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Question 1Geotechnical engineering models and applications

In a geotechnical site investigation per GB 50021-2001, an SPT test is conducted in medium sand at a test depth of 12.0 m using a drill rod length L = 13.5 m. The measured raw field blow count is N = 24. If the drill rod length correction coefficient is α = 0.79, what is the corrected penetration blow count N'?

  1. 15.8
  2. 19.0 Correct answer.
  3. 24.0
  4. 30.4

Explanation

According to GB 50021-2001 (Section 10.5), when standard penetration test blow counts require drill rod length correction, the corrected blow count is calculated as N' = α · N. Here, N' = 0.79 × 24 = 18.96 ≈ 19.0 blows.

Question 2Geotechnical engineering models and applications

During a static double-bridge Cone Penetration Test (CPT, 双桥静力触探), the measured cone tip resistance is qc = 2.40 MPa and the sleeve friction is fs = 48.0 kPa. What is the friction ratio Rf (%) of this soil stratum?

  1. 0.50%
  2. 1.20%
  3. 2.00% Correct answer.
  4. 4.80%

Explanation

Per GB 50021-2001 (Section 10.4), the CPT friction ratio Rf is defined as Rf = (fs / qc) × 100%. Converting units: fs = 48.0 kPa = 0.048 MPa; qc = 2.40 MPa. Thus, Rf = (0.048 / 2.40) × 100% = 2.00%.

Question 3Geotechnical engineering models and applications

A cross-hole seismic shear wave test (GB 50021-2001) is conducted between two vertical boreholes spaced L = 4.0 m apart. At a depth of 15.0 m, the travel time of the shear wave from source to receiver is measured as t = 16.0 ms (0.016 s). What is the shear wave velocity vs of the soil stratum?

  1. 125 m/s
  2. 250 m/s Correct answer.
  3. 320 m/s
  4. 500 m/s

Explanation

Shear wave velocity is computed directly as vs = L / t. Here, vs = 4.0 m / 0.016 s = 250 m/s.

Question 4Geotechnical engineering models and applications

A pumping test in an unconfined aquifer (GB 50021-2001) achieves steady state. The initial groundwater table is H0 = 12.0 m, the steady well water depth is hw = 8.0 m, the well radius is rw = 0.15 m, the radius of influence is R = 60.0 m, and the steady discharge is Q = 360 m³/d (4.167×10⁻³ m³/s). Using Dupuit's formula k = Q · ln(R/rw) / [π(H0² - hw²)], what is the hydraulic conductivity k of the aquifer?

  1. 1.25×10⁻⁴ m/s (10.8 m/d)
  2. 3.12×10⁻⁴ m/s (27.0 m/d)
  3. 9.94×10⁻⁵ m/s (8.59 m/d) Correct answer.
  4. 5.85×10⁻⁵ m/s (5.05 m/d)

Explanation

Applying Dupuit's formula: ln(R/rw) = ln(60.0 / 0.15) = ln(400) = 5.9915. H0² - hw² = 12.0² - 8.0² = 144 - 64 = 80 m². With Q = 360 m³/d: k = [360 × 5.9915] / [π × 80] = 2156.93 / 251.33 = 8.582 m/d ≈ 8.59 m/d. In SI units: 8.582 / 86400 s = 9.933×10⁻⁵ m/s ≈ 9.94×10⁻⁵ m/s.

Question 5Geotechnical engineering models and applications

A flat dilatometer test (DMT, 扁铲侧胀试验) conducted at a depth of 8.0 m yields an initial lift-off pressure p0 = 220 kPa and a membrane expansion pressure p1 = 480 kPa. The in-situ initial total vertical stress is σv0 = 150 kPa and pore water pressure is u0 = 40 kPa. What is the dilatometer modulus ED (assume membrane constant ΔA = ΔB = 0, dimension coefficient E = 34.7(p1 - p0))?

  1. 4.52 MPa
  2. 9.02 MPa Correct answer.
  3. 13.52 MPa
  4. 18.04 MPa

Explanation

According to the DMT formulation in GB 50021-2001 (Appendix C) and standard dilatometer specifications, the dilatometer modulus is ED = 34.7 × (p1 - p0). Substituting: ED = 34.7 × (480 - 220) kPa = 34.7 × 260 kPa = 9022 kPa ≈ 9.02 MPa.

Question 6Geotechnical engineering models and applications

A soil specimen has a natural moisture content w = 25.0%, specific gravity of soil solids Gs = 2.70, and wet unit weight γ = 19.0 kN/m³. Taking the unit weight of water γw = 10.0 kN/m³, what is the void ratio e and degree of saturation Sr of this soil?

  1. e = 0.776, Sr = 87.0% Correct answer.
  2. e = 0.650, Sr = 95.2%
  3. e = 0.820, Sr = 72.5%
  4. e = 0.550, Sr = 100.0%

Explanation

Dry unit weight γd = γ / (1 + w) = 19.0 / (1 + 0.25) = 15.20 kN/m³. Void ratio e = (Gs · γw / γd) - 1 = (2.70 × 10.0 / 15.20) - 1 = 1.7763 - 1 = 0.7763 ≈ 0.776. Degree of saturation Sr = (w · Gs) / e = (0.25 × 2.70) / 0.7763 = 0.675 / 0.7763 = 0.8695 ≈ 87.0%.

Question 7Geotechnical engineering models and applications

A soil layer of saturated sand has specific gravity Gs = 2.65 and void ratio e = 0.65. Taking γw = 10.0 kN/m³, what is the critical hydraulic gradient (icr, 临界水力比降) for upward seepage boiling/piping?

