CELE Reinforced & Prestressed Concrete — Reinforced Concrete ColumnsMemory Anchors

Memory aid

Remember: 'Point-Eight-Five Steel Net Plus Steel Yield' → P_o = 0.85 f'c (A_g − A_st) + fy A_st. Use the acronym 'PENS': P-o, Eight-five concrete on Net area, Steel yield times Steel area. Picture a PENS (ballpen) standing upright like a column — it holds its shape because the outer plastic (concrete net area) and inner ink tube (steel) work together.

Why it works

Breaking the formula into two clear parts (concrete contribution + steel contribution) with a physical image of a pen as a column makes the additive structure unforgettable.

Example usage

Exam asks: 'Find the nominal axial strength of a 400×400 column with 8-25mm bars, f'c=28 MPa, fy=415 MPa.' Recall PENS → Po = 0.85(28)(160000−3927) + 415(3927) = 5344 kN.

Recall trigger

Imagine a ballpen standing vertically like a tiny column on your exam paper.

Memory aid

Tied column = '80-65': the cap is 0.80 and φ is 0.65. Multiply: 0.80 × 0.65 = 0.52. Chunk it as 'EIGHT-ZERO, SIX-FIVE' like a Philippine area code you dial for a tied column. 0.52 Po is less than 0.64 (spiral) — because ties provide less confinement.

Why it works

Chunking two numbers together and linking them to the familiar concept of dialing a phone number creates an instant retrieval path for both the cap factor and φ.

Example usage

Given Po = 5344 kN (tied), φPn,max = 0.52 × 5344 = 2779 kN. Check: you dialed 80-65, product is 0.52.

Recall trigger

Tied column → dial '0.80 – 0.65' like a phone area code.

Memory aid

Spiral column = '85-75': the cap is 0.85 and φ is 0.75. Chunk it as 'EIGHTY-FIVE, SEVENTY-FIVE' — both numbers end in FIVE. Product = 0.6375 ≈ 0.64. Remember: 'Spirals are FIVEY' (both end in 5) and BIGGER than tied (0.64 > 0.52). The spiral is like a coiled spring — it gives more, so it gets more.

Why it works

The pattern that both numbers end in 5 is a self-reinforcing chunk. Connecting it to the physical superiority of spiral confinement adds meaning.

Example usage

Same Po = 5344 kN but spiral: φPn,max = 0.6375 × 5344 = 3407 kN — about 23% more than tied (2779 kN).

Recall trigger

Spiral → both factors end in FIVE → 85-75 → product ≈ 0.64.

Memory aid

Think of a tied column as a box of marbles held by rubber bands — when the outer concrete spalls, the bands snap and marbles scatter (brittle failure). A spiral column is like a slinky — even when the outer shell breaks, the coiled spring holds everything together (ductile failure). More ductility = more warning before collapse = higher reward from the code (bigger φ and bigger cap). NSCP/ACI rewards ductility with generosity.

Why it works

Contrasting brittle (rubber bands snap) vs. ductile (slinky holds) behavior creates an emotional and visual distinction that maps directly to the numerical difference in code factors.

Example usage

Exam question asks why spiral columns have φ=0.75: recall the slinky — ductility means warning before failure, so code grants higher strength reduction factor.

Recall trigger

Slinky = spiral = ductile = higher φ and higher cap.

Memory aid

Rhyme: 'One percent to eight percent — column steel must stay content. Less than one, the steel is vain; more than eight, the bars complain.' Picture the steel bars as workers in a column-shaped office: too few (below 1%) and the building relies on concrete alone; too many (above 8%) and bars are so crowded they can't be properly embedded. Practically, keep ρg ≤ 4% (0.04) where bars lap-splice to avoid congestion.

Why it works

Rhymes activate the brain's auditory memory loop, making the 1% and 8% limits pop up automatically during recall.

Example usage

ρg = 3927/160000 = 0.0245. Is 0.01 ≤ 0.0245 ≤ 0.08? Yes — steel is content. ✓

Recall trigger

Sing the rhyme mentally: 'One to eight — column steel must stay content.'

Memory aid

Visualize a TABLE (4 legs = 4 bars for a rectangular tied column) and a BICYCLE WHEEL (6 spokes = 6 bars for a circular spiral column). A table stands on 4 legs — if you remove one, it still barely stands (rectangular/tied). A wheel needs at least 6 spokes to be round and balanced (circular/spiral). Both images are things you see every day in the Philippines.

