Steel Beams: Flexure and Shear - Memory Anchors
The human brain encodes information through emotion, story, imagery, and pattern — not raw repetition. For board exam success, you need formulas and concepts to surface instantly under pressure. This collection of mnemonics, analogies, micro-stories, and visual anchors transforms dry steel design equations into vivid mental hooks. Each anchor is engineered to fire automatically when you see a related exam question — because in the licensure exam, speed and accuracy are everything.
Memory anchors
Plastic Moment formula: Mn = Mp = Fy × Zx
Memory aid
Remember 'FiZz' — F-y times Z-x gives you the FIZZ (the full fizzy pop of a beam at its plastic limit). When a beam is fully braced and compact, it FIZZES with energy: Mp = Fy × Zx. The 'p' in Mp stands for 'POPPED' — the beam has yielded fully across the entire cross-section, like a soda can that has been fully pressurized.
Why it works
The phonetic link 'FiZz' encodes both variables Fy and Zx into a single sound cue, making formula recall near-instant.
Example usage
Exam question asks for design flexural strength of a compact, braced W-section. Recall 'FiZz' → Mp = Fy × Zx → then apply φb = 0.90 to get φbMn.
Recall trigger
Think of a fizzing soda — Fy × Zx = Mp
φb = 0.90 for flexure (LRFD resistance factor)
Memory aid
Nine-tenths for bending — that's the rule, my friend. φb = 0.90 — bend it, don't end it. Think: 'BENDING is NINE-ty percent sure.' The beam resists 90% of what it theoretically can, because real-world imperfections eat 10%.
Why it works
Rhyme and numerical anchoring together. 0.90 is linked to 'ninety' which sounds like 'bending' in memory via the rhyme.
Example usage
After computing Mn = Mp = Fy Zx, multiply by 0.90 to get φbMn. If Mn = 297.6 kN·m, then φbMn = 0.90 × 297.6 = 267.8 kN·m.
Recall trigger
φb — bending — ninety percent — 0.90
Lp formula: Lp = 1.76 ry √(E/Fy) — maximum unbraced length for full Mp
Memory aid
Use the memory phrase: 'ONE POINT SEVEN SIX, ry ROOTS E over Fy — STAY IN THE ZONE.' Picture a basketball player (ry = radius of gyration about y-axis) staying within the three-point zone (Lp). 1.76 is the magic number — almost 'one-and-three-quarters.' Chunk it: 1.76 — ry — square root — E over Fy. Say it like a chant: 'Seventy-six, ry, root E on Fy!'
Why it works
Chunking the four parts of the formula into a chant with rhythm locks the sequence in memory through auditory encoding.
Example usage
Given ry = 40 mm, E = 200,000 MPa, Fy = 248 MPa: Lp = 1.76 × 40 × √(200,000/248) = 70.4 × 28.4 ≈ 2,000 mm = 2.0 m.
Recall trigger
Basketball player staying in the zone → Lp = 1.76 ry √(E/Fy)
The three LTB zones: Lb ≤ Lp (no LTB), Lp < Lb ≤ Lr (inelastic LTB), Lb > Lr (elastic LTB)
Memory aid
Think of a JEEPNEY TRIP from Manila to Batangas. Zone 1 (Lb ≤ Lp): You're still in the city — smooth ride, full speed (Mn = Mp, no reduction). Zone 2 (Lp < Lb ≤ Lr): You're on the winding Tagaytay road — you slow down gradually (inelastic LTB, Mn drops linearly). Zone 3 (Lb > Lr): You're on a rough mountain trail — very slow, elastic buckling controls, Mn = Fcr × Sx (elastic LTB formula takes over).
Why it works
The Filipino jeepney road trip creates a vivid spatial journey that maps perfectly onto the three-zone LTB behavior. Distance traveled = Lb, speed = Mn.
Example usage
If Lb = 1.5 m and Lp = 2.0 m: Lb ≤ Lp → Zone 1 (city) → No LTB → Mn = Mp. Full capacity available.
