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Memory AnchorsCELE · Steel & Timber DesignReal content

CELE Steel & Timber DesignSteel ConnectionsMemory Anchors

If you keep missing Steel Connections items on your CELE mocks despite having read the notes, the gap is usually recall speed. Memory anchors close that gap. These Steel Connections mnemonics have been tuned to the kinds of triggers Professional Regulation Commission (PRC) — Board of Civil Engineering builds into CELE Steel & Timber Design questions.

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 Connections in the 4th 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 Connections - Memory Anchors

Memory techniques can increase long-term retention by up to 400% compared to passive re-reading. When you encode a formula as a vivid story, a strange analogy, or a catchy rhyme, your brain creates multiple retrieval pathways — so even under exam pressure, at least one pathway fires. This set of 18 memory anchors covers every key concept in Steel Connections: bolt shear, bearing, block shear, fillet welds, and the critical phi = 0.75 limit state. Use these anchors the night before the board exam and during timed mock drills. The more absurd and sensory-rich the image, the longer it sticks.

Anchors

Tags

  • phi factor
  • LRFD
  • definition
  • formula

Topic

General LRFD Connection Philosophy

Concept

φ = 0.75 for ALL connection limit states (bolt shear, bearing, block shear, welds)

Anchor Id

A1

Difficulty

easy

Memory Aid

Think of a BASKETBALL COURT: Three-quarters (¾) of the court is safe territory for connections. The remaining quarter is your safety margin. Whenever you see a bolt, a weld, or a bearing plate, shout internally: 'THREE-QUARTERS!' φ = 0.75 = ¾.

Anchor Type

mnemonic

Why It Works

The fraction ¾ is universally recognizable. Associating it with a basketball court (hugely popular in the Philippines) creates a vivid, culturally relevant mental image.

Example Usage

Board question asks for design bearing strength. Before computing: 'Basketball court — φ = 0.75.' Multiply your nominal Rn by 0.75.

Recall Trigger

Basketball + connections → ¾ court → φ = 0.75

Tags

  • formula
  • bolt shear
  • acronym

Topic

Bolt Shear Strength

Concept

Bolt Shear Formula: φRn = 0.75 × Fnv × Ab

Anchor Id

A2

Difficulty

easy

Memory Aid

Remember 'FAB': F(nv) × A(b) gives the bolt shear, then multiply by the basketball factor 0.75. FAB = Fnv × Ab. The bolt is FABulous — it resists shear with its full cross-section area and stress rating.

Anchor Type

acronym

Why It Works

FAB is a common Filipino slang word for 'fabulous,' creating an emotional hook. The acronym directly maps to the formula variables.

Example Usage

20 mm A325-N bolt: 'FAB' → Fnv = 372 MPa, Ab = π/4(20²) = 314.2 mm². φRn = 0.75 × 372 × 314.2 = 87.7 kN.

Recall Trigger

Bolt shear → 'FAB' → Fnv × Ab × 0.75

Tags

  • double shear
  • bolt shear
  • analogy

Topic

Bolt Shear — Single vs. Double

Concept

Double shear doubles the bolt shear strength (two shear planes)

Anchor Id

A3

Difficulty

easy

Memory Aid

Imagine cutting a hotdog with one knife vs. two knives attacking from both sides simultaneously. One knife (single shear) takes the full load. Two knives (double shear) share the work — each takes half, so the hotdog resists TWICE as much total force before it breaks. Count your shear PLANES like you count your knife cuts.

Anchor Type

analogy

Why It Works

Hotdog (longganisa) is a familiar Filipino food. The tactile image of knives cutting from both sides is vivid and easy to recall under pressure.

Example Usage

A bolt in double shear: φRn (double) = 2 × 0.75 × Fnv × Ab. If single-shear strength = 87.7 kN, double-shear = 175.4 kN.

Recall Trigger

Double shear → two knives on a hotdog → multiply bolt shear by 2

Tags

  • A325
  • bolt type
  • Fnv
  • classification

Topic

Bolt Types and Nominal Shear Stress

Concept

A325-N vs A325-X: threads IN vs OUT of the shear plane, Fnv = 372 vs 469 MPa

Anchor Id

A4

Difficulty

medium

Memory Aid

STORY: Pedro (A325-N) forgot to comb his hair before the board exam — his 'threads' were in his face (IN the shear plane). He scored only 372. His classmate Xander (A325-X) was clean-cut — threads EXCLUDED, combed back neatly. He scored 469. N = Normal (threads in), lower score. X = eXcluded threads, higher score. N<X, 372<469.

