Reinforced Concrete Footings, Bond and Development - Memory Anchors
Instead of re-reading the same formulas repeatedly, you attach each concept to a vivid image, story, or sound pattern already wired into your brain. For the PRC Civil Engineer board exam, where you must juggle dozens of formulas under time pressure, these anchors act like mental shortcuts — one trigger word unlocks the entire formula, procedure, or concept. Use these anchors during your review sessions, then test yourself with the quick-recall chains and revision game. The goal: see 'punching shear' on the board exam and instantly visualize the right formula without hesitation.
Memory anchors
Footing sizing uses SERVICE load; structural design uses FACTORED load
Memory aid
Think of buying a dining table (footing). You choose the table size based on how many guests normally come (service load = normal operation). But when you actually bolt the table legs to the floor (structural design = factored load), you use the maximum possible force so the bolts never fail. Two different jobs: CHOOSING SIZE vs. MAKING IT STRONG.
Why it works
The analogy maps a familiar Filipino household scenario (preparing for a family gathering) onto a two-step engineering process, making the distinction between service and factored loads intuitive.
Example usage
When the exam gives a column load and asks for footing size, use P_service / q_a. When it asks for shear or flexure, switch to P_u. Remember: SIZE = service, STRENGTH = factored.
Recall trigger
Dining table purchase → two steps: pick the size, then bolt it down
Footing plan area formula: A_req = P_service / q_a
Memory aid
Remember 'A = P over Q' as 'Apat na Piso sa Queso' (Four Pesos for Cheese). Area equals P (load) divided by Q (allowable pressure). The word 'queso' reminds you of q_a, the allowable bearing pressure of the soil — the soil's 'price per square meter.'
Why it works
Tagalog food reference creates a culturally vivid mental image. 'Queso' sounds exactly like 'q' and anchors the denominator as the soil's 'cost per area.'
Example usage
Given P_service = 1200 kN, q_a = 200 kPa → A = 1200/200 = 6.0 m². Side B = √6 = 2.45 m → round up to 2.5 m.
Recall trigger
'Queso' → q_a → A = P/q_a
Two-way punching shear critical perimeter: b_o = 4(c + d) for a square column
Memory aid
Imagine the column is a boxer's fist punching downward through the footing. The punch doesn't just hit the column footprint — it spreads out by d/2 on all four sides before it breaks. So you draw a square frame, each side = (c + d), and the total perimeter of that frame = 4(c + d). The column is the fist; the footing is the face; the critical perimeter is the bruise outline.
Why it works
The boxing image makes 'punching shear' literal — the fist creates a square bruise outline around the column, etching the formula into spatial memory.
Example usage
Column c = 400 mm, d = 500 mm → b_o = 4(400 + 500) = 4(900) = 3600 mm.
Recall trigger
Boxer's fist punching through concrete → square bruise → b_o = 4(c + d)
Punching shear capacity: V_c = 0.33 λ √f'c · b_o · d (governing for square columns)
Memory aid
Remember the coefficient as '0.33' by saying 'One-Third Punch.' 0.33 ≈ 1/3. The formula chunks as: ONE-THIRD × LAMBDA × ROOT-F'C × PERIMETER × DEPTH. Say it out loud: 'One-third lambda root-fc B-naught d.' The 'one-third' sticks because punching uses 0.33 (roughly 1/3) while beam shear uses 0.17 (roughly 1/6) — punching is TWICE as strong per unit because the slab wraps around on all sides.
Why it works
Chunking the formula into spoken syllables creates an auditory memory trace. The contrast with 0.17 (beam shear) reinforces the difference between the two shear types.
Example usage
V_c = 0.33(1.0)√28 × 3600 × 500 = 0.33 × 5.292 × 1,800,000 = 3,143,000 N = 3143 kN. Then φV_c = 0.75 × 3143 = 2357 kN.
Recall trigger
'One-Third Punch' → 0.33 λ √f'c b_o d
One-way beam shear capacity: V_c = 0.17 λ √f'c · B · d
Memory aid
Remember 0.17 as 'One-Way = One-Seven.' One direction, one seven: 0.17. The full formula: 0.17 λ √f'c × WIDTH × DEPTH. Vs punching (0.33), beam shear is HALF — because the slab only resists in one direction, not wrapped around all four sides.
