CELE Engineering Mechanics — Analysis of TrussesDetailed Explanation
Want to really understand Analysis of Trusses before tackling CELE Engineering Mechanics questions? This detailed explanation breaks down every key concept, shows you why it matters for the CELE 2026, and walks through the reasoning Professional Regulation Commission (PRC) — Board of Civil Engineering expects on high-difficulty questions.
Exam context
Professional Regulation Commission (PRC) — Board of Civil Engineering runs the Civil Engineer Licensure Examination on May and November 2026. Its Engineering Mechanics section sits under a "Core" weighting, and Analysis of Trusses is the 3rd chapter in the 8-chapter CELE Engineering Mechanics rotation. The CELE passing mark is 70% weighted average, no sub-test below 50%, and the most recent 2026 paper drew about a meaningful share of questions from Engineering Mechanics.
Analysis of Trusses - Detailed Explanation
Trusses are among the most frequently tested structures in the PRC Civil Engineer Licensure Examination. A truss is an assemblage of straight, slender members joined at their ends by frictionless pins to form a stable, rigid framework. Because loads are applied only at the joints and members are pin-connected, every member acts as a pure axial two-force member — carrying either tension or compression with no bending. Real-world applications include roof trusses, bridge trusses, transmission towers, and long-span steel frames governed by NSCP 2015 and AISC 360-16. This chapter develops three core analytical tools — the determinacy check, the method of joints, and the method of sections — along with the rapid identification of zero-force members. Master these and you will solve virtually every truss problem that appears on the board exam.
Concepts
Truss Assumptions and the Two-Force Member Principle
A truss is idealized under three classical assumptions that are the foundation of all subsequent analysis. First, members are connected by frictionless pins at the joints; this means joints can transmit force in any direction but cannot resist or transmit moment. Second, all external loads and support reactions act exclusively at the joints; distributed loads on members (e.g., roof cladding weight) are first converted to equivalent joint loads before analysis. Third, members are straight, and their self-weight is negligible — or, if considered, half is lumped to each end joint. Under these assumptions, the resultant force in any member must be directed along the member's longitudinal axis, making every truss member a two-force member. A two-force member is in pure tension if it pulls on its connecting joints (member is being stretched) or in pure compression if it pushes on its connecting joints (member is being shortened). The standard sign convention adopted throughout this chapter — and on the board exam — is: tension is positive (+) and compression is negative (−). When analyzing a joint, we always assume unknown member forces are in tension, pointing away from the joint along the member. If the computed value is positive, the assumption is correct (tension); if negative, the member is actually in compression.
Examples
The force arrow points away from joint D toward E, consistent with tension. On a free-body diagram of joint D, the 20 kN and 15 kN components act away from D and must be balanced by other member forces or the external reaction at D.
Scenario
A diagonal member DE of a roof truss connects joint D(0,0) to joint E(4,3) m. It is subjected to a resultant force of 25 kN directed from D to E. Is the member in tension or compression, and what are the x- and y-components of the force it exerts on joint D?
Solution
Step 1: Compute the member length. L = √(4² + 3²) = √(16 + 9) = 5.0 m. Step 2: Determine the direction cosines. cos θ = 4/5 = 0.80 (horizontal); sin θ = 3/5 = 0.60 (vertical). Step 3: The force is directed from D toward E (away from D), which means the member pulls on joint D. A member that pulls on its joints is in TENSION. F_DE = +25 kN. Step 4: Components at joint D: F_x = +25(0.80) = +20 kN (→); F_y = +25(0.60) = +15 kN (↑).
Applications
- Roof trusses in industrial buildings (NSCP 2015 Section 204 — load combination at joints).
- Bridge truss chords and web members in DPWH road bridge design.
- Transmission tower diagonals and cross-bracing members.
- Long-span steel trusses in sports arenas and hangars analyzed per AISC 360-16 Chapter D (tension) and Chapter E (compression).
Misconceptions
- MISCONCEPTION: 'A pin joint can resist moment.' CORRECTION: An ideal pin transmits force only; moment = 0 at pin joints, which is exactly why no bending occurs in truss members.
- MISCONCEPTION: 'A member with larger cross-section carries more load.' CORRECTION: In truss analysis, member forces depend solely on geometry and joint loads, not on cross-sectional area.
- MISCONCEPTION: 'Tension always acts upward and compression always acts downward.' CORRECTION: Tension or compression is about the axial state of the member, not its orientation.
Related Concepts
- Free-body diagrams
- Equilibrium of a particle
- Two-force members vs. three-force members
- Axial stress: f_a = P/A (NSCP 2015 / AISC 360)
Common Exam Questions
Example
Board exam stem: 'Member BC of a truss has a computed force of −18 kN. The member is in ___.' Answer: Compression (18 kN C).
Approach
Draw the member isolated as a free body. If the force arrows point inward (toward the member), it is compression. If they point outward (away from the member/toward the joints), it is tension.
Question Type
Identification of tension vs. compression
Example
If a member spans Δx = 6 m, Δy = 4 m, then L = √52 = 7.211 m; sin θ = 4/7.211 = 0.5547; cos θ = 6/7.211 = 0.8321.
