand–silt–clay triangle) label zones like sandy clay loam, silty clay, etc., derived purely from PSD percentages. Why Other Options Are Wrong: USCS and HRB/AASHTO require plasticity data in addition to PSD. “Particle classification” lacks a recognized standard definition as a system name. Common Pitfalls: Assuming USCS is PSD-only; ignoring that plasticity controls whether fines are silt or clay in USCS. Final Answer: Textural classification

Question 2

Phase-relationship identity: Select the correct fundamental relationship linking dry unit weight (gamma_d), bulk unit weight (gamma), and water content (w) for a soil mass.

Accepted Answer

Introduction / Context: Routine field and lab work constantly switches between bulk (wet) and dry unit weights using the measured water content. Remembering the equivalent identities avoids algebra mistakes when interpreting compaction and in-situ density results. Given Data / Assumptions: gamma is the total (bulk) unit weight of the moist specimen. gamma_d is the unit weight of solids per total volume (i.e., dried basis). w is water content expressed as a decimal (not percent). Concept / Approach: Start with the mass partition: total mass M = Ms + Mw with Mw = w * Ms by definition of water content. Substituting into unit weight definitions yields a simple proportionality between gamma and gamma_d. Rearrangement then produces equivalent forms useful for conversions in either direction. Step-by-Step Solution: Write total mass: M = Ms + Mw = Ms * (1 + w).Define unit weights: gamma = M * g / V and gamma_d = Ms * g / V.Divide: gamma / gamma_d = (M / Ms) = (1 + w).Hence gamma = gamma_d * (1 + w) and gamma_d = gamma / (1 + w).Rearrange to find w: w = (gamma - gamma_d) / gamma_d. Verification / Alternative check: Example: if gamma = 20 kN/m^3 and w = 0.10, then gamma_d = 20/1.10 ≈ 18.18 kN/m^3. Substituting back: 18.18*(1+0.10) ≈ 20, confirming consistency. Why Other Options Are Wrong: Individually, Options A–C are each correct statements; however, none alone is “the only” relation. Option D correctly recognizes that they are equivalent identities derived from the same definition. Common Pitfalls: Using w in percent instead of decimal; mixing mass density with unit weight; rounding too early when converting. Final Answer: All of the above are equivalent identities

Question 3

Consolidation states and lab simulation: Identify the correct statements about isotropic consolidation, normally consolidated clay, and overconsolidated clay.

Accepted Answer

Introduction / Context: Consolidation history governs stiffness, strength, and compressibility of fine-grained soils. Recognizing whether a deposit is normally consolidated (NC) or overconsolidated (OC) is crucial for settlement prediction and undrained shear strength correlations. Lab apparatus can reproduce stress paths to study behavior. Given Data / Assumptions: Triaxial apparatus permits application of isotropic confining pressure. Effective stress concept is used throughout. Definitions of NC and OC clays follow consolidation history relative to present effective stress. Concept / Approach: NC clay: current effective vertical stress equals the historical maximum; void ratio–log sigma′ plots show the current point on the virgin compression line. OC clay: soil has been unloaded from a higher past stress; it lies on the recompression line and exhibits higher stiffness at small strains. In triaxial labs, isotropic consolidation is achieved by equal all-round pressure (sigma1′ = sigma3′ initially). Step-by-Step Solution: Relate NC definition: sigma′present = sigma′max, past.Relate OC definition: sigma′max, past > sigma′present (overconsolidation ratio > 1).Confirm isotropic consolidation setup: apply equal cell pressure in all directions.Thus all three statements are correct. Verification / Alternative check: Oedometer curves show a break at the preconsolidation pressure; triaxial tests reproduce NC and OC responses consistent with these definitions. Why Other Options Are Wrong: Any option excluding one of the correct statements would be incomplete. Common Pitfalls: Confusing overconsolidation with cementation; misinterpreting preconsolidation pressure from Casagrande construction; ignoring anisotropy in natural deposits even when isotropic lab consolidation is applied. Final Answer: All of the above

Question 4

Plasticity index in Atterberg limits: The plasticity index (PI) of a fine-grained soil is defined as the numerical difference between which two consistency limits?

