Stresses in Beams - Memory Anchors
Instead of memorizing dry formulas, your brain latches onto stories, images, and emotions. This set of 20 memory anchors transforms every key concept in Stresses in Beams — from the flexure formula to shear flow in built-up sections — into vivid, unforgettable mental hooks. Filipino engineering reviewees who use these anchors during study sessions consistently recall formulas faster under exam pressure. Use each recall trigger as a mental shortcut: when you see a beam problem, the right anchor fires automatically, and the formula follows.
Memory anchors
Flexure formula: σ = My/I — bending stress varies linearly with distance y from neutral axis
Memory aid
Imagine a LAYERED BIBINGKA fresh from the oven. The top crust is darkest (most stressed — maximum compression), the bottom crust is also dark (maximum tension), but the soft middle rice layer is pale and unstressed — that is the neutral axis. The further you are from the middle, the more 'cooked' (stressed) you get. Just like σ = My/I: the further y is from the neutral axis, the higher the stress.
Why it works
Familiar Filipino food creates a vivid spatial image of stress distribution through depth. The visual gradient from unstressed center to stressed fibers mirrors the linear variation in the formula.
Example usage
Board exam asks: 'Where is bending stress maximum in a rectangular beam?' Recall bibingka: darkest at top and bottom fibers (y = c), zero at the soft middle (neutral axis). Apply σ_max = Mc/I.
Recall trigger
Think BIBINGKA — dark crust top and bottom, soft unstressed middle
Shear stress formula: τ = VQ/(Ib) — maximum at neutral axis, zero at top and bottom
Memory aid
Think of a HALO-HALO glass. The sweetest, most concentrated layer is in the MIDDLE where all the ingredients meet — that is the neutral axis where shear stress is maximum. The plain shaved ice at the very top and crushed ice at the very bottom are bland and unstressed. Shear stress is the exact OPPOSITE of bending stress — maximum in the middle, zero at the edges.
Why it works
The contrast with bibingka (bending) reinforces memory through opposition. Filipino students immediately associate halo-halo's layered middle concentration with shear stress peaking at the neutral axis.
Example usage
If asked 'At which fiber of a rectangular beam is shear stress zero?', recall halo-halo: bland at top and bottom edges → τ = 0 at extreme fibers. Maximum is at the neutral axis.
Recall trigger
Think HALO-HALO — sweetest in the middle, bland at the top and bottom edges
Section modulus S = I/c, and σ_max = M/S
Memory aid
Remember the phrase: 'I over C gives S, and M over S gives Stress.' Shorten it to the slogan: 'ICE Makes Stress' → I/c = S, then M/S = σ. The word ICE helps: I (moment of inertia) over C (extreme fiber distance) equals S (section modulus). And M/S gives you the maximum stress — like ice cracking under maximum pressure.
Why it works
The acronym ICE is a single memorable word that encodes two formula steps in order. The image of ice cracking under pressure links S to maximum stress.
Example usage
Problem gives M = 50 kN·m and asks for σ_max of a W-section with I = 200×10⁶ mm⁴, d = 400 mm. Recall ICE: S = I/c = 200×10⁶/200 = 1×10⁶ mm³. Then σ_max = M/S = 50×10⁶/1×10⁶ = 50 MPa.
Recall trigger
ICE CRACKS — I/c = S, then M/S = σ_max
Section modulus of a rectangle: S = bh²/6
Memory aid
Chant this out loud: 'B times H-squared, divided by SIX — that's the Section Modulus of a rectangle, FIX IT IN YOUR BRAIN!' Or simply: 'B-H-squared over 6 — quick, quick, FIXED!' The word SIX rhymes with FIX, making it a two-beat memory hook. Alternatively, remember that 6 = 2×3, and a rectangle has 2 dimensions (b and h) while 3 refers to h cubed (I = bh³/12) divided in half (÷2).
Why it works
Rhythmic repetition activates phonological memory loops. The formula structure bh²/6 is chunked as 'b, h-squared, six' — three beats that mirror the three variables.
Example usage
Given a 100 mm × 300 mm rectangular timber beam: S = (100)(300)²/6 = 1.5×10⁶ mm³. Then σ_max = M/S directly.