Approach
Always compute the member length first, then use L to find sin θ and cos θ. Never use angles from memory unless the geometry is a standard 3-4-5 or 30-60-90 triangle.
Question Type
Component calculation
Key Points To Remember
- Frictionless pin joints → no moment transfer; only force is transmitted.
- All loads and reactions act at joints only.
- Every member is a two-force member: force is axial (along the member).
- Assume unknown forces as tension (pointing away from the joint); negative result = compression.
- Self-weight, if given, is split equally to the two end joints of the member.
Static Determinacy and Stability of Trusses
Before solving a truss, you must verify that it is statically determinate and stable. The governing equation for a plane (2D) truss is: m + r = 2j, where m = number of members, r = number of external reaction components, and j = number of joints. Derivation: at each joint we have two equilibrium equations (ΣFx = 0 and ΣFy = 0), giving 2j equations total. The unknowns are m member forces plus r reaction components. For a unique solution, unknowns must equal equations, hence m + r = 2j. Three outcomes exist: (1) m + r = 2j → statically determinate and (potentially) stable; (2) m + r > 2j → statically indeterminate (redundant members or reactions exist, requiring compatibility equations); (3) m + r < 2j → mechanism (insufficient members, will collapse). Critical warning: m + r = 2j is necessary but NOT sufficient for stability. A truss satisfying this equation can still be geometrically unstable if members are improperly arranged — for example, if all members meeting at a joint are concurrent through a single point on the same side, or if an entire panel has members that are nearly collinear. Always inspect the geometry. Standard stable trusses are built by the simple truss rule: start with a basic triangle (3 members, 3 joints) and add 2 new members plus 1 new joint at each step. This guarantees m = 2j − 3, and with r = 3 (pin + roller), m + r = 2j − 3 + 3 = 2j ✓.
Examples
This is a board-exam-style quick check. Always state the numerical comparison explicitly in your solution to earn full marks.
Scenario
A plane truss has m = 13 members, r = 3 reaction components (one pin + one roller), and j = 8 joints. Is it determinate, indeterminate, or unstable?
Solution
Check: m + r = 13 + 3 = 16. 2j = 2(8) = 16. Since 16 = 16, the truss satisfies the determinacy condition. Provided the geometry is proper (no concurrent-member instability), the truss is STATICALLY DETERMINATE.
The truss has 2 fewer unknowns than equations, meaning two equilibrium equations cannot be satisfied — it will collapse under general loading.
Scenario
A truss has m = 9, r = 3, j = 7. Classify the truss.
Solution
m + r = 9 + 3 = 12. 2j = 2(7) = 14. Since 12 < 14, the truss is a MECHANISM (geometrically unstable — insufficient members to maintain rigidity).
Applications
- Pre-analysis check before setting up joint or section equations in any board problem.
- Checking submitted structural drawings for proper truss layout (RA 544 — professional responsibility of a licensed civil engineer).
- Determining if a truss bridge panel requires additional redundant members for fatigue or load redistribution.
- NSCP 2015 Section 407 — structural systems must be adequately braced and stable.
Misconceptions
- MISCONCEPTION: 'If m + r = 2j, the truss is always stable.' CORRECTION: Geometric instability can still exist; check the physical arrangement of members.
- MISCONCEPTION: 'r is always 3.' CORRECTION: A fixed support contributes r = 3 (Ax, Ay, M), but this would make the joint a rigid connection, violating truss assumptions. A pin gives r = 2; a roller gives r = 1; special supports may differ.
- MISCONCEPTION: 'A truss with more members is always stronger.' CORRECTION: Redundant members make the truss indeterminate, changing the analysis method entirely — not simply 'stronger.'
Related Concepts
- Degree of static indeterminacy
- Geometric vs. static instability
- Simple vs. compound vs. complex trusses
- Compatibility equations for indeterminate structures
Common Exam Questions
Example
m = 21, r = 4, j = 12: m + r = 25; 2j = 24; 25 > 24 → indeterminate to the 1st degree.
Approach
Compute m + r and 2j. Compare. State result clearly: determinate, indeterminate (specify degree = m + r − 2j), or unstable.
Question Type
Classify truss type given m, r, j
Example
j = 10 joints, r = 3: m = 2(10) − 3 = 17 members needed for determinacy.
Approach
Use m = 2j − r. For a standard pin-roller support (r = 3): m = 2j − 3.
Question Type
Find the minimum number of members for a stable determinate truss with given joints
Key Points To Remember
- Determinacy equation: m + r = 2j (plane truss).
- m + r < 2j → unstable mechanism; m + r > 2j → indeterminate.
- m + r = 2j is necessary but not sufficient — geometry must also be proper.
- Simple truss: built from a triangle by successive addition of 2 members and 1 joint.
- Typical supports: pin = 2 reactions (Ax, Ay); roller = 1 reaction (normal to surface).
