CELE Geotechnical Engineering — Shear Strength of SoilsDetailed Explanation
Detailed explanations for CELE Geotechnical Engineering — Shear Strength of Soils. This page treats you like a serious reviewer: we unpack the concepts thoroughly, show worked examples of how Professional Regulation Commission (PRC) — Board of Civil Engineering frames Shear Strength of Soils questions, and explain the underlying reasoning that gets you to the right answer every time.
Exam context
Professional Regulation Commission (PRC) — Board of Civil Engineering runs the Civil Engineer Licensure Examination on May and November 2026. Its Geotechnical Engineering section sits under a "Core" weighting, and Shear Strength of Soils is the 7th chapter in the 11-chapter CELE Geotechnical Engineering rotation. The CELE passing mark is 70% weighted average, no sub-test below 50%, and the most recent 2026 paper drew about a meaningful share of questions from Geotechnical Engineering.
Shear Strength of Soils - Detailed Explanation
Shear strength is the most critical mechanical property of soil in geotechnical engineering. It defines the maximum resistance a soil mass can offer against sliding or shearing failure — the very phenomenon responsible for foundation bearing failures, landslides, slope collapses, and retaining wall overturns. For the PRC Civil Engineer Licensure Examination, shear strength problems consistently appear in both the morning (mathematics) and afternoon (design) sessions of the Geotechnical Engineering component. Mastery of the Mohr-Coulomb failure criterion, the three triaxial drainage conditions (UU, CU, CD), the unconfined compression test, and the drained-versus-undrained distinction is non-negotiable for passing this examination. This chapter presents these concepts with board-exam precision, SI-unit worked examples, and strategic exam guidance aligned with the NSCP 2015 and standard geotechnical practice.
Concepts
The Mohr-Coulomb Failure Criterion
The Mohr-Coulomb failure criterion is the fundamental equation governing soil shear strength. It states that failure on any plane occurs when the shear stress on that plane reaches a critical value that is a linear function of the normal stress acting on the same plane. The equation takes two forms depending on whether total or effective stresses are used: **Total Stress Form:** τ_f = c + σ·tan(φ) **Effective Stress Form:** τ_f = c' + σ'·tan(φ') Where: - τ_f = shear stress at failure (kPa) - c = cohesion intercept in total stress (kPa) - c' = effective cohesion (kPa) - σ = total normal stress on failure plane (kPa) - σ' = effective normal stress on failure plane (kPa) - φ = angle of internal friction in total stress (degrees) - φ' = effective friction angle (degrees) **Physical Meaning:** Think of the Mohr-Coulomb equation as analogous to sliding friction on a surface. The cohesion 'c' is like the adhesive bond between soil particles (present even at zero normal stress), while σ·tan(φ) is the frictional resistance that increases with the normal compression on the failure plane. For clean sands, c ≈ 0 and resistance is purely frictional. For saturated clays under rapid (undrained) loading, φ = 0 and resistance is purely cohesive with strength = c_u. **The Mohr Circle:** At failure, the stress state is represented by a Mohr circle tangent to the failure envelope (the line τ_f = c + σ·tan(φ)). The center of the Mohr circle is at ((σ_1 + σ_3)/2, 0) and its radius is (σ_1 - σ_3)/2. The failure plane makes an angle θ = 45° + φ/2 with the major principal plane. **Key relationship for principal stresses at failure (general case):** σ_1 = σ_3 · tan²(45 + φ/2) + 2c · tan(45 + φ/2) **For cohesionless soil (c = 0), the friction angle from principal stresses:** sin(φ) = (σ_1 - σ_3) / (σ_1 + σ_3) **Critical Rule:** ALWAYS use effective stresses with c' and φ' for drained analysis. Total stresses are used only in undrained (φ = 0, c = c_u) analysis.
Examples
This is the most direct application of the Mohr-Coulomb equation. Note: tan(30°) = 1/√3 ≈ 0.5774. The cohesion contributes 20 kPa regardless of normal stress, while the frictional component adds 86.6 kPa. This problem appears frequently on PRC board exams as a straightforward computation.
Scenario
A soil sample has c = 20 kPa and φ = 30°. Calculate the shear strength on a plane where the effective normal stress is 150 kPa.
Solution
τ_f = c + σ'·tan(φ) τ_f = 20 + 150 × tan(30°) τ_f = 20 + 150 × 0.5774 τ_f = 20 + 86.6 τ_f = 106.6 kPa
The formula sin(φ) = (σ_1 - σ_3)/(σ_1 + σ_3) is derived by noting that the Mohr circle of diameter (σ_1 - σ_3) must be tangent to the failure envelope passing through the origin (c = 0). The radius of the Mohr circle divided by the distance from origin to circle center equals sin(φ). Always double-check: φ for dense sand typically ranges 35°–45°, loose sand 28°–35°.
Scenario
The principal stresses at failure in a drained triaxial test on sand (c = 0) are σ_1 = 400 kPa and σ_3 = 150 kPa. Find the friction angle φ' and the angle of the failure plane.
