CELE Geotechnical Engineering — Lateral Earth Pressure and Retaining StructuresExam Answer Templates

Marks

1

Topic

Earth-Pressure Coefficients (Rankine)

Difficulty

easy

Template Id

T1

Examiner Tip

Examiners accept either equivalent form of K_a. Writing both forms (the fraction and the tan² form) in a 1-mark question shows mastery and protects you if one form is transcribed with a minor error.

Model Answer

The coefficient of active earth pressure K_a (Rankine) is: K_a = (1 − sin φ) / (1 + sin φ) = tan²(45° − φ/2) where φ is the effective angle of internal friction of the cohesionless backfill, assuming a smooth vertical wall and horizontal ground surface.

Question Type

very_short_answer

Answer Structure

• Line 1: State the formula with correct notation [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing K_p formula instead of K_a (sign confusion in the angle).
• Omitting the condition of smooth vertical wall and horizontal backfill.
• Using degree symbol without converting — always keep φ in degrees for tan function.

Key Phrases To Include

• K_a
• tan²(45° − φ/2)
• (1 − sin φ)/(1 + sin φ)
• angle of internal friction
• Rankine
• smooth vertical wall

Marks

2

Topic

Earth-Pressure Coefficients (Rankine)

Difficulty

easy

Template Id

T2

Examiner Tip

The relationship K_p = 1/K_a is a quick self-check. Always verify: if K_a × K_p ≠ 1.0, you have an arithmetic error. Examiners note this check favorably.

Model Answer

For φ = 30°: At-rest: K_0 = 1 − sin 30° = 1 − 0.500 = 0.500 Active: K_a = tan²(45° − 30°/2) = tan²(30°) = (0.5774)² = 0.333 Passive: K_p = tan²(45° + 30°/2) = tan²(60°) = (1.7321)² = 3.000 Relationship: K_a < K_0 < K_p (0.333 < 0.500 < 3.000) ✓ Note: K_p = 1/K_a = 1/0.333 = 3.00 (valid for Rankine with φ only).

Question Type

short_answer

Answer Structure

• Line 1–3: Compute K_0, K_a, K_p with substitution shown [1 mark]
• Line 4: State the inequality K_a < K_0 < K_p and verify with computed values [1 mark]

Scoring Breakdown

Common Mark Deductions

• Correct values but inequality written backwards (K_p < K_0 < K_a).
• Not showing the substitution step — just writing the answer without formula use loses the method mark.
• Rounding K_a to 0.34 or K_p to 2.99 without acknowledging it is an approximation.

Key Phrases To Include

• K_0 = 1 − sin φ
• K_a = tan²(45° − φ/2)
• K_p = tan²(45° + φ/2)
• K_a < K_0 < K_p
• K_p = 1/K_a

Marks

2

Topic

Resultant Active Thrust

Difficulty

easy

Template Id

T3

Examiner Tip

Write 'per unit length of wall' alongside your P_a answer. Board exam questions specify 'per meter of wall' in the answer choices — matching this phrasing avoids confusion.

Model Answer

Step 1 — Compute K_a: K_a = tan²(45° − 28°/2) = tan²(31°) = (0.6009)² = 0.361 Step 2 — Compute active thrust: P_a = ½ K_a γ H² P_a = ½ × 0.361 × 17 × (4)² P_a = ½ × 0.361 × 17 × 16 P_a = 49.1 kN/m Point of application: H/3 = 4/3 = 1.33 m above the base.

Question Type

numerical

Answer Structure

• Step 1: K_a formula and value [0.5 mark]
• Step 2: P_a formula with correct substitution and units [1 mark]
• Final: Numeric answer in kN/m + point of application [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using γ = 9.81 (confusing unit weight of soil with unit weight of water).
• Forgetting the ½ factor in the thrust formula — this is the most common arithmetic error on board exams.
• Not stating the point of application (H/3) — partial credit is lost.
• Leaving the answer in kN instead of kN/m (per unit length of wall).

Key Phrases To Include

• K_a = tan²(45° − φ/2)
• P_a = ½ K_a γ H²
• kN/m
• H/3 above the base
• triangular pressure distribution

Marks

3

Topic

Earth-Pressure States

Difficulty

easy

Template Id

T4

Examiner Tip

For a 'differentiate' question worth 3 marks, a table format (two columns: Active | Passive; three rows: movement, soil state, magnitude) is perfectly acceptable and often faster to write than prose.