Why it works
The word play 'one-way = one-seven (0.17)' creates a direct linguistic link. The comparison to punching (double the coefficient) embeds relational understanding.
Example usage
For a 3.0 m wide footing with d = 500 mm, f'c = 28 MPa: V_c = 0.17(1.0)√28 × 3000 × 500 = 0.17 × 5.292 × 1,500,000 = 1,349,460 N ≈ 1349 kN.
Recall trigger
'One-way, one-seven' → 0.17 λ √f'c B d
One-way shear critical section is at distance d from the column face
Memory aid
Think of a karate chop on a plank. You don't measure the breaking point AT the hand — you measure it ONE ARM-LENGTH (d) away from the hand. The concrete arch action carries shear diagonally from the support, so the diagonal crack forms at angle near 45°, landing at distance d from the face. The 'arm' of the karate master = effective depth d.
Why it works
The karate chop image is action-packed and memorable. It physically explains WHY the critical section is at d (the diagonal strut length), not at the face.
Example usage
Column face at x = 0, effective depth d = 550 mm → critical section for one-way shear is at x = 550 mm from column face.
Recall trigger
Karate chop → one arm-length away → critical section at d from face
Punching shear demand: V_u = q_u [A_footing − (c + d)²]
Memory aid
Imagine a trampoline (the footing) with a square hole cut in the middle where the column sits. The upward soil pressure pushes on the WHOLE trampoline, but the column handles the middle square (c+d by c+d). The part of the trampoline still pushing up and threatening to punch through is TOTAL AREA minus that MIDDLE SQUARE. That difference × q_u = the punching threat. 'The trampoline outside the hole fights the column.'
Why it works
The trampoline story makes the subtraction intuitive — students visualize the area outside the critical perimeter as the 'enemy' of punching resistance.
Example usage
q_u = 272 kPa, A = 6.25 m², (c+d)² = (0.4+0.5)² = 0.81 m² → V_u = 272(6.25 − 0.81) = 272 × 5.44 = 1480 kN.
Recall trigger
Trampoline with middle hole → V_u = q_u × (total − middle square)
Flexural critical section is at the FACE of the column (not at d from face)
Memory aid
Picture a diving board. The moment is largest at the ROOT of the diving board where it meets the pool deck — not somewhere along the board. The column face IS the root. The footing cantilevers out from the column face like a diving board cantilevering from the pool deck. Maximum bending = at the root = at the column face.
Why it works
The diving board is a universally understood cantilever. It viscerally shows that bending peaks at the fixed end (column face), not at some point away from it.
Example usage
For a 2.5 m footing with a 400 mm column: projection ℓ = (2500 − 400)/2 = 1050 mm = 1.05 m. M_u = q_u × B × ℓ²/2.
Recall trigger
Diving board → maximum bend at root → flexure critical section at column face
Flexural moment at column face: M_u = q_u · B · ℓ²/2
Memory aid
Say it as 'Moo = Q-Be-L-squared-half.' M (moo, like a cow) = q_u (Q) × B (Be) × ℓ² (L-squared) / 2 (half). The cow 'moos' at the column face because that's where the bending is greatest. Picture a cow standing at the column face, mooing loudly. Every time you see 'moment at column face,' you hear the cow.
Why it works
The silly cow image (Filipino students often see cows in the provinces) creates a phonetic anchor. 'Moo' = M_u; 'Q-Be' = q_u × B; 'half' = /2.
Example usage
q_u = 272 kPa, B = 2.5 m, ℓ = 1.05 m → M_u = 272 × 2.5 × 1.05²/2 = 272 × 2.5 × 0.5513 = 375 kN·m.
Recall trigger
Cow at column face mooing → M_u = q_u · B · ℓ²/2
Net factored soil pressure: q_u = P_u / A_footing
Memory aid
Rhyme it: 'P-sub-u on Area true, gives you q_u — the pressure that's due.' Short, rhythmic, and direct. q_u is what the soil pushes back up with after you've loaded it with P_u. It's the factored upward pressure per square meter of footing.
Why it works
Rhymes create phonological loops in working memory, making the formula retrievable through sound even under exam stress.
Example usage
P_u = 1700 kN, A = 2.5² = 6.25 m² → q_u = 1700/6.25 = 272 kPa (upward factored soil pressure).
