Reinforced Concrete Beams: Shear and Torsion - Memory Anchors
When preparing for the PRC Civil Engineer board exam, you are not just memorizing formulas; you are building a mental scaffolding that lets you retrieve the right equation under exam pressure within seconds. The anchors in this module attach each key concept to something you already know emotionally or visually, creating a durable 'hook' in long-term memory. Use these tools actively: say them out loud, sketch the diagrams, and test yourself with the recall chains before every mock exam.
Memory anchors
Total shear capacity Vn = Vc + Vs (concrete + stirrups)
Memory aid
Think of shear resistance like a barangay tanod team guarding a gate. The concrete (Vc) is the permanent guard — always on duty, rain or shine. The stirrups (Vs) are the volunteer tanods called in during an emergency (high shear). Together, they secure the gate. If the emergency is too big (Vu too large), you call in more volunteers or build a bigger gate (enlarge the section).
Why it works
The 'permanent guard + volunteer' analogy maps directly to the additive nature of Vn = Vc + Vs, making the two-component model intuitive and hard to forget.
Example usage
When asked 'What is the total nominal shear strength?', visualize the gate: Vn = Vc (permanent guard) + Vs (volunteer tanods).
Recall trigger
Barangay tanod gate guard
Strength reduction factor φ = 0.75 for shear and torsion
Memory aid
SHEAR = 3/4. Remember: '3-quarters for shear.' Alternatively, think of a 75 centavo coin — shear always pays at 75% of face value. If you ever mix this up with flexure (φ = 0.90), remember shear is MORE dangerous (brittle failure), so the code is MORE conservative — it pays you LESS (75 centavos, not 90).
Why it works
Anchoring φ = 0.75 to a 75-centavo coin makes it tangible and distinguishes it from the flexure phi-factor through a logical safety argument.
Example usage
When computing φVc, immediately picture a 75-centavo coin: φ = 0.75. So φVc = 0.75 × Vc.
Recall trigger
75 centavo coin
Vc = 0.17 λ √f'c bw d (NSCP 2015 simplified formula)
Memory aid
REMEMBER: '0.17 is the concrete's birthday — January 7 (1/7 ≈ 0.143... no, but 0.17 sounds like Oh-seventeen).' More precisely, use the phrase: 'Point-one-seven LABS' — Lambda, Area-of-web (bw), Bottom-distance (d), Square-root-of-f'c. The order in the formula is: constant (0.17) → lambda → root f'c → bw → d.
Why it works
The acronym LABS (Lambda, Area-web, Bottom-d, Square-root-f'c) gives the four multipliers in order after 0.17, preventing term omissions during calculation.
Example usage
Exam asks for Vc with bw=300 mm, d=500 mm, f'c=28 MPa, normal weight: Vc = 0.17 × 1.0 × √28 × 300 × 500. Recall LABS to confirm all four terms are included.
Recall trigger
0.17 LABS
λ = 1.0 for normal-weight concrete, λ < 1.0 for lightweight
Memory aid
Lambda (λ) is like a loyalty discount for concrete quality. Normal-weight concrete gets FULL loyalty points: λ = 1.0 (no discount on strength). Lightweight concrete gets a discount because it is weaker in tension — so λ is reduced below 1.0. A lighter material = a lighter lambda.
Why it works
The loyalty discount analogy captures the direction of the effect (lighter → lower λ) without requiring memorization of numerical values for lightweight.
Example usage
Problem says 'normal-weight concrete' — full loyalty, λ = 1.0. Problem says 'all-lightweight' — apply a discount, λ < 1.0 (check NSCP table for exact value).
Recall trigger
Loyalty discount for concrete weight
Critical section for shear is at distance d from the support face, not at the support itself
Memory aid
Imagine a tricycle driver delivering goods. The moment he leaves the terminal (support face), he travels a distance d before he can be stopped by a traffic enforcer (shear crack). Checking shear AT the terminal is like fining him before he even moves — not fair, and not how the code works. NSCP says: check shear at distance d FROM the terminal face.
Why it works
The micro-story creates a spatial journey (terminal → d → enforcer check), embedding the concept of 'd from support face' as a logical process, not an arbitrary rule.
Example usage
If a support reaction is at x=0, the design shear Vu is taken at x = d (the effective depth) from the support face, not at x = 0.
