CELE Geotechnical Engineering — Shear Strength of SoilsRevision Notes
Revision notes for CELE Geotechnical Engineering — Shear Strength of Soils. Short, focused, and designed for the week before exam day. Use these when you are already familiar with the chapter and need a quick refresh on the high-yield items Professional Regulation Commission (PRC) — Board of Civil Engineering tests.
Exam context
The Civil Engineer Licensure Examination is conducted by Professional Regulation Commission (PRC) — Board of Civil Engineering and is scheduled for May and November 2026. The Geotechnical Engineering subtest is marked as "Core" in the official pattern, and Shear Strength of Soils appears in position 7th of 11 in the CELE Geotechnical Engineering review rotation. Passing mark: 70% weighted average, no sub-test below 50%. Recent CELE 2026 papers have drawn roughly a meaningful share of questions from this subject.
Shear Strength of Soils - Revision Notes
Shear strength is the most critical soil property in geotechnical engineering. It governs the safety of every foundation, embankment, slope, and retaining wall. In PRC board examinations, shear strength problems consistently appear — expect 3 to 6 items per sitting covering the Mohr-Coulomb criterion, triaxial test interpretation, unconfined compression, and the drained-vs-undrained distinction. Master the formulas, understand which test applies to which drainage condition, and avoid the classic pitfall of confusing qu with cu. This chapter equips you with every formula, concept, and exam strategy you need.
Sections
Formulas
Example
Soil: c = 20 kPa, φ = 30°, σ = 150 kPa. τf = 20 + 150 × tan 30° = 20 + 150 × 0.5774 = 20 + 86.6 = 106.6 kPa.
Formula
τf = c + σ tan φ
Variables
τf = shear strength at failure (kPa); c = cohesion intercept (kPa); σ = total normal stress on failure plane (kPa); φ = angle of internal friction (degrees)
Application
Total stress analysis — used in UU triaxial results and undrained clay problems where pore pressure is not measured.
Example
c' = 10 kPa, φ' = 28°, σ = 200 kPa, u = 60 kPa. σ' = 200 − 60 = 140 kPa. τf = 10 + 140 × tan 28° = 10 + 74.5 = 84.5 kPa.
Formula
τf = c' + σ' tan φ'
Variables
τf = shear strength at failure (kPa); c' = effective cohesion (kPa); σ' = effective normal stress = σ − u (kPa); φ' = effective friction angle (degrees); u = pore water pressure (kPa)
Application
Effective stress analysis — used in CD and CU (with pore-pressure measurement) triaxial tests; governs long-term stability.
Example
Sand: σ3 = 100 kPa, deviator stress = 200 kPa → σ1 = 300 kPa. sin φ = (300 − 100)/(300 + 100) = 200/400 = 0.5 → φ = 30°.
Formula
sin φ = (σ1 − σ3) / (σ1 + σ3) [for c = 0]
Variables
φ = friction angle; σ1 = major principal stress at failure (kPa); σ3 = minor principal stress (confining pressure) (kPa); (σ1 − σ3) = deviator stress at failure (kPa)
Application
Used exclusively for cohesionless soils (c = 0) such as sand in CD or CU triaxial tests to directly compute φ from principal stresses.
Example
c = 25 kPa, φ = 20°, σ3 = 100 kPa. tan(45 + 10°) = tan 55° = 1.428. σ1 = 100 × (1.428)² + 2 × 25 × 1.428 = 100 × 2.04 + 71.4 = 204 + 71.4 = 275.4 kPa.
Formula
σ1 = σ3 tan²(45 + φ/2) + 2c tan(45 + φ/2)
Variables
σ1 = major principal stress at failure (kPa); σ3 = confining pressure (kPa); c = cohesion (kPa); φ = friction angle (degrees); tan²(45 + φ/2) = Kp = passive earth pressure coefficient
Application
Relates principal stresses to shear strength parameters; used to find σ1 when σ3, c, and φ are known — common in triaxial problem types.
Example
φ = 30°. θf = 45 + 15 = 60° — the failure plane inclines 60° from horizontal (where σ1 acts vertically).
Formula
θf = 45° + φ/2
Variables
θf = angle of failure plane measured from the plane of the major principal stress (degrees); φ = friction angle (degrees)
Application
Determines the orientation of the failure plane inside a triaxial or direct shear specimen.