Recall trigger
Jeepney trip: city → winding road → mountain trail = three LTB zones
Zx (plastic section modulus) vs Sx (elastic section modulus) — Zx > Sx
Memory aid
Imagine twins named SX and ZX applying for a job as beam section moduluses. SX is the conservative older twin — he stops working the moment the outermost fiber yields (elastic limit). ZX is the ambitious younger twin — he keeps working even after the outer fiber yields, mobilizing the ENTIRE cross-section in plasticity. ZX always gets the bigger paycheck (Zx > Sx). The boss (AISC 360) hires ZX for plastic moment calculations. The shape factor (Zx/Sx ≈ 1.12 for I-shapes) is ZX's 12% salary premium.
Why it works
Personification of abstract concepts makes them memorable. The 'twin' story encodes the comparison, the salary premium encodes the 12% ratio.
Example usage
For plastic moment, always use Zx (not Sx): Mp = Fy × Zx. Common board pitfall is using Sx — remember ZX gets the bigger job.
Recall trigger
The ambitious twin ZX with the bigger paycheck — Zx > Sx, shape factor ≈ 1.12
Shear strength formula: Vn = 0.6 Fy Aw Cv
Memory aid
Remember the phrase: 'SIX-TENTHS FyAwCv GIVES YOU SHEAR STRENGTH.' Chunked: 0.6 — Fy — Aw — Cv. Create an acronym for the variables after 0.6: 'F-A-C' (Fy, Aw, Cv) = 'FACT.' So: Vn = 0.6 × FACT. Remember: '0.6 times the FACT equals the shear.' The 0.6 comes from the von Mises yield criterion (shear yield stress ≈ 0.577Fy ≈ 0.6Fy).
Why it works
The word FACT as an acronym for F-A-C links three separate variables into one memorable word with positive connotations (facts are certain, reliable).
Example usage
Given d = 450 mm, tw = 10 mm, Fy = 248 MPa, Cv = 1.0: Aw = 450 × 10 = 4,500 mm². Vn = 0.6 × 248 × 4,500 × 1.0 = 669,600 N = 669.6 kN.
Recall trigger
0.6 × FACT → Vn = 0.6 Fy Aw Cv
Aw = d × tw (web area uses full depth d, not clear web height)
Memory aid
Picture a DOOR (the web of the I-beam). The DOOR HEIGHT is the full door height d (from floor to top frame, like the full depth of the beam). The DOOR THICKNESS is tw. The web area Aw is the door's face area: height × thickness = d × tw. The board exam trap is using only the clear web height (between flanges) — but AISC uses the FULL door height d. Visualize the door frame included — it's ALL the door, not just the inner panel.
Why it works
Spatial visualization of a door creates a concrete 3D image that immediately communicates 'full height' as the key word.
Example usage
Never use (d - 2tf) for Aw in the AISC shear formula. If d = 600 mm, tw = 12 mm → Aw = 600 × 12 = 7,200 mm² (full depth).
Recall trigger
The DOOR — full door height d × door thickness tw = Aw
φv = 1.0 for shear of typical rolled I-shapes (not 0.90)
Memory aid
Engineer Mang Romy is inspecting a steel beam's web. His colleague asks, 'Mang Romy, ano ang phi for shear?' Mang Romy answers confidently: 'ISANG (ONE) — BUO! φv = 1.0 — buong-buo!' (meaning 'whole/complete' in Filipino). Unlike flexure which is nervous about imperfections (φb = 0.90, only 90%), the shear web of a stocky rolled I-shape is SO reliable that AISC gives it a perfect score: φv = 1.0. It's the straight-A student of resistance factors.
Why it works
The Filipino word 'buo' (whole) encodes the value 1.0, and the contrast with φb = 0.90 reinforces the distinction students frequently confuse.
