CELE Engineering Mechanics — Equilibrium of Particles and Rigid BodiesSummary

A sketch isolating the body of interest, showing every external force acting on it: applied loads, self-weight, and support reactions. The FBD is the critical first step in any equilibrium problem. Replace each support type with its correct reaction: a roller exerts a single normal force perpendicular to the surface; a pin exerts two orthogonal force components (typically horizontal and vertical); a fixed end exerts two force components plus a reaction moment. A correct FBD is 80% of the solution—errors here propagate through all subsequent calculations.

For a body in static equilibrium, the three independent conditions are: (1) ΣFx = 0 (sum of horizontal forces), (2) ΣFy = 0 (sum of vertical forces), (3) ΣM = 0 (sum of moments about any point). These three equations can be used to solve for at most three unknowns. The moment equation is particularly powerful: choosing the moment center to pass through unknown forces eliminates them, often solving a reaction in one step. For 3D systems, six equations apply (three force, three moment).

A particle (treated as having no rotational inertia, with all forces concurrent at a point) requires only ΣFx = 0 and ΣFy = 0. A rigid body in general may require the moment equation as well. Particle equilibrium problems are common in suspension systems, cable-stayed structures, and joint analysis in trusses.

When exactly three concurrent forces act on a particle in equilibrium, Lami's theorem provides a direct solution: F₁/sin(α₁) = F₂/sin(α₂) = F₃/sin(α₃), where each αᵢ is the angle between the other two forces (not the angle between that force and a reference axis). This is faster than component methods for problems with three forces. Example: a weight hanging from two cables—identify the three forces (two tensions and the weight), measure angles between them, apply the law of sines directly.

A member loaded only at two points (with no other forces applied along its length) must be in equilibrium under forces acting only at those two points. The force must act along the line joining the two points—resulting in pure tension or compression. All truss members are two-force members. The direction of the internal force is therefore known (along the member axis), reducing unknowns in the problem.

A rigid body in equilibrium under three forces can be in equilibrium only if (1) the three forces are concurrent (all meet at a single point), or (2) all three are parallel. If a member is loaded at three points, the directions of two forces are known, and the third force must pass through the intersection of the lines of action of the first two. This property is used to find unknown force directions without solving equations.

Supports provide reactions that prevent motion. A roller (or rocker) provides one reaction perpendicular to the surface. A pin provides two orthogonal reactions (e.g., Hx and Vy). A fixed end (or built-in support) provides two force reactions plus a reaction moment. A free end provides no reaction. The number and types of reactions determine the degree of static determinacy: a simply supported beam (two supports) is statically determinate; additional supports make it indeterminate (redundant), requiring compatibility of deformation equations from structural analysis.

A uniformly distributed load (UDL) of intensity w over length L is equivalent to a single concentrated force of magnitude w × L acting at the centroid of the distribution (at midspan for uniform load). Before applying equilibrium equations, always replace the distributed load with its resultant. Example: a UDL of 20 kN/m over 5 m = 100 kN acting at 2.5 m from the start. Failure to do this is a common exam error.

The moment of a force about a point is M = F × d, where d is the perpendicular distance from the point to the line of action. For a coplanar system, take moments about a point in the plane, using the right-hand rule (counterclockwise positive, clockwise negative) or simply tracking sign carefully. The power of the moment equation lies in choosing the moment center to eliminate unknown forces. Example: to find a roller reaction, take moments about the other support, which eliminates that reaction and the vertical reaction at the chosen point.

The FBD sets up the problem by identifying all external forces and support reactions. The three equilibrium equations (ΣFx = 0, ΣFy = 0, ΣM = 0) are then applied directly to the FBD to solve for unknowns. A correct FBD is necessary; the equations alone cannot fix an incorrect diagram.

Particle equilibrium is a special case of rigid body equilibrium where all forces are concurrent (meet at a point). When forces are concurrent, the moment equation is automatically satisfied, and only two force equations (ΣFx = 0, ΣFy = 0) are needed. For general rigid bodies, all three equations are required.

Lami's theorem and the component method (resolving forces into x and y) are two routes to solve three concurrent-force problems. Lami is faster for three forces; components are more systematic and work for any number of forces. Choice depends on problem geometry and personal preference.

Recognizing a two-force member (such as a truss bar) immediately tells you the force is along the member axis. This reduces unknowns in the problem, because the direction is known; only magnitude remains.

