CELE Engineering Mechanics — Analysis of TrussesRevision Notes

Exam Tips

• Sketch the truss and label all joints with coordinates — geometry is the foundation of everything.
• Always state 'assume tension' at the start of your solution; board examiners award marks for consistent sign convention.
• Identify truss type (Pratt, Howe, Warren, Fink) early — it helps you predict which members are tension and which are compression before calculating.

Key Points

• A truss is composed of straight members connected at their ends by frictionless pin joints (nodes/joints).
• Loads and support reactions act ONLY at the joints — never along the length of a member.
• Member self-weight, if accounted for, is split equally between its two end joints.
• Because of pin connections and joint-only loading, every member is a TWO-FORCE MEMBER: the resultant force acts purely along the member's axis.
• A member pulling on its end joints is in TENSION (positive, T); one pushing on its joints is in COMPRESSION (negative, C).
• The standard sign convention: assume all unknown member forces are TENSION when writing equilibrium equations; a negative result confirms compression.
• Common truss types encountered in PRC boards: Pratt, Howe, Warren, Fink (roof), K-truss, and simple triangular trusses.
• Trusses appear in Philippine practice as roof structures (NSCP 2015 Section 204 — structural loads on roofs), bridges, transmission towers, and long-span floor systems.

Definitions

Section Title

Truss Fundamentals and Assumptions

Common Mistakes

• Applying a distributed load along a member length instead of converting it to joint forces first.
• Assuming a rigid joint (like a welded connection) instead of a pin — which would introduce moments and invalidate pure two-force member behavior.
• Forgetting that the direction of the assumed tension force on a free-body diagram must point AWAY from the joint for each member.
• Mixing up which direction a force acts on the joint versus how the joint pulls/pushes the member.

Formulas

Exam Tips

• Do the m + r = 2j check as Step 1 — if the truss is indeterminate, the board problem will either ask you to recognize it or will provide additional compatibility data.
• For a simple truss (triangular base + 2 members/joint), use m = 2j − 3 as a quick internal check.
• Space trusses (3D) appear occasionally — remember the equation changes to m + r = 3j.

Key Points

• For a PLANE (2D) truss, apply the determinacy equation: m + r = 2j.
• m = number of members, r = number of external reaction components, j = number of joints.
• If m + r = 2j: statically DETERMINATE (solvable by statics alone).
• If m + r > 2j: statically INDETERMINATE (redundant members; statics alone is insufficient — compatibility equations needed).
• If m + r < 2j: UNSTABLE mechanism (not a valid rigid truss).
• WARNING: m + r = 2j is NECESSARY but NOT SUFFICIENT for determinacy. Geometric instability can still occur if members are improperly arranged (e.g., a rectangular panel without a diagonal is a mechanism even if the count is satisfied).
• A simple truss is built by starting with a triangle (3 members, 3 joints) and adding 2 members per new joint: m = 2j − 3 for an internally determinate simple truss with 3 reaction components.
• For space (3D) trusses, the determinacy equation becomes: m + r = 3j.

Definitions

Section Title

Determinacy and Stability

Common Mistakes

• Counting a roller support as providing 2 reaction components — a roller gives only 1 (perpendicular to the rolling surface).
• Counting a pin-connected internal hinge as a joint but forgetting to subtract the additional equations of condition (each internal hinge provides one additional condition equation).
• Concluding a truss is stable simply because m + r = 2j without checking geometric arrangement.
• For a truss with m = 13, r = 3, j = 8: 13 + 3 = 16 = 2(8) = 16. Correct count. But if two of those members are coplanar and concurrent in a panel without triangulation, it is still geometrically unstable.

Formulas

Exam Tips

• In board problems, look for joints at the supports first (they already have 1 known reaction) and joints at the free end of a cantilever (often 2 unknowns immediately).
• If the truss is symmetric and the loading is symmetric, use symmetry to halve your work — only solve half the truss and state the symmetric member forces.
• The slope ratio method (Δx/L and Δy/L) is faster and less error-prone than computing arctan in time-pressured board exams.
• Write F_AC = ? (T assumed) explicitly — examiners follow your sign convention to award partial marks.

