CELE Geotechnical Engineering — Shear Strength of SoilsMemory Anchors

Memory aid

Think of a bangus (milkfish) sliding on a banana leaf. The banana leaf has a natural stickiness (that's cohesion 'c') AND roughness (that's friction angle 'φ'). The heavier the fish (normal stress σ), the more friction resists sliding. The total resistance to sliding = stickiness + (weight × roughness). That IS the Mohr-Coulomb equation. τ_f = c + σ·tan(φ): cohesion + normal stress × tangent of friction angle.

Why it works

The analogy maps each variable to a sensory, everyday object. The bangus is familiar to every Filipino student. Stickiness → c, roughness → tan(φ), fish weight → σ.

Example usage

Exam problem: c = 20 kPa, φ = 30°, σ' = 150 kPa. Trigger: bangus on leaf → τ_f = c + σ'tan(φ) = 20 + 150·tan30° = 20 + 86.6 = 106.6 kPa.

Recall trigger

Picture a bangus sliding on a banana leaf.

Memory aid

Use the phrase: 'Sand is Silly — it has Zero Cohesion.' → Sand ≈ c = 0 (purely frictional, φ > 0). Then: 'Clay is Clingy — it Clings with Cohesion.' → Saturated clay under undrained loading: φ = 0, strength comes purely from c_u. Acronym: SC-ZC-CC → Sand = c=0; Clay (undrained) = c_u only.

Why it works

Alliteration and contrasting character descriptions (silly vs. clingy) create a strong dual-coding effect — verbal + personality image.

Example usage

Board exam asks: 'A saturated clay is loaded quickly. What are the shear strength parameters?' → Clay is Clingy → φ = 0, strength = c_u. Use τ = c_u.

Recall trigger

Sand is Silly (zero c), Clay is Clingy (c_u only).

Memory aid

Chant this: 'q_u divided by two, that's what c_u is due!' Unconfined compressive strength (q_u) is TWICE the undrained shear strength. So always halve it: c_u = q_u ÷ 2. Think of cutting a bibingka (rice cake) exactly in half — the undrained strength is just one half of the unconfined strength.

Why it works

Rhyme encodes the formula in phonological memory (auditory loop), which is highly robust under stress. The bibingka image reinforces the 'half' concept visually.

Example usage

Problem: q_u = 120 kPa. → Chant the rhyme → c_u = 120/2 = 60 kPa. Common pitfall: do NOT use q_u directly as c_u — always divide by 2.

Recall trigger

Chant 'q_u divided by two' OR picture cutting bibingka in half.

Memory aid

Remember the three barangay officials at a meeting: Uncle Ulysses (UU = Unconsolidated-Undrained), Cousin Umberto (CU = Consolidated-Undrained), and Chief Doming (CD = Consolidated-Drained). Each level of drainage = more time for water to leave = more realistic long-term behavior. Uncle (UU) rushed in without preparing; Cousin (CU) prepared but did not drain; Chief (CD) was fully prepared AND drained — the most realistic for long-term.

Why it works

Narrative + character personalities make the sequence memorable. Filipino students relate to barangay community structures.

Example usage

Board question: 'Which triaxial test gives effective parameters c' and φ'?' → Chief CD (fully drained) → CD test gives effective c', φ'.

Recall trigger

Three barangay officials: Uncle UU, Cousin CU, Chief CD.

Memory aid

It's the first day of a big typhoon in Manila. A contractor needs to build an emergency embankment on soft clay RIGHT NOW — no time to wait for pore pressures to dissipate. The soil engineer shouts: 'Use UU! The clay has NO time to drain — φ is ZERO, use only c_u!' The typhoon = emergency loading = short-term = UU. Every time you see 'short-term stability of clay,' picture that typhoon scene.

Why it works

Emotional context (typhoon emergency, urgency) dramatically increases encoding strength. Filipino students have direct experience with typhoons, making this highly relatable.

