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CELE Steel & Timber Design — Steel Compression MembersMemory Anchors

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Memory anchors and mnemonic tricks for Steel Compression Members. If you find yourself forgetting key facts from this chapter during CELE mocks, these anchors are your fix. Built for Professional Regulation Commission (PRC) — Board of Civil Engineering's question style and the time pressure of the CELE 2026.

Exam context

For the Civil Engineer Licensure Examination, Professional Regulation Commission (PRC) — Board of Civil Engineering tests Steel & Timber Design under a "Core" label, with Steel Compression Members in the 2nd slot across 5 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 Steel & Timber Design questions. Date to watch: May and November 2026.

Steel Compression Members - Memory Anchors

Memory techniques can increase long-term recall by up to 400% compared to passive re-reading. For Steel Compression Members, the challenge is keeping straight the two Fcr formulas, the transition slenderness, the 0.658 exponent, and the phi factor. Each anchor below encodes a critical concept into a vivid image, story, or sound pattern so that when you see a board exam problem, the right formula fires instantly — no blank-staring at the page. Use these anchors actively: say them aloud, sketch the images, and quiz yourself with the recall triggers. Combine them with the Mermaid diagrams to build a complete mental map of the chapter.

Anchors

Topic

Elastic Buckling Stress

Concept

Euler elastic buckling stress formula: Fe = π²E / (KL/r)²

Anchor Id

Difficulty

medium

Memory Aid

Remember 'PIE over KLoR-squared' — imagine slicing a PIE (π²) and dividing it over a KLoR (KL/r) that is squared. Every time you see Fe, picture cutting a pie and placing it over a squared kilo. The 'E' stands for 'Euler's Engine' powering the formula.

Anchor Type

mnemonic

Why It Works

The phonetic hook 'PIE over KLoR-squared' maps directly to the formula structure. Pie = π², KLoR = KL/r, and squaring is the last operation. The visual of cutting pie reinforces the division.

Example Usage

Board exam gives KL/r = 70, E = 200,000 MPa. Think PIE over KLoR-squared: Fe = π²(200,000)/70² = 402.8 MPa.

Recall Trigger

Think: 'What is Fe?' → Visualize cutting a PIE → PIE over KLoR-squared → Fe = π²E/(KL/r)²

Topic

Transition Slenderness

Concept

Transition slenderness: KL/r = 4.71√(E/Fy)

Anchor Id

Difficulty

medium

Memory Aid

Use '4.71 — FOR SEVENTY-ONE percent of columns are inelastic!' The number 4.71 sounds like 'four-seventy-one.' Pair it with the image of a traffic boundary sign: on one side is a short, stocky jeepney (inelastic/stocky column) and on the other side is a tall, slender bamboo pole (elastic/slender column). The sign reads '4.71√(E/Fy)' — cross this and you enter elastic territory.

Anchor Type

mnemonic

Why It Works

The jeepney-vs-bamboo contrast creates a vivid boundary image. 4.71 is an unusual number, so the traffic-sign metaphor anchors it spatially. Filipino students instantly recognize a jeepney as short and robust.

Example Usage

For Fy = 248 MPa: 4.71√(200,000/248) = 133.7. If KL/r = 70 < 133.7 → stocky jeepney side → inelastic formula.

Recall Trigger

Think: 'Boundary between inelastic and elastic' → Jeepney vs. bamboo pole → 4.71√(E/Fy)

Topic

Inelastic Buckling — Fcr

Concept

Inelastic buckling formula: Fcr = [0.658^(Fy/Fe)] × Fy

Anchor Id

Difficulty

hard

Memory Aid

Think of 0.658 as a 'salary reduction factor.' Imagine Fy is your full salary (MPa). When a column is stocky but not perfectly straight, it doesn't get its full salary — it gets penalized by the factor 0.658 raised to the power (Fy/Fe). The harder the column tries to work (higher Fy/Fe), the bigger the penalty. The base 0.658 is fixed — memorize it as 'sixty-five-point-eight, never sixty-eight.'

Anchor Type

analogy

Why It Works

Salary-reduction analogy is relatable to Filipino graduates entering the workforce. The warning 'never sixty-eight' directly targets the common board-exam error of writing 0.685 or 0.680.

Example Usage

Fy/Fe = 0.616 → Fcr = 0.658^0.616 × 248. Compute 0.658^0.616 = 0.773, then 0.773 × 248 = 191.7 MPa.

