CELE Engineering Mechanics — Analysis of TrussesStudy Notes

A truss is an assemblage of straight, slender members joined together at points called joints. The defining characteristic of a truss is that **members are connected by frictionless pins** and **loads and reactions act only at the joints**. These assumptions have profound consequences: **Why These Assumptions Matter:** - Pin connections transmit force but **not moment** (torque). Each joint acts as a hinge. - Because loads act only at joints, not along member length, each member experiences force only at its ends. - Combining these conditions: **every truss member is a two-force member**, meaning it carries force purely along its longitudinal axis—either pulling (tension) or pushing (compression). **Two-Force Member Principle:** If a member carries force only at two points and those forces must be in equilibrium, the two forces must be equal in magnitude, opposite in direction, and collinear (along the member axis). This is the cornerstone of truss analysis: the force in any member **acts along the member**. There is no bending in ideal trusses. **Sign Convention:** - **Tension (positive):** Member pulls on the joints; the member is being stretched. - **Compression (negative):** Member pushes on the joints; the member is being squeezed. In board examinations, it is standard practice to **assume every unknown member force is in tension** (positive). If the calculation yields a negative value, the member is actually in compression. This removes the guesswork and standardizes the solution. **Self-Weight Treatment:** In practice, members have weight. The code (NSCP 2015, Section 2.2.1) requires that self-weight and distributed dead load be assigned to the joints. Typically, half the member weight goes to each end joint. This converts the distributed load into joint loads, preserving the two-force member assumption.

Before solving a truss, confirm it is **statically determinate**—that is, the support reactions and member forces can be found using equilibrium equations alone. An indeterminate truss requires additional equations from material behavior (strain compatibility). An unstable truss is a mechanism and cannot support load. **Determinacy Equation (Plane Trusses):** $$m + r = 2j$$ where: - m = number of members - r = number of reaction components (horizontal and vertical at supports) - j = number of joints **Interpretation:** - If m + r = 2j: the truss is **statically determinate** (typically). - If m + r > 2j: the truss is **statically indeterminate** (redundant members). - If m + r < 2j: the truss is **unstable** (a mechanism with more degrees of freedom than constraints). **Critical Caveat:** The equation m + r = 2j is a **necessary but not sufficient** condition. A truss can satisfy this equation and still be **geometrically unstable** if the members are arranged improperly. For example: - Three parallel members cannot stabilize a joint against all directions of load. - Members arranged concentrically (all meeting at one point) provide no lateral bracing. **Checking Geometric Stability:** 1. Ensure the support reactions form a stable restraint (e.g., a pin at one point and a roller at another, properly oriented, for a plane truss). 2. Verify that member groups are not concurrent (not all meeting at one point) or collinear (all in a line). 3. Use engineering judgment: a truss should appear "stiff" and rigid when drawn. **Example Determinacy Checks:** *Simple triangular truss:* 3 members (two legs + base), 3 joints, 3 reaction components (one pin + one roller). m + r = 3 + 3 = 6; 2j = 2(3) = 6. ✓ Determinate. *Pratt truss (5-panel bridge):* 17 members, 12 joints, 3 reactions. m + r = 17 + 3 = 20; 2j = 2(12) = 24. NOT determinate; this equation gives 20 ≠ 24. **This truss is unstable.** (Correcting the Pratt: 2 diagonals added → 19 members → m + r = 22 ≠ 24, still unstable. Add more members carefully.)

