Statics

A. Resultants of force systems

Resultants and Equivalent Force Systems

Combine forces into a single resultant, take moments with M=Fd and r×F, recognize couples, and reduce any system to an equivalent force-couple at a point.

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C. Equilibrium of rigid bodies

Equilibrium, Trusses, and Frames

Draw free-body diagrams, solve support reactions with ΣF=0 and ΣM=0, and analyze trusses by joints and sections, frames and machines, and zero-force members.

Equilibrium is the heart of statics and the most heavily tested idea on the FE Civil exam: a body at rest has zero net force and zero net moment, full stop. Every reaction, every member force, every "is this beam balanced" question reduces to writing

\sum\mathbf{F}=0

and

\sum\mathbf{M}=0

and solving. The skill that separates fast solvers from slow ones is the free-body diagram — isolate the right body, show every force once, and the equations almost write themselves. The FE Reference Handbook's Statics chapter gives the equilibrium requirements, the truss methods, and the zero-force-member rules; this concept turns those bare statements into a working procedure.

The free-body diagram

A free-body diagram (FBD) isolates a single body and shows every external force and couple acting on it: applied loads, the weight at the center of gravity, and reactions where supports were removed. Replace each support by the forces it can transmit — a roller gives one force perpendicular to its surface, a pin gives two perpendicular force components, and a fixed support gives two force components plus a couple. The FBD is not optional scratch work; an honest FBD with consistent positive directions is what makes the sign of your answer trustworthy.

\text{roller}:1\text{ reaction}\quad \text{pin}:2\text{ reactions}\quad \text{fixed}:2\text{ forces}+1\text{ couple}

Equilibrium equations

For a 2-D rigid body, equilibrium gives three independent scalar equations — two force sums and one moment sum — so you can solve for at most three unknown reactions on a determinate body. The moment equation may be taken about any point; choosing a point where two unknowns intersect drops them out and isolates the third. In 3-D there are six equations: three force sums and three moment sums.

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

Trusses and the method of joints

An ideal truss carries load only through two-force members — straight, pin-connected, and loaded only at the joints — so every member is in pure tension or compression along its axis. The method of joints applies

\sum F_x=0

and

\sum F_y=0

at one joint at a time; with only two equations per joint, start at a joint with two or fewer unknown members (often a support) and march through the truss. Assume every unknown member is in tension; a negative result simply means compression.

\text{at each joint:}\quad \sum F_H=0,\qquad \sum F_V=0

The method of sections

When you need the force in just one or two interior members, the method of sections is far faster than chaining through joints. Cut the truss with an imaginary line that passes through no more than three members of unknown force, draw the FBD of either side, and apply the three equilibrium equations to the whole section at once. The trick is the moment equation: take moments about the point where two of the cut members intersect, and the third member's force falls out directly.

Zero-force members

Spotting zero-force members before you compute saves real time. Two rules cover almost every case: (1) if exactly two non-collinear members meet at a joint that carries no external load or reaction, both members are zero-force; and (2) if three members meet at an unloaded joint and two of them are collinear, the odd (non-collinear) member is zero-force. These members are not useless — they brace against buckling and handle alternate load cases — but for the given loading they carry nothing, and the exam loves to ask you to identify them by inspection.

\text{2 members, no load}\Rightarrow \text{both }=0;\quad \text{3 members, 2 collinear, no load}\Rightarrow \text{3rd}=0

Frames and machines

Frames and machines contain at least one multi-force member — a member loaded at more than two points, so its internal force is not a simple axial pull — which means the truss shortcut does not apply. The strategy is to dismember: draw an FBD of the whole structure to find the external reactions, then separate it into individual members and apply equilibrium to each, invoking Newton's third law so that the force one member exerts on a pin is equal and opposite to the force the pin exerts back. Machines differ from frames only in that they transmit and modify forces (think pliers or a bolt cutter), but the analysis is identical.