  1. 0.50
  2. 0.80
  3. 1.00 Correct answer.
  4. 1.65

Explanation

Critical hydraulic gradient under upward seepage is given by Terzaghi's formula: icr = (Gs - 1) / (1 + e) = γ' / γw. Substituting values: icr = (2.65 - 1) / (1 + 0.65) = 1.65 / 1.65 = 1.00.

Question 8Geotechnical engineering models and applications

A consolidated-drained (CD) triaxial test is carried out on a normally consolidated clay specimen. At failure, the effective confining pressure is σ'3 = 100 kPa and the axial deviator stress is (σ1 - σ3)f = 200 kPa. Assuming the effective cohesion c' = 0, what is the effective internal friction angle φ'?

  1. 19.5°
  2. 30.0° Correct answer.
  3. 36.9°
  4. 45.0°

Explanation

Major effective principal stress at failure is σ'1 = σ'3 + (σ1 - σ3)f = 100 + 200 = 300 kPa. For c' = 0, Mohr-Coulomb failure criterion is σ'1 / σ'3 = tan²(45° + φ'/2) = (1 + sin φ') / (1 - sin φ'). Here, σ'1 / σ'3 = 300 / 100 = 3.0. Thus, (1 + sin φ') / (1 - sin φ') = 3.0 ⇒ 1 + sin φ' = 3 - 3 sin φ' ⇒ 4 sin φ' = 2 ⇒ sin φ' = 0.500 ⇒ φ' = 30.0°.

Question 9Geotechnical engineering models and applications

A saturated clay layer has thickness H = 6.0 m, initial void ratio e0 = 0.90, and compression index Cc = 0.30. The initial effective overburden stress at the layer mid-depth is σ'0 = 100 kPa. A surface fill creates a uniform vertical stress increment Δσ = 100 kPa throughout the layer. What is the primary consolidation settlement Sc of this clay layer (assume normally consolidated soil)?

  1. 95 mm
  2. 190 mm
  3. 285 mm Correct answer.
  4. 380 mm

Explanation

Primary consolidation settlement of normally consolidated clay is calculated using: Sc = [Cc · H / (1 + e0)] · log10[(σ'0 + Δσ) / σ'0]. Substituting values: Sc = [0.30 × 6.0 m / (1 + 0.90)] · log10[(100 + 100) / 100] = (1.80 / 1.90) · log10(2) = 0.94737 × 0.30103 m = 0.28518 m ≈ 285 mm.

Question 10Geotechnical engineering models and applications

A rock core sample undergoes point load strength testing (GB/T 50218). The core diameter is D = 50 mm and the failure load is P = 15.0 kN. What is the equivalent point load strength index Is(50)?

  1. 3.0 MPa
  2. 6.0 MPa Correct answer.
  3. 12.0 MPa
  4. 24.0 MPa

Explanation

Point load strength index Is is defined as Is = P / De², where equivalent core diameter De² = D² for diametral test. For D = 50 mm = 0.050 m: De² = 0.0025 m². Is(50) = 15.0 kN / 0.0025 m² = 6000 kPa = 6.0 MPa.

Question 11Geotechnical engineering models and applications

A soil profile consists of 4.0 m of sand (total unit weight γ = 18.0 kN/m³) overlying 6.0 m of saturated clay (saturated unit weight γsat = 20.0 kN/m³). The groundwater table is located at a depth of 4.0 m from the ground surface. Taking γw = 10.0 kN/m³, what is the total vertical stress σv, pore water pressure u, and effective vertical stress σ'v at a depth of 7.0 m?

  1. σv = 132 kPa, u = 30 kPa, σ'v = 102 kPa Correct answer.
  2. σv = 140 kPa, u = 70 kPa, σ'v = 70 kPa
  3. σv = 132 kPa, u = 70 kPa, σ'v = 62 kPa
  4. σv = 120 kPa, u = 30 kPa, σ'v = 90 kPa

Explanation

Total vertical stress at 7.0 m depth: σv = (γsand × 4.0 m) + (γsat,clay × 3.0 m) = (18.0 × 4.0) + (20.0 × 3.0) = 72.0 + 60.0 = 132.0 kPa. Pore water pressure (groundwater table at 4.0 m, depth of water column hw = 7.0 - 4.0 = 3.0 m): u = γw × hw = 10.0 × 3.0 = 30.0 kPa. Effective vertical stress: σ'v = σv - u = 132.0 - 30.0 = 102.0 kPa (or σ'v = 18.0×4.0 + (20.0-10.0)×3.0 = 72.0 + 30.0 = 102.0 kPa).

Question 12Geotechnical engineering models and applications

In a direct shear test on a silty clay, normal stresses of 100 kPa, 200 kPa, and 300 kPa produce peak shear strengths of 65 kPa, 105 kPa, and 145 kPa, respectively. What are the cohesion c and internal friction angle φ of this soil?

  1. c = 25.0 kPa, φ = 21.8° Correct answer.
  2. c = 20.0 kPa, φ = 26.6°
  3. c = 30.0 kPa, φ = 18.4°
  4. c = 15.0 kPa, φ = 30.0°

Explanation

The Mohr-Coulomb shear strength equation is τ = c + σ · tan φ. Using linear regression or slope between points: tan φ = Δτ / Δσ = (105 - 65) / (200 - 100) = 40 / 100 = 0.400. Thus, φ = arctan(0.400) = 21.80° ≈ 21.8°. Cohesion c = τ - σ · tan φ = 65 - 100 × 0.400 = 65 - 40 = 25.0 kPa.