Recall trigger
'P on A gives q_u' rhyme
Development length formula for large bars (>20 mm): ℓ_d = [f_y · ψ_t · ψ_e / (1.7 λ √f'c)] · d_b
Memory aid
Remember the coefficient '1.7' for large bars using: 'BIG bars need MORE time to embed — 1.7.' And recall the full formula with 'FED': F = f_y (yield strength of bar), E = ψ (modification factors, psi looks like E), D = d_b (bar diameter). 'FED into the concrete over length 1.7.' Big bars use 1.7; small bars (≤20 mm) use 2.1 (higher coefficient = MORE development for same bar size? No — higher denominator = SHORTER length. Big bars → 1.7 → LONGER ℓ_d. Small bars → 2.1 → shorter ℓ_d because smaller bars grip better per diameter.
Why it works
The 'FED' acronym chunks the numerator. The size comparison (1.7 vs 2.1) inverts the intuition, making students think carefully — and that cognitive effort improves retention.
Example usage
25 mm bar, f_y = 415 MPa, f'c = 28 MPa, ψ_t = ψ_e = 1.0, λ = 1.0: ℓ_d = [415 × 1 × 1 / (1.7 × 1 × √28)] × 25 = [415/8.996] × 25 = 46.1 × 25 = 1153 mm ≥ 300 mm ✓
Recall trigger
'FED into concrete' → ℓ_d = [f_y · ψ_t · ψ_e / (1.7 λ √f'c)] · d_b for large bars
Development length formula for small bars (≤20 mm): coefficient is 2.1 (shorter ℓ_d)
Memory aid
Think of two people gripping a rope: a child (small bar, ≤20 mm) has tiny hands but grips RELATIVELY more surface per diameter — so it needs less rope length to hold on. An adult (large bar, >20 mm) has big hands but proportionally less grip per diameter — needs more rope. Small bar → 2.1 denominator → shorter ℓ_d. Large bar → 1.7 denominator → longer ℓ_d. The bigger the bar, the more concrete it needs to grip.
Why it works
The rope-gripping analogy maps relative bond strength to bar size in a physically intuitive way. The child/adult contrast is vivid and easy to remember.
Example usage
20 mm bar (≤20, so use 2.1): ℓ_d = [f_y · ψ_t · ψ_e / (2.1 λ √f'c)] × 20 mm. Note: 20 mm is the boundary — use 2.1.
Recall trigger
Child gripping rope (small bar, 2.1) vs adult (large bar, 1.7)
Minimum development length: ℓ_d ≥ 300 mm always
Memory aid
300 mm = 30 cm = roughly one school ruler. Picture a standard 30-cm ruler buried inside the footing alongside the bar. No matter how short your formula gives you, the bar must at least span one ruler length into the concrete. 'Always plant at least one ruler deep.'
Why it works
The ruler is a universal, physically graspable reference. Students literally 'see' the minimum length as a familiar object.
Example usage
Computed ℓ_d = 250 mm → not acceptable. Must use ℓ_d = 300 mm (one ruler). Always check the computed value against 300 mm floor.
Recall trigger
School ruler = 300 mm → minimum ℓ_d
ψ_t = 1.3 for top bars (horizontal bars with more than 300 mm of fresh concrete below during casting)
Memory aid
Story: 'The Top Bar feels lonely at the top. Concrete bleeds and settles BELOW it during pouring, leaving tiny voids under the bar. Those voids weaken bond. So the Top Bar must work 30% harder to compensate — it gets a penalty multiplier of 1.3.' Picture a bar with a tiny gap under it because all the wet concrete sank away. That 30% penalty = ψ_t = 1.3.
Why it works
The story anthropomorphizes the top bar and explains the physics (bleed water and settlement) in narrative form. Understanding WHY the factor exists makes it unforgettable.
Example usage
Top bar 20 mm, ψ_t = 1.3 → ℓ_d = [415 × 1.3 × 1.0 / (2.1 × 1.0 × √35)] × 20 = [539.5 / 12.42] × 20 = 43.44 × 20 = 869 mm.