Accepted Answer

Introduction / Context: The Atterberg limits quantify consistency changes of fine-grained soils with water content. The plasticity index is widely used for classification (USCS, AASHTO) and correlating engineering behavior such as compressibility and strength. Given Data / Assumptions: Standard definitions: LL, PL, and SL are water contents defining transitions among liquid, plastic, semi-solid, and solid states. PI is a scalar difference expressed in percentage points (or decimal fraction if preferred). Concept / Approach: PI measures the width of the plastic range: PI = LL − PL. A larger PI indicates greater plasticity and usually more clay-sized, active minerals. LL − SL or PL − SL are other indices (e.g., toughness index, shrinkage index) but are not the plasticity index itself. Step-by-Step Solution: State the definition: PI = LL − PL.Relate meaning: span of water content where soil can be molded without cracking or flow.Use PI along with LL on the plasticity chart to distinguish clays from silts.Conclude Option A is correct. Verification / Alternative check: USCS boundaries (A-line) use PI and LL; road specifications often set PI limits for subgrade materials. Why Other Options Are Wrong: PL − SL defines a shrinkage index component, not PI. LL − SL is not the plastic range; it exceeds the true plastic window. “None of these” is incorrect because PI is clearly LL − PL. Common Pitfalls: Confusing PI with liquidity index or consistency index; mixing percentage points with decimals without unit clarity. Final Answer: Liquid limit (LL) and plastic limit (PL)

Question 5

Pile spacing guideline – friction piles: For friction piles of 1.0 m diameter, what is the typical minimum centre-to-centre spacing recommended in practice?

Accepted Answer

Introduction / Context: Pile spacing affects group efficiency, construction feasibility, and soil–pile interaction. Friction piles transfer load largely through shaft resistance, so keeping adequate spacing minimizes overlap of stress bulbs and simplifies reinforcement and head connection details. Given Data / Assumptions: Pile type: friction piles (skin-friction dominant). Pile diameter: 1.0 m. General practice values without site-specific group interaction analysis. Concept / Approach: Rules of thumb: end-bearing piles are often spaced at about 2.5d to 3.5d; friction piles typically need slightly larger spacing to reduce interaction but still often fall within 2d to 3d minimums for constructability and load transfer, unless rigorous analysis dictates otherwise. Many codes and handbooks cite minimum center-to-center spacing ≥ 2.5d, but ranges of 2d to 3d are commonly accepted for preliminary layout. Step-by-Step Solution: Compute spacing range in meters: for d = 1.0 m, 2d to 3d → 2.0 m to 3.0 m.Check constructability: allows head connection and reinforcement cages.Check interaction: limits excessive overlap of shear zones along pile shafts.Select the practical range option provided. Verification / Alternative check: Detailed group efficiency analyses (e.g., Converse–Labarre) or numerical models often validate preliminary spacings in the 2d–3d range for friction piles unless constrained by geometry or high loads. Why Other Options Are Wrong: 2.0 m (exactly 2d) may be acceptable as an absolute minimum but lacks the typical range context. 3.0–4.0 m and 5.0 m are not “minimum” values; they may be used but are not minimal practice guidelines. Common Pitfalls: Ignoring group effects in soft clays; using too-tight spacing that complicates construction or reduces efficiency; forgetting to check pilecap geometry and reinforcement congestion. Final Answer: 2.0 m to 3.0 m

Question 6

Coulomb’s wedge theory – core assumptions for earth pressure on retaining walls: Which of the following assumptions are inherent in Coulomb’s earth pressure theory for planar failure wedges?