Model Answer

Active Earth Pressure (K_a): • Occurs when the wall moves AWAY from the backfill, allowing the soil to expand laterally. • The soil is on the verge of shear failure in the active (stretching) state. • K_a = tan²(45° − 35°/2) = tan²(27.5°) = 0.271 → smallest earth pressure. Passive Earth Pressure (K_p): • Occurs when the wall is pushed INTO the soil (e.g., at the toe of a retaining wall), compressing it. • The soil resists the wall movement — shear failure in the passive (compressing) state. • K_p = tan²(45° + 35°/2) = tan²(62.5°) = 3.690 → largest earth pressure. Key Comparison: Active: K_a = 0.271 (drives the wall) → DESTABILIZING force Passive: K_p = 3.690 (resists the wall) → STABILIZING force Ratio: K_p/K_a = 3.690/0.271 ≈ 13.6× — passive is over 13 times larger than active for φ = 35°.

Question Type

short_answer

Answer Structure

• Part 1: Define active pressure — wall movement direction + soil state [1 mark]
• Part 2: Define passive pressure — wall movement direction + soil state [1 mark]
• Part 3: Compute K_a and K_p for φ = 35°, compare magnitudes and engineering significance [1 mark]

Scoring Breakdown

Common Mark Deductions

• Confusing which direction of wall movement triggers active vs. passive.
• Stating K_p = 1/K_a without verifying: for φ = 35°, 1/0.271 = 3.690 ✓ — skipping this check looks careless.
• Not addressing the engineering significance (destabilizing vs. stabilizing) — this is what examiners expect in a 'differentiate' question.

Key Phrases To Include

• wall moves away
• soil expands laterally
• active state — destabilizing
• wall pushed into soil
• passive resistance — stabilizing
• K_p = 1/K_a (Rankine)
• tan²(45° ± φ/2)

Marks

3

Topic

Resultant Thrust with Surcharge

Difficulty

medium

Template Id

T5

Examiner Tip

Draw a quick pressure diagram on your answer sheet: triangle (soil) + rectangle (surcharge). This sketch earns a diagram mark if available and prevents the H/2 vs. H/3 confusion.

Model Answer

Given: H = 5 m, γ = 18 kN/m³, φ = 30°, q = 20 kPa Step 1 — Compute K_a: K_a = tan²(45° − 30°/2) = tan²(30°) = 0.333 Step 2 — Pressure components at depth H = 5 m: Soil pressure at base: σ_a = K_a γ H = 0.333 × 18 × 5 = 29.97 kPa ≈ 30.0 kPa Surcharge pressure: σ_q = K_a q = 0.333 × 20 = 6.67 kPa (uniform over full height) Step 3 — Resultant thrusts: P_soil = ½ K_a γ H² = ½ × 0.333 × 18 × 25 = 75.0 kN/m (at H/3 = 1.67 m from base) P_surcharge = K_a q H = 0.333 × 20 × 5 = 33.3 kN/m (at H/2 = 2.50 m from base) Step 4 — Total active thrust: P_a,total = 75.0 + 33.3 = 108.3 kN/m ∴ Total active thrust = 108.3 kN/m acting as two resultants: triangular (at 1.67 m) + rectangular (at 2.50 m).

Question Type

numerical

Answer Structure

• Step 1: K_a computed correctly [0.5 mark]
• Step 2: Separate pressure diagrams for soil and surcharge identified [0.5 mark]
• Step 3: Individual thrusts P_soil and P_surcharge with correct formulas and points of application [1 mark]
• Step 4: Total thrust summed with units [1 mark]

Scoring Breakdown

Common Mark Deductions

• Adding surcharge thrust to soil thrust without showing the two separate pressure components — loses the method marks.
• Using P_surcharge acting at H/3 instead of H/2 — surcharge is rectangular (uniform), not triangular.
• Forgetting to multiply K_a into the surcharge: writing P_surcharge = qH = 100 kN/m instead of K_a×q×H = 33.3 kN/m.