Recall Trigger

Think: 'Inelastic column, stocky' → salary reduction → 0.658^(Fy/Fe) × Fy

Topic

Elastic Buckling — Fcr

Concept

Elastic buckling formula: Fcr = 0.877 × Fe

Anchor Id

Difficulty

medium

Memory Aid

Remember '0.877 = 87.7% of Euler.' Think of an 87.7% exam score — you almost got perfect Euler, but you lost 12.3% because the column was not perfectly straight from the start (initial out-of-straightness). Say it as: 'Elastic column? Take 87.7% of Euler.' The 0.877 factor has exactly three significant digits — '8-7-7, not 8-8-8.'

Anchor Type

mnemonic

Why It Works

Framing 0.877 as an exam score percentage is instantly relatable. The '87.7% of Euler' phrase gives physical meaning (imperfection penalty). The digit pattern 8-7-7 is easy to recall versus 8-8-8.

Example Usage

KL/r = 150, Fe = 87.73 MPa → Fcr = 0.877 × 87.73 = 76.9 MPa. (Slender column, Euler minus imperfection penalty.)

Recall Trigger

Think: 'Elastic column, slender' → 87.7% exam score → 0.877 Fe

Topic

Design Strength

Concept

Design strength: φcPn = 0.90 × Fcr × Ag

Anchor Id

Difficulty

easy

Memory Aid

Use 'PHI NINE-ZERO FAR AGain' — φ (phi) = 0.90, FAR = Fcr, AG = Ag. Say it fast: 'Phi nine-zero, Fcr, Ag.' Note: φc = 0.90 for steel columns — NOT 0.65 or 0.75 (those are for RC columns). Remember: 'Steel columns are 90-percenters — they give 90% of their nominal capacity.'

Anchor Type

mnemonic

Why It Works

The contrast with RC column phi factors (0.65/0.75) prevents the very common cross-subject confusion. '90-percenter' is a Filipino cultural expression meaning someone who gives near-full effort, which makes it sticky.

Example Usage

Fcr = 191.7 MPa, Ag = 8000 mm² → φcPn = 0.90 × 191.7 × 8000 = 1,380,240 N ≈ 1380 kN.

Recall Trigger

Think: 'Steel column design strength' → 90-percenter → φc = 0.90 → φcPn = 0.90 FcrAg

Topic

Effective Slenderness — Weak Axis

Concept

Always use the LARGEST KL/r (weakest axis governs)

Anchor Id

Difficulty

easy

Memory Aid

A chain breaks at its WEAKEST link. A steel column buckles about its WEAKEST axis — the one with the SMALLEST radius of gyration (r), giving the LARGEST KL/r ratio. Remember: 'LARGEST KL/r = WEAKEST axis = SMALLEST r.' Think of a flat ruler standing upright: it buckles easily about the thin direction (weak axis) not the wide direction. Always pick the largest KL/r for design.

Anchor Type

analogy

Why It Works

The flat-ruler image is tactile and universal. The chain analogy reinforces the 'weakest governs' principle that appears across multiple CE topics (shear, torsion, connections), making it cross-contextually reinforced.

Example Usage

If rx = 80 mm and ry = 50 mm, use KL/ry = KL/50 (larger ratio) for Fcr computation.

Recall Trigger

Think: 'Which KL/r to use?' → weakest chain link → largest KL/r governs

Topic

Slenderness Limit

Concept

Recommended slenderness limit: KL/r ≤ 200

Anchor Id

Difficulty

easy

Memory Aid

Sing to yourself: 'Two hundred is the max for steel, beyond that the column has no appeal!' KL/r > 200 means the column is too slender to be practical — it will shake and vibrate even under its own weight. In Philippine construction, think of a 200-story limit: no building in the Philippines reaches 200 stories (yet), and no compression member should reach KL/r of 200.

Anchor Type

rhyme

Why It Works

Rhymes are processed in the musical memory area of the brain, improving recall. The Philippine context (tallest buildings) makes the number 200 spatially anchored.

Example Usage

KL = 6 m = 6000 mm, r = 25 mm → KL/r = 240 > 200 → redesign required; increase section size.

Recall Trigger

Think: '200' → rhyme 'two hundred is the max' → KL/r ≤ 200 for compression members

Topic

Effective Length Factor K

Concept

K factor — effective length factor for buckling

Anchor Id

Difficulty

medium

Memory Aid

Visualize the 'K-flag positions': (1) Both ends pinned — the column buckles in a full sine wave, K = 1.0. (2) One end fixed, one end free (flagpole) — it buckles in a quarter wave, K = 2.0. (3) Both ends fixed — double curvature, K = 0.5. (4) Fixed-pinned — K = 0.7. Draw these as stick figures: (a) soldier at attention, (b) flagpole at the Mall of Asia, (c) bowed spring, (d) one knee bent.