Accepted Answer

Introduction / Context: Coulomb’s earth pressure theory provides active and passive pressures on retaining structures by considering a rigid wedge of backfill sliding along a planar surface. Understanding its assumptions clarifies applicability and limitations versus Rankine’s simpler theory and more advanced wedge or numerical methods. Given Data / Assumptions: Backfill is idealized as dry, cohesionless, homogeneous, isotropic (basic Coulomb case). Planar failure surface intersects the wall, typically near the heel. Wall friction angle and backface inclination are known inputs. Concept / Approach: Coulomb analyzes equilibrium of a rigid wedge bounded by the wall backface and a trial plane. The resultant earth force acts at a known inclination to the wall normal depending on wall friction; magnitude comes from maximizing or minimizing the thrust with respect to the plane angle for active or passive conditions. Variants extend the method to c–φ soils, sloping backfills, and submerged conditions but maintain the same core assumptions of planarity and rigid wedge behavior. Step-by-Step Solution: List the assumptions (dry φ-soil, planar slip, rigid wedge, specified resultant direction).Recognize that solution seeks limiting thrust via extremum of potential planes.Conclude that all listed statements are consistent with Coulomb’s framework. Verification / Alternative check: Comparisons with Rankine (no wall friction, vertical wall, horizontal backfill) show differences arising precisely from these assumptions. Why Other Options Are Wrong: Any single statement alone is incomplete; the theory needs all these assumptions to be well-defined. Common Pitfalls: Applying Coulomb to cohesive backfills without modifications; assuming a nonplanar slip surface; ignoring wall friction or backface batter effects. Final Answer: All of the above

Question 7

Quick condition and piping – critical hydraulic gradient: Under what conditions can the critical hydraulic gradient (quick condition) occur in soils subjected to seepage?

Accepted Answer

Introduction / Context: Boiling, quicksand, or piping occurs when seepage forces counteract the submerged weight of soil grains, reducing effective stress to zero. Recognizing the conditions that lead to the critical hydraulic gradient is essential near excavations, sheet piles, and hydraulic structures. Given Data / Assumptions: Steady seepage governed by Darcy’s law. Effective stress principle: sigma′ = sigma − u. Upward gradients are especially dangerous beneath floors or at downstream toes. Concept / Approach: The critical gradient icr is reached when the upward seepage force equals the submerged unit weight of the soil skeleton, driving sigma′ → 0. Typically icr ≈ (Gs − 1) / (1 + e). When upward flow raises pore pressure, effective stress diminishes; at zero, particles can become mobile, causing boils and piping channels. Step-by-Step Solution: Note direction: upward hydraulic head loss creates upward seepage forces.Recognize that upward seepage pressure increases pore pressure u.Apply effective stress: when u approaches sigma, sigma′ becomes zero → quick condition.Therefore, all listed statements describe the same onset condition. Verification / Alternative check: Field indicator: boiling at the base of an excavation with upward gradients; preventive measures include relief wells, cutoffs, or lowering the water table to keep i < icr. Why Other Options Are Wrong: Each alone is a partial description. The complete understanding requires acknowledging all three simultaneously. Common Pitfalls: Ignoring that even moderate upward gradients can be critical in loose silts; overlooking factor of safety against piping defined as icr / i. Final Answer: All of the above

Question 8

Active earth pressure in cohesionless backfill (horizontal ground): In the active state of plastic equilibrium for a cohesionless soil with a horizontal ground surface, which principal stress is vertical?

Accepted Answer

Introduction / Context: Active earth pressure corresponds to the state where lateral confinement is reduced until the soil mass reaches limiting equilibrium and lateral stress becomes the minor principal stress. Understanding the orientation of principal stresses clarifies wall design and interpretation of K-active relationships. Given Data / Assumptions: Backfill is cohesionless with a level surface. Retaining wall yields sufficiently to mobilize active conditions. Gravity loading yields vertical overburden stress sigma_v = gamma * z. Concept / Approach: Under active conditions, the horizontal earth pressure equals sigma_h = Ka * sigma_v, where Ka < 1. Thus sigma_h is the minor principal stress, and sigma_v is the major principal stress. Conversely, in passive conditions, sigma_h becomes the major principal stress. For horizontal ground and no surcharge, principal stresses align with vertical and horizontal directions. Step-by-Step Solution: Establish overburden: vertical stress increases with depth due to self-weight.At active limit, sigma_h reduces to Ka * sigma_v and is the minimum principal stress.Therefore, sigma_v > sigma_h and is the major principal stress.So the major principal stress is vertical and the minor is horizontal. Verification / Alternative check: From Mohr–Coulomb, active condition occurs when the Mohr circle is tangent to the failure envelope on the “active” side, giving sigma_1 = sigma_v and sigma_3 = sigma_h. Why Other Options Are Wrong: Option A reverses roles of sigma_v and sigma_h. Option B mislabels the minor principal stress as vertical. Option D is incorrect for horizontal backfill; principal directions align with vertical and horizontal axes. Common Pitfalls: Confusing active and passive states; forgetting that sufficient wall movement is required to mobilize active pressure; mixing sign conventions. Final Answer: Major principal stress is vertical