Question 13Geotechnical engineering models and applications

An oedometer test on an undisturbed clay sample yields an in-situ effective overburden stress σ'0 = 120 kPa and a preconsolidation pressure pc = 240 kPa determined via Casagrande's construction method. What is the Overconsolidation Ratio (OCR) and the consolidation state of this soil?

  1. OCR = 0.50, Under-consolidated (欠固结土)
  2. OCR = 1.00, Normally consolidated (正常固结土)
  3. OCR = 2.00, Over-consolidated (超固结土) Correct answer.
  4. OCR = 4.00, Heavily over-consolidated (强超固结土)

Explanation

The Overconsolidation Ratio is defined as OCR = pc / σ'0. Substituting values: OCR = 240 kPa / 120 kPa = 2.00. Since OCR > 1.0, the clay is classified as over-consolidated (超固结土).

Question 14Geotechnical engineering models and applications

A soft organic clay layer has thickness H = 5.0 m, void ratio at the end of primary consolidation ep = 1.20, and secondary compression index Cα = 0.022. The time at the completion of primary consolidation is t1 = 2.0 years. What is the secondary consolidation settlement Ss from year 2 to year 20 (t2 = 20 years)?

  1. 25 mm
  2. 50 mm Correct answer.
  3. 100 mm
  4. 220 mm

Explanation

Secondary consolidation settlement is given by Ss = [Cα · H / (1 + ep)] · log10(t2 / t1). Substituting: Ss = [0.022 × 5.0 m / (1 + 1.20)] · log10(20 / 2) = [0.110 / 2.20] · log10(10) = 0.050 m × 1.0 = 0.050 m = 50 mm.

Question 15Geotechnical engineering models and applications

In the p'-q effective stress space (where p' = (σ'1 + 2σ'3)/3 and q = σ'1 - σ'3 for triaxial compression), a soil has an effective critical state internal friction angle φ' = 30.0° and c' = 0. What is the slope M of the Critical State Line (CSL, q = M·p')?

  1. M = 0.577
  2. M = 0.866
  3. M = 1.200 Correct answer.
  4. M = 1.732

Explanation

In standard Cambridge triaxial stress path notation, the Critical State Line slope M under triaxial compression is M = (6 sin φ') / (3 - sin φ'). For φ' = 30.0°, sin φ' = 0.500: M = (6 × 0.500) / (3 - 0.500) = 3.0 / 2.5 = 1.200.

Question 16Geotechnical engineering models and applications

A rock mass is characterized using the Barton Q-system (NGI). The parameters are determined as: RQD = 80%, Joint Set Number Jn = 4 (two joint sets plus random), Joint Roughness Number Jr = 2.0 (rough/irregular, undulating), Joint Alteration Number Ja = 1.0 (unaltered joint walls), Joint Water Reduction Factor Jw = 1.0 (dry / minor inflow), and Stress Reduction Factor SRF = 2.5 (medium stress). What is the rock mass quality index Q?

  1. 4.0
  2. 16.0 Correct answer.
  3. 40.0
  4. 160.0

Explanation

Barton's Q-system formula is Q = (RQD / Jn) · (Jr / Ja) · (Jw / SRF). Substituting: Q = (80 / 4) × (2.0 / 1.0) × (1.0 / 2.5) = 20 × 2.0 × 0.40 = 16.0. A Q-value of 16.0 represents 'Good' quality rock mass.

Question 17Geotechnical engineering models and applications

Under GB 50007-2011 (Section 5.2.2), when the eccentricity of the vertical load resultant on a rectangular foundation (length l, width b) exceeds one-sixth of the base dimension (e = Mk / (Fk + Gk) > b/6), part of the foundation detaches from the soil. What formula is used to calculate the maximum contact edge pressure pkmax?

  1. pkmax = (Fk + Gk) / (l · b) · (1 + 6e/b)
  2. pkmax = 2 · (Fk + Gk) / [3 · l · (b/2 - e)] Correct answer.
  3. pkmax = 1.2 · (Fk + Gk) / (l · b)
  4. pkmax = (Fk + Gk) / [l · (b - 2e)]

Explanation

Per GB 50007-2011 (Formula 5.2.2-4), when e > b/6, tension occurs and the contact zone reduces to a triangular pressure distribution with effective length 3a' = 3(b/2 - e). The maximum edge pressure is pkmax = 2(Fk + Gk) / [3 · l · (b/2 - e)].

Question 18Geotechnical engineering models and applications

A rigid rectangular spread footing with b = 2.0 m and l = 3.0 m is subjected to biaxial eccentric vertical load (Fk + Gk = 900 kN) with eccentricities ex = 0.15 m (along width b) and ey = 0.20 m (along length l). Assuming no detachment occurs, what is the maximum corner contact pressure pkmax (GB 50007-2011)?

  1. 150.0 kPa
  2. 217.5 kPa
  3. 277.5 kPa Correct answer.
  4. 350.0 kPa

Explanation

Average contact pressure pk = (Fk + Gk) / (b · l) = 900 / (2.0 × 3.0) = 150.0 kPa. Biaxial contact pressure formula: pkmax = pk · [1 + (6ex / b) + (6ey / l)]. Substituting values: 6ex / b = 6 × 0.15 / 2.0 = 0.45; 6ey / l = 6 × 0.20 / 3.0 = 0.40. Thus, pkmax = 150.0 × [1 + 0.45 + 0.40] = 150.0 × 1.85 = 277.5 kPa.