Example usage

Exam: 'What test is used for short-term stability of a newly constructed clay embankment?' → Typhoon = urgent = UU → answer: UU test, φ = 0.

Recall trigger

Typhoon emergency → no time to drain → UU → φ = 0, c_u.

Memory aid

Years after the typhoon, the embankment has been standing for a long time. The pore water has fully drained, the soil is consolidated. The old engineer checks stability using c' and φ' — effective parameters from the CD test. Long time passing = pore pressure gone = drained = CD = effective stress parameters. Picture a dried-out bangus under the sun after many years — fully drained, fully consolidated.

Why it works

Contrast with A5 (typhoon = short term) creates a paired memory. The dried bangus image completes the story arc from wet to dry, reinforcing drainage concept.

Example usage

Problem asks for long-term slope stability analysis → use c', φ' from CD test. Paired with UU (short-term) for comparison.

Recall trigger

Dried bangus under the sun = long-term = CD = c', φ' effective.

Memory aid

Visualize a Mohr circle drawn on a whiteboard. The circle has DIAMETER = (σ₁ - σ₃) and CENTER at (σ₁ + σ₃)/2 from the origin. The sine of the friction angle = opposite over hypotenuse = RADIUS over DISTANCE-FROM-ORIGIN = [(σ₁-σ₃)/2] / [(σ₁+σ₃)/2] = (σ₁-σ₃)/(σ₁+σ₃). Picture a basketball (Mohr circle) where sin φ is simply the ratio of half-diameter to half-span. Every time you see a Mohr circle, recall: sin φ = difference ÷ sum.

Why it works

Geometric visualization of the Mohr circle is the CORRECT mathematical basis of the formula. This anchor teaches the formula by deriving it visually, making forgetting almost impossible.

Example usage

Triaxial test: σ₃ = 100 kPa, σ₁ = 300 kPa. → sin φ = (300-100)/(300+100) = 200/400 = 0.5 → φ = 30°.

Recall trigger

Mohr circle on whiteboard → sin φ = difference over sum.

Memory aid

Break the formula into two CHUNKS. CHUNK 1: σ₃·tan²(45 + φ/2) — this is the CONFINEMENT contribution. CHUNK 2: 2c·tan(45 + φ/2) — this is the COHESION contribution. Memory phrase: 'Confinement SQUARED, Cohesion DOUBLED — both times N.' Where N = tan(45 + φ/2) is the flow number (Rankine's coefficient Nφ). So: σ₁ = σ₃·N² + 2c·N. Two chunks: confinement × N², cohesion × 2N.

Why it works

Chunking reduces cognitive load by grouping the complex formula into two meaningful parts with a shared factor N. Once N is memorized, both terms are easy to reconstruct.

Example usage

Given: c = 20 kPa, φ = 30°, σ₃ = 100 kPa. N = tan(60°) = 1.732. σ₁ = 100·(1.732)² + 2·20·1.732 = 100·3 + 69.28 = 300 + 69.28 = 369.28 kPa.

Recall trigger

Confinement (σ₃·N²) + Cohesion (2c·N) = σ₁. Where N = tan(45 + φ/2).

Memory aid

Think of the direct shear test as the pandesal-cutting test. You have a stack of pandesal (soil sample). You push down on top (normal load σ), then push horizontally from the side (shear force T) until the bread splits on a horizontal plane. Plot how much sideways force you needed (τ_f) for each different downward force (σ). The graph gives you a straight line — the y-intercept is cohesion c, and the slope angle is φ. Simple, direct, on a predetermined failure plane.

Why it works

Pandesal is iconic Filipino bread. The physical cutting action directly mirrors the mechanics of the direct shear apparatus — horizontal splitting plane, normal force, shear force.

Example usage

Exam: 'What does a direct shear test plot look like?' → Pandesal image → τ vs σ plot → y-intercept = c, slope = tan(φ).

Recall trigger

Pandesal-cutting test: push down (σ), push sideways (τ), plot the line → c and φ.

