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CELE Reinforced & Prestressed Concrete — Reinforced Concrete Beams: FlexureMemory Anchors

Summary Concepts Notes Revision Deep Dive Slides Flashcards Mistakes Cheat Mnemonics Templates

Memory anchors for Reinforced Concrete Beams: Flexure — mnemonic devices, acronyms, and tricks that make the CELE Reinforced & Prestressed Concrete syllabus stick. Use these when a concept just will not stay in your head.

Exam context

For the Civil Engineer Licensure Examination, Professional Regulation Commission (PRC) — Board of Civil Engineering tests Reinforced & Prestressed Concrete under a "Core" label, with Reinforced Concrete Beams: Flexure in the 2nd slot across 7 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 Reinforced & Prestressed Concrete questions. Date to watch: May and November 2026.

Reinforced Concrete Beams: Flexure - Memory Anchors

Memory techniques transform abstract formulas into vivid mental images that survive the stress of board-exam day. Research in cognitive psychology (Paivio's dual-coding theory) shows that pairing verbal information with a strong image or story increases recall by up to 40%. For RC beam flexure — a topic guaranteed to appear in every PRC Civil Engineer board exam — you need instant, error-free retrieval of stress-block formulas, steel-ratio limits, and the ductility check. The anchors below exploit mnemonics, analogies, micro-stories, visual cues, and rhymes so that even at 2 a.m. the night before the exam, every formula snaps back into focus. Work through each anchor actively: close your eyes, replay the story or image, then write the formula from memory. Do this three times per anchor and it will stick for life.

Anchors

Topic

Singly Reinforced Beam — Stress Block

Concept

Equivalent stress-block depth: a = As·fy / (0.85·f'c·b)

Anchor Id

Difficulty

easy

Memory Aid

Imagine a SARI-SARI STORE counter (the compression zone). The stress block 'a' is how HIGH the cashier stacks the goods (compression force). The goods on the shelf are As·fy (all your steel selling power). The counter can only hold 0.85·f'c worth of goods per unit width b. So the stack height 'a' = (what steel sells) ÷ (what the counter can carry per mm). More steel → taller stack; wider counter → shorter stack. It balances perfectly — that's force equilibrium.

Anchor Type

analogy

Why It Works

The sari-sari store is a universal Filipino reference. Mapping 'stack height' to stress-block depth makes the formula feel physical and intuitive rather than abstract.

Example Usage

Exam asks for 'a' given As = 1473 mm², fy = 415 MPa, f'c = 28 MPa, b = 300 mm. Recall the counter: a = 1473×415 / (0.85×28×300) = 85.6 mm.

Recall Trigger

Think: sari-sari store counter, stack of goods

Topic

Nominal Moment Capacity

Concept

Nominal moment: Mn = As·fy·(d − a/2)

Anchor Id

Difficulty

easy

Memory Aid

Think of a LEVER at a palengke (wet market): the steel bars are a MARKET VENDOR pulling DOWN with force As·fy. The arm of the lever is the distance from where the vendor pulls (centroid of steel at depth d) to the middle of the stack of goods (at a/2 from the top). Moment = Force × Arm = As·fy × (d − a/2). The vendor's pull times the lever arm gives you the torque — that's your Mn.

Anchor Type

analogy

Why It Works

The palengke lever grounds the abstract lever-arm concept in a concrete, familiar Filipino setting, making the (d − a/2) term instantly visualizable.

Example Usage

After computing a = 85.6 mm: Mn = 1473×415×(500 − 42.8) = 279.4 kN·m. The vendor pulls with 1473×415 N; arm is (500 − 42.8) mm.

Recall Trigger

Think: palengke vendor pulling a lever arm

Topic

Ductility Check and φ-factor

Concept

Tension-controlled limit: εt ≥ 0.005, then φ = 0.90

Anchor Id

Difficulty

medium

Memory Aid

Remember: '5 for FULL PAY, less gets a CUT.' If the steel strain εt ≥ 0.005 (five thousandths), the section is tension-controlled and you get FULL PAY — the full φ = 0.90. If εt < 0.005, the section is in the transition or compression-controlled zone and φ takes a CUT (goes below 0.90). The number 5 = full strength reduction factor. Less than 5 = penalty.

Anchor Type

mnemonic

Why It Works

The pay-cut analogy ties the abstract φ-factor to a relatable concept of earning full or reduced compensation, making the threshold 0.005 memorable.