Before analyzing member forces, find the truss support reactions using global equilibrium. Treat the entire truss as a free body and apply: $$\sum F_x = 0, \quad \sum F_y = 0, \quad \sum M = 0$$ **Common Support Types (Plane Truss):** - **Pin Support (or Hinge):** Provides 2 reaction components (Rx, Ry). Prevents translation in both directions; allows rotation. - **Roller Support:** Provides 1 reaction component, perpendicular to the support surface. Allows translation along the surface and rotation. - **Fixed Support (Encastré):** Provides 2 reaction components + 1 moment reaction. Prevents translation and rotation. Rare in trusses (which are pinned structures); more common in continuous beams. **Procedure:** 1. Draw a free-body diagram of the entire truss with all external loads and support reactions labeled. 2. Choose a convenient origin and coordinate system (typically, x horizontal to the right, y vertical upward). 3. Write equilibrium equations. For a plane truss with one pin and one roller (determinate), three equations suffice: - ΣFx = 0 (usually gives the horizontal reaction directly) - ΣFy = 0 (usually gives the vertical reaction at the roller) - ΣM = 0 (taken about any point; often used to find vertical reaction at the pin) 4. Solve algebraically. **Sign Convention:** - Assume reaction components point in the positive direction (right for Rx, up for Ry). - A negative result means the reaction actually points opposite. **Key Insight for Exam Problems:** Symmetry often appears in exam trusses. If the truss geometry and loading are symmetric about a vertical centerline, the vertical reactions at the two supports are equal, and any horizontal reaction is zero. Exploit symmetry to simplify calculations.

The **method of joints** isolates each joint as a particle in equilibrium. At each joint, the sum of forces in x and y directions equals zero. Procedure: **Step-by-Step:** 1. Find support reactions for the entire truss (see Section 3). 2. Select a joint with at most **two unknown member forces**. (If three or more members meet at a joint and all are unknown, you cannot solve with two equilibrium equations.) 3. Draw a free-body diagram of the selected joint, showing: - All member forces acting on the joint (assume tension; arrows pull away from the joint). - Any external load or reaction at the joint. 4. Write equilibrium equations: ΣFx = 0, ΣFy = 0. 5. Solve for the two unknowns. 6. **Mark these forces as known** and move to an adjacent joint with now-only-two unknowns. 7. Repeat until all members are solved. **Assumption Convention:** Assuming every unknown member is in **tension** means drawing member force vectors **pulling away from the joint**. If the calculation gives a negative value, the member is in compression (it actually pushes on the joint). **Selection Strategy:** Always start at a support joint. Supports provide known reaction forces, reducing the number of unknowns at that joint. Then work methodically inward. **When Method of Joints Works Best:** Use when you need **all or most member forces**. It is systematic but can be tedious for large trusses. For a few specific interior members, the method of sections (Section 5) is faster. **Common Pitfall:** In a truss with multiple loading cases or unequal geometry, do not assume member forces "look" small. Compute them; a vertical member may carry substantial compression despite appearing slender.

The **method of sections** is an elegant alternative to the method of joints, especially useful for finding a few specific member forces deep inside a truss. Instead of isolating each joint, you **cut the truss** through the members of interest and apply equilibrium to one side as a rigid body. **Principle:** When you cut a truss, you expose the internal member forces at the cut. Treating one cut section as a free body in equilibrium gives you equations to solve for those forces. **Procedure:** 1. Find support reactions for the entire truss. 2. Decide which members you need to find. Plan a cut that **passes through at most three members with unknown forces**. (If you cut through four or more unknowns, you only have three equilibrium equations and cannot solve.) 3. Draw a free-body diagram of one side of the cut (usually the side with fewer external loads or a simpler geometry). 4. On the cut surface, draw arrows representing the member forces, **assuming tension** (forces point away from the section, toward the cut). 5. Write three equilibrium equations: ΣFx = 0, ΣFy = 0, ΣM = 0. 6. **Key trick:** Take moments about the intersection point of the other two cut members. This eliminates two unknowns, leaving one equation in one unknown. Solve. 7. Repeat for the other cut members if needed. **Why the Moment Trick Works:** If you take moments about point P where two members intersect, their moment arms are zero, and their forces drop out of the moment equation. You are left with a single unknown and one equation. **When to Use Method of Sections:** - Finding 1–3 specific member forces, especially interior members far from joints. - Cross-checking results from the method of joints. - When the truss is large and solving all joints is tedious. **Limitation:** You must be able to cut through the members of interest with a single line and expose at most three unknowns. Some truss geometries make this impossible; then you need method of joints or a multi-step approach. **Coordinate System:** Choose axes aligned with the truss orientation (horizontal for chords, vertical for verticals) to minimize coordinate transformations and angle calculations.