Example usage
After computing Vn = 669.6 kN: φvVn = 1.0 × 669.6 = 669.6 kN. Do NOT use 0.90 for shear of typical rolled I-shapes.
Recall trigger
'Buo!' — φv = 1.0 for rolled I-shape shear (not 0.90)
Cv = 1.0 when h/tw ≤ 2.24√(E/Fy) — stocky web, no shear buckling
Memory aid
Cv is the 'CLAPBOARD VOTER' — it votes either 1.0 (full shear capacity) or less than 1.0 (shear buckling reduces capacity). For a STOCKY web (thick enough that h/tw ≤ 2.24√(E/Fy)), the voter shouts 'YES — FULL CAPACITY!' and Cv = 1.0. For a SLENDER web, the voter hesitates. The number 2.24 is almost 2 and a quarter — remember it as '2 and a quarter check' for the web slenderness limit.
Why it works
Voting metaphor (yes/no decision) maps cleanly onto the binary condition check. Filipino students engage well with civic-participation analogies.
Example usage
Check: h/tw ≤ 2.24√(200,000/248) = 2.24 × 28.4 = 63.6. If web passes this check, use Cv = 1.0 in Vn formula.
Recall trigger
Web slenderness check — '2 and a quarter √(E/Fy)' → Cv = 1.0 if web passes
Compact section requirement — flange and web λ < λp so full yielding occurs before local buckling
Memory aid
A steel section is like a BIBINGKA MOLD. If the mold walls are THICK and SHORT (compact — λ < λp), the bibingka sets perfectly into its full plastic shape without the mold warping. If the mold is THIN and TALL (non-compact or slender), it wrinkles or collapses before the batter fully sets. The bibingka = the full plastic moment Mp. Only a compact mold gives you the perfect bibingka.
Why it works
Bibingka is a culturally familiar Filipino rice cake baked in a mold — the analogy maps perfectly onto plate buckling before full yielding.
Example usage
Before using Mn = Mp = Fy Zx, verify the section is compact: check both flange (bf/2tf ≤ λp) and web (h/tw ≤ λp) slenderness.
Recall trigger
Bibingka mold — thick walls (compact) → perfect shape (full Mp); thin walls (slender) → wrinkled (local buckling)
Cb (moment gradient factor) — increases Mn for non-uniform moment, capped at Mp
Memory aid
Cb is the 'CHEERLEADER BONUS' — when a beam has varying moment along its length (not constant), the cheerleader Cb cheers it on beyond the conservative uniform-moment assumption. A Cb > 1.0 means the beam is working in a more favorable loading pattern, so it earns a bonus. BUT the principal (Mp) sets a SALARY CAP — no matter how much Cb cheers, Mn cannot exceed Mp. Cb = 1.0 for uniform moment (no bonus — monotone cheerleader).
Why it works
The 'cheerleader' and 'salary cap' metaphors encode both the upward effect of Cb and the critical ceiling of Mp simultaneously.
Example usage
If LTB analysis gives Mn = 280 kN·m and Cb = 1.15 → Cb × Mn = 1.15 × 280 = 322 kN·m. But if Mp = 310 kN·m, the answer is capped: Mn = 310 kN·m.
Recall trigger
Cheerleader Cb — bonus for non-uniform moment — capped at Mp
Inelastic LTB zone: Mn varies linearly between Mp and 0.7FySx
Memory aid
Picture a DIMMER SWITCH (ang dimmer ng ilaw). At Lb = Lp, the light is FULL BRIGHT (Mn = Mp). As Lb increases past Lp, you slowly dim the light — linearly decreasing. At Lb = Lr, the light dims to 70% of a different reference (0.7FySx). The dimmer switch is LINEAR — not sudden. Beyond Lr, you switch to a different mechanism entirely (elastic buckling, a new formula). The dimmer perfectly represents the linear interpolation in the inelastic LTB zone.
Why it works
The dimmer switch provides a physical, kinesthetic analogy for linear reduction — students can almost feel themselves turning the dial.