Question 19Geotechnical engineering models and applications

A raft foundation for a 20-story building has planar dimensions B = 20.0 m, L = 40.0 m. The average contact net pressure is p0 = 220 kPa. Using the layer-wise summation method (GB 50007-2011), the equivalent modulus of compressibility is calculated as Es = 15.0 MPa. If the theoretical calculated settlement is s' = 60.0 mm, and the empirical settlement coefficient table gives ψs = 0.85 for p0 = 220 kPa and Es = 15.0 MPa, what is the predicted final settlement s?

  1. 51.0 mm Correct answer.
  2. 60.0 mm
  3. 70.6 mm
  4. 102.0 mm

Explanation

In GB 50007-2011 (Formula 5.3.5-1), the final foundation settlement is s = ψs · s'. Substituting values: s = 0.85 × 60.0 mm = 51.0 mm.

Question 20Geotechnical engineering models and applications

A combined rectangular footing of length L = 6.0 m supports two interior columns C1 and C2 spaced 3.6 m apart. Column C1 carries vertical load F1k = 800 kN and Column C2 carries F2k = 1200 kN. To achieve a perfectly uniform contact pressure distribution under concentric loading, where must the centroid of the footing base be located relative to Column C1?

  1. 1.44 m from C1 towards C2
  2. 2.16 m from C1 towards C2 Correct answer.
  3. 3.00 m from C1 towards C2
  4. 3.60 m from C1 towards C2

Explanation

For uniform pressure, the geometric centroid of the footing must coincide with the resultant of the column loads. Taking moments about Column C1: Total load R = F1k + F2k = 800 + 1200 = 2000 kN. Distance from C1 to resultant: xR = (F2k × 3.6 m) / R = (1200 × 3.6) / 2000 = 4320 / 2000 = 2.16 m from C1.

Question 21Geotechnical engineering models and applications

A high-strain dynamic pile test (JGJ 106-2014) is performed using the CASE method. The measured impact force and particle velocity at time t1 (impact peak) and t1 + 2L/c (wave return from pile toe) yield a total dynamic resistance R_CASE = 3200 kN. If the damping coefficient is Jc = 0.35 and the dynamic damping resistance is computed as Rdamp = 600 kN, what is the static ultimate compressive capacity estimate Rst?

  1. 2080 kN
  2. 2600 kN Correct answer.
  3. 3200 kN
  4. 3800 kN

Explanation

In high-strain CASE method analysis (JGJ 106-2014), the static capacity Rst is obtained by subtracting the dynamic damping resistance from the total dynamic resistance: Rst = R_CASE - Rdamp = 3200 kN - 600 kN = 2600 kN.

Question 22Geotechnical engineering models and applications

In a deep soft clay deposit overlying dense gravel (JGJ 94-2008), a heavy site fill of 2.0 m is placed after driving end-bearing piles. As the soft clay consolidates, Negative Skin Friction (NSF, 负摩阻力) develops. What is the location of the Neutral Point (中性点) along the pile shaft?

  1. At the pile head
  2. At the depth where pile settlement equals the surrounding soil settlement (relative displacement S_soil - S_pile = 0) Correct answer.
  3. At the pile toe resting on the dense gravel
  4. At exactly 0.5 times the pile length regardless of soil stratigraphy

Explanation

According to JGJ 94-2008 (Section 5.4.2), the Neutral Point (中性点) is the cross-section along the pile shaft where the relative displacement between the pile and the surrounding consolidating soil is zero (S_soil = S_pile). Above the neutral point, the soil settles more than the pile, exerting downward negative skin friction; below the neutral point, the pile settles more than the soil, mobilizing positive upward skin friction.

Question 23Geotechnical engineering models and applications

Under JGJ 94-2008 (Appendix C), the 'm-method' (m法) is used to compute the lateral deflection and bending moment of piles under horizontal loads. What fundamental assumption does the m-method make regarding the coefficient of horizontal subgrade reaction Cz?

  1. Cz is constant with depth (Cz = k0)
  2. Cz increases linearly with depth z from zero at ground surface (Cz = m · z) Correct answer.
  3. Cz increases quadratically with depth (Cz = k · z²)
  4. Cz decreases exponentially with depth

Explanation

The Chinese standard 'm-method' (JGJ 94-2008 Appendix C) assumes that the coefficient of horizontal subgrade reaction Cz is proportional to depth z below the ground surface (or dredge line): Cz = m · z, where m is the coefficient of horizontal subgrade reaction proportionality (kN/m⁴).

Question 24Geotechnical engineering models and applications

A 9-pile cap (3×3 square layout, center-to-center spacing sa = 3d) is subjected to an eccentric vertical load with overturning moment. Which piles experience the highest axial compressive reaction under standard load combinations (JGJ 94-2008)?

  1. The central interior pile
  2. The four edge midpoint piles
  3. The corner pile on the side of load eccentricity Correct answer.
  4. All 9 piles carry exactly identical axial reactions

Explanation

Under eccentric vertical loading with moments Mx and My, the pile reaction formula is Ni = (F + G)/n ± (Mx · yi / Σy²) ± (My · xi / Σx²). The corner pile located at the maximum coordinate distances (+xmax, +ymax) experiences the maximum combined compressive load.