Exam Tips
- Identify whether c = 0 first — if yes, use sin φ formula directly. If c ≠ 0, use the principal stress–Mohr-Coulomb relationship.
- Label your stress state clearly: is this total or effective? Check if pore pressure u is given — if so, it is an effective stress problem.
- The tan²(45 + φ/2) term equals Kp (passive coefficient) — recognizing this speeds up calculations significantly.
- Quick check: for φ = 30°, sin 30° = 0.5, so (σ1 − σ3)/(σ1 + σ3) = 0.5 means σ1 = 3σ3. Memorize this ratio for common angles.
Key Points
- Soil fails by shearing along a plane when the applied shear stress reaches the shear strength of that plane.
- The Mohr-Coulomb criterion defines shear strength as a linear function of normal stress: τf = c + σ tan φ.
- Two forms exist: TOTAL STRESS form (uses total normal stress σ, total cohesion c, total friction angle φ) and EFFECTIVE STRESS form (uses effective normal stress σ' = σ − u, effective cohesion c', effective friction angle φ').
- The effective stress form is fundamentally correct for all soils; total stress is a shortcut valid only for undrained loading of saturated clay.
- For SAND (cohesionless soil): c ≈ 0, so τf = σ tan φ — purely frictional resistance.
- For SATURATED CLAY under UNDRAINED loading: φ = 0, so τf = cu — purely cohesive resistance.
- The failure plane makes an angle θ = 45° + φ/2 with the major principal stress plane.
- Typical φ values: clean sand 28°–36°, gravel 35°–45°, soft clay φ' ≈ 20°–30°.
- Typical cu values: very soft clay < 12.5 kPa, stiff clay 50–100 kPa.
Definitions
Term
Cohesion (c or c')
Definition
The shear strength intercept at zero normal stress. Represents inter-particle bonding (true cohesion) or apparent cohesion from capillary suction or cementation.
Importance
Clays have significant c'; clean sands have c = 0. In PRC problems, the value of c signals whether a simple sin φ formula is applicable.
Term
Angle of Internal Friction (φ or φ')
Definition
The angle that the Mohr-Coulomb failure envelope makes with the normal stress axis; represents frictional resistance between soil particles.
Importance
Higher φ means stronger soil under confining pressure. For undrained saturated clay, φu = 0, meaning strength is independent of confining pressure.
Term
Mohr's Circle
Definition
A graphical representation of the state of stress at a point; at failure, the largest Mohr's circle is tangent to the failure envelope τ = c + σ tan φ.
Importance
The geometry of Mohr's circle directly yields the sin φ = (σ1 − σ3)/(σ1 + σ3) formula and the failure plane orientation.
Term
Deviator Stress (Δσ or qf)
Definition
The difference between major and minor principal stresses at failure: Δσ = σ1 − σ3. It is the axial load per unit area applied in a triaxial test beyond the cell pressure.
Importance
Frequently given in board problems — always add σ3 to get σ1 before applying failure criteria.
Section Title
The Mohr-Coulomb Failure Criterion
Common Mistakes
- Using total stress parameters (c, φ) when the problem asks for effective stress analysis (use c', φ' and subtract pore pressure from σ to get σ').
- Forgetting to add σ3 to the deviator stress: σ1 = σ3 + (σ1 − σ3); a common arithmetic error in triaxial problems.
- Applying sin φ = (σ1 − σ3)/(σ1 + σ3) when c ≠ 0 — this formula is ONLY valid for cohesionless soils (c = 0).
- Confusing the failure plane angle with 45° (correct only when φ = 0); for φ > 0 the failure plane is steeper: θf = 45 + φ/2.
Formulas
Example
UC test gives qu = 120 kPa. cu = 120/2 = 60 kPa. This soil classifies as medium stiff clay.
Formula
cu = qu / 2
Variables
cu = undrained shear strength (kPa); qu = unconfined compressive strength (kPa) = peak axial load / cross-sectional area
Application
Unconfined Compression Test on saturated clay; σ3 = 0, φ = 0, so the Mohr's circle has diameter qu and radius = cu.
