Free PE Civil Structural Practice Test

Correct answer: C — 2.17 kip/ft

The beam collects half of each adjoining bay: 8/2 + 12/2 = 10 ft of tributary width. Dead load = (70 × 10 + 40)/1,000 = 0.740 kip/ft. Live load = 80 × 10/1,000 = 0.800 kip/ft. Factored: 1.2(0.740) + 1.6(0.800) = 2.168 kip/ft ≈ 2.17 kip/ft.

Why the other answers miss: A is the unfactored service total (0.740 + 0.800). B forgets the beam's own 40 lb/ft. D gives the beam both full bays (20 ft) instead of half of each. This checks the one combination given; it isn't a search for the governing combination.

Fix it: Write units next to the tributary width, each area load and each line load before you apply factors. Add the beam's self-weight once, as a line load.

Source: Engineering Statics, Sec. 7.8, distributed loads (Secs. 7.8.1–7.8.2: resultant = area under the load curve, acting at its centroid)

Correct answer: B — 22.5 kips

K_LL·A_T = 4 × 900 = 3,600 ft², so √3,600 = 60. L = 50(0.25 + 15/60) = 50(0.50) = 25 psf, which equals the 0.50Lo floor (25 psf), so 25 psf stands. Column load = 25 × 900/1,000 = 22.5 kips.

Why the other answers miss: A uses the unreduced 50 psf. C applies the 0.50 reduction twice. D applies a 0.40Lo floor, which isn't the limit given for this member. Use the limit stated for the case in front of you.

Fix it: Always compute the reduced value, then compare it with the stated minimum before multiplying by area.

Source: Engineering Statics, Sec. 7.8, distributed loads (Secs. 7.8.1–7.8.2: resultant = area under the load curve, acting at its centroid)

Correct answer: A — 11.5 psf uplift

With uplift, dead load is the only thing holding the deck down, so the smaller dead-load factor governs. 0.9(15) − 25 = −11.5 psf, meaning 11.5 psf net uplift. The other combination gives 1.2(15) − 25 = −7.0 psf, which is less severe.

Why the other answers miss: B uses unfactored dead load (15 − 25). C uses 1.2D, which helps resist uplift and so isn't the critical case. D adds the wind as if it pushed down.

Fix it: Ask which load resists the effect you're checking. For the two combinations here, the lower dead-load factor controls.

Source: worked directly from the criteria and definitions stated in the question.

Correct answer: C — 2.15 kip

Try both internal signs. With GCpi = +0.18: p = 35(−1.10 − 0.18) = −44.8 psf. With GCpi = −0.18: p = 35(−1.10 + 0.18) = −32.2 psf. Positive internal pressure pushes the panel out from inside, adding to the external suction. Force = 44.8 × 48/1,000 = 2.15 kip outward.

Why the other answers miss: A uses the smaller suction case. B leaves out internal pressure entirely. D doubles the internal coefficient. The gust effect is already inside the combined coefficients, so don't multiply by a separate gust factor.

Fix it: Write both signed equations before you take a magnitude. Internal pressure that pushes outward adds to roof suction.

Source: NIST Technical Note 1415 (1995) (Sec. 3.2.2, Eq. (6), PDF p. 18: net pressure from external and internal coefficients)

Correct answer: 267 kips

Cs = 0.80/(6/1.0) = 0.1333. V = (0.80/6) × 2,000 = 266.7 kips, which rounds to 267 kips.

Why the other answers miss: 1,600 kips forgets to divide by R. 44 kips divides by R twice. R reduces the elastic demand; Ie (1.0 here) increases it for more important buildings.

Fix it: Write Cs as a dimensionless number first and sanity-check it. For ordinary buildings, values well under 1.0 are normal.

Source: FEMA 451, NEHRP Recommended Provisions: Design Examples (2006) (Sec. 6.2.5, printed p. 6-14 (PDF p. 365): Cs and V = CsW; historical method example, with coefficients and applicability supplied here)

Correct answer: 77.5 kips

Σw·h = 200(12) + 180(24) + 160(36) = 12,480 kip-ft. Level forces: F1 = 96(2,400)/12,480 = 18.46 kips; F2 = 96(4,320)/12,480 = 33.23 kips; Froof = 96(5,760)/12,480 = 44.31 kips. They add back to 96 kips. A cut through the story between 12 ft and 24 ft carries everything above it: 33.23 + 44.31 = 77.5 kips.

