CELE Engineering Mechanics — Analysis of TrussesSummary

Under the assumptions of truss theory (pin joints, loads at joints only, straight weightless members), every truss member is a two-force member — meaning the internal force acts along the member axis, either pulling (tension) or pushing (compression). This simplification eliminates bending moments from axial loads. If a member has a force at one end, equilibrium requires an equal, opposite force at the other end, both directed along the member centerline. This is central to understanding why truss analysis is simpler than general frame analysis.

Truss joints are idealized as frictionless pins that transmit force but NOT moment. In reality, connections are bolted or welded, but the pin assumption is valid when members are straight, loads are applied only at joints, and the centerlines of members intersect at a point. A pin joint allows rotation without friction, ensuring each member experiences only axial force. This contrasts with rigid frames where joints transfer both force and moment.

For a plane truss, the equation m + r = 2j determines whether the structure is statically determinate (one unique solution), indeterminate (infinite solutions, requires material properties), or unstable (no solution, mechanism). Here: m = number of members, r = number of reaction components (pin = 2, roller = 1), j = number of joints. The criterion arises from the fact that each joint furnishes 2 equilibrium equations (ΣF_x = 0, ΣF_y = 0), and each unknown (member force and reaction) requires one equation. If m + r = 2j, the system is determinate; if m + r > 2j, indeterminate; if m + r < 2j, unstable. However, this is a NECESSARY but not SUFFICIENT condition — geometric arrangement matters. A truss can satisfy m + r = 2j and still be unstable if members are collinear or otherwise improperly arranged (e.g., all diagonals meeting at one point).

A systematic approach treating each joint as a particle in equilibrium, applying ΣF_x = 0 and ΣF_y = 0. Procedure: (1) Find support reactions for the entire truss using global equilibrium. (2) Identify a joint with at most 2 unknown member forces. (3) Apply joint equilibrium to solve for those forces (assume tension; negative result = compression). (4) Move to an adjacent joint, carrying known forces forward. Continue until all members are solved. Advantages: yields all member forces; creates an audit trail; straightforward for hand calculation. Disadvantages: must solve all joints even if only a few forces are needed; accumulated rounding errors if many joints; prone to sign errors if conventions drift.

An alternative approach where the entire truss (or a portion) is cut through selected members (typically ≤3 unknowns), and one side is treated as a rigid body in equilibrium. After cutting, apply ΣF_x = 0, ΣF_y = 0, and ΣM = 0 (moment about any point). The trick is to take moments about the intersection point of two of the cut members, eliminating them and leaving one unknown to solve directly. Then use force equilibrium to find the others. Advantages: finds specific interior members without solving all joints; fewer equations; useful for checking Method of Joints results. Disadvantages: requires visualizing the intersection point (may require slope/angle calculation); cannot find forces in members not cut; if more than 3 unknowns are cut, the system is unsolvable.

Members carrying no internal force under a particular loading condition. Identification rules: (1) At a joint with 2 non-collinear members and NO external load, BOTH are zero-force. (2) At a joint with 3 members where 2 are collinear and NO external load, the third (non-collinear) member is zero-force. Example: in a roof truss, a vertical web member between two horizontal chords (no load at the joint) is zero-force. These members are not useless — they stabilize against buckling, carry load under different scenarios, and are mandated by design codes (e.g., NSCP 2015 and AISC 360 for lateral bracing). Recognizing zero-force members simplifies hand calculation by allowing temporary removal during analysis.

In truss analysis, a positive member force denotes TENSION (the member pulls inward on both joints; it is 'stretched'). A negative force denotes COMPRESSION (the member pushes outward; it is 'squeezed'). Consistent sign convention is crucial: assume every unknown member force is tension. If the solution yields a negative value, the member is in compression. This convention prevents confusion when writing equilibrium equations and interpreting results. On free-body diagrams, draw tension forces pointing AWAY from the joint (pulling), and compression forces pointing TOWARD the joint (pushing).

Before analyzing joints or sections, find the support reactions using global equilibrium of the entire truss. For a 2D planar truss: ΣF_x = 0 (horizontal equilibrium), ΣF_y = 0 (vertical equilibrium), ΣM = 0 (rotational equilibrium about any point). A pin support provides 2 reaction components (H, V); a roller provides 1 (perpendicular to the surface). Once reactions are known, they become known forces at the boundary joints, enabling downstream joint analysis. Failure to correctly find reactions invalidates the entire analysis.

A truss satisfying m + r = 2j is statically determinate only if it is also GEOMETRICALLY STABLE — i.e., the member arrangement prevents collapse into a mechanism. Examples of geometric instability: (1) Three non-concurrent, non-parallel reaction forces (stable), but if all diagonals meet at one point, the chord may bend (unstable). (2) A truss where all members pass through a single point (concurrent members). (3) Collinear members in a region without lateral support. The determinacy equation guarantees enough equations to solve; geometric stability ensures a unique, rigid solution. A structure can be determinate but unstable if members are improperly arranged.

The two-force-member assumption is valid ONLY because of pin-jointed connections. If joints were rigid and transmitted moments, members could be bent by moments, and forces would not be purely axial. The combination of pin joints + loads at joints only + straight members = every member is a two-force member. Conversely, if a member is loaded at midspan or connected rigidly, it is NOT a two-force member and requires bending analysis.

If m + r = 2j (determinate), the Method of Joints (or Method of Sections) yields a unique solution using only equilibrium equations. If m + r > 2j (indeterminate), equilibrium alone is insufficient; additional compatibility and material-property equations are needed (beyond the scope of this chapter, but essential for advanced courses and professional design). If m + r < 2j (unstable), no equilibrium solution exists — the structure is a mechanism.

Both methods use the same fundamental principle: equilibrium. Method of Joints applies it locally (each joint as a particle); Method of Sections applies it globally (a section as a rigid body). For any determinate truss, both yield identical answers for the same member. The choice depends on the problem goal: all forces → Method of Joints; specific interior forces → Method of Sections. On exams, using one method to verify the other demonstrates mastery.

A member is zero-force not because of weak loading, but because of JOINT GEOMETRY. If a joint has 2 non-collinear members and no load, the geometry forces both forces to be zero (the joint has no way to transmit load). This is a direct consequence of equilibrium and geometry, not a design choice. Recognizing zero-force members requires geometric insight, not just arithmetic.

Support reactions are the boundary conditions for the entire truss. Any error in reactions propagates through all subsequent joint and section calculations, invalidating the entire solution. Thus, reactions must be found with care and often verified using multiple moment centers or alternative equilibrium equations.

Assuming tension (positive) and interpreting negative as compression must be applied consistently. If one joint is analyzed assuming tension and another assuming compression, the results cannot be compared or compiled into a final member list. The convention is arbitrary (one could assume compression instead), but consistency is essential.