Analysis of Determinate Structures - Memory Anchors
For the PRC Civil Engineer board exam, you will face problems under time pressure — your brain needs instant, reliable recall of formulas, sign conventions, and procedures. This collection uses mnemonics, analogies, micro-stories, and vivid visual anchors to wire every key concept of Analysis of Determinate Structures into long-term memory. Think of each anchor as a mental 'hook' — the stranger, funnier, or more emotionally charged, the harder it is to forget. Pair these with practice problems and you will approach the board exam with genuine confidence.
Memory anchors
Degree of Indeterminacy formula for frames: DI = (3m + r) - (3n + ec)
Memory aid
Use the phrase: '3 Members plus Reactions, MINUS 3 Nodes plus Conditions' → '3MR minus 3NC'. Imagine a balance scale: the LEFT side piles up '3 Members + Reactions' (unknowns), and the RIGHT side piles up '3 Nodes + Conditions' (equations). If left outweighs right, the structure is OVER-constrained (indeterminate). If balanced → determinate (DI=0). If right outweighs → unstable.
Why it works
The balance-scale image gives a physical meaning to the arithmetic, and the 3MR-3NC phrase has a rhythm that makes it easy to chant mentally during the exam.
Example usage
For a portal frame: m=3, r=6, n=4, ec=0. Left pan: 3(3)+6=15. Right pan: 3(4)+0=12. DI = 15-12 = 3 → indeterminate to 3rd degree.
Recall trigger
Picture a Filipino 'timbangan' (scale) with members and reactions on the left pan.
Degree of Indeterminacy for beams: DI = r - (3 + ec)
Memory aid
Chant: 'R minus three plus EC, that is DI for a beam, you see!' (r − (3 + ec) = DI). The '3' stands for the three classic equilibrium equations (ΣFx, ΣFy, ΣM). Every internal hinge you ADD increases ec by 1, giving you ONE MORE equation and making the structure EASIER to solve — like adding a helper to your team.
Why it works
Rhymes create phonological loops in working memory, enabling effortless retrieval under stress. The helper analogy adds conceptual grounding.
Example usage
Beam with 5 reaction components, 2 internal hinges: DI = 5 − (3+2) = 0 → determinate. Perfect for a Gerber beam!
Recall trigger
Humming 'R minus three plus EC' while looking at a beam problem.
DI = 0 → determinate; DI > 0 → indeterminate; DI < 0 → unstable
Memory aid
Think of a BASKETBALL TEAM analogy: DI=0 means your team has EXACTLY 5 players on court — perfect game (determinate). DI>0 means you have TOO MANY players (too constrained, you need extra compatibility equations). DI<0 means you have FEWER than 5 players — the team COLLAPSES (unstable structure). The PBA referee (equilibrium) can only manage 5 at a time!
Why it works
Filipinos are passionate about basketball. Mapping an abstract number to a familiar sports rule makes the classification instantly memorable.
Example usage
If DI = -1, your 'team' is short one player — the structure will collapse under any load (unstable). Check if reactions are concurrent or parallel.
Recall trigger
Picture 5 PBA players on a court whenever you see a DI calculation.
Stability vs. Determinacy — a structure can be geometrically unstable even if the reaction count is sufficient
Memory aid
Imagine Engineer Rico builds a bridge and counts his supports: 'I have exactly 3 reaction components — it must be determinate and stable!' But he forgot to check the arrangement. All three reactions are PARALLEL (all vertical) — the bridge slides horizontally on the first wind. Rico's boss scolds him: 'Counting is not enough — POSITION matters!' Rico never forgets: check concurrent/parallel reaction lines, not just the total count.
Why it works
Narrative memory (episodic encoding) is among the most durable forms. Engineer Rico's embarrassing mistake creates a vivid emotional anchor.
Example usage
Three rollers all pointing vertically under a beam: r=3, DI=0 mathematically, but the structure is still unstable horizontally. Always sketch the reaction lines.
Recall trigger
Think of Rico's parallel-reaction bridge whenever a problem has 'just enough' reactions.