Key Phrases To Include

• P_soil = ½ K_a γ H²
• P_surcharge = K_a q H
• triangular pressure distribution at H/3
• rectangular surcharge block at H/2
• total active thrust = soil + surcharge components

Marks

2

Topic

Active Pressure in Cohesive Soils

Difficulty

medium

Template Id

T6

Examiner Tip

Memorize z_c = 2c/(γ√K_a) as a single formula rather than deriving it from the Rankine active pressure equation during the exam — this saves 1–2 minutes under time pressure.

Model Answer

Given: c = 18 kPa, φ = 20°, γ = 19 kN/m³ Step 1 — Compute K_a: K_a = tan²(45° − 20°/2) = tan²(35°) = 0.490 √K_a = √0.490 = 0.700 Step 2 — Tension crack depth: z_c = 2c / (γ √K_a) z_c = 2 × 18 / (19 × 0.700) z_c = 36 / 13.30 z_c = 2.71 m Physical meaning: No net active pressure acts above z_c = 2.71 m. The soil can theoretically stand unsupported (or a crack opens) to this depth.

Question Type

numerical

Answer Structure

• Step 1: K_a and √K_a computed [0.5 mark]
• Step 2: z_c formula applied with correct substitution and units [1 mark]
• Physical meaning stated [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using z_c = 2c/γ (forgetting the √K_a term) — this is a very common error.
• Using K_a directly instead of √K_a in the denominator.
• Not stating the physical meaning of z_c — 'depth of tension crack' alone is insufficient; explain that no active pressure acts above it.

Key Phrases To Include

• z_c = 2c / (γ √K_a)
• tension crack
• no lateral pressure above z_c
• √K_a
• cohesive soil / clay

Marks

2

Topic

Retaining Wall Stability — Sliding

Difficulty

medium

Template Id

T7

Examiner Tip

Board exam questions often phrase this as 'Why is passive earth pressure neglected?' Accept only two clear engineering reasons. The phrase 'conservative design practice' alone earns no credit without justification.

Model Answer

Passive earth pressure at the toe of a retaining wall is conservatively neglected in sliding calculations for two engineering reasons: 1. Mobilization requirement: Passive resistance requires significant lateral displacement of the wall into the soil in front of the toe — typically 3–5% of the wall height. This displacement may be unacceptable in service. 2. Reliability of passive zone: The passive soil in front of the wall may be removed (by erosion, excavation, or utility trenching) during the life of the structure, making it an unreliable source of resistance. Conclusion: Including P_p is non-conservative design practice. The sliding check FS = μΣW / P_a,H ≥ 1.5 (without P_p) gives a conservative (safe-side) result that is preferred in Philippine practice.

Question Type

short_answer

Answer Structure

• Reason 1: Mobilization/displacement requirement [1 mark]
• Reason 2: Reliability/risk of removal of passive zone [1 mark]

Scoring Breakdown

Common Mark Deductions

• Saying only 'it is conservative' without explaining WHY passive is mobilized differently from active.
• Not citing the displacement incompatibility (active mobilized at small strains; passive requires large strains).
• Confusing passive toe resistance with friction at the base — these are two separate resisting mechanisms.

Key Phrases To Include

• large lateral displacement required
• conservative design
• erosion or excavation of passive zone
• unreliable
• safe-side
• FS = μΣW / P_a,H

Marks

2

Topic

Retaining Wall Stability — Overturning

Difficulty

easy

Template Id

T8

Examiner Tip

Overturning moments are always taken about the TOE of the wall. State this in your solution. If the question says 'about the base corner,' confirm it is the toe side — this avoids confusion in asymmetric wall problems.

Model Answer

(a) Overturning check: FS_OT = ΣM_R / M_O FS_OT = 480 / 180 FS_OT = 2.67 (b) Minimum required: FS_OT ≥ 1.5 to 2.0 (typically 2.0 for retaining walls in Philippine practice). Conclusion: FS_OT = 2.67 > 2.0 ∴ Wall is SAFE against overturning. ✓

Question Type

numerical

Answer Structure

• Part (a): Apply FS_OT = ΣM_R/M_O and compute [1 mark]
• Part (b): State minimum FS criterion and conclude safe/unsafe [1 mark]

Scoring Breakdown

Common Mark Deductions

• Inverting the ratio: FS = M_O/ΣM_R — always resisting over overturning.
• Not writing the conclusion (SAFE/UNSAFE) — examiners specifically look for this decision statement.
• Stating FS ≥ 1.0 as the criterion — this is technically correct but unprofessional; always use the practical FS ≥ 2.0 for overturning.

