Free PE: Principles and Practice of Engineering Exam Questions and Answers
The professional engineering licensure examination, taken in a chosen discipline — civil, mechanical, electrical and others — after the FE exam and the required experience.
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Last updated: October 25, 2026
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Question #1
PE Civil: Structural. A rectangular spread footing is 6 ft wide (B) by 8 ft long (L). The column transmits a service axial load of 180 kips, which includes the weight of the footing and soil above it, together with a service moment of 120 ft-kips acting about the axis parallel to the 6 ft dimension, so the eccentricity lies along the 8 ft length. Assume the footing remains in full contact with the soil. The maximum soil bearing pressure (ksf) is most nearly: (Enter your answer to the nearest tenth.)
Type your answer, then reveal.
Correct answer: 5.6
Explanation
The eccentricity is e = M/P = 120/180 = 0.667 ft, which is less than L/6 = 1.33 ft, so the full base is in compression. Then q_max = (P/BL)(1 + 6e/L) = (180/48)(1 + 6(0.667)/8) = 3.75(1.5) = 5.6 ksf. Using P/A alone gives 3.75 ksf and ignores the moment.
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Question #2
PE Civil: Structural. A 16 in. by 16 in. tied reinforced concrete column is reinforced with six No. 9 bars (A_st = 6.00 in.^2). The concrete has f'c = 5 ksi and the bars have fy = 60 ksi. The column is subject to axial load only. Using ACI 318 strength design, the maximum design axial compressive strength, φPn,max (kips), is most nearly:
Correct answer: A
Explanation
Po = 0.85 f'c(Ag - Ast) + fy Ast = 0.85(5)(256 - 6) + 60(6) = 1,062.5 + 360 = 1,422.5 kips. For a tied column, φPn,max = 0.80(0.65)(1,422.5) = 740 kips. Omitting the 0.80 factor gives 925 kips, omitting φ gives 1,140 kips, and using Po alone gives 1,420 kips.
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Question #3
PE Civil: Structural. A simply supported steel beam spans 20 ft. It carries a uniformly distributed load of 1.5 kips/ft over its full length and a concentrated load of 10 kips located 5 ft from the left support. The beam's self-weight is included in the distributed load. The maximum bending moment (ft-kips) is most nearly:
Correct answer: C
Explanation
Taking moments gives a left reaction of 1.5(20)/2 + 10(15)/20 = 22.5 kips. Shear just right of the point load is 22.5 - 7.5 - 10 = 5 kips, and it reaches zero 5/1.5 = 3.33 ft farther along, at x = 8.33 ft. The moment there is 22.5(8.33) - 10(3.33) - 1.5(8.33)^2/2 = 102 ft-kips. The moment under the point load is only 94 ft-kips, which is the common trap of assuming the maximum occurs at the concentrated load.
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Question #4
PE Civil: Structural. A 30 ft long steel bar (A = 2.0 in.^2, E = 29,000 ksi, α = 6.5 x 10^-6 per °F) is fixed to a rigid support at one end. The other end is installed with a 0.10 in. gap to a second rigid abutment. The bar is then heated uniformly by 70°F. Assume the abutment does not move and the bar does not buckle. The axial compressive force in the bar (kips) is most nearly:
Correct answer: B
Explanation
Free thermal expansion is 6.5e-6(70)(360 in.) = 0.1638 in., so only 0.1638 - 0.10 = 0.0638 in. is restrained. P = AEδ/L = 2.0(29,000)(0.0638)/360 = 10.3 kips. Ignoring the gap gives 26.4 kips, and 5.1 kips is the stress in ksi, which omits the cross-sectional area.
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Question #5
PE Civil: Structural. W16x40 beams were ordered to ASTM A992 for a floor system. A review of the mill test reports found that four heats of steel were supplied. ASTM A992 requires a yield strength of 50 to 65 ksi, a minimum tensile strength of 65 ksi, and a maximum yield-to-tensile ratio of 0.85. Which heat does NOT conform to the specification?
Correct answer: B
Explanation
The heat with Fy = 58 ksi and Fu = 66 ksi meets the yield range and the minimum tensile strength but has a yield-to-tensile ratio of 58/66 = 0.88, which exceeds 0.85. The other heats have ratios of 0.76, 0.79 and 0.83, and all lie within the yield and tensile limits. The ratio limit is what ensures ductility, so the yield strength alone cannot be used to judge conformance.