Why the other answers miss: 33.2 kips is the level-2 floor force, not a story shear. 96 kips is the shear in the bottom story. Distributing by weight alone ignores the height term in the formula.

Fix it: Draw a horizontal cut through the story you want, then add only the lateral forces above the cut.

Source: FEMA 451, NEHRP Recommended Provisions: Design Examples (2006) (Sec. 6.2.6, printed p. 6-15, development of equivalent lateral forces)

Correct answer: B — 240 lb/ft

A flexible diaphragm acts like a simple beam between the end walls. Each end reaction = 240 × 100/2 = 12,000 lb. That reaction is spread along the 50-ft wall: 12,000/50 = 240 lb/ft. (Matching the line load is a coincidence of the 2:1 plan shape.)

Why the other answers miss: C divides the whole 24,000-lb load by one 50-ft wall instead of splitting it between two walls. A halves the correct answer again. D is the end reaction, a force, not a unit shear in lb/ft.

Fix it: Follow the load path in order: wall → diaphragm (beam) → reactions → shear walls. Check units at each step.

Source: Engineering Statics, Sec. 7.8, distributed loads (Secs. 7.8.1–7.8.2: resultant = area under the load curve, acting at its centroid); APA, Shear Walls and Diaphragms ('What Are Diaphragms?': diaphragm as a deep, thin beam transferring load to shear walls)

Correct answer: 47.02 kip-ft

The uniform load's resultant is 10.8 kips at x = 6 ft. Right reaction = [10.8(6) + 12(15)]/18 = 13.6 kips, so left reaction = 22.8 − 13.6 = 9.2 kips. Inside the loaded length, V(x) = 9.2 − 0.90x, which is zero at x = 10.22 ft. There M = 9.2²/(2 × 0.90) = 47.02 kip-ft. Check the other candidates: M(12) = 45.6 and M(15) = 40.8 kip-ft, both lower.

Why the other answers miss: Midspan (x = 9 ft) gives only 46.35 kip-ft. The point-load location isn't automatically critical; here it's 40.8 kip-ft. Using wL²/8 over the full span changes the loading.

Fix it: Find the reactions, mark where shear crosses zero, and compare that moment with the moments at load points.

Source: MIT 2.001, Lecture 5 notes (Livermore) (PDF pp. 5–8, reactions, cuts and internal shear and moment)

Correct answer: B — 14.06 kips tension

Reactions: By = 30(9)/24 = 11.25 kips; Ay = 18.75 kips. Member AC is 15 ft long (9 across, 12 up). At joint A, vertical equilibrium: F_AC(12/15) = −18.75, so F_AC = −23.44 kips (compression). Horizontal equilibrium: F_AB = 23.44(9/15) = 14.06 kips tension.

Why the other answers miss: A is the roller reaction. C is the pin reaction, with a sense attached. D is the force in AC, not AB.

Fix it: Draw the joint free-body diagram, assume every member in tension, and let the sign tell you the answer.

Source: Engineering Statics, Sec. 6.4, method of joints (Sec. 6.4.1, joint equilibrium of ideal trusses)

Correct answer: A — 1.69 mm

For a uniformly loaded beam with both ends fixed, δmax = wL⁴/(384EI), at midspan. Work in N and mm: w = 12 N/mm, L = 6,000 mm, E = 200,000 N/mm². δ = 12(6,000)⁴/[384(200,000)(120 × 10⁶)] = 1.69 mm.

Why the other answers miss: B treats 12 kN/m as 0.012 N/mm (it's 12 N/mm). C uses the simply supported coefficient, 5/384. D uses the cantilever coefficient, 1/8.

Fix it: Match the end restraints to the formula first, then convert everything into one unit system before you substitute.

Source: USDA Forest Products Laboratory, Wood Handbook (FPL-GTR-282, 2021), Ch. 9 (Eq. 9-2, printed p. 9-2; Table 9-1, printed p. 9-3: uniformly distributed load, both ends clamped, kb = 1/384)

Correct answer: A — 0 ksi

Axial stress P/A = 300/576 = 0.521 ksi. Bending: M = 300 × 4 = 1,200 kip-in and S = 24(24²)/6 = 2,304 in³, so M/S = 0.521 ksi. Minimum = 0.521 − 0.521 = 0 ksi. An eccentricity of h/6 = 4 in puts the load exactly at the edge of the kern; any further out and that face would go into tension.