A **zero-force member** carries no internal force under a given loading. Identifying zero-force members simplifies analysis—you can remove them (conceptually) and reduce the problem size. They are not useless, however; they are essential for resisting other load cases and for lateral stability and bracing. **Rules for Identifying Zero-Force Members:** **Rule 1: Two Non-Collinear Members at an Unloaded Joint** If a joint has exactly two members and no external load (and no reaction) applied to it, both members are zero-force. Why? With no external force, ΣFx = 0 and ΣFy = 0 are satisfied only if both member forces are zero (assuming the members are not collinear; if they are, the joint is degenerate and should be removed). **Rule 2: Three Members, Two Collinear, at an Unloaded Joint** If a joint connects three members, two of which are collinear (forming a straight line), and there is no external load at the joint, then the **non-collinear (odd) member is zero-force**. The collinear pair carries force along the line; the odd member has no role in equilibrium. **Why These Rules?** Forces at a joint must sum to zero. If you have a restricted set of members and no external load, the only way to achieve equilibrium is for some members to carry zero force. The geometry determines which ones. **Practical Use:** Before solving a truss: 1. Scan all unloaded joints. 2. Apply the two rules to identify zero-force members. 3. Mark them on your diagram. 4. **Either remove them (for analysis) or skip them (when solving).** They do not affect the primary force path under the given loading. 5. When you report final member forces, include the zero-force members with force = 0 kN. **Important Caveat:** Zero-force members under one loading may carry significant force under a different loading (e.g., wind, lateral load, different live-load pattern). In design, you must check multiple load cases. A "useless" diagonal in one case might be essential in another. Trusses are over-braced with some zero-force members under typical loading to ensure strength under all conditions (as required by NSCP 2015, Chapter 2: Load Cases). **Example from Philippine Practice:** In a roof truss under vertical gravity loads, many diagonals might be zero-force. However, when the truss is subjected to wind uplift or lateral wind pressure, those diagonals activate and resist the induced moments and shears. Removing them would make the truss unstable under wind. This is why roof trusses appear "over-braced": extra members for robustness.

**Problem Statement (Exam Level):** A simply supported roof truss spans 10 m horizontally and rises 2 m vertically. The truss is pin-supported at the left end (point A) and roller-supported at the right end (point B). The geometry is: - Lower left support: A(0, 0) - Lower right support: B(10, 0) - Mid-span peak: C(5, 2) - Lower chord joints: A(0,0), D(2.5, 0), E(7.5, 0), B(10, 0) - Upper chord joint: C(5, 2) Members: - Lower chord: AD, DE, EB (three segments) - Upper chord: AC, CB (two segments) - Verticals: none in this simple truss - Diagonals: DC (connecting D to C), EC (connecting E to C) Loading: - Downward load of 20 kN applied at C (mid-span peak). - Self-weight: distributed among the joints as follows: 2 kN at each of A and B (end reactions account for half), 1 kN at D, 1 kN at E, 1.5 kN at C (in addition to the 20 kN external load). Total downward load = 2 + 2 + 1 + 1 + 1.5 + 20 = 27.5 kN, but the 2 kN at A and B are partial end reactions. Net load = 1 + 1 + 1.5 + 20 = 23.5 kN applied at interior points, plus support reactions balance the ends. (Simplification for this example: ignore self-weight to focus on the 20 kN load at C. Total external load = 20 kN.) **Given Data:** - External load at C: 20 kN downward - Truss height: 2 m; half-span: 5 m - Geometry: AC and CB each have length √(5² + 2²) = √29 ≈ 5.385 m - Angle of AC and CB with horizontal: α = tan⁻¹(2/5) ≈ 21.8° (sin α ≈ 0.371, cos α ≈ 0.928) - DC and EC: D(2.5, 0), C(5, 2) → DC has length √((5−2.5)² + 2²) = √(6.25 + 4) = √10.25 ≈ 3.202 m; angle with horizontal: β = tan⁻¹(2/2.5) = 38.66° (sin β ≈ 0.625, cos β ≈ 0.781). By symmetry, EC has the same length and angle. **Analysis:** **Step 1: Determinacy Check** m = 7 members (AD, DE, EB, AC, CB, DC, EC) r = 3 (A is pin: 2 components; B is roller: 1 component) j = 5 joints (A, D, E, B, C) m + r = 7 + 3 = 10; 2j = 10. ✓ Determinate. **Step 2: Support