Key Phrases To Include

• FS_OT = ΣM_R / M_O
• FS ≥ 2.0
• moment about the toe
• SAFE against overturning
• resisting moment vs. overturning moment

Marks

2

Topic

Retaining Wall Stability — Sliding

Difficulty

easy

Template Id

T9

Examiner Tip

When FS is very close to the minimum (e.g., 1.516 vs. 1.50), add a professional comment: 'The wall is marginally safe; a key or passive resistance at the toe is recommended.' This demonstrates engineering judgment beyond mere calculation.

Model Answer

Given: ΣW = 320 kN/m, μ = 0.45, P_a,H = 95 kN/m Sliding resistance: F_R = μ × ΣW = 0.45 × 320 = 144 kN/m Factor of safety against sliding: FS_slide = F_R / P_a,H = 144 / 95 = 1.516 Minimum required: FS_slide ≥ 1.5 Conclusion: FS_slide = 1.516 > 1.5 ∴ Wall is SAFE against sliding (marginally). ✓ Note: Passive resistance at the toe has been conservatively neglected in this calculation.

Question Type

numerical

Answer Structure

• Step 1: Compute sliding resistance F_R = μΣW [0.5 mark]
• Step 2: Apply FS_slide = F_R / P_a,H [0.5 mark]
• State FS criterion and conclusion [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using ΣW without multiplying by μ — writing FS = ΣW/P_a,H = 320/95 = 3.37 is a completely wrong formula.
• Not noting that passive earth pressure was neglected.
• Not writing 'marginally safe' or commenting on how close FS is to the minimum limit — this shows engineering judgment.

Key Phrases To Include

• FS_slide = μΣW / P_a,H
• FS ≥ 1.5
• sliding resistance
• coefficient of friction μ
• passive neglected (conservative)
• SAFE against sliding

Marks

5

Topic

Retaining Wall Stability — Complete Analysis

Difficulty

hard

Template Id

T10

Examiner Tip

In 5-mark questions, structure your answer with clearly labeled STEPS. Examiners award marks per step, not per equation. A neat, organized solution where each step is identifiable is worth significantly more than a jumbled correct answer.

Model Answer

Given: H = 6 m, γ = 18 kN/m³, φ = 32°, B = 3 m, ΣW = 280 kN/m, μ = 0.50 ━━━━━━━━━━━━━━━━━━━━━━━━━━ STEP 1 — Active Earth Pressure Coefficient ━━━━━━━━━━━━━━━━━━━━━━━━━━ K_a = tan²(45° − 32°/2) = tan²(29°) tan 29° = 0.5543 K_a = (0.5543)² = 0.307 ━━━━━━━━━━━━━━━━━━━━━━━━━━ STEP 2 — Total Active Thrust ━━━━━━━━━━━━━━━━━━━━━━━━━━ P_a = ½ K_a γ H² P_a = ½ × 0.307 × 18 × (6)² P_a = ½ × 0.307 × 18 × 36 P_a = 99.5 kN/m Point of application: H/3 = 6/3 = 2.0 m above the base ━━━━━━━━━━━━━━━━━━━━━━━━━━ STEP 3 — Overturning Check ━━━━━━━━━━━━━━━━━━━━━━━━━━ M_O = P_a × (H/3) = 99.5 × 2.0 = 199.0 kN·m/m For a simple gravity wall, resisting moment: ΣM_R = ΣW × (B/2) [approximation: resultant at center of base] ΣM_R = 280 × (3/2) = 280 × 1.5 = 420 kN·m/m FS_OT = ΣM_R / M_O = 420 / 199 = 2.11 Required: FS_OT ≥ 2.0 ∴ FS_OT = 2.11 > 2.0 → SAFE against overturning ✓ ━━━━━━━━━━━━━━━━━━━━━━━━━━ STEP 4 — Sliding Check ━━━━━━━━━━━━━━━━━━━━━━━━━━ F_R = μ × ΣW = 0.50 × 280 = 140 kN/m FS_slide = F_R / P_a = 140 / 99.5 = 1.407 Required: FS_slide ≥ 1.5 ∴ FS_slide = 1.407 < 1.5 → UNSAFE against sliding ✗ ━━━━━━━━━━━━━━━━━━━━━━━━━━ STEP 5 — Engineering Recommendation ━━━━━━━━━━━━━━━━━━━━━━━━━━ The wall fails the sliding check. Remedial measures: • Increase base width B to increase ΣW. • Provide a shear key at the base to increase resistance. • Use roughened base surface to increase μ.