Anchor Type

visual_association

Why It Works

Visual-spatial encoding with familiar Philippine landmarks (Mall of Asia flagpole) creates long-term memory hooks. The stick-figure mnemonic can be quickly sketched during exam review.

Example Usage

Column braced top and bottom (pinned-pinned) → K = 1.0 → KL = 1.0 × L. Cantilever column (fixed-free) → K = 2.0 → KL = 2.0 × L.

Recall Trigger

Think: 'K = ?' → visualize the flagpole at MOA vs. soldier vs. spring → K values

Topic

Inelastic Fcr — Exponent Base

Concept

Exponent base is 0.658, NOT 0.685 (common typo)

Anchor Id

Difficulty

hard

Memory Aid

Story: 'Engineer Carlo was rushing in his board exam and wrote 0.685 instead of 0.658. He transposed the last two digits — 5 and 8. He failed by 2 points. When he retook the exam, he wrote on his palm (mentally): SIX-FIVE-EIGHT, not EIGHT-FIVE — 0.658. He passed.' The digits go in ASCENDING order: 6 → 5 → 8? Wait — think of it as '6 is the first digit, 5 is the second, 8 is the last: 0.6-5-8.' The 8 comes at the END.

Anchor Type

micro_story

Why It Works

The cautionary tale creates an emotional anchor (fear of failing the board). Personal narrative engages episodic memory. The explicit warning against the specific typo directly addresses the most common error cited in board review centers.

Example Usage

When computing Fcr inelastic: write 0.658^(Fy/Fe) × Fy. Double-check: the exponent base ends in 8, NOT 5. So 0.658, not 0.685.

Recall Trigger

Think: 'Carlo's mistake' → 0.685 is WRONG → correct is 0.658

Topic

Buckling Modes

Concept

Buckling modes: flexural, torsional, flexural-torsional

Anchor Id

A10

Difficulty

medium

Memory Aid

Think of three ways a student can fail a plank walk: (1) FLEXURAL — the plank bends sideways in one plane (doubly symmetric sections, like W-shapes, standard failure mode). (2) TORSIONAL — the plank twists along its own axis without bending (cruciform or built-up sections with coincident centroid and shear center). (3) FLEXURAL-TORSIONAL — the plank bends AND twists simultaneously (channels, tees, angles — singly symmetric or unsymmetric sections, the most dangerous mode).

Anchor Type

analogy

Why It Works

The plank-walk analogy provides a physical action (bending vs. twisting vs. both) that maps to each buckling mode. The three-category classification appears frequently in board exams for section selection.

Example Usage

W-section column (doubly symmetric): check flexural buckling only. Channel section (singly symmetric): must check flexural-torsional buckling — it may govern.

Recall Trigger

Think: 'Three ways a plank can fail' → bend / twist / bend+twist → flexural / torsional / flexural-torsional

Topic

Local Buckling and Q Factor

Concept

Local buckling — slender elements reduce capacity (Q factor)

Anchor Id

A11

Difficulty

hard

Memory Aid

Think of a drinking straw (thin wall = slender element). A short straw can carry load, but if you press on it, the wall crinkles locally before the whole straw buckles — that is LOCAL buckling. A thick-walled bamboo section (compact element) won't crinkle. NSCP 2015 uses a Q factor (Q ≤ 1.0) to reduce Fy for slender elements — the thinner the straw wall, the smaller Q, the lower the strength.

Anchor Type

analogy

Why It Works

Straws and bamboo are everyday Filipino objects. The crinkling image of a drinking straw is viscerally familiar and provides an immediate physical intuition for local buckling.

Example Usage

If a W-section flange width-to-thickness ratio exceeds the limit λr, the flange is slender: use Q < 1.0 to reduce effective Fy in the Fcr formula.

Recall Trigger

Think: 'Drinking straw crinkling' → local buckling → Q factor → reduces capacity

Topic

ASD Design — Ωc

Concept

ASD safety factor for compression: Ωc = 1.67

Anchor Id

A12

Difficulty

easy

Memory Aid

Remember: 'LRFD steel compression: φc = 0.90 (a 90-percenter). ASD: Ωc = 1.67 (one-sixty-seven).' Chunk the pair: '90 and 167 — the steel compression duo.' If you know 1/0.90 = 1.111 and you double-check: Pallow = Pn/1.67 in ASD. Think of '1.67 ≈ 5/3,' a fraction that is easy to compute: 5/3 = 1.6667 ≈ 1.67. So ASD allowable = (5/3)Pn divided... no — Pa = Pn/Ωc = Pn × (3/5) × (1/Fcr × ...) — just remember Ωc = 1.67 = 5/3.