Question 25Geotechnical engineering models and applications

A precast driven concrete square pile (0.40 m × 0.40 m) of length 20.0 m is driven into saturated silty sand. What phenomenon typically causes the pile bearing capacity measured immediately after driving (end-of-driving EOD) to be significantly lower than that measured after a 28-day rest period (set-up / 桩的休止期)?

  1. Rapid degradation of pile concrete compressive strength over time
  2. Dissipation of excess pore water pressure generated during driving and thixotropic soil structure recovery (土体触变恢复与超静孔隙水压力消散) Correct answer.
  3. Thermal contraction of the concrete pile core
  4. Chemical dissolution of quartz sand grains by groundwater

Explanation

During pile driving into saturated fine soils, intense shear remolding generates high positive excess pore water pressures, drastically reducing effective stresses and shear strength. During the statutory rest period (休止期, 14–28 days), excess pore pressures dissipate, effective stresses rise, and thixotropic recovery restores soil shear strength, resulting in significant soil setup (bearing capacity increase).

Question 26Geotechnical engineering models and applications

A vertical smooth retaining wall of height H = 6.0 m supports a dry homogeneous sand backfill with horizontal surface (unit weight γ = 18.0 kN/m³, internal friction angle φ = 30.0°, cohesion c = 0). Using Rankine's earth pressure theory, what is the coefficient of active earth pressure Ka and the total active earth thrust Ea per unit length of wall?

  1. Ka = 0.333, Ea = 108.0 kN/m Correct answer.
  2. Ka = 0.500, Ea = 162.0 kN/m
  3. Ka = 0.333, Ea = 216.0 kN/m
  4. Ka = 3.000, Ea = 972.0 kN/m

Explanation

Rankine's active earth pressure coefficient: Ka = tan²(45° - φ/2) = tan²(45° - 15°) = tan²(30°) = (1/√3)² = 1/3 = 0.3333. The total active earth pressure thrust on a wall of height H is: Ea = 1/2 · γ · H² · Ka = 0.5 × 18.0 kN/m³ × (6.0 m)² × (1/3) = 0.5 × 18.0 × 36.0 × (1/3) = 108.0 kN/m.

Question 27Geotechnical engineering models and applications

A smooth vertical retaining wall of height H = 5.0 m supports a cohesive backfill with unit weight γ = 19.0 kN/m³, cohesion c = 15.0 kPa, and internal friction angle φ = 20.0° (Ka = tan²(45° - 10°) = 0.490, √Ka = 0.700). According to Rankine's theory, what is the depth of the tension crack z0 where active lateral earth pressure is zero?

  1. z0 = 0.79 m
  2. z0 = 1.58 m
  3. z0 = 2.26 m Correct answer.
  4. z0 = 3.16 m

Explanation

Rankine's active earth pressure in cohesive soil is σa = γ · z · Ka - 2c · √Ka. At the bottom of the tension crack, σa = 0: γ · z0 · Ka = 2c · √Ka ⇒ z0 = (2c) / (γ · √Ka). Substituting values: z0 = (2 × 15.0 kPa) / (19.0 kN/m³ × 0.700) = 30.0 / 13.30 = 2.256 m ≈ 2.26 m.

Question 28Geotechnical engineering models and applications

A deep excavation in soft clay is supported by a strutted diaphragm wall. According to JGJ 120-2012 (Section 4.10), what failure mechanism must be checked at the excavation base to prevent upward plastic soil flow and base heave into the pit?

  1. Prandtl / Terzaghi basal heave stability (基坑底抗隆起稳定性验算) Correct answer.
  2. Creep rupture of steel reinforcement
  3. Thermal expansion buckling of diaphragm concrete
  4. Dynamic liquefaction under zero seismic excitation

Explanation

JGJ 120-2012 (Section 4.10.1) dictates that deep excavations in soft cohesive soils must undergo basal heave stability verification (基坑底抗隆起稳定性验算) using Prandtl's bearing capacity mechanism or circular slip surface methods to prevent upward extrusion of the bottom clay.

Question 29Geotechnical engineering models and applications

A soil nailing wall (土钉墙) is constructed to support an 8.0 m deep excavation in silty clay (JGJ 120-2012). A bored grouted soil nail of hole diameter d = 110 mm has an effective bond length la = 9.0 m behind the potential slip surface. The characteristic bond strength at the soil-grout interface is qsik = 35.0 kPa. What is the ultimate pullout resistance Tujk of this soil nail?

  1. 34.6 kN
  2. 72.8 kN
  3. 108.9 kN Correct answer.
  4. 152.4 kN

Explanation

According to JGJ 120-2012 (Formula 5.2.2), the ultimate pullout resistance of a grouted soil nail is Tujk = π · d · la · qsik. Substituting: Tujk = 3.1416 × 0.110 m × 9.0 m × 35.0 kPa = 3.1416 × 0.99 × 35.0 = 108.85 kN ≈ 108.9 kN.

Question 30Geotechnical engineering models and applications

A vertical retaining wall of height H = 5.0 m retains dry sand (γ = 18.0 kN/m³, φ = 30.0°, c = 0, Ka = 0.333). A uniform surcharge q = 18.0 kPa acts on the entire backfill surface. What is the total active earth thrust Ea on the wall including the surcharge effect?

  1. 75.0 kN/m
  2. 105.0 kN/m Correct answer.
  3. 135.0 kN/m
  4. 180.0 kN/m

Explanation

Active thrust from soil self-weight: Ea1 = 1/2 · γ · H² · Ka = 0.5 × 18.0 × 5.0² × (1/3) = 0.5 × 18.0 × 25.0 × (1/3) = 75.0 kN/m. Active thrust from uniform surface surcharge: Ea2 = q · H · Ka = 18.0 kPa × 5.0 m × (1/3) = 30.0 kN/m. Total active thrust: Ea = Ea1 + Ea2 = 75.0 + 30.0 = 105.0 kN/m.