Recall trigger
Chant: 'B-H-squared over SIX'
Maximum shear stress in a rectangular section: τ_max = 1.5 V/A (or 3V/2A)
Memory aid
Remember 'ONE AND A HALF AVERAGE' — the maximum shear stress in a rectangle is exactly 1.5 times the average shear stress (V/A). Think of it as: you ordered one cup of rice but the canteen scoops one and a half cups — 50% more than average, always. The number 1.5 is the magic multiplier for rectangles. Associate: RECTANGLE → 1.5 × (V/A).
Why it works
The everyday canteen overpour image makes the 1.5 multiplier concrete and memorable. It also helps distinguish rectangles (1.5) from circles (4/3 ≈ 1.33), which is a classic exam trap.
Example usage
V = 60 kN, b = 150 mm, h = 300 mm. A = 45,000 mm². τ_max = 1.5 × (60,000/45,000) = 2.0 MPa. Recall: canteen gives 1.5× the average.
Recall trigger
Canteen overpour — 1.5 times the average
Maximum shear stress in a solid circular section: τ_max = 4V/(3A)
Memory aid
For a CIRCLE, remember '4 over 3' — the same fraction as the volume of a sphere (4/3 π r³)! The geometry connection: a circle's shear factor shares the '4/3' with the sphere volume formula — both involve circular geometry. Say: 'Circle shear = 4V/3A. Four-thirds, like a sphere's volume.' This cross-formula link burns '4/3' into memory for circular sections.
Why it works
Cross-linking to the already-memorized sphere volume formula (4/3 πr³) creates an associative hook. The fraction 4/3 appears in both formulas — exploiting existing memory traces reduces new learning load.
Example usage
A solid circular shaft (d = 100 mm) carries V = 30 kN. A = π(100)²/4 = 7,854 mm². τ_max = 4(30,000)/[3(7,854)] = 5.09 MPa. Recall: 4/3 like a sphere.
Recall trigger
Think SPHERE VOLUME → 4/3 → τ_max = 4V/3A for circles
Sagging (positive) moment → bottom fiber in tension, top in compression
Memory aid
Picture a BAMBOO BRIDGE over a river in the province. When you walk across, the bamboo sags in the middle — a SAGGING beam. Look at the bamboo: the bottom fibers are being STRETCHED (tension — they try to pull apart). The top fibers are being SQUISHED (compression — the bamboo is bent over them). SAG = bottom STRETCHES, top SQUISHES. The bridge sags, so the bottom goes to tension — just like a smile: the bottom lip stretches outward.
Why it works
The bamboo bridge is a vivid rural Filipino image. The physical intuition of stretching (tension) at the bottom of a sagging member is reinforced by the smile analogy — bottom fiber curves outward and stretches.
Example usage
A simply supported beam with UDL has a sagging BMD. The bottom fiber is in tension → reinforcement goes at the bottom in RC beams (ACI 318 / NSCP 2015 Section 406 principle).
Recall trigger
Sagging bamboo bridge — bottom stretches (tension), top squishes (compression)
First moment of area Q — the correct area to use in τ = VQ/(Ib)
Memory aid
Imagine you are a SECURITY GUARD (Q) at the boundary level where you want to find shear stress. You count EVERYONE ABOVE YOU (the area above that level) and record how far their center of gravity is from the neutral axis. Q is the total 'report' from everyone above: Q = A_above × ȳ_above. If the stress is at the web-flange junction of an I-beam, Q includes only the flange area times its centroid distance from NA. The guard only counts the people BEYOND the checkpoint, not everyone in the building.
Why it works
The security guard checkpoint metaphor clearly demarcates which area contributes to Q. It prevents the classic mistake of using the total section area. The story creates a spatial mental model.
Example usage
To find τ at the neutral axis of a rectangle (b × h): Q = (b × h/2) × (h/4) = bh²/8. The guard counts the top half of the rectangle and finds its centroid at h/4 from the NA. Then τ = VQ/Ib.