Key Points

• Isolate each joint as a concurrent force system (all forces meet at the pin); apply ΣFx = 0 and ΣFy = 0.
• At each joint you can solve for AT MOST 2 unknown member forces — therefore, always start at a joint with only 2 unknowns.
• The support reactions MUST be found first (ΣFx = 0, ΣFy = 0, ΣM = 0 for the whole truss) before applying the Method of Joints.
• Assume every unknown member force is in TENSION (force directed away from the joint along the member). A positive result confirms tension; negative confirms compression.
• After solving one joint, treat those member forces as KNOWN when you move to the next adjacent joint.
• The method of joints is most efficient when ALL member forces are required.
• Work systematically: label joints, set up a consistent x-y coordinate system, and resolve each diagonal force into components using the member's angle or slope ratio.
• Always resolve angled members using: Fx = F·cosθ, Fy = F·sinθ, or using the slope ratio (Δx/L, Δy/L) when coordinates are given.
• Final check: apply ΣFx = 0 and ΣFy = 0 at the last joint to verify your solution — if they are satisfied, the analysis is correct.

Definitions

Section Title

Method of Joints

Common Mistakes

• Starting at a joint with 3 or more unknown member forces — you cannot solve it with only 2 equations.
• Forgetting to find support reactions before applying the method of joints.
• Drawing the assumed tension force pointing TOWARD the joint (which is compression direction) — this reverses the sign and causes systematic errors.
• Using the wrong angle for a diagonal member — always recompute θ from the joint coordinates of that specific member.
• Not carrying the solved force (with its correct sign) to the next joint — reintroducing it as unknown again.

Formulas

Exam Tips

• In board exams, 'find the force in member DF' (interior member) almost always calls for Method of Sections — recognize this pattern.
• Identify the moment center FIRST (where 2 of 3 cut members meet) before writing any equation — this is the key step.
• If all three cut members are non-concurrent and non-parallel, you need all three equations (ΣFx, ΣFy, ΣM) — but often one equation suffices when the geometry is favorable.
• For Pratt trusses (vertical web members + diagonal members sloping inward from ends to center): top chord is compression, bottom chord is tension, end diagonals are tension, intermediate diagonals are compression.

Key Points

• Pass an imaginary cut through the truss, cutting through the members whose forces you want to find. Isolate either the LEFT or RIGHT portion as a free body.
• The cut must pass through AT MOST 3 members with unknown forces (since you have 3 equilibrium equations for a rigid body: ΣFx = 0, ΣFy = 0, ΣM = 0).
• The cut does NOT need to be straight — it can be curved or stepped to isolate specific members.
• POWER MOVE: Take moments about the point where TWO of the three cut members INTERSECT. This eliminates those two unknowns and gives the third member force directly.
• If two of the three cut members are PARALLEL, they have no moment about any point on their common line of action — take moments about a point on their line to solve the third member.
• The Method of Sections is most efficient when only a FEW specific interior member forces are needed — it avoids solving many joints sequentially.
• Support reactions must still be computed for the whole truss before applying sections.
• The assumed-tension convention applies here too: draw each cut member force pointing AWAY from the cut section (tension assumed). Negative result means compression.
• You may use either the left or right portion — choose the side with FEWER external loads for simpler computation.

Definitions

Section Title

Method of Sections

Common Mistakes

• Cutting through MORE than 3 members with unknown forces — 3 equations cannot solve 4 unknowns.
• Taking moments about the wrong point — the moment center should eliminate 2 unknowns, leaving only 1.
• Forgetting to include ALL external forces on the chosen portion (missing a load or a reaction).
• Drawing the cut member forces pointing toward the section (compression assumed) inconsistently — pick one convention and stick with it.
• Using the moment arm of the force's line of action incorrectly — always measure the perpendicular distance from the moment center to the LINE OF ACTION of the force.

Exam Tips

• On the board exam, a truss diagram question may ask 'how many zero-force members are there?' — scan all joints systematically using Rules 1 and 2.
• ZFMs typically appear at interior panel joints of roof trusses (Fink, Pratt) where web members meet without load.
• In a truss loaded only at bottom-chord joints, top-chord-only joints (like apex joints with no external load) are prime candidates for ZFM identification.
• Marking ZFMs with a circle or X on your diagram helps prevent accidentally including them in subsequent joint/section calculations.

Key Points

• Zero-force members (ZFMs) carry NO axial force under a SPECIFIC loading condition. Identifying them early simplifies the analysis significantly.
• RULE 1 (Two-Member Joint, No Load): If only TWO members meet at a joint AND there is NO external load or reaction at that joint, BOTH members are zero-force members.
• RULE 2 (Three-Member Joint, Two Collinear, No Load): If THREE members meet at an unloaded joint AND TWO of them are collinear (form a straight line), then the THIRD (non-collinear) member is zero-force. The two collinear members carry equal and opposite forces.
• RULE 2 EXTENSION: If an external load acts along the collinear line of the two straight members, the third member is still zero-force. If the load is in another direction, Rule 2 does not apply.
• Zero-force members are NOT useless — they serve to: (1) brace compression members against buckling (provide lateral support), (2) carry forces under DIFFERENT load combinations, and (3) maintain the geometric rigidity of the truss.
• After identifying ZFMs, remove them from the truss model and re-apply Rules 1 and 2 — sometimes their removal creates new ZFMs at adjacent joints (a cascade effect).
• ZFM identification is a direct, fast question type on PRC boards — recognize both rules instantly.