Question 31Geotechnical engineering models and applications

Under the Technical Code for Ground Treatment of Buildings (JGJ 79-2012), a sand-gravel cushion replacement (砂石垫层) of thickness z = 1.5 m is placed under a strip footing of width b = 2.0 m. The stress diffusion angle of the compacted granular cushion is θ = 30.0° (tan 30° = 0.577). What is the expanded bottom width b' of the cushion over which vertical stress is distributed to the underlying soft soil?

  1. 2.00 m
  2. 2.87 m
  3. 3.73 m Correct answer.
  4. 5.00 m

Explanation

For a strip footing, the dispersed width at the cushion base is b' = b + 2 · z · tan θ. Substituting values: b' = 2.0 m + 2 × 1.5 m × tan(30.0°) = 2.0 + 3.0 × 0.5774 = 2.0 + 1.732 = 3.732 m ≈ 3.73 m.

Question 32Geotechnical engineering models and applications

In composite ground improvement design (JGJ 79-2012), column piles of diameter d = 0.50 m are installed in an equilateral triangular pattern with center-to-center spacing s = 1.50 m. What is the area replacement ratio m (面积置换率) of this composite foundation?

  1. 0.089
  2. 0.101 Correct answer.
  3. 0.111
  4. 0.157

Explanation

For an equilateral triangular layout, the effective influence diameter of one column is de = 1.05 · s = 1.05 × 1.50 m = 1.575 m. The area replacement ratio is m = d² / de² = (0.50 / 1.575)² = (0.3175)² = 0.1008 ≈ 0.101 (or m = (π d² / 4) / ((√3 / 2) · s²) = (0.19635) / (0.8660 × 2.25) = 0.19635 / 1.9485 = 0.1008 ≈ 0.101).

Question 33Geotechnical engineering models and applications

A composite foundation treated with cement-soil deep mixing columns has an area replacement ratio m = 0.15. The characteristic bearing capacity of a single column is fpk = 600 kPa, and the characteristic bearing capacity of the natural soil between columns is fsk = 80 kPa. The pile-soil bearing capacity mobilization coefficient is β = 0.90 (JGJ 79-2012). What is the characteristic bearing capacity fspk of the composite subgrade?

  1. 124.0 kPa
  2. 151.2 kPa Correct answer.
  3. 180.0 kPa
  4. 224.0 kPa

Explanation

According to JGJ 79-2012 (Formula 7.1.5-2), the composite ground characteristic bearing capacity is fspk = m · fpk + β · (1 - m) · fsk (or fspk = λ·m·fpk + β·(1-m)·fsk with λ=1.0). Substituting values: fspk = 0.15 × 600 kPa + 0.90 × (1 - 0.15) × 80 kPa = 90.0 + 0.90 × 0.85 × 80 = 90.0 + 61.2 = 151.2 kPa.

Question 34Geotechnical engineering models and applications

When designing preloading consolidation with vertical drains (sand drains / PVDs) per JGJ 79-2012, Carrillo's theorem is used to compute the combined average degree of consolidation Ut. If the radial degree of consolidation is Ur = 0.80 and vertical degree of consolidation is Uz = 0.30, what is the combined degree of consolidation Ut?

  1. 0.56
  2. 0.86 Correct answer.
  3. 0.94
  4. 1.10

Explanation

Carrillo's formula for combined radial and vertical consolidation (JGJ 79-2012 Formula 5.2.14-1) is: (1 - Ut) = (1 - Ur) · (1 - Uz). Substituting: 1 - Ut = (1 - 0.80) × (1 - 0.30) = 0.20 × 0.70 = 0.14. Thus, Ut = 1 - 0.14 = 0.86 (86%).

Question 35Geotechnical engineering models and applications

In dynamic compaction design (强夯法, JGJ 79-2012 Section 6.2), Menard's empirical formula H = α · √(M · h / 10) is used to estimate the effective reinforcement depth H (m), where M is tamper mass (kN), h is drop height (m), and α is an empirical coefficient. If tamper mass is M = 200 kN, drop height is h = 15.0 m, and α = 0.60 for gravelly sand, what is the estimated reinforcement depth H?

  1. 5.2 m
  2. 10.4 m Correct answer.
  3. 17.3 m
  4. 30.0 m

Explanation

Energy level = M · h / 10 = 200 × 15.0 / 10 = 300 (or nominal energy 3000 kN·m). Effective depth: H = α · √(M · h / 10) = 0.60 × √(300) = 0.60 × 17.32 m = 10.39 m ≈ 10.4 m.

Question 36Geotechnical engineering models and applications

In vibro-replacement stone column improvement (振冲碎石桩) of saturated silty sand ground (JGJ 79-2012), what are the dual engineering functions achieved by the installed gravel columns?

  1. Vibration compaction / densification of the surrounding sand and provision of high-permeability vertical gravel drains for rapid pore pressure dissipation during earthquakes Correct answer.
  2. Total chemical solidification and permanent groundwater contamination
  3. Thermal insulation and permafrost protection
  4. Increasing ground electrical conductivity for lightning arrestors

Explanation

In cohesionless and silty sands, vibro-replacement densifies the surrounding soil matrix through lateral vibratory compaction while the aggregate column functions as a rigid reinforcement element and a continuous vertical gravel drain, preventing excess pore water pressure buildup and mitigating liquefaction during seismic events.