Anchor Type

micro_story

Why It Works

Filipino board exam culture makes the test-score analogy instantly relatable. The names Pedro and Xander start with N and X respectively, reinforcing the variable names.

Example Usage

Problem says 'A325-N bolt' → Fnv = 372 MPa. Problem says 'A325-X bolt' → Fnv = 469 MPa.

Recall Trigger

N = threads in shear plane = 372 MPa; X = threads excluded = 469 MPa

Tags

  • bearing
  • formula
  • min function

Topic

Bearing Strength at Bolt Holes

Concept

Bearing strength formula: φRn = 0.75 × min(1.2 lc t Fu, 2.4 db t Fu)

Anchor Id

A5

Difficulty

medium

Memory Aid

The SMALLER TWIN WINS. Imagine twins: Twin 1.2 (using lc) and Twin 2.4 (using db). They fight over who gets to be the bearing strength. The SMALLER twin wins (governs). The '2.4 twin' is always the CAP — it stops Twin 1.2 from going too high. Remember: '1.2 uses lc, 2.4 uses db, MINIMUM governs.'

Anchor Type

mnemonic

Why It Works

The twin competition metaphor makes the min() function memorable. The labeling of each twin with its coefficient directly maps to the formula.

Example Usage

lc = 35 mm, db = 20 mm, t = 10 mm, Fu = 400 MPa: Twin 1.2 = 1.2(35)(10)(400) = 168 kN. Twin 2.4 = 2.4(20)(10)(400) = 192 kN. Smaller = 168 kN. φRn = 0.75(168) = 126 kN.

Recall Trigger

Bearing → SMALLER TWIN WINS → min(1.2 lc t Fu, 2.4 db t Fu)

Tags

  • bearing
  • rhyme
  • formula components

Topic

Bearing Strength at Bolt Holes

Concept

In bearing formula: 1.2 uses lc (clear distance); 2.4 uses db (bolt diameter)

Anchor Id

A6

Difficulty

medium

Memory Aid

'ONE-POINT-TWO times the CLEAR, TWO-POINT-FOUR times the DIAMETER — pick the one that's smaller.' Sing it to the tune of a simple counting song. 1.2 → lc (Clear distance), 2.4 → db (Diameter of bolt). The rhyme 'clear' and 'diameter' lock the pairing.

Anchor Type

rhyme

Why It Works

Rhyme and rhythm activate the brain's auditory-sequential memory, distinct from rote verbal memory. The tune makes it reproducible during silent exam conditions.

Example Usage

Immediately before substituting values: recite the rhyme quietly → confirm: 1.2 × lc × t × Fu, and 2.4 × db × t × Fu, take min.

Recall Trigger

'One-point-two times the clear' → start the rhyme, recall both terms

Tags

  • weld
  • throat
  • formula
  • geometry

Topic

Fillet Weld Geometry

Concept

Fillet weld throat = 0.707 × a (leg size); use throat NOT leg in strength formula

Anchor Id

A7

Difficulty

easy

Memory Aid

Picture a right triangle — the weld cross-section. The two legs (equal length 'a') are the visible weld size. The THROAT is the hypotenuse divided by √2 = 0.707. Visualize a right-triangle sandwich: the bread slices are the legs, but you chew through the DIAGONAL filling (the throat). You always bite through 0.707a, never through just 'a'.

Anchor Type

visual_association

Why It Works

The right-triangle geometry is universally recognized from trigonometry. The sandwich analogy adds a sensory (taste/bite) dimension, making it multi-modal.

Example Usage

6 mm fillet weld: throat = 0.707(6) = 4.24 mm. Use 4.24 mm in the formula, NOT 6 mm.

Recall Trigger

Fillet weld → right-triangle sandwich → bite through 0.707a (throat, not leg)

Tags

  • weld
  • formula
  • fillet
  • sequence

Topic

Fillet Weld Strength

Concept

Fillet weld design strength: φRn = 0.75 × 0.60 × FEXX × 0.707a × L

Anchor Id

A8

Difficulty

medium

Memory Aid

THE WELD CHANT: '75-60-FEXX-707-L.' Chant it like a basketball play call: '75! 60! FEX! 707! L!' Each number is a factor in sequence: φ=0.75, shear stress fraction=0.60, electrode strength FEXX, throat factor=0.707, length L. Five factors, five chants.

