Frames, Machines and Cables - Memory Anchors
Memory techniques transform passive reading into active recall. For the PRC Civil Engineer Licensure Examination, where you must recall formulas and procedures under pressure, these anchors give your brain a 'mental hook' — a shortcut that retrieves the full concept in seconds. Each anchor here is tied to a specific board-exam concept in Frames, Machines, and Cables. Read each one, close your eyes, visualize it, then test yourself using the recall trigger. With repetition, recall becomes automatic — exactly what you need on exam day.
Memory anchors
Frames have multi-force members (not two-force like trusses)
Memory aid
Think of a TRUSS as a Filipino 'bilao' tray with only bamboo sticks — each stick is pulled or pushed from both ends only (two-force). A FRAME is like a 'bahay-kubo' — the posts and beams are nailed together and carry loads from windows, walls, and people standing in the middle. The bahay-kubo posts bend and shear; the bilao sticks only stretch or compress. Multi-force = multi-load, like a busy bahay-kubo.
Why it works
Connecting abstract structural mechanics to a familiar Filipino dwelling makes the distinction visceral and easy to picture during exams.
Example usage
When a problem says 'frame,' immediately think: 'bahay-kubo post — multi-force — I must dismember and apply full equilibrium, not just axial force.'
Recall trigger
Picture a bahay-kubo vs a bilao tray of bamboo sticks.
Analysis procedure for frames: (1) External reactions, (2) Dismember, (3) Pin forces by equilibrium
Memory aid
Remember the acronym E-D-P: 'Every Dismembered Part.' Step 1 = External reactions (whole frame FBD). Step 2 = Dismember (separate each member). Step 3 = Pin forces (apply ΣFx, ΣFy, ΣM to each part). 'Every Dismembered Part' — like dissecting a frog in NSCP lab — you always check the whole frog first, then cut it open, then analyze each organ.
Why it works
The acronym E-D-P gives a three-step ordered sequence. The frog dissection analogy reinforces the 'whole before parts' logic.
Example usage
Exam question asks for pin force at C in a frame. Recall E-D-P: first find support reactions at A and B (whole frame), then dismember at pin C, then apply ΣM=0 on one member.
Recall trigger
Say 'E-D-P' and picture cutting open a frog.
Newton's Third Law at pins — equal and opposite forces on the two members sharing a pin
Memory aid
Imagine two engineering students arm-wrestling at a pin joint. If Student A pushes Student B's arm to the RIGHT with 50 N, then Student B is pushing Student A's arm to the LEFT with exactly 50 N. Neither can cheat — the pin forces must be equal and opposite on both FBDs. If you draw the force going right on member 1, it MUST go left on member 2. Forgetting this is the #1 frame mistake in board exams — 'the arm-wrestling rule.'
Why it works
A competitive physical scenario creates an emotional memory. The arm-wrestling image makes Newton's Third Law concrete and hard to forget.
Example usage
When drawing the dismembered FBDs, after assigning pin force direction on member AB, immediately flip the arrow for member BC — arm-wrestling rule.
Recall trigger
Two students arm-wrestling at a pin.
Two-force member: force acts along the line connecting the two pin ends
Memory aid
Visualize a tug-of-war rope — two teams pull from both ends. The rope (two-force member) can ONLY be pulled along its own length. No matter how hard you try, you cannot pull a tug-of-war rope sideways. A two-force member is that rope: force is always axial (along the member), never transverse. Spot it by checking: is it loaded ONLY at two pin ends, with no loads in between? If yes — it's a tug-of-war rope!
Why it works
Tug-of-war is universally understood. The visual of a rope that can only transmit axial force is physically intuitive.
Example usage
In a frame problem, scan members first. Find the strut with pins only at both ends and no intermediate loads — label it as two-force, draw the force along the member, and immediately reduce the unknowns from 2 to 1.
Recall trigger
Tug-of-war rope — force only along its length.
Machines transmit or modify force — input effort produces amplified output load
Memory aid
A machine is like a BOTE-OPENER (bottle opener) at a fiesta. You push down gently on the long handle (effort arm), and the short end under the bottle cap pops it open with a huge force (load arm). The machine doesn't create energy — it trades DISTANCE for FORCE. Long effort arm = big mechanical advantage = big output force. This is the essence of every machine problem: find the ratio of effort arm to load arm.