Solution
Step 1 — Friction angle: sin(φ') = (σ_1 - σ_3)/(σ_1 + σ_3) sin(φ') = (400 - 150)/(400 + 150) sin(φ') = 250/550 = 0.4545 φ' = arcsin(0.4545) = 27.04° ≈ 27° Step 2 — Failure plane angle: θ = 45° + φ'/2 = 45° + 27°/2 = 45° + 13.5° = 58.5° (measured from the major principal plane, which is horizontal in a triaxial test)
The term tan²(45+φ/2) is the passive pressure coefficient K_p from Rankine theory. This dual appearance — in shear strength AND in lateral earth pressure — is a favorite board exam connection. Always compute tan(45+φ/2) first, then square it. Do not confuse: the deviator stress = σ_1 - σ_3 = 340.6 - 100 = 240.6 kPa, which is what the load cell measures in a triaxial test.
Scenario
A soil has c = 30 kPa and φ = 25°. If the confining pressure in a triaxial test is σ_3 = 100 kPa, find the major principal stress σ_1 at failure.
Solution
σ_1 = σ_3·tan²(45 + φ/2) + 2c·tan(45 + φ/2) Compute tan(45 + 25/2) = tan(45 + 12.5°) = tan(57.5°) tan(57.5°) = 1.5697 tan²(57.5°) = 2.4640 σ_1 = 100 × 2.4640 + 2 × 30 × 1.5697 σ_1 = 246.40 + 94.18 σ_1 = 340.6 kPa
Applications
- Bearing capacity analysis of shallow and deep foundations — shear failure of soil beneath footing
- Slope stability analysis — determining factor of safety against sliding along a failure surface
- Lateral earth pressure on retaining walls and basement walls
- Sheet pile and braced excavation design
- Embankment and dam stability analysis for Philippine infrastructure projects
Misconceptions
- WRONG: Using total normal stress with effective strength parameters (c', φ') in a drained analysis — always pair effective stresses with effective parameters
- WRONG: Assuming c and φ are fixed material constants — they change with drainage condition and test type (UU gives different values than CD)
- WRONG: Thinking the failure plane is at 45° regardless of φ — the correct angle is 45° + φ/2
- WRONG: Using σ_1 = σ_3 + deviator stress only for sand — this is general; for clay with c ≠ 0, the relationship changes
Related Concepts
- Effective stress principle (Terzaghi)
- Pore water pressure and pore pressure parameters (A and B)
- Rankine earth pressure theory
- Bearing capacity equations (Terzaghi, Meyerhof)
- Slope stability — infinite slope and circular slip methods
Common Exam Questions
Example
Given c = 15 kPa, φ = 25°, σ' = 200 kPa — find τ_f. Answer: 15 + 200×tan25° = 15 + 93.26 = 108.26 kPa
Approach
Identify c, φ, and the normal stress σ (or σ'). Substitute directly into τ_f = c + σ·tan(φ). Ensure stress units are consistent (kPa).
Question Type
Direct computation of τ_f
Example
σ_3 = 80 kPa, deviator stress = 180 kPa → σ_1 = 260 kPa; sin(φ) = 180/340 = 0.529; φ = 31.9°
Approach
Use sin(φ) = (σ_1 - σ_3)/(σ_1 + σ_3). Note: σ_1 = σ_3 + Δσ where Δσ is the deviator stress from the test.
Question Type
Finding φ from principal stresses (sand, c=0)
Example
σ_3 = 120 kPa, c = 0, φ = 30° → σ_1 = 120×tan²(60°) = 120×3 = 360 kPa
Approach
Use σ_1 = σ_3·tan²(45+φ/2) + 2c·tan(45+φ/2). Compute the bracket quantity once, then apply.