Recall trigger
Top bar feeling lonely → concrete sank away → 30% penalty → ψ_t = 1.3
ϕ = 0.75 for shear (both punching and beam shear in footings)
Memory aid
Remember '0.75 for shear' with the phrase 'SHEAR = SEVEN-FIVE.' Shear → 0.75. Contrast: flexure (bending) uses ϕ = 0.90 ('NINETY for NICE bending'). Shear is less ductile and more unpredictable → lower ϕ. 'Seven-five for shear, ninety for bending, sixty-five for columns.'
Why it works
Rhyming the phi values with their magnitudes creates a simple lookup table in memory. The reason (ductility) reinforces understanding.
Example usage
V_c = 3143 kN (punching). ϕV_c = 0.75 × 3143 = 2357 kN. Check: ϕV_c = 2357 ≥ V_u = 1480 kN → OK.
Recall trigger
'Seven-five for shear' → ϕ = 0.75
Standard hooks reduce required development length where straight embedment is insufficient
Memory aid
A hook is like a fish hook in concrete. A straight bar is like a straight wire — it can pull out if not long enough. But bend the end into a hook (90° or 180°) and it catches the concrete like a fishhook catches a fish. When you don't have space for a full straight ℓ_d, throw a 'fishhook' on the end. The hook develops the bar mechanically, not just by bond.
Why it works
The fishhook analogy is universally understood and perfectly captures the mechanical anchorage principle of a standard hook.
Example usage
In a shallow footing where ℓ_d = 1153 mm but available concrete cover only gives 500 mm, specify a standard 90° hook to develop the 25 mm bar within the available length.
Recall trigger
Fishhook in concrete → hook anchorage → use when straight ℓ_d doesn't fit
Square footing side: B = √(P_service / q_a)
Memory aid
Rhyme: 'When the load lands on the ground, take the square root of P over Q to find B — that's the length all around.' B is each side of the square. Square root of area = side length. 'P over Q under the root — that's B, the boot!'
Why it works
Short rhymes activate phonological memory. The 'boot' (B = boot) is a quirky Filipino-English rhyme that sticks.
Example usage
P_service = 900 kN, q_a = 150 kPa → A = 900/150 = 6.0 m² → B = √6 = 2.449 → use 2.5 m × 2.5 m.
Recall trigger
'Square root of P over Q = B the boot' → B = √(P_service/q_a)
Two shear checks in footings: punching (two-way) AND beam (one-way) — both must pass
Memory aid
Story: 'Footing Inspector Mang Pedro has TWO checkpoints. First checkpoint: he walks around the column perimeter (two-way shear check). Second checkpoint: he walks across the full width at distance d from the column (one-way beam shear check). A footing can only open if it PASSES BOTH guards.' Mang Pedro is thorough — he never skips either checkpoint.
Why it works
The inspector/checkpoint narrative makes it memorable that BOTH checks are mandatory. Filipino students relate to the 'Mang' persona (a familiar authority figure).
Example usage
In any footing problem: Step 1 — check punching shear (ϕV_c ≥ V_u,punching). Step 2 — check beam shear (ϕV_c ≥ V_u,beam). Do NOT skip Step 2 even if Step 1 passes.
Recall trigger
Mang Pedro's two checkpoints → punching AND beam shear — both required
λ = 1.0 for normal-weight concrete; λ = 0.75 for lightweight concrete
Memory aid
Normal concrete is FULL WEIGHT — so lambda is FULL VALUE: λ = 1.0. Lightweight concrete is like using pumice (volcanic rock from Pinatubo!) instead of gravel — lighter, but weaker in tension → lambda drops to 0.75. Picture a heavy concrete block (λ = 1.0) and a Pinatubo pumice block (λ = 0.75). The Philippine volcanic reference makes it culturally vivid.
Why it works
The Pinatubo pumice reference is uniquely Filipino and scientifically accurate (pumice IS volcanic lightweight aggregate). It anchors the concept to local geography.
Example usage
Unless specifically stated as lightweight, always use λ = 1.0 in board exam problems. If the problem says 'lightweight concrete,' use λ = 0.75 in all shear and development length formulas.
Recall trigger
Pinatubo pumice = lightweight = λ = 0.75; regular gravel = normal weight = λ = 1.0
The three NSCP punching shear expressions — the 0.33 value governs for compact/square columns
Memory aid
Remember there are THREE formulas for punching V_c (NSCP/ACI 318) using '3-B-A': (1) Beta formula (involving β_c = column aspect ratio), (2) Bo formula (involving α_s and b_o), (3) Alpha-simple formula: V_c = 0.33√f'c · b_o · d. For square columns, β_c = 1.0, and the 0.33 form governs — it gives the SMALLEST result and is most critical. Always compute all three and use the minimum, but for square columns in board exams, 0.33 almost always controls.