Internal hinge adds one equation of condition (ec = 1 per hinge)
Memory aid
An internal hinge is like a BALIKBAYAN BOX HINGE on a suitcase lid: at the hinge point, the lid can rotate freely, so it CANNOT transfer moment across it. Mathematically, ΣM = 0 at that point for either side — that is your extra equation. Each hinge = one 'extra receipt' (equation) you can use at checkout (solving the system).
Why it works
Balikbayan boxes are deeply familiar to Filipinos. The physical inability to transfer moment is captured by the 'rotating lid' image.
Example usage
A Gerber beam with 1 internal hinge: ec=1. Use ΣM=0 at the hinge (take one side of the cut) as your 4th equation alongside 3 global equilibrium equations.
Recall trigger
Visualize a balikbayan box hinge that cannot resist being twisted.
Procedure for Gerber (compound) beam — solve suspended span FIRST
Memory aid
Remember: 'SIP then POUR' — Solve the Inner (suspended) Part first, then POUR its reaction onto the outer span as an applied load. Like making 'sinigang': you must first prepare the tamarind base (inner span) before adding it to the main pot (outer span). Getting the sequence wrong ruins the whole dish!
Why it works
Filipino cooking analogies are culturally resonant. The sequential dependency (inner→outer) mirrors the dependency in Gerber beam analysis.
Example usage
Identify the suspended (released) span. Apply ΣFy=0 and ΣM=0 for that span to get its reactions. Transfer the reaction at the connection point as a downward load onto the main span, then solve the main span normally.
Recall trigger
Sinigang base first, then the main pot — SIP then POUR.
Sign convention for internal forces: positive shear (V) and positive moment (M)
Memory aid
For SHEAR: imagine a HAPPY FACE (→←) — left side pushes right, right side pushes left → positive shear (counter-clockwise couple). For MOMENT: think of a SMILING BEAM — positive moment causes SAGGING (concave up, like a smile ☺). Negative moment causes HOGGING (concave down, like a frown ☹). Filipino engineers use 'ngiti = pababa' (smile = sag = positive) to remember this instantly.
Why it works
Associating a mathematical sign with a facial expression ties emotion to abstraction, making the convention virtually impossible to forget.
Example usage
When computing M at a section and the left side produces sagging (bottom fiber in tension), report M as positive. If your calculation gives a negative number, the beam hogs at that section.
Recall trigger
Ngiti (smile) = sag = positive moment. Frown = hog = negative moment.
The three internal forces at any section: N (axial), V (shear), M (moment)
Memory aid
Remember 'NVM' — 'Never Void the Method' (N=axial, V=shear, M=moment). To find each: take ONE SIDE of the cut, then: N = sum of forces ALONG the member axis; V = sum of forces PERPENDICULAR to the axis; M = sum of moments about the cut point. NVM — Never forget to check all three!
Why it works
NVM is a familiar internet acronym (Never Mind), repurposed here. The repurposing itself creates a memorable 'twist' that aids encoding.
Example usage
At any section in a frame member, isolate one side, compute ΣF along axis (=N), ΣF perpendicular (=V), ΣM at cut (=M). Always report all three.
Recall trigger
NVM: axial Never hides, shear is Visible transversely, Moment sums about the cut.
Three-hinged arch is statically determinate (4 unknowns, 4 equations)
Memory aid
Chunk it as '4-4 LOCKDOWN': 4 UNKNOWNS (Ax, Ay, Bx, By) and 4 EQUATIONS (ΣFx=0, ΣFy=0, ΣM_A=0, PLUS the crown condition ΣM_C=0 for one side). It's like a championship game that goes EXACTLY 4 quarters with a decisive score — no overtime needed (no extra compatibility equations). '4 unknowns meet 4 equations = LOCKDOWN determinate.'
Why it works
Chunking reduces cognitive load by grouping related items (4+4). The sports metaphor reinforces that the match is perfectly balanced.
Example usage
For a three-hinged arch, immediately write 4 equations. The 4th equation (crown condition) is ΣM_C = 0 applied to EITHER the left or right portion — use whichever has fewer loads.
Recall trigger
4-4 LOCKDOWN — 4 unknowns, 4 equations, zero overtime.