Question Type
Finding σ_1 given σ_3, c, φ
Key Points To Remember
- τ_f = c + σ·tan(φ) for total stress; τ_f = c' + σ'·tan(φ') for effective stress
- Failure plane angle with respect to major principal plane: θ = 45° + φ/2
- For c = 0 (sand): sin(φ) = (σ_1 - σ_3)/(σ_1 + σ_3)
- General: σ_1 = σ_3·tan²(45+φ/2) + 2c·tan(45+φ/2)
- The term tan²(45+φ/2) is called the Rankine active/passive coefficient; it appears again in lateral earth pressure problems
- c and φ are NOT soil constants — they depend on drainage condition, stress history, and test type
- Effective stress = Total stress − Pore water pressure: σ' = σ − u
Laboratory Shear Strength Tests
Three major laboratory tests are used to determine shear strength parameters. Each has specific drainage conditions that produce different stress parameters applicable to different field scenarios. **1. DIRECT SHEAR TEST** The sample is placed in a split box. The lower half is fixed while the upper half is pushed horizontally. The normal load N is applied vertically, producing σ = N/A on the failure plane. The horizontal force at failure gives τ_f. Multiple specimens are tested at different normal stresses to plot the failure envelope τ_f vs σ, from which c (y-intercept) and φ (slope angle) are obtained by regression. Advantages: Simple, fast, inexpensive. Limitations: Failure plane is FORCED (horizontal), not the weakest plane; drainage control is poor; stress distribution is non-uniform. Applicable for: Determining φ' of sands and gravels (drain quickly); getting preliminary c and φ for clays. **2. TRIAXIAL COMPRESSION TEST** A cylindrical soil sample (usually 38 mm dia × 76 mm height or 50 mm × 100 mm) is encased in a rubber membrane, placed in a cell filled with water, and subjected to an all-around confining pressure σ_3. An axial load is then applied through a piston until failure. The maximum deviator stress (σ_1 - σ_3) at failure defines one Mohr circle. Multiple tests at different σ_3 values define multiple Mohr circles; the common tangent line is the failure envelope. Three drainage types: **UU (Unconsolidated-Undrained):** - No drainage allowed during consolidation OR shearing stages - Pore pressure NOT measured - Results in φ = 0 for saturated clay (all Mohr circles have the same diameter, envelope is horizontal) - Gives total stress parameter c_u (undrained shear strength) - Use for: SHORT-TERM (end-of-construction) stability of saturated clay - Philippine application: Immediate bearing capacity of soft clay deposits in Metro Manila and coastal areas **CU (Consolidated-Undrained):** - Sample consolidated under σ_3 (drainage allowed) → sheared without drainage but WITH pore pressure measurement - Gives BOTH total stress parameters (c, φ) AND effective stress parameters (c', φ') from pore pressure correction - Most common test for clay in practice - Use for: Analyzing stability where consolidation has occurred under one loading and rapid additional loading is applied **CD (Consolidated-Drained):** - Sample consolidated and then sheared SLOWLY with full drainage - No excess pore pressure builds up - Gives directly c' and φ' (effective parameters) - Most time-consuming test - Use for: LONG-TERM stability analysis, stability of slopes in drained condition, analysis of sands (fast drainage) **3. UNCONFINED COMPRESSION (UC) TEST** Special case of UU triaxial with σ_3 = 0. The cylindrical sample stands on its own (for cohesive soil only) and is loaded axially to failure. The unconfined compressive strength q_u = failure axial load / cross-sectional area. Since σ_3 = 0 and φ = 0 for saturated clay: σ_1 = q_u, σ_3 = 0 The Mohr circle center is at (q_u/2, 0), radius = q_u/2 The failure envelope (horizontal line) is tangent at τ = c_u **Therefore: c_u = q_u/2** (the undrained shear strength equals HALF the unconfined compressive strength) This is one of the most frequently tested formulas in the PRC board examination.
Examples
Note the conversion: force in N, area in m², gives Pa; divide by 1000 for kPa. The formula c_u = q_u/2 applies because in the UC test, σ_3 = 0, and for saturated clay φ = 0, making the Mohr circle tangent point exactly at τ = q_u/2. Board exam shortcut: if only the failure load and diameter are given, compute q_u first, then halve it.
Scenario
An unconfined compression test on a clay sample gives a failure load of 250 N on a sample with diameter 38 mm. Find the undrained shear strength c_u.
Solution
Step 1 — Cross-sectional area: A = π/4 × (0.038)² = π/4 × 0.001444 = 1.134 × 10⁻³ m² Step 2 — Unconfined compressive strength: q_u = F/A = 250/(1.134 × 10⁻³) = 220,460 Pa ≈ 220.5 kPa Step 3 — Undrained shear strength: c_u = q_u/2 = 220.5/2 = 110.2 kPa
In board exams, you are often given three data pairs and asked to find c and φ. The fastest method: use two extreme points to find the slope (= tan φ), then back-calculate c from any one point. ALWAYS verify with the third point. If all three don't lie perfectly on a line, use the average or least-squares fit.
Scenario
Three direct shear tests on a clay give the following results: Test 1: σ = 50 kPa, τ_f = 48.2 kPa; Test 2: σ = 100 kPa, τ_f = 75.8 kPa; Test 3: σ = 150 kPa, τ_f = 103.4 kPa. Determine c and φ.
Solution
The failure envelope is τ_f = c + σ·tan(φ), a linear regression through the three points. Using two points (Test 1 and Test 3): Slope = tan(φ) = (103.4 - 48.2)/(150 - 50) = 55.2/100 = 0.552 φ = arctan(0.552) = 28.9° ≈ 29° Using Test 1 for intercept: c = τ_f - σ·tan(φ) = 48.2 - 50 × 0.552 = 48.2 - 27.6 = 20.6 kPa Verify with Test 2: τ_f = 20.6 + 100 × 0.552 = 20.6 + 55.2 = 75.8 kPa ✓ Result: c ≈ 20.6 kPa, φ ≈ 29°
CU tests with pore pressure measurement are powerful because they give BOTH total-stress (c, φ) and effective-stress (c', φ') parameters from a single test. The board exam may give total stresses + pore pressure and ask for effective parameters. The procedure is always: subtract pore pressure from both principal stresses to get effective values, then apply the Mohr-Coulomb equations.
