Steel Connections - Memory Anchors
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.
Memory anchors
φ = 0.75 for ALL connection limit states (bolt shear, bearing, block shear, welds)
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.
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
Bolt Shear Formula: φRn = 0.75 × Fnv × Ab
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.
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
Double shear doubles the bolt shear strength (two shear planes)
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.
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
A325-N vs A325-X: threads IN vs OUT of the shear plane, Fnv = 372 vs 469 MPa
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.
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
Bearing strength formula: φRn = 0.75 × min(1.2 lc t Fu, 2.4 db t Fu)
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.'
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)
In bearing formula: 1.2 uses lc (clear distance); 2.4 uses db (bolt diameter)
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.
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
Fillet weld throat = 0.707 × a (leg size); use throat NOT leg in strength formula
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'.
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)
Fillet weld design strength: φRn = 0.75 × 0.60 × FEXX × 0.707a × L
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.
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
E70 electrode: FEXX = 482 MPa (≈ 70 ksi)
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.
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
Block shear formula: Rn = 0.6Fu×Anv + Ubs×Fu×Ant ≤ 0.6Fy×Agv + Ubs×Fu×Ant
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).
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
Ubs = 1.0 for uniform tension; 0.5 for non-uniform tension in block shear
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.
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
The governing connection strength is the MINIMUM across all limit states
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.
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
Slip-critical connections resist by FRICTION (no slip at service load)
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.
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
Groove (full-penetration) welds develop FULL base-metal strength — no throat reduction
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.
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
Common pitfall: using weld leg 'a' instead of throat '0.707a' in weld formula
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.
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
Single vs. double shear: always COUNT the shear planes on the bolt
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.
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
Net area Ant and Anv in block shear: subtract hole area (db + 2 mm for standard hole)
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.'
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
Five connection limit states to check in order (systematic approach)
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.'
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
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?
Answer
φ = 0.75 (LRFD resistance factor for connections)
Memory link
A1 — Basketball court, ¾ territory
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?
Answer
The weld throat = 0.707 × a (leg size)
Memory link
A7 — Right-triangle sandwich, bite through the diagonal
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?
Answer
A325-X has higher Fnv = 469 MPa; A325-N = 372 MPa
Memory link
A4 — Pedro (N=372) vs Xander (X=469) board exam scores
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?
Answer
Ubs = 1.0 (uniform tension; non-uniform = 0.5)
Memory link
A11 — Group project grading: uniform = full credit 1.0
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?
Answer
BBBWN: Bolt Shear, Bearing, Block Shear, Weld, Net-Section Rupture
Memory link
A18 — BBBWN Radio Station: tune into all channels before broadcasting
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?
Answer
The MINIMUM of the two governs (Smaller Twin Wins)
Memory link
A5 — Smaller Twin Wins: min(1.2 lc t Fu, 2.4 db t Fu)
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?
Answer
Two shear planes (double shear): multiply single-shear bolt capacity by 2
Memory link
A3 — Two knives on a hotdog: double shear doubles resistance
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.)
Answer
FEXX = 482 MPa (PIN: 4-8-2)
Memory link
A9 — E70 PIN code: 4-8-2 → 482 MPa
Formula mnemonics
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
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
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
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
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
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)
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
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
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
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)
Diagrams
Steel Connections — Complete Concept Map
A comprehensive mind map showing all Steel Connections topics branching from bolted to welded connections, with every key formula variable and governing principle labeled. Use this as a pre-study orientation and post-study self-test tool.
Step-by-Step Connection Design Strength Check (LRFD)
A decision-tree flowchart showing the complete systematic procedure for determining connection design strength under AISC 360 LRFD. Follow this flow on every board problem involving connections: start at the top, compute every applicable limit state, and always report the minimum as the governing capacity.
Bearing Strength: The Smaller Twin Decision
A focused flowchart for the bearing strength computation showing the two candidate expressions (1.2 lc t Fu and 2.4 db t Fu), the minimum selection step, and the final phi-factor application. Use this diagram as a mental checklist whenever bearing strength appears in a problem.
Block Shear: Pizza Escape — 8-Step Procedure
An 8-step flowchart for the complete block shear computation procedure. The pizza-escape metaphor maps to this flow: sketch the block (pizza), compute shear and tension areas (edges and front), decide Ubs (topping uniformity), compute both rupture and yield paths, take the lesser, then apply phi = 0.75.