Why it works

Associating abstract numbers with familiar everyday objects (table, bike wheel) gives the brain a concrete image to retrieve during the exam.

Example usage

Design circular spiral column → must have at least 6 longitudinal bars. Remember the bike wheel.

Recall trigger

Tied rectangular → 4-legged table. Spiral circular → 6-spoke bike wheel.

Memory aid

Story: Engineer Ana is designing a spiral column. She asks: 'How much spiral steel do I need?' Her mentor says: 'Ana, think of it as COMPENSATION — the core (Ach) is smaller than the gross section (Ag). The spiral must compensate for the lost concrete outside the core. The ratio Ag/Ach − 1 tells you how much you lost; multiply by the material ratio f'c/fyt to get the minimum spiral. And always start with 0.45 — that's the code's safety deposit.' Ana never forgot: 0.45 × (how much you lost) × (material ratio) = minimum spiral.

Why it works

Embedding the formula in a character-driven story (Engineer Ana and her mentor) creates an episodic memory trace, which is far more durable than rote memorization.

Example usage

400mm circular column, 40mm cover: Ach = π/4(320²)=80,425 mm², Ag=π/4(400²)=125,664 mm². ρs,min = 0.45(125664/80425 − 1)(f'c/fyt). Recall Ana's story → apply the formula.

Recall trigger

Think of Engineer Ana asking about compensation for lost concrete outside the core.

Memory aid

Remember '700 is the CEILING for spiral yield.' Visualize a low ceiling marked '700 MPa' — no matter how strong your spiral wire is, it can't jump higher than this ceiling in the ρs formula. This prevents underestimating the required spiral by assuming unrealistically high yield strength. The number 700 → think '7' is the highest single digit, but it's capped there.

Why it works

The physical image of a ceiling cap directly encodes the concept of an upper bound, preventing the common mistake of using fyt > 700 MPa in the spiral ratio equation.

Example usage

If spiral wire has fyt = 800 MPa, use fyt = 700 MPa in the spiral ratio formula — the ceiling applies.

Recall trigger

Spiral formula → low ceiling at 700 MPa for fyt.

Memory aid

Imagine you're filling a glass with water (concrete). But some space is taken by straws (steel bars). The water only fills the space NOT occupied by straws. So, concrete stress × (glass area − straw area) = 0.85f'c × (Ag − Ast). Don't charge the concrete for the space the steel already occupies. This is the most common board-exam pitfall — using Ag instead of the net area.

Why it works

The water-glass-straw analogy is physically intuitive and immediately explains WHY the net area (Ag − Ast) is used, not just what it is.

Example usage

A_g = 160,000 mm², A_st = 3,927 mm². Concrete term = 0.85(28)(160,000 − 3,927) = 0.85(28)(156,073). Do NOT use 160,000 for the concrete term.

Recall trigger

Glass with straws → water fills the net space → concrete acts on (Ag − Ast).

Memory aid

Visualize the interaction diagram as a MANGO (Filipino fruit): the TIP of the mango at the top is pure axial load (Po) — tall and pointy, all compression. The FATTEST MIDDLE of the mango is the balanced point — biggest moment capacity. The FLAT BOTTOM of the mango is pure flexure (P = 0). Any design point (Pu, Mu) must be INSIDE the mango. If it falls outside, the column fails — like biting into a rotten mango.

Why it works

The mango shape is culturally familiar to Filipino students and perfectly mirrors the elongated drop shape of an RC column interaction diagram, with the three key points mapping naturally.

Example usage

Exam plots a point (Pu, Mu) → check if it's inside the mango (φ-reduced diagram). If outside = column inadequate.

Recall trigger

Interaction diagram → mango shape. Top = pure axial, middle fat = balanced, bottom = pure flexure.

Memory aid

Story: Two Olympic sprinters race side by side. One is named 'Concrete Crush' (εc = 0.003) and the other is 'Steel Yield' (εs = fy/Es). At the BALANCED POINT, they both cross the finish line at EXACTLY the same time. If Concrete Crush wins (crosses first), the failure is brittle compression-controlled. If Steel Yield wins (crosses first), it's ductile tension-controlled. The balanced point is the PHOTO FINISH — both cross simultaneously — and it gives the maximum moment on the interaction diagram.

Why it works

The sprint race story makes the simultaneous failure conditions memorable and also correctly encodes the compression-controlled vs. tension-controlled distinction.