Recall trigger
Tricycle leaving the terminal
Stirrup spacing formula: s = Av fy d / Vs
Memory aid
Use the phrase 'AFD over S' backward: 'S equals AFD/Vs.' Remember the word 'AFFORDS' — A stirrup AFFORDS spacing: s = A(v) × F(y) × D / V(s). Visualize: the stirrup can 'afford' to be spaced further apart when Av, fy, or d is large, but must be spaced closely when Vs demand is high.
Why it works
The word AFFORDS encodes Av, fy, d in the numerator and creates a cause-and-effect mental model for why larger values of each term allow wider spacing.
Example usage
Given Av=157 mm², fy=415 MPa, d=500 mm, Vs=198,400 N: s = (157)(415)(500)/198,400 = 164 mm. Recall AFFORDS: A×F×D on top, Vs on bottom.
Recall trigger
AFFORDS spacing
Vs must not exceed 0.66 √f'c bw d (web crushing limit)
Memory aid
Stirrups are like jeepney passengers — you can load more, but there's a maximum load limit (66) before the jeepney chassis (concrete web) collapses. The limit Vs ≤ 0.66 √f'c bw d is the chassis rating. Exceed it and you don't add more stirrups — you BUILD A BIGGER JEEPNEY (enlarge the beam section).
Why it works
The overloaded jeepney analogy captures both the limit and the correct response (enlarge section, not add stirrups), preventing a common design error.
Example usage
If computed Vs = 600 kN and 0.66√f'c bw d = 524 kN, the section is INADEQUATE — you must enlarge bw or d, not simply add more stirrups.
Recall trigger
Overloaded jeepney chassis — 0.66 limit
Maximum stirrup spacing: d/2 or 600 mm (when Vs ≤ 0.33 √f'c bw d), halves to d/4 or 300 mm when Vs > 0.33 √f'c bw d
Memory aid
Remember the NUMBER PAIRS: '2–600 becomes 4–300 when you cross the HALF-WAY LINE.' The halfway line is 0.33 √f'c bw d (which is exactly HALF of the crushing limit 0.66). Crossing it HALVES your maximum spacing. Visualize a road speed limit: below 50 kph you can space checkpoints 600 m apart; above 50 kph (more dangerous), you halve the spacing to 300 m and cut the fraction from d/2 to d/4.
Why it works
The 'doubling/halving' symmetry is memorable: 600→300 and d/2→d/4 are both halved at the same threshold, so one number (HALF) unlocks both relationships.
Example usage
Compute Vs; compute 0.33√f'c bw d. If Vs is below this value: smax = min(d/2, 600 mm). If Vs is above: smax = min(d/4, 300 mm).
Recall trigger
Cross the half-way line, halve everything
Minimum stirrups required when Vu > 0.5 φ Vc
Memory aid
Picture a thermometer with 'φVc' as the boiling point. At HALF the boiling point (0.5 φVc), the 'warning light' turns on — minimum stirrups are MANDATORY. Below 0.5 φVc, the thermometer is in the green zone and stirrups are optional (except in specific member types). Above φVc, you compute Vs and design fully. The thermometer has three zones: Green (no stirrups needed) → Yellow at ½ mark (minimum stirrups) → Red above φVc (full design).
Why it works
The three-zone thermometer creates a visual gradient that maps directly to the three shear design regimes, making the transition points impossible to confuse.
Example usage
Given Vu = 55 kN, φVc = 101 kN: Is Vu > 0.5 × 101 = 50.5 kN? Yes (55 > 50.5), so minimum stirrups are REQUIRED even though Vu < φVc.
Recall trigger
Three-zone thermometer at φVc
Minimum Av = 0.062 √f'c (bw s / fy) ≥ 0.35 (bw s / fy)
Memory aid
MINIMUM STIRRUPS have TWO FLOORS: the concrete floor (0.062 √f'c bw s/fy) and the absolute floor (0.35 bw s/fy). Think of a building with two minimum floor heights — you must satisfy BOTH and use the LARGER value. The 0.062 floor depends on f'c (material quality), while 0.35 is the absolute ground floor that never changes.
Why it works
The 'two-floor building' analogy captures the dual-limit nature and the requirement to take the larger of the two, a common source of error in board exams.