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Question #6
PE Civil: Structural. A statement of special inspections is being prepared for a two-story office building. A steel support frame for suspended mechanical equipment will be attached to the underside of a concrete slab using adhesive anchors installed in drilled holes oriented vertically upward. The weight of the equipment will place sustained tension on the anchors. The anchors have an evaluation report that permits their use in cracked concrete. Which level of special inspection should the engineer of record specify for installation of these anchors?
Correct answer: C
Explanation
Adhesive anchors installed horizontally or upwardly inclined that resist sustained tension are sensitive to workmanship, so the installation must be continuously inspected. Periodic inspection is acceptable only for other anchor orientations and loading conditions. An evaluation report that permits cracked-concrete use does not remove the inspection requirement, and keeping instructions on site is not a substitute for inspection.
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Question #7
PE Civil: Structural. A rectangular timber beam has actual dimensions of 3.5 in. wide by 11.25 in. deep (I = 415 in.^4). At a given section the internal shear force is 4,200 lb. The shear stress (psi) at a point 2.0 in. above the neutral axis is most nearly:
Correct answer: C
Explanation
Q for the area above the point is 3.5(5.625 - 2.0)[(5.625 + 2.0)/2] = 48.4 in.^3. Then τ = VQ/(Ib) = 4,200(48.4)/(415)(3.5) = 140 psi. The value 160 psi is the maximum shear stress at the neutral axis, 107 psi is the average V/A, and 53 psi results from dividing the shear force by twice the area.
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Question #8
PE Civil: Structural. A simply supported reinforced concrete beam has b = 12 in., d = 21.5 in., and A_s = 3.00 in.^2 of tension reinforcement. The concrete strength is f'c = 4,000 psi and the reinforcement yield strength is f_y = 60,000 psi. The section is tension-controlled. Using ACI 318-19, the design flexural strength, φM_n (ft-kips), is most nearly:
Correct answer: B
Explanation
a = A_s f_y /(0.85 f'c b) = 180/(0.85 × 4 × 12) = 4.41 in. M_n = 180 × (21.5 − 4.41/2) = 3,473 in.-kips = 289 ft-kips. The net tensile strain is about 0.0094, which exceeds 0.005, so φ = 0.90 and φM_n = 261 ft-kips. 289 omits φ, and 323 uses the full depth d as the lever arm.
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Question #9
PE Civil: Structural. A crane bridge applies two wheel loads of 24 kips and 12 kips, spaced 8 ft apart, to a simply supported runway girder with a 40 ft span. Neglect impact and girder self-weight. The maximum vertical support reaction (kips) as the wheels move across the span is most nearly:
Correct answer: C
Explanation
The maximum reaction occurs when the heavier wheel sits directly over the support with both wheels on the span. R = 24 + 12 × (40 − 8)/40 = 33.6 kips. 36.0 kips assumes both loads act at the support, which is not possible with an 8 ft wheel spacing.
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Question #10
PE Civil: Structural. Each beam below has constant EI and length L. Match each loading case to the expression for its maximum deflection. Each item may be used once. Targets: Target 1: Simply supported beam with a uniformly distributed load w over the full span Target 2: Simply supported beam with a single concentrated load P at midspan Target 3: Cantilever beam with a concentrated load P at the free end Target 4: Cantilever beam with a uniformly distributed load w over the full length Items: Item A: PL^3/(3EI) Item B: 5wL^4/(384EI) Item C: wL^4/(8EI) Item D: PL^3/(48EI)
Drag or tap each item onto its target, then reveal.
- Target 1Drop here
- Target 2Drop here
- Target 3Drop here
- Target 4Drop here
Correct answer: Target 1 = Item B; Target 2 = Item D; Target 3 = Item A; Target 4 = Item C
Explanation
These are the standard elastic-curve results found in the reference handbook. The simply supported cases carry the larger denominators (384 and 48), while the cantilever cases have denominators of 3 and 8. The uniform load cases contain wL^4 and the point load cases contain PL^3.
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FAQ
Learn More: https://ncees.org/engineering/pe/
- Q1: What is the PE: Principles and Practice of Engineering exam?
- A: PE: Principles and Practice of Engineering is a professional examination offered by NCEES.
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- A: Yes. The first ten are free to read with their answers and no account is needed; the rest of the bank and the practice test come with access to the exam.
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- A: Start from the official NCEES exam objectives, which define what is actually tested, and practise against questions written to them rather than to a syllabus someone has guessed at.
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- A: Registration is handled by the awarding body — see https://ncees.org/engineering/pe/ for the current fee, the testing options and the booking process.