Method of Joints
The method of joints treats each joint of the truss as a concurrent force system in equilibrium. At each joint, we apply the two particle-equilibrium equations: ΣFx = 0 and ΣFy = 0. Since only two independent equations are available, each joint must have at most two unknown member forces for a direct solution. Procedure: (1) Solve for all support reactions using the entire truss as a free body (ΣFx = 0, ΣFy = 0, ΣM = 0). (2) Identify a joint with at most two unknown members (often a support joint after reactions are known, or a free end with one member). (3) Draw the FBD of that joint showing the known reaction or applied load and assuming all unknown member forces as tension (arrows pointing away from the joint). (4) Apply ΣFx = 0 and ΣFy = 0 to solve for the two unknowns. (5) Transfer the now-known forces to adjacent joints (reverse the arrow direction on the adjacent joint's FBD — Newton's third law). (6) Repeat until all member forces are found. The method of joints is best when you need all member forces. It can be tedious for large trusses — in those cases, combine with zero-force-member detection to reduce the number of joints to analyze.
Examples
Note that at joint A, F_AC was assumed pointing from A toward C (tension). The negative result tells us C actually points from C toward A — the member pushes on joint A, confirming compression. Always verify at the last joint for error-checking — a crucial board-exam habit.
Scenario
Triangular truss: A(0,0) pin support, B(6,0) m roller support, apex C(3, 2.5) m. Downward load P = 20 kN at C. Find all member forces.
Solution
STEP 1 — REACTIONS. ΣMA = 0 (taking moments about A): RB(6) − 20(3) = 0 → RB = 10 kN ↑. ΣFy = 0: RAy + RB − 20 = 0 → RAy = 10 kN ↑. ΣFx = 0: RAx = 0. STEP 2 — GEOMETRY. Member AC: Δx = 3, Δy = 2.5, L_AC = √(9 + 6.25) = √15.25 = 3.905 m; cos θ_AC = 3/3.905 = 0.7682; sin θ_AC = 2.5/3.905 = 0.6401. Member BC: Δx = 3, Δy = 2.5 (from B(6,0) to C(3,2.5)), L_BC = 3.905 m; same angles by symmetry. Member AB: horizontal, L = 6 m. STEP 3 — JOINT A (unknowns: F_AC and F_AB). Assume both in tension (F_AC directed from A toward C; F_AB directed from A toward B). ΣFy = 0: RAy + F_AC sin θ_AC = 0 → 10 + F_AC(0.6401) = 0 → F_AC = −15.62 kN (COMPRESSION, 15.62 kN C). ΣFx = 0: RAx + F_AC cos θ_AC + F_AB = 0 → 0 + (−15.62)(0.7682) + F_AB = 0 → F_AB = +12.00 kN (TENSION, 12.00 kN T). STEP 4 — JOINT B (by symmetry): F_BC = −15.62 kN (15.62 kN C). STEP 5 — VERIFY at Joint C: ΣFy: −20 + F_AC sin θ + F_BC sin θ = −20 + (−15.62)(−0.6401) + (−15.62)(−0.6401) = −20 + 10 + 10 = 0 ✓. ΣFx: F_AC(−cos θ) + F_BC(cos θ) = (−15.62)(−0.7682) + (−15.62)(0.7682) = 12 − 12 = 0 ✓.
Applications
- Complete force analysis of all members in a simply supported roof truss.
- Determining which chord members of a bridge truss are in tension (AISC 360-16 Chapter D design) vs. compression (Chapter E).
- Sizing of truss members in steel construction after obtaining axial forces.
- Verification of computer output from structural analysis software in engineering practice.
Misconceptions
- MISCONCEPTION: 'I can start at any joint.' CORRECTION: Start only at a joint with at most 2 unknowns; otherwise you have more unknowns than equations.
- MISCONCEPTION: 'I reverse the arrow for the same joint.' CORRECTION: You reverse the arrow only when drawing the force on the ADJACENT joint (Newton's 3rd law — equal and opposite).
- MISCONCEPTION: 'I can ignore the x-component of a diagonal because it looks small.' CORRECTION: Always resolve all forces; even small horizontal components contribute to equilibrium and change results significantly.
Related Concepts
- Method of sections
- Zero-force members
- Particle equilibrium
- Free-body diagram construction
Common Exam Questions
Example
A Pratt truss with 4 panels, span 16 m, height 3 m — find the end diagonal force at joint A. Set up geometry, get reactions, apply ΣF at joint A.
Approach
Set up the geometry table first (Δx, Δy, L, sin θ, cos θ for every member). Then solve joints sequentially. Tabulate answers clearly: member name, force magnitude, T or C.
Question Type
Find specific member force using method of joints
Example
At apex joint C with 20 kN downward load, substitute computed F_AC and F_BC and verify ΣFy = 0 and ΣFx = 0.
Approach
Substitute computed forces into ΣFx and ΣFy. If both equal zero (or nearly zero, within rounding), equilibrium is satisfied.
Question Type
Verify equilibrium at a specific joint
Key Points To Remember
- Apply ΣFx = 0 and ΣFy = 0 at each joint (2 equations, max 2 unknowns).
- Always find support reactions FIRST using the whole truss.
- Assume all unknown member forces are tension (arrows away from joint).