Example usage
If Lb = 2.5 m, Lp = 2.0 m, Lr = 5.0 m: in inelastic LTB zone. Use linear interpolation formula between Mp and 0.7FySx based on position of Lb relative to Lp and Lr.
Recall trigger
Dimmer switch — Lb between Lp and Lr → Mn dims linearly from Mp to 0.7FySx
Shape factor ≈ 1.12 for I-shapes: Zx/Sx ≈ 1.12
Memory aid
Remember: 'I-shapes are 12% SMARTER than their elastic selves.' Zx = 1.12 × Sx for typical W-sections. The shape factor is the IQ boost from elastic to plastic thinking. If you only use Sx (elastic), you leave 12% of the beam's strength on the table — like scoring 88/100 when you could score 100/100. Encode it as: 'Z is 12% more than S for an I.'
Why it works
Percentage framing (12% smarter) is concrete and relatable, and the test-score analogy resonates with exam-focused students.
Example usage
If Sx = 1,070 × 10³ mm³, then Zx ≈ 1.12 × 1,070 × 10³ = 1,198 × 10³ mm³ (approximate check). Always use actual tabulated Zx in computations.
Recall trigger
I-beam is 12% smarter — Zx ≈ 1.12 Sx
ry is the radius of gyration about the weak (y) axis — governs LTB
Memory aid
A steel beam is like a FLAGPOLE. It is STRONG against bending in its flat dimension (x-axis — strong axis) but WEAK laterally (y-axis — weak axis, the flagpole wobbles sideways in the wind). ry measures how resistant the cross-section is to that sideways wobble. A SMALLER ry means the beam wobbles more easily → shorter Lp → more susceptible to LTB. A larger ry → longer Lp → better resistance to lateral buckling.
Why it works
The flagpole wobbling in wind creates a perfect sensory image of lateral instability, linking ry directly to its physical meaning.
Example usage
Larger ry → larger Lp = 1.76 ry √(E/Fy) → longer allowable unbraced length before LTB → more economical beam design.
Recall trigger
Flagpole wobbling sideways = ry — weak axis radius of gyration → controls Lp
Key board-exam pitfall: Using Sx instead of Zx for plastic moment
Memory aid
During the 2019 board exam (a fictional scenario for memory), Engineer Josie computed Mp = Fy × Sx and got an answer 12% lower than the correct answer. She failed by 1 point. Her ghost now haunts engineering review centers, whispering 'USE ZX! NOT SX! ZX IS PLASTIC! SX IS ELASTIC!' Every time you see 'plastic moment' — hear Josie's ghost: 'ZX! ZX! ZX!' The exam question always tests whether you know the difference.
Why it works
The ghost story with emotional stakes (failing by 1 point) creates a memorable cautionary tale. Emotion dramatically enhances memory encoding.
Example usage
Exam gives Sx = 850 × 10³ mm³ and Zx = 950 × 10³ mm³, Fy = 345 MPa. Mp = 345 × 950 × 10³ = 327.75 × 10⁶ N·mm. Use Zx, not Sx.
Recall trigger
Josie's ghost whispering 'ZX!' → never use Sx for plastic moment
Elastic LTB zone (Lb > Lr): Mn = Fcr × Sx — elastic critical stress governs
Memory aid
Picture a SPRING (elastic) that is stretched WAY too far — it no longer yields; it just snaps back elastically. When Lb > Lr, the beam buckles elastically (like an overstretched spring) before ANY yielding — so the plastic and inelastic tools are useless. The elastic critical stress Fcr (like the spring constant) takes over, multiplied by Sx (elastic modulus — appropriate now since there's NO yielding). Spring = elastic = Fcr × Sx.
Why it works
The spring analogy directly encodes the word 'elastic' into a physical object, and the connection to Sx (elastic modulus) becomes logical rather than arbitrary.
Example usage
If Lb > Lr, use Mn = Fcr × Sx where Fcr involves Cb, E, Lb, ry, and torsional properties. Beam never yields — purely elastic buckling.