Why the other answers miss: B is P/A alone. D is the maximum stress, 0.521 + 0.521. C keeps the tensile M/S term but leaves out the uniform compression P/A.

Fix it: Superpose P/A and M/S on each face separately. For rectangles, remember the kern limit e = h/6.

Source: D. Roylance, MIT 3.11, Stresses in Beams (flexure formula σ = My/I and shear stress τ = VQ/Ib)

Correct answer: B — 0.693 ksi compression

Take compression as positive. P/A = 240/250 = 0.960. Prestress eccentricity adds bottom compression: Pe/Sb = 240(3)/900 = 0.800. Sagging gravity moment causes bottom tension: M/Sb = 80(12)/900 = 1.067. Bottom stress = 0.960 + 0.800 − 1.067 = 0.693 ksi compression.

Why the other answers miss: A adds all three terms as compression. C keeps only the direct prestress. D keeps only the gravity bending.

Fix it: Sketch the sign of each contribution at the bottom fiber before adding. Prestress below the centroid and sagging moment work against each other there.

Source: D. Roylance, MIT 3.11, Stresses in Beams (flexure formula σ = My/I and shear stress τ = VQ/Ib); FHWA LRFD prestressed girder example (Design Step 5.6.4.2: direct prestress, eccentric prestress and gravity-moment stress terms; historical method support only)

Correct answer: A — 11.3 ksi compression

Free expansion would be a strain of αΔT = 6.5 × 10⁻⁶ × 60 = 3.9 × 10⁻⁴. The supports push it back to zero length change, so σ = EαΔT = 29,000 × 3.9 × 10⁻⁴ = 11.3 ksi compression.

Why the other answers miss: B has the wrong sense: a heated bar that can't grow is squeezed. C is true only for a bar free to expand. D is a factor-of-ten error in the thermal strain, using 3.9 × 10⁻³ instead of 3.9 × 10⁻⁴.

Fix it: Ask what the restraint prevents. Prevented expansion produces compression; prevented contraction produces tension.

Source: OpenStax, University Physics Vol. 2, Sec. 1.3 (linear thermal expansion and thermal stress)

Correct answer: C — 42 kips

An axle P at distance x from the left support adds P(1 − x/42) to the left reaction. Put the heavy 30-kip axle right at the left support and the 18-kip axle 14 ft in: R = 30 + 18(28/42) = 42 kips. The reverse order gives 18 + 30(28/42) = 38 kips.

Why the other answers miss: A counts only the heavy axle. B is the other driving direction. D gives both full axle weights to one support, which only happens if both sit on it.

Fix it: Sketch the reaction influence line (a triangle from 1 to 0) and test both axle orders.

Source: Engineering Statics, Sec. 7.8, distributed loads (Secs. 7.8.1–7.8.2: resultant = area under the load curve, acting at its centroid)

Correct answer: 451 kip-ft

The maximum occurs under one axle when midspan sits halfway between that axle and the resultant of both loads. The resultant is at the middle of the pair, 2 ft from each axle, so place one axle 1 ft past midspan, at x = 21 ft, with the other at 17 ft. Left reaction = 25(19/40) + 25(23/40) = 26.25 kips. Moment under the 21-ft axle, from the right: right reaction = 50 − 26.25 = 23.75 kips; M = 23.75 × 19 = 451.25 kip-ft, entered as 451.

Why the other answers miss: 450 kip-ft comes from centering the pair on the span (axles at 18 and 22 ft), which is close but not the maximum. 500 kip-ft treats both axles as one 50-kip load at midspan, which they can't physically occupy. 250 kip-ft uses one axle only.

Fix it: For these two equal axles, place their resultant and either axle equally far from midspan on opposite sides. For a general axle group, compare the candidate maxima under the different axles and check which loads remain on the span.

Source: Felix Udoeyo, Structural Analysis, Chapter 9 (Sec. 9.4.4, Eq. 9.10 and Example 9.14: absolute moment maxima and comparison under different axles)

Correct answer: A — 9.0 in

Shear flow q = VQ/I = 40(100)/2,000 = 2.0 kip/in. Each row of two fasteners resists 18 kips, so spacing s = 18/2.0 = 9.0 in.

