Statics is worth 8 to 12 questions on the FE Civil exam, and it is the knowledge area most likely to reward pure speed. The algebra is short, the geometry is familiar, and almost every problem reduces to summing forces and moments. The problems below are the kind you should be finishing in under two minutes each.
Each problem here is followed by a full solution that shows the free-body diagram, the equilibrium equations actually written, and the sign convention used. Where the NCEES FE Reference Handbook supplies a formula or a table value, the solution names it, because on exam day you will be pulling that value from a search box rather than memory.
Exam weight: NCEES lists Statics at 8-12 questions (7-11%) of the 110-question FE Civil exam. Work each problem below on paper first, then reveal the worked solution — reading a solution you have not attempted builds recognition, not recall.
What NCEES Tests in Statics
The NCEES specification puts 8 to 12 statics questions on the 110-question FE Civil exam, roughly 7 to 11 percent. Expect resultants of two- and three-dimensional force systems, concurrent and non-concurrent systems, equilibrium of a rigid body with pins, rollers and fixed supports, and equivalent force-couple systems replacing a distributed load with a single resultant at the centroid of the load diagram.
The rest of the area splits between trusses and frames, centroids and moments of inertia, and friction. Truss questions ask for one member force, so method of sections usually beats method of joints; zero-force members appear often enough to be worth recognizing on sight. Friction items test whether the block is on the verge of sliding or tipping, not just F equals mu times N.
5 Free Statics Practice Problems
Each problem below comes from the PECivilClick FE Civil question bank and matches the style, difficulty and format of the real exam. Attempt each one under a three-minute limit — roughly the pace the exam demands.
Problem 1
Which type of load is NOT resisted by an ideal pin support in 2D?
A) Axial force
B) Shear force
C) Moment (couple)
D) Combination of axial and shear
Answer: C) Moment (couple)
Step 1 – Identify given information: An ideal pin support in 2D is described. We must determine which load type it cannot resist.
Step 2 – Support reaction rules (FE Handbook): An ideal pin (hinge) provides two reaction force components (\(A_x\) and \(A_y\)) but zero moment resistance. A roller provides one force (perpendicular to the surface). A fixed support provides two forces and one moment.
Step 3 – Evaluate each option:
| Option | Load Type | Resisted by Pin? |
|---|---|---|
| A | Axial force | Yes (\(A_x\)) |
| B | Shear force | Yes (\(A_y\)) |
| C | Moment (couple) | No |
| D | Axial + Shear | Yes (both \(A_x\) and \(A_y\)) |
Step 4 – Answer: A pin support cannot resist a couple moment because it allows free rotation at the joint. Only a fixed (cantilever) support can resist moments.
Correct answer: (C)
Problem 2
A \( 400\,\mathrm{N} \) force lies between two nonperpendicular lines P and Q that form a \( 50^{\circ} \) angle. The force makes a \( 20^{\circ} \) angle with line P toward Q. Resolve the force into components along P and Q. What are \( (F_P,\, F_Q) \)?
A) (261 N, 179 N)
B) (279 N, 143 N)
C) (179 N, 261 N)
D) (307 N, 105 N)
Answer: A) (261 N, 179 N)
Step 1 – Identify given information: A \(400\,\mathrm{N}\) force lies between lines P and Q that form a \(50^{\circ}\) angle. The force makes a \(20^{\circ}\) angle with line P (so \(30^{\circ}\) with line Q).
Step 2 – Sine rule for non-perpendicular components: From the FE Handbook, when resolving a force along two non-orthogonal directions, use the law of sines:
\(\frac{F}{\sin(\angle PQ)} = \frac{F_P}{\sin(\angle FQ)} = \frac{F_Q}{\sin(\angle FP)}\)
Step 3 – Solve for each component:
\(F_P = F \cdot \frac{\sin(\angle FQ)}{\sin(\angle PQ)} = 400 \cdot \frac{\sin 30^{\circ}}{\sin 50^{\circ}} = 400 \cdot \frac{0.5}{0.7660} \approx 261\,\mathrm{N}\)
\(F_Q = F \cdot \frac{\sin(\angle FP)}{\sin(\angle PQ)} = 400 \cdot \frac{\sin 20^{\circ}}{\sin 50^{\circ}} = 400 \cdot \frac{0.3420}{0.7660} \approx 179\,\mathrm{N}\)
Step 4 – Answer: \((F_P, F_Q) = (261\,\mathrm{N},\, 179\,\mathrm{N})\). Option (B) reverses the angle assignments; (C) swaps the two components; (D) uses incorrect angles. The key is that the component along P uses the angle to Q, not P.