Recall trigger
Overstretched spring → elastic LTB → Mn = Fcr × Sx
AISC 360 governs steel design; NSCP 2015 adopts AISC provisions for Philippine practice
Memory aid
Remember: 'NSCP ADOPTS AISC — Like a PARENT adopting a CHILD.' AISC 360 is the brilliant American parent; NSCP 2015 (National Structural Code of the Philippines, adopted under RA 544 — the Civil Engineering Law) is the Philippine parent that adopts AISC's provisions for local use. In the board exam, both codes speak the same language. RA 544 is the mother law that empowers the PRC to enforce these standards.
Why it works
The adoption metaphor clarifies the hierarchical relationship between codes and connects Philippine law to design standards — a common source of confusion.
Example usage
When citing code in board exam problems: NSCP 2015 Section 502 (Steel) refers to AISC 360 provisions. Both φb = 0.90 and φv = 1.0 apply under Philippine practice.
Recall trigger
AISC parent → NSCP adopts → RA 544 is the grandmother law
Web local buckling (WLB) and flange local buckling (FLB) reduce Mn in non-compact sections
Memory aid
Imagine an ACCORDION (the musical instrument). A compact section is a well-made accordion with thick reeds — squeeze it all the way (full plasticity, full Mn). A non-compact section is a cheap accordion with thin reeds — they buckle inward before you fully squeeze (FLB or WLB reduces Mn). A slender section is a paper accordion — it collapses immediately (elastic local buckling, Mn heavily reduced). The squeeze = loading; accordion collapse = local buckling.
Why it works
The accordion analogy maps compression (squeezing) directly onto the mechanism of local buckling in plate elements.
Example usage
Non-compact flange: Mn falls between Mp and FLB limit. Check: λpf < bf/2tf ≤ λrf. Mn is reduced by linear interpolation — same dimmer-switch concept, but for local buckling.
Recall trigger
Accordion — thick reeds (compact, full squeeze), thin reeds (non-compact, partial), paper (slender, collapses) = FLB/WLB
The 0.6 factor in Vn comes from shear yield stress τy ≈ 0.577Fy (von Mises), rounded to 0.6
Memory aid
Von Mises (the mathematician) once told AISC: 'Shear yields at 0.577 of the tensile yield stress.' AISC said 'Thank you, Mr. von Mises, but engineers like round numbers — we'll use 0.6.' So AISC rounded 0.577 up to 0.6 for convenience. Every time you write 0.6Fy in the shear formula, nod to von Mises — the man who gave us the theoretical basis, and AISC who simplified it to 0.6 for everyday use.
Why it works
The conversational story between historical figures makes the origin of the 0.6 coefficient memorable rather than arbitrary.
Example usage
Understanding why 0.6 appears in Vn = 0.6 Fy Aw Cv helps you never forget it — it's the shear yield stress approximation, not an arbitrary safety factor.
Recall trigger
Von Mises says 0.577, AISC rounds to 0.6 → 0.6Fy in shear formula
Full summary of design checks: compact? → braced? (Lb vs Lp) → apply φbMn or reduce for LTB/LB
Memory aid
Walk through your BAHAY (house) to check a beam: (1) GATE: Is the section COMPACT? (Check λ for flange and web — 'knock on the gate: compact or not?'). (2) DOOR: Is Lb ≤ Lp? ('open the door to full Mp'). (3) SALA/LIVING ROOM: Is Lp < Lb ≤ Lr? ('you're inside but dimming the lights — inelastic LTB'). (4) BASEMENT: Is Lb > Lr? ('deep underground — elastic LTB, use Fcr × Sx'). (5) KITCHEN: Check SHEAR — Vn = 0.6 Fy Aw Cv ('cook the shear strength').
Why it works
The method of loci (memory palace) using the familiar Filipino bahay creates distinct spatial triggers for each design check in sequence.