Why the other answers miss: B counts only one fastener per row. C inverts the ratio. D reports the shear flow itself, which is in kip/in, not inches.

Fix it: Shear flow is force per unit length. Divide the strength of one row by it to get the spacing.

Source: D. Roylance, MIT 3.11, Stresses in Beams (flexure formula σ = My/I and shear stress τ = VQ/Ib)

Correct answer: C — Keep the shores until adequate strength is established

OSHA's rule says forms and shores (other than slabs on grade and slip forms) stay in place until the employer determines the concrete has gained sufficient strength to support its weight and superimposed loads. Nothing in the facts shows that yet, so C. A separate paragraph, (e)(2), applies the same idea to reshoring.

Why the other answers miss: A, B and D substitute a calendar age or an appearance for the strength determination the rule requires.

Fix it: Name the evidence (specified conditions or test results) and the loads it has to cover before deciding on removal.

Source: 29 CFR 1926.703(e), July 1, 2020 edition (govinfo) (§1926.703(e)(1)–(2), printed pp. 382–383 (PDF pp. 3–4))

Correct answer: A and B

The limited access zone is the wall height plus 4 ft (10 + 4 = 14 ft), runs the entire length of the wall, sits on the unscaffolded side, and must be in place before construction starts. A wall over 8 ft tall that isn't otherwise supported must be braced, and the bracing stays until permanent supporting elements are in place. So A and B.

Why the other answers miss: C drops the wall height from the zone width. D replaces the access restriction with PPE; the zone is limited to employees actively building the wall. Scoring is all-or-nothing: both correct choices and nothing else.

Fix it: Treat the zone and the bracing as two separate requirements with different triggers.

Source: 29 CFR 1926.706, July 1, 2020 edition (govinfo) (§1926.706(a)(1)–(5) and (b), printed pp. 385–386 (PDF pp. 6–7))

Correct answer: A and B

§1926.502(b)(1) sets the top-rail height at 42 in ± 3 in (39 to 45 in). §1926.502(b)(3) requires the top edge to withstand at least 200 lb applied within 2 in of the top edge, outward or downward. So A and B. When conditions warrant, (b)(1) allows a top edge above 45 in if the other criteria are met; when employees use stilts, their height is added to the top-rail height. Those cases are excluded here.

Why the other answers miss: C is below the 39-in minimum. D understates the 200-lb requirement. (Midrails, screens and similar members are a separate 150-lb requirement in (b)(5); don't mix the two.) All-or-nothing scoring.

Fix it: Memorize these as a pair: 42 ± 3 in and 200 lb. Then remember midrails are a different, lower load.

Source: 29 CFR 1926.502(b), July 1, 2020 edition (govinfo) (§1926.502(b)(1), (3)–(5), printed pp. 299–300 (PDF pp. 1–2))

Correct answer: A — Written notice of 75% strength (or sufficient strength)

§1926.752(a)(1) puts the duty on the controlling contractor, who must notify the steel erector in writing. The concrete has to reach either 75% of its intended minimum compressive design strength or enough strength for the erection loads, based on an appropriate ASTM test method of field-cured samples. So A. Paragraph (a)(2) separately requires written notification about any anchor-bolt repairs, replacements or modifications.

Why the other answers miss: B names the wrong party and the wrong form of notice. C sets a higher bar than the rule and uses lab-cured, not field-cured, samples. D swaps a strength basis for a calendar.

Fix it: Remember three pieces: who notifies (controlling contractor), how (in writing), and the threshold (75% or sufficient strength).

Source: 29 CFR 1926.752(a), July 1, 2020 edition (govinfo) (§1926.752(a)(1)–(2), printed p. 390 (PDF p. 1))

Correct answer: 0.914 ksf

γsat = 62.4(2.68 + 0.72)/1.72 = 123.35 pcf. The submerged (buoyant) unit weight is 123.35 − 62.4 = 60.95 pcf. σ′ = 60.95 × 15/1,000 = 0.914 ksf. Check: total stress 1.850 ksf minus pore pressure 0.936 ksf = 0.914 ksf.

Why the other answers miss: 1.850 ksf is total stress (pore pressure not subtracted). 0.936 ksf is the water pressure alone. Using a dry unit weight contradicts the saturated condition.