Question 9

Triaxial shear testing modes (concept check): If drainage is permitted continuously during the application of both the normal stress (cell pressure) and the shear (deviator) stress so that full consolidation occurs and no excess pore-water pressure …

Accepted Answer

Introduction / Context: Understanding the drainage and consolidation conditions in laboratory shear tests is fundamental for interpreting soil strength parameters. The three classical triaxial modes are UU (quick), CU, and CD, each producing different pore pressure responses and effective stresses. Given Data / Assumptions: Drainage is allowed at all stages: during application of cell pressure and during shearing. No excess pore-water pressure builds up at any time. Full consolidation is achieved before and maintained during shear. Concept / Approach: In a consolidated–drained (CD) test, the specimen first consolidates under the applied all-round pressure with drainage open. During subsequent shearing, drainage remains open and the piston is advanced slowly so pore pressures do not develop. Shear strength is evaluated directly in terms of effective stresses; c′ and φ′ are obtained for design under long-term drained conditions. Step-by-Step Solution: Identify drainage condition during consolidation: open → consolidation complete.Identify drainage condition during shearing: open → no excess pore pressure.Conclusion: The described procedure corresponds to CD (drained) triaxial testing. Verification / Alternative check: In UU, drainage is closed at all times → excess pore pressure develops. In CU, drainage is open during consolidation but closed during shear → excess pore pressure develops only during shearing. Only CD has no excess pore pressure at any stage. Why Other Options Are Wrong: UU (quick): always undrained, hence excess pore pressure occurs. CU: undrained during shear; pore pressures are generated. “None of these” and “partially drained direct shear”: do not match the fully drained definition. Common Pitfalls: Confusing CU with CD because both consolidate under open drainage; the key difference is drainage during shearing. Final Answer: Drained test (CD: consolidated–drained)

Question 10

Void ratio from saturation condition: A saturated soil has water content w = 40% and specific gravity of solids G = 2.7. Determine the void ratio e (assume full saturation).

Accepted Answer

Introduction / Context: For fully saturated soils, a very useful identity links void ratio e, water content w, and specific gravity G. This provides a quick check on laboratory data and helps initialize consolidation and seepage analyses. Given Data / Assumptions: Saturation: S_r = 1.0 (fully saturated). Water content w = 40% = 0.40 (decimal). Specific gravity of solids G = 2.7. Concept / Approach: Under full saturation, the fundamental phase-relationship reduces to e = w * G (with w in decimal form). This comes from S_r = (w * G)/e and S_r = 1 at saturation, giving e = w * G directly. Step-by-Step Solution: Use S_r = (w * G) / e with S_r = 1.Therefore e = w * G.Substitute: e = 0.40 * 2.7 = 1.08. Verification / Alternative check: Porosity n = e / (1 + e) = 1.08 / 2.08 ≈ 0.519 (~52%), a realistic value for saturated fine-grained soils at this water content and mineral density. Why Other Options Are Wrong: 0.40 or 0.52 or 0.80: Do not satisfy e = w * G for the given data. “None of these”: incorrect because a valid computed value exists in the options. Common Pitfalls: Using w as a percent without converting to decimal; confusing G (dimensionless) with unit weights. Final Answer: 1.08

Question 11

Historical note in soil mechanics: Who is credited with coining and popularizing the term “soil mechanics” and establishing its modern scientific basis?