Reactions** By symmetry of geometry and loading: - RA,y = RB,y = 10 kN (upward) - RA,x = 0 (no horizontal loads) Verification: ΣFy = 10 + 10 − 20 = 0 ✓; ΣM about A = 10(10) − 20(5) = 100 − 100 = 0 ✓. **Step 3: Method of Joints** Start at joint A (left support). **At Joint A:** Members: AD (lower chord, horizontal) and AC (diagonal, sloping up-right). External: Reaction RA,x = 0, RA,y = 10 kN (upward). Free-body diagram: RA,y pointing up; FAD pointing right (assume tension); FAC pointing up-right (assume tension; this means the force on the joint is in the direction of AC from A toward C). Wait, if FAC is tension in member AC, the force on joint A is *pulling* A toward C, i.e., up and to the right. Unit vector along AC (from A toward C): (5/5.385, 2/5.385) ≈ (0.928, 0.371). Equilibrium at A: ΣFx: FAD + FAC(0.928) = 0 ΣFy: 10 + FAC(0.371) = 0 → FAC = −10/0.371 ≈ −26.95 kN (compression, magnitude ≈ 27 kN) From ΣFx: FAD = −FAC(0.928) = −(−26.95)(0.928) ≈ 25.01 kN (tension) So: FAD ≈ 25 kN (T), FAC ≈ 27 kN (C). **At Joint D:** Members: AD (from left), DE (to right), DC (sloping up-right toward C). External: none (assuming no load at D in this simplified version). Force from AD on D: equal and opposite to force on A, so −FAD, pointing left (compression reaction), i.e., FD,left = 25 kN pointing left. But wait, we found FAD = 25 kN tension in AD. Member AD pulls both of its ends; so it pulls A to the right and pulls D to the left. The force on joint D from member AD is pointing left (in the negative x-direction): (−25, 0). Assuming unknown members are tension (FDE, FDC): - FDE points right: (FDE, 0) - FDC points up-right toward C: FDC(cos β, sin β) = FDC(0.781, 0.625) External load at D: none. Equilibrium: ΣFx: −25 + FDE + FDC(0.781) = 0 ... (i) ΣFy: FDC(0.625) = 0 → FDC = 0 kN (zero-force!) From (i): FDE = 25 kN (tension). So: FDE ≈ 25 kN (T), FDC ≈ 0 kN. **At Joint E:** By symmetry: FEB ≈ 25 kN (T), FEC ≈ 0 kN. **At Joint C (Peak):** Members: AC (from left), CB (to right), DC (from left), EC (from right). External: 20 kN downward. Force from AC: member is in compression (−27 kN), so it pushes on C. Direction from C toward A: (−0.928, −0.371). Force on C: −(−27)(−0.928, −0.371) = −(+25.06, +10.02) ≈ (−25, −10). Wait, let me be more careful. If FAC = −27 kN (compression), the magnitude of compression is 27 kN. At joint C, the member AC pushes on C, i.e., the force points from C away from A (outward). Direction from A to C: (0.928, 0.371). Force on C from AC: 27(0.928, 0.371) ≈ (25, 10). By symmetry, FCB (compression in CB) exerts force on C: 27(−0.928, 0.371) ≈ (−25, 10). Forces from diagonals: FDC = 0, FEC = 0, so no contribution. External load at C: (0, −20). Equilibrium at C: ΣFx: 25 − 25 + 0 + 0 = 0 ✓ ΣFy: 10 + 10 + 0 + 0 − 20 = 0 ✓ Verification complete. **Summary of Member Forces:** - AD: 25 kN tension - DE: 25 kN tension - EB: 25 kN tension (by symmetry) - AC: 27 kN compression - CB: 27 kN compression (by symmetry) - DC: 0 kN (zero-force) - EC: 0 kN (zero-force) **Step 4: Verification Using Method of Sections** Cut through AC, DC, and DE. Take moments about the intersection of AC and DC. But AC and DC meet at C(5,2). Moment arm of DE about C: y-distance from DE to C is 2 m. DE carries 25 kN tension (pointing right). Force on the left section from the cut: DE pulls to the right with 25 kN, so on the cut face (at the boundary), there is a 25 kN force pointing left (internal action-reaction). Moment of this force about C: 25 × 2 = 50 kN·m (clockwise). On the left section: RA,y = 10 kN at A, external load at C = 20 kN down. Net external moment about C: 10(5) − 20(0) = 50 kN·m (counterclockwise). This is balanced by the internal member moment. Forces from AC and DC also contribute. AC is at C, so zero moment arm about C. DC carries 0 kN, so no contribution. Moment equation balances: 50 (external) = 50 (internal from DE) ✓. (This section is a quick check; full resolution would compute all three cut member forces.) **Key Takeaway:** In this symmetric truss under a central load, the lower chord carries tension uniformly (resisting bending), the upper chord carries compression, and the diagonals DC and EC carry zero force. This is characteristic of symmetric load cases where the truss "acts like a beam" in bending, with the lower chord as the tension flange and upper chord as the compression flange. Diagonals become active under other loading (e.g., if load were off-center).