Example
Specimen 60 mm × 60 mm. Normal load = 360 N, Shear force at failure = 200 N. σ = 360/(0.06 × 0.06) = 100 kPa. τf = 200/(0.06 × 0.06) = 55.6 kPa.
Formula
τf = (P/A) for direct shear at failure
Variables
τf = shear stress at failure (kPa); P = horizontal shear force at failure (kN); A = cross-sectional area of specimen (m²)
Application
Direct shear test — record the peak shear force and divide by cross-sectional area to get τf; plot three (σ, τf) points to draw Mohr-Coulomb envelope.
Example
UU test on saturated clay: σ3 = 50 kPa, deviator stress at failure = 100 kPa. cu = 100/2 = 50 kPa. Another test at σ3 = 100 kPa gives same deviator ≈ 100 kPa — confirms φ = 0.
Formula
σ1 (at failure, UU test on saturated clay) = σ3 + qu_equivalent = 2cu
Variables
For saturated clay UU test: φ = 0, so Mohr's circle has diameter = σ1 − σ3 = deviator stress at failure; cu = (σ1 − σ3)/2
Application
Multiple UU tests at different σ3 produce circles of the same diameter — the envelope is horizontal (φ = 0) at height cu.
Exam Tips
- Memorize the drainage table: UU = no drainage anywhere; CU = consolidation drained + shear undrained; CD = drained throughout.
- For any problem stating 'saturated clay, UU test': immediately set φ = 0 and use τf = cu.
- For any problem stating 'sand, CD test': immediately set c = 0 and use sin φ = (σ1−σ3)/(σ1+σ3).
- When three Mohr's circles are plotted and you need to find c and φ, use two tangent points to form two simultaneous equations in c and φ.
- In CU test with pore-pressure data: effective σ3' = σ3 − u; effective σ1' = σ1 − u. Apply effective stress formula.
Key Points
- Three main laboratory tests measure shear strength: Direct Shear Test, Triaxial Compression Test, and Unconfined Compression Test (UC).
- The DIRECT SHEAR TEST shears soil on a predetermined horizontal plane; plots τf vs σ to obtain c and φ directly.
- The TRIAXIAL TEST is the most versatile; it measures shear strength under controlled drainage and confining pressure conditions.
- Three drainage conditions in triaxial testing: UU (Unconsolidated-Undrained), CU (Consolidated-Undrained), CD (Consolidated-Drained).
- The UNCONFINED COMPRESSION TEST (UC) is a special case of UU triaxial with σ3 = 0; rapid and simple but only for cohesive soils.
- For the UC test: undrained shear strength cu = qu/2, where qu is the unconfined compressive strength.
- In direct shear, the normal load is applied first, then shear force is gradually increased until failure; shear stress at failure and normal stress are recorded for each test.
- At least three triaxial specimens are tested at different σ3 values to draw the failure envelope.
- Direct shear is simple but the failure plane is forced — triaxial allows the natural failure plane to develop.
Definitions
Term
UU Test (Unconsolidated-Undrained)
Definition
No drainage allowed during either consolidation stage or shearing stage. Tests soil at its in-situ water content and void ratio. For saturated clay: gives φ = 0 and total stress cu.
Importance
Models the short-term, rapid-loading condition (e.g., sudden embankment construction on soft clay). Most conservative for immediate stability.
Term
CU Test (Consolidated-Undrained)
Definition
Sample is consolidated under σ3 (drainage allowed) then sheared without drainage. With pore-pressure measurement, yields both total-stress and effective-stress parameters.
Importance
Most commonly used test in practice. If pore pressures are measured, it gives effective φ' and c' — needed for long-term analysis of clay slopes.
Term
CD Test (Consolidated-Drained)
Definition
Sample is consolidated under σ3 then sheared so slowly that no excess pore pressure develops (drainage fully allowed throughout shearing). Directly gives effective c' and φ'.
Importance
Gold standard for effective strength parameters but very slow for clays (days to weeks). Essential for long-term drained stability of embankments and slopes.
Term
Unconfined Compressive Strength (qu)
Definition
The axial stress at failure when σ3 = 0; equal to twice the undrained shear strength for saturated clay: qu = 2cu.
Importance
Quick field classification: very soft clay qu < 25 kPa; stiff clay qu = 100–200 kPa. Board exam problems frequently give qu and ask for cu — always divide by 2.