Example usage

At the balanced point: ε_c = 0.003 (concrete crushes), ε_s = f_y/E_s (steel just yields). This is where M_n is maximum on the interaction diagram.

Recall trigger

Photo finish race — Concrete Crush and Steel Yield cross together = balanced point.

Memory aid

Above the balanced point, the column is like a KARAOKE MACHINE set too loud — it crushes (distorts) before you get the best musical moment (max moment). Below the balanced point, it's like a guitar string — it yields and stretches (ductile), giving you warning. High axial load + small moment = compression-controlled = dangerous brittle. Low axial + high moment = tension-controlled = safe ductile. Remember: UP = UGLY (brittle compression), DOWN = DUCTILE (tension).

Why it works

The UP/DOWN spatial association maps to the vertical axis of the interaction diagram, while the familiar (and funny) karaoke reference makes the brittle/ductile distinction culturally sticky for Filipino students.

Example usage

If Pu > Pb (above balanced), column is compression-controlled, φ=0.65 (tied). If Pu < Pb, it transitions toward tension-controlled, φ→0.90 for pure flexure.

Recall trigger

UP on diagram = UGLY brittle compression. DOWN = Ductile tension.

Memory aid

Mnemonic: '34 MINUS 12-to-ONE, cap at FORTY, then you're done.' The ratio M1/M2 is the ratio of smaller to larger end moment (positive if bent in double curvature). The limit = 34 − 12(M1/M2) but never more than 40. Think: '34 is the BASE, 12 is the ADJUSTMENT for curvature.' If both moments bend the same way (single curvature), M1/M2 is negative, making the limit smaller — tighter. Safer column, lower limit.

Why it works

The rhyme '34 minus 12-to-one, cap at forty, then you're done' encodes all three numbers (34, 12, 40) in a single sentence with a clear endpoint.

Example usage

Braced column: M1/M2 = +0.5 (double curvature). Limit = 34 − 12(0.5) = 34 − 6 = 28. If kℓu/r = 25 < 28 → column is short. ✓

Recall trigger

'34 minus 12-to-one, cap at 40, then you're done' — braced frame slenderness.

Memory aid

Unbraced = sway = DANGER. The limit is only 22 — the lowest slenderness limit. Remember: 'TWENTY-TWO is TOO FEW for sway' — if slenderness exceeds 22 in a sway frame, the column is slender and needs moment magnification. The number 22 is also the age most CE graduates take the board exam — if you're 22 and sway, you need more support (moment magnification).

Why it works

Connecting 22 to the typical age of board exam takers creates a personal, emotionally resonant hook that makes the number impossible to forget.

Example usage

Sway frame column with kℓu/r = 28 > 22 → column is slender → must apply moment magnification (P-Δ effect).

Recall trigger

Sway frame → age 22 → limit is 22 → slender if above 22.

Memory aid

A slender column under axial load is like a fishing rod (bending wand) with a fish at the tip. As the fish pulls down (P), the rod bends laterally (δ), which creates an additional moment (P×δ). This moment bends the rod more, increasing δ, increasing the moment — a feedback loop. The moment magnifier δ is like an 'amplifier' on a stereo: it takes the first-order moment M2 and cranks it up: Mc = δ × M2. The heavier the fish (bigger P) and the more slender the rod (bigger ℓu/r), the louder the amplification.

Why it works

The fishing rod analogy physically demonstrates the P-δ feedback mechanism, while the stereo amplifier image immediately conveys what the magnifier δ does mathematically.

Example usage

Slender column: Mc = δns × M2 (braced) or Mc = δs × M2 (sway). If δ = 1.25 and M2 = 200 kN·m, design moment = 1.25 × 200 = 250 kN·m.

Recall trigger

Slender column = fishing rod with fish. Amplifier knob = moment magnifier δ.

Memory aid

Acronym: '16-48-LEAST' → Tie spacing must be the smallest of: 16 × longitudinal bar diameter, 48 × tie bar diameter, or the LEAST column dimension. Remember '16-48-LEAST' like a sports jersey number sequence: '16 is the quarterback (longitudinal), 48 is the linebacker (tie), and LEAST wins the game.' Always take the minimum of the three.

Why it works

Acronyms with numbers are effective for code provisions. The sports metaphor adds a competitive angle that reinforces the 'take the minimum' rule.