Why it works
The bottle opener is a familiar Filipino social tool. It perfectly illustrates the lever principle and mechanical advantage in a tactile, everyday context.
Example usage
For a machine problem, identify the pivot, measure effort arm and load arm, then M.A. = effort arm / load arm. Load = Effort × M.A.
Recall trigger
Fiesta bote-opener — long handle, big pop.
Mechanical Advantage formula: M.A. = effort arm / load arm
Memory aid
Short rhyme to remember the formula:
'Effort over load? No — that's not right.
Arm over arm gives you the might.
Effort ARM on top, load ARM below —
Multiply by effort, and watch the load grow!'
Why it works
Rhymes engage the auditory memory system and are naturally repetitive. The correction in the first line prevents the common mistake of using forces instead of arms.
Example usage
Exam problem: effort = 80 N, effort arm = 0.6 m, load arm = 0.04 m. Recall the rhyme: arm over arm → M.A. = 0.6/0.04 = 15. Load = 80 × 15 = 1200 N.
Recall trigger
Recite the first line: 'Effort over load? No — that's not right.'
Cable carries only tension — no compression, no shear, no bending
Memory aid
A cable is like a KULOG-AT-KIDLAT (thunder and lightning) safety guideline: it only works ONE WAY. A clothesline (sampayan) can only be pulled taut — it cannot push clothes apart, cannot twist, cannot bend itself. The moment you try to push a clothesline, it goes slack. A cable is always in TENSION, never anything else. If a calculated cable tension comes out negative, the cable is slack — physically impossible in the model.
Why it works
The clothesline/sampayan is a daily household object every Filipino student has used. The one-way behavior is instantly relatable.
Example usage
Before solving any cable problem, write at the top: 'T > 0 always.' If your answer is negative, recheck your geometry or load direction.
Recall trigger
Picture a sampayan (clothesline) — pull only, never push.
Horizontal component H is constant throughout a cable
Memory aid
Imagine you are sending a 'tabo' sliding across a table on a string. No matter how steep or shallow the string's angle at any point, the horizontal pull at your end never changes — you always exert the same sideways tug. That steady horizontal pull IS the constant H. The vertical pull changes because the angle changes, but horizontal? Constant. Always. Even when the cable dips and rises under multiple loads, H stays the same from support to support.
Why it works
The tabo-on-string mental image creates a tangible sense of the horizontal component being 'locked in.' The micro-story replaces an abstract equation with a physical intuition.
Example usage
In a cable with three point loads, find H from one segment using ΣM=0 at a point, then use the same H to find tension in every other segment: T = H/cosθ.
Recall trigger
Tabo sliding across a table — the horizontal tug never changes.
Parabolic cable formula: H = wL²/8d
Memory aid
Remember the formula H = wL²/8d using the phrase: 'Weightlifter LOADS up at EIGHT Decks.' W = load per meter (w), L = span (L²), 8 = eight, d = sag/deck depth. Or use the visual: picture the number '8' lying on its side — that's the infinity symbol of a cable sagging under uniform load. The '8' is in the DENOMINATOR with d (sag). Bigger sag = lower H (less horizontal pull needed). Bigger span or load = higher H.
Why it works
The phrase encodes the order of variables. The visual of '8 on its side' connects the denominator to the cable's sinuous shape.
Example usage
Cable spans 40 m, sag 4 m, w = 6 kN/m. Recall: H = wL²/8d = 6(40²)/[8(4)] = 6(1600)/32 = 9600/32 = 300 kN.
Recall trigger
Say 'Weightlifter Loads at Eight Decks' → w, L², 8, d → H = wL²/8d.
Maximum tension in a parabolic cable occurs at the supports (steepest angle)
Memory aid
Visualize a roller-coaster track (like Enchanted Kingdom's Space Shuttle). The cable is steepest where it connects to the towers — like the steep climb of a roller coaster at the sides. At the bottom (midspan), the ride is flat and easy; at the towers, the slope is maximum and so is the tension. MAX TENSION = MAX ANGLE = AT THE SUPPORTS. Draw a smiley-face parabola and mark 'DANGER: MAX T' at both ends, 'CALM: H only' at the bottom.