Term
Failure Envelope
Definition
The straight line τ = c + σ tan φ drawn tangent to the family of Mohr's circles at failure from multiple tests; its intercept is c and slope angle is φ.
Importance
Three test points (three Mohr's circles) are needed to establish a reliable failure envelope — this is the basis of multi-specimen triaxial testing.
Section Title
Laboratory Shear Strength Tests
Common Mistakes
- Using cu = qu (not dividing by 2) — the most frequent arithmetic error in UC test problems. Always: cu = qu/2.
- Thinking UU test gives φ ≠ 0 for a saturated clay — it does not; the envelope is horizontal and φu = 0 regardless of confining pressure.
- Confusing deviator stress with σ1: the cell pressure (σ3) must be added. If σ3 = 100 kPa and deviator = 180 kPa, then σ1 = 280 kPa.
- Applying the UC test to sands — the UC test is only valid for cohesive soils; a sand specimen with σ3 = 0 would simply fall apart.
- Reporting CD test results as total stress parameters — CD always gives effective parameters directly (no excess pore pressure exists at failure).
Formulas
Example
Saturated clay (B = 1), A = 0.40, Δσ3 = 0, Δσ1 = 100 kPa. u = 1 × [0 + 0.40 × 100] = 40 kPa.
Formula
u = B[Δσ3 + A(Δσ1 − Δσ3)]
Variables
u = excess pore water pressure generated (kPa); B = Skempton's B parameter (= 1.0 for saturated soil); A = Skempton's A parameter (varies: −0.5 to 1.0 for clays); Δσ1, Δσ3 = changes in principal stresses
Application
Predicts pore pressure generated in CU triaxial test or under undrained field loading. At failure, Af = undrained A value used in slope stability.
Example
Total stress on failure plane = 200 kPa, pore pressure = 80 kPa. σ' = 200 − 80 = 120 kPa. Use 120 kPa in τf = c' + σ' tan φ'.
Formula
σ' = σ − u (Terzaghi's effective stress principle)
Variables
σ' = effective stress (kPa); σ = total stress (kPa); u = pore water pressure (kPa)
Application
Converts total to effective stresses for drained analysis. Foundation-level principle — appears in every geotechnical problem involving pore pressure.
Exam Tips
- Board exam trigger words: 'immediately after construction' or 'rapid loading on clay' → undrained, use cu, φ = 0.
- Board exam trigger words: 'long-term stability', 'after consolidation', 'pore pressures dissipated' → drained, use c', φ'.
- Sand problems: always effective/drained unless explicitly stated otherwise — sand has negligible drainage time.
- If a CU test is described 'with pore pressure measurement', you can compute BOTH total and effective envelope — a frequent board exam bonus scenario.
- Memorize: embankment on soft clay → undrained governs short-term; cut slope in stiff clay → drained governs long-term.
Key Points
- The key question in any geotechnical stability problem is: 'Has enough time passed for pore pressures to dissipate?'
- UNDRAINED condition: load applied rapidly; excess pore pressures cannot dissipate; use total stress parameters cu and φu = 0 for saturated clay.
- DRAINED condition: load applied slowly or enough time has passed; excess pore pressures = 0; use effective stress parameters c' and φ'.
- For SANDS and GRAVELS: drainage occurs almost instantly due to high permeability; stability analysis is always effectively drained.
- For CLAYS: drainage is very slow; short-term (undrained) stability governs immediately after construction; long-term (drained) governs years later.
- Critical scenario for CLAY CUT SLOPES: long-term drained stability is more critical (φ' > φu = 0 but negative pore pressures during excavation dissipate over time reducing strength).
- Critical scenario for CLAY EMBANKMENTS/FOUNDATIONS: short-term undrained stability governs (pore pressures build up during rapid loading).
- The Skempton pore pressure parameters A and B relate stress changes to pore pressure changes in undrained loading; B = 1 for fully saturated soils.
- Terzaghi's effective stress principle (σ' = σ − u) underpins all drained analysis.
Definitions
Term
Short-term (Undrained) Stability
Definition
Stability evaluated immediately after or during construction, before pore pressures equalize. Governed by undrained shear strength cu. Uses total stress analysis (φ = 0 for saturated clay).