Memory aid

Remember: 'DEVIATE from confinement to reach the peak.' The deviator stress (Δσ) is the EXTRA axial load beyond the confining pressure σ₃. Think of σ₁ as the total vertical push = base confinement (σ₃) + extra deviator kick (Δσ). Board exam pitfall: if the problem gives you 'deviator stress = 200 kPa' and σ₃ = 100 kPa, then σ₁ = 100 + 200 = 300 kPa — do NOT use 200 as σ₁ directly!

Why it works

The word 'deviate' is phonetically and semantically linked to 'deviator stress.' The warning about the common mistake makes the anchor doubly useful — it prevents the #1 board exam error.

Example usage

Problem: σ₃ = 80 kPa, deviator = 180 kPa. → σ₁ = 80 + 180 = 260 kPa. Then sin φ = (260-80)/(260+80) = 180/340 = 0.529 → φ ≈ 32°.

Recall trigger

DEVIATE = extra push above σ₃. σ₁ = σ₃ + deviator. Never use deviator as σ₁.

Memory aid

Meet Cousin Umberto (CU). He consolidates first (waits for drainage during consolidation), but during shearing, he plugs his ears and refuses to drain. However, his friends measure BOTH his emotional reaction (total stress envelope) AND his true feelings (effective stress with pore pressure measured). CU test is the 'middle child' — partially drained during consolidation, undrained during shear, but with pore pressure measurement, you get both total (c, φ) and effective (c', φ') parameters.

Why it works

Character narrative (Cousin Umberto) from the A4 anchor extends the story, building a connected memory palace. The 'middle child' analogy clarifies the intermediate nature of CU.

Example usage

Exam: 'Which test gives both total and effective shear parameters?' → Cousin Umberto (CU) → CU test with pore pressure measurement.

Recall trigger

Cousin CU: consolidates yes, drains during shear NO, measures pore pressure YES → both total and effective parameters.

Memory aid

Total stress is like your gross salary — what you earn on paper. Pore water pressure u is like tax. Effective stress σ' is your NET take-home pay — what actually controls your buying power (shear strength). τ_f (effective) = c' + σ'·tan(φ'). Always subtract the 'tax' (pore pressure) before calculating real shear strength in drained analyses.

Why it works

Salary-tax-net pay is a universal analogy that every working professional immediately understands. It correctly maps the mathematical relationship σ' = σ - u.

Example usage

Problem: σ = 200 kPa, u = 50 kPa, c' = 10 kPa, φ' = 28°. σ' = 200 - 50 = 150 kPa. τ_f = 10 + 150·tan28° = 10 + 79.7 = 89.7 kPa.

Recall trigger

Gross salary (σ) minus tax (u) = net pay (σ'). Use net pay for drained shear strength.

Memory aid

Picture a clay cylinder standing ALONE on a table — no chamber, no confining pressure, nothing squeezing it from the sides. It's naked (σ₃ = 0). You just push down on top until it fails. That's the UC test. It's the 'cheapest triaxial' — UU test with zero confinement. The sample stands on its own because clay has cohesion; sand would fall apart immediately (no c). This also explains why UC test ONLY works for cohesive soils.

Why it works

The visual image of a naked, unconfined cylinder is striking and unique. The 'cheapest triaxial' label correctly positions the UC test in the hierarchy of shear tests.

Example usage

Exam: 'Why can't you perform UC test on sand?' → Naked cylinder → sand has no c → it collapses without confinement → UC only for cohesive soils.

Recall trigger

Naked clay cylinder on a table — no confinement, σ₃ = 0 = UC test.

Memory aid

Remember: '45 PLUS HALF the friction.' The failure plane in soil is NOT at 45° like pure shear — it tilts by an extra φ/2 due to friction. So it's steeper than 45°. Think of it as: 'Friction adds a lean — half of φ beyond 45.' For φ = 30°: failure plane = 45 + 15 = 60° from the σ₃ plane (or equivalently, 30° from the σ₁ plane).