Horizontal thrust H in a three-hinged arch reduces bending compared to a beam
Memory aid
Imagine a JEEPNEY LOAD: A flat beam bridge is like a jeepney with passengers sitting ON TOP — all weight pushes DOWN, creating huge bending. A three-hinged arch is like the jeepney's arched chassis — loads are redirected into the wheel axles as THRUST (compression along the arch). The horizontal thrust H is the secret weapon that turns bending into compression, making arches efficient for long spans. That is why arches appear in Rizal Park's monument gateway!
Why it works
Jeepneys are quintessentially Filipino. The contrast between flat load path and arched load path makes the structural efficiency of arches tangible.
Example usage
For an arch, M_arch at any section = M_beam − H·y, where y is the arch height at that section. The larger H, the smaller the net moment — the arch is efficient.
Recall trigger
Picture a jeepney chassis arch redirecting load to the wheels (supports) via thrust.
Crown condition for three-hinged arch: ΣM_C = 0 for one portion
Memory aid
Engineer Liza is at the crown of the arch, sitting on the hinge pin. 'I cannot resist rotation!' she shouts. 'So the moments about ME from EITHER SIDE must sum to zero!' She holds up a sign: 'ΣM_C = 0 — left side only!' Her colleague on the right side does the same: 'ΣM_C = 0 — right side only!' Both equations give the same result. Liza's zero-moment rule at the crown is how you always find the horizontal thrust H.
Why it works
Personifying the hinge as a character who 'refuses to resist rotation' makes the physical meaning of an internal hinge viscerally clear.
Example usage
After finding vertical reactions by global equilibrium, cut the arch at the crown and write ΣM_C = 0 for the LEFT portion. Solve for H. This is the unique step that distinguishes arch analysis from beam analysis.
Recall trigger
Liza sitting on the crown hinge, shouting 'I cannot resist rotation — ΣM_C = 0!'
How to find reactions using three equilibrium equations (ΣFx=0, ΣFy=0, ΣM=0)
Memory aid
The three equations spell 'FFM': Force-x, Force-y, Moment. Remember: 'FiRST Find Moments about the pin support to kill two unknowns at once.' This is the expert boarder's shortcut — taking moments about a support point eliminates the reactions AT that point from the equation. 'FiRST take Moments, then FFs (Forces)' → FFM solved cleanly.
Why it works
Providing a practical sequence ('take moments first') avoids simultaneous equations and is the actual technique used by top reviewees.
Example usage
For a simply supported beam with a pin at A and roller at B, take ΣM_A=0 first → solves for V_B directly. Then ΣFy=0 → V_A. Then ΣFx=0 → H_A (if any horizontal load).
Recall trigger
FFM: ΣFx, ΣFy, ΣM — always take the MOMENT equation about a pin support first.
Equations of condition for a hinge shared by k members: ec = k − 1
Memory aid
Think of a POTLUCK DINNER with k families sharing one table. Each extra family brings ONE extra dish (equation) beyond the first. So k families → k−1 extra dishes → ec = k−1. If only 2 families (2 members at a hinge), only 1 extra condition. If 3 members share a hinge, ec = 2. The first family is the 'host' (no extra equation), each additional family adds one.
Why it works
Filipino potluck (handaan) culture is universal. Counting 'extra dishes' maps directly to counting 'extra conditions,' with the k−1 formula arising naturally.
Example usage
A joint where 3 members meet at a hinge: ec = 3−1 = 2. This is important in complex Gerber beams with multiple internal hinges at a single point.
Recall trigger
Potluck: k families, k−1 extra dishes = k−1 equations of condition.
For frames, draw N, V, M diagrams member by member
Memory aid
Use the mental image of WALKING THROUGH A BAHAY-KUBO: First, walk through the VERTICAL POSTS (columns) — draw their N, V, M as you climb from base to top. Then walk through the HORIZONTAL BEAMS (rafters) — draw their diagrams left to right. At every CORNER (joint), forces and moments MUST BALANCE — check equilibrium at each joint before moving to the next member. Your mental walk through the house = your solution walk through the frame.
Why it works
The method of loci (memory palace) uses spatial navigation to encode procedural knowledge. Bahay-kubo is an iconic Filipino image, making the palace vivid.
Example usage
For an L-frame with column AB and beam BC: analyze AB first (isolate as free body), then BC. At joint B, the forces transferred from AB become loads on BC. Equilibrium at B must be satisfied.