Scenario
A CU triaxial test on clay gives: σ_3 = 100 kPa, σ_1 = 280 kPa at failure, with measured pore pressure u = 40 kPa. Find the effective principal stresses and check if they are consistent with φ' = 28°, c' = 10 kPa.
Solution
Step 1 — Effective stresses: σ_1' = σ_1 - u = 280 - 40 = 240 kPa σ_3' = σ_3 - u = 100 - 40 = 60 kPa Step 2 — Check sin(φ') assuming c' = 0 (quick check for sand-like behavior): sin(φ') = (240 - 60)/(240 + 60) = 180/300 = 0.60 → φ' = 36.9° Actual check with c' = 10 kPa and φ' = 28°: σ_1' expected = σ_3'·tan²(45+14°) + 2c'·tan(57°) = 60 × tan²(57°) + 2 × 10 × tan(57°) = 60 × 2.3699 + 20 × 1.5399 = 142.2 + 30.8 = 173 kPa ≠ 240 kPa Conclusion: The given c' and φ' parameters do NOT match this test result. The effective parameters for this sample are φ' ≈ 36.9° (with c' ≈ 0). Board exam lesson: always verify by substitution.
Applications
- Design of foundation systems on soft Manila clay — UU parameters for immediate bearing capacity
- Long-term slope stability of cut slopes in mountainous regions (Benguet, Bukidnon) — CD parameters
- Embankment construction on soft ground — staged construction analysis using CU parameters
- Liquefaction assessment for sands in seismic zones (Metro Manila, Western Visayas) — uses triaxial cyclic tests
- Soil improvement verification — before and after compaction or grouting
Misconceptions
- WRONG: Thinking c_u = q_u (it is HALF of q_u, not equal to q_u)
- WRONG: Applying UC test to sands — sands have no cohesion and the sample collapses without confinement; UC test is only valid for cohesive (clayey) soils
- WRONG: Saying 'CD test gives c and φ' without specifying 'effective' — CD directly gives c' and φ'; emphasize effective
- WRONG: Confusing deviator stress with σ_1 — deviator stress = σ_1 - σ_3; always add σ_3 to get σ_1
- WRONG: Saying UU test always gives φ = 0 — this is only true for SATURATED clays; partially saturated soils may show φ > 0 in UU
Related Concepts
- Consolidation and settlement (links to drained/undrained behavior)
- Pore water pressure coefficients A and B (Skempton)
- Sensitivity of clays — ratio of undisturbed to remolded c_u
- Thixotropy — regain of strength after remolding
- Vane shear test — in-situ measurement of c_u
Common Exam Questions
Example
q_u = 96 kPa → c_u = 48 kPa. If q_u = 0.096 MPa, same answer: 0.096 × 1000/2 = 48 kPa
Approach
Calculate q_u = failure load/area. Then c_u = q_u/2. Watch units: if force is in N and area in mm², q_u is in MPa; convert to kPa.
Question Type
Compute c_u from UC test data
Example
A clay embankment is to be constructed rapidly. Which test? → UU, because drainage cannot occur during rapid construction → undrained condition governs short-term stability
Approach
Ask: Is loading rapid (undrained) or slow (drained)? Is consolidation needed? Short-term → UU. Long-term → CD. Intermediate with pore pressure → CU.
Question Type
Identify correct test type for a scenario
Example
σ_3 = 200 kPa, deviator = 300 kPa, u = 80 kPa → σ_1' = 420 kPa, σ_3' = 120 kPa → sin(φ') = 300/540 = 0.556 → φ' = 33.8°
Approach
Given σ_3, σ_1, and u: compute σ_3' = σ_3 - u and σ_1' = σ_1 - u. Then apply Mohr-Coulomb in terms of effective stresses.