Why it works

The phrase '45 PLUS HALF' is a clean verbal chunk that encodes both the base angle and the modifier. It also immediately alerts students to the common error of using 45° alone.

Example usage

Exam: φ = 30°. Failure plane angle from horizontal = 45 + 30/2 = 45 + 15 = 60° from the minor principal plane.

Recall trigger

Failure plane = 45 + φ/2. Friction adds a lean.

Memory aid

A student named Carlo memorized 'c_u equals q_u' and used q_u = 120 kPa directly. He got 120 kPa as c_u — and failed that board question. His classmate Maria remembered 'cut the bibingka in half' and wrote c_u = 60 kPa. Maria passed. Every board review group in Manila tells Carlo's story: 'Don't be a Carlo — always HALVE the q_u.'

Why it works

Social consequences (passing vs. failing) create emotional stakes that dramatically improve memory encoding. Named characters make the story personal. The 'Don't be a Carlo' warning is quotable and repeatable.

Example usage

q_u = 90 kPa → Don't be Carlo → c_u = 90/2 = 45 kPa. Never use 90 kPa directly as c_u.

Recall trigger

Don't be Carlo — always halve q_u. c_u = q_u/2.

Memory aid

Saturated clay under fast (undrained) loading is like a water balloon. No matter how much you squeeze it (increase confining pressure), the water inside transfers the load — the effective stress barely changes. Shear strength does NOT increase with more confinement. It's like all Mohr circles for UU tests have the same diameter — they all touch the same horizontal failure envelope (φ = 0, a flat line at height c_u). The balloon can't get 'firmer' — it just redistributes water.

Why it works

The water balloon is a tactile, sensory analogy. Students can feel the concept. The visual of same-diameter Mohr circles all touching a horizontal line is a powerful geometric memory.

Example usage

UU triaxial tests on saturated clay at σ₃ = 50, 100, 200 kPa all give same failure stress difference → Mohr circles same size → φ_u = 0 → c_u = constant.

Recall trigger

Saturated clay = water balloon. More pressure = same strength. φ_u = 0, horizontal envelope at c_u.

Memory aid

Pour sugar on a plate — it forms a cone (the angle of repose). Sugar grains don't stick together (c = 0); they just pile up based on friction (φ). Dry sand behaves exactly like sugar: no cohesion, pure friction. If you remove the confining stress, the pile collapses. This is why σ₃ > 0 in triaxial tests matters for sand — without confinement, strength drops to zero for c = 0 soils.

Why it works

Sugar and sand are physically similar granular materials. Students can picture pouring sugar and see the angle of repose forming. The sugar analogy makes c = 0 intuitive.

Example usage

Triaxial on sand with σ₃ = 0: σ₁ = σ₃·tan²(45+φ/2) = 0 → sand has zero strength with zero confinement, confirming c = 0.

Recall trigger

Sand = sugar pile: c = 0, pure friction, strength depends on confinement.

Memory aid

In a CU test with pore pressure measurement, imagine drawing TWO Mohr-Coulomb envelopes on the same graph: a YELLOW line (total stress, using σ) and a GREEN line (effective stress, using σ' = σ - u). The green line (c', φ') is steeper and shifted right — it shows TRUE soil behavior. The yellow line (c_T, φ_T) is the apparent envelope. Board exams often ask which one to use for long-term → always GREEN (effective). For immediate loading → can use YELLOW (total).

Why it works

Color-coding two lines on one graph creates a distinctive visual memory. The GREEN = go (long-term, effective) mnemonic adds a traffic-light anchor.

Example usage

Exam: CU test data with pore pressure. Long-term analysis → use GREEN line parameters c', φ'. Short-term → use YELLOW (total) c, φ.

Recall trigger

Two lines on CU plot: YELLOW (total) vs. GREEN (effective). Go green for long-term.