**Mistake 1: Sign Confusion in Member Forces** - **Error:** Inconsistent convention for tension vs. compression, or mixing directions. - **Fix:** Adopt the convention early: assume all unknown members are in tension (forces point away from the joint, pulling on the joint). Negative result = compression. Stick to it throughout. - **Exam Consequence:** Losing points for incorrect signs or inconsistency, even if the magnitude is correct. **Mistake 2: Cutting Through More Than Three Unknowns** - **Error:** In the method of sections, cutting through four or more members with unknown forces. - **Fix:** Always plan your cut to expose at most three unknown member forces. If needed, use two or more cuts in sequence. - **Exam Consequence:** Unsolvable system of equations; incomplete or incorrect answer. **Mistake 3: Forgetting Support Reactions** - **Error:** Jumping to member force analysis without first computing support reactions. - **Fix:** Always find RA, RB, etc., before solving joints or sections. They are essential boundary conditions. - **Exam Consequence:** Wrong external forces at joints; cascading errors in member force calculations. **Mistake 4: Wrong Geometry or Angle Calculation** - **Error:** Miscomputing member length, slope angle, or unit vectors, especially in complex geometries. - **Fix:** Redraw the truss geometry carefully. Compute Δx, Δy, and length = √(Δx² + Δy²) explicitly. Double-check sin and cos of the angle. Use a calculator or trigonometric tables if unsure. - **Exam Consequence:** Correct method but wrong numerical answer due to geometric error. **Mistake 5: Assuming a Joint Has Only Two Unknowns When It Has Three or More** - **Error:** At a joint with three unknown members, writing two equilibrium equations and expecting to solve all three. - **Fix:** Always identify joints with exactly two unknowns before starting the method of joints. Skip joints with three or more unknowns initially; solve them later once you have more known forces from adjacent joints. - **Exam Consequence:** Incomplete or contradictory system; stuck in analysis. **Mistake 6: Misidentifying Zero-Force Members** - **Error:** Applying the zero-force rules incorrectly (e.g., calling a member zero-force when it has external load or when the geometry doesn't match the rule). - **Fix:** Carefully re-read the two rules. A joint must be unloaded. Members must be non-collinear (Rule 1) or form a 2-collinear-plus-1 pattern (Rule 2). If the joint has external load or the member configuration doesn't match, do not apply the rules; solve normally. - **Exam Consequence:** Incorrect member forces, cascading to later joints. **Mistake 7: Confusing Determinacy with Stability** - **Error:** Checking m + r = 2j and concluding the truss is determinate and stable, without verifying geometric stability (no concurrent or collinear member groups). - **Fix:** Use m + r = 2j as a necessary condition, then inspect the truss layout for proper member arrangement. Ask: "Can this truss resist loads in all directions?" If members are clustered or poorly arranged, it may be unstable despite satisfying the equation. - **Exam Consequence:** Attempting to solve an unstable (indeterminate or mechanism) truss, leading to inconsistent results or realizing mid-analysis that the problem is flawed. **Mistake 8: Inconsistent Free-Body Diagrams** - **Error:** Drawing the free-body diagram of a joint with unclear or contradictory force directions, leading to confusion in equilibrium equations. - **Fix:** For each joint, carefully draw all member forces. If assuming tension, draw all member forces pointing *away* from the joint (pulling). Include external loads and reactions with correct directions and magnitudes. Label everything. - **Exam Consequence:** Difficult to follow your work; risk of algebraic errors in equilibrium equations. **Mistake 9: Rounding Errors Accumulating Over Multiple Joints** - **Error:** Rounding intermediate results (e.g., member