Accepted Answer

Introduction / Context: Soil mechanics emerged as a modern discipline in the early 20th century. Knowing key contributors helps contextualize foundational theories used every day in geotechnical practice. Given Data / Assumptions: Question seeks the principal figure associated with the term and field of soil mechanics. Emphasis is on the origin and consolidation of the discipline as a coherent science. Concept / Approach: Karl Terzaghi is widely regarded as the father of soil mechanics. His pioneering work on effective stress, consolidation theory, and earth pressure laid the foundations of geotechnical engineering. His 1925 book “Erdbaumechanik” unified disparate observations into a systematic framework and popularized the terminology and methods. Step-by-Step Solution: Identify the seminal contributor: Karl Terzaghi.Connect core concepts (effective stress, consolidation) to Terzaghi’s publications and lectures.Conclude he coined and popularized the modern notion of soil mechanics. Verification / Alternative check: Textbooks routinely begin with Terzaghi’s contributions; professional societies (e.g., ASCE) honor him through Terzaghi Lectures and awards. Why Other Options Are Wrong: Kray, Leygue: lesser-known figures without foundational roles in defining the field. Fellenius: notable for pile capacity methods and slope stability developments, but not credited with originating “soil mechanics.” Casagrande: made major advances in classification and consolidation, yet Terzaghi remains the primary originator. Common Pitfalls: Confusing early contributors; remembering that “effective stress” is central to Terzaghi’s legacy. Final Answer: Karl Terzaghi

Question 12

Seepage through earth dams: The shape of the phreatic line (upper seepage line) within a homogeneous earth dam with a horizontal drainage filter is typically:

Accepted Answer

Introduction / Context: Estimating seepage and pore-pressure distribution in earth dams is vital for stability and filter design. The phreatic line represents the top boundary of saturated flow, separating saturated and unsaturated zones inside the embankment. Given Data / Assumptions: Homogeneous earth dam with a drainage filter (toe drain). Steady-state seepage conditions. Darcian flow and Dupuit assumptions commonly used. Concept / Approach: Under Dupuit–Forchheimer assumptions for homogeneous isotropic media and horizontal drainage, the free seepage surface (phreatic line) becomes approximately parabolic. Graphical methods such as Casagrande’s construction are based on this property, facilitating routine design checks. Step-by-Step Solution: Model seepage with a flow net or Dupuit approach.Apply boundary conditions at upstream water level and at the drain.Construct the phreatic line; standard solution yields a parabola terminating at the drainage filter. Verification / Alternative check: Finite element seepage analyses for homogeneous sections reproduce a near-parabolic free surface for moderate slopes and steady states. Why Other Options Are Wrong: Circular/Elliptical/Straight line: not consistent with theory or practice for homogeneous dams with toe drains. “Depends only on upstream slope”: many factors influence the line; and the canonical shape under these assumptions is parabolic. Common Pitfalls: Ignoring anisotropy and zoning; in zoned dams or anisotropic conditions, the shape deviates from a simple parabola. Final Answer: Parabolic

Question 13

Purely cohesive (φ = 0) slope — critical circle: In a purely cohesive soil (φ = 0) slope, the critical center of the circular slip surface lies at the intersection of:

Accepted Answer

Introduction / Context: For homogeneous slopes, Swedish circle (method of slices) and stability charts provide critical slip surfaces. In the special case of purely cohesive soil (φ = 0), classical results locate the critical circle center using simple geometrical rules that aid preliminary design checks. Given Data / Assumptions: Homogeneous soil with φ = 0 (purely cohesive). Planar ground surface forming a slope with a defined toe and crest. Classical circular slip surface assumption. Concept / Approach: For φ = 0 slopes, the most critical (minimum factor of safety) circle is found at the intersection of the locus of potential centers and a perpendicular erected at one-third of the slope length measured from the toe. This geometric rule stems from analyses that balance driving and resisting moments for fully cohesive materials. Step-by-Step Solution: Mark the slope toe and measure the slope length along the face.At a point one-third of this length from the toe, draw a perpendicular to the slope face.Determine the locus of possible circle centers (standard construction line).Their intersection gives the center of the critical slip circle for φ = 0. Verification / Alternative check: Stability charts (Taylor’s stability numbers for φ = 0) and trial slip-circle analyses confirm this location for the critical circle in cohesive slopes. Why Other Options Are Wrong: Perpendicular bisector or two-thirds position: do not match the established φ = 0 rule. Directional angles with slope normal: not the classical geometric construction for cohesive slopes. Common Pitfalls: Applying this geometric rule to c–φ soils with φ > 0 where the critical center location differs; ignoring groundwater effects that alter stability. Final Answer: A perpendicular drawn at one-third of the slope length from the toe and the locus of centers

