3.1 Statics I: Resultants, Equilibrium & Trusses

Forces, Components & Resultants

Statics is the study of bodies in equilibrium — bodies at rest or moving at constant velocity, so the net force and net moment are zero. On the Fundamentals of Engineering (FE) Civil exam, Statics is one of the largest knowledge areas (about 7–11 of the 110 questions). All formulas used here appear in the NCEES FE Reference Handbook under Statics; your job on the open-resource exam is to find and apply them quickly, not to memorize them.

A force is a vector. To work with several forces you resolve each into rectangular components. For a force F at angle θ from the positive x-axis:

• Fx = F·cos θ, Fy = F·sin θ (units: pound-force, lbf, or newtons, N)

The resultant R of a coplanar force system is found by summing components:

• Rx = ΣFx, Ry = ΣFy
• R = √(Rx² + Ry²), direction θ_R = arctan(Ry / Rx)

A common trap is mixing degrees and radians, or measuring θ from the wrong axis. Always sketch the angle. In 3D, a force along a line is found from the position vector between two points: F = F·(d/|d|), where d = ⟨x₂−x₁, y₂−y₁, z₂−z₁⟩ and |d| = √(dx² + dy² + dz²).

Equilibrium & Free-Body Diagrams

The heart of statics is the free-body diagram (FBD): isolate the body, draw every external force and reaction, and apply equilibrium. In two dimensions there are three independent scalar equations:

• ΣFx = 0, ΣFy = 0, ΣM = 0 (moment about any point)

In three dimensions there are six: ΣFx = ΣFy = ΣFz = 0 and ΣMx = ΣMy = ΣMz = 0. A moment of a force about a point is M = F·d, where d is the perpendicular distance from the point to the force's line of action (units: lbf·ft or N·m). Vector form: M = r × F.

Support reactions (2D)

A structure is statically determinate when the number of unknown reactions equals the number of equilibrium equations (3 in 2D). Count reactions first: a simple beam (pin + roller) has 3 unknowns and 3 equations — solvable. When unknowns exceed equations, the structure is statically indeterminate and needs compatibility (deflection) equations beyond statics. Choose your moment center at a pin to eliminate two unknowns at once, then back-substitute into the force equations.

The sign convention matters: pick positive directions for the reaction arrows on your free-body diagram and keep them consistent. A negative answer simply means the reaction acts opposite to your assumed direction — it is not an error. For distributed loads, replace the load with its resultant (area under the load curve) acting at the load's centroid before taking moments; a uniform load w over length a has resultant w·a acting at a/2 from its start.

Worked Example — Beam Reactions

A simply supported beam spans L = 10 ft. A 4 kip downward point load sits 3 ft from the left pin A; the right end B is a roller. Find the reactions.

Take moments about A (clockwise positive):

• ΣM_A = 0: (4 kip)(3 ft) − R_B(10 ft) = 0 → R_B = 12/10 = 1.2 kip (up)

Sum vertical forces:

• ΣFy = 0: R_A + R_B − 4 = 0 → R_A = 4 − 1.2 = 2.8 kip (up)

Check with ΣM_B: R_A(10) − 4(7) = 28 − 28 = 0. ✓ The load closer to A throws more reaction onto A, which matches R_A > R_B. Always verify with a second, independent equation — it catches sign and arithmetic errors.

Truss Analysis

A truss is an assembly of straight two-force members joined at pins; each member carries only axial tension (+) or compression (−). Two solution techniques are tested:

• Method of joints: isolate one joint; apply ΣFx = 0 and ΣFy = 0 (two equations, so solve a joint with ≤2 unknowns). Work joint to joint.
• Method of sections: cut through ≤3 members, treat one side as a free body, and use ΣM about a strategic point to solve a single member force directly — ideal when you need only one or two interior members.

Zero-force members

Spot them by inspection to save time:

1. At a joint with two non-collinear members and no external load, both are zero-force.
2. At a joint with three members where two are collinear and no load, the odd (non-collinear) member is zero-force.

Zero-force members are not useless — they brace other members against buckling and handle load under different loading cases.

Sign convention and procedure

By convention, draw all unknown member forces in tension (arrows pulling away from the joint). A positive result confirms tension; a negative result means the member is in compression. For the method of joints, start at a support or a loaded joint with no more than two unknown members so the two equilibrium equations are solvable, then march through the truss joint by joint, carrying forward each solved member force.

For the method of sections, the cut must pass through the member(s) you want and divide the truss into two complete parts. Apply all three equilibrium equations to either part. A smart choice of moment center — the intersection of two of the three cut members — isolates the third member force in a single ΣM equation, which is far faster than chasing it joint by joint across the whole truss. This is why the method of sections is the go-to technique when a question asks for one specific interior member.