Memory aid

Place yourself in a Filipino wet market (palengke). STALL 1 (Consolidation): The soil is squeezed under the confining pressure — like pressing water out of a wet towel at the first stall. STALL 2 (Shearing): The sample is now sheared — like slicing through the pressed towel at the second stall. In CU — you squeeze at Stall 1 (drain open), then slice at Stall 2 (drain CLOSED). In CD — you squeeze at Stall 1, then slice at Stall 2 with drain STILL OPEN. The difference is just whether the drain valve stays open during slicing.

Why it works

Method of loci (memory palace) using a familiar Philippine setting. Each stall represents a test phase. The wet towel gives a tangible feel for consolidation and drainage.

Example usage

Exam asks to distinguish CU from CD: → palengke image → CU closes drain valve at Stall 2 (no drainage during shear); CD keeps it open → pore pressures dissipate during shear in CD.

Recall trigger

Palengke: Stall 1 = consolidate (squeeze), Stall 2 = shear (slice). CU: Stall 2 drain CLOSED; CD: Stall 2 drain OPEN.

Memory aid

Imagine two engineers, Ate Dina (short-term) and Kuya Ernesto (long-term). Ate Dina is worried about the day the embankment is built — pore pressures are high, no drainage yet — she uses UU parameters (c_u, φ=0). Kuya Ernesto checks back 5 years later — pore pressures have dissipated, soil is drained — he uses CD parameters (c', φ'). Both must be checked. In Philippine practice (DPWH embankment design), BOTH stability conditions are evaluated because each can govern depending on soil type.

Why it works

Named Filipino characters with clear roles are memorable. The timeline (day of construction vs. 5 years later) gives a concrete temporal anchor. Professional practice context adds exam relevance.

Example usage

Board question: 'Which stability condition governs just after construction of a clay embankment?' → Ate Dina (short-term) → UU → φ = 0, c_u controls.

Recall trigger

Ate Dina (short-term, UU, c_u) vs. Kuya Ernesto (long-term, CD, c' φ').

Answer

Unconfined Compression (UC) Test — σ₃ = 0

Memory link

A13 — Naked clay cylinder on a table (visual association)

Answer

One-half (1/2) — c_u = q_u / 2

Memory link

A3 — 'q_u divided by two, that's what c_u is due!' (rhyme) and A15 — Don't be Carlo!

Answer

σ₁ = 300 kPa; sin φ = (300-100)/(300+100) = 200/400 = 0.5; φ = 30°

Memory link

A7 — Mohr circle on whiteboard: sin φ = difference over sum; A10 — DEVIATE: σ₁ = σ₃ + deviator

Answer

UU test parameters — c_u with φ_u = 0 (undrained, short-term)

Memory link

A5 — Typhoon emergency micro-story; A20 — Ate Dina (short-term, UU)

Answer

Chief Doming — the CD (Consolidated-Drained) test

Memory link

A4 — Three barangay officials: Uncle UU, Cousin CU, Chief CD

Answer

τ_f = 15 + 200·tan(25°) = 15 + 200(0.4663) = 15 + 93.26 = 108.26 kPa

Memory link

A1 — Bangus on banana leaf: stickiness (c) + weight × roughness (σ·tan φ)

Answer

σ' = 200 - 50 = 150 kPa. Use in τ_f = c' + σ'·tan(φ') for drained/long-term analysis.

Memory link

A12 — Salary-tax-net pay analogy for effective stress (σ' = σ - u)

Answer

Sand is Silly → c ≈ 0 (no cohesion), φ > 0 (frictional). Clay is Clingy (undrained) → c_u > 0, φ_u = 0 (purely cohesive in short-term).

Memory link

A2 — SC-ZC-CC acronym and personality analogy

Mnemonic

CST: Cohesion + Sigma Times tangent-phi. 'C plus STan-phi.' Think: Every COST has a fixed base (c) plus a variable rate (σ·tan φ). The shear strength COST = fixed cohesion + variable friction.

When to use

Always, for any soil shear strength calculation. Use total stress (σ, c, φ) for total stress analysis (UU conditions). Use effective stress (σ', c', φ') for drained or long-term analysis (CD conditions).