\mathbf{F}_{AB}=-\mathbf{F}_{BA}\quad(\text{Newton's third law at every connection})

Exam strategy

Always start with an FBD and a sign convention — tension-positive for members, CCW-positive for moments. For reactions, take moments about a support pin to kill two unknowns at once and solve a single equation for one reaction. For trusses, scan for zero-force members first, then pick the tool: method of joints when you need many member forces or the whole truss, method of sections when you need one or two interior members fast (cut through at most three unknowns, take moments at the intersection of two of them). For frames, get external reactions first, then dismember. If a member crosses your section as a slope, resolve it into horizontal and vertical components using the member's own geometry (run, rise, length).

Key equations

2-D equilibrium

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

Three independent equations for a planar rigid body; solves up to three unknown reactions.

3-D equilibrium

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

Six equations for a spatial body; needed for 3-D supports and space trusses.

Method of joints

\sum F_H=0,\qquad \sum F_V=0

Applied at each pin; start where two or fewer members are unknown. Tension assumed positive.

Member force components

F_x=\frac{x}{L}F,\qquad F_y=\frac{y}{L}F

Resolve an axial member force along its geometry (run $x$ , rise $y$ , length $L=\sqrt{x^2+y^2}$ ).

Determinacy of a plane truss

m+r=2j

$m$ = members, $r$ = reactions, $j$ = joints. Equality → statically determinate (necessary condition).

Zero-force member rules

\text{2 non-collinear, unloaded}\Rightarrow 0;\ \ \text{3 with 2 collinear, unloaded}\Rightarrow \text{odd}=0

Identify members carrying no force by inspection before computing.

Newton's third law at a connection

\mathbf{F}_{AB}=-\mathbf{F}_{BA}

For frames/machines: force of member A on a pin equals minus the force of the pin on A.

Two-force member

\mathbf{F}_1=-\mathbf{F}_2,\ \text{collinear}

A member loaded only at two points carries a force directed along the line joining them (truss member).

Worked examples

Reactions of a loaded simple beam

Problem. A simply supported beam spans $8\,\text{m}$ with a pin at $A$ (left) and a roller at $B$ (right). It carries a uniform load of $5\,\text{kN/m}$ over the full span plus a $20\,\text{kN}$ downward point load at $6\,\text{m}$ from $A$ . Find the support reactions.

Solution. Replace the uniform load by its resultant:

W=5\times8=40\,\text{kN}

acting at midspan,

x=4\,\text{m}

. Moments about

A

(CCW positive), solving for

R_B

:

\sum M_A=R_B(8)-40(4)-20(6)=0\Rightarrow R_B=(160+120)/8=35\,\text{kN}

. Vertical equilibrium:

R_A=40+20-R_B=60-35=25\,\text{kN}

. Horizontal:

A_x=0

(no horizontal loads).

R_A=25\,\text{kN}\uparrow

,

R_B=35\,\text{kN}\uparrow

. Sanity check: total up

=25+35=60\,\text{kN}

equals total down

=40+20=60\,\text{kN}

; the heavier reaction is at

B

, nearer the point load — consistent.

R_B=\frac{40(4)+20(6)}{8}=35\ \text{kN}

Truss member forces by the method of joints

Problem. In the figure's truss, panels are $4\,\text{m}$ wide and the height is $3\,\text{m}$ ( $A$ , $B$ , $C$ , $D$ along the bottom at $0,4,8,12\,\text{m}$ ; $E$ and $F$ on top above $B$ and $C$ ). A single $9\,\text{kN}$ downward load acts at $B$ ; $A$ is a pin, $D$ a roller. Find the forces in members $AE$ and $AB$ .

Solution. Reactions:

\sum M_A=R_D(12)-9(4)=0\Rightarrow R_D=3\,\text{kN}

;

R_A=9-3=6\,\text{kN}\uparrow

. Isolate joint

A

. Member

AE

runs from

(0,0)

to

(4,3)

: length

5

, so its unit direction is

(0.8,0.6)

. Member

AB

is horizontal. Assume both in tension.

\sum F_y=0:\ 6+0.6\,F_{AE}=0\Rightarrow F_{AE}=-10\,\text{kN}

(negative → compression).