Example usage
For f'c = 28 MPa: 0.062√28 = 0.328 < 0.35. So the controlling minimum is 0.35 bw s / fy (the absolute floor governs).
Recall trigger
Two-floor minimum building: 0.062 floor vs 0.35 floor
A U-stirrup has TWO legs, so Av = 2 × (single bar area)
Memory aid
A U-stirrup looks like the letter U — it crosses the crack with its LEFT leg and RIGHT leg. Always count both. Picture two legs of a person straddling a crack in the floor. If you only count one leg, the person falls (design fails). Av = 2 × Ab for a standard U-stirrup.
Why it works
The mental image of straddling legs anchors the 'two-leg' rule viscerally. This is the single most common arithmetic error in stirrup design problems.
Example usage
10 mm dia stirrup: Ab = π/4 × 10² = 78.5 mm². Av = 2 × 78.5 = 157 mm². NEVER use 78.5 directly in the spacing formula.
Recall trigger
Two legs straddling a crack
Vs = (Vu − φVc) / φ — divide the WHOLE expression by φ, not just Vu
Memory aid
A board examinee named 'Cardo' always forgets to divide by φ. He writes Vs = Vu − φVc. His professor (also named φ = 0.75) says: 'Cardo, I am in BOTH terms — I control the whole bracket (Vu − φVc), not just my own term!' So the correct formula is Vs = (Vu − φVc) / φ. Every time you see that formula, remember: 'Don't be like Cardo.'
Why it works
The story gives the error a face and a name, creating emotional memory. 'Don't be like Cardo' becomes a self-monitoring prompt during exams.
Example usage
Vu = 250 kN, φVc = 101.2 kN: Vs = (250 − 101.2) / 0.75 = 198.4 kN. NOT Vs = 250 − 101.2 = 148.8 kN (Cardo's mistake).
Recall trigger
Don't be like Cardo
Torsion threshold formula: Tu < φ (0.083) λ √f'c (Acp² / pcp)
Memory aid
For torsion, remember the phrase: 'ACP-squared over PCP — if Tu is BELOW, you can SLEEP.' The threshold uses the OUTSIDE dimensions (Acp = full cross-section area, pcp = full outside perimeter). The constant is 0.083, and it is multiplied by φ = 0.75, λ, and √f'c. Recall: '0.083 is like a stop-sign for torsion — if you haven't reached it, don't design for torsion.'
Why it works
The rhyme 'ACP-squared over PCP — if below, you sleep' encodes the formula components and the decision outcome simultaneously.
Example usage
Compute φ × 0.083 × λ × √f'c × Acp²/pcp. If Tu < this value, skip torsion design. If Tu ≥ this value, design closed stirrups and longitudinal bars for torsion.
Recall trigger
ACP over PCP — below it, sleep (no torsion design needed)
Torsion uses OUTSIDE perimeter (Acp, pcp) for threshold, NOT stirrup centerline
Memory aid
Torsion threshold is measured from the OUTSIDE of the building (the property line), not from the inside wall (stirrup centerline). When the whole building is in danger of twisting, you measure its FULL footprint, not just the rooms inside. Acp = full cross-section area, pcp = outer perimeter = 2(b+h). This is the opposite of the torsion design itself (which uses Aoh, the stirrup-enclosed area).
Why it works
The 'property line vs. interior wall' analogy creates a clear spatial distinction between threshold (outside = Acp, pcp) and design (inside = Aoh, ph), preventing formula mix-ups.
Example usage
For a 300×500 mm beam: Acp = 300×500 = 150,000 mm², pcp = 2(300+500) = 1,600 mm. Use these FULL outer dimensions in the threshold formula only.
Recall trigger
Property line (outside) for threshold; interior wall (inside) for design
Torsion requires CLOSED stirrups + longitudinal bars (not open U-stirrups)
Memory aid
Imagine a bottle of water being twisted. If the cap is open (U-stirrup), the water spills — torsion escapes. You need a CLOSED cap (closed stirrup) to contain the twisting stress. Add longitudinal bars along the full height to 'brace' the bottle at every corner. Open stirrups = open bottle = torsion fails.
Why it works
The closed-bottle analogy captures the structural logic: torsion creates a circulating shear flow around the cross-section, which requires a closed path (closed stirrup) to be effective.