- Positive result = tension (T); negative result = compression (C).
- When carrying forces to the next joint, reverse the arrow (Newton's 3rd law).
- Work systematically joint by joint, never skip to a joint with more than 2 unknowns.
Method of Sections
The method of sections is the most efficient technique when you need the force in one or a few specific interior members — especially those deep inside a large truss where working joint by joint would require many steps. The method treats a portion of the truss as a rigid body in equilibrium under the applied forces, reactions, and the cut member forces. Procedure: (1) Determine support reactions for the entire truss. (2) Pass an imaginary cut through the truss, slicing through the members whose forces you want. CUT THROUGH AT MOST THREE MEMBERS with unknown forces (otherwise you have more unknowns than the three rigid-body equilibrium equations: ΣFx = 0, ΣFy = 0, ΣM = 0). (3) Choose the simpler of the two portions (usually the one with fewer forces). (4) Assume unknown cut-member forces are in tension (arrows pointing away from the cut surface). (5) Write equilibrium equations. THE KEY STRATEGY: take moments about the intersection point of two of the three cut members. This eliminates those two unknowns and leaves a single equation in one unknown — solving directly. (6) If the result is positive, the member is in tension; if negative, in compression. Important: for a chord member, take moments about the joint where the other chord intersects the cut plane; for a diagonal, take moments about a point on the bottom or top chord. Practice identifying these strategic moment points — they are the heart of the method of sections.
Examples
The moment of F_EF about C: EF acts at y = 3, and C is at y = 0, so the moment arm is 3 m. The sign is correctly handled by noting that F_EF assumed in tension points to the LEFT on the left free-body (away from the cut surface), creating a clockwise (negative) moment about C — hence the equation gives a negative F_EF confirming compression.
Scenario
Parallel-chord truss: joints A(0,0), B(4,0), C(8,0), D(12,0) along the bottom chord; E(4,3), F(8,3) along the top chord. Members: bottom chord AB, BC, CD; top chord EF; verticals BE, CF; diagonals AE, EC, FD. Supports: pin at A, roller at D. Downward loads of 12 kN at B and at C. Find the force in top-chord member EF using the method of sections.
Solution
STEP 1 — REACTIONS. By symmetry: RA = RD = (12 + 12)/2 = 12 kN ↑. STEP 2 — CUT. Pass a vertical cut through members EF, EC (diagonal), and BC. Retain the LEFT portion: forces acting on it are RA = 12 kN ↑ at A, the 12 kN load ↓ at B, and the three cut-member forces F_EF (horizontal, in top chord), F_EC (diagonal), and F_BC (horizontal, in bottom chord). STEP 3 — STRATEGIC MOMENT POINT. EF is horizontal at y = 3; EC goes from E(4,3) to C(8,0); BC is horizontal at y = 0. The lines of action of EC and BC both pass through point C(8,0). Take ΣMc = 0 to eliminate F_EC and F_BC. ΣMc = 0 (positive counterclockwise): RA(8) − 12(8 − 4) + F_EF(3) = 0 → 12(8) − 12(4) + 3F_EF = 0 → 96 − 48 + 3F_EF = 0 → 3F_EF = −48 → F_EF = −16 kN. ANSWER: F_EF = 16 kN COMPRESSION (top chord of simply supported truss is always in compression ✓). STEP 4 — CHECK. By taking ΣME = 0 on the left portion: RA(4) − 12(4 − 4) + F_BC(−3) = 0 → 48 + F_BC(−3) = 0 → F_BC = +16 kN T (bottom chord in tension ✓).
Applications
- Rapid determination of chord forces in a long multi-panel bridge truss without analyzing every joint.
- Checking only the most critically loaded members in a preliminary structural design.
- Board exam problems where the question asks for one specific interior member force.
- Retrofit analysis: verifying adequacy of a specific diagonal in an existing structure.
Misconceptions
- MISCONCEPTION: 'I can cut through any number of members.' CORRECTION: Cut through at most 3 members with unknown forces; otherwise you have more unknowns than equations.
- MISCONCEPTION: 'The strategic moment point must be on the truss.' CORRECTION: The point can be anywhere in the plane — even off the truss — as long as it is the intersection of the lines of action of two cut forces.
- MISCONCEPTION: 'Take the moment about a support joint.' CORRECTION: Take moments about the intersection of the OTHER TWO cut members' lines of action, not about the support.
Related Concepts
- Method of joints
- Rigid body equilibrium (ΣF = 0, ΣM = 0)
- Strategic moment point selection
- Parallel-chord vs. Pratt vs. Howe truss geometries
Common Exam Questions
Example
Board question: 'Find the force in member FG of a Pratt truss.' Cut through FG, the adjacent diagonal, and the opposite chord. Take moments about the intersection of the diagonal and opposite chord.
Approach
1. Get reactions. 2. Identify the cut (usually a vertical or diagonal cut through 3 members). 3. Identify the strategic moment point. 4. Write ΣM = 0 about that point. 5. State T or C.