For a three-force member in equilibrium, if two force directions are known, the third force must pass through the intersection of the first two (or all must be parallel). This geometric constraint determines the third force direction without calculation, allowing a graphical or analytical solution.

The type of support (roller, pin, fixed) determines how many and which reactions the support exerts. Correct FBD requires correct identification of support type. Statically determinate systems have reactions equal in number to equilibrium equations (3 in 2D); indeterminate systems have more.

A distributed load is not directly used in equilibrium equations. It must be replaced by a concentrated resultant force (magnitude = intensity × length, location = centroid of the distribution) before applying ΣF and ΣM.

The moment equation ΣM = 0 can be applied about any point in the plane. Choosing the moment center through unknown forces eliminates them, reducing the number of equations to solve simultaneously. This is a key strategy for efficient problem solving.

When designing a reinforced concrete beam (per ACI 318 and NSCP 2015), the first step is always to find the support reactions using equilibrium equations. A cantilever beam requires both a shear reaction and a moment reaction at the fixed end; a simply supported beam requires vertical reactions at both ends. These reactions determine the shear and moment diagrams, which in turn dictate reinforcement design. Example: a 6 m simply supported beam with 50 kN point load at midspan produces reactions of 25 kN at each end. This forms the basis of all subsequent strength and serviceability checks.

Trusses consist of two-force members (bars in tension or compression). To find member forces, first determine support reactions using equilibrium (treating the entire truss as a rigid body), then use method of joints (analyzing each joint as a particle in equilibrium) or method of sections (cutting the truss and applying equilibrium to isolated parts). Example: a roof truss supporting a 10 kN load per panel. Equilibrium at the support determines external reactions; then each joint equation (ΣFx = 0, ΣFy = 0) finds the bar forces without knowing any internal moments (because all members are two-force members).

Cables in suspension bridges, guy-wires for towers, and hanging loads are particle-equilibrium problems. The cable tensions are found by resolving forces at the junction point: ΣFx = 0 and ΣFy = 0 at the load point. If exactly three forces act (two cable tensions and the load), Lami's theorem gives a direct answer. Example: a 600 N lamp hangs from two cables at 30° and 50° from the ceiling. Using Lami or component equations, find both cable tensions instantly. This is common in design of roof hangers and suspended structures.

A column carries a building load down to the foundation. The base support (pile cap, footing) must be designed to resist the column reactions: vertical load and possibly bending moment. Using equilibrium of the foundation element (free-body diagram of the footing), the designer determines bearing pressure distribution and checks against soil capacity. Example: a 2 m × 3 m footing under a 1500 kN column load with 200 kN·m moment. Equilibrium equations determine how the bearing pressure varies across the footing, which must not exceed soil bearing capacity (per NSCP 2015, Chapter 2).

Rigid frames (building frames, gantry cranes, portal frames) have both force and moment reactions at supports. Finding these reactions requires all three equilibrium equations. Example: a gantry crane spans 10 m between two pin supports, carries a 50 kN hoisted load at midspan. Taking moments about one pin gives the reaction at the other; force balance gives remaining reactions. These reactions then load the frame members, which must be checked for bending stress per AISC 360.

A cantilever is fixed at one end and free at the other. The fixed support exerts both a shear reaction and a reaction moment. The reaction moment is often larger than the shear force, governing the design. Example: a concrete balcony overhanging 2 m with live load 5 kN/m produces: shear = 10 kN, moment = 20 kN·m at the fixed end. This moment must be resisted by the connection and the beam section, per ACI 318 and NSCP 2015 design criteria.

Problems with inclined planes (ramps, sloped roofs, retaining walls) require decomposition of forces. A load on a 30° slope has components parallel and perpendicular to the slope. Equilibrium equations are written in slope-normal and slope-parallel directions, or in horizontal-vertical directions with care for angles. Example: a 100 kN load on a 20° slope resting on two supports at different elevations. Resolve the load into components, draw the FBD, and solve for the two support reactions. This is common in analyzing bridge approaches and foundation structures.

Under RA 544 (Hydraulics and Hydrostatics Law), civil engineers may supervise load tests on structures. The test loads are applied at known locations, and reactions are measured at supports. Comparing measured reactions to calculated reactions (using equilibrium equations) verifies structural integrity. Discrepancies may indicate damage or deterioration. Example: a test load of 100 kN applied at 3 m from support A on a 10 m span should produce reactions calculable by equilibrium; if measured values differ significantly, further investigation is warranted.