Definitions

Section Title

Zero-Force Members

Common Mistakes

• Concluding a ZFM is permanently unloaded — ZFMs carry force under DIFFERENT load cases (e.g., dead load only vs. dead + live load). Never remove them from the physical structure.
• Applying Rule 2 when an external load acts at the joint — the load direction must be checked carefully. If the load is NOT along the collinear members, Rule 2 still applies only if the load component perpendicular to the collinear direction is zero.
• Missing cascade ZFMs — after removing one ZFM, recheck adjacent joints for new instances of Rules 1 or 2.
• Confusing Rule 2 with a reaction at the joint — if there is a support reaction at the joint, the rules do NOT directly apply unless the reaction is collinear with the two straight members.

Exam Tips

• In a 5-item truss problem set on PRC boards (typical format), allocate: 1 item for determinacy, 1 for reactions, 2 for member forces (one joints, one sections), 1 for ZFM identification.
• Show your free body diagrams — partial marks are awarded even for incorrect final answers if the FBD and equations are set up correctly.
• Round only at the final answer, not at intermediate steps, to avoid accumulated rounding error.

Key Points

• Example 1 (Method of Joints — Triangular Truss): Joints A(0,0), B(6,0), C(3,2.5); pin at A, roller at B; 20 kN downward at C.
• Step 1 — Reactions: ΣM_A = 0 → R_B(6) − 20(3) = 0 → R_B = 10 kN↑. ΣFy = 0 → R_A + 10 − 20 = 0 → R_A = 10 kN↑. ΣFx = 0 → R_Ax = 0.
• Step 2 — Geometry of AC: Δx = 3, Δy = 2.5, L_AC = √(9 + 6.25) = 3.905 m; cosθ = 0.768; sinθ = 0.640.
• Step 3 — Joint A (unknowns: F_AC, F_AB): ΣFy = 0 → F_AC(0.640) + 10 = 0 → F_AC = −15.63 kN (C). ΣFx = 0 → −15.63(0.768) + F_AB = 0 → F_AB = +12.0 kN (T). [Note: F_AC enters at A pointing away from A toward C; its y-component is +0.640·F_AC pointing upward when tension assumed — since reaction is upward and load is downward, the member is found to be compressive.]
• Step 4 — By symmetry: F_BC = −15.63 kN (C). Verify at joint C: ΣFy = F_AC·sinθ_CA + F_BC·sinθ_CB − 20 = (−15.63)(0.640)·(−1) + (−15.63)(0.640)·(−1) − 20 = 10 + 10 − 20 = 0 ✓.
• Example 2 (Method of Sections — Parallel Chord Truss): Joints A(0,0), B(4,0), C(8,0), D(12,0), E(4,3), F(8,3); loads 12 kN↓ at B and C; pin at A, roller at D.
• Reactions (symmetric loading): R_A = R_D = (12 + 12)/2 = 12 kN↑.
• Find F_EF: Cut through EF, EC, BC. Left portion (A-side). Point of moment: C(8,0) — both BC and EC lines of action pass through C. ΣM_C = 0: +R_A(8) − 12(8−4) + F_EF(3) = 0 → 12(8) − 12(4) + 3F_EF = 0 → 96 − 48 + 3F_EF = 0 → F_EF = −16 kN (C).
• Find F_BC: Take moments about E(4,3) — EF and EC pass through E. ΣM_E = 0: +R_A(4) − 12(4−4) − F_BC(3) = 0 → 12(4) − 0 − 3F_BC = 0 → F_BC = +16 kN (T). [Bottom chord in tension — expected for simply supported truss.]
• Determinacy check: m = 9, r = 3, j = 6; 9 + 3 = 12 = 2(6) ✓ DETERMINATE.

Section Title

Comprehensive Worked Examples — Board Exam Style

Common Mistakes

• In Example 1, a common error is writing F_AC·sinθ = −10 (mixing signs) — be explicit that ΣFy includes the REACTION as a positive upward force.
• In Example 2, a common error is using the wrong moment arm for F_EF — the perpendicular distance from C to the line of action of EF (which is horizontal at height 3 m) is simply 3 m.
• Not verifying the last joint or last equation — always check equilibrium at one more joint/equation than strictly necessary.