Recall trigger
SECURITY GUARD at the level — count only the area BEYOND that point
Shear flow q = VQ/I — used for connector spacing in built-up beams
Memory aid
Think of WATER FLOWING between two glued planks of wood. The shear flow q (in N/mm) is like the water pressure per unit length trying to slide the planks apart. The nails or bolts are like DAMS spaced at distance s — each dam (nail) has a holding capacity F (N). The spacing s = F/q is how far apart you can place the dams before the water pressure (shear flow) breaks through. More shear flow → closer nail spacing.
Why it works
Flow is an intuitive physical metaphor for a force distributed per unit length. The dam-spacing analogy directly models the design equation s = F/q, making it self-explanatory.
Example usage
V = 5 kN, Q = 1.5×10⁵ mm³, I = 4×10⁷ mm⁴. q = VQ/I = 5000(1.5×10⁵)/4×10⁷ = 18.75 N/mm. Nail capacity F = 400 N. s = 400/18.75 = 21.3 mm → use 20 mm spacing.
Recall trigger
Water flowing between planks — nail spacing s = F/q (dam capacity over flow rate)
Combined axial and bending stress: σ = P/A ± Mc/I
Memory aid
Remember 'PAM IS COMBINED': P/A is the axial (uniform) stress — it's the same everywhere like a flat palm. M/c/I is the bending stress — it varies. Together: PALM flat + BENDING = combined. The plus-or-minus (±) means one fiber gets the addition, the other gets the subtraction. Think of it as: your hand (palm = P/A) + bending your fingers (±Mc/I) = total finger stress. Compressed side: palm + bend. Tension side: palm − bend.
Why it works
The hand gesture is a body-linked memory anchor — highly durable. The PAM IS COMBINED phrase encodes the formula structure. Physical enactment during study strengthens kinesthetic memory.
Example usage
Short column 200×200 mm, P = 300 kN, e = 40 mm. σ = P/A ± Mc/I = 7.5 ± 9.0 MPa → σ_max = 16.5 MPa, σ_min = −1.5 MPa. Recall: palm ± bent fingers.
Recall trigger
PALM IS COMBINED — P/A (flat palm) ± Mc/I (bent fingers)
Middle-third rule (kern): load must act within h/6 from center to avoid tension
Memory aid
Visualize a JEEPNEY that can only safely pick up passengers within the MIDDLE THIRD of its roof — if the load (passenger) sits on the edge beyond h/6, the jeepney tips (tension develops). Draw a rectangle: divide it into thirds vertically. Only the middle third is the 'safe zone' for eccentric loads. The kern is that middle-third diamond. If e > h/6, you have gone beyond the jeepney's safe loading zone → tension appears.
Why it works
The jeepney is deeply familiar to Filipino students. The tipping metaphor physically represents the onset of tension. Drawing the rectangle in thirds creates a vivid mental diagram.
Example usage
Column 300×300 mm. Kern limit = h/6 = 50 mm. If e = 60 mm > 50 mm → tension occurs. Recall: jeepney tipped beyond the middle third.
Recall trigger
JEEPNEY MIDDLE THIRD — safe loading zone is ±h/6 from center
For an unsymmetric section (like a T-beam), the neutral axis is NOT at mid-depth; top and bottom fiber stresses are different
Memory aid
Think of a BALANCING SCALE (timbangan) at the palengke. A T-beam is like a scale where one side (the flange) is much heavier than the other (the web). The balance point (neutral axis) is NOT at the geometric center — it shifts toward the heavier (flange) side. Because the NA is closer to the flange, c_top < c_bot, so the BOTTOM FIBER gets more stress. In a concrete T-beam, the bigger stress at the bottom means more tension reinforcement is needed there.
Why it works
The timbangan (weighing scale) is a culturally resonant image for the centroid shift. The asymmetry of a T-section is naturally modeled by the unequal pan weights. Practical link to RC design makes it exam-relevant.
Example usage
T-beam: flange 200×50 mm top + web 50×200 mm. ȳ from top = 87.5 mm → c_bot = 162.5 mm > c_top. σ_bot > σ_top for same M. Recall: the heavier flange pan pulls the NA upward.