Question 37Structural engineering models and applications

A simply supported reinforced concrete beam has an effective span of L = 6.0 m and carries a total factored uniform ultimate load of w = 40 kN/m. What is the maximum ultimate bending moment M_u acting at mid-span?

  1. 120 kN·m
  2. 180 kN·m Correct answer.
  3. 240 kN·m
  4. 360 kN·m

Explanation

For a simply supported beam subject to a uniform load w, the maximum ultimate bending moment occurs at mid-span and is calculated using M_u = (w * L^2) / 8. Substituting w = 40 kN/m and L = 6.0 m yields M_u = (40 * 6.0^2) / 8 = (40 * 36) / 8 = 180 kN·m. This fundamental structural analysis formula is standard in HK structural concrete design.

Question 38Structural engineering models and applications

A cantilever retaining wall stem of effective height L = 4.0 m experiences a uniform ultimate lateral earth pressure load of w = 25 kN/m. What is the ultimate bending moment M_u at the base of the cantilever?

  1. 100 kN·m
  2. 200 kN·m Correct answer.
  3. 400 kN·m
  4. 800 kN·m

Explanation

For a cantilever beam subjected to a uniform distributed load w over length L, the maximum bending moment occurs at the fixed support (base) and equals M_u = (w * L^2) / 2. Substituting w = 25 kN/m and L = 4.0 m gives M_u = (25 * 4.0^2) / 2 = (25 * 16) / 2 = 200 kN·m. This governs the main vertical reinforcement required in cantilever retaining walls.

Question 39Structural engineering models and applications

In elastic uncracked reinforced concrete elastic analysis, what is the modular ratio α_e defined as?

  1. The ratio of concrete compressive strength to steel yield strength (f_cu / f_yk)
  2. The ratio of steel modulus of elasticity to concrete modulus of elasticity (E_s / E_c) Correct answer.
  3. The ratio of concrete modulus of elasticity to steel modulus of elasticity (E_c / E_s)
  4. The ratio of steel yield strain to concrete ultimate compressive strain (ε_y / ε_cu)

Explanation

The modular ratio α_e is defined as the ratio of the modulus of elasticity of reinforcing steel to that of concrete, α_e = E_s / E_c. In transformed section analysis, it is used to convert the steel area into an equivalent area of concrete. Typical values for short-term loading range between 6 and 10 depending on concrete grade.

Question 40Structural engineering models and applications

A pin-ended steel column of length L = 4.0 m has flexural rigidity EI = 2,000 kN·m². According to Euler's column formula, what is the critical elastic buckling load P_cr?

  1. 1,233.7 kN Correct answer.
  2. 2,467.4 kN
  3. 4,934.8 kN
  4. 9,869.6 kN

Explanation

Euler's critical buckling load formula for a pinned-pinned column is P_cr = (π² * EI) / L^2. Substituting EI = 2,000 kN·m² and L = 4.0 m gives P_cr = (9.8696 * 2,000) / (4.0^2) = 19,739.2 / 16 = 1,233.7 kN. This represents the theoretical elastic stability limit before material yielding occurs.

Question 41Structural engineering models and applications

What is the primary structural function of transverse link reinforcement (shear stirrups) in a reinforced concrete beam?

  1. To resist longitudinal flexural tension stresses at mid-span
  2. To resist diagonal tension forces resulting from shear and control shear cracking Correct answer.
  3. To increase the modulus of elasticity of concrete in compression
  4. To prevent drying shrinkage cracking of unreinforced concrete cover

Explanation

Transverse links (stirrups) cross potential diagonal tension cracks caused by shear forces near beam supports. They transfer vertical shear loads across cracks, provide mechanical confinement to compression concrete, and secure longitudinal reinforcement against buckling. Flexural tension at mid-span is resisted primarily by main longitudinal steel.

Question 42Structural engineering models and applications

In reinforced concrete beam analysis, what defines the balanced strain condition?

  1. Concrete reaches ultimate compressive strain ε_cu = 0.0035 exactly as tension steel reaches its yield strain ε_y Correct answer.
  2. Concrete compression stress reaches 0.67 f_cu while steel tension stress reaches zero
  3. The neutral axis depth x equals the total beam depth h
  4. Tension reinforcement yields before concrete compressive strain reaches 0.0010

Explanation

The balanced strain condition occurs when the extreme fiber of concrete in compression reaches its maximum design strain (ε_cu = 0.0035 under HK Concrete Code) simultaneously as the outermost tension reinforcement reaches its design yield strain (ε_y = f_yk / (γ_s * E_s)). Under-reinforced sections yield steel first, ensuring ductile failure warnings.

Question 43Structural engineering models and applications

A rectangular solid timber beam section has width b = 200 mm and total depth h = 400 mm. What is its second moment of area (moment of inertia) I about the major centroidal axis?

  1. 5.33 × 10^7 mm^4
  2. 1.07 × 10^9 mm^4 Correct answer.
  3. 2.13 × 10^9 mm^4
  4. 4.27 × 10^9 mm^4

Explanation

The second moment of area for a rectangular section about its centroidal axis is given by I = (b * h^3) / 12. Substituting b = 200 mm and h = 400 mm gives I = (200 * 400^3) / 12 = (200 * 64,000,000) / 12 = 12,800,000,000 / 12 = 1,066,666,667 mm^4 = 1.07 × 10^9 mm^4. This value is critical for bending stress and deflection checks.

