Statics

B. Force systems

Vectors, Force Systems, and Resultants

Resolve forces into components, build unit vectors in two and three dimensions, and combine any force system into a single resultant.

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C. Force couple systems

Moments and Force-Couple Systems

Compute moments by lever arm and by cross product, recognize couples, and reduce any load set to an equivalent force-couple at a chosen point.

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D. Equilibrium of rigid bodies

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

. Verify by checking

\Sigma M = 0

about a second, independent point — if it does not close, you dropped a force or a sign. Before solving, count unknowns against equations to confirm the problem is determinate.

Key equations

2-D force equilibrium

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

Net force is zero along two perpendicular axes; signs follow your chosen positive directions.

2-D moment equilibrium

\sum M_O = 0

Net moment about any point $O$ is zero. Pick $O$ to eliminate unknowns from the equation.

Three independent 2-D equations

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

Solves up to three unknowns for a single rigid body in a plane.

3-D equilibrium

\sum \mathbf{F} = 0, \qquad \sum \mathbf{M} = 0

Six scalar equations (three force, three moment); solves up to six unknown components.

Two-force member condition

\mathbf{F}_A = -\mathbf{F}_B \ \text{(equal, opposite, collinear)}

Loaded at only two points; force direction is along the line connecting them.

Three-force member condition

\text{three forces concurrent or all parallel}

Equilibrium of a body under exactly three forces requires their lines of action to meet at a point (or be parallel).

Roller / smooth-surface reaction

R \perp \text{surface},\ \ 1\ \text{unknown}

Single reaction normal to the contact surface or the roller's travel direction.

Pin (hinge) reaction

A_x,\ A_y\ \ (2\ \text{unknowns, no moment})

Two force components; allows free rotation, so it transmits no reaction moment.

Determinacy criterion (single body)

r = 3 \ (\text{2-D}), \quad r = 6 \ (\text{3-D})

$r$ = number of unknown reaction components for a statically determinate rigid body.

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 $3\,\text{m}$ from $A$ and a $6\,\text{kN}$ downward load $6\,\text{m}$ from $A$ . Find the support reactions.

Solution. Unknowns:

A_x

,

A_y

(pin) and

B_y

(roller) — three unknowns, three equations, determinate.

\sum M_A = 0

(CCW +):

B_y(8) - 10(3) - 6(6) = 0 \Rightarrow B_y = (30 + 36)/8 = 8.25\,\text{kN}

up.

\sum F_y = 0

:

A_y + B_y - 10 - 6 = 0 \Rightarrow A_y = 16 - 8.25 = 7.75\,\text{kN}

up.

\sum F_x = 0

:

A_x = 0

(no horizontal loads). Check

\sum M_B = 0

:

-A_y(8) + 10(5) + 6(2) = -62 + 50 + 12 = 0

✓. Reactions:

A_x = 0

,

A_y = 7.75\,\text{kN}

,

B_y = 8.25\,\text{kN}

.

B_y = \frac{10(3) + 6(6)}{8} = 8.25\ \text{kN}, \qquad A_y = 16 - 8.25 = 7.75\ \text{kN}

Beam with an inclined load

Problem. A beam $4\,\text{m}$ long, pinned at $A$ (left) and on a roller at $B$ (right, vertical reaction only), carries a single $20\,\text{kN}$ force at midspan directed $30^\circ$ above the horizontal (pointing up and to the right). Find all reactions.

Solution. Load components:

F_x = 20\cos 30^\circ = 17.32\,\text{kN}

(right),

F_y = 20\sin 30^\circ = 10.0\,\text{kN}

(up). Take all reactions positive upward and CCW positive; the upward load at

x = 2\,\text{m}

from

A

makes a positive (CCW) moment about

A

.

\sum M_A = 0

:

B_y(4) + 10.0(2) = 0 \Rightarrow B_y = -5.00\,\text{kN}

; the negative sign means

B_y

acts downward,

5.00\,\text{kN}

. (The horizontal component passes through the beam axis, contributing no moment about

A

.)

\sum F_y = 0

:

A_y + B_y + 10.0 = 0 \Rightarrow A_y = -10.0 - (-5.00) = -5.00\,\text{kN}

; the negative sign means

A_y

also acts downward,

5.00\,\text{kN}

.

\sum F_x = 0

:

A_x + 17.32 = 0 \Rightarrow A_x = -17.3\,\text{kN}

(i.e.

17.3\,\text{kN}

to the left). Check

\sum M_B = 0

(CCW +):

-A_y(4) - 10.0(2) = -(-5.00)(4) - 20.0 = 20.0 - 20.0 = 0

✓. Sanity: the load pushes up at midspan, so both supports must pull the beam down to hold it, and because the load sits exactly at midspan the two vertical reactions are equal by symmetry —

A_y = B_y = -5.00\,\text{kN}

(each

5.00\,\text{kN}

down) is exactly that.

B_y = -\frac{10.0(2)}{4} = -5.00\ \text{kN}, \quad A_y = -5.00\ \text{kN}, \quad A_x = -17.3\ \text{kN}

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

E. Internal forces in rigid bodies

Trusses, Frames, and Machines

Find member forces by the method of joints and the method of sections, spot zero-force members, and tell trusses, frames, and machines apart.

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F. Area properties

Centroids and Area Moments of Inertia

Locate centroids of areas and composite shapes, compute area moments of inertia, and shift axes with the parallel-axis theorem.

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G. Static friction

Static Friction

Apply Coulomb's dry-friction law, find impending motion on inclines and wedges, use belt friction, and distinguish tipping from sliding.

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I. Weight and mass computations

Weight, Mass, and the gc Factor

Keep mass and force units straight across SI and USCS, apply W=mg correctly, and use the gc conversion factor to dodge the classic lbm/lbf trap.

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