Anchor Type

mnemonic

Why It Works

Sequential chanting mimics how athletes memorize plays. The five distinct numbers create a chunked sequence that is easy to reproduce under stress.

Example Usage

E70 weld, a=6mm, L=200mm: φRn = 0.75 × 0.60 × 482 × 0.707(6) × 200 = 184 kN.

Recall Trigger

'75-60-FEXX-707-L' chant → write out the fillet weld formula

Tags

  • weld
  • electrode
  • FEXX
  • chunking

Topic

Electrode Strength

Concept

E70 electrode: FEXX = 482 MPa (≈ 70 ksi)

Anchor Id

A9

Difficulty

easy

Memory Aid

E70 → '70 ksi' → convert: 70 × 6.895 ≈ 482 MPa. Remember 'E70 = 482' using chunking: 4-8-2. Think: 4 floors, 8 rooms, 2 exits. Or simpler: '482 = 4 times 120 + 2' is not easy, so use the PIN: 482. Drill it as a PIN number. Every time you think 'E70,' mentally type your PIN: 4-8-2.

Anchor Type

chunking

Why It Works

PIN/passcode memory is one of the most over-trained memory circuits in modern life. Hijacking that circuit for FEXX = 482 makes recall automatic.

Example Usage

Problem gives E70 electrodes → automatically write FEXX = 482 MPa before computing weld strength.

Recall Trigger

E70 electrode → PIN 4-8-2 → FEXX = 482 MPa

Tags

  • block shear
  • formula
  • limit state
  • rupture

Topic

Block Shear

Concept

Block shear formula: Rn = 0.6Fu×Anv + Ubs×Fu×Ant ≤ 0.6Fy×Agv + Ubs×Fu×Ant

Anchor Id

A10

Difficulty

hard

Memory Aid

STORY: A pizza slice (the block) tries to escape the box (the connection). It can fail in TWO ways: (1) RUPTURE PATH — the edges (shear) and front (tension) both tear at Fu (ultimate strength, brutal failure), capped by (2) YIELD PATH — the edges yield first at Fy (milder failure). The pizza slice takes the SMALLER of the two escape routes. Shear edges = Anv or Agv. Tension front = Ant. Ubs = topping uniformity (1.0 if evenly loaded).

Anchor Type

micro_story

Why It Works

Pizza is universally loved in Filipino college culture. The physical image of a slice escaping a box maps to the block shear tear-out path geometry.

Example Usage

Compute both: Rupture = 0.6Fu×Anv + Ubs×Fu×Ant; Yield = 0.6Fy×Agv + Ubs×Fu×Ant. φRn = 0.75 × (whichever is smaller).

Recall Trigger

Block shear → pizza slice escaping → rupture path (Fu) ≤ yield path (Fy), pick smaller

Tags

  • block shear
  • Ubs
  • classification

Topic

Block Shear — Ubs Coefficient

Concept

Ubs = 1.0 for uniform tension; 0.5 for non-uniform tension in block shear

Anchor Id

A11

Difficulty

medium

Memory Aid

Ubs is like a GRADING FACTOR for how evenly distributed the tension load is. If all students in a group (bolts) carry equal tension — UNIFORM — everyone gets full credit: Ubs = 1.0 (100%). If some carry more and some less — NON-UNIFORM — the group gets partial credit: Ubs = 0.5 (50%). In most board problems (single row of bolts, coped beams), tension is uniform: Ubs = 1.0.

Anchor Type

analogy

Why It Works

Filipino students relate strongly to group project grading dynamics. The grading analogy makes the abstract coefficient immediately intuitive.

Example Usage

Standard bracket with one bolt row → Ubs = 1.0. Problem states non-uniform tension → Ubs = 0.5.

Recall Trigger

Block shear Ubs → group grade → uniform = 1.0, non-uniform = 0.5

Tags

  • limit state
  • governing
  • sequence
  • process

Topic

Governing Limit State

Concept

The governing connection strength is the MINIMUM across all limit states

Anchor Id

A12

Difficulty

medium

Memory Aid

The connection is like a CHAIN with 5 links: (1) Bolt Shear, (2) Bearing, (3) Block Shear, (4) Net-Section Rupture, (5) Weld Strength. A chain ALWAYS breaks at its WEAKEST link. No matter how strong the other four are, the weakest one controls. The design strength = the smallest computed value across all applicable limit states.