Question Type
Find the force in a specific chord or diagonal using sections
Example
In a Howe truss cut through U2L2, U2U3 (top chord), and L2L3 (bottom chord): to find U2U3, take ΣM at L2 (intersection of the diagonal and bottom chord lines).
Approach
For a cut through top chord, diagonal, and bottom chord: the strategic point for finding the top chord is where the diagonal and bottom chord meet (or their extensions meet); for the bottom chord, it is where the top chord and diagonal meet.
Question Type
Determine the strategic moment point for a given cut
Key Points To Remember
- Cut through AT MOST 3 members with unknown forces.
- Isolate one portion (left or right, whichever is simpler).
- Assume cut-member forces are tension (pointing away from cut surface).
- To find a member force directly: take ΣM = 0 about the point where the other two cut members (or their lines of action) INTERSECT.
- Three equilibrium equations: ΣFx = 0, ΣFy = 0, ΣM = 0 for the isolated portion.
- Method of sections is best for finding forces in a FEW SPECIFIC members.
Zero-Force Members
A zero-force member carries no axial force under a specific loading condition. Identifying zero-force members before starting joint-by-joint analysis dramatically reduces the number of equations to solve — a critical time-saving skill on the board exam. There are two standard rules for identifying zero-force members, both derived directly from the equilibrium of a joint with no external load: RULE 1 — Two-Member Joint (No Load): If only two non-collinear members meet at an unloaded joint, BOTH members are zero-force members. Proof: ΣF along one member → the other member has zero force; ΣF along the other → the first also has zero force. RULE 2 — Three-Member Joint with Two Collinear (No Load): If three members meet at an unloaded joint, two of which are collinear (i.e., they form a straight line), then the third (non-collinear or 'odd') member is a zero-force member. Proof: ΣF perpendicular to the collinear pair → the odd member = 0; the two collinear members can then have equal and opposite forces. Important caveat: zero-force members are zero under the CURRENT loading only. Under a different load case, they will carry force. They also serve structural functions: they reduce the unsupported length of compression members (preventing buckling) and provide stability during construction. Never recommend removing them from the truss in practice.
Examples
Proof: resolve forces perpendicular to JK-JL (i.e., in the vertical direction at J). The only force in the vertical direction is F_JM sin 90° = F_JM. With no external vertical load at J, ΣFy = F_JM = 0. Then ΣFx along the chord gives F_JK = F_JL.
Scenario
A roof truss has joint J where three members meet: members JK and JL are along the same horizontal bottom-chord line, and member JM goes vertically upward. No external load is applied at J. Identify any zero-force members.
Solution
Identify the configuration: two members (JK and JL) are COLLINEAR (both horizontal), and the third member JM is NON-COLLINEAR (vertical). No load at J. Applying RULE 2: the non-collinear member JM is a ZERO-FORCE MEMBER. Members JK and JL carry equal forces (they are collinear and form the chord).
ΣF along PQ direction: F_PQ + F_PR cos(90°) = 0 → F_PQ = 0. ΣF along PR direction: F_PR + 0 = 0 → F_PR = 0.
Scenario
At joint P of a truss, only two members meet: PQ (going up-right at 45°) and PR (going up-left at 135° from horizontal). No load at P. Are these zero-force?
Solution
Two members at an unloaded joint (RULE 1). Since PQ and PR are NOT collinear (they make a 90° angle with each other), both members PQ and PR are ZERO-FORCE MEMBERS.
Applications
- Rapid simplification of complex trusses before starting joint analysis.
- Identification of redundant anti-buckling members in long-span steel trusses.
- Understanding why certain members in a bridge truss appear to carry no load under symmetric loading but are essential for asymmetric (live) loads.
- Pre-check in computer structural analysis to validate output (zero-force members should show near-zero axial force).
Misconceptions
- MISCONCEPTION: 'Zero-force members can be removed from the truss.' CORRECTION: Remove them only from the analysis of the current load case, never from the physical structure.
- MISCONCEPTION: 'Rule 2 applies when two of three members are close in direction but not exactly collinear.' CORRECTION: The two members must be EXACTLY collinear (on the same straight line) for Rule 2 to apply.
- MISCONCEPTION: 'A zero-force member has zero stress.' CORRECTION: Zero axial force → zero axial stress under the current load case. Under a different load or during construction, it carries stress.
Related Concepts
- Method of joints
- Equilibrium at a particle
- Buckling length of compression members (NSCP 2015 / AISC 360 Chapter E)
- Load combinations and load cases (NSCP 2015 Section 203)
Common Exam Questions
Example
In a symmetrically loaded Pratt truss, the center vertical member under a symmetric load at the center joint is often a zero-force member — verify by applying Rule 2 or by inspection of ΣFy at the center top joint.
Approach
Survey every joint. At each unloaded joint, check Rule 1 (two members) and Rule 2 (three members, two collinear). List all zero-force members and state which rule applies.
Question Type
Identify all zero-force members in a given truss under a specified loading
Example
Board exam: 'Member MN is a zero-force member. If it is removed, what happens to the truss under the current loading?' Answer: The truss remains in equilibrium under the same loading, but becomes potentially unstable under different load cases.