Recall trigger
TIMBANGAN at the palengke — heavy flange pulls the balance point up
I-beam efficiency: more depth → much more section modulus because S ∝ h² for rectangles
Memory aid
Imagine two BAMBOO POLES — one thin and one thick. If you double the DEPTH of a rectangular beam, the section modulus S = bh²/6 goes up by a factor of FOUR (because h is squared). It is like upgrading from a single bamboo rafter to one twice as deep — it can carry four times the bending moment! This is why steel I-beams are tall and thin — maximum depth for minimum weight, squeezing every bit out of h².
Why it works
The bamboo analogy ties the abstract h² relationship to a physical building material familiar in the Philippines. The 4× improvement from doubling depth is a concrete, surprising fact that sticks.
Example usage
A 100×200 mm beam has S = 100(200²)/6 = 666,667 mm³. Doubling depth to 400 mm: S = 100(400²)/6 = 2,666,667 mm³ — exactly 4×. Recall: bamboo upgrade gives 4× capacity.
Recall trigger
Double the depth → FOUR TIMES the section modulus (h² effect)
Unit consistency: N·mm with mm⁴ gives MPa (N/mm²)
Memory aid
Remember 'kN·m is DANGEROUS in the formula' — always convert to N·mm before plugging into σ = My/I or τ = VQ/Ib. The mantra: 'N dot mm, mm to the fourth, gives N per mm-squared = MPa.' Write it as an equation: [N·mm] / [mm⁴] = [N/mm²] = MPa. Think of it as: NEWTON-MILLIMETER divided by MM-FOURTH = MEGAPASCAL. The memory hook: 'kN·m times a million becomes N·mm' (multiply by 10⁶).
Why it works
Unit errors are the #1 source of mistakes in beam stress calculations in Philippine board exams. A dedicated mnemonic for this procedural step prevents careless errors.
Example usage
M = 12 kN·m = 12×10⁶ N·mm. S = 2.25×10⁶ mm³. σ = 12×10⁶/2.25×10⁶ = 5.33 N/mm² = 5.33 MPa. Recall: always convert to N·mm first.
Recall trigger
DANGEROUS kN·m — convert to N·mm first! N·mm ÷ mm⁴ = MPa
Bending stress vs. shear stress: opposite distributions through the depth
Memory aid
Draw two triangles pointing in OPPOSITE DIRECTIONS, one on top of the other — like an HOURGLASS (or a double-arrowhead). The top triangle represents bending stress: maximum at top (σ_max), zero at middle (NA). The bottom triangle represents shear stress: maximum at middle (NA), zero at top and bottom. The HOURGLASS image: the wide ends of bending are at the fibers, the wide end of shear is at the middle. They are MIRROR IMAGES of each other's zero-and-maximum locations.
Why it works
The hourglass is a single image encoding both stress distributions simultaneously and their opposition. Visual-spatial memory is extremely durable, especially for distribution shapes.
Example usage
Exam asks: 'At the neutral axis of a loaded beam, which stress is zero?' Recall hourglass: bending stress is zero at the NA, but shear stress is maximum there.
Recall trigger
HOURGLASS — bending wide at ends, shear wide at middle
Short beams are shear-governed; long beams are flexure/deflection-governed
Memory aid
Think of a BENCH PRESS vs. a DIVING BOARD. A SHORT, THICK bench press beam is limited by shear — it can barely handle the concentrated load without splitting. A LONG, SLENDER diving board flexes greatly — deflection and bending govern its design. Short = Shear, Long = bending (flexure). The alliteration helps: Short → Shear; Long → bending (Letter L for both Long and fLexure).
Why it works
Physical sports images create strong emotional-motor memory. The alliterative Short-Shear and Long-bending pattern uses phonological similarity as a memory aid.
Example usage
Design a 1 m heavily loaded bracket (short): check shear first. Design a 10 m floor beam: check bending and deflection first. Recall: bench press vs. diving board.
Recall trigger
SHORT bench press = SHEAR; LONG diving board = bending/deflection
Plane sections remain plane — the fundamental assumption of beam bending theory
Memory aid
Imagine a STACK OF PANCAKES (hotcakes) with a stripe drawn vertically down the stack. When you bend the stack, the stripe remains straight (planar) — it just tilts. Each pancake layer moves a different amount, but no individual pancake crumbles or warps independently. This is 'plane sections remain plane' — the Bernoulli-Euler hypothesis. It is why strain varies linearly with depth, and therefore stress does too (for elastic material). The pancake stack confirms: straight line before bending, straight line after.