Anchor Type

analogy

Why It Works

The 'weakest link' metaphor is universally known. It maps perfectly to the engineering concept of governing limit state, making it both intuitive and memorable.

Example Usage

After computing bolt shear = 350 kN, bearing = 280 kN, block shear = 320 kN — the connection capacity is 280 kN (bearing governs, weakest link).

Recall Trigger

Connection strength → chain → weakest link = governing limit state = minimum value

Tags

  • slip-critical
  • friction
  • bolt tension
  • definition

Topic

Slip-Critical Connections

Concept

Slip-critical connections resist by FRICTION (no slip at service load)

Anchor Id

A13

Difficulty

medium

Memory Aid

Picture two rough SANDPAPER SHEETS clamped tightly together by a pre-tensioned bolt. They don't slip because friction holds them — not the bolt shank in bearing. Slip-critical = SANDPAPER FRICTION CONNECTION. Used where: (a) FATIGUE loads (vibrating machines — like generators at a power plant), or (b) OVERSIZED holes. As long as the clamping force (Tb) is maintained, no slip.

Anchor Type

visual_association

Why It Works

The tactile sensation of sandpaper is instantly imaginable. The clamping image directly maps to the pre-tension (Tb) in the slip-critical formula.

Example Usage

Board question: 'Connection is subject to fatigue — what type?' → Slip-critical. Formula uses friction coefficient μ, pre-tension Tb, number of slip planes ns.

Recall Trigger

Slip-critical → sandpaper clamped tight → friction = Rn = μ × Du × hf × Tb × ns

Tags

  • weld
  • groove
  • CJP
  • base metal

Topic

Groove Welds

Concept

Groove (full-penetration) welds develop FULL base-metal strength — no throat reduction

Anchor Id

A14

Difficulty

medium

Memory Aid

A fillet weld is like a MANGO HALF glued to a plate — the attachment area is only at the surface (throat). A full-penetration groove weld is like two pieces of metal FUSED COMPLETELY into one — they become one material. You cannot tell where the weld is. Strength = base metal strength, no separate weld check needed for CJP groove welds.

Anchor Type

analogy

Why It Works

The mango analogy (Filipino fruit) vs. total fusion creates a memorable contrast. The fused-metal concept matches the engineering reality of CJP groove welds.

Example Usage

Problem specifies CJP groove weld: design strength = φ × Fy × Ag (base metal governs, not weld).

Recall Trigger

CJP groove weld → fully fused metal → use base-metal strength, no throat reduction

Tags

  • weld
  • throat
  • common mistake
  • pitfall

Topic

Common Pitfalls — Weld Throat

Concept

Common pitfall: using weld leg 'a' instead of throat '0.707a' in weld formula

Anchor Id

A15

Difficulty

easy

Memory Aid

STORY: Engr. Reyes failed his first mock board because he used '6 mm' directly in the weld formula instead of '0.707 × 6 = 4.24 mm.' He thought: 'the weld IS 6 mm, why reduce it?' His professor told him: 'The weld fails diagonally through the THROAT, not through the fat leg. Always slice diagonally — use 0.707a.' Now every time Reyes sees a fillet weld size, he automatically multiplies by 0.707 first.

Anchor Type

micro_story

Why It Works

A story of a student making a mistake and learning from it is extremely relatable to board reviewees. Narrative encoding is among the most powerful for error-avoidance.

Example Usage

Given: 8 mm fillet weld → throat = 0.707(8) = 5.66 mm. Use 5.66 in φRn = 0.75(0.60)(482)(5.66)(L).

Recall Trigger

Fillet weld problem → remember Engr. Reyes → multiply leg by 0.707 FIRST

Tags

  • bolt shear
  • shear planes
  • visual
  • process

Topic

Shear Plane Counting

Concept

Single vs. double shear: always COUNT the shear planes on the bolt

Anchor Id

A16

Difficulty

easy

Memory Aid

VISUALLY SCAN the connection: How many plates are trying to slide PAST the bolt? In a lap joint (2 plates, 1 interface) → SINGLE shear (1 plane). In a double-lap or clevis (3 plates, 2 interfaces) → DOUBLE shear (2 planes). Count the CUTS. Draw an imaginary knife line at every plane where failure would occur. Number of knife lines = number of shear planes.