forces at early joints) and using those rounded values in later joints, causing small errors to compound. - **Fix:** Carry extra significant figures in intermediate calculations. Only round the final answer to a reasonable precision (usually 0.1 kN or 0.01 kN in exam problems). - **Exam Consequence:** Final answers slightly off, potentially marked incorrect if the examiner expects exact numerical results. **Mistake 10: Neglecting to Check Equilibrium at the Last Joint** - **Error:** Solving all but one joint and assuming the last one automatically balances. If it doesn't, it reveals a calculation error, but you've run out of time to fix it. - **Fix:** After solving all members, select one joint you haven't explicitly solved (usually an interior joint with several known members) and verify equilibrium. This catches mistakes early. - **Exam Consequence:** Undetected error; wrong final answer. **Mistake 11: Misinterpreting Support Conditions** - **Error:** Treating a pin support as a roller, or vice versa, affecting the number of reaction components and thus determinacy. - **Fix:** Read problem statements carefully. Pin = 2 reactions (Rx, Ry). Roller = 1 reaction (perpendicular to surface). Fixed support (encastré) = 2 reactions + moment. If the diagram is unclear, ask or re-read the problem. - **Exam Consequence:** Wrong determinacy assessment; misleading analysis from the start. **Mistake 12: Forgetting Vertical or Horizontal Components** - **Error:** In equilibrium equations, omitting the y-component (for instance) and solving an incomplete system. - **Fix:** Always write out both ΣFx = 0 and ΣFy = 0, even if one seems obviously zero. This ensures completeness and catches oversights. - **Exam Consequence:** Missing a solution; incomplete answer.

**Philippine Standards and Codes:** - **NSCP 2015 (National Structural Code of the Philippines, 2nd Edition).** Section 2: Loads and Load Cases. Specifies load combinations, self-weight treatment, and load assumptions for trusses in buildings and bridges. - **RA 544 (Republic Act No. 544, "An Act to Regulate the Practice of Profession of Civil Engineering in the Philippines").** Mandates that only licensed civil engineers may design and certify structural calculations, including truss analysis, ensuring professional competence. **International Standards (for reference and advanced study):** - **AISC 360 (American Institute of Steel Construction, Specification for Structural Steel Buildings).** Chapter J: Connections, Joints, and Fasteners. Details pin connections, frictionless joint assumptions, and member behavior in steel trusses. - **ACI 318 (American Concrete Institute, Building Code Requirements for Structural Concrete).** While primarily for concrete, it addresses timber and composite truss design and load factors in reinforced structures. **Mechanics Texts:** - **Hibbeler, R.C., "Structural Analysis", 8th Edition.** Comprehensive coverage of truss analysis methods, determinacy, and worked examples. - **Leet, K.M., Uang, C.M., and Gilbert, A.M., "Fundamentals of Structural Analysis", 5th Edition.** Clear exposition of method of joints, method of sections, and zero-force members. **Exam Preparation:** - **PRC Board of Civil Engineers Review Materials.** Official exam syllabi and past exam questions often feature truss problems at varying difficulty levels. - **UPCA (UP College of Architecture) and Other Universities' Review Notes.** Pedagogical approaches and problem sets from leading Philippine engineering programs. **Note:** For the PRC Licensure Examination (both the Civil Engineer and Structural Engineer exams), truss analysis typically comprises 5–10% of the Engineering Mechanics/Structural Analysis section. Mastery of the method of joints, method of sections, and zero-force identification is essential. The exam often presents trusses under multiple load cases, requiring checklist-based determinacy verification and careful sign convention.