Correct answer: (A)
Problem 3
A \( 200\,\mathrm{mm} \times 100\,\mathrm{mm} \) rectangle has a circular hole of radius \( 25\,\mathrm{mm} \) centered at \( (150\,\mathrm{mm},\, 50\,\mathrm{mm}) \). Origin is at the rectangle's lower-left corner. What is the x-coordinate of the centroid of the remaining area?
A) 84.6 mm
B) 94.6 mm
C) 100.0 mm
D) 105.4 mm
Answer: B) 94.6 mm
Step 1 – Identify given information:
| Shape | Dimensions | Centroid \(x\) | Area |
|---|---|---|---|
| Rectangle | \(200 \times 100\,\mathrm{mm}\) | \(100\,\mathrm{mm}\) | \(20{,}000\,\mathrm{mm^2}\) |
| Circular hole | \(r = 25\,\mathrm{mm}\) | \(150\,\mathrm{mm}\) | \(\pi(25)^2 = 1{,}963.5\,\mathrm{mm^2}\) |
Step 2 – Composite centroid formula (FE Handbook): For a shape with a hole removed:
\(\bar{x} = \frac{A_{\text{rect}} \bar{x}_{\text{rect}} - A_{\text{hole}} \bar{x}_{\text{hole}}}{A_{\text{rect}} - A_{\text{hole}}}\)
Step 3 – Substitute and solve:
\(\bar{x} = \frac{20{,}000(100) - 1{,}963.5(150)}{20{,}000 - 1{,}963.5} = \frac{2{,}000{,}000 - 294{,}525}{18{,}036.5} = \frac{1{,}705{,}475}{18{,}036.5} \approx 94.6\,\mathrm{mm}\)
Step 4 – Answer: \(\bar{x} \approx 94.6\,\mathrm{mm}\). The centroid shifts left from the rectangle center (\(100\,\mathrm{mm}\)) because material is removed from the right side. Option (A) over-shifts; (C) ignores the hole; (D) shifts right instead of left.
Correct answer: (B)
Problem 4
A cantilever beam is fixed at the right end O. Downward loads of \( 50\,\mathrm{N} \) at \( 0.5\,\mathrm{m} \) from O and \( 80\,\mathrm{N} \) at \( 1.5\,\mathrm{m} \) from O act on the beam. A counterclockwise couple of \( 60\,\mathrm{N{\cdot}m} \) is applied at O. What upward force \( F \) at the free end (\( 2.0\,\mathrm{m} \) from O) is required for equilibrium?
A) 35 N
B) 42.5 N
C) 55 N
D) 85 N
Answer: B) 42.5 N
Step 1 – Identify given information:
| Load | Value | Distance from O |
|---|---|---|
| Downward force 1 | \(50\,\mathrm{N}\) | \(0.5\,\mathrm{m}\) |
| Downward force 2 | \(80\,\mathrm{N}\) | \(1.5\,\mathrm{m}\) |
| CCW couple at O | \(60\,\mathrm{N{\cdot}m}\) | at O |
| Upward force \(F\) | Unknown | \(2.0\,\mathrm{m}\) |
Step 2 – Moment equilibrium about O (FE Handbook): For static equilibrium, \(\sum M_O = 0\) (taking CCW as positive).
Step 3 – Substitute and solve:
\(\sum M_O = 0: \quad 60 + F(2.0) - 50(0.5) - 80(1.5) = 0\)
\(60 + 2F - 25 - 120 = 0\)
\(2F = 85 \quad \Rightarrow \quad F = 42.5\,\mathrm{N}\)
Step 4 – Answer: \(F = 42.5\,\mathrm{N}\) upward. Option (A) \(35\,\mathrm{N}\) neglects the couple; (C) \(55\,\mathrm{N}\) uses incorrect moment arms; (D) \(85\,\mathrm{N}\) forgets to divide by 2. Always check sign convention consistency.