Example usage

Column with 25mm long. bars and 10mm ties, 400mm column width. s ≤ min(16×25=400, 48×10=480, 400) = 400 mm. Use s = 400 mm max.

Recall trigger

'16-48-LEAST' → three competing limits → smallest controls.

Memory aid

Story: In the PE Review Center, a nervous reviewee writes φ = 0.90 for a column problem. The reviewer shakes his head: 'Hoy! Columns are NOT beams!' The 0.90 is for BEAMS under flexure — they are tension-controlled and ductile. Columns are compression-controlled (upper part of interaction diagram), so they get the stricter 0.65 (tied) or 0.75 (spiral). The reviewee learned: 0.90 belongs to the beam next door, not in the column family.

Why it works

The classroom micro-story dramatizes the most common board-exam mistake (using φ=0.90 for columns) and creates a strong emotional memory through embarrassment and correction.

Example usage

Column problem: always check — tied or spiral? Then apply 0.65 or 0.75. Never 0.90 for columns.

Recall trigger

Columns are NOT beams — φ=0.65 tied, φ=0.75 spiral. NOT 0.90.

Memory aid

Chunk: 'Spiral earns a TWENTY-THREE PERCENT BONUS.' From the reference example: tied gives 2779 kN, spiral gives 3407 kN — ratio ≈ 1.23. The extra 23% is the reward for the spiral's ductility and confinement. In the Philippines, a 23% salary bonus sounds very attractive — just like the spiral column's bonus capacity. Remember this number for comparison problems.

Why it works

Linking a technical percentage to the universally exciting concept of a salary bonus creates an instant emotional hook, especially relevant to young professionals.

Example usage

If tied φPn = 2779 kN, expect spiral φPn ≈ 2779 × 1.23 = 3419 kN ≈ 3407 kN. Quick sanity check for spiral vs. tied comparison problems.

Recall trigger

Spiral = 23% salary bonus over tied. Same materials, 23% more capacity.

Memory aid

Imagine trying to squeeze into a packed jeepney (Filipino public transport). The code says the column can officially hold up to 8% steel (8 passengers per row), but in practice, when bars lap-splice, you're doubling the steel locally — like two sets of passengers crammed in the same spot. So the practical limit drops to 4% (4 passengers) to avoid steel congestion and ensure proper concrete consolidation. 'Lap-splice = doubling = half the room = 4% max.'

Why it works

The jeepney analogy is culturally iconic for Filipino students and perfectly illustrates why the practical limit (4%) is half the code maximum (8%) — because lap-splicing doubles local steel density.

Example usage

When detailing column lap splices, keep ρg ≤ 4% (0.04) to avoid bar congestion and ensure concrete can be properly placed and consolidated.

Recall trigger

Lap-splice = two sets of passengers in one jeepney = congestion → limit to 4%.

Memory aid

Visualize the number 0.85 as an 85% efficiency rating on an appliance label (like an inverter aircon). Concrete doesn't develop its full cylinder strength (f'c) in a real beam or column — it achieves about 85% because of shrinkage, loading rate differences, and the shape of the stress distribution. The 0.85 factor is the 'energy efficiency label' stamped on all concrete in the code. It appears in BOTH the Po formula (concrete contribution) and the stress block depth factor β1.

Why it works

The appliance energy label is a familiar modern reference that encodes both the value (0.85) and the reason (not 100% efficient) in a single image.

Example usage

In Po formula: 0.85 f'c × (Ag − Ast). In Whitney stress block: a = 0.85f'c × b. Both use 0.85 as the concrete efficiency factor.

Recall trigger

0.85 = concrete's efficiency rating — appears wherever concrete compression is involved.

Answer

Po = 0.85 f'c (Ag − Ast) + fy Ast

Memory link

A1 — PENS mnemonic. The ballpen standing upright: outer plastic (concrete net) + inner ink tube (steel).

Answer

0.52 (from 0.80 × 0.65)

Memory link

A2 — Phone area code '80-65'. Product = 0.52.

Answer

700 MPa

Memory link

A8 — The low ceiling at 700 MPa. No spiral yield can jump higher than this ceiling.

Answer

Balanced point (Pb, Mb): εc = 0.003 AND εs = fy/Es simultaneously

Memory link

A11 — The photo-finish race: Concrete Crush and Steel Yield cross the line together.

Answer

Sway limit = 22. Yes, 30 > 22 → column is SLENDER. Moment magnification required.