Fix it: Keep three columns: total stress, pore pressure, effective stress. The last is always the first minus the second.

Source: FHWA NHI-05-037, Geotechnical Aspects of Pavements Reference Manual, Ch. 5 (Sec. 5.3.1, Table 5-8: saturated unit weight); FHWA NHI-06-088, Soils and Foundations, Vol. I (2006) (Secs. 2.2–2.3, Eq. 2-13, printed pp. 2-19–2-20 (PDF pp. 75–76): effective stress equals total stress minus pore-water pressure)

Correct answer: C — 0.336 in

With everything else the same, deflection is inversely proportional to E. δB = 0.240(4.2/3.0) = 0.336 in. The softer mix deflects more.

Why the other answers miss: A flips the ratio. B assumes equal strength means equal stiffness; it doesn't. D adds the two beams' deflections.

Fix it: Use the measured property the calculation actually needs. Equal compressive strength doesn't guarantee equal modulus.

Source: Noguchi et al., ACI Structural Journal 106(5), 2009 (abstract) (public abstract: concrete modulus depends on more than compressive strength); USDA Forest Products Laboratory, Wood Handbook (FPL-GTR-282, 2021), Ch. 9 (Eq. 9-2, printed p. 9-2: bending deflection is proportional to 1/E for fixed geometry, loading and supports; shear deflection is a separate term)

Correct answer: B — Not acceptable: a running three-test average is below f′c

Criterion (2) passes; the lowest test, 3,600 psi, is exactly 400 psi low. But the running averages are 3,933, 3,867 and 4,100 psi. The first two fall below 4,000 psi, so criterion (1) fails and the concrete is not acceptable under these criteria. B.

Why the other answers miss: A checks only one of the two criteria. C invents a rule; a single test below f′c doesn't fail criterion (2) unless it's more than 500 psi low. D averages all five tests, which isn't one of the stated criteria.

Fix it: Write out every consecutive group of three and check both criteria every time.

Source: worked directly from the criteria and definitions stated in the question.

Correct answer: A — 99.4%, passes

Convert to dry unit weight first: γd = γ/(1 + w) = 128.0/1.12 = 114.3 pcf. Relative compaction = 114.3/115.0 = 99.4%, which meets the 95% requirement.

Why the other answers miss: B divides the moist unit weight by the lab maximum; the lab value is a dry unit weight, so that compares unlike things. C inverts the ratio and uses the moist weight (115.0/128.0). D assumes the minimum without calculating.

Fix it: Compare dry with dry. Convert the field reading with γd = γ/(1 + w) before dividing.

Source: FHWA NHI-05-037, Geotechnical Aspects of Pavements Reference Manual, Ch. 5 (Sec. 5.3.2, Eq. 5.1 (moist-to-dry unit weight) and Eq. 5.4 (relative compaction))

Correct answer: A — 24.2 in²

Modular ratio n = E_s/E_c = 29,000/3,600 = 8.06. For tension steel in a cracked section, the transformed area is n·A_s = 8.06 × 3.00 = 24.2 in².

Why the other answers miss: B uses (n − 1)A_s. In a cracked section the concrete on the tension side is ignored, so nothing is counted twice and the full n·A_s applies. C forgets to transform. D divides by n instead of multiplying.

Fix it: The stiffer material gets multiplied by n. Before adjusting by one, ask whether the concrete around that steel is actually being counted.

Source: I. Almohanna, CE 370 Tutorial Note 2, King Saud University (2021) (modular ratio n = Es/Ec and transformed tension steel area n·As in cracked-section examples)

Correct answer: B and D

Small clear specimens show what defect-free wood can do. Structural lumber has knots, slope of grain and other features, and its strength varies, so design values come from grading and the applicable design specification, with adjustments for size and conditions. So B and D.

Why the other answers miss: A treats an average failure value as an allowable stress. C invents a universal factor. E swaps in a different property. All-or-nothing scoring.

Fix it: Label what a test number is (specimen type, statistic) before deciding whether it can be used as a design value.

Source: USDA Forest Products Laboratory, Wood Handbook (FPL-GTR-282, 2021), Ch. 5 (printed pp. 5-1–5-2, clear-wood properties and variability)

Correct answer: C — 116.0 kips, net-section rupture; not adequate

Yielding: 0.90(50)(3.60) = 162.0 kips. Effective net area Ae = U·An = 0.85(2.80) = 2.38 in². Rupture: 0.75(65)(2.38) = 116.0 kips, which governs. Pu = 120 kips is more than 116.0 kips, so the member fails these checks.