Why it works
The '3-B-A' acronym (Three Formulas, Beta, Alpha-simple) provides a framework for remembering that three formulas exist and that 0.33 is typically the minimum/governing one.
Example usage
Square column (β_c = 1): all three NSCP expressions yield V_c ≥ 0.33√f'c · b_o · d, so 0.33 formula controls. Use it directly in board exam calculations.
Recall trigger
'3-B-A' → three punching formulas → 0.33 governs for square columns
Revision game
I am the boundary between service and factored. Cross me and you switch from sizing to strength. What am I?
Answer
The decision to use service load for footing area vs. factored load for shear and flexure design
Memory link
A1 — Dining table analogy: pick the size (service), then bolt it down (factored)
I am 0.33 × λ × √f'c × b_o × d. I resist the column trying to drop through the footing like a fist through paper. Name my check.
Answer
Two-way punching shear capacity (V_c for punching)
Memory link
A4 — 'One-Third Punch' chunking mnemonic
My critical section is exactly ONE effective depth away from the column face — not at the face, not at the edge. Which shear check am I?
Answer
One-way beam shear (the critical section for beam shear is at d from the column face)
Memory link
A6 — Karate chop: one arm-length (d) away from the column
I am a 30-centimeter school ruler buried in the concrete. No bar may be anchored in less length than my full body. Who am I?
Answer
Minimum development length of 300 mm (ℓ_d ≥ 300 mm always)
Memory link
A13 — The school ruler visual association
I am 1.3, and I apply to bars that were cast at the TOP while wet concrete settled and bled away below me, leaving voids that weakened my bond. Who am I?
Answer
Top bar modification factor ψ_t = 1.3
Memory link
A14 — 'Top bar feels lonely' micro-story
I am what you SUBTRACT from the total footing area to find the punching shear demand. I am a square of side (c + d). What is my area?
Answer
(c + d)² — the area inside the critical punching perimeter
Memory link
A7 — Trampoline with middle hole cut out
I am a fishhook in concrete. When the bar is too short to develop by bond alone, engineers bend my end so I grip the concrete mechanically. What am I called?
Answer
Standard hook (90° or 180° hook) for development/anchorage
Memory link
A16 — Fishhook in concrete analogy
I am the maximum product of two modification factors. No matter how bad a top epoxy-coated bar is, my value caps the combined penalty. What is my value?
Answer
ψ_t × ψ_e ≤ 1.7 (product cap for development length modification factors)
Memory link
Quick-recall chain 3: 'Even troublemakers have limits — capped at 1.7'
Formula mnemonics
A_req = P_service / q_a
Mnemonic
Queso Price: Area = Load divided by the Soil's Price (q_a). 'Apat na Piso sa Queso' — four pesos per queso = area per unit load.
When to use
FIRST step in any footing problem — before any shear or flexure calculation. Use service (unfactored) load only.
What each part means
A_req = required footing plan area (m²); P_service = total service (unfactored) column load (kN); q_a = allowable soil bearing capacity (kPa = kN/m²)
q_u = P_u / A_footing
Mnemonic
P on A gives q_u. Factored load on actual chosen area = net upward design pressure. 'P-you on A = Q-you.'
When to use
After sizing the footing, compute q_u for use in shear demand (V_u) and flexural moment (M_u) calculations.
What each part means
q_u = net factored upward soil pressure (kPa); P_u = factored column load (kN); A_footing = actual footing plan area chosen (m²)
b_o = 4(c + d)
Mnemonic
Boxer's Square Bruise: perimeter of the punching square = 4 sides × (column size c + depth d). 'Four sides of the punch.'
When to use
Two-way punching shear check only. Critical perimeter is located at d/2 from all column faces.
What each part means
b_o = critical punching perimeter (mm); c = column side dimension (mm); d = effective depth of footing (mm). The critical perimeter is at d/2 from column face on all four sides.
V_c = 0.33 λ √f'c · b_o · d (punching)
Mnemonic
'One-Third Punch': 0.33 ≈ 1/3. Multiply lambda, root-f'c, perimeter, depth. The 'wrapped-around' shear resistance — all four sides contribute.