Anchor Type

visual_association

Why It Works

The knife-line visualization is a concrete, teachable technique. Counting is a reliable low-cognitive-load task, even under exam stress.

Example Usage

Sketch shows bolt through 3 plates (outer-inner-outer): 2 knife lines → double shear → φRn = 2 × 0.75 × Fnv × Ab.

Recall Trigger

Bolt shear → draw knife lines → count planes → multiply bolt shear capacity accordingly

Tags

  • net area
  • hole deduction
  • block shear
  • formula

Topic

Net Area Computation

Concept

Net area Ant and Anv in block shear: subtract hole area (db + 2 mm for standard hole)

Anchor Id

A17

Difficulty

medium

Memory Aid

HOLE DEDUCTION RULE: 'ADD 2, THEN SUBTRACT.' Standard hole diameter = bolt diameter + 2 mm (for punching tolerance). When computing net areas for block shear (or net section): hole diameter to deduct = db + 2 mm. Mnemonic: 'The HOLE is ALWAYS 2 mm WIDER than the bolt that fills it — it's like buying pants one size up.'

Anchor Type

mnemonic

Why It Works

The pants/clothing analogy is universally relatable. The 'add 2' rule is easily reproduced because it is linked to a familiar overcorrection habit.

Example Usage

20 mm bolt, standard hole → hole for area deduction = 22 mm. Net width = gross width − 22 mm × (number of holes in path).

Recall Trigger

'ADD 2 THEN SUBTRACT' → hole diameter for deduction = bolt diameter + 2 mm

Tags

  • limit state
  • sequence
  • acronym
  • process

Topic

Systematic Connection Design

Concept

Five connection limit states to check in order (systematic approach)

Anchor Id

A18

Difficulty

medium

Memory Aid

Use the acronym 'BBBWN' — 'Bolts Bring Blocks, Welds, and Net-sections.' B = Bolt Shear, B = Bearing, B = Block Shear, W = Weld Strength (if welded), N = Net-section Rupture. Check ALL that apply, then take the MINIMUM. 'BBBWN sounds like a radio station — tune into ALL channels before broadcasting the final capacity.'

Anchor Type

acronym

Why It Works

Acronyms reduce multi-item lists to a single retrievable code. The radio-station metaphor adds a behavioral instruction: do not broadcast (report capacity) until all channels are checked.

Example Usage

Start board solution: List BBBWN. Compute each applicable one. φRn (connection) = min of all computed values.

Recall Trigger

New connection problem → tune to BBBWN radio → check all five limit states

Revision Game

φ = 0.75 (LRFD resistance factor for connections)

Clue

I am the Greek letter that equals three-quarters. I apply to EVERY connection limit state in LRFD. Without me, you overestimate strength. Who am I?

Memory Link

A1 — Basketball court, ¾ territory

The weld throat = 0.707 × a (leg size)

Clue

I am the part of a fillet weld that actually fails in shear. I am shorter than the leg by a factor of 0.707. Engineers who forget me overestimate weld strength. What am I?

Memory Link

A7 — Right-triangle sandwich, bite through the diagonal

A325-X has higher Fnv = 469 MPa; A325-N = 372 MPa

Clue

A325-N says 'my threads are IN the shear plane.' A325-X says 'my threads are OUT.' Which has higher Fnv, and what are the exact values in MPa?

Memory Link

A4 — Pedro (N=372) vs Xander (X=469) board exam scores

Ubs = 1.0 (uniform tension; non-uniform = 0.5)

Clue

In block shear, I am called when tension stress is UNIFORM across the tension area. My value is 1.0. When tension is non-uniform, my twin is used at 0.5. What coefficient am I?

Memory Link

A11 — Group project grading: uniform = full credit 1.0

BBBWN: Bolt Shear, Bearing, Block Shear, Weld, Net-Section Rupture

Clue

I am an acronym for a radio station. I list all five connection limit states you must check before reporting a final design capacity. What do my letters stand for?