Why the other answers miss: A ignores the lower rupture value. B forgets the shear lag factor (uses An). D applies the yielding φ to rupture.

Fix it: Put each limit state on its own line, with its own area, stress and φ. Then take the smaller.

Reference check: On the exam, these checks are in AISC 360 Chapter D (Secs. D2–D3).

Source: ANSI/AISC 360-16, July 7, 2016 (AISC-authored, University of Maryland–hosted copy) (Secs. D2–D3, Eqs. D2-1, D2-2 and D3-1, printed pp. 16.1-28–16.1-29 (PDF pp. 84–85): the lesser of 0.90FyAg and 0.75FuAe, with Ae = UAn)

Correct answer: A — 139 kips

The effective throat of an equal-leg fillet is 0.707 × leg = 0.707 × 0.3125 = 0.221 in. Strength per inch = 0.75 × 0.60 × 70 × 0.221 = 6.96 kip/in. Total length 20 in gives 139 kips.

Why the other answers miss: B uses the leg size as the throat. C counts only one weld line. D leaves out φ.

Fix it: Weld strength lives on the throat, not the leg. Multiply by 0.707 for equal-leg fillets.

Reference check: On the exam, fillet weld strength is in AISC 360 Chapter J. This exercise uses the verified 2016 Specification's LRFD weld-metal form, 0.45F_EXX·Awe, for the stated loading and geometry.

Source: ANSI/AISC 360-16, July 7, 2016 (AISC-authored, University of Maryland–hosted copy) (Sec. J2.2, printed pp. 16.1-119–120 (PDF pp. 175–176); Sec. J2.4(a), Eq. J2-3, printed p. 16.1-122 (PDF p. 178); Table J2.5, printed p. 16.1-124 (PDF p. 180))

Correct answer: 19.70 kips

M = 40 × 6 = 240 kip-in. Each bolt is r = 5 in from the centroid, so Σr² = 4(25) = 100 in². Direct shear = 40/4 = 10 kips, down. The moment component is 240(5)/100 = 12 kips, which splits into 240(4)/100 = 9.6 kips horizontal and 240(3)/100 = 7.2 kips vertical. At the right-hand bolts the vertical parts add: √[9.6² + (10 + 7.2)²] = 19.70 kips. (The left-hand bolts get √[9.6² + 2.8²] = 10.0 kips.)

Why the other answers miss: 10 kips is direct shear only. 12 kips is the moment part only. 22 kips adds the two magnitudes as if they were parallel, but they aren't.

Fix it: Draw the moment force perpendicular to each bolt's radius, then add x and y components before taking a magnitude.

Source: NASA RP-1228, Fastener Design Manual (1990) (printed p. 17, 'Finding Shear Loads on Fastener Group')

Correct answer: A — 192 kips

Aw = 18.0 × 0.355 = 6.39 in². Vn = 0.6(50)(6.39)(1.0) = 191.7 kips. With φv = 1.00, φvVn = 192 kips.

Why the other answers miss: B uses φ = 0.90, the general value, instead of the 1.00 the stem says this web qualifies for. C leaves out d when calculating Aw, using the web thickness as if it were the web area. D drops the 0.6, treating shear yield as tensile yield.

Fix it: Check which φ applies before you multiply. For rolled shapes, Aw uses the overall depth.

Reference check: On the exam, member shear is in AISC 360 Chapter G. Confirm the web-slenderness condition for φv = 1.00 there.

Source: ANSI/AISC 360-16, July 7, 2016 (AISC-authored, University of Maryland–hosted copy) (Secs. G1 and G2.1, Eqs. G2-1–G2-2, printed pp. 16.1-70–71 (PDF pp. 126–127): Aw = d·tw and the qualifying rolled-I web case)

Correct answer: A — 298 kip-ft

Mp = FyZx = 50(101) = 5,050 kip-in = 420.8 kip-ft. 0.7FySx = 0.7(50)(88.9) = 3,111.5 kip-in = 259.3 kip-ft. The interpolation fraction is (12 − 5.83)/(17.0 − 5.83) = 0.552. Mn = 420.8 − (420.8 − 259.3)(0.552) = 331.6 kip-ft. φbMn = 0.90 × 331.6 = 298 kip-ft.