Correct answer: (B)
Problem 5
Two equal and opposite \( 120\,\mathrm{N} \) forces separated by \( 0.25\,\mathrm{m} \) form a couple. What is the couple moment magnitude?
A) 15 N·m
B) 30 N·m
C) 48 N·m
D) 60 N·m
Answer: B) 30 N·m
Step 1 – Identify given information: Two equal and opposite forces: \(F = 120\,\mathrm{N}\), separated by \(d = 0.25\,\mathrm{m}\).
Step 2 – Couple moment formula (FE Handbook): A couple consists of two equal, opposite, non-collinear forces. The moment of a couple is:
\(M = F \cdot d\)
where \(d\) is the perpendicular distance between the force lines of action.
Step 3 – Substitute and solve:
\(M = 120 \times 0.25 = 30\,\mathrm{N{\cdot}m}\)
Step 4 – Answer: The couple moment is \(30\,\mathrm{N{\cdot}m}\). Option (A) \(15\) incorrectly halves the result; (C) \(48\) multiplies \(120 \times 0.4\); (D) \(60\) uses \(d = 0.5\,\mathrm{m}\). A couple moment is the same about any point, which distinguishes it from a single-force moment.
Correct answer: (B)
Using the FE Reference Handbook for Statics
Do not derive centroids or moments of inertia. Search the handbook for the shape name and pull Ixc, Iyc and the centroid location straight from the table, then apply the parallel axis theorem yourself. Bookmark that table plus the friction and belt-friction relations before your first practice exam so the search is a reflex, not a hunt.
Four Mistakes That Cost Points
- Skipping the free-body diagram On a two-minute question the diagram feels like wasted time, but most wrong answers here come from a forgotten reaction component or a force drawn in the wrong direction. Sketch the body, mark every support reaction, then write equations.
- Choosing the wrong moment point Summing moments about an arbitrary origin leaves two unknowns in one equation. Take moments about a pin or about the intersection of two unknown lines of action and the target reaction usually falls out in a single line of algebra.
- Placing a distributed load resultant incorrectly A uniform load resultant sits at midspan; a triangular load resultant sits at one third of the base from the tall end. Reversing that third catches people every time, and the wrong answer choices are built to match the mistake.
- Forgetting the parallel axis term Table values for I are about the shape's own centroid. For a built-up section you still need A times d squared for each piece, measured from the composite centroid, not from the edge of the section. Locate the composite centroid first.
Frequently Asked Questions
How many statics questions are on the FE Civil exam?
The NCEES specification lists 8 to 12 statics questions out of 110, about 7 to 11 percent. That makes it one of the two largest mechanics areas alongside mechanics of materials. Because the problems are short and formulaic, statics is usually the best return on study hours of any area on the exam.
Should I use method of joints or method of sections on the exam?
Sections, almost always. FE truss questions ask for the force in one named member, and a single cut through three members with one moment equation gets you there. Method of joints only wins when the member you want touches a support joint, or when you are checking for zero-force members.
Do I need to memorize moment of inertia formulas for statics?
No. The on-screen NCEES FE Reference Handbook contains centroid and area moment of inertia tables for rectangles, triangles, circles and semicircles, and you can search it during the exam. What you do need from memory is the parallel axis theorem and the habit of finding the composite centroid before shifting any axes.
How hard are FE Civil statics questions compared to a university statics course?
The mechanics is easier but the clock is tighter. Exam items are single-concept: one truss member, one reaction, one friction check, no multi-page frame analyses. The difficulty is recognizing the setup in ten seconds and executing cleanly, which is why timed practice matters more than reworking long textbook problems.
Keep Going
Work through the other FE Civil knowledge areas with truss analysis practice problems, more free practice problems, review the full FE Civil exam topic breakdown, or plan your preparation with our FE exam study guide and study timeline.