What each part means

τ_f = shear stress at failure (kPa); c = cohesion intercept (kPa), soil's inherent 'stickiness'; σ = normal stress on failure plane (kPa); tan(φ) = tangent of friction angle, the 'slope' of the failure envelope; φ = angle of internal friction (degrees).

Mnemonic

HALF THE UCS. 'c_u = HALF of q_u.' Chant: 'Undrained strength is half the unconfined — always cut it in two, you'll be fine.' The bibingka rule: cut the unconfined strength in half.

When to use

Only for saturated clay under undrained conditions (φ = 0) from unconfined compression test results. Do NOT use for drained conditions or for sands.

What each part means

c_u = undrained shear strength (kPa); q_u = unconfined compressive strength (kPa) from UC test; the factor 1/2 arises because at failure in the UC test, the Mohr circle has diameter q_u and the shear stress at the top of the circle (representing the failure plane for φ = 0) is q_u/2.

Mnemonic

SINE = DIFF over SUM. 'Sin phi — difference over sum.' Like a fraction: top = how far apart the principals are; bottom = how far they span total. Picture the Mohr circle: sin(φ) = radius / center-distance = [(σ₁-σ₃)/2] / [(σ₁+σ₃)/2].

When to use

For sands (c = 0) in CD or drained triaxial tests. Also works for any c = 0 soil. NOT directly applicable when c ≠ 0 — must use graphical Mohr circle or the general tangent-line method.

What each part means

φ = friction angle (degrees); σ₁ = major principal stress at failure (kPa); σ₃ = minor principal stress / confining pressure (kPa); (σ₁ - σ₃) = deviator stress at failure = diameter of Mohr circle; (σ₁ + σ₃) = twice the center location of Mohr circle from origin. Valid only when c = 0.

Mnemonic

SIGMA-ONE = CONFINEMENT-N-squared PLUS TWO-C-N. Where N = tan(45 + φ/2). 'Two chunks: σ₃·N² + 2c·N.' The confinement chunk uses N², the cohesion chunk uses 2N.

When to use

When you know σ₃, c, and φ and need to find the major principal stress at failure. Common in triaxial test problems and bearing capacity checks. For c = 0 soils, the formula simplifies to σ₁ = σ₃·tan²(45 + φ/2).

What each part means

σ₁ = major principal stress at failure (kPa); σ₃ = confining (minor principal) stress (kPa); c = cohesion (kPa); φ = friction angle (degrees); N = tan(45 + φ/2) = Rankine's passive pressure coefficient (√K_p) = flow factor. The term tan²(45 + φ/2) is sometimes written as K_p (passive coefficient) or N_φ².

Mnemonic

THETA = 45 PLUS HALF-PHI. 'Start at 45, add half the friction.' The failure plane is always tilted more than 45° from the σ₃ plane (horizontal in standard triaxial setup) by an extra φ/2.

When to use

When locating the failure plane in a triaxial specimen or when drawing the Mohr-Coulomb envelope graphically. Essential for understanding why triaxial specimens fail at an angle rather than at 45°.

What each part means

θ_f = angle of failure plane measured from the minor principal stress plane (degrees); 45° = base angle for pure shear; φ/2 = additional tilt due to internal friction. Note: measured from σ₃ plane. If measured from σ₁ plane, the angle is (45 - φ/2).

Mnemonic

EFFECTIVE STRENGTH = subtract the tax first. 'Take away pore pressure u (the tax), then apply Mohr-Coulomb.' c' and φ' are effective (true) soil parameters; σ' is net (take-home) stress.

When to use

For all drained analyses (CD test, long-term stability, sand in any condition). The fundamental equation for realistic soil behavior. Always preferred over total stress when pore pressures are known.

What each part means

c' = effective cohesion (kPa) — true cohesion after drainage; φ' = effective friction angle (degrees) — true friction; σ = total normal stress (kPa); u = pore water pressure (kPa); σ' = effective normal stress = σ - u (kPa). This is Terzaghi's effective stress principle applied to shear strength.