Approach
State that removal is valid for the current load case ONLY. Under other load cases, the member will carry force. In practice, removal is never recommended.
Question Type
Effect of removing a zero-force member
Key Points To Remember
- Rule 1: Two non-collinear members at an unloaded joint → both are zero-force.
- Rule 2: Three members at an unloaded joint, two collinear → the odd (non-collinear) one is zero-force.
- Zero-force ≠ useless — they serve anti-buckling and load-redistribution functions.
- Zero-force condition is load-case-specific.
- After identifying zero-force members, remove them from the FBDs to simplify analysis.
Practice Problems
The degree of indeterminacy equals (m + r) − 2j. Each redundant member requires one additional compatibility equation (e.g., virtual work or force method) to solve. On the board exam, always state the degree of indeterminacy explicitly.
Problem
PROBLEM 1 (Determinacy Check). A plane truss has 17 members, a pin support (2 reactions), and a roller support (1 reaction), with 10 joints. (a) Check if the truss is statically determinate. (b) If two more members are added without adding joints, reclassify the truss.
Solution
(a) m = 17, r = 2 + 1 = 3, j = 10. m + r = 17 + 3 = 20. 2j = 2(10) = 20. Since 20 = 20, the truss is STATICALLY DETERMINATE (assuming proper geometric arrangement). (b) Adding 2 members: m = 19, r = 3, j = 10. m + r = 19 + 3 = 22. 2j = 20. Since 22 > 20, the truss is STATICALLY INDETERMINATE to the 2nd degree (two redundant members).
The horizontal load shifts the reactions asymmetrically, making BC carry significantly more compression than AC. Always verify equilibrium at the last joint — this single check catches most computational errors in board exam solutions.
Problem
PROBLEM 2 (Method of Joints — Full Truss). A simple triangular truss has: A(0,0) pin, B(8,0) m roller, and C(4, 3) m apex. A 30 kN downward load acts at C and a 10 kN horizontal load (pointing right) acts at C. Find all member forces.
Solution
STEP 1 — REACTIONS. ΣFx = 0: RAx + 10 = 0 → RAx = −10 kN (i.e., 10 kN ←). ΣMA = 0: RB(8) − 30(4) − 10(3) = 0 → 8RB = 120 + 30 = 150 → RB = 18.75 kN ↑. ΣFy = 0: RAy + RB − 30 = 0 → RAy = 30 − 18.75 = 11.25 kN ↑. STEP 2 — GEOMETRY. AC: Δx=4, Δy=3, L=5 m; cos θ=0.8, sin θ=0.6. BC: Δx=4 (from B(8,0) to C(4,3), so Δx=−4, Δy=3), L=5 m; cos θ=0.8, sin θ=0.6. AB: horizontal, L=8 m. STEP 3 — JOINT A. ΣFy=0: 11.25 + F_AC(0.6) = 0 → F_AC = −18.75 kN → 18.75 kN C. ΣFx=0: −10 + F_AC(0.8) + F_AB = 0 → −10 + (−18.75)(0.8) + F_AB = 0 → F_AB = 10 + 15 = 25.0 kN T. STEP 4 — JOINT B. At B, reaction RB = 18.75 kN ↑. Member BC goes from B(8,0) toward C(4,3): direction cosines cos θ = −4/5 = −0.8 (leftward), sin θ = 3/5 = 0.6 (upward). ΣFy=0: 18.75 + F_BC(0.6) = 0 → F_BC = −31.25 kN → 31.25 kN C. ΣFx=0: F_BC(−0.8) + (−F_AB) = 0? Check: (−31.25)(−0.8) − 25 = 25 − 25 = 0 ✓. STEP 5 — VERIFY at Joint C. ΣFx: 10 + F_AC(−0.8) + F_BC(0.8)? F_AC acts from C toward A (compression means pushing), so direction from A to C reversed → force on C: F_AC pointing from C toward A, i.e., (−0.8, −0.6)×(−18.75) = (15, 11.25). F_BC pointing from C toward B: (0.8, −0.6)×(−31.25) = (−25, 18.75). External at C: (10, −30). ΣFx = 10 + 15 − 25 = 0 ✓; ΣFy = −30 + 11.25 + 18.75 = 0 ✓. RESULTS: F_AC = 18.75 kN C; F_BC = 31.25 kN C; F_AB = 25.0 kN T.
The strategic moment points U1 and L2 each eliminate two of the three unknowns, allowing direct solution for the third. The diagonal is found last using ΣFy because it is the only force with a vertical component not yet used. The equilibrium check ΣFx = 0 confirms all three values simultaneously.
Problem
PROBLEM 3 (Method of Sections). A simply supported Pratt truss has 4 panels: bottom joints L0(0,0), L1(4,0), L2(8,0), L3(12,0), L4(16,0) m; top joints U1(4,4), U2(8,4), U3(12,4) m. Verticals: L1U1, L2U2, L3U3. Diagonals: L0U1 (end rafter), U1L2 (falling Pratt diagonal), U2L3, U3L4. Loads: 20 kN ↓ at L1, 20 kN ↓ at L2, 20 kN ↓ at L3. Supports: pin at L0, roller at L4. Using the method of sections, find the forces in members U1U2 (top chord), L1L2 (bottom chord), and U1L2 (diagonal) simultaneously.