Question 14

Phase relationship definition: The ratio of the volume of water present in a soil mass to the total volume of its voids is called:

Accepted Answer

Introduction / Context: Phase relationships define how water, air, and solids occupy space in soils. Correct terminology is essential for interpreting compaction, seepage, and consolidation behavior. Given Data / Assumptions: Vw = volume of water, Va = volume of air, Vv = Vw + Va = total void volume. V = total volume of soil; Vs = volume of solids. Concept / Approach: Degree of saturation S_r is defined as Vw / Vv. Porosity n is Vv / V; void ratio e is Vv / Vs; percentage of voids expresses n as a percent. Air content a_c is Va / Vv. Only S_r uses water volume over void volume as requested in the stem. Step-by-Step Solution: Recognize required ratio: Vw / Vv.Map to standard term: S_r (degree of saturation) = Vw / Vv.Conclude correct option: Degree of saturation. Verification / Alternative check: At full saturation, S_r = 1.0; when the soil is dry, S_r = 0, matching intuitive end states. Why Other Options Are Wrong: Porosity (Vv/V) and void ratio (Vv/Vs) have different denominators. Percentage of voids is n expressed in percent, not Vw/Vv. Air content is Va/Vv, not Vw/Vv. Common Pitfalls: Confusing S_r with moisture content w (a mass-based ratio, Mw/Ms). Final Answer: Degree of saturation

Question 15

Shear strength of sands — effect of particle shape and density: For which soil is the angle of internal friction (φ) typically the smallest?

Accepted Answer

Introduction / Context: In granular soils, shear strength arises mainly from interlocking and dilation. Particle shape (angular vs rounded) and relative density (dense vs loose) strongly influence the friction angle φ measured in drained shear tests. Given Data / Assumptions: Clean sands with varying particle shape and density. Effective-stress, drained behavior governs φ. No significant cementation or cohesion. Concept / Approach: Angular particles interlock and resist sliding better than rounded ones, yielding a higher φ. Dense packs dilate upon shearing, further increasing peak φ compared with loose packs. Hence the smallest φ occurs when both interlocking and dilation are minimal: rounded particles in a loose state. Step-by-Step Solution: Rank by shape: angular > rounded (in φ).Rank by density: dense > loose (in φ).Combine: minimum φ → round-grained + loose state. Verification / Alternative check: Typical peak φ values may be 40–45° for angular dense sands, 35–40° for rounded dense sands, 30–34° for angular loose sands, and as low as high-20s to low-30s for rounded loose sands (ranges indicative). Why Other Options Are Wrong: Angular dense or angular loose: angularity increases φ via interlock. Round dense: density raises φ via dilation. Silty sand with fines may reduce φ, but not as consistently minimal as round loose sands under clean conditions. Common Pitfalls: Comparing clays to sands; φ in clays is a different mechanism and often treated via effective-stress parameters with possible apparent cohesion. Final Answer: Round-grained loose sand

Question 16

Consolidation terminology: The reduction in soil volume caused by squeezing out water from the voids under sustained load is termed:

Accepted Answer

Introduction / Context: Settlements in saturated fine-grained soils manifest as immediate (elastic) settlement, primary consolidation, and secondary compression. Correctly naming the mechanism guides prediction and mitigation measures in design. Given Data / Assumptions: Sustained static load applied to saturated soil with drainage paths available. Time-dependent dissipation of excess pore pressure. Volume change due to water expulsion, not elastic distortion alone. Concept / Approach: Primary consolidation is the process by which excess pore pressure dissipates and effective stress increases, causing a reduction in void ratio. Secondary compression then continues at approximately constant effective stress due to viscous adjustment of the soil skeleton. Immediate settlement is nearly instantaneous elastic strain without significant pore-pressure dissipation in coarse soils or at very small strains. Step-by-Step Solution: Apply load → generate excess pore pressure u in saturated clay.Drainage allows u to dissipate with time → σ′ increases.Void ratio decreases → volume reduces → settlement accumulates.This time-dependent phase is primary consolidation. Verification / Alternative check: Oedometer curves show a characteristic primary consolidation segment followed by a near-linear secondary compression tail in e–log t space. Why Other Options Are Wrong: Secondary compression: occurs after primary consolidation, not the initial water expulsion phase. Immediate settlement: largely elastic and rapid. Elastic rebound: recovery upon unloading, opposite effect. Creep under constant effective stress: describes secondary time effects, not primary consolidation. Common Pitfalls: Using “primary compression” loosely; precise term in soil mechanics is “primary consolidation.” Final Answer: Primary consolidation

Question 17

Compaction behavior — key statements: Which of the following statements about dry density versus water content behavior is/are correct? Select the best combined answer.

Accepted Answer

Introduction / Context: Laboratory compaction tests (Standard/Modified Proctor) produce a characteristic dry density–water content curve. Recognizing its features is crucial for field compaction control and specification compliance. Given Data / Assumptions: Soil is compacted at different water contents with fixed compactive effort. Dry density γd is measured after compaction. Zero air-voids (ZAV) line represents the theoretical upper limit of γd at a given water content. Concept / Approach: As water content increases from very dry conditions, a thin water film lubricates particles, aiding rearrangement and increasing γd, reaching a peak at OMC. Further addition of water fills voids without improving particle interlock, reducing γd. The ZAV line (na = 0) is given by γd(ZAV) = γw / (w + 1/G) and is an asymptotic upper bound rarely achieved in practice. Step-by-Step Solution: Confirm A: Peak γd at OMC is a defining feature of the curve.Confirm B: Dry soil lacks lubrication → high stiffness → low achievable density.Confirm C: Beyond OMC, added water displaces air but reduces interparticle friction benefits → γd falls.Confirm D: ZAV condition provides the maximum theoretical γd for the given w. Verification / Alternative check: Compare Standard vs Modified Proctor: higher effort shifts the curve up and left (higher γd, lower OMC) but preserves these qualitative statements. Why Other Options Are Wrong: Each individual statement A–D is correct; therefore the combined correct choice is “All the above”. Common Pitfalls: Confusing ZAV line with air-void lines of constant nonzero air content; assuming γd keeps increasing with water content indefinitely. Final Answer: All the above

Question 18

Geotechnical Laboratory Practice – Purpose of the Pycnometer (Soil Testing) In routine soil mechanics testing, a glass pycnometer (or density bottle) is primarily used to determine which property from a soil sample, especially for coarse-grained soi…

Accepted Answer

Introduction / Context: The pycnometer is a simple but powerful tool in geotechnical laboratories. While it is widely known for aiding measurements related to water and volume, students often confuse what exact index property it directly determines versus what requires additional tests. This question checks whether you can distinguish the pycnometer’s primary role in routine moisture determination, especially for coarse soils. Given Data / Assumptions: Standard geotechnical lab equipment includes a pycnometer (density bottle) with stopper. Tests follow conventional procedures used for index properties. Focus is on what the pycnometer is primarily used to determine directly and efficiently. Concept / Approach: A pycnometer allows rapid and fairly accurate calculation of the water (moisture) content of coarse-grained soils by mass–volume bookkeeping. The wet sample is placed in the bottle and topped with water, and companion readings (including the bottle filled only with water) enable separation of the mass of solids and mass of water within the specimen. Although a pycnometer also assists in determining the specific gravity of soil solids, among the options listed here the direct, routine outcome emphasized is the moisture content percentage, especially when oven-drying is impractical for large coarse particles. Step-by-Step Solution: Weigh the empty pycnometer and record M1.Place moist soil in the pycnometer, add water to expel air, and record combined mass M2.Fill the pycnometer with water only and record M3.Use the standard pycnometer relations to find mass of water in the specimen and then compute w (%) = (mass of water in soil / mass of dry soil) * 100. Verification / Alternative check: Oven-dry determination (105–110 °C) remains the reference method; pycnometer results agree closely if air is fully expelled and the procedure is followed carefully. Why Other Options Are Wrong: Voids ratio and density index require additional gradation/volume data. Dry density requires field/lab compaction or core cutter data, not just a pycnometer reading. Common Pitfalls: Trapped air bubbles, incorrect temperature assumptions for water density, and poor cleaning/drying of the bottle lead to errors. Final Answer: Water (moisture) content (%)