Example Usage

Computed εt = 0.0119 > 0.005. Student recalls '5 for full pay' → tension-controlled → φ = 0.90 → φMn = 0.90 × 279.4 = 251.5 kN·m.

Recall Trigger

Think: '5 for full pay' — εt = 0.005

Topic

Strain Compatibility

Concept

Net tensile strain formula: εt = 0.003·(d − c)/c

Anchor Id

Difficulty

medium

Memory Aid

Picture a BAMBOO POLE leaning against a wall (the neutral axis). The TOP of the pole is pinned at concrete crush strain 0.003 (like the wall). The BOTTOM tip is the steel at depth d. The neutral axis is at height c from the top. As the bamboo bends, the tip (steel strain) moves proportionally: εt = 0.003 × (how far steel is from top) / (how far the crush point is from the top) = 0.003 × (d − c)/c. Similar triangles on the strain diagram — just like the bamboo bending shape.

Anchor Type

visual_association

Why It Works

Bamboo is a culturally resonant Filipino material. The visual of a bamboo bending encodes the similar-triangles geometry of the strain diagram directly into long-term memory.

Example Usage

c = 100.7 mm, d = 500 mm. Recall bamboo: εt = 0.003 × (500 − 100.7)/100.7 = 0.0119.

Recall Trigger

Think: bamboo pole bending, tip-to-wall ratio

Topic

Stress-Block Factor β₁

Concept

β₁ factor: 0.85 for f'c ≤ 28 MPa, decreases by 0.05 per 7 MPa above 28, minimum 0.65

Anchor Id

Difficulty

medium

Memory Aid

BETA STARTS HIGH, GETS SHY: β₁ = 0.85 (proud and tall) when concrete is normal (≤ 28 MPa). Every time concrete gets STRONGER by 7 MPa, beta gets SHY and drops 0.05. But beta has a FLOOR of 0.65 — it refuses to go any lower, like a shy person who still won't leave the room. Remember: 85, shy by 5, floor at 65.

Anchor Type

mnemonic

Why It Works

The personification of β₁ as 'getting shy' makes the piecewise rule memorable. The three numbers (85, 5, 65) rhyme rhythmically.

Example Usage

f'c = 35 MPa: β₁ = 0.85 − 0.05×(35−28)/7 = 0.85 − 0.05 = 0.80. f'c = 56 MPa: β₁ = 0.85 − 0.05×4 = 0.65 (floor).

Recall Trigger

Think: 'Beta starts high, gets shy — 85, shy by 5, floor at 65'

Topic

Balanced Steel Ratio

Concept

Balanced steel ratio: ρb = 0.85·β₁·(f'c/fy)·[600/(600+fy)]

Anchor Id

Difficulty

hard

Memory Aid

Story: 'CAPTAIN BALANCE commanded his two soldiers — CONCRETE COLONEL (who crushes at 0.003) and STEEL SERGEANT (who yields at fy/Es). On the fateful day of balanced failure, both soldiers hit their limit at EXACTLY the same time. Captain Balance's order was: ρb = 0.85·β₁·(f'c/fy) × [600/(600+fy)]. The 600 comes from Es = 200,000 MPa × 0.003 = 600 MPa — the concrete's dying breath. When both soldiers fall simultaneously, that's the balanced condition.' Every time you see 600 in the formula, remember: it's the concrete's last breath.

Anchor Type

micro_story

Why It Works

The military story with two characters failing simultaneously encodes the physical meaning of the balanced condition (simultaneous crushing and yielding) alongside the origin of the 600 constant.

Example Usage

f'c = 28, fy = 415: ρb = 0.85×0.85×(28/415)×[600/1015] = 0.0288. Both soldiers fall together at this ratio.

Recall Trigger

Think: Captain Balance — both soldiers fall at the same time; 600 = concrete's last breath

Topic

Maximum Steel Ratio

Concept

Maximum steel ratio (tension-controlled): ρmax = 0.85·β₁·(f'c/fy)·(0.375)

Anchor Id

Difficulty

medium

Memory Aid

CHUNK IT as: ρmax = [0.85·β₁·(f'c/fy)] × 0.375. The first part [0.85·β₁·(f'c/fy)] is the 'concrete-over-steel fraction.' The 0.375 is the magic multiplier from setting εt = 0.005: c/d = 0.003/(0.003+0.005) = 3/8 = 0.375. Remember '3-8-MAX': three (0.003) over eight (0.003+0.005) gives MAX ratio. Three eighths = 0.375. Say it: 'Three-eighths is my max!'