Anchor Type

chunking

Why It Works

Pairing φc and Ωc as a duo reduces memory load by grouping related information. Converting 1.67 to the fraction 5/3 gives a simpler number to recall and compute with.

Example Usage

ASD allowable axial load: Pa = Pn/Ωc = FcrAg/1.67. Or equivalently, Pa ≈ 0.60 × FcrAg (since 1/1.67 ≈ 0.60).

Recall Trigger

Think: 'ASD compression' → 5/3 → Ωc = 1.67

Topic

Transition — Fy/Fe = 2.25

Concept

Transition condition: Fy/Fe = 2.25 is equivalent to KL/r = 4.71√(E/Fy)

Anchor Id

A13

Difficulty

hard

Memory Aid

Remember '2.25 = the magic ratio.' When Fy/Fe equals exactly 2.25, you are right at the inelastic-elastic boundary. Check: 0.658^2.25 × Fy vs. 0.877 × Fe — they give the same Fcr at the boundary (both equal 0.39Fy approximately). Think '2.25 = nine-quarters = 9/4.' Memorize: 'If Fy/Fe ≤ 2.25, use 0.658; if Fy/Fe > 2.25, use 0.877.' The number 2.25 = 2¼ — picture two-and-a-quarter cups of rice: below 2.25 cups you cook with 0.658, above 2.25 cups you cook with 0.877.

Anchor Type

mnemonic

Why It Works

The fraction 9/4 is easier to recall than 2.25. The cooking-rice analogy is deeply Filipino and provides a concrete threshold image. The decision rule (≤ 2.25 vs. > 2.25) is embedded in the analogy.

Example Usage

Fy = 345, Fe = 197.4 → Fy/Fe = 1.748 < 2.25 → inelastic → Fcr = 0.658^1.748 × 345.

Recall Trigger

Think: '2.25 cups of rice' → the boundary ratio → Fy/Fe ≤ 2.25 → inelastic (0.658); > 2.25 → elastic (0.877)

Topic

Nominal Strength

Concept

Pn = Fcr × Ag — nominal compressive strength

Anchor Id

A14

Difficulty

easy

Memory Aid

Visualize a STAMP pressing down on a plate: the STAMP face area is Ag (gross area), and the pressure of the stamp is Fcr (critical stress). The total FORCE = stress × area = Fcr × Ag = Pn. Every time you see Pn, picture a huge stamp pressing down on a steel plate. The stamp face area is Ag, the stamp pressure is Fcr.

Anchor Type

visual_association

Why It Works

Stress × area = force is a fundamental mechanics concept. The stamp image makes it tactile and visual, preventing the common error of using net area (An) instead of gross area (Ag) for compression.

Example Usage

Fcr = 166.0 MPa, Ag = 6000 mm² → Pn = 166.0 × 6000 = 996,000 N = 996 kN → φcPn = 0.90 × 996 = 896.4 kN.

Recall Trigger

Think: 'Pn = ?' → stamp pressing on plate → force = pressure × area → Fcr × Ag

Topic

0.877 Factor — Physical Meaning

Concept

The 0.877 factor accounts for initial out-of-straightness in elastic range

Anchor Id

A15

Difficulty

medium

Memory Aid

Story: 'In a steel mill in Batangas, a long slender column was rolled and cooled — but due to imperfections in the process, it came out 1% curved (not perfectly straight). When engineers tested it, it buckled at only 87.7% of the theoretical Euler load, not 100%. From then on, AISC adopted the 0.877 factor to account for real-world columns that are never perfectly straight.' The story makes 0.877 feel like a historical accident — it was measured from real steel, not invented from math.

Anchor Type

micro_story

Why It Works

Origin stories are highly memorable (narrative memory). Localizing the factory to Batangas (a real Philippine industrial zone) grounds the abstract factor in a Filipino context. The physical explanation (imperfect column) gives the factor meaning beyond a mere number.

Example Usage

Elastic range (KL/r > 4.71√(E/Fy)): Fcr = 0.877 Fe. The column is imperfect → cannot reach full Euler load.