Why the other answers miss: B is φMp, which ignores the unbraced length. C is φ × 0.7FySx, the value at Lr, not at 12 ft. D is the nominal Mn without φ.

Fix it: Place Lb in its zone (plastic, inelastic or elastic) before choosing an equation, then apply φ last.

Reference check: On the exam, Lp, Lr and the flexural equations come from AISC 360 Chapter F (and the Manual's tables for real shapes).

Source: ANSI/AISC 360-16, July 7, 2016 (AISC-authored, University of Maryland–hosted copy) (Secs. F1(a), F2.1 and F2.2(b), Eqs. F2-1–F2-2, printed pp. 16.1-46–47 (PDF pp. 102–103))

Correct answer: 692 kips

Lc/r = 180/3.00 = 60. The limit is 4.71√(29,000/50) = 113.4, so 60 is below it and the inelastic equation applies. Fe = π²(29,000)/60² = 79.505 ksi. Critical stress Fcr = 0.658^(50/79.505) × 50 = 38.4286 ksi. Keeping full precision, φcPn = 0.90 × 38.4286 × 20.0 = 691.714 kips, which rounds to 692 kips.

Why the other answers miss: About 1,255 kips comes from using 0.877Fe, the elastic equation, outside its range. 900 kips is the squash load φFyAg, which ignores buckling. 769 kips leaves out φ.

Fix it: Compute Lc/r and compare it with the 4.71√(E/Fy) limit before you pick a buckling equation.

Reference check: On the exam, column strength is in AISC 360 Chapter E.

Source: ANSI/AISC 360-16, July 7, 2016 (AISC-authored, University of Maryland–hosted copy) (Secs. E1 and E3, Eqs. E3-1–E3-4, printed pp. 16.1-33 and 16.1-35–36 (PDF pp. 89 and 91–92))

Correct answer: 470.6 kip-ft

Tension T = 4.00 × 60 = 240 kips. If the block stays in the flange, its depth a = 240/(4.25 × 60) = 0.941 in, which is less than the 4-in flange, so that assumption holds. Mn = T(d − a/2) = 240(24 − 0.471)/12 = 470.6 kip-ft.

Why the other answers miss: Using the 12-in web width makes the block about 4.7 in deep and wrongly pushes it into the web. Using d as the lever arm gives 480 kip-ft. Applying a strength reduction factor answers a different question; this asks for the nominal value.

Fix it: Balance compression and tension first, confirm the block's location, then take moments about the compression resultant.

Source: FHWA LRFD prestressed girder example, Design Step 5.6.4 (compression-block force equilibrium and nominal moment couple)

Correct answer: B — Member B only

M = 150(18²)/8 = 6,075 lb-ft = 72,900 lb-in. Limit L/360 = 216/360 = 0.600 in. Member A: S = 49.9 in³, so fb = 1,461 psi > 1,200 (fails); I = 230.8 in⁴, so live-load Δ = 5(100/12)(216⁴)/[384(1.6 × 10⁶)(230.8)] = 0.640 in > 0.600 (fails). Member B: S = 73.8 in³, so fb = 987 psi (passes); I = 415.3 in⁴, so Δ = 0.355 in (passes). Member B only.

Why the other answers miss: A picks the member that fails both checks. C misses A's failures. D rejects B, which passes both; a common cause is entering 100 lb/ft as lb/in in the deflection formula.

Fix it: Use total load for strength and live load for the live-load deflection limit. Keep loads in lb/in when E is in psi and lengths in inches.

Source: USDA Forest Products Laboratory, Wood Handbook (FPL-GTR-282, 2021), Ch. 9 (Eq. 9-2, printed p. 9-2; Table 9-1, printed p. 9-3: uniformly distributed load, both ends simply supported, kb = 5/384; Eq. 9-13, printed p. 9-7: bending stress M/S)

Correct answer: 5.21 ft

Total P = 400 kips. Resultant location: [160(0) + 240(18)]/400 = 10.8 ft. For uniform bearing, the footing centroid sits there: −2 + L/2 = 10.8, so L = 25.6 ft. Required area = 400/3.00 = 133.3 ft². Width B = 133.3/25.6 = 5.21 ft.