Anchor Type

chunking

Why It Works

Chunking separates the formula into two digestible parts. The fraction 3/8 = 0.375 derived from strain compatibility makes the multiplier logical, not arbitrary.

Example Usage

ρmax = 0.85×0.85×(28/415)×0.375 = 0.0183. Verify: 0.375 = 0.003/(0.003+0.005).

Recall Trigger

Think: '3-8-MAX — three over eight is my max ratio'

Topic

Minimum Steel Ratio

Concept

Minimum steel ratio: ρmin = max(1.4/fy, √f'c / 4fy)

Anchor Id

Difficulty

medium

Memory Aid

RMIN = 'Take the MAX of the MIN pair' — sounds ironic but that's exactly it! Two candidates fight for minimum title: CANDIDATE A = 1.4/fy (the flat one, doesn't change with concrete strength) and CANDIDATE B = √f'c/(4fy) (the root one, grows with concrete strength). The WINNER (the larger one) becomes ρmin. Remember: 'One-point-four or root over four — whichever is MORE, that's the floor!'

Anchor Type

mnemonic

Why It Works

The rhyme 'one-point-four or root over four, whichever is more, that's the floor' encodes both expressions and the selection rule in one memorable line.

Example Usage

fy = 415, f'c = 28: Candidate A = 1.4/415 = 0.00337; Candidate B = √28/(4×415) = 5.292/1660 = 0.00319. Winner = 0.00337. ρmin = 0.00337.

Recall Trigger

Think: '1.4 or root-over-4, whichever is more, that's the floor'

Topic

Design — Coefficient of Resistance

Concept

Coefficient of resistance: Rn = Mu / (φ·b·d²)

Anchor Id

Difficulty

easy

Memory Aid

Rn is like a LAND AREA PRICE per square meter: you have a budget (Mu = your total money), you divide by the capacity factor (φ = like a discount), and then by the floor area (b·d²). What you get is the price per mm² of beam — that's Rn in MPa. Think: 'How much moment can I squeeze out of each square millimeter of beam cross-section?' More Rn = more stressed beam — you're pushing it harder.

Anchor Type

analogy

Why It Works

Real-estate price-per-sqm is a concept every Filipino student understands viscerally (especially in Metro Manila). Mapping Rn to price per unit area makes the formula intuitively obvious.

Example Usage

Mu = 200 kN·m, φ = 0.90, b = 300, d = 450: Rn = 200×10⁶/(0.90×300×450²) = 3.66 MPa.

Recall Trigger

Think: land price per sqm — Rn = Mu/(φ·b·d²)

Topic

Design — Required Steel Ratio

Concept

Steel ratio from Rn: ρ = (0.85·f'c/fy)·[1 − √(1 − 2Rn/(0.85·f'c))]

Anchor Id

A10

Difficulty

hard

Memory Aid

Story: 'ENGINEER RONA wanted to find how much steel she needed. She computed the STRESS RATIO = 0.85·f'c/fy (how strong the concrete team is versus the steel team). Then she computed the DAMAGE FACTOR under the square root: how much of the 0.85·f'c capacity has the load Rn used up? She subtracted the remaining undamaged fraction from 1 (because she needs only the USED part), then multiplied by the stress ratio. Like subtracting what's LEFT on the plate — you eat (use) the portion ρ, the leftovers under the root are what concrete still has.' The formula is: stress ratio × (1 − undamaged fraction).

Anchor Type

micro_story

Why It Works

The eating analogy maps the abstract quadratic formula derivation into a familiar act (finishing food on a plate), encoding both the structure and meaning of the formula.

Example Usage

Rn = 3.66, f'c = 28, fy = 415: ρ = (0.85×28/415)×[1−√(1−2×3.66/(0.85×28))] = 0.05735×(1−√0.6926) = 0.05735×0.1678 = 0.00962.

Recall Trigger

Think: Engineer Rona eating — stress ratio × (1 − what's left on the plate)

Topic

Analysis Procedure

Concept

Analysis procedure: given section → find φMn (5-step process)

Anchor Id

A11

Difficulty

medium

Memory Aid

Use the acronym A-C-E-C-M: A = compute 'a' (stress block depth). C = compute 'c' (neutral axis: c = a/β₁). E = check εt (strain: εt = 0.003(d−c)/c). C = check φ (is εt ≥ 0.005? → φ = 0.90). M = compute φMn (φ × As×fy×(d−a/2)). Say: 'ACE-CM — Ace your Capacity Moment analysis!'