Question Type

numerical

Answer Structure

• Step 1: K_a formula + correct value (0.307) [1 mark]
• Step 2: P_a formula + correct value (99.5 kN/m) + point of application [1 mark]
• Step 3: M_O and ΣM_R computed; FS_OT = 2.11 > 2.0 → SAFE [1.5 marks]
• Step 4: FS_slide = 1.407 < 1.5 → UNSAFE conclusion [1 mark]
• Step 5: Engineering recommendation for sliding failure [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Stopping after FS_OT and not checking sliding — the question explicitly asks for BOTH checks.
• Not writing SAFE or UNSAFE conclusions — just computing numbers without engineering judgment.
• Omitting the engineering recommendation for the failing check — this is always expected in a 5-mark answer.
• Using the wrong lever arm for M_O: P_a acts at H/3 from the base, not at H/2.

Key Phrases To Include

• K_a = tan²(45° − φ/2)
• P_a = ½K_aγH²
• H/3 from base
• FS_OT = ΣM_R / M_O ≥ 2.0
• FS_slide = μΣW / P_a ≥ 1.5
• SAFE / UNSAFE
• shear key
• engineering recommendation

Marks

5

Topic

Lateral Pressure with Water Table

Difficulty

hard

Template Id

T11

Examiner Tip

The water table behind a retaining wall is a classic board exam trap. Always split the problem into two forces: effective soil pressure (with γ') + hydrostatic water pressure (with γ_w). Never use γ_sat in the K_a formula when a water table is present.

Model Answer

Given: H = 5 m, γ_sat = 19.5 kN/m³, φ = 28°, γ_w = 9.81 kN/m³ Water table at top of backfill → use submerged (buoyant) unit weight throughout. ━━━━━━━━━━━━━━━━━━━━━━━━━━ STEP 1 — Submerged Unit Weight ━━━━━━━━━━━━━━━━━━━━━━━━━━ γ' = γ_sat − γ_w = 19.5 − 9.81 = 9.69 kN/m³ ━━━━━━━━━━━━━━━━━━━━━━━━━━ STEP 2 — Active Earth Pressure Coefficient ━━━━━━━━━━━━━━━━━━━━━━━━━━ K_a = tan²(45° − 28°/2) = tan²(31°) = (0.6009)² = 0.361 ━━━━━━━━━━━━━━━━━━━━━━━━━━ STEP 3 — Effective Active Thrust (soil skeleton only) ━━━━━━━━━━━━━━━━━━━━━━━━━━ P_a (effective) = ½ K_a γ' H² P_a = ½ × 0.361 × 9.69 × (5)² P_a = ½ × 0.361 × 9.69 × 25 P_a = 43.8 kN/m (acts at H/3 = 1.67 m from base) ━━━━━━━━━━━━━━━━━━━━━━━━━━ STEP 4 — Hydrostatic Water Pressure ━━━━━━━━━━━━━━━━━━━━━━━━━━ P_w = ½ γ_w H² = ½ × 9.81 × (5)² P_w = ½ × 9.81 × 25 = 122.6 kN/m (acts at H/3 = 1.67 m from base) ━━━━━━━━━━━━━━━━━━━━━━━━━━ STEP 5 — Total Lateral Force ━━━━━━━━━━━━━━━━━━━━━━━━━━ P_total = P_a (effective) + P_w P_total = 43.8 + 122.6 = 166.4 kN/m Compare with dry backfill: P_a,dry = ½ × 0.361 × 18 × 25 = 81.2 kN/m The presence of the water table MORE THAN DOUBLES the lateral force (166.4 vs. 81.2 kN/m). This illustrates why drainage behind retaining walls is critical.