Why it works
Roller coasters are exciting and emotionally charged. The spatial mapping of steepness to tension creates a geometric memory that transfers directly to cable diagrams.
Example usage
For T_max: always compute at the support. T_max = √(H² + (wL/2)²). The wL/2 term is the vertical reaction at each support for a symmetric parabolic cable.
Recall trigger
Enchanted Kingdom roller coaster — steepest at the towers, calmest at the bottom.
Parabola vs Catenary: uniform horizontal load → parabola; uniform load along cable length → catenary
Memory aid
Think of two types of bridges in the Philippines:
- SUSPENSION BRIDGE (like those in Samar or the Passi bridge) carries a road deck — the load is uniform PER HORIZONTAL METER of deck → PARABOLA. The bridge deck straightens the load horizontally.
- A HANGING CHAIN (like a rosary or kwintas hung between two fingers) carries its own weight along its LENGTH → CATENARY. The chain follows its own arc.
Memory: DECK = PARABOLA (D and P share the same 'engineering drawing' feel). CHAIN = CATENARY (both start with 'C-A').
Why it works
Filipino infrastructure examples ground the abstract math. The alliterative links (D-P, C-C) reinforce the pairing.
Example usage
Problem says 'cable carries a uniformly distributed load of 5 kN/m along the horizontal.' Recall: horizontal load = deck = PARABOLA → use H = wL²/8d.
Recall trigger
DECK → Parabola. CHAIN → Catenary. (D-P, C-C)
Cable with concentrated loads — cable is straight between load points (funicular polygon)
Memory aid
Picture a PAROL (Philippine Christmas lantern) being held by two strings from its tips, with ornament balls hanging from the middle joints. The strings between each ornament are perfectly straight — no curves. The entire cable shape looks like a series of straight sticks arranged in a V-shape or W-shape. This is the funicular polygon. 'Funicular' = rope-like. Between any two adjacent loads, the cable is a straight line because there is no distributed load to curve it.
Why it works
The parol is a deeply cultural Filipino image. The ornament weights create an instantly visualized funicular polygon with straight segments between loads.
Example usage
Cable with two point loads: split into three straight segments. Find H from ΣM=0 at a point on the cable, then use T = H/cosθ for each segment's tension.
Recall trigger
Parol with hanging ornaments — straight segments between each ball.
Dismembering a frame — separate each member and show pin forces with Newton's 3rd Law
Memory aid
Engineer Maria is analyzing a frame for a footbridge in her hometown in Batangas. She imagines she has a magical hacksaw — she cuts through every pin in the frame and separates the members into individual pieces. At each cut, she draws two equal-but-opposite arrows: one on each piece. She calls it her 'EQUAL AND OPPOSITE ACCOUNTING SYSTEM.' Like a sari-sari store ledger — every debit on one member's account is a credit on the other. The total of the system is zero. Maria never forgets Newton's Third Law because she remembers her ledger.
Why it works
The sari-sari store ledger metaphor connects Newton's Third Law to a familiar financial system. The story has a named character (relatable) and a concrete setting.
Example usage
After finding that pin C exerts 30 kN upward on member AB, immediately write: pin C exerts 30 kN DOWNWARD on member BC — ledger balanced.
Recall trigger
Maria's magical hacksaw + sari-sari store ledger.
Identifying two-force members before solving — reduces unknowns
Memory aid
Before solving, do a TWO-FORCE PATROL. Ask three questions using the acronym P-I-N: P = Pinned at both ends? I = Is it straight? N = No intermediate loads? If all three are YES → TWO-FORCE MEMBER. Mark it, assign its force along its axis, and reduce unknowns. Think of a PIN as the keyword — a pure pin-to-pin member with nothing else is always two-force.
Why it works
The acronym P-I-N creates a checklist that uses the word 'pin,' which is already relevant to the topic. This prevents accidentally treating a beam with a midspan load as a two-force member.
Example usage
Before solving, scan all members. For each: P — yes (pin at C and D). I — yes (straight). N — yes (no other loads). → Two-force member. Force along CD. Reduces unknowns from 2 to 1.