Anchor Type

acronym

Why It Works

The acronym ACE-CM ties to the idea of 'acing' the exam, making it motivationally as well as mnemonically effective. The sequence maps exactly to the five computation steps.

Example Usage

Given b=300, d=500, As=1473, f'c=28, fy=415: A→a=85.6; C→c=100.7; E→εt=0.0119; C→φ=0.90; M→φMn=251.5 kN·m.

Recall Trigger

Think: ACE-CM — Ace your Capacity Moment

Topic

Design Procedure

Concept

Design procedure: given Mu → find As (4-step process)

Anchor Id

A12

Difficulty

medium

Memory Aid

Use R-ρ-A-CHECK: R = compute Rn = Mu/(φ·b·d²). ρ = compute steel ratio from the Rn formula. A = compute As = ρ·b·d. CHECK = verify ρmin ≤ ρ ≤ ρmax. Pronounce it 'R-ROW-A-CHECK' — like a rowing boat race where you row to victory and then CHECK if you crossed the line within limits!

Anchor Type

acronym

Why It Works

The rowing race metaphor adds kinetic energy to a four-step sequence, and 'CHECK' at the end reminds students not to skip the critical code-compliance verification.

Example Usage

Mu=200 kN·m, b=300, d=450, f'c=28, fy=415: R→3.66; ρ→0.00962; A→1299 mm²; CHECK→ρmin=0.00337≤0.00962≤ρmax=0.0183 ✓.

Recall Trigger

Think: rowing race — R-ROW-A-CHECK

Topic

Over-Reinforced Beams and Code Rationale

Concept

Over-reinforced beam: concrete crushes before steel yields → brittle failure

Anchor Id

A13

Difficulty

medium

Memory Aid

Story: 'GREEDY GENERAL CONCRETE had too many soldiers (too much steel) so he hoarded all the compression. When the load came, the CONCRETE SIDE collapsed suddenly — BOOM — without warning, because the steel (the flexible, ductile part) never got a chance to yield and warn everyone. People died without notice. That's why NSCP 2015 BANS over-reinforced sections for new designs — no greedy generals allowed. Always let the steel yield first (tension-controlled) so you get VISIBLE sagging before collapse.'

Anchor Type

micro_story

Why It Works

The story of a greedy general hoarding resources and causing sudden collapse makes the engineering consequence of over-reinforcement vivid and emotionally memorable, linking the concept to why the code exists.

Example Usage

Exam question: 'Why does NSCP 2015 require εt ≥ 0.005?' Answer: To ensure tension-controlled ductile behavior — steel yields first, giving visible deflection warning before failure (no Greedy General Concrete).

Recall Trigger

Think: Greedy General Concrete — sudden collapse, no warning

Topic

T-Beam Analysis

Concept

T-beam: if a ≤ tf, treat as rectangular beam of width bf

Anchor Id

A14

Difficulty

medium

Memory Aid

Imagine a HAMBURGER (the T-beam cross-section). The BUN (flange, width bf) is on top; the PATTY (web, width bw) is below. When you compress the burger, if you only squish the bun (a ≤ tf), you only need to consider the BUN width — treat it as a rectangular burger of full bun-width bf. If you squish SO HARD that the compression goes into the patty (a > tf), now you must account for the patty separately. Rule: Check the bun first — if the stress block fits in the bun, you're done (rectangular analysis with bf)!

Anchor Type

analogy

Why It Works

Every Filipino student knows a burger. The bun = flange and patty = web is a perfect visual analogy for the T-beam two-region analysis. 'Check the bun first' is an unforgettable rule.

Example Usage

bf=800, tf=100, As=2500, fy=415, f'c=28: a=2500×415/(0.85×28×800)=54.4 mm < tf=100 mm → compression in bun only → treat as b=800 mm rectangular beam. ✓

Recall Trigger

Think: T-beam = hamburger — does compression stay in the bun (flange)?

Topic

Doubly Reinforced Beams

Concept

Doubly reinforced beam: use when Mu exceeds tension-controlled Mn of singly reinforced

Anchor Id

A15

Difficulty

hard

Memory Aid

Imagine a JEEPNEY that's overloaded. The single engine (singly reinforced) can't handle the load (Mu too large). Solution: add a SECOND ENGINE at the back (compression steel A's). Now the jeepney has two couples working: (1) concrete compression vs. tension steel, and (2) compression steel vs. extra tension steel. That's why doubly reinforced beams are modeled as two superimposed couples. Add compression steel when the load exceeds what your single engine (singly reinforced) can deliver.