Question Type

numerical

Answer Structure

• Step 1: Compute γ' = γ_sat − γ_w [0.5 mark]
• Step 2: K_a = 0.361 [0.5 mark]
• Step 3: Effective active thrust P_a = 43.8 kN/m using γ' [1 mark]
• Step 4: Hydrostatic thrust P_w = 122.6 kN/m [1.5 marks]
• Step 5: Total = 166.4 kN/m + comparison/engineering insight [1.5 marks]

Scoring Breakdown

Common Mark Deductions

• Using γ_sat directly in the active thrust formula instead of γ' — this is the most critical error; it double-counts the water pressure.
• Not separating the soil effective stress and water pressure into two distinct forces.
• Missing the engineering insight on the importance of drainage — often tested as a follow-up in board exams.

Key Phrases To Include

• γ' = γ_sat − γ_w
• effective active thrust using γ'
• separate hydrostatic force P_w = ½γ_wH²
• drainage behind retaining walls
• water table doubles lateral force

Marks

1

Topic

At-Rest Earth Pressure

Difficulty

easy

Template Id

T12

Examiner Tip

K_0 questions in 1-mark format only require the formula + one application. Do not over-explain — keep it to two lines maximum.

Model Answer

K_0 = 1 − sin φ (Jaky's formula for normally consolidated soils) K_0 is used when the wall does not move (rigid basement walls, braced excavations) so neither active nor passive conditions develop.

Question Type

very_short_answer

Answer Structure

• Line 1: K_0 formula [0.5 mark]
• Line 2: Condition for use — no wall movement [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Writing K_0 = 1 − sin φ without stating the condition for its use — gains only 0.5 mark.
• Confusing K_0 with K_a — K_0 > K_a always.

Key Phrases To Include

• K_0 = 1 − sin φ
• Jaky's formula
• no lateral movement
• rigid basement walls
• braced excavations

Marks

2

Topic

Retaining Wall Bearing Stability

Difficulty

medium

Template Id

T13

Examiner Tip

Draw a quick sketch of the base pressure distribution for two cases: e ≤ B/6 (trapezoidal — all compressive) and e > B/6 (triangular — toe side only). A 2-second sketch earns visual marks and clarifies your answer.

Model Answer

Middle-Third Rule: The resultant of all vertical forces (ΣW) must fall within the middle third of the base width B, i.e., the eccentricity e ≤ B/6. Significance: • If e ≤ B/6: The pressure distribution under the entire base is compressive — no tension between soil and wall base. This is the safe design condition. • If e > B/6: Tension develops under part of the base (soil cannot take tension), causing the effective bearing area to reduce and toe pressure to increase sharply, risking bearing capacity failure. Check: e = B/2 − x̄, where x̄ is the distance of the resultant from the toe. Allowable: e ≤ B/6

Question Type

short_answer

Answer Structure

• Part 1: State the rule with formula e ≤ B/6 [1 mark]
• Part 2: Explain significance — no tension when e ≤ B/6; bearing failure risk when e > B/6 [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating the rule without the formula e ≤ B/6.
• Not explaining what happens when e > B/6 (tension → reduced bearing area → high toe pressure).
• Confusing the middle-third rule with the overturning check — they are related but distinct stability criteria.

Key Phrases To Include

• eccentricity e ≤ B/6
• middle third of base
• no tension under base
• compressive pressure distribution
• e = B/2 − x̄
• bearing capacity failure

Marks

3

Topic

Earth-Pressure Coefficients (Rankine)

Difficulty

easy

Template Id

T14

Examiner Tip

This is a common Exercise 1 in the chapter — treat it as a guaranteed easy 3-mark question. Practice computing tan²(45° ± φ/2) quickly with a calculator to save time on board exam day.

Model Answer

Given: φ = 34° Step 1 — At-Rest: K_0 = 1 − sin 34° = 1 − 0.5592 = 0.441 Step 2 — Active: K_a = tan²(45° − 34°/2) = tan²(28°) tan 28° = 0.5317 K_a = (0.5317)² = 0.283 Step 3 — Passive: K_p = tan²(45° + 34°/2) = tan²(62°) tan 62° = 1.8807 K_p = (1.8807)² = 3.538 Step 4 — Verification: 1/K_a = 1/0.283 = 3.534 ≈ 3.538 ✓ (difference due to rounding) K_a × K_p = 0.283 × 3.538 ≈ 1.001 ≈ 1.0 ✓ Summary: K_a = 0.283 < K_0 = 0.441 < K_p = 3.538

Question Type

numerical

Answer Structure

• Step 1: K_0 = 0.441 [0.5 mark]
• Step 2: K_a = 0.283 with substitution shown [1 mark]
• Step 3: K_p = 3.538 with substitution shown [1 mark]
• Step 4: Verification K_p = 1/K_a confirmed [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Rounding intermediate tan values too early — keep at least 4 significant figures to avoid rounding error in the squared result.
• Skipping the verification step — K_p = 1/K_a check is explicitly asked and earns 0.5 mark.
• Not writing the inequality K_a < K_0 < K_p in the summary.