Memory link

A14 — '22 is the board exam age — sway frame limit is 22, very strict.'

Answer

6 bars minimum (6-spoke bike wheel)

Memory link

A6 — Bike wheel with 6 spokes = 6 bars for spiral/circular columns.

Answer

0.01 ≤ ρg ≤ 0.08; practical maximum at lap splices = 0.04

Memory link

A5 — 'One to eight — column steel must stay content.' Jeepney analogy (A19): lap-splice = 4% practical max.

Answer

φ (strength reduction factor). For columns: φ = 0.65 (tied) or φ = 0.75 (spiral). NOT 0.90.

Memory link

A17 — 'Columns are NOT beams!' story. The reviewer corrects the student who wrote φ = 0.90.

Mnemonic

PENS: P-o, Eight-five on Net area, Steel Yield times Steel area. Two parts: concrete NET + steel YIELD. Like a pen: outer plastic (concrete net) + inner ink tube (steel).

When to use

To find the nominal pure-axial (zero eccentricity) strength of any RC column before applying the cap factor and φ. This is always Step 1 in column axial capacity problems.

What each part means

0.85 f'c = effective concrete compressive stress; (Ag − Ast) = net concrete area (gross minus steel); fy = longitudinal steel yield strength; Ast = total longitudinal steel area. Both terms add together because concrete and steel both carry axial load.

Mnemonic

TIED: 'EIGHTY-ZERO, SIX-FIVE' like a phone area code. Product = 0.52. Tied columns = 0.52 Po. Easy: 0.52 is slightly more than HALF of Po.

When to use

For the MAXIMUM design axial load a tied column can carry. Apply this after computing Po. This is the allowable limit for pure or near-pure axial loading.

What each part means

0.80 = NSCP cap factor for tied columns accounting for unavoidable eccentricity; φ = 0.65 = strength reduction factor for compression-controlled tied columns; Po = nominal pure-axial strength from the Po formula.

Mnemonic

SPIRAL: 'EIGHTY-FIVE, SEVENTY-FIVE' — both end in FIVE. Product ≈ 0.64. Spiral columns ≈ 0.64 Po — significantly more than tied (0.52 Po).

When to use

For the MAXIMUM design axial load a spiral column can carry. Use whenever the problem specifies a spiral (circular) column.

What each part means

0.85 = NSCP cap factor for spiral columns (higher than 0.80 for tied due to greater ductility); φ = 0.75 = strength reduction factor for spiral columns (also higher than 0.65 for tied); Po = nominal pure-axial strength.

Mnemonic

'One percent to eight percent — column steel must stay content.' ρg is simply total steel over gross area. Limits: 1% minimum (avoid plain concrete behavior), 8% maximum (avoid bar congestion).

When to use

Always check ρg after designing or analyzing a column. Required verification in any column design problem.

What each part means

Ast = total area of all longitudinal bars; Ag = gross cross-sectional area of column; ρg = longitudinal reinforcement ratio. The 1% minimum prevents columns from acting like plain concrete; the 8% maximum prevents congestion.

Mnemonic

SPIRAL COMPENSATION: '0.45 × LOSS RATIO × MATERIAL RATIO.' The loss ratio (Ag/Ach − 1) quantifies concrete lost outside the core. The material ratio (f'c/fyt) scales by material strengths. 0.45 is the code's safety deposit. fyt ceiling = 700 MPa (the low ceiling).

When to use

When designing spiral reinforcement for a circular column. Compute Ach based on core diameter (column diameter minus 2 × cover), then solve for the minimum ρs.

What each part means

0.45 = empirical code constant; Ag = gross section area; Ach = core area measured to outside of spiral; f'c = concrete compressive strength; fyt = yield strength of spiral reinforcement (max 700 MPa). The formula ensures the spiral can replace the lost concrete toughness.

Mnemonic

'34 MINUS 12-to-ONE, cap at FORTY — braced is done. TWENTY-TWO for SWAY, anything more needs P-delta pay.'

When to use

Before any column design: compute kℓu/r and compare to the appropriate limit. If slenderness ratio exceeds the limit, the column is SLENDER and moment magnification is required.

What each part means

k = effective length factor; ℓu = unsupported column length; r = radius of gyration (= 0.3h for rectangular, 0.25D for circular); M1/M2 = ratio of smaller to larger factored end moment (positive = double curvature); The limits determine if slenderness effects can be ignored.