Why it works
The pancake stack is a hands-on image students can physically recreate. The stripe-tilting visualization directly demonstrates why strain distribution is linear, grounding the assumption physically.
Example usage
If asked: 'What assumption justifies the linear strain distribution in the flexure formula?', recall: pancake stripe stays straight → plane sections remain plane → ε = y/ρ → σ = My/I.
Recall trigger
PANCAKE STACK with a stripe — stripe stays straight even when bent
Design steps for a beam: (1) Find M_max & V_max, (2) Flexure check: S_req = M/σ_allow, (3) Shear check: τ_max ≤ τ_allow, (4) Deflection check
Memory aid
Remember the acronym 'FIND-S-T-D': Find (M_max, V_max from SFD/BMD) → S (Section modulus design: S_req = M/σ_allow) → T (Tau check: τ_max ≤ τ_allow) → D (Deflection serviceability check). Pronounce it 'FIND-STD' — as in 'Find the STanDard' for your beam. Each letter is one step. Short beams may skip to T first; long beams may be governed by D.
Why it works
A 4-step acronym condenses the entire beam design procedure into a single word. 'Find the STD' is a memorable phrase that respects the logical sequence engineers follow.
Example usage
Board problem: design a timber beam for given loads. Step 1: Find M_max = wL²/8, V_max = wL/2. Step 2: S_req = M/σ_allow → pick dimensions. Step 3: τ_max = 1.5V/A ≤ τ_allow. Step 4: Check δ_max ≤ L/360 (NSCP 2015 Table 406.2.1).
Recall trigger
FIND-STD: Find → S (section) → T (shear/tau) → D (deflection)
For I-beams (wide-flange), nearly all shear is in the web: τ_max ≈ V/A_web
Memory aid
Think of a HIGHWAY OVERPASS. The wide top and bottom flanges are like the WIDE LANES where cars (bending stress) travel most. The thin web in the middle is like the NARROW BRIDGE COLUMN that handles the VERTICAL SHEAR FORCE from the traffic. Almost all shear squeezes through the thin web column, so τ_max ≈ V/A_web (the web's cross-sectional area). Flanges are for bending; the web is for shear.
Why it works
The highway/overpass analogy separates flange and web roles clearly. Infrastructure imagery is directly relevant to Filipino CE students who study bridges and apply NSCP and AISC 360.
Example usage
W310×97 with A_web = t_w × d_w: τ_max ≈ V/A_web. This approximation is used in AISC 360 web shear checks and NSCP 2015 steel provisions.
Recall trigger
HIGHWAY OVERPASS — wide lanes (flanges) for bending, narrow column (web) for shear
Section modulus of a circle: S = πd³/32
Memory aid
Chunk it as 'PIE-D-CUBE-32'. Say out loud: 'PIE (π) × D-CUBED, all over 32.' Remember the sequence: π on top, d³ (d cubed) on top, 32 on the bottom. The number 32 is 2⁵ — think of it as a '32-count' pack: 32 = 2 × 16 = 2 × (diameter-to-radius conversion chain). Alternatively: I for a circle = πd⁴/64; S = I/c = (πd⁴/64)/(d/2) = πd³/32. The derivation itself is a memory chain.
Why it works
Chunking 'PIE-D-CUBE-32' into four syllables creates a phonological memory unit. Deriving S from I = πd⁴/64 also reinforces both formulas simultaneously through logical connection.
Example usage
Size a circular timber shaft for M = 9 kN·m, σ_allow = 8 MPa. S_req = 9×10⁶/8 = 1.125×10⁶ mm³. πd³/32 = 1.125×10⁶ → d³ = 11.46×10⁶ → d = 225.6 mm. Use d = 230 mm.
Recall trigger
PIE-D-CUBE-32 → S = πd³/32
Revision game
I am the stress that is ZERO in the middle and MAXIMUM at the top and bottom. What am I?