Recall trigger
Do the P-I-N check: Pinned both ends? Is it straight? No intermediate loads?
Sum of moments to find strut force in a frame (avoid solving two simultaneous equations)
Memory aid
The PRC board exam trick: 'TAKE MOMENTS AT THE UNKNOWN PIN.' When you have two unknown pin components at point A (Ax and Ay) and one known strut force angle, take ΣM=0 about point A. The two unknowns (Ax, Ay) create ZERO moment because they pass through A. Only the strut force and external load appear — one equation, one unknown. This is called 'killing two unknowns with one moment.' Remember: 'TAKE MOMENTS AT THE MESSY PIN — it kills the mess instantly.'
Why it works
The phrase 'kills the mess' creates a vivid action image. This technique is the single most important frame-solving shortcut and deserves a dedicated anchor.
Example usage
Frame with pin at A (unknown Ax, Ay) and strut force Fs at C. Take ΣM_A = 0: Fs × (perpendicular distance) − P × (distance) = 0. Solve Fs directly.
Recall trigger
Messy pin with two unknowns → take ΣM at that pin → unknowns disappear.
Tension in each cable segment: T = H / cosθ
Memory aid
Think of H (horizontal component) as the 'backbone' of the cable — it never changes. The tension T in any segment is like the 'hypotenuse' of a right triangle where H is the base and V is the height. From basic trigonometry, cos θ = H/T, so T = H/cosθ. Visualize a right triangle at every joint: the bottom side is always H (constant), the slanted side is T (varies with angle). Steeper angle → longer hypotenuse → bigger T. This is why maximum tension is at the steepest segment (at supports).
Why it works
Connecting T = H/cosθ to the constant-base right triangle makes the formula derivable from first principles, reducing memorization to understanding.
Example usage
Segment at angle 20° to horizontal, H = 40 kN. T = 40/cos20° = 40/0.940 = 42.6 kN. If angle is 35°, T = 40/cos35° = 48.8 kN. Steeper = bigger T. ✓
Recall trigger
Right triangle with constant base H — T is always the hypotenuse.
Machines — same procedure as frames: dismember, apply equilibrium, find force ratio
Memory aid
Short board-exam rhyme:
'Frame or machine — the rule's the same:
Disassemble, draw, and equilibrium claim.
For machines, find what effort achieves —
The load it lifts, the force it heaves.
Arm over arm, that's your M.A. gain —
Multiply by effort, and the load you'll name!'
Why it works
The rhyme reinforces that machines and frames share the same analysis method, preventing students from thinking machines require a special formula beyond lever M.A.
Example usage
Pliers problem: treat as a machine. Dismember at the pivot pin. Apply ΣM=0 on one handle: (Hand force × effort arm) = (Grip force × load arm). Solve for grip force.
Recall trigger
Recite: 'Frame or machine — the rule's the same.'
For a cable with a single central point load, vertical equilibrium: 2T sinθ = P
Memory aid
Remember '2T sinθ = P' as 'TWO TALL SOLDIERS hold up the PRISONER.' Two cable segments (2T) each pull upward (sinθ component) to hold up the load P (the prisoner). Since the cable is symmetric, both sides have equal tension T. The '2' comes from two symmetrical segments. This only works when the load is exactly at midspan (symmetric). For off-center loads, the two tensions are different — the 'soldiers' are unequal.
Why it works
The soldier analogy gives the '2' a physical meaning (two segments). The prisoner = load P creates a memorable image of force balance.
Example usage
Central load P = 10 kN, θ = 14°. T = P/(2sinθ) = 10/(2×0.2425) = 20.6 kN. Quick check: 2 × 20.6 × 0.2425 ≈ 10 kN ✓
Recall trigger
'Two tall soldiers (2T sinθ) holding the prisoner (P).' → 2T sinθ = P.
Common board-exam pitfall: treating a multi-force member as two-force
Memory aid
Engineering student Carlos was in a hurry during the board exam. He saw a horizontal beam with a pin at each end — and immediately labeled it 'two-force.' But he forgot: the beam had a 20 kN load applied at its midpoint by a pulley! Carlos computed a wrong answer and lost 1 mark. The lesson: even if a member LOOKS like a two-force member, CHECK for intermediate loads first. 'Carlos's costly shortcut' — never skip the P-I-N check (Pinned both ends? Is it straight? No intermediate loads?).