Solution
STEP 1 — REACTIONS. Total load = 60 kN. By symmetry, RL0 = RL4 = 30 kN ↑. STEP 2 — CUT. Pass a vertical cut between panels 1 and 2, slicing through U1U2, U1L2, and L1L2. Retain the LEFT portion: forces acting on it are RL0 = 30 kN ↑ at L0, and the 20 kN ↓ at L1, plus the three cut forces F_U1U2, F_U1L2, and F_L1L2. STEP 3 — GEOMETRY OF DIAGONAL U1L2. From U1(4,4) to L2(8,0): Δx = 4, Δy = −4, L = 4√2 = 5.657 m; cos θ = 4/5.657 = 0.7071; sin θ = 4/5.657 = 0.7071. The diagonal slopes downward to the right at 45°. STEP 4 — FIND F_U1U2 (top chord). Take ΣML1 = 0 (L1(4,0) is the intersection of diagonal U1L2 — note: U1 is directly above L1, so U1L2 extended does NOT pass through L1 exactly, but the bottom chord L1L2 and the vertical L1U1 are at L1, so the strategic point for F_U1U2 is actually the bottom chord level point directly below U1 — let's use ΣML2 instead). ΣML2 = 0: RL0(8) − 20(8−4) + F_U1U2(4) = 0 → 30(8) − 20(4) + 4F_U1U2 = 0 → 240 − 80 + 4F_U1U2 = 0 → F_U1U2 = −40 kN → 40 kN C. (At L2(8,0): diagonal U1L2 passes through L2, so F_U1L2 has zero moment about L2; bottom chord L1L2 also passes through L2, zero moment. Only F_U1U2 remains.) STEP 5 — FIND F_L1L2 (bottom chord). Take ΣMU1 = 0 (U1(4,4): the top chord U1U2 and the diagonal U1L2 both meet AT U1, so F_U1U2 and F_U1L2 have zero moment arm about U1). ΣMU1 = 0: RL0(4) − 20(4−4) − F_L1L2(4) = 0 → 30(4) − 0 − 4F_L1L2 = 0 → F_L1L2 = +30 kN → 30 kN T. STEP 6 — FIND F_U1L2 (diagonal). ΣFy = 0 on left portion: RL0 − 20 + F_U1L2(−sin 45°) = 0 [diagonal goes down-right, so its vertical component on the cut face points downward for tension assumption]. 30 − 20 − F_U1L2(0.7071) = 0 → F_U1L2 = 10/0.7071 = +14.14 kN → 14.14 kN T. STEP 7 — VERIFY ΣFx on left portion: F_U1U2 + F_U1L2 cos 45° + F_L1L2 = 0? −40 + 14.14(0.7071) + 30 = −40 + 10 + 30 = 0 ✓. RESULTS: U1U2 = 40 kN C; L1L2 = 30 kN T; U1L2 = 14.14 kN T.
The cascade effect is a hallmark of zero-force member problems on the board exam: removing one zero-force member (BF) reveals another (BG), and removing CG reveals CH. Always re-examine joints after each zero-force member is identified. The final answer should list all zero-force members in order of discovery.
Problem
PROBLEM 4 (Zero-Force Members). In the truss below, identify all zero-force members under the given loading. The truss has top joints A(0,4), B(3,4), C(6,4), D(9,4); bottom joints E(0,0) pin, F(3,0), G(6,0), H(9,0) roller. Members: top chord AB, BC, CD; bottom chord EF, FG, GH; verticals AF, BF, BG, CG, CH; diagonals AE (left rafter), DH (right rafter), and EG. External loads: 15 kN ↓ at A and 15 kN ↓ at D. No loads at B, C, F, G, H.
Solution
STEP 1 — Examine joint F (bottom joint at (3,0)). Members at F: EF (bottom chord, going left), FG (bottom chord, going right), BF (vertical, going up). No external load at F. EF and FG are COLLINEAR (both horizontal). BF is non-collinear (vertical). → RULE 2: BF is a ZERO-FORCE MEMBER. STEP 2 — Examine joint G (bottom joint at (6,0)). Members at G: FG (bottom chord, going left), GH (bottom chord, going right), CG (vertical, going up), EG (diagonal). External load at G? No. But G has FOUR members: FG, GH, CG, EG. Four members means we cannot apply the simple two-rule identification directly without writing equations. ΣFy at G: CG(sin90°) + EG(sinθ_EG) = 0; θ_EG: from E(0,0) to G(6,0) — wait, EG is horizontal (both at y=0)! EG goes from E(0,0) to G(6,0), so it is HORIZONTAL. That means EF, FG, GH, EG are all horizontal and CG is vertical. So at G, members EG (part of bottom chord line), FG, GH, and also EF are in play — actually only those connected to G: FG, GH, EG, CG. Among these, FG, GH, and EG are all collinear (horizontal). And CG is vertical (non-collinear). With no load at G: ΣFy = F_CG = 0. → CG is a ZERO-FORCE MEMBER. ΣFx = F_FG + F_GH + F_EG = 0 (three horizontals must sum to zero, but this doesn't separately zero them out — need more equations). STEP 3 — Examine joint B (top joint at (3,4)). Members at B: AB (horizontal left), BC (horizontal right), BF (now known = 0, vertical down), BG (diagonal, going down-right to G(6,0)). With BF = 0: effectively at B, two non-zero members remain: AB, BC (collinear top chord), and BG (non-collinear diagonal), with no external load at B. → RULE 2: BG is a ZERO-FORCE MEMBER. STEP 4 — Examine joint C (top joint at (6,4)). Members: BC (horizontal left), CD (horizontal right), CG (now known = 0, vertical), CH (diagonal going down-right to H). With CG = 0: at C, effectively BC, CD (collinear top chord), and CH (non-collinear), no load at C. → RULE 2: CH is a ZERO-FORCE MEMBER. SUMMARY OF ZERO-FORCE MEMBERS: BF, CG, BG, CH.