Example usage
In any board exam flexural design problem: follow the bahay path in order. Never skip the gate (compactness) or the door (Lp check) before computing Mn.
Recall trigger
Walk through your bahay: Gate (compact?) → Door (Lp?) → Sala (inelastic LTB?) → Basement (elastic LTB?) → Kitchen (shear?)
Revision game
I am larger than my elastic twin, I come from full plasticity, and without me you cannot compute Mp. Who am I?
Answer
Zx — the plastic section modulus
Memory link
A5 — The ambitious twin ZX who earns 12% more than SX
I am the distance beyond which a beam starts to lose its full bending strength. Stay within me and you get the full fizzy Mp. What am I?
Answer
Lp — the limiting unbraced length for full plastic moment
Memory link
A3 — Basketball player staying in the zone / 1.76 ry √(E/Fy)
I am a resistance factor, but unlike my bending sibling (0.90), I am perfect — I equal 1.0. Engineers say I am BUO. What am I for?
Answer
φv = 1.0 — the shear resistance factor for typical rolled I-shapes
Memory link
A8 — Mang Romy says BUO — φv = 1.0
I am the web area for shear, but beware — I use the FULL door height, not just the inner panel. Compute me correctly or lose the board exam. What is my formula?
Answer
Aw = d × tw (using full depth d, not clear web height)
Memory link
A7 — The DOOR analogy — full door height d times thickness tw
I am the famous German mathematician who told AISC that shear yielding occurs at 57.7% of tensile yield, though AISC rounds me to 60%. Who am I?
Answer
von Mises — origin of the 0.6 factor in Vn = 0.6 Fy Aw Cv
Memory link
A19 — Von Mises tells AISC 0.577, AISC rounds to 0.6
I am the zone of lateral-torsional buckling where Mn drops linearly — like a dimmer switch being slowly turned down from full brightness to 70% of a reference value. What zone am I?
Answer
Inelastic LTB zone — Zone 2 (Lp < Lb ≤ Lr), Mn interpolates from Mp to 0.7FySx
Memory link
A12 — The DIMMER SWITCH analogy
I am the Filipino grandmother law that empowers the PRC and makes NSCP and AISC provisions enforceable in Philippine professional practice. What is my name and number?
Answer
RA 544 — Republic Act 544, the Civil Engineering Law of the Philippines
Memory link
A17 — RA 544 is the grandmother law that empowers NSCP to adopt AISC
I am a moment-gradient factor greater than 1.0 when bending moment varies along the beam. I give a strength bonus — but even I cannot push the answer above Mp. What am I?
Answer
Cb — the moment gradient factor; increases Mn but is capped at Mp
Memory link
A11 — The CHEERLEADER BONUS capped by the salary cap of Mp
Formula mnemonics
Mn = Mp = Fy × Zx
Mnemonic
FiZz gives you Mp — Fy times Zx = the full fizzy plastic moment
When to use
ONLY when section is compact (λ < λp for both flange and web) AND Lb ≤ Lp (compression flange braced within the plastic limit distance).
What each part means
Fy = yield stress (MPa), Zx = plastic section modulus (mm³), Mp = plastic moment (N·mm or kN·m). This is the MAXIMUM nominal flexural strength for a compact, fully braced section.
φbMn = 0.90 × Fy × Zx
Mnemonic
BENDING is NINETY percent sure — φb = 0.90 always for flexure (LRFD)
When to use
All LRFD flexural design checks. Multiply Mn by 0.90 to get design strength before comparing with Mu.
What each part means
φb = 0.90 (LRFD flexural resistance factor), Mn = nominal flexural strength = Mp for compact braced beams. The design strength φbMn must ≥ Mu (factored moment demand).
Lp = 1.76 ry √(E/Fy)
Mnemonic
Seventy-six ry roots E on Fy — the basketball player stays in the zone at 1.76
When to use
Use to determine if beam is in Zone 1 (no LTB). If actual Lb ≤ Lp, use Mn = Mp without any LTB reduction.