Key Phrases To Include

• K_0 = 1 − sin φ
• K_a = tan²(45° − φ/2)
• K_p = tan²(45° + φ/2)
• K_p = 1/K_a (verification)
• K_a < K_0 < K_p

Marks

5

Topic

Resultant Thrust with Surcharge — Location

Difficulty

hard

Template Id

T15

Examiner Tip

The resultant point of application of combined loads is found using the same principle as finding the centroid of a force system: Σ(Force × moment arm) / ΣForce. Write it as a moment equation — this is the professionally correct format and earns full method marks even if arithmetic is slightly off.

Model Answer

Given: H = 6 m, γ = 19 kN/m³, φ = 32°, q = 10 kPa ━━━━━━━━━━━━━━━━━━━━━━━━━━ STEP 1 — K_a ━━━━━━━━━━━━━━━━━━━━━━━━━━ K_a = tan²(45° − 32°/2) = tan²(29°) tan 29° = 0.5543 K_a = (0.5543)² = 0.307 ━━━━━━━━━━━━━━━━━━━━━━━━━━ STEP 2 — Pressure Components at Base (z = 6 m) ━━━━━━━━━━━━━━━━━━━━━━━━━━ Surcharge pressure: σ_q = K_a × q = 0.307 × 10 = 3.07 kPa (uniform) Soil pressure at base: σ_soil = K_a × γ × H = 0.307 × 19 × 6 = 35.0 kPa ━━━━━━━━━━━━━━━━━━━━━━━━━━ STEP 3 — Resultant Forces ━━━━━━━━━━━━━━━━━━━━━━━━━━ P_surcharge = σ_q × H = 3.07 × 6 = 18.4 kN/m [rectangular; acts at H/2 = 3.0 m] P_soil = ½ × σ_soil × H = ½ × 35.0 × 6 = 105.0 kN/m [triangular; acts at H/3 = 2.0 m] Total: P_total = 18.4 + 105.0 = 123.4 kN/m ━━━━━━━━━━━━━━━━━━━━━━━━━━ STEP 4 — Resultant Point of Application (ȳ from base) ━━━━━━━━━━━━━━━━━━━━━━━━━━ Taking moments about the base: P_total × ȳ = P_surcharge × (H/2) + P_soil × (H/3) 123.4 × ȳ = 18.4 × 3.0 + 105.0 × 2.0 123.4 × ȳ = 55.2 + 210.0 = 265.2 ȳ = 265.2 / 123.4 = 2.15 m from the base ━━━━━━━━━━━━━━━━━━━━━━━━━━ FINAL ANSWERS: ━━━━━━━━━━━━━━━━━━━━━━━━━━ Total active thrust P_total = 123.4 kN/m Resultant acts at ȳ = 2.15 m above the base

Question Type

numerical

Answer Structure

• Step 1: K_a = 0.307 [0.5 mark]
• Step 2: Pressure diagram with both components identified [0.5 mark]
• Step 3: P_surcharge = 18.4 kN/m (at H/2) and P_soil = 105.0 kN/m (at H/3) [2 marks]
• Step 4: Moment calculation for resultant location ȳ = 2.15 m [1.5 marks]
• Final statement with correct units and answer [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Applying surcharge thrust at H/3 (triangular) instead of H/2 (rectangular) — costs 1 mark.
• Not computing the resultant point of application — the question explicitly asks for it.
• Forgetting to multiply by K_a when computing surcharge pressure: σ_q = K_a × q, not just q.
• Computing P_surcharge = q × H = 10 × 6 = 60 kN/m without applying K_a — a full-step error.

Key Phrases To Include

• P_soil = ½ K_a γ H² at H/3
• P_surcharge = K_a q H at H/2
• moment about the base
• P_total × ȳ = ΣM
• resultant location from base
• 123.4 kN/m
• ȳ = 2.15 m