Question Type
Find effective stress parameters from CU test
Key Points To Remember
- Direct shear: forced failure plane, gives c and φ directly from τ-σ plot; best for sands
- Triaxial: most versatile; three types — UU, CU, CD — each for a specific drainage condition
- UU → φ = 0, c = c_u → short-term, saturated clay, end-of-construction
- CD → c', φ' → long-term, drained condition, sands always effectively CD
- CU with pore pressure → both total and effective parameters
- UC test: c_u = q_u/2 (memorize this — it appears in almost every board exam set)
- σ_1 = σ_3 + deviator stress (deviator stress is what the load cell measures, NOT σ_1)
Drained vs Undrained Behavior and Practical Application
The distinction between drained and undrained behavior is perhaps the most conceptually important topic in soil shear strength. It determines which set of parameters to use in design and which test type is appropriate. **Why Drainage Matters:** When a load is applied to a saturated soil, the water in the pores must carry the load initially (as excess pore water pressure). Over time, water drains out, the effective stress increases, and the soil consolidates. This time-dependent process is what distinguishes drained from undrained conditions. **UNDRAINED (Short-Term) Condition:** - Loading is faster than drainage can occur - Excess pore pressure u develops and does NOT dissipate - Volume stays constant (no drainage) - Parameters used: φ = 0, c = c_u (total stress analysis) - τ_f = c_u (constant regardless of confining pressure!) - Critical for: End-of-construction stability of clay foundations, rapid loading - Governs: SHORT-TERM safety (during and immediately after construction) **DRAINED (Long-Term) Condition:** - Loading is slow OR sufficient time has passed for excess pore pressure to dissipate - Effective stresses are fully established - Parameters used: c', φ' (effective stress analysis) - τ_f = c' + σ'·tan(φ') - Critical for: Long-term stability of slopes, retaining walls, embankments - Governs: LONG-TERM safety (years after construction) **SANDS vs CLAYS:** Sands are highly permeable (k ≈ 10⁻³ to 10⁻¹ m/s). Any load applied drains almost instantly. Therefore: - Sands are ALWAYS effectively drained under static loading - Only exception: Seismic/cyclic loading (liquefaction analysis) - Sand parameters: c' ≈ 0, φ' = 28°–45° (effective, always) Clays are low-permeability (k ≈ 10⁻⁸ to 10⁻¹⁰ m/s). Drainage takes months to decades. - Undrained governs SHORT-TERM, drained governs LONG-TERM - For overconsolidated clays (OC), c' > 0 and φ' is relatively high - For normally consolidated clays (NC), c' ≈ 0 and φ' is the primary parameter **CRITICAL STATE:** At large strains, all soils approach a critical state where shear can continue at constant volume, constant stress, and constant pore pressure. The critical state friction angle φ_cs is a fundamental property used in advanced analysis. **Rule of Thumb for Board Exams:** - Problem says 'immediate' or 'just after construction' or 'rapid loading' → UNDRAINED → use c_u, φ = 0 - Problem says 'long-term' or 'after consolidation' or 'slow' → DRAINED → use c', φ' - Problem involves sand → DRAINED → use φ' (no c) - Problem gives UU test results → SHORT-TERM analysis - Problem gives CD test results → LONG-TERM analysis
Examples
This is a typical board exam scenario. Note that for purely cohesive soil (φ = 0), the undrained bearing capacity does NOT increase with depth of footing (only q surcharge term changes), because the Mohr-Coulomb envelope is horizontal — normal stress has no effect on undrained strength. This is a key insight that differentiates undrained from drained analysis.
Scenario
A 2 m wide strip footing on saturated soft clay (c_u = 60 kPa, φ = 0) is being designed. The geotechnical engineer wants to check (a) immediate stability after construction and (b) long-term stability. Which parameters apply for each case?
Solution
(a) Immediate (end-of-construction): Use UNDRAINED parameters φ = 0, c = c_u = 60 kPa For φ = 0 clay, Terzaghi's bearing capacity: q_ult = c_u × N_c + q With N_c = 5.14 for φ = 0 (from Terzaghi) q_ult = 60 × 5.14 = 308.4 kPa (approximate) (b) Long-term: Use DRAINED parameters Need c' and φ' from CD or CU-with-pore-pressure test. Typically for soft NC clay: c' ≈ 0, φ' = 20°–28° The long-term capacity is often HIGHER than immediate (as consolidation increases effective stress and strength).
This comparison shows that for NC clays, long-term drained strength can exceed short-term undrained strength because consolidation under the applied load increases effective stress. The undrained condition represents the CRITICAL (most dangerous) short-term state. After drainage is complete, the soil is actually stronger. This is why construction on soft clay is checked for short-term (undrained) stability first.
Scenario
A saturated NC clay has c_u = 45 kPa (from UU test) and c' = 0, φ' = 26° (from CD test). At a point in the soil, total vertical stress σ = 200 kPa and pore pressure u = 80 kPa. Calculate shear strength using (a) undrained parameters and (b) drained parameters.
Solution
(a) Undrained (short-term): τ_f = c_u = 45 kPa (independent of normal stress for φ = 0) (b) Drained (long-term, effective stress): σ' = σ - u = 200 - 80 = 120 kPa τ_f = c' + σ'·tan(φ') = 0 + 120 × tan(26°) τ_f = 0 + 120 × 0.4877 = 58.5 kPa Here, drained strength (58.5 kPa) > undrained strength (45 kPa), which is typical for NC clay under these conditions.
Applications
- Staged construction of embankments on soft clay — undrained analysis at each stage, with strength gain between stages
- Temporary vs permanent cuts in clay — temporary (undrained) may be safe; permanent (drained) may require flatter slopes
- Rapid drawdown in earth dams — water level drops quickly; pore pressures remain high → undrained instability
- Foundation design on soft clay — bearing capacity for immediate loading uses c_u
- NSCP 2015 Section 301 — foundation design must consider both short-term and long-term conditions for clay soils
Misconceptions
- WRONG: Thinking undrained is always more conservative (critical) than drained — for heavily overconsolidated clay with high c', undrained may actually give HIGHER strength than drained at low confining stresses
- WRONG: Applying undrained analysis to sands for static loading — sands are always effectively drained under static conditions
- WRONG: Thinking drainage condition depends on the test, not the field condition — you choose the test to MATCH the expected field drainage condition
- WRONG: Confusing 'undrained' with 'no pore pressure' — undrained means excess pore pressure IS present and NOT draining; it is the pore pressure that makes it undrained
Related Concepts
- Permeability and Darcy's Law
- Consolidation theory (Terzaghi) and time rate of consolidation
- Pore pressure parameters A and B (Skempton's equation: Δu = B[Δσ_3 + A(Δσ_1 - Δσ_3)])
- Over-consolidation ratio (OCR) and its effect on pore pressure response
- Liquefaction potential of saturated sands under cyclic loading
Common Exam Questions
Example
A highway fill is placed rapidly on soft clay. For immediate stability analysis, which parameter? → c_u from UU test, φ = 0
Approach
Identify soil type (sand/clay), loading rate (rapid/slow), and time frame (short/long term). Map to UU (short-term clay), CD (long-term clay or sand), CU (intermediate).