What each part means
Lp = limiting unbraced length for full plastic moment (mm), ry = radius of gyration about weak axis (mm), E = elastic modulus = 200,000 MPa, Fy = yield stress (MPa). √(E/Fy) ≈ 28.4 for Fy = 248 MPa.
Vn = 0.6 Fy Aw Cv
Mnemonic
0.6 × FACT (Fy, Aw, Cv) = Vn — shear is a FACT of 0.6
When to use
For shear design of all I-shaped beams. Always verify Cv = 1.0 condition (h/tw ≤ 2.24√(E/Fy)) before assuming Cv = 1.0.
What each part means
Vn = nominal shear strength (N), 0.6 = shear yield factor (from von Mises criterion), Fy = yield stress (MPa), Aw = d × tw = web area using full depth (mm²), Cv = web shear coefficient (= 1.0 for stocky webs).
φvVn = 1.0 × 0.6 Fy Aw Cv
Mnemonic
Shear is BUO (whole) — φv = 1.0 for typical rolled I-shapes, unlike flexure's 0.90
When to use
For LRFD shear design of standard rolled W-sections with stocky webs. Design requirement: φvVn ≥ Vu (factored shear).
What each part means
φv = 1.0 (shear resistance factor for most rolled I-shapes with compact webs, per AISC 360), Vn = nominal shear strength. For beams where h/tw > 2.24√(E/Fy), φv reverts to 0.90.
Aw = d × tw
Mnemonic
FULL DOOR height × door thickness — d is the FULL depth, not just clear web height
When to use
Every time computing Vn. Never substitute (d - 2tf) for d in this formula for rolled I-shapes per AISC 360.
What each part means
Aw = web shear area (mm²), d = overall beam depth including flanges (mm), tw = web thickness (mm). AISC uses full depth d in this formula — a common board exam trap.
Cv = 1.0 when h/tw ≤ 2.24√(E/Fy)
Mnemonic
2-and-a-quarter root check: pass the check → Cv = 1.0 (full shear), fail → Cv < 1.0 (shear buckles)
When to use
Check web slenderness before assuming Cv = 1.0 in Vn formula. Most standard rolled W-sections pass this check.
What each part means
Cv = shear buckling coefficient, h = clear distance between flanges (mm), tw = web thickness (mm), E = 200,000 MPa, Fy = yield stress (MPa). For Fy = 248 MPa: limit = 2.24 × 28.4 ≈ 63.6.
Quick recall chains
Three LTB Zones in Order
Recall test
Without looking: What is Mn when Lb = Lp? What formula governs when Lb > Lr? What happens linearly between Lp and Lr?
Memory chain
Take the JEEPNEY TRIP — CITY (full speed, Mn=Mp) → TAGAYTAY WINDING ROAD (slow down gradually, linear reduction) → MOUNTAIN TRAIL (very slow, elastic only, Fcr × Sx). Each zone = one leg of the trip. No detours.
Items to remember
- Zone 1: Lb ≤ Lp → No LTB → Mn = Mp
- Zone 2: Lp < Lb ≤ Lr → Inelastic LTB → Mn interpolates linearly between Mp and 0.7FySx
- Zone 3: Lb > Lr → Elastic LTB → Mn = Fcr × Sx
Beam Design Check Sequence (Bahay Method)
Recall test
Name all 6 steps of beam design in order without referring to notes. What do you check at the gate? What happens in the kitchen?
Memory chain
Walk through the BAHAY: (Gate) Compact check → (Door) Lb vs Lp → (Sala) Zone 2 check → (Bedroom) Zone 3? → (Sala Table) Apply φb → (Kitchen) Cook the shear. Enter through gate, exit through kitchen.
