Statics

9% of exam

Resultants and concurrent force systems, equilibrium of rigid bodies, frames and trusses, centroids and moments of inertia, and static friction.

4 concepts

Exam areas

C. Equilibrium of rigid bodies

Free-Body Diagrams, Resultants and Static Equilibrium

Resolve and combine forces, take moments and couples, draw an honest free-body diagram, and solve 2-D and 3-D equilibrium with ΣF=0 and ΣM=0.

Statics is the quiet foundation under every mechanical problem you will solve on the FE: before a beam can bend, a shaft can twist, or a bolt can fail, the loads on it must first be sorted out by equilibrium. A body at rest has zero net force and zero net moment — that is the whole subject, stated in one breath. What separates a fast, reliable solver from someone who churns is not cleverness with the equations but discipline with the free-body diagram (FBD): isolate exactly one body, show every force acting on it once and only once, pick a positive sign convention, and the equations almost write themselves. The FE Reference Handbook's Statics chapter lays out the force, resultant, moment, and equilibrium relations in bare form; this concept turns those statements into a procedure you can run under time pressure without second-guessing a sign.

Forces, components, and resultants

A force is a vector — it needs a magnitude, a direction, and a point of application before it means anything. The first move in almost every problem is to resolve each force into rectangular components, because components add as plain scalars while the original vectors do not. When the geometry is given as a run, a rise, and a length rather than an angle, project along the geometry directly:

F_x=(x/L)F

. The resultant of a force system is the single force whose components equal the sums

\sum F_x

and

\sum F_y

; its magnitude is the Pythagorean combination and its direction is the arctangent of the component ratio. Always keep the quadrant in mind —

\arctan

on a calculator collapses two quadrants into one.

R=\sqrt{\left(\sum F_x\right)^2+\left(\sum F_y\right)^2},\qquad \theta=\arctan\!\frac{\sum F_y}{\sum F_x}

Concurrent force systems

When the lines of action of every force pass through one common point, the system is concurrent and there is no tendency to rotate — the moment equation is satisfied automatically, so equilibrium collapses to the two (or three, in 3-D) force-sum equations. This is the regime of a single loaded pin, a knot where cables meet, or a particle. Because there is no moment bookkeeping, concurrent-force problems reduce to

\sum F_x=0

and

\sum F_y=0

, which you can solve for two unknowns — a cable tension and an angle, two strut forces, and so on.

\text{concurrent at a point}:\quad \sum F_x=0,\qquad \sum F_y=0

Moments and couples

A moment measures the turning effect of a force about a point and is the cross product

\mathbf{M}=\mathbf{r}\times\mathbf{F}

, where

\mathbf{r}

runs from the moment center to any point on the force's line of action. In 2-D this reduces to magnitude (force) times the perpendicular distance (the moment arm), with a sign for sense — adopt counter-clockwise positive and stay with it. A couple is a special pair: two forces equal in magnitude, opposite in direction, and parallel. Their force sum is zero but their moment is not — a couple is a pure moment

M=Fd

that is the same about every point, which is exactly why a fixed support can transmit a reaction couple.

\mathbf{M}=\mathbf{r}\times\mathbf{F},\qquad |\mathbf{M}_{2D}|=F\,d_\perp,\qquad M_{\text{couple}}=Fd

Building the free-body diagram

The FBD is where points are won or lost. Mentally cut the body free of everything touching it and, at each cut, draw the force(s) that the removed support could transmit. A roller (or a smooth surface, or a cable) gives a single reaction along one known line; a pin (or hinge) gives two unknown perpendicular components; a fixed support gives two force components plus a reaction couple. Add the applied loads and the weight at the center of gravity. Replace a distributed load by its resultant — for a uniform load

w

over length

L

the resultant is

W=wL

acting at the centroid of the load diagram (midspan for a rectangle). The figure shows a simply supported beam reduced to a pin, a roller, and the equivalent point load.

Equilibrium in two and three dimensions

For a planar rigid body there are exactly three independent scalar equations — two force sums and one moment sum — so a statically determinate 2-D body has at most three unknown reactions. The moment equation may be taken about any point, and choosing the intersection of two unknowns deletes them, isolating the third in a single equation. In three dimensions the count doubles to six: three force sums and three moment sums, needed for ball-and-socket joints, thrust bearings, and space frames. Count unknowns against available equations first — if reactions exceed equations the body is statically indeterminate and equilibrium alone will not finish it.