Why it works
A cautionary story with a named character creates an emotional memory. 'Carlos's costly shortcut' becomes a memorable warning phrase.
Example usage
Before analyzing any frame: list ALL members, apply P-I-N check to each. Only those passing all three criteria are two-force. All others are multi-force → need full equilibrium.
Recall trigger
'Don't be like Carlos!' Always do the P-I-N check before classifying members.
Cable sag relationship: larger sag → smaller H; smaller sag → larger H (and larger T_max)
Memory aid
Think of a HAMMOCK strung between two coconut trees. When you loosen the ropes and let it sag deeply, you can relax easily — the horizontal pull on the trees is LOW. When you tighten the ropes so the hammock is almost flat (small sag), the trees are pulled together with enormous force — you can feel the horizontal tension is HUGE. This is why suspension bridge cables must have a minimum sag — making them too flat would require impossibly large horizontal tension forces and anchors.
Why it works
The hammock is a universally understood rest object. The physical feeling of tightening ropes creating larger horizontal force makes H = wL²/8d immediately intuitive.
Example usage
Problem: same cable, same load, sag reduced from 4 m to 2 m. H doubles: H₁ = wL²/8(4), H₂ = wL²/8(2) = 2H₁. T_max also increases. Confirm with formula.
Recall trigger
Tight hammock (small d) = trees pulled hard (big H). Saggy hammock (big d) = trees relaxed (small H).
Revision game
I am a structure that stands still, but my members feel more than just a push or pull — they bend, they shear, and they carry loads from multiple directions. What am I?
Answer
Frame
Memory link
A1 (bahay-kubo analogy) — a frame is like a bahay-kubo, not a bilao tray of bamboo sticks.
I am the one force component in a cable that NEVER changes, no matter how steep or shallow the cable becomes. What am I?
Answer
Horizontal component H
Memory link
A8 (tabo-on-string micro-story) — the horizontal tug of the string never changes, like H in a cable.
A cable carries a uniform load of 10 kN/m on a 50 m span with a sag of 5 m. I am the formula you use to find H. State me.
Answer
H = wL²/8d = 10(50²)/[8(5)] = 10(2500)/40 = 625 kN
Memory link
A9 (Weightlifter Loads at Eight Decks mnemonic) — w, L², 8, d → H = wL²/8d.
I am the three-step acronym for solving frames. Say my letters and what each one stands for.
Answer
E-D-P: External reactions → Dismember → Pin forces by equilibrium
Memory link
A2 (E-D-P acronym + frog dissection analogy)
A lever has an effort arm of 0.5 m and a load arm of 0.025 m. A 40 N effort is applied. What load is balanced? I am the formula and the answer.
Answer
M.A. = 0.5/0.025 = 20. Load = 40 × 20 = 800 N
Memory link
A5 (bote-opener analogy) and A6 (rhyme: arm over arm gives the might)
I am the shape a cable takes when it carries a uniform load per meter of HORIZONTAL distance (like a bridge deck). Name me.
Answer
Parabola
Memory link
A11 (DECK = PARABOLA, D-P alliteration; suspension bridge deck carries horizontal load)
I am the mistake Carlos made: he called a beam with a midpoint load a two-force member. What is the correct classification, and what check should he have done?
Answer
Multi-force member. He should have done the P-I-N check: Pinned both ends? Straight? No intermediate loads? The beam fails 'N' (has an intermediate load) → multi-force.
Memory link
A19 (Carlos's costly shortcut micro-story) + A14 (P-I-N patrol)
I am the location in a cable where the tension is maximum. In a parabolic cable, where am I?
Answer
At the supports — where the cable angle is steepest.
Memory link
A10 (Enchanted Kingdom roller coaster — steepest at the towers, T_max at the towers)
Formula mnemonics
H = wL² / 8d
Mnemonic
Weightlifter LOADS at EIGHT Decks. w (weight/load), L² (span squared), 8 (eight), d (deck/sag). Or: 'Eight sags below, span squared above, load on the side — H is what you derive.'