Recall test

Without looking, name the three triaxial test types from least-drained to fully-drained, and state what parameters each gives. Time yourself: 30 seconds.

Memory chain

Three barangay officials at increasing levels of experience: UNCLE (UU) barged in with zero preparation and zero drainage. COUSIN (CU) prepared (consolidated) but panicked and closed the drain during shearing. CHIEF (CD) was calm, fully prepared, and kept everything open — the most experienced and most drained. Uncle → Cousin → Chief = UU → CU → CD. Each step adds MORE drainage.

Items to remember

• UU — Unconsolidated-Undrained (no drainage at all)
• CU — Consolidated-Undrained (drain during consolidation, close during shear)
• CD — Consolidated-Drained (drain throughout — fully drained)

Recall test

A CD test on sand: σ₃ = 80 kPa, deviator stress = 180 kPa. Run through DAVETs and find φ. Answer: σ₁ = 260 kPa; sin φ = 180/340 = 0.529; φ ≈ 31.9°.

Memory chain

DEVIATE to SUCCESS: Diagnose σ₃ → Add deviator to get σ₁ → Verify with SINE formula → Evaluate φ by arcsin → Take the answer. Short: DAVETs — Diagnose, Add, Verify, Evaluate, Take. Picture Dave taking a triaxial test step by step.

Items to remember

• Step 1: Identify σ₃ (confining pressure)
• Step 2: Calculate σ₁ = σ₃ + deviator stress
• Step 3: Apply sin φ = (σ₁ - σ₃)/(σ₁ + σ₃)
• Step 4: Find φ = arcsin of result
• Step 5: Verify using τ_f = σ·tan φ if needed

Recall test

A UC test on clay gives q_u = 150 kPa. What is c_u and φ? Answer: c_u = 75 kPa, φ = 0.

Memory chain

NAKED CLAY HALF STRENGTH: Naked (σ₃=0) Clay (cohesive only) HAlves the strength to give c_u. N-C-H-S. Or: 'Naked Clay Halves Strength.' Picture a naked clay cylinder on a table being cut in half to reveal c_u.

Items to remember

• UC test: σ₃ = 0 (no confinement)
• Failure load gives q_u (unconfined compressive strength)
• φ_u = 0 for saturated clay
• c_u = q_u / 2 (halve the UCS)
• Only valid for cohesive soils

Recall test

Describe a Mohr-Coulomb plot: what are the axes, what does the y-intercept represent, what does the slope represent, and write the equation of the failure envelope.

Memory chain

Graph story: 'I STAND on the NORMAL, I RISE with SHEAR. My starting height is C (cohesion), my rising angle is PHI (friction). The line I trace is Mohr-Coulomb.' Standing on x-axis (normal), rising on y-axis (shear), starting at c, rising at angle φ.

Items to remember

• X-axis: Normal stress σ (or σ')
• Y-axis: Shear stress τ
• Y-intercept: Cohesion c
• Slope angle: Friction angle φ
• Equation of line: τ_f = c + σ·tan φ
• Points on/below envelope: stable; above: failure

Recall test

A clay slope is built during rainy season. Which parameters govern stability (a) on the day of construction and (b) after 10 years? Answer: (a) UU — c_u, φ=0; (b) CD — c', φ'.

Memory chain

Timeline: TYPHOON DAY (UU) → CALM AFTER STORM (CU) → SUNNY YEARS LATER (CD). Typhoon = emergency short-term = UU = c_u. After storm = medium-term = CU. Sunny long-term = CD = c', φ'. Remember: Philippines has typhoon seasons (short-term) and dry seasons (long-term). Both must be designed for.

Items to remember

• Short-term (just after construction): use UU test → c_u, φ = 0
• Intermediate: use CU test → total or effective parameters
• Long-term (years after construction): use CD test → c', φ'
• Sand: always drained → always use φ' (CD parameters)
• Saturated clay: check both short-term (UU) and long-term (CD)