\text{2-D}:\ \sum F_x=\sum F_y=\sum M_O=0\qquad \text{3-D}:\ \sum F_{x,y,z}=0,\ \sum M_{x,y,z}=0

Exam strategy

Start every problem with an FBD and one declared sign convention (CCW-positive moments, up/right-positive forces). To find reactions, take moments about a support pin so its two unknown components drop out, leaving one equation in one unknown; then use

\sum F_y=0

and

\sum F_x=0

for the rest. Collapse distributed loads to a resultant before writing equations. Resolve sloped forces along the member's own run-rise-length so you never trust a memorized angle. Finish with the cheapest possible check: total up-forces should equal total down-forces, and the larger reaction should sit nearer the heavier load. If the unknown count exceeds three (2-D) or six (3-D), stop — the body is indeterminate and the FE answer will hinge on recognizing that, not on grinding.

Key equations

Force components

F_x=F\cos\theta_x,\quad F_y=F\cos\theta_y,\qquad F_x=\tfrac{x}{L}F,\ \ F_y=\tfrac{y}{L}F

Resolve a force by angle or by geometry (run $x$ , rise $y$ , length $L=\sqrt{x^2+y^2}$ ). Components in N or lbf.

Resultant of a force system

R=\sqrt{\left(\textstyle\sum F_x\right)^2+\left(\textstyle\sum F_y\right)^2},\quad \theta=\arctan\frac{\sum F_y}{\sum F_x}

Single equivalent force. Watch the quadrant when taking the arctangent.

Moment of a force

\mathbf{M}=\mathbf{r}\times\mathbf{F},\qquad M_z=xF_y-yF_x

$\mathbf{r}$ from the moment center to the line of action. In 2-D, magnitude $=F\,d_\perp$ (N·m or ft·lbf).

Couple (pure moment)

M=Fd

Two equal, opposite, parallel forces a distance $d$ apart; same moment about every point, zero net force.

2-D equilibrium

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

Three 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 (ball-and-socket, bearings, space frames).

Resultant of a uniform distributed load

W=wL\ \text{at midspan};\quad W=\tfrac{1}{2}w_0 L\ \text{(triangular, at }L/3)

Replace a line load by its area resultant acting at the load-diagram centroid. $w$ in N/m or lbf/ft.

Support reactions transmitted

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

Count of unknowns each idealized support adds to the FBD.

Worked examples

Reactions of a loaded simply supported beam

Problem. A beam spans $8\,\text{m}$ with a pin at $A$ (left) and a roller at $B$ (right). It carries a uniform load of $6\,\text{kN/m}$ over the full span plus a $24\,\text{kN}$ downward point load $6\,\text{m}$ from $A$ . Find the support reactions.

Solution. Replace the uniform load by its resultant:

W=wL=6\times8=48\,\text{kN}

acting at midspan,

x=4\,\text{m}

. Moments about

A

(CCW positive), solving for

R_B

\sum M_A=R_B(8)-48(4)-24(6)=0

R_B=\dfrac{192+144}{8}=42\,\text{kN}

. Vertical equilibrium:

R_A=48+24-R_B=72-42=30\,\text{kN}

. Horizontal:

A_x=0

(no horizontal loads).

R_A=30.0\,\text{kN}\uparrow

R_B=42.0\,\text{kN}\uparrow

. Sanity check: total up

=30+42=72\,\text{kN}

equals total down

=48+24=72\,\text{kN}

; the larger reaction is at

B

, nearer the point load — consistent.

R_B=\frac{48(4)+24(6)}{8}=42.0\ \text{kN}

Resultant of concurrent forces

Problem. Three forces act through one point: $200\,\text{N}$ at $30^\circ$ , $150\,\text{N}$ at $120^\circ$ , and $100\,\text{N}$ at $250^\circ$ (angles measured CCW from the $+x$ axis). Find the magnitude and direction of the resultant.

Solution. Resolve each into components and sum.