Question Type
Select appropriate test/parameter for given scenario
Example
σ = 150 kPa, u = 60 kPa, c_u = 50 kPa, φ' = 30°, c' = 0. Undrained: τ = 50 kPa. Drained: τ = 90 × tan30° = 52 kPa. Drained > Undrained slightly in this case.
Approach
Compute τ_f for both conditions. Note whether pore pressure is positive (reduces effective stress, reduces drained strength) or negative (increases effective stress).
Question Type
Compare undrained vs drained strength
Example
In a board exam essay/explanation question: 'Sands have high permeability (k > 10⁻³ m/s), so any load-induced pore pressure dissipates almost immediately. The drained condition (c' = 0, φ') is always used for static design.'
Approach
Permeability argument: sands drain near-instantaneously under static load, so excess pore pressures cannot build up. Effective stresses are always fully developed. Exception: seismic/dynamic loading (liquefaction).
Question Type
Explain why sands don't require undrained analysis (static loading)
Key Points To Remember
- Undrained = short-term; Drained = long-term — this is the single most important practical distinction
- For saturated clay, undrained analysis: φ = 0, τ_f = c_u (strength independent of confining pressure)
- Sands: always treat as drained under static loading (high permeability)
- Clays: undrained for rapid loading, drained for long-term assessment
- Over-consolidation ratio (OCR) affects strength: OC clays are generally stronger than NC clays at the same effective stress
- Pore pressure in undrained loading of NC clay is POSITIVE (reduces effective stress, reduces strength) — this is why undrained can be more critical
- Drained strength of NC clay is HIGHER than undrained strength at the same total confining stress
Practice Problems
Key technique: Always verify the computed c and φ against ALL test points, not just the ones used to calculate them. If the third point doesn't match, you may need to use a least-squares regression. In board exams, problems are typically set so all three points lie on the same line — use extreme values to find slope for best accuracy.
Problem
Problem 1 (Direct Shear): Three direct shear tests on a silty clay gave: Test 1: σ = 50 kPa, τ_f = 47 kPa; Test 2: σ = 100 kPa, τ_f = 73 kPa; Test 3: σ = 200 kPa, τ_f = 125 kPa. (a) Determine c and φ by plotting the best-fit line. (b) Find the shear strength at σ = 175 kPa.
Solution
(a) Using regression through the three points: Slope = tan(φ): Using Tests 1 and 3: tan(φ) = (125 - 47)/(200 - 50) = 78/150 = 0.520 φ = arctan(0.520) = 27.5° Using Test 1 for intercept: c = 47 - 50 × 0.520 = 47 - 26.0 = 21.0 kPa Verification: Test 2: τ = 21.0 + 100 × 0.520 = 21.0 + 52.0 = 73.0 kPa ✓ Test 3: τ = 21.0 + 200 × 0.520 = 21.0 + 104.0 = 125.0 kPa ✓ Result: c = 21.0 kPa, φ = 27.5° (b) At σ = 175 kPa: τ_f = 21.0 + 175 × tan(27.5°) = 21.0 + 175 × 0.520 = 21.0 + 91.0 = 112.0 kPa
Note that the Mohr circle center x-coordinate = radius = q_u/2 = c_u. This elegant geometric result — that the circle is tangent to a horizontal line at the top of the circle — is the proof that c_u = q_u/2 for φ = 0 soil. The bulging at mid-height confirms a valid failure mode (barrel-shaped failure). Board exam tip: if the problem gives dimensions and load, always compute q_u as an intermediate step.
Problem
Problem 2 (UC Test): A clay sample with diameter 50 mm and height 100 mm is tested in unconfined compression. The load at failure is 310 N. The sample bulges at mid-height. Find: (a) unconfined compressive strength q_u, (b) undrained shear strength c_u, (c) the Mohr circle parameters at failure.