Answer
Bending stress (σ = My/I)
Memory link
A1 — Bibingka analogy: dark crust at top and bottom, soft unstressed middle (neutral axis)
I am the stress that is ZERO at the top and bottom fibers and MAXIMUM in the middle. What am I?
Answer
Shear stress (τ = VQ/Ib)
Memory link
A2 — Halo-halo analogy: sweetest concentration in the middle, bland at the edges
I am a single number equal to I/c that tells you how efficient a cross-section is in bending. What am I?
Answer
Section modulus S
Memory link
A3 — ICE CRACKS mnemonic: I/c = S (ICE), then M/S = σ (cracks)
If you DOUBLE the DEPTH of a rectangular beam while keeping width constant, by what factor does the section modulus increase?
Answer
4 times (because S = bh²/6 and h is squared)
Memory link
A13 — Bamboo upgrade analogy: double the depth → four times the bending capacity
A short column carries a load that is 60 mm off-center. The column is 300×300 mm. Will tension develop somewhere in the section? (Hint: check the kern!)
Answer
Yes, because e = 60 mm > h/6 = 50 mm — the load is outside the kern (middle third)
Memory link
A11 — Jeepney middle-third rule: loading beyond ±h/6 from center → jeepney tips → tension appears
I am the shear stress shape factor for a RECTANGLE. Say my value as a fraction AND a decimal.
Answer
3/2 = 1.5 (one and a half)
Memory link
A5 — Canteen overpour: always 1.5 times the average shear stress V/A
A built-up wood beam has shear flow q = 25 N/mm. Each nail can carry 500 N. What is the maximum nail spacing?
Answer
s = F/q = 500/25 = 20 mm
Memory link
A9 — Water flow between planks: nail spacing s = F/q (dam capacity over flow rate)
In the formula τ = VQ/(Ib), what AREA do you use to compute Q when finding shear stress at mid-depth of a rectangular beam?
Answer
The area of the UPPER HALF of the rectangle (the area ABOVE the level of interest, not the whole section)
Memory link
A8 — Security guard protocol: count only the people BEYOND your checkpoint (above the cut level)
Formula mnemonics
σ = My/I
Mnemonic
MY EYES (My/I) see the stress — the bending stress σ equals M·y divided by I, like your eyes (I) reading from the neutral axis (y distance).
When to use
Any point inside a beam cross-section under a known bending moment M. Use y = c (extreme fiber) for maximum stress.
What each part means
σ = bending stress at distance y (MPa); M = bending moment (N·mm); y = distance from neutral axis to point of interest (mm); I = moment of inertia of section about NA (mm⁴)
σ_max = M/S where S = I/c
Mnemonic
ICE CRACKS under max stress — I/c = S (ICE), then M/S = σ_max (cracks). Three letters: I, C, S → formula chain.
When to use
Design problems where you need σ_max quickly, or when sizing a section given allowable stress: S_req = M/σ_allow.
What each part means
S = section modulus (mm³); I = moment of inertia (mm⁴); c = distance from NA to extreme fiber (mm); M = bending moment (N·mm); σ_max = maximum bending stress (MPa)
S_rect = bh²/6
Mnemonic
B-H-squared over SIX — chant it: 'bh-squared-six, bh-squared-six.' Six is 2×3, and the formula comes from I = bh³/12 divided by c = h/2.
When to use
Any rectangular beam cross-section: timber, plain concrete, or rectangular RC section in flexure design.
What each part means
b = width of rectangular section (mm); h = total depth of rectangular section (mm); 6 is the constant from the derivation (12 ÷ 2 = 6)
τ = VQ/(Ib)
Mnemonic
Very Quiet In Bed — V·Q over I·b. 'Very Quiet' = VQ (top); 'In Bed' = Ib (bottom). The formula is VQ over Ib.
When to use
Finding shear stress at any level in a beam cross-section under a known shear force V.