When to use
When the cable supports a load uniformly distributed per horizontal meter (e.g., a suspension bridge deck load). This is the primary cable formula in PRC board exams.
What each part means
H = horizontal component of cable tension (kN), constant throughout cable. w = uniform load per unit horizontal length (kN/m). L = total horizontal span (m). d = sag at midspan (m). The formula applies ONLY to a parabolic cable with uniform horizontal load.
T_max = √(H² + (wL/2)²)
Mnemonic
PYTHAGORAS at the TOWER: T_max is the hypotenuse of the right triangle formed by H (horizontal) and V = wL/2 (vertical reaction at support). 'Hypotenuse = Tower Tension.' Remember: V at each support = wL/2 for symmetric loading.
When to use
After computing H for a parabolic cable, use this formula to find the maximum tension (which occurs at the steepest point — the supports). Always the last step in a cable problem.
What each part means
T_max = maximum cable tension, at the supports (kN). H = horizontal component (kN). wL/2 = vertical reaction at each support for symmetric uniform load (kN). This is a direct Pythagorean identity: T² = H² + V².
T = H / cosθ (or T = √(H² + V²) at any point)
Mnemonic
H is the 'horizontal hero' — always present. θ is the cable's slope angle at the point of interest. 'Hero over cosine = Tension.' Since cos0° = 1 at midspan (flat), T = H at midspan for a shallow cable. At supports, θ is maximum, cosθ is smallest, T is largest.
When to use
For finding tension in any specific segment of a cable (both concentrated-load and parabolic cases). Start from H (constant), then use the angle of that segment.
What each part means
T = cable tension at any specific point (kN). H = constant horizontal component (kN). θ = angle of cable segment to horizontal at that point (degrees). V = vertical component of tension at that point (= wL/2 at supports for parabolic cable).
M.A. = effort arm / load arm; Load = Effort × M.A.
Mnemonic
'BOTE-OPENER RULE': Long arm (effort arm) on top, short arm (load arm) on bottom = big M.A. = big load. Or: 'Effort Arm Always goes on TOP of M.A. fraction.' Never confuse arms with forces when computing M.A.
When to use
For lever-type machines (pliers, crowbar, clamp, wheelbarrow). In board exam machine problems, identify pivot first, then measure both arms, then apply this formula.
What each part means
M.A. = mechanical advantage (dimensionless). Effort arm = distance from pivot (fulcrum) to where effort force is applied (m). Load arm = distance from pivot to where load acts (m). Load = output force (N or kN). Effort = input force (N or kN).
2T sinθ = P (symmetric cable with single central point load)
Mnemonic
'TWO TALL SOLDIERS (2T sinθ) HOLD THE PRISONER (P).' The '2' is because both cable segments share the load equally by symmetry. Rearranged: T = P / (2 sinθ).
When to use
When a single concentrated load acts at the midspan of a cable. First find θ from the sag geometry (tanθ = sag / half-span), then use T = P / (2sinθ).
What each part means
T = cable tension in each segment (kN) — equal by symmetry. θ = angle of each cable segment to horizontal (degrees). P = concentrated point load at midspan (kN). sinθ = vertical component factor. Note: H = T cosθ (computed after T is found).
Quick recall chains
Frame Analysis Steps (E-D-P)
Recall test
Close your eyes. A frame problem appears. What are the three steps in order? Say 'E-D-P' and list what each letter means.
Memory chain
Story chain: 'Engineer Eva ENTERS the frame (E = External reactions), then DISSECTS it with her magic hacksaw (D = Dismember), then PAYS each pin its due force (P = Pin forces by equilibrium).' E → D → P: 'Eva Dissects, Pays.' Alternatively: 'Every Dismembered Part' — a simple three-word phrase that spells E-D-P.
Items to remember
- Step 1: External reactions — analyze the WHOLE frame as a single FBD
- Step 2: Dismember — separate each member at the pins
- Step 3: Pin forces — apply ΣFx, ΣFy, ΣM to each individual member
P-I-N Check for Two-Force Members
Recall test
Name the three criteria for a two-force member using the P-I-N acronym. Then apply it: a horizontal beam is pinned at A and B with a downward load at midpoint C — does it pass P-I-N? (Answer: Fails N → multi-force.)