Example usage
When Tu exceeds the threshold, switch from open U-stirrups (shear only) to CLOSED stirrups, and add longitudinal torsion reinforcement along the perimeter.
Recall trigger
Closed bottle cap — closed stirrups for torsion
Shear failure is brittle and sudden; hence φ = 0.75 (more conservative than flexure)
Memory aid
Think of flexural failure as a slowly sinking ship — you see cracks forming gradually, have time to evacuate. Shear failure is a lightning strike — no warning, immediate collapse. Because shear is the 'lightning strike' failure mode, the code punishes it harder: φ = 0.75 (only 75 cents on the peso) versus flexure's φ = 0.90 (90 cents). The more dangerous the failure mode, the smaller the phi-factor.
Why it works
The contrast between 'sinking ship' (ductile, slow) and 'lightning strike' (brittle, sudden) encodes the WHY behind the phi-factor hierarchy, not just the numbers.
Example usage
Flexure: φ = 0.90 (gradual warning). Shear/Torsion: φ = 0.75 (sudden failure). Always apply the correct phi before computing design capacity.
Recall trigger
Lightning strike vs. sinking ship
Shear design checklist sequence: Compute Vc → Check 0.5φVc → Compute Vs → Check Vs,max → Compute spacing s → Check smax
Memory aid
Use the acronym 'CV-CV-SS' — Compute-Vc, Check-0.5φVc, Compute-Vs, check-Vs,max (Vs safety check), Spacing-s, Smax-check. Or remember: 'Very Cool Civil Volunteers Solve Systematically.' Each word's first letter: V(c) C(heck) C(ompute) V(s) S(pacing) S(max).
Why it works
Sequential acronyms prevent step-skipping under exam time pressure. 'Very Cool Civil Volunteers Solve Systematically' creates a sentence that is easy to recall even when anxious.
Example usage
When solving any stirrup design problem, mentally run through: Vc → 0.5φVc threshold → Vs → 0.66 limit → s formula → smax check. Tick off each step.
Recall trigger
Very Cool Civil Volunteers Solve Systematically
The constant 0.17 in the Vc formula comes from the ACI empirical derivation (approximately 1/6 in USC units converted to SI)
Memory aid
Chunk the Vc formula into three pieces: [0.17 λ √f'c] × [bw] × [d]. The first chunk [0.17 λ √f'c] is the 'material strength chunk' — it depends only on what the concrete IS. The second [bw] is the 'width chunk' — how wide the web is. The third [d] is the 'depth chunk' — how deep the beam is. Always compute the material chunk first, then multiply by area (bw × d).
Why it works
Chunking the three-term multiplication into a logical sequence (material × width × depth) reduces cognitive load and makes unit checking easier.
Example usage
f'c=28 MPa: [0.17×1.0×√28] = [0.17×5.292] = [0.900 MPa material chunk] × 300 mm × 500 mm = 134,933 N = 134.9 kN.
Recall trigger
Material × Width × Depth chunks
When Vs > 0.33 √f'c bw d, reduce max spacing to d/4 (≤ 300 mm)
Memory aid
Rhyme: 'When Vs is HIGH — 0.33 is WHY — d/4 and three-hundred apply!' When Vs exceeds the halfway threshold (0.33 × √f'c × bw × d), you've crossed into high-shear territory. The maximum stirrup spacing is cut in HALF: from d/2 down to d/4, and from 600 mm down to 300 mm. High shear → half the space.
Why it works
The rhyme provides a rhythmic recall cue, and the 'half the space' rule is logically consistent (both d/2→d/4 and 600→300 are halvings), making the rule self-reinforcing.
Example usage
Vs=300 kN; 0.33√28×300×500 = 261.7 kN. Since 300 > 261.7, spacing is reduced: smax = min(d/4, 300 mm) = min(125, 300) = 125 mm.
Recall trigger
High Vs = half the spacing
Torsion and shear use combined closed stirrups — torsion adds to the shear stirrup demand
Memory aid
Think of a sari-sari store with two delivery schedules: shear deliveries and torsion deliveries. The same truck (closed stirrups) handles BOTH deliveries. When torsion is significant, the truck makes MORE trips (closer stirrup spacing or larger Av). You combine the two demands — you don't get two separate trucks. One closed stirrup resists both shear and torsion when properly sized.