Recall trigger
Walking through a Bahay-kubo: posts first, then beams, equilibrium at every corner.
Concurrent or parallel reactions → geometric instability
Memory aid
CONCURRENT reactions: Imagine three soldiers POINTING RIFLES at the SAME TARGET. If someone pushes the structure perpendicular to their aim, the soldiers cannot help — they all shoot in the same direction. PARALLEL reactions: Three soldiers standing in a LINE, all facing UP. A sideways shove and the structure slides — no one is facing sideways to stop it. If your reactions all meet at one point (concurrent) or all face the same direction (parallel) → UNSTABLE.
Why it works
Military imagery with soldiers creates a strong visual-spatial memory. The 'can't help if push is sideways' logic makes geometric instability intuitively clear.
Example usage
A three-roller beam where all rollers are vertical: reactions are parallel → structure slides horizontally. Fix it by replacing one roller with a pin to add a horizontal reaction.
Recall trigger
Soldiers pointing the same direction = parallel reactions = unstable against lateral load.
The moment at a fixed support includes both a force reaction and a moment reaction
Memory aid
A FIXED support is like a SENATORIAL SEAT — it can resist EVERYTHING: horizontal push (Ax), vertical push (Ay), AND rotation (MA). A PIN support is like a MAYOR's seat — resists horizontal and vertical but can be ROTATED OUT of office (no moment resistance). A ROLLER is like a BARANGAY KAGAWAD — resists only the perpendicular direction, everything else slides away.
Why it works
Philippine political hierarchy is familiar to every Filipino. Mapping support types to government ranks creates a memorable, culturally resonant classification system.
Example usage
For a cantilever beam (fixed at A, free at B): r = 3 (Ax, Ay, MA). DI = 3−(3+0) = 0 → determinate. Find all three reactions from ΣFx=0, ΣFy=0, ΣM_A=0.
Recall trigger
Fixed = Senator (3 powers), Pin = Mayor (2 powers), Roller = Kagawad (1 power).
Relationship between load, shear, and moment: dV/dx = −w; dM/dx = V
Memory aid
Picture a WATER CHANNEL (kanal) with rain falling in at rate w: every meter you travel, the flow (shear V) DECREASES by w — that is dV/dx = −w, the load 'drains' the shear. Now, the VOLUME of water collected (moment M) grows with the flow rate V — that is dM/dx = V. A heavy downpour (large w) quickly reduces shear; a strong flow (large V) quickly builds up the moment. The kanal remembers both rules simultaneously.
Why it works
The water-channel analogy links the abstract differential relations to a physical, everyday Filipino experience (monsoon flooding). The imagery is vivid and sequential.
Example usage
For a UDL of w kN/m over length L, shear decreases linearly (slope = −w) and moment is parabolic (dM/dx = V which is linear). Use these relationships to sketch SFD and BMD without computing every section.
Recall trigger
Kanal: rain w drains the shear; shear flow V fills the moment M.
Axial force N in a frame column = sum of vertical forces from the beam(s) it supports
Memory aid
A frame column is like a PORTER (kargador) at the port. The beams dump their vertical reactions onto the porter's shoulders — these vertical beam reactions become the porter's axial load. The horizontal loads create shear in the porter's legs (columns). The porter must carry ALL vertical loads delivered from the beams above. Count every vertical force that 'lands' on the column's head to get N.
Why it works
The kargador image is visceral and familiar. It correctly maps vertical force transfer from beams to columns as axial compression.
Example usage
For an L-frame: the beam BC carries a vertical load of 15 kN at C. The beam transfers V=15 kN to the column AB at joint B. Therefore N at the column base = 15 kN (compression).
Recall trigger
Kargador (porter) carrying the vertical loads from beams = column axial force N.
Board-exam pitfall: mistaking determinacy for stability
Memory aid
Chant: 'Determinacy counts, stability SITS — check the lines where each reaction HITS!' Counting reactions tells you the degree (DI), but stability depends on WHERE reactions act. A structure with DI=0 can still collapse if reactions are badly positioned. After counting, always SKETCH the reaction lines and ask: 'Can this structure slide or rotate without resistance from any of my reactions?'
