FE Civil Structural Engineering practice problems
6 free FE Civil Structural Engineering practice problems with answers and full worked solutions. Structural Engineering carries 10–15 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 Structural Engineering problems
Question 1 of 6
A simply supported beam spans 20 ft under a uniformly distributed load of 2.5 kip/ft. The maximum shear is most nearly:
Simply supported beam of span 20 ft carrying a uniform load of 2.5 kip per foot over its full length, with a pin at A and a roller at B. Show the answer and the reasoning
Why B
Given: Simple span L = 20 ft, w = 2.5 kip/ft. Relation: Shear peaks at the supports, each carrying half the load: V = wL/2. Substitute: V = 2.5(20)/2 Result: V = 25.0 kips.
Why A is tempting
Divided by 8, which belongs to the moment expression.
Why C is tempting
Gave the total load rather than the shear at one support.
Why D is tempting
Computed the maximum moment. The units alone rule this out.
Source NCEES FE Reference Handbook 10.6 — Civil Engineering, Structural Design
Question 2 of 6
A cantilever 20 ft long carries a load P = 13 kips at its free end. With E = 29,000 ksi and I = 900 in^4, the tip deflection is most nearly:
Cantilever of length 20 ft built in at A, carrying a 13 kip load at the free end. Show the answer and the reasoning
Why C
Given: Cantilever L = 20 ft = 240 in, tip load P = 13 kips, E = 29,000 ksi, I = 900 in^4. Relation: delta = PL^3/(3EI). Substitute: delta = 13(240)^3/(3 x 29000 x 900) Result: delta = 2.30 in.
Why A is tempting
Used the simply supported central point load formula PL^3/48EI.
Why B is tempting
Used 5PL^3/48EI, the tip deflection for a load at mid-length rather than at the tip.
Why D is tempting
Multiplied the tip slope PL^2/2EI by the length. Integrating the slope brings in a factor of 1/3, not 1/2.
Source NCEES FE Reference Handbook 10.6 — Civil Engineering, Structural Design
Question 3 of 6
A slender steel column has a slenderness ratio of 150. With E = 29,000 ksi, the Euler critical stress is most nearly:
Slender column with a slenderness ratio of 150, failing by elastic buckling rather than by yielding. Show the answer and the reasoning
Why D
Given: KL/r = 150, E = 29,000 ksi. Relation: Fe = pi^2 E/(KL/r)^2. Substitute: Fe = pi^2(29000)/(150)^2 Result: Fe = 12.7 ksi.
Why A is tempting
Dropped pi squared entirely.
Why B is tempting
Used pi rather than pi squared.
Why C is tempting
Introduced a factor of two that the Euler relation does not carry.
Source NCEES FE Reference Handbook 10.6 — Civil Engineering, Structural Design
Question 4 of 6
A continuous planar beam has 6 reaction components and 1 internal hinge. Its degree of static indeterminacy is most nearly:
Continuous planar beam with 6 reaction components and 1 internal hinge, drawn over its supports so the released and restrained points can be counted. Show the answer and the reasoning
Why B
Given: r = 6 reactions, 1 internal hinge. Relation: Degree = r - 3 - c, each hinge adding one condition equation (zero moment). Substitute: 6 - 3 - 1 Result: 2.
Why A is tempting
Reversed the sign of the expression.
Why C is tempting
Used two equilibrium equations. A planar rigid body has three.
Why D is tempting
Omitted the three global equilibrium equations.
Source NCEES FE Reference Handbook 10.6 — Civil Engineering, Structural Design
Question 5 of 6
A beam of span 22 ft is built in at both ends and carries an 8-kip load at midspan. The magnitude of the moment at each fixed end is most nearly:
Beam of span 22 ft built in at both ends with a 8 kip load at midspan. Show the answer and the reasoning
Why C
Given: Fixed at both ends, L = 22 ft, P = 8 kips at midspan. Relation: Fixed-end moment PL/8 (equal and opposite at midspan). Substitute: M = 8(22)/8 Result: M = 22.00 ft-kips.
Why A is tempting
Halved the correct denominator.
Why B is tempting
Used PL/12, which belongs to a uniform load on a fixed-fixed beam.
Why D is tempting
Omitted the denominator entirely.
Source NCEES FE Reference Handbook 10.6 — Civil Engineering, Structural Design
Question 6 of 6
A 6-in normal-weight concrete slab has a unit weight of 145 pcf and carries 20 psf of superimposed finishes. Its dead load is most nearly:
Slab strip 6 in thick in normal-weight concrete at 145 pounds per cubic foot, carrying 20 pounds per square foot of superimposed finishes. Show the answer and the reasoning
Why A
Given: t = 6 in, unit weight 145 pcf, finishes 20 psf. Relation: Dead load = (t in ft)(unit weight) + finishes. Substitute: (6/12)(145) + 20 = 72.5 + 20 Result: 92.50 psf.
Why B is tempting
Omitted the thickness entirely, treating the unit weight as an area load.
Why C is tempting
Left the slab thickness in inches. It must be converted to feet before multiplying by a unit weight in pcf.
Why D is tempting
Multiplied by the finish load instead of adding it; the two have different units.
Source NCEES FE Reference Handbook 10.6 — Civil Engineering, Structural Design
Going further
Frequently asked questions
How many Structural Engineering questions are on the FE Civil exam?
Between 10 and 15 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 Structural Engineering 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.