Why it works
The 'one truck, two deliveries' analogy captures the combined design approach without requiring formula memorization at the introduction level, reinforcing the concept of combined action.
Example usage
When torsion Tu exceeds the threshold, increase closed stirrup area to carry BOTH Av (shear) and At (torsion) components. The stirrup does double duty.
Recall trigger
One truck (closed stirrup) for two deliveries (shear + torsion)
Revision game
I am the formula that tells you how much shear the plain concrete can carry, before any stirrup lifts a finger. I have four multipliers after a constant of 0.17. Who am I?
Answer
Vc = 0.17 λ √f'c bw d
Memory link
Anchor A3 — 0.17 LABS mnemonic
I am the phi-factor for shear. Think of me as 75 centavos — you get less than a peso because shear failure is dangerous and sudden. What is my value?
Answer
φ = 0.75
Memory link
Anchor A2 — 75 centavo coin analogy
Cardo forgot to include me in his Vs formula. Without me in the denominator, his stirrup demand is underestimated and the beam is unsafe. Who am I?
Answer
The second φ = 0.75 in Vs = (Vu − φVc) / φ
Memory link
Anchor A12 — Don't be like Cardo micro-story
I am the maximum value of Vs. Exceed me and no amount of stirrups can save the beam — you must build a bigger section. My constant is 0.66. What is my full expression?
Answer
Vs,max = 0.66 √f'c bw d
Memory link
Anchor A7 — Overloaded jeepney chassis analogy
I am the boundary that divides normal stirrup spacing from tighter stirrup spacing. Cross me and your maximum spacing is halved — from d/2 to d/4, and from 600 mm to 300 mm. What is my value in terms of beam properties?
Answer
0.33 √f'c bw d
Memory link
Anchor A8 — Cross the halfway line; Anchor A19 — rhyme
A U-stirrup crosses a shear crack with both of my arms. Count them — if you count only one, your stirrup design is 50% under-reinforced! How many legs does a standard U-stirrup contribute to Av?
Answer
Two legs; Av = 2 × Ab
Memory link
Anchor A11 — Two legs straddling a crack visual
I am the threshold below which torsion can be peacefully ignored. I use Acp and pcp — the OUTSIDE dimensions of the beam, not the stirrup centerline. Tu must be BELOW me for you to SLEEP. What is my formula?
Answer
Tu,threshold = φ × 0.083 × λ × √f'c × (Acp² / pcp)
Memory link
Anchor A13 — ACP over PCP sleep test; Anchor A14 — property line analogy
I am the location where shear is checked on a simply supported beam. I am NOT at the centerline of the support, and NOT at the face of the support either — I am a distance d AWAY from the face. Where am I?
Answer
At distance d from the support face (the critical section for shear)
Memory link
Anchor A5 — Tricycle leaving the terminal micro-story
Formula mnemonics
Vn = Vc + Vs
Mnemonic
TOTAL = CONCRETE + STEEL. Think of it as your total budget: concrete is your savings (always available), stirrups are your emergency fund (activated when needed). Together they form the total budget (Vn).
When to use
Always — this is the fundamental shear capacity equation. Apply φ to get design capacity: φVn ≥ Vu
What each part means
Vn = nominal shear strength (total capacity); Vc = concrete contribution (from aggregate interlock, dowel action, compression zone); Vs = steel stirrup contribution
Vc = 0.17 λ √f'c bw d
Mnemonic
0.17 LABS — Lambda, Area-web, Bottom-depth, Square-root-f'c. Compute the four factors and multiply them by 0.17.
When to use
For non-prestressed beams under simplified method (NSCP 2015 Section 422). Always check units: N, MPa, mm.
What each part means
0.17 = empirical constant (SI units); λ = lightweight modification factor (1.0 for normal weight); √f'c = square root of compressive strength in MPa; bw = web width in mm; d = effective depth in mm. Result is in Newtons.
Vs = (Vu − φVc) / φ
Mnemonic
Don't be like Cardo — DIVIDE the whole bracket by φ. Vs = (Vu minus φVc) ALL divided by φ.
When to use
When Vu > φVc, i.e., stirrups are needed beyond the minimum. This gives the required stirrup contribution.