Importance
Critical for soft clay foundations (bearing capacity) and embankments built on soft clay. PRC problems involving 'immediate settlement' or 'rapid construction' signal undrained conditions.
Term
Long-term (Drained) Stability
Definition
Stability evaluated after pore pressures have fully dissipated and effective stresses govern. Uses effective strength parameters c' and φ' in effective stress analysis.
Importance
Critical for permanent slopes, retaining walls, and foundations after consolidation is complete. For excavated slopes in stiff clay, long-term is MORE critical than short-term.
Term
Pore Water Pressure (u)
Definition
Pressure in the water filling the soil pores. Hydrostatic pore pressure is normal; excess pore pressure is generated by loading or unloading and dissipates with time.
Importance
Positive excess u reduces effective stress and reduces shear strength — it is the mechanism by which undrained loading weakens saturated clay.
Term
Normally Consolidated Clay (NC Clay)
Definition
A clay currently under its maximum ever-experienced stress. NC clay has c' ≈ 0 and φ' > 0. Positive excess pore pressure builds up during undrained shearing.
Importance
Most marine and alluvial clays in the Philippines (e.g., Metro Manila soft clay, Pasig River alluvium) are normally consolidated — expect low cu values.
Term
Overconsolidated Clay (OC Clay)
Definition
A clay that has been loaded and then unloaded (stress reduced below its past maximum). OC clay has measurable c' > 0 and higher shear strength than NC clay at same current stress.
Importance
Stiff fissured OC clays can be deceptive — fissures act as planes of weakness and long-term drained strength along fissures uses residual φ', not peak φ'.
Section Title
Drained vs Undrained Behaviour and Stability Applications
Common Mistakes
- Applying undrained (cu, φ = 0) parameters to a long-term stability problem involving clay — after pore pressures dissipate, effective parameters c', φ' must be used.
- Assuming sands are undrained — sands drain essentially immediately; always use drained (effective) parameters for sand.
- Forgetting that for cut slopes in OC clay, the long-term drained factor of safety is LOWER than the short-term undrained FS — the opposite of embankments.
- Using effective cohesion c' for normally consolidated clay as a large value — for NC clay, c' ≈ 0 is the correct assumption.
Formulas
Example
c = 15 kPa, φ = 25°, σ' = 200 kPa. τf = 15 + 200 × tan 25° = 15 + 93.3 = 108.3 kPa.
Formula
τf = c + σ tan φ [TOTAL STRESS] or τf = c' + σ' tan φ' [EFFECTIVE STRESS]
Variables
Master formula — see Section 1 for variable definitions.
Application
Applied whenever a normal stress on the failure plane is given; determine whether total or effective by checking for pore pressure data.
Example
σ3 = 80 kPa, σ1 = 260 kPa. sin φ = (260−80)/(260+80) = 180/340 = 0.529 → φ = 31.9° ≈ 32°.
Formula
φ = arcsin[(σ1 − σ3)/(σ1 + σ3)] [c = 0 only]
Variables
Use only for cohesionless soil (c = 0); σ1 and σ3 are principal stresses at failure.
Application
CD or CU triaxial test on sand — most common board problem type for φ determination.
Example
qu = 90 kPa → cu = 45 kPa.
Formula
cu = qu / 2
Variables
cu = undrained shear strength; qu = unconfined compressive strength.
Application
Unconfined Compression Test — the single most tested formula in the shear strength section.
Exam Tips
- In a 5-minute board problem, draw a quick Mohr's circle sketch — it takes 30 seconds and prevents formula confusion.
- Key angle: tan²(45 + φ/2) = Kp. Memorise: φ = 30° → Kp = 3.0; φ = 20° → Kp ≈ 2.04; φ = 0° → Kp = 1.0.
- If the problem gives σ3 and qu (not deviator stress), check: is this a UC test (σ3 = 0) or a triaxial test? The word 'unconfined' means σ3 = 0.
- Consistency check: for φ = 30° and c = 0, the Mohr's circle formula gives σ1/σ3 = 3. Use this as a rapid sanity check.