Why it works
The rhyme creates a mental checklist: count THEN check positions. The word 'SITS' (sits in place = stable) versus 'HITS' (where it applies) creates an auditory hook.
Example usage
If three rollers are parallel, DI=0 (determinate by count) but the structure is unstable (all reaction lines parallel — can't resist horizontal loads). Always sketch before concluding stable.
Recall trigger
Determinacy counts; stability SITS — check the lines where reactions HITS.
Free body diagram (FBD) as the foundation of every structural analysis
Memory aid
A great structural engineer named Ate Mira told her board-exam students: 'The FBD is your CONFESSION BOOTH. You go inside it with ONE part of the structure, confess ALL the forces acting on it, and then find peace (equilibrium). Never skip the FBD — structures that skip confession collapse!' She made every student draw the FBD before writing a single equation. All of them passed the board.
Why it works
The confession booth is a culturally resonant Catholic reference. The humor and the authoritative character (Ate Mira) make the procedural advice memorable and emotionally charged.
Example usage
Before writing ΣFx, ΣFy, ΣM for any structure or cut section, draw the FBD first. Show all known loads, reaction arrows with correct directions, and the internal forces at cuts (N, V, M).
Recall trigger
Ate Mira's confession booth: isolate the body, list all forces, find equilibrium.
Revision game
I am the hinge that sits at the top of an arch. Moments must sum to zero at my location — from either side. What single force does this condition let you find?
Answer
The horizontal thrust H of a three-hinged arch
Memory link
A11 (Liza sitting on the crown hinge, shouting 'ΣM_C = 0!')
I am the number you get when you subtract (3n + ec) from (3m + r). If I equal zero, you can solve the structure using equilibrium alone. What am I?
Answer
The Degree of Indeterminacy (DI) for a frame
Memory link
A1 (Timbangan / balance scale — 3MR minus 3NC)
A Kagawad, a Mayor, and a Senator walk into a structural analysis exam. The Kagawad can resist 1 force, the Mayor can resist 2, and the Senator resists 2 forces plus 1 moment. What are these three characters representing?
Answer
Roller (1 reaction), Pin/Hinge (2 reactions), Fixed Support (3 reactions)
Memory link
A16 (Senator/Mayor/Kagawad analogy for support types)
I am the step you must NEVER skip before writing your equilibrium equations. Engineers who skip me often get the wrong answer. My full name involves isolating a body and marking all the forces on it.
Answer
The Free Body Diagram (FBD)
Memory link
A20 (Ate Mira's confession booth analogy)
A beam with 5 reaction components and 2 internal hinges. Am I determinate, indeterminate, or unstable? Show your work using the beam formula.
Answer
DI = r − (3 + ec) = 5 − (3 + 2) = 0 → Statically Determinate
Memory link
A2 (rhyme: R minus three plus EC, that is DI for a beam, you see!)
I solve sinigang before the main pot. I am the span you MUST analyze first in a Gerber compound beam before moving on to the main span. What type of span am I?
Answer
The suspended (released) span — SIP then POUR
Memory link
A6 (Sinigang analogy — SIP then POUR for Gerber beams)
A portal frame fixed at both bases: m=3 members, r=6 reaction components, n=4 nodes, no internal hinges. How indeterminate is this structure? Name the type of structure.
Answer
DI = (3×3 + 6) − (3×4 + 0) = 15 − 12 = 3, indeterminate to the 3rd degree (a fixed-base portal frame)
Memory link
A1 (3MR minus 3NC mnemonic with the timbangan balance scale)
Engineer Rico counted exactly 3 reactions for his beam — two vertical rollers and one inclined roller pointing in the same direction as the first. He declared it determinate and stable. What dangerous mistake did Rico make?
Answer
Rico confused determinacy with stability. Three parallel reactions make the structure geometrically unstable (no horizontal resistance), even though DI = 0.
Memory link
A4 (Rico's parallel-reaction bridge micro-story) and A15 (soldiers pointing the same direction)
Formula mnemonics
DI = (3m + r) − (3n + ec) [frames]
Mnemonic
3MR minus 3NC: 'Three Members plus Reactions, minus Three Nodes plus Conditions.' Left of minus = unknowns; right = equations. Surplus unknowns = degree of indeterminacy.