What each part means
Vs = required stirrup shear strength; Vu = factored shear at section (kN or N); φVc = reduced concrete capacity; φ = 0.75 (appears in BOTH the subtraction and the denominator)
s = Av fy d / Vs
Mnemonic
AFFORDS spacing — A stirrup AFFORDS wider spacing when Av, fy, or d is large. s = A(v)×F(y)×D ÷ V(s).
When to use
After computing Vs, use this to find the required spacing. Then compare to smax and use the smaller value.
What each part means
s = stirrup spacing (mm); Av = total stirrup area crossing the crack (mm², = 2×Ab for U-stirrup); fy = stirrup yield strength (MPa); d = effective beam depth (mm); Vs = required stirrup shear force (N)
Vs,max = 0.66 √f'c bw d
Mnemonic
The 66 Chassis Limit — if Vs exceeds 0.66 √f'c bw d, the concrete web crushes. You cannot add more stirrups; you must enlarge the section. '66 is the overload limit.'
When to use
Immediately after computing Vs. If Vs exceeds this limit, STOP — enlarge the beam before proceeding.
What each part means
0.66 = upper-bound constant (= 4× the lower band constant of approximately 0.17); √f'c = concrete tensile indicator; bw d = shear area of the web. Units: N (if mm and MPa used consistently).
Av,min = max(0.062 √f'c × bw s / fy , 0.35 × bw s / fy)
Mnemonic
Two Floors of Minimum — the 0.062 concrete floor and the 0.35 absolute floor. Always use the HIGHER floor (larger Av,min).
When to use
When Vu > 0.5φVc but stirrups are not fully designed. Also check that your computed Av ≥ Av,min for any spacing chosen.
What each part means
0.062 √f'c bw s/fy = f'c-dependent minimum (governs for higher-strength concrete); 0.35 bw s/fy = absolute minimum (governs for low f'c when 0.062√f'c < 0.35, i.e., when f'c < 31.9 MPa approximately)
Tu,threshold = φ × 0.083 × λ × √f'c × (Acp² / pcp)
Mnemonic
ACP over PCP — if Tu is BELOW, you SLEEP (no torsion design). Constant = 0.083, same φ = 0.75, same λ. Use OUTSIDE dimensions only.
When to use
Given a factored torque Tu, compute this threshold. If Tu < threshold → ignore torsion. If Tu ≥ threshold → design for combined shear and torsion with closed stirrups and longitudinal bars.
What each part means
φ = 0.75; 0.083 = empirical SI constant; λ = lightweight factor; √f'c in MPa; Acp = gross area of entire cross-section (mm²); pcp = outer perimeter of entire cross-section (mm). Result in N·mm — convert to kN·m by dividing by 10⁶.
Quick recall chains
Shear Design Step-by-Step (The 6-Step Chain)
Recall test
Without looking: What are the 6 steps of stirrup design in order? Can you state what you compute or check at each step?
Memory chain
Use the sentence: 'Very Cautious Civil Volunteers Solve Systematically.' V=Vc computation, C=Check 0.5φVc, C=Check φVc, V=Vs computation, S=Section adequacy (0.66 limit), S=Spacing and smax. Run through this chain on every shear problem before writing any formula.
Items to remember
- Step 1: Compute Vc = 0.17 λ √f'c bw d
- Step 2: Check if Vu > 0.5 φVc (minimum stirrups needed?)
- Step 3: Check if Vu > φVc (full stirrup design needed?)
- Step 4: Compute Vs = (Vu − φVc) / φ
- Step 5: Check Vs ≤ 0.66 √f'c bw d (section adequacy)
- Step 6: Compute spacing s = Av fy d / Vs; check smax
Maximum Spacing Decision Tree (The Halving Rule)
Recall test
Vs = 200 kN, 0.33√f'c bw d = 262 kN, d = 500 mm. What is smax? (Answer: d/2 = 250 mm or 600 mm → smax = 250 mm)
Memory chain
Remember: '0.33 is the JUDGE. Below the judge: d/2 and 600. Above the judge: d/4 and 300. The judge HALVES your allowance when you misbehave (Vs too high).' Vs ≤ 0.33 = good citizen (d/2, 600). Vs > 0.33 = misbehaving (d/4, 300).