Anchor Type

analogy

Why It Works

The jeepney is the quintessential Filipino vehicle reference. The image of adding a second engine to handle overload perfectly captures the structural logic of doubly reinforced beams.

Example Usage

If computed ρ > ρmax for a singly reinforced section, add compression steel A's to form a doubly reinforced beam — two jeepney engines sharing the load.

Recall Trigger

Think: overloaded jeepney — add a second engine (compression steel)

Topic

Effective Depth vs. Total Depth

Concept

Common pitfall: using full depth h instead of effective depth d

Anchor Id

A16

Difficulty

easy

Memory Aid

Picture a SWIMMING POOL (the beam cross-section). The TOTAL DEPTH h is from the top of the pool to the bottom. But the effective depth d is to the CENTER OF THE STEEL BARS — not the bottom tile. The clear cover + stirrup + half bar diameter eat up typically 60–80 mm. If you use h instead of d, you're measuring to the pool tile, not to where the swimmer (steel) actually is. Always ask: 'Where is the swimmer?' d = h − cover − stirrup − db/2.

Anchor Type

visual_association

Why It Works

The swimming pool image with a swimmer at a specific depth (not the pool bottom) is vivid and spatially intuitive, encoding the cover deduction into a memorable scene.

Example Usage

Board exam gives h = 600 mm with 40 mm cover, 12 mm stirrups, 25 mm bars. d = 600 − 40 − 12 − 25/2 = 535.5 mm. Use d = 535.5, NOT h = 600!

Recall Trigger

Think: swimming pool — where is the swimmer? That's d, not h.

Topic

Whitney Stress Block — 0.85 Factor

Concept

The 0.85 in the stress block (0.85·f'c) is NOT negotiable — always present

Anchor Id

A17

Difficulty

easy

Memory Aid

Rhyme: 'Point-eight-five stays alive — in the stress block it will thrive. Never drop it, never hide — 0.85·f'c, take it in stride!' The 0.85 accounts for the difference between cylinder strength and in-situ concrete: real concrete in the beam is slightly weaker than the lab cylinder. It's the Whitney stress-block correction and it appears in EVERY flexure formula involving concrete compression.

Anchor Type

rhyme

Why It Works

Rhymes exploit phonological loop memory. The short, rhythmic verse encodes both the value 0.85 and its physical justification, preventing the common error of using f'c alone.

Example Usage

Computing a: always write 0.85·f'c in the denominator. Never write just f'c. Stress block = 0.85·f'c, not f'c. Rhyme triggers the correction.

Recall Trigger

Rhyme: 'Point-eight-five stays alive'

Topic

Neutral Axis Depth

Concept

c = a / β₁ (relationship between neutral axis depth and stress block depth)

Anchor Id

A18

Difficulty

easy

Memory Aid

Picture a BLANKET (stress block, depth a) covering only PART of a SLEEPING PERSON (the full compression zone, depth c). The blanket is 85% of the person's covered area (for normal concrete: β₁ = 0.85). The person's length c is bigger than the blanket length a. So c = a/β₁ — you need to DIVIDE by beta to find the full person from the blanket. The blanket (a) is always shorter than the person (c).

Anchor Type

visual_association

Why It Works

The blanket-person visual makes the c > a relationship immediately intuitive. The act of 'unwrapping' the blanket to find the full person encodes the division by β₁.

Example Usage

a = 85.6 mm, β₁ = 0.85: c = 85.6/0.85 = 100.7 mm. The person (neutral axis) is longer than the blanket (stress block).

Recall Trigger

Think: blanket (a) covering a person (c) — person is bigger → c = a/β₁

Topic

Strength Reduction Factor

Concept

φ = 0.90 applies only to tension-controlled flexure (NSCP 2015 / ACI 318)

Anchor Id

A19

Difficulty

medium

Memory Aid

Remember: '90% is an A in flexure — but only if you're tension-controlled.' In Philippine grading, 90 is an excellent grade. φ = 0.90 is the BEST reduction factor for flexure — you give up only 10% of your strength. But you only EARN that A if εt ≥ 0.005 (you studied hard = ductile section). If you're transition or compression-controlled, you get a lower grade (lower φ). Always CHECK your grade before claiming 90!

Anchor Type

mnemonic

Why It Works

The Philippine grading system (90 = excellent) is culturally resonant and ties the abstract 0.90 to a concrete academic reference every student knows emotionally.