When to use
Use for any planar frame (portal frames, multi-story frames, L-frames) where you have multiple members connected at joints.
What each part means
m = number of members; r = total external reaction components; n = number of joints/nodes; ec = equations of condition (1 per internal hinge for 2-member connection). Result: DI=0 determinate, DI>0 indeterminate, DI<0 unstable.
DI = r − (3 + ec) [beams]
Mnemonic
'R minus the crew (3) plus helpers (ec).' The crew of 3 are ΣFx, ΣFy, ΣM. Each internal hinge adds a helper (extra equation), making ec larger, reducing DI.
When to use
Use for beams and simple structures (not complex frames). Faster than the general frame formula when dealing with beams only.
What each part means
r = total reaction components from all supports; 3 = three global equilibrium equations; ec = number of internal hinge conditions (1 per hinge connecting 2 members).
ΣM_C = 0 (one side) → H for three-hinged arch
Mnemonic
'CROWN CUTS: Cut at Crown, take One side, Write ΣM=0, get H.' The letter C appears in Crown, Cut, and Condition — triple-C rule.
When to use
Use exclusively for three-hinged arches after finding vertical reactions from global equilibrium. This is the FOURTH equation unique to arch analysis.
What each part means
ΣM_C = sum of all moments about the crown hinge C, taken for EITHER the left or right free-body portion of the arch. Setting this to zero uses the internal hinge condition (the hinge transmits no moment) to solve for the horizontal thrust H.
M_arch = M_beam − H · y
Mnemonic
'Arch Moment = Beam Moment minus Thrust times Height.' The arch LIFTS the moment diagram DOWN by H·y. Think of H as a 'moment reducer' — the higher the arch (larger y), the more it reduces bending.
When to use
Use to find internal moments at any arch section after H is determined. Particularly useful for comparing arch efficiency against an equivalent flat beam.
What each part means
M_arch = bending moment at a section in the arch; M_beam = bending moment at the same horizontal position if the arch were a flat beam of the same span and loading; H = horizontal thrust; y = rise of the arch at that section above the chord.
dV/dx = −w; dM/dx = V
Mnemonic
'Load Drains Shear; Shear Fills Moment.' Load (w) drains shear at rate w (negative slope). Shear (V) fills moment (positive slope equals shear). L-D-S-F-M: Load Drains Shear, (Shear) Fills Moment.
When to use
Use to sketch shear and moment diagrams without computing forces at every section. Discontinuities in SFD occur at concentrated forces; discontinuities in BMD slope occur at concentrated moments.
What each part means
dV/dx = rate of change of shear along beam = −w (distributed load intensity). dM/dx = rate of change of moment = V (shear value at that point). Positive w is downward; these relations enable rapid SFD and BMD sketching.
ec = k − 1 (hinge shared by k members)
Mnemonic
'k members, k minus 1 equations — the first member is the HOST, rest are GUESTS who each bring one equation.' One hinge, two members → 1 equation. One hinge, three members → 2 equations.
When to use
Use when multiple members converge at a single internal hinge, which is common in complex trusses, arched structures, or multi-span Gerber beams.
What each part means
k = number of members meeting at one internal hinge; ec = equations of condition provided by that joint. Total ec for a structure = sum of (k−1) over all internal hinges.
Quick recall chains
Steps to Analyze a Three-Hinged Arch
Recall test
Without looking: what is the 4th step in arch analysis, and WHY is that step unique to arches?
Memory chain
The story of a Climb to the Crown: START at support A (FBD). Take MOMENTS at A to find your buddy at B (V_B). Sum FORCES vertically to find what you carry at A (V_A). Reach the CROWN C — there, the hinge shouts 'no moment from either side!' — that gives you the THRUST H. CHECK: horizontal forces balance (H_A = H_B). Finally, CUT anywhere to find the internal forces.
Items to remember
- Draw FBD and identify all loads and supports
- Apply ΣM_A = 0 to find V_B
- Apply ΣFy = 0 to find V_A
- Apply ΣM_C = 0 (left portion) to find H
- Apply ΣFx = 0 to confirm H_A = H_B
- Find internal N, V, M at any section by cutting
Support Types and Their Reaction Components
Recall test
A beam has two pins and one roller. What is r? (Answer: 2+2+1 = 5 reaction components.)