Recall Trigger

Think: 'Batangas steel mill, curved column' → 87.7% of Euler → 0.877 Fe

Topic

Problem-Solving Sequence

Concept

Step sequence for solving a compression member problem

Anchor Id

A16

Difficulty

medium

Memory Aid

Use the acronym 'SKETCH-FC-PHI': S = Solve KL/r (find largest), K = Know the transition (4.71√(E/Fy)), E = Evaluate Fe = π²E/(KL/r)², T = Test Fy/Fe or compare KL/r to transition, C = Choose formula (0.658 or 0.877), H = Hit Fcr (compute it), F = Find Pn = FcrAg, C = Cap with φc = 0.90, P = Produce φcPn, H = Highlight units (N or kN), I = Inspect KL/r ≤ 200. Say it: 'SKETCH-FC-PHI — and you'll never miss a step.'

Anchor Type

acronym

Why It Works

Acronyms encode sequences into retrievable chunks. SKETCH-FC-PHI is phonetically memorable and covers every step in the correct order. The final word 'PHI' links to the phi factor, reinforcing it at the end of the sequence.

Example Usage

Board exam: Given KL, r, Ag, Fy. Start: S — compute KL/r = KL/r; K — compute transition = 4.71√(E/Fy); E — compute Fe; T — test; C — choose formula; H — compute Fcr; FCP — compute φcPn; HI — check units and KL/r ≤ 200.

Recall Trigger

Think: 'Compression member problem' → SKETCH-FC-PHI → 10-step solution path

Topic

Inelastic vs. Elastic Buckling

Concept

Inelastic buckling occurs in STOCKY columns (short and stout); elastic in SLENDER columns

Anchor Id

A17

Difficulty

easy

Memory Aid

Think of a BARANGAY CAPTAIN (stocky, authority figure) vs. a BAMBOO POLE (tall and slender). The barangay captain is compact and fails inelastically — he resists the load but still yields a little before collapsing (inelastic buckling). The bamboo pole is slender and snaps elastically without yielding — it buckles suddenly while still below yield (elastic buckling, governed by Euler). In design: stocky captain → 0.658 formula; slender bamboo → 0.877Fe formula.

Anchor Type

analogy

Why It Works

Filipino cultural figures (barangay captain) and natural imagery (bamboo) are immediately relatable. The physical contrast between robust and slender is universally understood. The pairing of each figure with the correct formula creates a direct retrieval link.

Example Usage

KL/r = 70 < 133.7 → barangay captain (stocky, inelastic) → use Fcr = 0.658^(Fy/Fe) × Fy. KL/r = 150 > 133.7 → bamboo (slender, elastic) → use Fcr = 0.877Fe.

Recall Trigger

Think: 'Barangay captain = inelastic (0.658); bamboo = elastic (0.877Fe)'

Topic

Gross Area in Compression

Concept

Gross area Ag is used for compression (not net area An)

Anchor Id

A18

Difficulty

easy

Memory Aid

For TENSION members, bolts punch holes — use NET area (An). For COMPRESSION members, the column is squeezed — it FILLS the holes! The compression force pushes the metal together, so all the material is active. Picture a fat sponge (gross area) being squeezed — all of it resists. Compare to a tension sponge being pulled apart where the holes matter. COMPRESSION → GROSS; TENSION → NET. Slogan: 'Compression is GROSS; Tension is NET (of holes).'

Anchor Type

visual_association

Why It Works

The contrast with tension members (a related chapter) deepens understanding and prevents cross-chapter errors. The sponge analogy is tactile. The slogan uses everyday financial language (gross vs. net income) which Filipino graduates understand.

Example Usage

A W-shape with bolt holes: for column compression design, use full Ag (holes don't reduce compressive area). For tension design of the same section, use An = Ag minus hole areas.

Recall Trigger

Think: 'Compression → GROSS area' vs. 'Tension → NET area'

Topic

Effect of Fy on Transition

Concept

High-strength steel (Fy = 345 or 415 MPa) has lower transition slenderness than Fy = 248 MPa

Anchor Id

A19

Difficulty

hard

Memory Aid

Think: 'Strong material, shorter patience.' High-strength steel (Fy = 345 MPa) transitions to elastic buckling at a LOWER KL/r = 113.4 compared to Grade 248 (transition at 133.7). It's like a strong but impatient student — at a shorter column length, it already switches to the elastic (Euler) behavior. The formula confirms: 4.71√(E/Fy) decreases as Fy increases. Stronger steel → lower transition KL/r → elastic zone starts earlier.

