FE Civil Statics practice problems
6 free FE Civil Statics practice problems with answers and full worked solutions. Statics carries 8–12 of the 110 questions on the FE Civil exam (NCEES specification effective July 2020). Pick an answer to check it, then read the worked solution and why each wrong choice is wrong.
6 worked Statics problems
Question 1 of 6
Two forces of 300 lb and 290 lb act at a common point with 30 degrees between them. The magnitude of their resultant is most nearly:
Forces of 300 lb and 290 lb acting at a common point with 30° between them. Show the answer and the reasoning
Why C
Given: F1 = 300 lb and F2 = 290 lb with 30 degrees between them. Relation: Law of cosines on the force parallelogram: R = sqrt(F1^2 + F2^2 + 2 F1 F2 cos(theta)). Substitute: R = sqrt(300^2 + 290^2 + 2(300)(290)cos(30)) Result: R = 569.9 lb. Check: Resolving each force into x and y components and recombining gives the same value.
Why A is tempting
Treated the forces as perpendicular regardless of the stated angle.
Why B is tempting
Used a sine where the law of cosines calls for a cosine.
Why D is tempting
Added the magnitudes, which holds only when the forces are collinear.
Question 2 of 6
Two parallel forces of 280 lb and 260 lb act in the same direction, 17 ft apart. Their resultant acts at a distance from the 280-lb force of most nearly:
Parallel forces of 280 lb and 260 lb acting in the same direction 17 ft apart; their resultant acts on the line that reproduces their combined moment. Show the answer and the reasoning
Why A
Given: F1 = 280 lb and F2 = 260 lb, parallel, 17 ft apart. Relation: Moments about F1: (F1 + F2) x = F2 d, so x = F2 d/(F1 + F2). Substitute: x = 260(17)/540 Result: x = 8.19 ft from the 280-lb force.
Why B is tempting
Divided by the first force alone rather than by the total.
Why C is tempting
Placed the resultant at the second force.
Why D is tempting
Inverted the expression.
Question 3 of 6
A 150-lb weight hangs from two cables that each make an angle of 75 degrees with the horizontal, arranged symmetrically. The tension in each cable is most nearly:
A 150 lb weight hanging from two cables that each make 75° with the horizontal, arranged symmetrically about the load. Show the answer and the reasoning
Why C
Given: W = 150 lb on two symmetric cables at 75 degrees to the horizontal. Relation: Vertical equilibrium: 2T sin(theta) = W. Substitute: T = 150/(2 sin(75)) Result: T = 77.65 lb in each cable.
Why A is tempting
Multiplied by the sine instead of dividing.
Why B is tempting
Ignored the cable inclination, which is only correct for vertical cables.
Why D is tempting
Used a cosine. The vertical components of the two cables carry the weight, so the sine of the angle from horizontal applies.
Question 4 of 6
Analysis of a frame gives a pin reaction with components of 17 kips horizontally and 5 kips vertically. The magnitude of the resultant reaction is most nearly:
Pin reaction on a frame resolved into a 17 kip horizontal component and a 5 kip vertical component acting at the pin. Show the answer and the reasoning
Why D
Given: Ax = 17 kips, Ay = 5 kips. Relation: A pin reaction combines its components as a vector: R = sqrt(Ax^2 + Ay^2). Substitute: R = sqrt(17^2 + 5^2) Result: R = 17.72 kips.
Why A is tempting
Averaged the two components.
Why B is tempting
Subtracted the components.
Why C is tempting
Reported the larger component alone.
Question 5 of 6
A T-section has an 8 in by 1 in flange sitting on top of a 0.375 in by 12 in web. Measured from the bottom of the web, the centroid lies at most nearly:
T-section with a 8 in by 1 in flange sitting on top of a 0.375 in by 12 in web, with the centroid measured from the bottom of the web. Show the answer and the reasoning
Why B
Given: Flange 8 x 1 in on a 0.375 x 12 in web; datum at the web bottom. Relation: y = sum(A y)/sum(A). Substitute: web 4.50 in^2 at 6.00 in, flange 8.00 in^2 at 12.50 in Result: y = 10.16 in.
Why A is tempting
Used mid-height of the overall section. The flange carries more area than the thin web, so the centroid sits above midpoint.
Why C is tempting
Reported the flange centroid alone.
Why D is tempting
Left the flange area out of the denominator.
Question 6 of 6
A solid circular shaft has a diameter of 12 in. Its polar moment of inertia is most nearly:
Solid circular shaft 12 in in diameter; the polar moment is taken about the axis through the centre, perpendicular to the section. Show the answer and the reasoning
Why B
Given: D = 12 in. Relation: J = pi D^4/32, twice the diametral moment (J = Ix + Iy). Substitute: J = pi(12)^4/32 Result: J = 2036 in^4.
Why A is tempting
Gave the moment about a single diameter. The polar moment is the sum about two perpendicular diameters, so it is twice as large.
Why C is tempting
Used a denominator of 16.
Why D is tempting
Used the solid-cylinder mass moment form.
Going further
Frequently asked questions
How many Statics questions are on the FE Civil exam?
Between 8 and 12 of the 110 questions, according to the NCEES FE Civil specification effective July 2020.
Are these real NCEES exam questions?
No. NCEES does not release its exam questions. These are original FE-style problems written to the NCEES specification and the FE Reference Handbook; every calculated key is checked by two independent solution methods.
Are these Statics practice problems free?
Yes. All 6 problems on this page, with answers and full worked solutions, are free and need no account.
Check where you stand, free
The quick diagnostic is eight FE Civil questions in about ten minutes, with no account. The free mock is a full 110-question, 5 hour 20 minute FE Civil paper with a worked solution for every question.