Recall test

Without looking, list all 8 steps to find φPn,max for a 350×350 tied column with 6-20mm bars. What formula do you use in Step 4?

Memory chain

PENS story: 'I IDENTIFY my pen type (tied or spiral). I GRAB the gross area. I ASSIGN steel area. I COMPUTE Po (the pen's raw strength). I CAP it with the code factor. I φ-REDUCE it. I CHECK the steel ratio is content (1% to 8%).' Remember: IGACC-Check or simply '1-2-3-Po-Cap-φ-Check.'

Items to remember

• 1. Identify column type: tied or spiral
• 2. Compute Ag (gross area)
• 3. Compute Ast (total steel area)
• 4. Compute Po = 0.85f'c(Ag − Ast) + fy Ast
• 5. Apply cap factor: 0.80 (tied) or 0.85 (spiral)
• 6. Apply φ: 0.65 (tied) or 0.75 (spiral)
• 7. φPn,max = cap × φ × Po
• 8. Check ρg = Ast/Ag is between 0.01 and 0.08

Recall test

What two conditions are simultaneously satisfied at the balanced point? What happens to φ as you move down from the balanced point toward pure flexure?

Memory chain

MANGO: TIP (pure axial at top), FAT MIDDLE (balanced, max moment), FLAT BOTTOM (pure flexure). The photo-finish race (Concrete Crush vs. Steel Yield) happens exactly at the fat middle. Design point (Pu, Mu) must fall INSIDE the mango.

Items to remember

• Top: Pure axial (P = Po, M = 0) — compression-controlled
• Middle: Balanced point (Pb, Mb) — εc = 0.003 AND εs = fy/Es simultaneously — maximum moment
• Transition: φ transitions from 0.65/0.75 toward 0.90
• Bottom: Pure flexure (P = 0, M = Mn) — tension-controlled, φ = 0.90

Recall test

A 500mm diameter spiral column: what is the minimum number of bars? What is the maximum ρg at lap-splice zones? What is the maximum fyt in the spiral ratio formula?

Memory chain

The 'Table-Wheel-Jeepney' chain: TABLE has 4 legs (min bars for tied), WHEEL has 6 spokes (min bars for spiral), JEEPNEY holds 4% max (practical ρg at lap), '16-48-LEAST' wins the tie spacing game. Then Engineer Ana's COMPENSATION formula for spirals with the 700 MPa ceiling.

Items to remember

• ρg: 0.01 ≤ ρg ≤ 0.08 (practical max 0.04 at lap splices)
• Min bars: 4 (tied rectangular), 6 (spiral circular)
• Tie spacing: s ≤ min(16db_long, 48db_tie, least dimension)
• Spiral ratio: ρs ≥ 0.45(Ag/Ach − 1)(f'c/fyt), fyt ≤ 700 MPa

Recall test

A braced-frame column has M1/M2 = −0.3 (single curvature) and kℓu/r = 36. Is it short or slender? What is the slenderness limit?

Memory chain

S-K-R-COMPARE-SHORT/SLENDER: 'SWAY or BRACED? Know k. Radius r. COMPARE to limit. SHORT means stop; SLENDER means amplify.' Sway = 22 (board exam age). Braced = 34-minus-12, cap 40.

Items to remember

• 1. Determine if frame is braced (non-sway) or unbraced (sway)
• 2. Find k (effective length factor)
• 3. Compute r = 0.3h (rectangular) or 0.25D (circular)
• 4. Compute kℓu/r
• 5. Compare: braced → 34−12(M1/M2) ≤ 40; sway → 22
• 6. If slenderness ratio ≤ limit → SHORT column (no magnification needed)
• 7. If slenderness ratio > limit → SLENDER (apply moment magnifier)

Recall test

Without looking: what is φPn,max for a spiral column if Po = 4000 kN? What about for a tied column? What percentage more does the spiral column carry?

Memory chain

PHONE CODE vs. BONUS: Tied = 'dial 80-65' (area code), product 0.52. Spiral = 'FIVEY numbers 85-75', product 0.64 — a 23% salary bonus. Slinky (spiral) beats rubber band (tied) in ductility every time.

Items to remember

• Tied: cap = 0.80, φ = 0.65, product = 0.52, min bars = 4
• Spiral: cap = 0.85, φ = 0.75, product ≈ 0.64, min bars = 6
• Spiral is ~23% stronger for same materials
• Spiral is more ductile (slinky), tied is more brittle (rubber bands)