Question 44Structural engineering models and applications

A symmetric steel UB section has second moment of area I_x = 3.0 × 10^8 mm^4 and depth D = 600 mm. If it experiences a design bending moment M = 450 kN·m, what is the maximum extreme fiber bending stress σ?

  1. 225.0 N/mm² Correct answer.
  2. 450.0 N/mm²
  3. 675.0 N/mm²
  4. 900.0 N/mm²

Explanation

The elastic bending stress formula is σ = (M * y) / I, where y is the distance from the neutral axis to the extreme fiber (y = D / 2 = 300 mm). Converting M = 450 kN·m = 4.5 × 10^8 N·mm: σ = (4.5 × 10^8 N·mm * 300 mm) / (3.0 × 10^8 mm^4) = 1,350 / 6 = 225 N/mm². This is within the allowable design strength of S275 or S355 structural steel.

Question 45Structural engineering models and applications

A prestressed concrete beam has cross-sectional area A = 2.0 × 10^5 mm², section modulus Z_bottom = 2.5 × 10^7 mm³, and prestress force P = 1,200 kN acting at an eccentricity e = 150 mm below the centroid. If the applied sagging bending moment is M = 300 kN·m, what is the net stress at the bottom fiber of the concrete?

  1. +1.2 N/mm² (compression) Correct answer.
  2. +1.2 N/mm² (tension)
  3. +25.2 N/mm² (compression)
  4. -5.2 N/mm² (tension)

Explanation

Bottom fiber stress formula: σ_bottom = P/A + (P * e) / Z_bottom - M / Z_bottom. Direct compression P/A = 1,200,000 / 200,000 = +6.0 N/mm². Prestress moment compression at bottom (P * e) / Z = (1,200,000 * 150) / 25,000,000 = +7.2 N/mm². Applied moment tension at bottom -M / Z = -(300,000,000) / 25,000,000 = -12.0 N/mm². Net bottom stress = +6.0 + 7.2 - 12.0 = +1.2 N/mm² (compression).

Question 46Structural engineering models and applications

A compact steel structural section has a plastic section modulus S_p = 1,200 cm³ and yield strength f_y = 355 N/mm². What is its ultimate plastic moment capacity M_p?

  1. 213.0 kN·m
  2. 426.0 kN·m Correct answer.
  3. 852.0 kN·m
  4. 1,278.0 kN·m

Explanation

Plastic moment capacity is calculated as M_p = f_y * S_p. Converting S_p = 1,200 cm³ = 1,200 × 10^3 mm³ = 1.2 × 10^6 mm³: M_p = 355 N/mm² * 1.2 × 10^6 mm³ = 426,000,000 N·mm = 426.0 kN·m. Plastic analysis allows full section yielding compared to elastic analysis.

Question 47Structural engineering models and applications

On a reinforced concrete column axial force-bending moment (P-M) interaction diagram, what signifies a tension-controlled failure region?

  1. Axial load P is above the balanced failure load P_b; steel strain is below yield while concrete crushes first
  2. Axial load P is below the balanced failure load P_b; tension steel yields before concrete reaches crushing strain Correct answer.
  3. Axial load P equals zero while bending moment M equals zero
  4. Pure axial compression with zero bending moment eccentricity

Explanation

In a P-M interaction diagram, the tension-controlled region occurs when the applied axial force P is less than the balanced axial capacity P_b. Under low axial load and high bending moment, tension reinforcement yields ductilely before concrete crushes, providing visible warning prior to structural failure.

Question 48Water Resources and Environmental engineering models and applications

For a hypothetical catchment, use Q=C*(I/1000)*A/3600 with I inmm/h and A inm^2. Given C=0.85,I=120mm/h and A=150000m^2, what modeled discharge inm^3/s follows?

  1. 4.25 m³/s Correct answer.
  2. 5.10 m³/s
  3. 18.36 m³/s
  4. 51.0 m³/s

Explanation

Convert120mm/h to0.12m/h. The defined model gives Q=0.85*0.12*150000/3600=4.25m^3/s. The supplied coefficient and intensity are exercise assumptions; rainfall duration, catchment applicability and actual drainage requirements remain separate.

Question 49Water Resources and Environmental engineering models and applications

What is the Manning equation formula for average flow velocity V (m/s) in an open channel under uniform flow conditions?

  1. V = (1 / n) * R^(2/3) * S^(1/2) Correct answer.
  2. V = n * R^(1/2) * S^(2/3)
  3. V = (1 / n) * R^(1/2) * S^(1/2)
  4. V = (1 / n) * R^(3/2) * S^(2)

Explanation

Manning's formula in SI units is V = (1 / n) * R^(2/3) * S^(1/2), where n is Manning's roughness coefficient, R is hydraulic radius (A / P), and S is the energy slope (bed slope for uniform flow). It is the fundamental equation specified in the DSD Stormwater Drainage Manual for gravity pipe and channel sizing.

Question 50Water Resources and Environmental engineering models and applications

What is the hydraulic radius R of a fully flowing circular pipe of internal diameter D?

  1. D / 2
  2. D / 4 Correct answer.
  3. π D / 4
  4. D / 8

Explanation

Hydraulic radius R is defined as cross-sectional flow area A divided by wetted perimeter P. For a circular pipe flowing full: A = (π * D²) / 4 and P = π * D. Therefore, R = [(π * D²) / 4] / (π * D) = D / 4. This key relationship simplifies pipe flow calculations.