Memory AnchorsCELE · Geotechnical EngineeringReal content

CELE Geotechnical Engineering — Shear Strength of SoilsMemory Anchors

Summary Concepts Notes Revision Deep Dive Slides Flashcards Mistakes Cheat Mnemonics Templates

Memory anchors for Shear Strength of Soils — mnemonic devices, acronyms, and tricks that make the CELE Geotechnical Engineering syllabus stick. Use these when a concept just will not stay in your head.

Exam context

For the Civil Engineer Licensure Examination, Professional Regulation Commission (PRC) — Board of Civil Engineering tests Geotechnical Engineering under a "Core" label, with Shear Strength of Soils in the 7th slot across 11 chapters. CELE candidates must clear the 70% weighted average, no sub-test below 50% cut on the 2026 paper, which draws about a meaningful share of Geotechnical Engineering questions. Date to watch: May and November 2026.

Shear Strength of Soils - Memory Anchors

Memory anchors work by connecting unfamiliar technical information to vivid, emotionally-charged, or story-driven mental images that your brain already knows how to store. Research in cognitive science shows that spaced repetition + memorable associations can boost long-term recall by up to 400% compared to re-reading notes. For the PRC Civil Engineer Licensure Examination, where Geotechnical Engineering questions are frequently numerical and concept-heavy, these anchors give you instant access to formulas, classifications, and procedures under exam pressure. Use each anchor as a mental 'hook' — when you see a problem, the hook fires automatically, pulling the right formula or concept into working memory. The anchors in this set cover every major concept in Shear Strength of Soils: Mohr-Coulomb criterion, triaxial test types, drainage conditions, and principal stress relationships.

Anchors

Topic

Mohr-Coulomb Failure Criterion

Concept

Mohr-Coulomb Failure Criterion: τ_f = c + σ·tan(φ)

Anchor Id

Difficulty

easy

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.

Anchor Type

analogy

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.

Topic

Soil Classification by Shear Behavior

Concept

Cohesion c vs. Friction angle φ — which soil has which

Anchor Id

Difficulty

easy

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.

Anchor Type

acronym

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).

Topic

Unconfined Compression Test

Concept

Undrained shear strength from UC test: c_u = q_u / 2

Anchor Id

Difficulty

easy

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.

Anchor Type

rhyme

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.

Topic

Triaxial Test Types

Concept

Triaxial Test Types: UU, CU, CD — what they mean

Anchor Id

Difficulty

medium

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.

Anchor Type

acronym

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.

Topic

Drainage Conditions — Short Term vs. Long Term

Concept

UU test → short-term (undrained) parameters: φ = 0, c = c_u

Anchor Id

Difficulty

medium

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.

Anchor Type

micro_story

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.

Topic

Drainage Conditions — Long Term

Concept

CD test → long-term (drained) parameters: c', φ' (effective)

Anchor Id

Difficulty

medium

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.

Anchor Type

micro_story

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.

Topic

Principal Stresses at Failure

Concept

Principal stress relationship at failure (c = 0 sand): sin φ = (σ₁ - σ₃)/(σ₁ + σ₃)

Anchor Id

Difficulty

hard

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.

Anchor Type

visual_association

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.

Topic

Principal Stress at Failure — General Formula

Concept

σ₁ at failure in terms of σ₃: σ₁ = σ₃·tan²(45 + φ/2) + 2c·tan(45 + φ/2)

Anchor Id

Difficulty

hard

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.

Anchor Type

chunking

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).

Topic

Direct Shear Test

Concept

Direct shear test — plots τ_f vs σ to get c and φ

Anchor Id

Difficulty

easy

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.

Anchor Type

analogy

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.

Topic

Triaxial Test — Deviator Stress

Concept

Deviator stress vs. major principal stress: σ₁ = σ₃ + Δσ (deviator)

Anchor Id

A10

Difficulty

medium

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!

Anchor Type

mnemonic

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.

Topic

CU Triaxial Test

Concept

CU test — gives BOTH total stress and effective stress parameters

Anchor Id

A11

Difficulty

medium

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.

Anchor Type

micro_story

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.

Topic

Effective Stress Principle

Concept

Effective vs. total stress: use σ' = σ - u for drained analyses

Anchor Id

A12

Difficulty

medium

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.

Anchor Type

analogy

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.

Topic

Unconfined Compression Test

Concept

Unconfined compression test (UC): σ₃ = 0, special case of UU

Anchor Id

A13

Difficulty

easy

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.

Anchor Type

visual_association

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.

Topic

Failure Plane Orientation

Concept

Failure plane orientation: 45 + φ/2 degrees from major principal plane

Anchor Id

A14

Difficulty

medium

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).

