FE Civil · Chapter 5 · 7–11 exam questions

FE Civil Statics

This chapter tests your ability to analyze forces, moments, and equilibrium in rigid body systems.

What the FE tests in Statics

Forces & Equilibrium

Every structural analysis starts with resolving forces into components and writing equilibrium equations. Whether you are finding the reactions of a simply supported bridge girder, combining dead load, live load, and wind on a building connection, or checking the overturning stability of a retaining wall, the process is the same: draw the free body diagram, resolve forces, and apply ΣF = 0 and ΣM = 0.

Trusses & Friction

Civil engineers analyze trusses when designing roof systems, pedestrian bridges, and transmission towers. Method of joints gives you all member forces; method of sections lets you cut straight to the one member you need to size. Friction governs the sliding stability of retaining walls, the traction of vehicles on road surfaces, and belt-driven equipment loads.

Section Properties

Centroids locate the neutral axis of composite beam cross-sections and the point of application of distributed loads. Moments of inertia control how much a beam bends and where it fails in flexure. You compute these for composite sections using the parallel axis theorem — built-up steel columns, reinforced concrete T-beams, and timber box beams all require combining individual shape properties about a common axis.

Key Statics formulas

$\sum F_x = 0,\quad \sum F_y = 0,\quad \sum M = 0$
Equilibrium EquationsFE Handbook p. 94
$\bar{x} = \frac{\sum A_i x_i}{\sum A_i}$
Composite CentroidFE Handbook p. 95
$I_x = \bar{I}_x + Ad^2$
Parallel Axis TheoremFE Handbook p. 95
$F \leq \mu_s N$
Static FrictionFE Handbook p. 96
$M = r \times F$
Moment of a ForceFE Handbook p. 94
$F_{out} = F_{in}\,\frac{a_{in}}{a_{out}}$
Lever / Machine (mechanical advantage)

Sample Statics problems

Q1. A guy wire anchoring a utility pole exerts a 2,600 N force directed along a line that runs 5 m horizontally and 12 m vertically. What is the horizontal component of the force?

Answer: 1,000 N

Explain it simply

When a force direction is given by geometry (rise and run), resolve it with that geometry: the hypotenuse is $R=\sqrt{5^2+12^2}=13$ m, so $F_x=(5/13)\times 2{,}600 = 1{,}000$ N. This is a 5-12-13 right triangle — a pattern the FE loves to reuse.

Q2. Two forces act at a gusset plate: $F_1 = 300$ N horizontal and $F_2 = 400$ N vertical. What is the magnitude of the resultant?

Answer: 500 N

Explain it simply

A 3-4-5 triangle in disguise. $R=\sqrt{300^2+400^2}=\sqrt{250{,}000}=500$ N, at $\theta=\arctan(400/300)=53.1^\circ$ above horizontal.

These are 2 of 1,126 problems across all 15 chapters. The full bank, lessons, mastery tracking, and timed exam simulation live inside the app.

Common Statics mistakes on the FE

Incorrect free body diagram — forgetting a reaction force or including an internal force on the wrong side of the cut.
Wrong sign convention for moments — pick CW or CCW positive and stick with it throughout.
In trusses, confusing method of joints (all forces at a joint) with method of sections (cutting through members).
Forgetting the Ad² term in the parallel axis theorem when computing composite moments of inertia.
Assuming friction force equals μN before verifying impending motion — it could be less than the maximum.
A two-force member carries force along the line joining its two pins; a multi-force member (in frames/machines) does not.
Analyze frames and machines by dismembering them — at a shared pin, the forces on the two members are equal and opposite.

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