Solution
(a) Cross-sectional area: A = π/4 × (0.050)² = π/4 × 0.0025 = 1.9635 × 10⁻³ m² q_u = 310 N / (1.9635 × 10⁻³ m²) = 157,920 Pa = 157.9 kPa ≈ 158 kPa (b) Undrained shear strength: c_u = q_u/2 = 158/2 = 79 kPa (c) Mohr circle parameters: - Major principal stress: σ_1 = q_u = 158 kPa - Minor principal stress: σ_3 = 0 (unconfined) - Center of Mohr circle: ((158 + 0)/2, 0) = (79, 0) - Radius of Mohr circle: (158 - 0)/2 = 79 kPa - Since φ = 0, the failure envelope is the horizontal line τ = c_u = 79 kPa - The circle is tangent to the failure envelope at τ = 79 kPa, confirming c_u = 79 kPa ✓
For cohesionless soil (c = 0), the formula sin(φ) = (σ_1 - σ_3)/(σ_1 + σ_3) applies independently to each test. The consistency of all three results confirms the assumption c = 0 is correct. In board exams, you may be given only one test for sand and asked to find φ. If two or three tests are given for a cohesive soil (c ≠ 0), you need Mohr circles — they will NOT give the same φ value unless you account for c.
Problem
Problem 3 (Triaxial, Sand): A CD triaxial test on dry sand gives: Test A: σ_3 = 50 kPa, deviator stress at failure = 119.4 kPa; Test B: σ_3 = 100 kPa, deviator stress = 238.8 kPa; Test C: σ_3 = 150 kPa, deviator stress = 358.2 kPa. Find the friction angle φ'.
Solution
For each test, σ_1 = σ_3 + deviator stress: Test A: σ_1 = 50 + 119.4 = 169.4 kPa Test B: σ_1 = 100 + 238.8 = 338.8 kPa Test C: σ_1 = 150 + 358.2 = 508.2 kPa Using sin(φ) = (σ_1 - σ_3)/(σ_1 + σ_3) for each test: Test A: sin(φ) = 119.4/219.4 = 0.5443 → φ = 32.97° Test B: sin(φ) = 238.8/438.8 = 0.5442 → φ = 32.96° Test C: sin(φ) = 358.2/658.2 = 0.5442 → φ = 32.96° All three give consistent results: φ' ≈ 33° Note: The ratio (deviator stress)/(σ_1 + σ_3) is constant for all three tests, confirming c = 0 (sand — passes through origin) and φ' = 33°.
This problem demonstrates why CU tests with pore pressure measurement are so valuable: from a single test, you obtain both total (φ_total = 22°) and effective (φ' = 37°) parameters. Note: the deviator stress (σ_1 - σ_3 = 240 kPa) is the SAME in both total and effective stress analysis because pore pressure affects both σ_1 and σ_3 equally — only the mean stress changes. This is a critical conceptual point often tested in the board exam.
Problem
Problem 4 (CU Test with pore pressure): A CU triaxial test on a clay sample gives: σ_3 = 200 kPa, σ_1 at failure = 440 kPa, and measured pore pressure at failure u = 120 kPa. Assuming c' = 0 for this normally consolidated clay, find: (a) total stress friction angle φ_total, (b) effective friction angle φ'.
Solution
(a) Total stress analysis: sin(φ_total) = (σ_1 - σ_3)/(σ_1 + σ_3) sin(φ_total) = (440 - 200)/(440 + 200) = 240/640 = 0.375 φ_total = arcsin(0.375) = 22.0° (b) Effective stress analysis: σ_1' = σ_1 - u = 440 - 120 = 320 kPa σ_3' = σ_3 - u = 200 - 120 = 80 kPa sin(φ') = (σ_1' - σ_3')/(σ_1' + σ_3') sin(φ') = (320 - 80)/(320 + 80) = 240/400 = 0.600 φ' = arcsin(0.600) = 36.87° ≈ 37° Note: φ' (37°) > φ_total (22°) because the positive pore pressure reduced effective confining stress, making the effective Mohr circle larger relative to its mean stress.
This result — that long-term drained strength (29.5 kPa) is LOWER than short-term undrained (54 kPa) — occurs when effective stresses are low (shallow depth, high water table). At greater depth with higher effective stresses, the drained strength increases and can exceed the undrained strength. This depth-dependency is why geotechnical engineers must evaluate BOTH conditions at every depth. This multi-part problem format is typical of PRC board exam difficulty level.
Problem
Problem 5 (Combined): A saturated clay has the following properties from laboratory tests: UU test: c_u = 55 kPa; UC test: q_u = 108 kPa; CD test: c' = 5 kPa, φ' = 28°. At a depth of 4 m in this clay deposit, the total vertical stress is 76 kPa and the pore water pressure is 30 kPa. Find: (a) undrained shear strength from UC test; (b) verify consistency between UU and UC results; (c) long-term drained shear strength on the horizontal plane at this depth.