\sum F_x=0:\ F_{AB}+0.8\,F_{AE}=0\Rightarrow F_{AB}=-0.8(-10)=+8\,\text{kN}

(positive → tension).

F_{AE}=10\,\text{kN (C)}

,

F_{AB}=8\,\text{kN (T)}

. Sanity check: the inclined member must push down-and-in to balance the upward reaction, hence compression; the bottom chord pulls back, hence tension.

F_{AE}=-\frac{R_A}{0.6}=-10\ \text{kN}\ (\text{C})

Interior members by the method of sections

Problem. For the same truss, find the forces in the top chord $EF$ , the bottom chord $BC$ , and the diagonal $EC$ using one section cut a–a through that panel.

Solution. Reactions as before:

R_A=6\,\text{kN}\uparrow

. Cut through

EF

,

EC

, and

BC

; analyze the left portion (joints

A

,

B

,

E

), which carries

R_A=6\,\text{kN}\uparrow

and the

9\,\text{kN}

load down at

B

. Diagonal

EC

from

(4,3)

to

(8,0)

has unit direction

(0.8,-0.6)

. Vertical equilibrium of the left part:

\sum F_y=6-9-0.6\,F_{EC}=0\Rightarrow F_{EC}=-5\,\text{kN}

(compression). Bottom chord: take moments about

E\,(4,3)

, where

EF

and

EC

intersect.

\sum M_E=-R_A(4)+F_{BC}(3)=0\Rightarrow F_{BC}=6(4)/3=+8\,\text{kN}

(tension). Top chord: take moments about

C\,(8,0)

, where

BC

and

EC

intersect.

\sum M_C=-R_A(8)+9(4)-F_{EF}(3)=0\Rightarrow F_{EF}=(-48+36)/3=-4\,\text{kN}

(compression).

F_{EF}=4\,\text{kN (C)}

,

F_{BC}=8\,\text{kN (T)}

,

F_{EC}=5\,\text{kN (C)}

. Sanity check:

\sum F_x

of the left part

=F_{EF}+F_{BC}+0.8F_{EC}=-4+8+0.8(-5)=0

— equilibrium confirmed.

F_{EC}=\frac{R_A-9}{0.6}=\frac{6-9}{0.6}=-5\ \text{kN}\ (\text{C})

Common pitfalls

•Modeling a pin as a roller (or vice versa). A roller gives one reaction normal to its surface; a pin gives two components. Miscounting reactions makes a determinate problem look unsolvable — or hides indeterminacy.
•Sign confusion on member forces. Assume tension positive everywhere; a negative answer means compression. Re-labeling mid-solution to 'fix' a sign corrupts later joints.
•Cutting more than three unknown members in the method of sections. With only three equations you cannot solve four unknown forces; choose a cut that exposes at most three.
•Applying truss equations (axial-only members) to a frame. Frames have multi-force members; you must dismember and use Newton's third law, not assume forces act along the members.
•Forgetting zero-force members and grinding through joints needlessly — or, worse, assuming a member is zero-force when an external load or reaction sits at that joint, which voids the rule.
•Resolving a diagonal member with the wrong triangle. Use the member's own run, rise, and length; mixing up which leg is horizontal flips the component ratios (0.8 vs 0.6 here).
•Taking the moment sum about a poorly chosen point. Summing moments about a point where an unknown's line of action passes eliminates it — choosing any other point leaves two unknowns in one equation.

References

NCEES FE Reference Handbook — Statics (Equilibrium, Method of Joints, Method of Sections, Zero-Force Members)
Hibbeler, Engineering Mechanics: Statics — FBDs, trusses, frames and machines
Beer & Johnston, Vector Mechanics for Engineers: Statics

E. Centroid of area

Centroids and Area Moments of Inertia

Locate centroids of composite areas, compute area moments of inertia with the parallel-axis theorem, and find radius of gyration and polar moment of inertia.

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G. Static friction

Static Friction

Apply Coulomb dry friction with μs, find impending motion on inclines, wedges, and belts, and distinguish tipping from sliding.

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