FE Civil Mechanics of Materials practice problems
6 free FE Civil Mechanics of Materials practice problems with answers and full worked solutions. Mechanics of Materials carries 7–11 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 Mechanics of Materials problems
Question 1 of 6
A simply supported beam spans 24 ft and carries a 32-kip load at midspan. The maximum bending moment is most nearly:
Simply supported beam of span 24 ft with a 32 kip load at midspan. Show the answer and the reasoning
Why B
Given: Simple span L = 24 ft, P = 32 kips at midspan. Relation: Each reaction is P/2, so at midspan M = (P/2)(L/2) = PL/4. Substitute: M = 32(24)/4 Result: M = 192.0 ft-kips.
Why A is tempting
Used a denominator with no standard basis.
Why C is tempting
Used PL/2, which corresponds to a cantilever.
Why D is tempting
Omitted the denominator.
Source NCEES FE Reference Handbook 10.6 — Mechanics of Materials
Question 2 of 6
A member with a cross-sectional area of 4.5 in^2 carries an axial load of 85 kips. The normal stress is most nearly:
Member with a cross-sectional area of 4.5 square inches carrying an axial load of 85 kips, uniformly distributed across the section. Show the answer and the reasoning
Why D
Given: P = 85 kips, A = 4.5 in^2. Relation: sigma = P/A; kips over in^2 gives ksi. Substitute: sigma = 85/4.5 Result: sigma = 18.89 ksi.
Why A is tempting
Gave P/(2A), the normal stress on a plane at 45 degrees to the axis. The normal stress on the cross section is P/A.
Why B is tempting
Divided by a safety factor of 1.67. That turns a strength into an allowable stress; the stress a load produces has no factor in it.
Why C is tempting
Applied a resistance factor of 0.90. Resistance factors reduce a capacity, not the stress a load produces.
Source NCEES FE Reference Handbook 10.6 — Mechanics of Materials
Question 3 of 6
A beam spanning 40 ft deflects 0.8 in under service load. Expressed as L/N, the value of N is most nearly:
Show the answer and the reasoning
Why C
Given: L = 40 ft = 480 in, delta = 0.8 in. Relation: N = L/delta, both in inches. Substitute: N = 480/0.8 Result: N = 600, i.e. L/600.
Why A is tempting
Left the span in feet while the deflection is in inches.
Why B is tempting
Halved the ratio, which would report the beam as twice as flexible relative to its span as it is.
Why D is tempting
Doubled the ratio, which would report the beam as twice as stiff relative to its span as it is.
Source NCEES FE Reference Handbook 10.6 — Mechanics of Materials
Question 4 of 6
A plane stress element has σx = 50 ksi, σy = 0 ksi and τxy = 16 ksi. The minimum principal stress is most nearly:
Plane stress element with a normal stress of 50 ksi on the x faces, 0 ksi on the y faces, and a shear stress of 16 ksi. Show the answer and the reasoning
Why A
Given: sigma_x = 50 ksi, sigma_y = 0 ksi, tau_xy = 16 ksi. Relation: sigma_2 = C - R, C = (sigma_x + sigma_y)/2, R = sqrt(((sigma_x - sigma_y)/2)^2 + tau_xy^2). Substitute: sigma_2 = 25.00 - 29.682 Result: sigma_2 = -4.682 ksi (negative is compression).
Why B is tempting
Took the smaller applied normal stress, ignoring the shear.
Why C is tempting
Subtracted the shear stress directly.
Why D is tempting
Gave the maximum principal stress.
Source NCEES FE Reference Handbook 10.6 — Mechanics of Materials
Question 5 of 6
A simply supported beam spans 28 ft under a uniform load of 1.5 kip/ft. The shear 9 ft from the left support is most nearly:
Simply supported beam of span 28 ft under a uniform load of 1.5 kip per foot, with a section marked 9 ft from the left support. Show the answer and the reasoning
Why A
Given: Simple span L = 28 ft, w = 1.5 kip/ft, section at x = 9 ft. Relation: V(x) = wL/2 - wx, falling linearly through zero at midspan. Substitute: V = 21.00 - 13.50 Result: V = 7.500 kips.
Why B is tempting
Counted only the load to the left of the section and omitted the reaction.
Why C is tempting
Reported the reaction at the support without deducting the load between the support and the section.
Why D is tempting
Used the full load as the reaction.
Source NCEES FE Reference Handbook 10.6 — Mechanics of Materials
Question 6 of 6
A 0.5-in thick plate is connected with 2 bolts of 8/8-in diameter, carrying 45 kips in bearing. The bearing stress is most nearly:
Show the answer and the reasoning
Why C
Given: P = 45 kips, 2 bolts, d = 1 in, t = 0.5 in. Relation: Bearing acts on the projected area n d t: sigma_b = P/(n d t). Substitute: n d t = 2(1)(0.5) = 1.0000 in^2; sigma_b = 45/1.0000 Result: sigma_b = 45.00 ksi.
Why A is tempting
Used the curved surface of the hole rather than its projected area.
Why B is tempting
Used the bolt cross-sectional area. Bearing acts on the projected rectangle of diameter times plate thickness.
Why D is tempting
Multiplied by the bolt count instead of dividing.
Source NCEES FE Reference Handbook 10.6 — Mechanics of Materials
Going further
Frequently asked questions
How many Mechanics of Materials questions are on the FE Civil exam?
Between 7 and 11 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 Mechanics of Materials 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.