Why the other answers miss: Dividing the area by the 18-ft column spacing uses the wrong length. Centering the footing between the columns ignores the unequal loads and creates eccentric bearing. Ignoring the fixed left edge changes the geometry.

Fix it: Find the resultant first. The footing length follows from it; the width follows from the area.

Source: Engineering Statics, Sec. 7.8, distributed loads (Secs. 7.8.1–7.8.2: resultant = area under the load curve, acting at its centroid); FHWA-HRT-14-094 (2015), Ch. 4 (eccentricity, bearing pressure diagram and footing liftoff)

Correct answer: A — 4.30 ksf

e = M/P = 0.5 ft, which is less than B/6 = 1.33 ft, so the whole base stays in compression. q_max = (P/A)(1 + 6e/B) = (200/64)(1 + 6 × 0.5/8) = 3.125 × 1.375 = 4.30 ksf. (q_min = 1.95 ksf.)

Why the other answers miss: B is P/A only. C is the minimum pressure. D computes the section modulus as B²/6, leaving out the footing width.

Fix it: Check e against B/6 first. Inside the kern, use P/A ± M/S; outside it, the pressure diagram changes shape.

Source: FHWA-HRT-14-094 (2015), Ch. 4 (eccentricity, bearing pressure diagram and footing liftoff); D. Roylance, MIT 3.11, Stresses in Beams (flexure formula σ = My/I and shear stress τ = VQ/Ib)

Correct answer: 71.25 kips

P/n = 360/6 = 60 kips. Σx² = 2(9) + 2(0) + 2(9) = 36 ft². At x = +3 ft: 135(3)/36 = 11.25 kips. Maximum = 60 + 11.25 = 71.25 kips. (The two piles at x = −3 ft carry 48.75 kips each.)

Why the other answers miss: 82.5 kips counts only one pile at each x position (Σx² = 18 ft²), which doubles the moment share. 60 kips ignores the moment.

Fix it: Build Σx² pile by pile. It's the same idea as the elastic bolt group, turned on its side.

Source: Missouri DOT Engineering Policy Guide, 751.36 Driven Piles (Sec. 751.36.5.8, 'Additional Provisions for Pile Cap Footings' → 'Pile Group Layout': rigid-cap axial load distribution; no state-specific capacities adopted)

Correct answer: D — 9.248 kip/ft

Soil thrust (triangle): ½(0.307)(118)(9²)/1,000 = 1.467 kip/ft. Surcharge thrust (rectangle): 0.307(250)(9)/1,000 = 0.691 kip/ft. Total driving force = 2.158 kip/ft. Sliding FS = μW/driving force, so W = 1.50(2.158)/0.35 = 9.248 kip/ft.

Why the other answers miss: A multiplies by μ instead of dividing. B leaves out the 1.50 safety factor. C leaves out the surcharge.

Fix it: Draw the two pressure diagrams separately. A uniform surcharge adds a rectangle, Ka·q·H, not a triangle.

Source: FHWA Rockery Design and Construction Guidelines, Ch. 4 ('Lateral Earth Pressures' (surcharge pressure Ka·q) and 'Sliding Resistance' (resisting ÷ driving forces))

Correct answer: B — 200.0 psi, passes

Net area = 620 + 4(28) = 732 in². Stress = 146,400/732 = 200 psi, below the 225-psi allowable, so it passes this check.

Why the other answers miss: A counts the grouted cells twice (844 in²). C leaves out the grout (620 in²). D reports the allowable stress instead of calculating the demand.

Fix it: Inventory the area piece by piece: solid masonry, plus grouted cells, minus nothing else. Only then divide.

Source: worked directly from the criteria and definitions stated in the question.

Correct answer: A — 75 kips

The brace runs diagonally: length = √(20² + 15²) = 25 ft. Its horizontal component has to carry the full 60 kips: F(20/25) = 60, so F = 75 kips.

Why the other answers miss: B ignores the brace angle. C multiplies by 20/25 instead of dividing. D splits the shear between both braces, but this is a tension-only system, so only one brace works at a time.

Fix it: Resolve the member force into components and match the horizontal one to the story shear.

Source: Engineering Statics, Sec. 6.4, method of joints (Sec. 6.4.1, joint equilibrium of ideal trusses)