When to use
Two-way punching shear capacity. Governs for square columns (β_c = 1.0). Apply φ = 0.75: check φV_c ≥ V_u.
What each part means
0.33 = punching shear coefficient (≈ 1/3); λ = concrete density factor (1.0 NW, 0.75 LW); f'c = concrete compressive strength (MPa); b_o = critical perimeter (mm); d = effective depth (mm). Result in N.
V_c = 0.17 λ √f'c · B · d (one-way beam shear)
Mnemonic
'One-Way One-Seven': 0.17. Multiply lambda, root-f'c, full width B, depth d. Half the punching coefficient because only one-directional resistance.
When to use
One-way shear check across a section at distance d from the column face. Apply φ = 0.75: check φV_c ≥ V_u.
What each part means
0.17 = beam shear coefficient (≈ 1/6); λ = concrete density factor; f'c = compressive strength (MPa); B = full footing width perpendicular to shear plane (mm); d = effective depth (mm). Result in N.
M_u = q_u · B · ℓ² / 2
Mnemonic
'Moo Q-Be-L-squared-Half': Cow at column face. M_u (moo) = q_u (Q) × B (Be) × ℓ² (L-squared) ÷ 2 (half).
When to use
Flexural design of footing reinforcement. Critical section is at the column face. ℓ is the cantilever arm.
What each part means
M_u = factored moment at column face (kN·m); q_u = factored upward soil pressure (kPa); B = footing width (m); ℓ = projection length from column face to footing edge = (B − c)/2 (m)
ℓ_d = [f_y · ψ_t · ψ_e / (1.7 λ √f'c)] · d_b (bars > 20 mm)
Mnemonic
'FED into concrete, 1.7 for big bars': F = f_y, E = ψ factors, D = d_b diameter. Big bars (>20 mm) use 1.7 in the denominator → longer ℓ_d.
When to use
Tension development length for bars larger than 20 mm diameter, with adequate clear spacing and cover. Minimum ℓ_d = 300 mm always applies.
What each part means
f_y = bar yield strength (MPa); ψ_t = top bar factor (1.3 if top bar, 1.0 otherwise); ψ_e = epoxy factor (1.5 if epoxy-coated, 1.0 otherwise); λ = density factor; f'c = compressive strength (MPa); d_b = bar diameter (mm). Always: ℓ_d ≥ 300 mm.
ℓ_d = [f_y · ψ_t · ψ_e / (2.1 λ √f'c)] · d_b (bars ≤ 20 mm)
Mnemonic
'Small bars, 2.1, shorter journey': ≤ 20 mm bars use 2.1 (larger denominator → shorter ℓ_d). Small child gripping rope needs less length.
When to use
Tension development length for bars ≤ 20 mm with adequate cover and spacing per NSCP/ACI 318. Minimum ℓ_d = 300 mm.
What each part means
Same variables as above. 2.1 coefficient applies ONLY when bar ≤ 20 mm diameter AND clear spacing ≥ d_b AND cover ≥ d_b. Otherwise use the less-favorable (smaller) coefficients 1.4 or 1.1.
Quick recall chains
Complete Footing Design Steps (7 Steps)
Recall test
Without looking, list the 7 steps of footing design in order. What load do you use for Step 1? What load for Steps 4–7?
Memory chain
Story: 'Architect Armand (A = area) Rounds up Quickly (q_u = P_u/A), then Punches the Board (punching shear), Beams at the Crowd (beam shear), Flexes His Muscles (flexure design), and Develops the Plan (development length).' A → Round → Q → Punch → Beam → Flex → Develop = 7 steps.
Items to remember
- 1. Compute required area: A = P_service / q_a
- 2. Round up footing dimensions (square: B × B)
- 3. Compute factored upward pressure: q_u = P_u / A
- 4. Check punching shear: V_u vs φV_c (0.33 formula)
- 5. Check beam shear: V_u vs φV_c (0.17 formula) at d from face
- 6. Design flexural steel: M_u at column face, solve for A_s
- 7. Check development length: ℓ_d available ≥ ℓ_d required
Key Phi Factors in Footing Design
Recall test
Quick: What is φ for punching shear? What is φ for flexural bar design in a footing? Name all three in order from largest to smallest.