Key Points
- Board problems on shear strength typically test: (1) direct application of τf = c + σ tan φ, (2) finding φ from triaxial data using sin φ formula, (3) finding σ1 from c, φ, σ3 using the tan² formula, (4) finding cu from UC test, and (5) identifying the correct test for a given scenario.
- Always write out the given information, identify the formula, substitute, and check units before finalising the answer.
- Multiple-choice boards often include a 'trap' answer that differs by factor of 2 (e.g., qu instead of cu) — be vigilant.
- In three-specimen triaxial problems, draw three Mohr's circles, draw the common tangent, read off c (y-intercept) and φ (slope angle).
- Worked Problem 1 — Mohr-Coulomb: c = 20 kPa, φ = 30°, σ' = 150 kPa. τf = 20 + 150 tan 30° = 20 + 86.6 = 106.6 kPa.
- Worked Problem 2 — UC Test: qu = 120 kPa. cu = qu/2 = 60 kPa.
- Worked Problem 3 — CD Triaxial on Sand: σ3 = 100 kPa, deviator stress = 200 kPa → σ1 = 300 kPa. sin φ = (300−100)/(300+100) = 0.5 → φ = 30°.
- Worked Problem 4 — Principal Stress: c = 25 kPa, φ = 20°, σ3 = 100 kPa. θ = tan(55°) = 1.4281. σ1 = 100(1.4281)² + 2(25)(1.4281) = 204.1 + 71.4 = 275.5 kPa.
- Worked Problem 5 — Failure Plane Angle: φ = 36°. θf = 45 + 18 = 63° from horizontal.
Definitions
Term
Shear Strength (τf)
Definition
The maximum shear stress a soil can sustain on a given plane before failure. It is the resistance to sliding or shearing along that plane.
Importance
The fundamental quantity in all geotechnical stability analyses. All bearing capacity, slope stability, and earth pressure problems ultimately depend on τf.
Term
Undrained Shear Strength (cu or Su)
Definition
The shear strength of a saturated soil under undrained conditions where φ = 0; numerically equal to half the unconfined compressive strength.
Importance
Used in φ = 0 analysis for short-term loading of saturated clays. Directly obtainable from UC test, field vane shear test, and UU triaxial.
Section Title
Board-Exam Worked Problems
Common Mistakes
- Forgetting to convert degrees to check reasonableness: φ for soil should be between 0° and 45°; a result outside this range signals an error.
- Using σ1 and σ3 from before the test (confining pressures) rather than at failure — always use stresses at failure point.
- In the σ1 formula with c and φ: computing tan(45 + φ/2) instead of tan²(45 + φ/2) for the σ3 coefficient.
- Rounding φ prematurely: carry at least three decimal places in sin φ before taking arcsin to avoid a 1–2° error that changes the answer choice.
Connections
- Shear strength parameters c and φ are directly used in Terzaghi's and Hansen's bearing capacity equations (Chapter: Bearing Capacity of Shallow Foundations) — cu governs undrained bearing capacity qu = 5.14 cu (for strip footing, φ = 0).
- Slope stability analysis (Chapter: Stability of Slopes) uses τf = c + σ tan φ along the assumed failure surface; the Factor of Safety equals available shear strength divided by mobilised shear stress.
- Lateral earth pressure (Chapter: Retaining Walls and Lateral Earth Pressure) uses φ to compute Rankine's Ka = tan²(45 − φ/2) and Kp = tan²(45 + φ/2); recognising Kp = tan²(45+φ/2) links directly to the triaxial failure formula.
- Consolidation theory (Chapter: Consolidation Settlement) explains WHY drainage takes time in clay — the same low permeability that governs consolidation rate also controls whether loading is drained or undrained.
- Effective stress principle (Chapter: Stresses in Soil Masses) — Terzaghi's σ' = σ − u is the theoretical foundation for all effective stress shear strength analysis.
- Soil classification (Chapter: Classification of Soils) — the Unified Soil Classification System (USCS) categories GW, SP, ML, CH directly predict expected ranges of c and φ; CH (fat clay) has high cu and low φ'; SW (well-graded sand) has φ' = 32°–40° and c' ≈ 0.
- Field testing (Chapter: Site Investigation) — the Standard Penetration Test (SPT) N-value correlates empirically with φ for sands (Peck, Hanson & Thornburn); the Vane Shear Test directly measures cu in soft clays in-situ.
