Statics

Statics · Study · FE Other Disciplines · FE → PE Prep

Equilibrium, Free-Body Diagrams, and Reactions

Draw correct free-body diagrams, identify support reactions, and apply the equilibrium equations to solve for unknown forces and reactions.

Equilibrium is the heart of statics: a body at rest has zero net force and zero net moment, and almost every statics problem on the FE is, underneath, an exercise in writing those conditions and solving them. The skill that separates fast solvers from slow ones is not the algebra — it is drawing a correct free-body diagram (FBD), because a force forgotten or a reaction drawn the wrong way poisons every equation that follows. The Statics chapter of the FE Reference Handbook states the equilibrium requirements in one line,

\Sigma\mathbf{F}=0

and

\Sigma\mathbf{M}=0

; this concept teaches you to set up those equations so they actually solve.

The free-body diagram

A free-body diagram isolates one body and shows every external force and reaction acting on it — applied loads, weights, and the reactions from each support, and nothing internal. Replace each support by the force(s) and moment(s) it can exert, include the body's weight at its center of gravity when it matters, and label every magnitude and direction. Guess a direction for each unknown; a negative answer simply means the true direction is opposite to your guess. The FBD is the contract: if it is right, the equations are routine.

Support types and their reactions

Each support is read by what it prevents. A roller or smooth surface resists only along the direction normal to the surface, giving one reaction; a link (or short connecting two-force member) resists only along its own axis, giving one reaction along that known line of action. A pin (or hinge) resists translation in two directions but allows rotation, giving two reaction components and no moment. A fixed (built-in) support resists translation and rotation, giving two force components plus a reaction moment in 2-D. Counting reactions correctly is the first half of every equilibrium problem; the rule of thumb is one reaction per degree of motion the support prevents.

\text{roller} \to 1\ \text{reaction}, \quad \text{pin} \to 2, \quad \text{fixed} \to 3\ (\text{2-D})

The 2-D equilibrium equations

A rigid body in a plane is in equilibrium when the sum of forces in two perpendicular directions and the sum of moments about any point are all zero — three independent equations, so three unknowns are solvable. The strategic move is to take moments about a point where unknown forces intersect, eliminating them from that equation. Choosing the moment center well often hands you one reaction directly, with no simultaneous equations.

\sum F_x = 0, \qquad \sum F_y = 0, \qquad \sum M_O = 0

Two-force and three-force members

A two-force member carries load at only two points and has no other forces; equilibrium then forces the two resultants to be equal, opposite, and collinear — the line joining the two points. This means its force direction is known from geometry alone, which is exactly why truss members are so tractable. A three-force member is in equilibrium only if the three forces are concurrent (their lines of action meet at one point) or all parallel; spotting concurrency can replace a moment equation with a quick geometric argument.

\text{Two-force member: } \mathbf{F}_A = -\mathbf{F}_B, \ \text{collinear along } AB

Three-dimensional equilibrium

In space, equilibrium requires six scalar equations: forces summing to zero along three axes and moments summing to zero about three axes. With six equations you can solve up to six unknown reaction components. Supports generalize accordingly — a ball-and-socket gives three force reactions, a journal bearing or hinge gives forces and possibly moments transverse to its axis. The vector forms

\Sigma\mathbf{F}=0

and

\Sigma\mathbf{M}=0

are most reliably applied through components.

\sum F_x = \sum F_y = \sum F_z = 0, \qquad \sum M_x = \sum M_y = \sum M_z = 0

Statical determinacy

A structure is statically determinate when the number of unknown reactions (and member forces) equals the number of independent equilibrium equations available — three per rigid body in 2-D, six in 3-D. Fewer unknowns than equations means a mechanism (unstable); more unknowns means statically indeterminate, requiring deformation compatibility beyond statics alone. The FE almost always presents determinate structures, but recognizing indeterminacy tells you immediately that

\Sigma F

and

\Sigma M

alone cannot finish the job.

r = 3 \;(\text{2-D, single body}) \Rightarrow \text{determinate}; \quad r > 3 \Rightarrow \text{indeterminate}

Exam strategy

Always draw the FBD first and never skip it — it is faster than the time it saves. Replace each support with its correct reactions, assume directions, and let the algebra report sign. Take moments about the point where the most unknowns intersect (usually a pin) so one reaction drops out and you solve it directly; then back-substitute into

\Sigma F_y = 0

and

\Sigma F_x = 0

Key equations

2-D force equilibrium

\sum F_x = 0, \qquad \sum F_y = 0

Worked examples

Reactions of a simply supported beam

Problem. A beam $8\,\text{m}$ long has a pin at the left end $A$ and a roller at the right end $B$ . It carries a $10\,\text{kN}$ downward load $3\,\text{m}$

Common pitfalls

•Skipping or mis-drawing the FBD. A forgotten weight or a reaction at the wrong support is the leading cause of wrong equilibrium answers — draw it every time.
•Wrong reaction set for a support. A roller gives one reaction, a pin two, a fixed support three (2-D). Over- or under-counting reactions makes the problem look indeterminate or unstable.
•Taking moments about a poorly chosen point. Choose the point where the most unknowns intersect so they drop out; otherwise you create needless simultaneous equations.
•Forgetting that a negative result just flips the assumed direction. Do not re-solve — report the magnitude with the corrected sense.
•Including internal forces on the FBD of a whole body. Only external loads and reactions belong; internal member forces appear only when you cut the body.
•Assuming a three-force body's forces are arbitrary. They must be concurrent or parallel — using concurrency can replace a moment equation entirely.
•Not checking determinacy. If reactions exceed the available equations, statics alone cannot solve it; recognize indeterminacy before grinding algebra.

References

NCEES FE Reference Handbook — Statics
Hibbeler, Engineering Mechanics: Statics — free-body diagrams, supports, equilibrium
Beer & Johnston, Vector Mechanics for Engineers: Statics

B. Force systems

Vectors, Force Systems, and Resultants

C. Force couple systems

Moments and Force-Couple Systems

D. Equilibrium of rigid bodies

Equilibrium, Free-Body Diagrams, and Reactions

The free-body diagram

Support types and their reactions

The 2-D equilibrium equations

Two-force and three-force members

Three-dimensional equilibrium

Statical determinacy

Exam strategy

Key equations

Worked examples

Common pitfalls

References

E. Internal forces in rigid bodies

Trusses, Frames, and Machines

F. Area properties

Centroids and Area Moments of Inertia

G. Static friction

Static Friction

I. Weight and mass computations

Weight, Mass, and the gc Factor