Solution
(a) Undrained shear strength from UC test: c_u(UC) = q_u/2 = 108/2 = 54 kPa (b) Consistency check: UU test gives c_u = 55 kPa UC test gives c_u = 54 kPa Difference = 1 kPa (about 1.8%) — essentially the same; the small difference is due to sample variability and test imperfections. Both tests are measuring the same undrained shear strength, and the values are consistent. ✓ (c) Long-term drained shear strength: Effective vertical stress: σ_v' = σ_v - u = 76 - 30 = 46 kPa On the horizontal plane, the normal stress is the vertical effective stress: τ_f = c' + σ_v'·tan(φ') τ_f = 5 + 46 × tan(28°) τ_f = 5 + 46 × 0.5317 τ_f = 5 + 24.5 τ_f = 29.5 kPa Note: Short-term undrained strength ≈ 54–55 kPa >> long-term drained strength = 29.5 kPa This means the LONG-TERM drained condition is MORE CRITICAL (lower strength) at this stress level.
Exam Preparation Tips
- MEMORIZE THE BIG THREE FORMULAS: τ_f = c + σ·tan(φ), sin(φ) = (σ_1-σ_3)/(σ_1+σ_3) for c=0, and c_u = q_u/2. These three appear in nearly every board exam set on shear strength.
- MASTER THE UU/CU/CD DECISION TABLE: When given a scenario, ask (1) Is it short-term or long-term? (2) Is it clay or sand? (3) Is there pore pressure measurement? This determines which test and parameters apply.
- DEVIATOR STRESS vs σ_1: σ_1 = σ_3 + deviator stress. Never plug the deviator stress directly into Mohr circle formulas as σ_1 — this is the most common arithmetic error in triaxial problems.
- EFFECTIVE STRESS IS KING: For any drained analysis, ALWAYS subtract pore pressure from total stresses before computing τ_f with c' and φ'. Mixing total stress with effective parameters is the most common conceptual error.
- TAN(45+φ/2) SHORTCUT FOR COMMON ANGLES: φ=30°: tan(60°)=√3=1.732, so K_p=tan²(60°)=3.0. φ=45°: tan(67.5°)=2.414. φ=0°: tan(45°)=1.0, K_p=1.0. Having these memorized saves critical time.
- c_u = q_u/2, NOT q_u: Half the unconfined compressive strength. This has been tested on the board exam at least once in every recent exam cycle. Write it 10 times if necessary.
- USE UNITS CONSISTENTLY: Work entirely in kPa. If given forces in kN and areas in m², the result is already in kPa. If given N and mm², convert: (N/mm²) × 1000 = kPa.
- DRAW THE MOHR CIRCLE: Even a quick sketch helps you visualize the problem, check if the circle is tangent to the failure envelope, and verify your answer is geometrically reasonable.
- KNOW THE TYPICAL RANGES: Dense sand: φ' = 38°–45°. Loose sand: φ' = 28°–35°. NC clay: φ' = 20°–28°, c' ≈ 0. OC clay: c' > 0, φ' similar to NC. Soft clay: c_u = 10–50 kPa. Stiff clay: c_u = 50–200 kPa. Unusual values should prompt you to recheck.
- PRACTICE ALL FIVE PROBLEM TYPES: (1) Find τ_f from c, φ, σ; (2) Find φ from principal stresses (sand); (3) Find σ_1 from σ_3, c, φ; (4) Find c_u from UC test; (5) Find effective parameters from CU test with pore pressure. These cover approximately 90% of board exam shear strength problems.
- LINK TO BEARING CAPACITY: The Terzaghi bearing capacity equation uses c and φ (or c_u and 0 for undrained). After mastering shear strength, immediately review Terzaghi and Meyerhof bearing capacity factors — the topics are inseparable in the board exam.
- TIME MANAGEMENT: A straightforward τ_f computation should take 30–45 seconds. A triaxial test analysis (find φ) should take 1–2 minutes. If a shear strength problem is taking more than 3 minutes, skip and return — do not let it eat into time for other questions.
In summary
Shear strength of soils is a cornerstone of geotechnical engineering and a guaranteed topic in the PRC Civil Engineer Licensure Examination. The three pillars of this chapter — the Mohr-Coulomb failure criterion, the laboratory test types (direct shear, triaxial UU/CU/CD, unconfined compression), and the drained-versus-undrained distinction — are interconnected: the test type determines which parameters you obtain, and the field drainage condition determines which parameters you use in design. For the board exam, these are your non-negotiables: (1) τ_f = c + σ·tan(φ) in effective stress form for drained analysis; (2) sin(φ) = (σ_1 - σ_3)/(σ_1 + σ_3) for cohesionless soil; (3) c_u = q_u/2 from the unconfined compression test; (4) UU → short-term clay, CD → long-term/sand, CU-with-pore-pressure → both; and (5) always distinguish deviator stress from σ_1. Beyond the examination, these concepts directly govern the safety of every foundation, slope, and retaining structure you will design as a licensed civil engineer under RA 544 (Civil Engineering Law of the Philippines). Soft clay deposits in the Philippine coastal lowlands, volcanic soils in Luzon and Mindanao, and residual soils in hilly terrain all present distinct shear strength challenges that require both textbook mastery and engineering judgment. Begin with the five practice problems in this chapter, then extend your practice to past board exam sets from 2018–2024, where shear strength problems appear in at least 3–5 items per examination. Consistent, disciplined review of this topic will yield reliable exam points and — more importantly — prepare you to protect public safety as a professional engineer.
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