Memory chain
The word P-I-N is itself the anchor — it's what a two-force member IS (pin-to-pin). Story: 'A PIN is pinned at both ends, is straight as an arrow, and carries no extra luggage.' If all three PINs light up green → two-force member → force acts along the member axis → one unknown only.
Items to remember
- P — Pinned at BOTH ends (no fixed connections)
- I — Is it straight? (no bends or offsets)
- N — No intermediate loads (nothing between the two pins)
Cable Problem Solving Sequence
Recall test
Walk through the 6 D's for this problem: w = 8 kN/m, L = 30 m, d = 3 m. What is H? What is T_max? (H = 8×900/24 = 300 kN; T_max = √(300²+120²) = 323.1 kN)
Memory chain
Story: 'Draw the DECK (geometry). DECIDE the shape (parabola/catenary). DETERMINE H first. DERIVE vertical reactions. DEDUCE segment tension. DECLARE T_max at supports.' Six D's: Draw, Decide, Determine, Derive, Deduce, Declare. Remember '6 D's to solve a cable.'
Items to remember
- Step 1: Draw the cable geometry (identify sag d, span L, load type)
- Step 2: Identify load type → horizontal load = parabola; along-length load = catenary
- Step 3: Compute H (horizontal component) using H = wL²/8d or ΣM at cut
- Step 4: Compute vertical reactions at supports
- Step 5: Find tension in required segment using T = H/cosθ or T = √(H²+V²)
- Step 6: Find T_max at the steepest point (supports for parabolic)
Three Types of Structural Members by Force Content
Recall test
Without looking: what internal forces can develop in a cable? In a truss member? In a frame member? Spell TAC and explain each letter.
Memory chain
'TAC = Two, All, Cables.' T = Truss (Two-force). A = A frame/machine has All three internal forces. C = Cables have one type only (tension). Visualize TAC as an acronym on a structural engineering textbook spine: T-A-C stacked vertically like members on a structure.
Items to remember
- TRUSS member: two-force only (axial: tension or compression)
- FRAME/MACHINE member: multi-force (axial + shear + bending)
- CABLE: one-force only (tension along cable length, no compression)
Common Board Exam Pitfalls (4 F's to AVOID)
Recall test
List the 4F pitfalls from memory. For each, state one check you can do on the exam to avoid it.
Memory chain
'Four Fatal Frame/Cable Fails = 4F.' F1: False two-force classification. F2: Forgotten Newton's Third. F3: Forgetting H is constant. F4: Finding T_max at the wrong location. Remember Carlos (A19) for F1, the arm-wrestling rule (A3) for F2, the tabo story (A8) for F3, and the roller coaster (A10) for F4.
Items to remember
- False two-force: calling a multi-force member a two-force member
- Forgotten Newton's 3rd Law: not flipping the pin force direction on the opposite FBD
- Forgetting H is constant: recalculating H for each segment (it never changes!)
- Finding T_max at midspan instead of at supports
Diagrams
Frames, Machines and Cables — Master Concept Map
A hierarchical mind map covering all three main topics (Frames, Machines, Cables) with their key subtopics, formulas, and pitfalls. Use this as a visual summary before attempting practice problems.
Frame and Cable Analysis Decision Flowchart
A comprehensive decision flowchart guiding students through identifying the structure type and applying the correct analysis method. Includes branching for frames, trusses, machines, and all three cable types. FontAwesome icons highlight key warnings and checkpoints.
Two-Force Member P-I-N Identification Flowchart
A focused flowchart for the P-I-N check procedure. Students follow the three yes/no questions to determine if a member is two-force or multi-force, then see the consequence of each classification.
Distribution of Cable Load Types in PRC Board Exam Problems
An approximate distribution of cable problem types commonly encountered in PRC Civil Engineer board examinations. Parabolic cable problems dominate, confirming that H = wL²/8d and T_max = √(H²+(wL/2)²) should be prioritized in review.
Step-by-Step Solution Timeline for a Parabolic Cable Problem
A step-by-step timeline showing the logical order of solution steps for parabolic cable problems. The linear timeline format reinforces the sequential nature of the solution and prevents skipping critical steps.