Anchor Type

mnemonic

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.

Topic

UC Test — Common Pitfall

Concept

Board exam pitfall: c_u = q_u/2, NOT q_u

Anchor Id

A15

Difficulty

easy

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.'

Anchor Type

micro_story

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.

Topic

φ = 0 Concept for Saturated Undrained Clay

Concept

Saturated clay undrained behavior: φ_u = 0 (φ = 0 concept)

Anchor Id

A16

Difficulty

medium

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.

Anchor Type

analogy

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.

Topic

Sand Shear Behavior

Concept

Sand behavior: purely frictional, c = 0, φ typically 28°–45°

Anchor Id

A17

Difficulty

easy

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.

Anchor Type

analogy

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.

Topic

CU Test Envelopes

Concept

CU test: total stress parameters vs. effective stress parameters

Anchor Id

A18

Difficulty

hard

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).

Anchor Type

visual_association

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.

Topic

CU vs. CD Test Procedure

Concept

Sequence: Consolidation then Shearing in CU and CD tests

Anchor Id

A19

Difficulty

medium

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.

Anchor Type

method_of_loci

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.

Topic

Short-term vs. Long-term Stability

Concept

Short-term vs. Long-term stability: which governs for clay embankments

Anchor Id

A20

Difficulty

medium

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.

Anchor Type

micro_story

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').

Revision Game

Unconfined Compression (UC) Test — σ₃ = 0

Clue

I am the test where the soil cylinder stands completely naked — no jacket, no confining pressure, just being pushed from above until I fail. What test am I?

Memory Link

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

One-half (1/2) — c_u = q_u / 2

Clue

I am the fraction that converts the unconfined compressive strength into undrained shear strength. Say my value and the formula.

Memory Link

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

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

Clue

In a triaxial test on sand, σ₃ = 100 kPa and the deviator stress = 200 kPa. What is φ? (Solve in your head using the SINE formula.)

Memory Link

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

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

Clue

A typhoon just hit and the contractor must check the clay embankment stability RIGHT NOW. Which test type parameters should the engineer use, and what is the friction angle value?

Memory Link

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

Chief Doming — the CD (Consolidated-Drained) test

Clue

I am the barangay official who is fully prepared, completely drained, and gives you the most realistic long-term effective parameters c' and φ'. Who am I?

Memory Link

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

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

Clue

A soil has c = 15 kPa and φ = 25°. The normal stress on the failure plane is 200 kPa. What is the shear strength? (Use the bangus-on-banana-leaf formula.)

Memory Link

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

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

Clue

My salary (total stress) is ₱200 kPa. I pay ₱50 kPa in tax (pore pressure). What is my take-home pay (effective stress), and how do I use it in the shear strength formula?

Memory Link

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

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).

Clue

Sand is Silly. Clay is Clingy. What do these phrases tell you about the shear parameters c and φ for each soil type?

Memory Link

A2 — SC-ZC-CC acronym and personality analogy

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)

Loading diagram…

Previous chapter

Consolidation and Settlement

Next chapter

Lateral Earth Pressure and Retaining Structures

Ready to practise for the CELE 2026?

Super Tutor's AI review plan adapts to your weak areas and builds a weekly practice schedule around your target CELE exam date.

Start free See the full CELE guide CELE preparation guide

← Back to Geotechnical Engineering hub

CELE Geotechnical Engineering — Shear Strength of SoilsMemory Anchors

Exam context

Shear Strength of Soils - Memory Anchors

Anchors

Tags

Topic

Concept

Anchor Id

Difficulty

Memory Aid

Anchor Type

Why It Works

Example Usage

Recall Trigger

Tags

Topic

Concept

Anchor Id

Difficulty

Memory Aid

Anchor Type

Why It Works

Example Usage

Recall Trigger

Tags

Topic

Concept

Anchor Id

Difficulty

Memory Aid

Anchor Type

Why It Works

Example Usage

Recall Trigger

Tags

Topic

Concept

Anchor Id

Difficulty

Memory Aid

Anchor Type

Why It Works

Example Usage

Recall Trigger

Tags

Topic

Concept

Anchor Id

Difficulty

Memory Aid

Anchor Type

Why It Works

Example Usage

Recall Trigger

Tags

Topic

Concept

Anchor Id

Difficulty

Memory Aid

Anchor Type

Why It Works

Example Usage

Recall Trigger

Tags

Topic

Concept

Anchor Id

Difficulty

Memory Aid

Anchor Type

Why It Works

Example Usage

Recall Trigger

Tags

Topic

Concept

Anchor Id

Difficulty

Memory Aid