Memory Link

A18 — BBBWN Radio Station: tune into all channels before broadcasting

The MINIMUM of the two governs (Smaller Twin Wins)

Clue

In the bearing formula, there are two expressions. One uses clear distance lc with coefficient 1.2; the other uses bolt diameter db with coefficient 2.4. Both are multiplied by t and Fu. Which one governs?

Memory Link

A5 — Smaller Twin Wins: min(1.2 lc t Fu, 2.4 db t Fu)

Two shear planes (double shear): multiply single-shear bolt capacity by 2

Clue

A bolt passes through three plates: an outer plate, a middle plate, and another outer plate. The middle plate is the connected member. How many shear planes are there, and how does this affect bolt shear capacity?

Memory Link

A3 — Two knives on a hotdog: double shear doubles resistance

FEXX = 482 MPa (PIN: 4-8-2)

Clue

I am the electrode classification strength for E70 electrodes in SI units. State my value in MPa. (Hint: I am also a memorable PIN code.)

Memory Link

A9 — E70 PIN code: 4-8-2 → 482 MPa

Formula Mnemonics

Formula

φRn = 0.75 × Fnv × Ab (bolt shear, single plane)

Mnemonic

FAB × 0.75: F(nv) × A(b) = FABulous bolt strength, then take three-quarters.

When To Use

Any bolt in direct shear (lap joints, bracing connections). Multiply by number of shear planes (1 or 2) and by number of bolts in group.

What Each Part Means

φ = 0.75 (LRFD resistance factor for connections); Fnv = nominal shear stress of bolt (A325-N: 372 MPa; A325-X: 469 MPa); Ab = nominal bolt cross-sectional area = π/4 × db²

Formula

φRn = 0.75 × min(1.2 lc t Fu, 2.4 db t Fu) — bearing per bolt

Mnemonic

SMALLER TWIN WINS: Twin 1.2 (lc) vs. Twin 2.4 (db) — minimum governs, both multiplied by t × Fu, then by 0.75.

When To Use

Every bolt in a bearing-type connection. Check each bolt separately if clear distances vary (edge bolt vs. interior bolt). Use 1.5 lc t Fu and 3.0 db t Fu only when hole deformation is NOT a design consideration (rare in board problems).

What Each Part Means

lc = clear distance in direction of force (edge distance to first hole, or hole-to-hole clear distance); t = thickness of plate being checked; Fu = ultimate tensile strength of the plate; db = nominal bolt diameter

Formula

φRn = 0.75 × (0.60 × FEXX) × (0.707a) × L — fillet weld total strength

Mnemonic

CHANT: '75-60-FEX-707-L' → multiply in order: 0.75 × 0.60 × FEXX × 0.707a × L

When To Use

All fillet weld strength checks. For welds on both sides of a plate (two lines), multiply L by 2. Per-unit-length strength = 0.75 × 0.60 × FEXX × 0.707a (N/mm).

What Each Part Means

0.75 = φ (connection); 0.60 = fraction of electrode strength for shear on weld throat (AISC J2.4); FEXX = electrode classification strength (E70 = 482 MPa); 0.707a = weld throat (a = leg size); L = effective weld length

Formula

Rn (block shear) = 0.6Fu×Anv + Ubs×Fu×Ant ≤ 0.6Fy×Agv + Ubs×Fu×Ant

Mnemonic

PIZZA ESCAPE: Rupture path (Fu both) ≤ Yield path (Fy on shear side). Shear × 0.6, Tension × Ubs, pick the LESSER of the two caps, then φ = 0.75.

When To Use

Angles bolted one leg, coped beams, gusset plates — any connection where a block of material can tear out. Always check alongside bolt shear and bearing.

What Each Part Means

Anv = net shear area (along shear failure planes); Agv = gross shear area; Ant = net tension area (perpendicular to load); Ubs = 1.0 uniform, 0.5 non-uniform; Fu = ultimate strength; Fy = yield strength of plate

Formula

Rn (slip-critical) = μ × Du × hf × Tb × ns

Mnemonic

MUDTHINS: μ (friction) × Du (ratio ~1.13) × hf (hole factor ≤1.0) × Tb (pre-tension) × ns (slip planes) = MUDTHINS resistance.

When To Use

Fatigue-loaded connections, connections with oversized or short-slotted holes perpendicular to load direction, or where slip would be functionally unacceptable.