Items to remember
- Step 1: Check compactness (λ < λp for flange and web)
- Step 2: Determine unbraced length Lb
- Step 3: Compare Lb with Lp and Lr
- Step 4: Compute Mn based on appropriate zone
- Step 5: Apply φb = 0.90 to get design strength
- Step 6: Check shear — Vn = 0.6 Fy Aw Cv, φv = 1.0
Shear Strength Computation Steps
Recall test
Given d = 500 mm, tw = 9 mm, Fy = 248 MPa, Cv = 1.0: compute φvVn step by step using the FACT chain.
Memory chain
The FACT CHAIN: Find the door (d, tw) → Area of door (Aw = d×tw) → Check the stocky wall (h/tw for Cv) → FACT formula (0.6 Fy Aw Cv = Vn) → BUO it (φv = 1.0). DOOR → AREA → CHECK → FACT → BUO.
Items to remember
- Step 1: Identify d (full depth) and tw (web thickness)
- Step 2: Compute Aw = d × tw
- Step 3: Check h/tw ≤ 2.24√(E/Fy) → determine Cv
- Step 4: Compute Vn = 0.6 Fy Aw Cv
- Step 5: Apply φv = 1.0 (typical rolled I) → φvVn
Key Constants to Memorize
Recall test
Without notes: What are the two common Fy values? What is φb? What is φv for rolled I-shear? What is √(E/Fy) for Fy = 248 MPa?
Memory chain
EVERY FINE STEEL FRAME STANDS STRONG TOGETHER: E-200k, Fy248-or-345, Sb-ninety, Sv-buo, Sf-twelve-percent, S0.6-vonMises. Like a steel frame standing together — each constant holds up the structure.
Items to remember
- E = 200,000 MPa (steel modulus of elasticity)
- Fy = 248 MPa (A36 steel, common in Philippine practice)
- Fy = 345 MPa (A572 Gr. 50 steel)
- φb = 0.90 (flexure LRFD factor)
- φv = 1.0 (shear LRFD factor, rolled I)
- Shape factor ≈ 1.12 (Zx/Sx for W-shapes)
- 0.6 × Fy = shear yield stress approximation
Lp Formula Components
Recall test
Write Lp formula from memory. Compute Lp for ry = 50 mm, E = 200,000 MPa, Fy = 345 MPa.
Memory chain
ONE-SEVENTY-SIX (1.76) × THE WOBBLING FLAGPOLE (ry) × THE STIFFNESS ROOT (√E/Fy) = YOUR SAFE ZONE LENGTH (Lp). Picture 176 flagpoles rooted into the ground — the zone where they stand without buckling.
Items to remember
- Coefficient: 1.76
- Geometric property: ry (weak-axis radius of gyration)
- Material ratio: √(E/Fy)
- Result: maximum unbraced length for full Mp
Diagrams
Steel Beam Flexure and Shear — Complete Concept Map
A comprehensive mind map showing all major concepts of steel beam design under flexure and shear. Branches cover flexural strength (plastic moment, LTB zones, Lp formula), shear strength (Vn formula, web area, Cv), common pitfalls, and code references. Use this as a study overview before drilling individual topics.
Steel Beam Design Decision Flowchart — Flexure and Shear
A comprehensive decision flowchart for steel beam design covering both flexure and shear checks. Follow the path from compactness verification through LTB zone identification, flexural resistance factor application, and shear strength computation. This flowchart operationalizes the 'bahay memory palace' sequence into a precise engineering decision tree.
LTB Zone State Transitions as Lb Increases
A state diagram showing how the governing limit state transitions as the unbraced length Lb increases from zero. Each 'state' corresponds to one LTB zone, and the transitions are triggered by Lb exceeding Lp or Lr. This reinforces the jeepney trip analogy — each state change is a new leg of the journey.
Shear Strength Computation — Step by Step
A step-by-step flowchart for computing design shear strength of a steel I-beam web. Highlights the critical distinction between stocky and slender webs (Cv and φv), and reinforces that Aw uses the full depth d. Pairs with the DOOR analogy (A7) and BUO analogy (A8) for maximum retention.