Memory chain
Remember '90-75-65' as a basketball jersey numbers countdown: Star player wears 90 (best = flexure), starter wears 75 (shear), sub wears 65 (columns). 'Ninety, Seventy-Five, Sixty-Five — the Footing Basketball Team.'
Items to remember
- φ = 0.90 for flexure (bending/tension-controlled)
- φ = 0.75 for shear and torsion
- φ = 0.65 for compression-controlled (columns)
Development Length Modification Factors (ψ values)
Recall test
What is ψ_t for a bottom bar? What is the maximum product of ψ_t × ψ_e? When does ψ_e = 1.5 apply versus 1.2?
Memory chain
Story: 'The TOP student (ψ_t = 1.3) got extra credit. The EPOXY-COATED student (ψ_e = 1.5) cheated with a thick coat and got the maximum penalty. But the total penalty is CAPPED at 1.7 — even troublemakers have limits.' TOP = 1.3, EPOXY = 1.5 or 1.2, CAP = 1.7.
Items to remember
- ψ_t = 1.3 for top bars (more than 300 mm fresh concrete below)
- ψ_t = 1.0 for all other bars
- ψ_e = 1.5 for epoxy-coated bars with cover < 3d_b or spacing < 6d_b
- ψ_e = 1.2 for other epoxy-coated bars
- ψ_e = 1.0 for uncoated bars
- ψ_t × ψ_e ≤ 1.7 (product cap)
Two Shear Check Formulas Side by Side
Recall test
Fill in the blanks: Punching V_c = ___ × λ × √f'c × ___ × d. Beam V_c = ___ × λ × √f'c × ___ × d. What is φ for both?
Memory chain
Punch uses ZERO-POINT-THIRTY-THREE with b_o (the boxing ring perimeter). Beam uses ZERO-POINT-SEVENTEEN with B (the full width bridge). 'Boxing Ring vs Bridge.' Punch demand = total minus middle hole. Beam demand = upward load on the strip beyond the cut.
Items to remember
- Punching: V_c = 0.33 λ √f'c · b_o · d (coefficient 0.33, use b_o = 4(c+d))
- Beam: V_c = 0.17 λ √f'c · B · d (coefficient 0.17, use full width B)
- Punching demand: V_u = q_u [A − (c+d)²]
- Beam demand: V_u = q_u × B × [(B−c)/2 − d]
- Both use φ = 0.75
Critical Sections for Each Check
Recall test
Where is the critical section for punching shear? For one-way shear? For flexure? All measured from which reference point?
Memory chain
Distance from column face for each check — HALF d, FULL d, ZERO, and ALL THE WAY to the end. Think of a runner: 'Half-step (d/2) punch, Full-step (d) beam, At the start line (0) flex, Run to the end (ℓ_d) to develop.' Half → Full → Zero → End.
Items to remember
- Punching shear: at d/2 from column face (all four sides)
- Beam shear: at d from column face (one plane across width)
- Flexure: at column face (zero distance from face)
- Development: from the point of maximum stress (column face) to the bar end
Diagrams
Reinforced Concrete Footings: Complete Concept Map
A hierarchical mind map showing how all topics in the chapter — sizing, design pressure, punching shear, beam shear, flexure, and development length — connect to each other. Use this as a mental framework before drilling individual formulas.
Step-by-Step Footing Design Procedure (Board Exam Decision Flowchart)
A decision flowchart for the complete footing design process. Shows the sequence of checks, pass/fail decision points, and remedial actions. Mirrors the board exam problem structure exactly. The two shear checks (Steps 4 and 5) are highlighted as sequential mandatory gates — both must pass before proceeding to flexure and development length design.
Development Length Selection Guide
A decision flowchart for selecting the correct development length formula and modification factors. Guides the student through bar size selection (1.7 vs 2.1 coefficient), top bar factor (ψ_t), epoxy factor (ψ_e), product cap check (1.7 maximum), and the 300 mm minimum floor. Use this as a checklist during board exam problems involving development length.
Distribution of Footing Design Checks by Board Exam Frequency
A pie chart showing the approximate relative frequency of each footing design topic in PRC board exam questions based on historical patterns. Punching shear and development length together account for over half of footing-related board exam items, making them the highest-priority topics for review.