What each part means
τ = shear stress at level of interest (MPa); V = shear force at cross-section (N); Q = first moment of area beyond cut level about NA (mm³); I = moment of inertia of whole section about NA (mm⁴); b = width of section at cut level (mm)
τ_max,rect = 1.5 V/A = 3V/(2A)
Mnemonic
ONE AND A HALF AVERAGE — τ_max for a rectangle is always 1.5 times the average shear stress V/A. Remember: 3/2 = 1.5.
When to use
Rectangular timber or concrete beams under transverse loading. Direct shortcut without computing Q and I separately.
What each part means
τ_max = maximum shear stress at neutral axis (MPa); V = shear force (N); A = total cross-sectional area bh (mm²); 3/2 is the shape factor for a rectangle
τ_max,circle = 4V/(3A)
Mnemonic
FOUR-THIRDS like a SPHERE — τ_max for a solid circle uses 4/3, the same fraction in the sphere volume formula (4/3 πr³). Circular geometry → 4/3.
When to use
Solid circular cross-sections (round timber logs, dowels, cylindrical shafts) under transverse shear.
What each part means
τ_max = maximum shear stress at NA of circular section (MPa); V = shear force (N); A = πd²/4 (mm²); 4/3 is the shape factor for a solid circle
q = VQ/I (shear flow)
Mnemonic
Shear Flow = VQ/I — drop the 'b' from the shear stress formula because flow is force PER UNIT LENGTH, not per area. q = VQ/I gives N/mm (flow rate of force along the beam length).
When to use
Built-up beams: plank-and-joist timber, plate girders, box sections. Use q = VQ/I then s = F/q for nail/bolt/weld spacing.
What each part means
q = shear flow (N/mm); V = shear force (N); Q = first moment of connected area about NA (mm³); I = moment of inertia of whole section about NA (mm⁴). Connector spacing: s = F/q where F is connector capacity (N).
σ = P/A ± Mc/I (combined axial and bending)
Mnemonic
PALM ± BEND — P/A is the flat palm (uniform axial stress), ±Mc/I is the bending (one side adds, other side subtracts). Always check BOTH extreme fibers.
When to use
Short columns with eccentric loads, masonry walls with wind moment, footings, any member with simultaneous axial force and bending moment.
What each part means
P/A = uniform axial stress (+ tension, − compression) (MPa); M = Pe = bending moment from eccentricity (N·mm); c = distance to extreme fiber (mm); I = moment of inertia (mm⁴). Sign: same side as load offset gets addition; opposite side gets subtraction.
Quick recall chains
Beam Design Checklist (FIND-STD)
Recall test
Without looking: list the 4 steps of beam design in order. What does each letter of FIND-STD stand for?
Memory chain
FIND-STD: 'FIND the problem, pick the right Standard size.' F = Find loads, S = Section modulus design, T = Tau (shear) check, D = Deflection (serviceability). Say it like a command: 'FIND STD!' every time you start a beam design problem.
Items to remember
- Find M_max and V_max from SFD/BMD
- Section modulus: compute S_req = M_max / σ_allow
- Tau (shear): verify τ_max ≤ τ_allow
- Deflection: check δ_max ≤ allowable (NSCP Table 406.2.1)
Shear Stress Maxima for Different Sections
Recall test
Name the shear stress shape factors for: (a) rectangle, (b) solid circle, (c) I-beam web. Which is largest? Which smallest?
Memory chain
Order the shape factors from smallest to largest: I-beam (≈1), Circle (4/3 ≈ 1.33), Rectangle (3/2 = 1.5), Tube (2). Memory story: 'The I-beam is the MOST EFFICIENT (factor 1), the circle is MIDDLE (4/3), the rectangle is ONE AND A HALF, and the thin tube is DOUBLE.' Think: I-C-R-T → '1, 4/3, 3/2, 2'
Items to remember
- Rectangle: τ_max = 1.5 V/A (factor = 3/2)
- Solid circle: τ_max = 4V/(3A) (factor = 4/3)
- Thin-walled circular tube: τ_max = 2V/A (factor = 2)
- Wide-flange (I-beam): τ_max ≈ V/A_web (factor ≈ 1)
Steps to Compute Shear Stress τ = VQ/(Ib) at a Specific Level
Recall test
Walk through computing τ at the neutral axis of a 100×200 mm rectangle with V = 20 kN. Use the 7-step protocol.