Items to remember
- Compute Vs (required stirrup strength)
- Compute the band limit: 0.33 √f'c bw d
- If Vs ≤ band limit: smax = min(d/2, 600 mm)
- If Vs > band limit: smax = min(d/4, 300 mm)
- Always use the smaller of computed s and smax
Torsion Go/No-Go Sequence
Recall test
300×500 mm beam, f'c=28 MPa, Tu=4 kN·m: Compute threshold and state whether torsion design is required. (Answer: threshold = 4.63 kN·m > 4 kN·m → SLEEP, torsion neglected)
Memory chain
The 'ACP-PCP SLEEP TEST': Compute Acp → Compute pcp → Compute the sleep threshold → Compare Tu → SLEEP or WAKE. Remember: Acp and pcp always use the OUTER boundary (property line), never the stirrup centerline.
Items to remember
- Compute Acp = b × h (outer area)
- Compute pcp = 2(b + h) (outer perimeter)
- Compute threshold = φ × 0.083 × λ × √f'c × Acp² / pcp
- Compare Tu to threshold
- If Tu < threshold: SLEEP (ignore torsion, design for shear only)
- If Tu ≥ threshold: WAKE UP (design closed stirrups + longitudinal torsion bars)
Stirrup Leg Count and Av (U-Stirrup vs. Multi-Leg)
Recall test
12 mm dia U-stirrup: what is Av? (Ab = π/4 × 144 = 113.1 mm², Av = 2 × 113.1 = 226.2 mm²)
Memory chain
Rhyme: 'Count the legs before you calculate — one per side keeps the beam straight.' U = 2 legs. Box = 4 legs. A person has 2 legs; a table has 4. A U-stirrup is a person (2 legs), a closed rectangular stirrup with a middle bar has 4 legs (table). Count your legs, then multiply by Ab.
Items to remember
- Single bar area: Ab = π/4 × db²
- U-stirrup (2 legs): Av = 2 × Ab
- 4-leg stirrup: Av = 4 × Ab
- Always identify number of legs from problem diagram
- Use total Av in the spacing formula
Key Constants in Shear and Torsion (The Number Chain)
Recall test
Without looking, list all 7 constants in descending order with their associated concept.
Memory chain
The Number Staircase from top to bottom: 0.75 (big phi at the top) → 0.66 (next: crushing limit) → 0.35 (absolute floor) → 0.33 (half-limit for spacing) → 0.17 (Vc formula) → 0.083 (torsion threshold) → 0.062 (Av,min concrete floor). Descending staircase: 0.75, 0.66, 0.35, 0.33, 0.17, 0.083, 0.062.
Items to remember
- φ = 0.75 (shear and torsion)
- 0.17 (Vc constant, SI)
- 0.33 (half of 0.66, spacing band boundary)
- 0.66 (Vs,max constant)
- 0.083 (torsion threshold constant)
- 0.062 (Av,min concrete-based constant)
- 0.35 (Av,min absolute constant)
Diagrams
Shear and Torsion in RC Beams — Master Concept Map
A comprehensive mind map showing all major topics in shear and torsion design: the Vn = Vc + Vs framework, stirrup design rules, spacing limits, minimum reinforcement, critical section location, and the torsion threshold/design pathway. Use this map to orient yourself before solving problems.
Shear Design Decision Flowchart — Step-by-Step Board Exam Process
A complete decision flowchart for stirrup design that walks through every decision node: minimum stirrup threshold (0.5φVc), full design trigger (φVc), web crushing check (0.66 limit), spacing band selection (0.33 limit), and final spacing comparison to smax. Follow this chart on every board exam shear problem.
Torsion Go/No-Go Decision Flowchart
A focused flowchart for the torsion threshold decision. Shows the computation path using outer dimensions (Acp, pcp), the comparison to Tu, and the two outcomes: ignore torsion (SLEEP) or proceed with closed stirrup and longitudinal bar design (WAKE UP). Reinforces the critical distinction between outer dimensions for threshold vs. stirrup centerline dimensions for design.
Three Shear Zones — State Diagram for Design Regime Selection
A state diagram showing the three design regimes (Green, Yellow, Red) determined by comparing Vu to 0.5φVc and φVc. Each state leads to a different design action: no stirrups, minimum stirrups, or full stirrup design. The diagram visually reinforces the thermometer analogy from Anchor A9.