\sum F_x=200\cos30^\circ+150\cos120^\circ+100\cos250^\circ=173.2-75.0-34.2=64.0\,\text{N}

\sum F_y=200\sin30^\circ+150\sin120^\circ+100\sin250^\circ=100.0+129.9-94.0=135.9\,\text{N}

R=\sqrt{64.0^2+135.9^2}=\sqrt{4096+18469}=150\,\text{N}

\theta=\arctan(135.9/64.0)=64.8^\circ

(both components positive → first quadrant).

R=150\,\text{N}

64.8^\circ

. Sanity check: both sums are positive and

F_y>F_x

, so the resultant should point up and to the right, steeper than

45^\circ

— it does.

R=\sqrt{(64.0)^2+(135.9)^2}=150\ \text{N}\ \angle\,64.8^\circ

Moment of a 3-D force about a point

Problem. A force $\mathbf{F}=(40\,\mathbf{i}-30\,\mathbf{j}+60\,\mathbf{k})\,\text{N}$ acts at a point located by $\mathbf{r}=(0.5\,\mathbf{i}+0.2\,\mathbf{k})\,\text{m}$ from the origin $O$ . Find the moment about $O$ and its magnitude.

Solution. Use

\mathbf{M}=\mathbf{r}\times\mathbf{F}

with

\mathbf{r}=(0.5,0,0.2)

and

\mathbf{F}=(40,-30,60)

M_x=r_yF_z-r_zF_y=(0)(60)-(0.2)(-30)=+6.0\,\text{N·m}

M_y=r_zF_x-r_xF_z=(0.2)(40)-(0.5)(60)=8-30=-22.0\,\text{N·m}

M_z=r_xF_y-r_yF_x=(0.5)(-30)-(0)(40)=-15.0\,\text{N·m}

|\mathbf{M}|=\sqrt{6.0^2+22.0^2+15.0^2}=\sqrt{36+484+225}=27.3\,\text{N·m}

\mathbf{M}=(6.0\,\mathbf{i}-22.0\,\mathbf{j}-15.0\,\mathbf{k})\,\text{N·m}

|\mathbf{M}|=27.3\,\text{N·m}

. Sanity check:

\mathbf{M}\cdot\mathbf{F}=6(40)-22(-30)-15(60)=240+660-900=0

, confirming

\mathbf{M}\perp\mathbf{F}

as a cross product must be.

\mathbf{M}=\mathbf{r}\times\mathbf{F}=(6.0\,\mathbf{i}-22.0\,\mathbf{j}-15.0\,\mathbf{k})\ \text{N·m}

Common pitfalls

•Modeling a pin as a roller (or vice versa). A roller/cable/smooth surface gives ONE reaction along a known line; a pin gives TWO components. Miscounting makes a determinate beam look unsolvable — or hides indeterminacy.
•Forgetting the moment arm is the PERPENDICULAR distance. $M=Fd$ needs $d$ measured perpendicular to the line of action, not the slant distance to the force's point of application.
•Using $\arctan$ blindly for resultant direction. A calculator returns one quadrant; check the signs of $\sum F_x$ and $\sum F_y$ to place the angle in the correct quadrant.
•Leaving a distributed load as $w$ in the equilibrium equations. Replace it by its resultant $W=wL$ at the load-diagram centroid before summing forces and moments.
•Treating a couple as a force. A couple has zero net force but a nonzero, position-independent moment — it shows up only in $\sum M=0$ , never in $\sum F=0$ .
•Mixing sign conventions mid-problem. Declare CCW-positive (or CW) and up/right-positive once, and a negative answer then honestly means the reaction points opposite to your assumed arrow.
•Trying to solve a statically indeterminate body with equilibrium alone. When reactions exceed 3 (2-D) or 6 (3-D), you need compatibility/material relations — recognize it rather than forcing a wrong split.

References

NCEES FE Reference Handbook — Statics
Hibbeler, Engineering Mechanics: Statics — force systems, moments, equilibrium of rigid bodies

D. Frames and trusses

Trusses and Dry/Belt Friction

Solve trusses by joints and sections, spot zero-force members by inspection, and apply Coulomb (dry), wedge, and belt friction at impending motion.

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Frames and Machines: Multi-Force Members

Identify two-force versus multi-force members, find external reactions, then dismember frames and machines to solve pin forces with Newton's third law.

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E. Centroids and moments of inertia

Centroids and Moments of Inertia

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

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