Question 19

Sedimentation Theory for Soil Particles – Lower Size Limit for Stokes’ Law Validity Stokes’ law is used in sedimentation analysis of fine soils. It does not hold good if the particle size becomes smaller than approximately which value (due to Browni…

Accepted Answer

Introduction / Context: Hydrometer and pipette analyses rely on Stokes’ law to relate particle size to settling velocity. However, Stokes’ law has bounds of applicability. At extremely small sizes, random thermal motion and non-ideal effects distort the laminar settling assumptions, invalidating direct use of the formula. Given Data / Assumptions: Lamina flow with Reynolds number < 1 is assumed for Stokes’ law. Particles are assumed rigid, smooth, and spherical, without mutual interference. Fluid properties are constant and temperature effects are controlled. Concept / Approach: Stokes’ law gives terminal settling velocity v = (g * (G_s − 1) * d^2) / (18 * ν) for a single, isolated sphere in laminar flow. The relationship fails at both extremes: (i) at larger sizes (higher Reynolds number) where turbulence and form drag increase; and (ii) at very tiny sizes (~sub-micron to a few tenths of micron) where Brownian motion dominates. In classical soil mechanics, a widely cited lower size threshold is about 0.0002 mm (0.2 micrometre). Below this, Stokes’ sedimentation assumptions break down. Step-by-Step Solution: Identify the lower applicability bound dominated by Brownian motion.Recall practical threshold: about 0.0002 mm.Select the option that matches this threshold. Verification / Alternative check: Laboratory observations show erratic settling for colloidal-sized particles; dispersion chemistry becomes more significant than gravity settling predictions. Why Other Options Are Wrong: 0.2 mm and 0.02 mm are much larger (where Stokes may fail due to turbulence, but the question asks about smaller than a limit). 0.002 mm is still too large for the Brownian-motion-driven lower limit. Common Pitfalls: Confusing upper and lower validity limits; ignoring flocculation that invalidates the single-particle assumption. Final Answer: 0.0002 mm

Question 20

Consistency Limits – Definition of Plasticity Index (Atterberg Limits) In soil consistency characterization, the plasticity index (PI) is defined as the numerical difference between which two Atterberg limits?

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Introduction / Context: Atterberg limits quantify water content thresholds at which fine-grained soils transition between states. The plasticity index (PI) is a foundational parameter used in classification systems (e.g., plasticity charts) and in predicting engineering behavior of clays and silty clays. Given Data / Assumptions: Standard Atterberg limits: liquid limit (LL), plastic limit (PL), and shrinkage (semi-solid) limit (SL). Water content is reported as percent of dry mass. Concept / Approach: The plasticity index (PI) expresses the width of the plastic range: PI = LL − PL. It measures the span of water contents over which soil exhibits plastic behavior (mouldable without cracking and without water bleeding). Higher PI usually indicates higher clay content and greater compressibility and swelling potential. Step-by-Step Solution: Recall definition: PI = LL − PL.Identify the correct pair among the options: liquid limit and plastic limit.Therefore choose the option stating 'Liquid limit and plastic limit'. Verification / Alternative check: Plasticity charts (Casagrande) plot PI versus LL to classify clays and silts; this usage confirms the definition. Why Other Options Are Wrong: Pairs involving the semi-solid or solid limits do not define PI; those limits are used for shrinkage analysis. Common Pitfalls: Using different moisture units or mixing up LL and PL procedures (cone penetrometer vs Casagrande device) does not alter the PI definition. Final Answer: Liquid limit and plastic limit

Soil Mechanics and Foundation Engineering Questions