Example Usage

After computing εt: if εt = 0.0119 ≥ 0.005 → 'I earned my A' → φ = 0.90. If εt = 0.003 < 0.005 → 'I failed the ductility check' → φ < 0.90.

Recall Trigger

Think: 90 is an A — but you must earn it with εt ≥ 0.005

Topic

Legal Framework — RA 544

Concept

RA 544 — Philippine Civil Engineering Act: basis for PRC CE licensure requirements

Anchor Id

A20

Difficulty

easy

Memory Aid

Story: 'In 1950, Republic Act 544 was born — the legal BACKBONE of Philippine civil engineering. It said: only LICENSED engineers may sign and seal structural designs (including RC beams!). The PRC board exam tests NSCP 2015 formulas precisely because RA 544 mandates competence before licensure. Every time you compute φMn and check steel ratios, you are performing the exact task that RA 544 says only a licensed CE can legally certify. Honor the law — master the formula.'

Anchor Type

micro_story

Why It Works

Connecting the technical formula work to the legal framework (RA 544) gives emotional weight and professional significance to what might otherwise feel like abstract number-crunching.

Example Usage

Board exam or interview: 'What law governs CE practice in the Philippines?' RA 544 — it mandates licensure for structural design sign-and-seal authority.

Recall Trigger

Think: RA 544 — the law that makes this formula work matter professionally

Revision Game

Stress block depth a = As·fy / (0.85·f'c·b)

Clue

I am the height of the Whitney stress block. I am computed by dividing the steel's total force by the concrete's resistance per unit width. What am I?

Memory Link

A1 — Sari-sari store counter: stack height = steel selling force ÷ counter carrying capacity

β₁ (beta-one), the stress-block factor

Clue

I am the factor that equals 0.85 when f'c ≤ 28 MPa, and I get smaller as concrete gets stronger. I have a minimum value of 0.65. What am I?

Memory Link

A5 — 'Beta starts high, gets shy — 85, shy by 5, floor at 65'

εt = 0.005

Clue

I am the minimum net tensile strain that a beam must achieve to be called 'tension-controlled' and earn φ = 0.90. What am I?

Memory Link

A3 — '5 for full pay' — five thousandths earns you φ = 0.90

ρmax = 0.85·β₁·(f'c/fy)·0.375

Clue

I am the upper steel ratio limit for tension-controlled beams per NSCP 2015. I involve the fraction 0.375 — which is 3/8 — derived from the strain compatibility condition at εt = 0.005. What formula gives me?

Memory Link

A7 — 'Three-eighths is my max!'

Rn = Mu / (φ·b·d²), the coefficient of resistance

Clue

I am the starting step in ANY beam design problem. I normalize the factored moment demand to the beam's cross-sectional capacity. I have units of MPa. What am I?

Memory Link

A9 — 'Land price per sqm' — Rn = Mu ÷ (φ × floor area)

If the stress-block depth a ≤ tf (flange thickness), treat as rectangular beam of width bf

Clue

I am the condition that allows you to treat a T-beam as a simple rectangular beam. I compare a calculated quantity to the flange thickness. Describe me.

Memory Link

A14 — 'Check the bun' — if compression stays in the bun (flange), the hamburger is rectangular

Republic Act 544 (RA 544) — the Philippine Civil Engineering Act of 1950

Clue

I am the Philippine law that makes the entire content of this chapter professionally significant. Without me, signing RC beam designs would be illegal. What law am I, and what year?

Memory Link

A20 — RA 544 story: the legal backbone of Philippine CE practice

Doubly reinforced beam — two internal couples: concrete-steel couple plus compression steel-tension steel couple

Clue

I am the reason compression steel (A's) is added to a beam. I happen when the factored moment exceeds the tension-controlled capacity of a singly reinforced section. What configuration do I create?

Memory Link

A15 — Overloaded jeepney: add a second engine (compression steel) when singly reinforced can't handle the load

0.85f'c/fy = concrete-to-steel stress ratio; 2Rn/(0.85f'c) = normalized demand ratio; [1−√(…)] = the 'consumed' fraction of concrete capacity; ρ = required steel ratio

Assume rectangular beam, width = bf (full flange)
Compute a = As·fy/(0.85·f'c·bf)
Compare a with tf (flange thickness)
If a ≤ tf: DONE — treat as rectangular beam of width bf
If a > tf: must use T-beam formula accounting for web

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CELE Reinforced & Prestressed Concrete — Reinforced Concrete Beams: FlexureMemory Anchors

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Reinforced Concrete Beams: Flexure - Memory Anchors

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