Exam Preparation Tips
- REACTIONS FIRST, ALWAYS. Never attempt joint or section equations without solving the global equilibrium (ΣFx = 0, ΣFy = 0, ΣM = 0) for the support reactions. Errors in reactions propagate to every member force.
- GEOMETRY TABLE. Before computing any equilibrium equation, tabulate all member geometries: Δx, Δy, L = √(Δx² + Δy²), cos θ = Δx/L, sin θ = Δy/L. Board exam time pressure causes geometry errors — a table prevents them.
- CHECK DETERMINACY FIRST. Compute m + r vs. 2j. If the truss is indeterminate, you cannot use the methods of joints or sections without modification — and the problem likely cannot be solved by statics alone.
- ZERO-FORCE MEMBER SCAN. Before starting the joint analysis, spend 60–90 seconds identifying all zero-force members using Rules 1 and 2. In a 15-member truss, finding 4 zero-force members reduces your work by more than 25%.
- CONSISTENT SIGN CONVENTION. Adopt one convention at the start and never deviate: tension (+), compression (−). Write 'T' or 'C' next to every computed force — board examiners deduct points if the nature (T or C) is not stated.
- METHOD SELECTION. Method of joints: use when all member forces are required, or when only 2–3 joints away from a known start. Method of sections: use when you need one or a few specific interior members, especially chord members of a parallel-chord truss.
- MOMENT POINT STRATEGY FOR SECTIONS. Memorize: to find force in a chord member, take moments about the joint where the diagonal and the other chord intersect. To find the diagonal force directly, take ΣFy (if the chord forces are horizontal) — this avoids the need for a special moment point.
- VERIFY AT THE LAST JOINT. After method of joints, always check equilibrium at the final joint. After method of sections, verify at least one unused equilibrium equation (e.g., after using ΣM and ΣFy, check ΣFx). This catches arithmetic errors before submitting.
- SYMMETRY SHORT-CUT. For symmetric trusses under symmetric loading, reactions are equal (RA = RB = P_total/2) and mirror-image members have equal magnitudes. Analyze only half the truss and mirror the results — halves the computational work.
- BOARD EXAM TIMING. A well-prepared examinee should solve a complete 7-member truss (method of joints) in about 6–8 minutes, and a method-of-sections problem (one member) in 3–4 minutes. Practice timed problems using past PRC board exam questions from 2015–2023.
- COMMON LOAD SETUP. The most frequent board exam truss configuration is the simply supported parallel-chord truss with 3–5 panels and equal point loads at the joints. Be thoroughly comfortable with this setup.
- UNITS. Work exclusively in kN and meters throughout. Mixing kN and N, or m and mm, is the single most common source of order-of-magnitude errors on the board exam.
In summary
The analysis of trusses is a foundational skill in Engineering Mechanics that appears consistently in the PRC Civil Engineer Licensure Examination. The four pillars of truss analysis are: (1) the truss assumptions — particularly the two-force member principle — which simplify every member force to a pure axial tension or compression; (2) the determinacy check m + r = 2j, which must be performed before any force calculation; (3) the method of joints — systematic, joint-by-joint application of ΣFx = 0 and ΣFy = 0, starting always from a support joint or a joint with at most two unknowns; and (4) the method of sections — isolating a portion of the truss with a strategic cut and using ΣM = 0 about the intersection of two cut members to find the force in the third member directly. Zero-force member identification using the two standard rules is an indispensable time-saving technique that every board examinee must master — it reduces the number of joints to analyze and reveals the overall force flow pattern in the structure. In professional practice governed by RA 544, NSCP 2015, and AISC 360-16, the member forces computed by these methods form the basis of member sizing, connection design, and stability checks. Build fluency through timed practice: aim to complete a standard 7-to-9-member truss problem in under 10 minutes, and a single method-of-sections problem in under 4 minutes. Mastery of truss analysis is not merely about passing the board exam — it is a core competency for every licensed Filipino civil engineer engaged in structural design, review, and construction supervision.
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