- Philippine context: NSCP 2015 Section 304 references allowable bearing pressures that indirectly rely on shear strength parameters for cohesive soils; geotechnical investigation requirements under NSCP guide which tests to specify.
Exam Strategy
Approach shear strength problems in four systematic steps: (1) IDENTIFY the soil type — sand (c=0) or clay (check if undrained/drained). (2) IDENTIFY the drainage condition — look for keywords: 'rapid loading', 'short-term', 'immediate' → undrained (UU, cu, φ=0 for saturated clay); 'long-term', 'drained', 'effective' → CD parameters. (3) SELECT the correct formula — single-point problems: τf = c + σ tanφ; triaxial principal stresses with c=0: sinφ = (σ1−σ3)/(σ1+σ3); with c≠0: σ1 = σ3 tan²(45+φ/2) + 2c tan(45+φ/2); UC test: cu = qu/2. (4) CHECK units (always kPa), check numerical reasonableness (φ between 0° and 45°, cu between 10 and 300 kPa for clays), and watch for the trap of using qu instead of cu. For the PRC board exam, allocate no more than 2–3 minutes per shear strength item. If the problem feels long, sketch a Mohr's circle — it resolves 90% of formula uncertainty in under 30 seconds. Prioritise memorising: cu = qu/2 (most tested), sinφ formula for sand triaxial, and the drainage condition table (UU/CU/CD). In the week before the exam, drill at least 10 problems of each type until the formula recall is automatic.
Quick Review Questions
A soil has c = 30 kPa and φ = 35°. The effective normal stress on the failure plane is 120 kPa. What is the shear strength?
Direct application of Mohr-Coulomb: τf = c + σ' tan φ. tan 35° = 0.7002. Multiply: 120 × 0.7002 = 84.0. Add cohesion: 30 + 84.0 = 114.0 kPa.
An unconfined compression test on a saturated clay gives qu = 144 kPa. What is the undrained shear strength?
For the UC test (σ3 = 0) on saturated clay with φ = 0, the undrained shear strength equals half the unconfined compressive strength. Never use qu directly as cu.
A CD triaxial test on sand (c = 0) gives σ3 = 150 kPa and a deviator stress at failure of 270 kPa. Find φ.
For c = 0: sin φ = (σ1 − σ3)/(σ1 + σ3). Key step: compute σ1 by adding deviator stress to σ3 before substituting. Deviator stress is NOT σ1.
For short-term stability immediately after construction of a clay embankment, which triaxial test type and which strength parameters should be used?
Rapid loading on saturated clay generates excess pore pressure with no time to drain. The UU test mimics this condition. The failure envelope is horizontal (φ = 0) and the undrained strength cu governs.
A triaxial test on clay uses c = 20 kPa and φ = 25°. The confining pressure σ3 = 100 kPa. Find σ1 at failure.
Use σ1 = σ3 tan²(45 + φ/2) + 2c tan(45 + φ/2). Compute (45 + φ/2) = 45 + 12.5 = 57.5°. Find tan(57.5°) = 1.5697. Square it: 2.464. Then substitute.
What is the inclination of the failure plane in a triaxial specimen if φ = 28°?
The failure plane makes an angle of 45 + φ/2 with the major principal stress plane. Since σ1 is vertical in a standard triaxial test, the failure plane inclines at 59° from horizontal.
A saturated clay is tested in three UU triaxial tests at σ3 = 50, 100, and 150 kPa. The deviator stresses at failure are approximately 96, 98, and 100 kPa respectively. What are c and φ?
For saturated clay UU, all Mohr's circles have essentially the same diameter regardless of σ3 — the envelope is horizontal (φ = 0). The undrained shear strength is the radius of any circle = (σ1 − σ3)/2 ≈ 50 kPa. This confirms φu = 0 for saturated clay UU.
Why can the formula sin φ = (σ1 − σ3)/(σ1 + σ3) NOT be applied to a clay with c = 15 kPa?
The geometric derivation sin φ = R/distance = (σ1−σ3)/2 ÷ (σ1+σ3)/2 assumes the failure envelope passes through the origin (c = 0). With c > 0, the centre of Mohr's circle shifts relative to the tangent point, and the simpler formula overestimates φ.
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