Memory chain
SECURITY GUARD PROTOCOL (7 steps): Level → Guard Post. Area above → Count the people. Centroid → Find their average height. Q = A×ȳ → Submit report. I of whole section → Building total inertia. b = width → Door width. τ = VQ/Ib → Stress per door. Recall each step by picturing the guard filing a report.
Items to remember
- Step 1: Identify the level (depth) where τ is needed
- Step 2: Identify the area ABOVE (or below) that level
- Step 3: Find centroid of that area (ȳ from NA)
- Step 4: Compute Q = A_above × ȳ
- Step 5: Get I of whole section about NA
- Step 6: Get b = width at the cut level
- Step 7: Apply τ = VQ/(Ib)
Extreme Fiber Stress Signs for Sagging and Hogging Beams
Recall test
A continuous beam has hogging moment over a support. Which fiber is in tension? Where should reinforcement go in an RC beam at that location?
Memory chain
SMILE vs. FROWN: A sagging beam SMILES → bottom is stretched (tension), top is squished (compression). A hogging beam FROWNS → top is stretched (tension), bottom is squished (compression). For kern: 'Stay in the middle or someone cries (tension appears).' Neutral axis: always peaceful, always zero.
Items to remember
- Sagging moment (+): top fiber = compression (−), bottom fiber = tension (+)
- Hogging moment (−): top fiber = tension (+), bottom fiber = compression (−)
- Neutral axis: zero bending stress always
- Eccentric compression load: both fibers start in compression; kern violation → one goes to tension
T-Beam Neutral Axis Location Procedure
Recall test
A T-beam has a 150×50 mm flange and 50×150 mm web (total depth 200 mm). Find ȳ from the top. Then find c_bot.
Memory chain
TIMBANGAN PROTOCOL: Divide → Weigh each pan (A_i). Measure moment arms (ȳ_i). Sum-of-moments over total weight gives balance point (ȳ). Then c_top and c_bot follow from geometry. Parallel axis: I = I_centroidal + Ad². Think: 'Balance the timbangan, then measure each arm length.'
Items to remember
- Step 1: Divide into rectangular sub-areas (flange + web)
- Step 2: Compute area of each sub-area (A_i)
- Step 3: Measure centroid of each sub-area from the top (ȳ_i)
- Step 4: ȳ_total = Σ(A_i × ȳ_i) / ΣA_i
- Step 5: c_top = ȳ_total; c_bot = total depth − ȳ_total
- Step 6: Use parallel axis theorem for I_total
Diagrams
Stresses in Beams — Master Concept Map
A hierarchical mind map showing all major topics in Stresses in Beams, their sub-concepts, and key formulas. Use this as a visual roadmap before and after studying each section to reinforce connections between concepts.
Beam Design Decision Flowchart (FIND-STD Protocol)
A step-by-step decision flowchart for designing beams against flexure, shear, and deflection. Follows the FIND-STD protocol (A18). The diamond decision nodes highlight where design checks can trigger iteration back to section selection. Use this during practice problems to train the habit of checking all three criteria.
Stress Distribution Comparison — Bending vs. Shear in a Rectangular Beam
Side-by-side comparison of bending stress (linear, maximum at fibers) and shear stress (parabolic, maximum at neutral axis) through the depth of a rectangular beam. This is the HOURGLASS visual from anchor A15. Reinforces that bending and shear distributions are opposite in nature — a perennial board exam topic.
Computing First Moment of Area Q — Security Guard Protocol
A procedural flowchart for computing the first moment of area Q at any level in a beam cross-section. Follows the Security Guard Protocol from anchor A8. Includes a special branch for the neutral axis case, which gives the maximum shear stress. Use this flowchart when practicing τ = VQ/(Ib) problems.
Distribution of Beam Problem Types in PRC CE Board Exams (Approximate)
An approximate distribution of beam stress problem types that appear in Philippine CE licensure exams based on typical syllabi. The largest slice (flexure formula and section modulus design combined = 50%) confirms that mastering σ = My/I and S = I/c is the highest-value study priority. Shear and combined stress problems each represent significant portions that should not be neglected.