What Each Part Means

μ = mean slip coefficient (Class A=0.35, Class B=0.50); Du = 1.13 (ratio of mean installed bolt pretension to specified); hf = factor for fillers and shims (≤1.0); Tb = minimum bolt pre-tension (tabulated, e.g., 142 kN for 20mm A325); ns = number of slip planes

Quick Recall Chains

Chain Title

BBBWN — Five Connection Limit States in Order

Recall Test

Without looking: name all 5 limit states using BBBWN, then write the formula for each from memory.

Memory Chain

Tune into BBBWN Radio Station: 'Bolts Bring Blocks, Welds, and Net-sections.' Each letter is a limit state. Broadcast (report) the MINIMUM after checking all channels.

Items To Remember

  • Bolt Shear (φRn = 0.75 Fnv Ab per plane)
  • Bearing (φRn = 0.75 min(1.2 lc t Fu, 2.4 db t Fu))
  • Block Shear (φRn = 0.75 [0.6Fu Anv + Ubs Fu Ant])
  • Weld Strength (φRn = 0.75 × 0.60 FEXX × 0.707a × L)
  • Net-Section Rupture (φRn = 0.75 Fu Ae)

Chain Title

Fillet Weld Formula Build-Up — 75-60-FEX-707-L

Recall Test

Cover the formula. Say '75-60-FEX-707-L' aloud, then reconstruct: φRn = 0.75 × 0.60 × FEXX × 0.707a × L. Check if correct.

Memory Chain

Chant: '75-60-FEX-707-L!' five times. Picture a basketball play being called: the coach shouts '75! 60! FEX! 707! L!' Each number is a factor. Multiply them left to right to get φRn.

Items To Remember

  • 0.75 — phi factor
  • 0.60 — shear fraction of electrode strength
  • FEXX — electrode nominal strength (E70 = 482 MPa)
  • 0.707a — weld throat (a = leg size)
  • L — effective weld length

Chain Title

Bearing Formula Components — TWIN CHECK

Recall Test

Write both bearing expressions, identify which uses lc and which uses db, then state which governs and why.

Memory Chain

1.2-lc and 2.4-db are TWINS. 'lc' has 2 letters → lower coefficient (1.2). 'db' has 2 letters too, but the bolt DIAMETER is the bigger physical dimension → bigger coefficient (2.4). Both × t × Fu. Min wins. × 0.75.

Items To Remember

  • Coefficient 1.2 pairs with lc (clear distance)
  • Coefficient 2.4 pairs with db (bolt diameter)
  • Both multiplied by t (plate thickness) and Fu (plate ultimate)
  • Take the MINIMUM of the two products
  • Then multiply by φ = 0.75

Chain Title

Block Shear Step-by-Step Sequence

Recall Test

Without notes, write the 8 steps for block shear and identify which two expressions are compared to find Rn.

Memory Chain

PIZZA SLICE ESCAPE in 8 moves: SKETCH → SHEAR AREAS → TENSION AREA → UBS GRADE → RUPTURE PATH → YIELD PATH → LESSER WINS → MULTIPLY 0.75. Draw the pizza slice first, then compute systematically.

Items To Remember

  • Step 1: Identify the tear-out block geometry (sketch it)
  • Step 2: Compute Agv, Anv (shear areas gross and net)
  • Step 3: Compute Ant (net tension area, deduct bolt holes)
  • Step 4: Determine Ubs (1.0 or 0.5)
  • Step 5: Compute Rupture path = 0.6Fu Anv + Ubs Fu Ant
  • Step 6: Compute Yield path = 0.6Fy Agv + Ubs Fu Ant
  • Step 7: Rn = lesser of Rupture and Yield paths
  • Step 8: φRn = 0.75 × Rn

Chain Title

A325 Bolt Fnv Values

Recall Test

Which bolt has higher shear stress: A325-N or A325-X? What are the exact Fnv values? Why is N lower?

Memory Chain

Pedro (N) = 372 board score. Xander (X) = 469 board score. Pedro's threads (hair) were in his face, reducing his performance. Xander neatly excluded them. N = 372, X = 469. Always: X > N in both alphabet and strength.

Items To Remember

  • A325-N (threads in shear plane): Fnv = 372 MPa
  • A325-X (threads excluded from shear plane): Fnv = 469 MPa
  • N < X, 372 < 469 (N is lower, threads weaken the shear plane)
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