Anchor Type

analogy

Why It Works

The 'strong but impatient' analogy gives intuitive directionality (higher Fy → lower threshold). This prevents the incorrect assumption that stronger steel is always better for slender columns — a nuanced board exam trap.

Example Usage

Fy = 415 MPa: transition = 4.71√(200,000/415) = 103.4. A column with KL/r = 110 is elastic for Fy = 415 but inelastic for Fy = 248 (transition = 133.7).

Recall Trigger

Think: 'High Fy = impatient = lower transition KL/r'

Topic

Material Property — Modulus of Elasticity

Concept

E = 200,000 MPa (modulus of elasticity for steel)

Anchor Id

A20

Difficulty

easy

Memory Aid

Remember: 'E for steel = 200 GPa = 200,000 MPa.' Chunk it as TWO-HUNDRED-THOUSAND. In the board exam, E is almost always given as 200,000 MPa. If not given, use this value. Cross-check: 200 GPa = 200 × 10³ MPa = 200,000 MPa. The chunk '200-200-200' reminds you: 200 GPa, 200,000 MPa, 200 kN/mm². All three are the same number in different units — steel's elastic modulus is always 200.

Anchor Type

chunking

Why It Works

Repetition of '200' in three unit systems creates a rhythmic pattern. Chunking reduces three separate facts into one memorable number with unit conversions attached.

Example Usage

Fe = π²(200,000)/(KL/r)² — always substitute E = 200,000 MPa when steel is the material.

Recall Trigger

Think: 'E for steel = 200' → 200 GPa = 200,000 MPa = 200 kN/mm²

Revision Game

4.71 — the transition slenderness constant (as in 4.71√(E/Fy))

Clue

I am the constant that separates the Jeepney column from the Bamboo column. What number am I?

Memory Link

Anchor A2 — Jeepney vs. Bamboo boundary sign reads 4.71√(E/Fy)

0.685 — the wrong value. The correct base is 0.658.

Clue

Carlo failed his board exam because he wrote me instead of the correct base for the inelastic Fcr formula. What wrong number did Carlo write?

Memory Link

Anchor A9 — Carlo's cautionary micro-story about transposing digits in 0.658

φc = 0.90

Clue

I am the phi factor for steel compression in LRFD. I am NOT 0.65 (that's for RC tied columns) and NOT 0.75 (RC spiral). What am I?

Memory Link

Anchor A5 — 'Steel columns are 90-percenters'

87.7% — expressed as the factor 0.877 in Fcr = 0.877 Fe

Clue

A slender column buckles before yielding and its Fcr is only this percentage of its Euler stress — accounting for initial out-of-straightness. What percentage?

Memory Link

Anchor A4 and A15 — '87.7% exam score' and 'Batangas steel mill imperfect column'

KL/r = 200

Clue

I am the maximum recommended slenderness ratio for steel compression members under NSCP 2015. Exceed me and your column is impractical.

Memory Link

Anchor A7 — 'Two hundred is the max for steel — beyond that the column has no appeal!'

The weak axis (axis with smallest radius of gyration r, giving largest KL/r)

Clue

For a W-shape column, I always check buckling about this axis — the one that gives the LARGEST KL/r ratio. Which axis am I?

Memory Link

Anchor A6 — 'A chain breaks at its weakest link — use the largest KL/r'

Ag — the gross cross-sectional area

Clue

When computing the nominal compressive strength, I use this area — not the net area, not the effective area (for compact sections). What area do I use?

Memory Link

Anchor A18 — 'Compression is GROSS; Tension is NET' and the sponge analogy

Flexural-torsional buckling (because a channel is singly symmetric — shear center does not coincide with centroid)

Clue

A channel section used as a compression strut must be checked for this more critical buckling mode beyond simple flexural buckling. Name it.

Memory Link

Anchor A10 — 'Three ways a plank can fail: bend / twist / bend+twist' — Channel = bend+twist

Ωc = 1.67 ≈ 5/3 = ASD safety factor for compression; Pa = allowable compressive strength; must satisfy P ≤ Pa where P is the service (unfactored) load

E (steel) = 200,000 MPa
φc (LRFD compression) = 0.90
Ωc (ASD compression) = 1.67
Inelastic exponent base = 0.658
Elastic reduction factor = 0.877
Transition constant = 4.71
Transition Fy/Fe ratio = 2.25
Max recommended slenderness = KL/r ≤ 200

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CELE Steel & Timber Design — Steel Compression MembersMemory Anchors

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Steel Compression Members - Memory Anchors

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