Memory chain
KAGAWAD-MAYOR-SENATOR: Kagawad (Roller) has 1 power, Mayor (Pin) has 2 powers, Senator (Fixed) has 3 powers. Each step up in 'political rank' adds one more reaction component. Counting the total reactions r is as easy as totalling the political power of all supports.
Items to remember
- Roller: 1 reaction (perpendicular to surface)
- Pin / Hinge: 2 reactions (Fx, Fy)
- Fixed: 3 reactions (Fx, Fy, M)
Procedure for Solving Determinate Beam Reactions
Recall test
If a simply supported beam has only vertical loads, which equation do you apply first, and why does it simplify the solution?
Memory chain
DTMFC: 'Draw Then Moment, Force, Check' — Draw FBD, Take moments first, then Forces (Fy then Fx), finally Check with a second moment equation. Pronounce it 'DiTiMFiC' as a Filipino nickname: 'Di ka mabibigo sa DiTiMFiC!'
Items to remember
- Draw the FBD with all loads and reactions labeled
- Take ΣM about one support (eliminate 2 unknowns at once)
- Use ΣFy = 0 to find the remaining vertical reaction
- Use ΣFx = 0 to find horizontal reaction (if any)
- Check with a moment equation about the other support
Determinacy Classification Steps
Recall test
A frame with m=4 members, r=9 reaction components, n=5 joints, no internal hinges: compute DI. (Answer: DI = (12+9)−(15+0) = 21−15 = 6, indeterminate to 6th degree.)
Memory chain
RMN-EC-DI-S: 'Radio-Manigong-Network: Every Crusade Delivers Instant Stability.' R (reactions), M (members), N (nodes), EC (equations of condition), DI (compute), S (stability check). Like tuning into RMN radio before starting your day — follow the sequence to tune your structure.
Items to remember
- Count all reaction components (r) — use Senator/Mayor/Kagawad rule
- Count members (m) and joints/nodes (n) for frames
- Count equations of condition (ec) — 1 per internal hinge (2-member case)
- Compute DI using the appropriate formula
- Interpret result: DI=0 determinate, DI>0 indeterminate, DI<0 unstable
- Additionally check stability (concurrent/parallel reactions)
Internal Force Sign Convention Summary
Recall test
At a section where the left portion exerts an upward shear and the beam sags: is shear V positive or negative? Is moment M positive or negative?
Memory chain
TAC-SMS: 'Tension, (clockwise) Shear, Smile (sag) = positive N, V, M.' The word TAC (short for 'tack,' to fix in place) reminds you these are the POSITIVE cases. Tension keeps the member together (T), clockwise shear (C), and the beam Smiles (S) when positive moment causes sag.
Items to remember
- Positive axial N: tension (member elongates)
- Positive shear V: left face down, right face up (or: tends to rotate clockwise)
- Positive moment M: sagging (concave up, bottom fiber in tension)
Diagrams
Master Concept Map: Analysis of Determinate Structures
A comprehensive mind map showing all major topics and their relationships in the Analysis of Determinate Structures chapter. Use this as a big-picture navigator before diving into individual memory anchors.
Decision Flowchart: Classifying a Structure (Determinacy and Stability)
Step-by-step decision tree for classifying any planar structure as stable/unstable, determinate/indeterminate. Follow the arrows: identify type, count components, compute DI, interpret, then check geometric stability. This flowchart embodies the complete decision logic in one visual.
Three-Hinged Arch Analysis Procedure
A complete procedural flowchart for three-hinged arch analysis. Highlights the unique 4th step (crown condition) that distinguishes arch analysis from beam analysis. The horizontal thrust H emerges from the crown hinge condition.
States of a Structure Based on Degree of Indeterminacy
State diagram showing the possible structural states and the transitions between them based on the DI calculation and geometric stability check. Illustrates clearly that determinacy and stability are two SEPARATE checks.
Distribution of Reaction Components by Support Type in Typical Board Problems
Pie chart illustrating the relative frequency and weight of each support type in typical structural analysis problems. Fixed supports (Senators) contribute the most reaction components per support; rollers (Kagawads) contribute the least.