60 problems covering all 14 sections, sized to the midpoints of the NCEES question ranges: six each from the three largest civil areas, five from transportation, construction, statics and math, down to two from ethics and dynamics. About half are calculations laid out as Given / Find with a worked solution; the rest test concepts and definitions. Units follow each problem, SI or USCS, as on the exam.
Updated Oct 29, 2026 · Not affiliated with NCEES.
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Question 1 of 60 · Materials · 5–8 items
In high-performance concrete mixes, which physical property of silica fume most significantly contributes to enhanced strength and durability?
Solution
Silica fume: amorphous SiO2, mean particle size ≈ 0.1 μm; typical dosage 5–10 % of cementitious materials.
Particles about 0.1 μm across fill the voids between cement grains and react fast, densifying the paste and the transition zone.
Silica fume is highly pozzolanic, not inert; it reacts with calcium hydroxide to form more C-S-H.
Its specific surface is extremely high, about 20,000 m²/kg, roughly 50 times that of cement.
Silica fume raises water demand, but water retention is not how it adds strength.
Question 2 of 60 · Materials · 5–8 items
Which test is used to determine the fineness of Portland cement?
Solution
Typical Type I cement Blaine fineness ≈ 350–400 m²/kg; finer cement gains strength faster.
Slump measures fresh-concrete consistency.
Vicat needle penetration measures normal consistency and setting time.
Le Chatelier tests soundness (expansion) of cement, or density with the flask; it does not measure fineness.
ASTM C204 measures specific surface (m²/kg) from the time air takes to pass through a compacted bed of cement.
Question 3 of 60 · Materials · 5–8 items
Which property of wood is most affected by moisture content?
Solution
Below FSP (≈ 28 %), strength increases roughly 4–6 % per 1 % drop in moisture content; NDS applies CM for wet service.
Below the fiber saturation point, strength and stiffness rise as wood dries; design values are adjusted for wet service.
Density changes a little with moisture, but 'only' misses the large effect on strength.
Grain pattern and figure are set by growth and sawing, not by moisture.
Color varies with species and exposure, not with moisture content in any design sense.
Question 4 of 60 · Fluid Mechanics · 6–9 items
A concrete-lined trapezoidal channel has a bottom width of 4 m, side slopes of 1:1 and Manning n = 0.02. What bed slope is needed for it to carry 22 m³/s at a uniform depth of 2 m?
Given
Bottom width b = 4 m
Side slopes 1:1 (z = 1)
Flow depth y = 2 m
Discharge Q = 22 m³/s
Manning n = 0.02
Find
Bed slope S (m/m)
Solution
A = (4 + 2)(2) = 12 m²; P = 4 + 2(2)√2 = 9.66 m; R = 1.243 m
V = Q/A = 1.833 m/s
S = (Vn/R2/3)² = (1.833 × 0.02/1.156)² = 0.0010
FE Reference Handbook, Fluid Mechanics — Manning's equation (SI).
0.00053 takes R equal to the 2-m depth instead of A/P = 1.243 m.
A = 12 m², P = 9.66 m, R = 1.243 m, V = 22/12 = 1.83 m/s, so S = (Vn/R2/3)² = (1.83 × 0.02/1.156)² = 0.0010.
0.0030 treats the section as a 4 × 2 m rectangle (A = 8 m², R = 1.0 m).
0.032 is Vn/R2/3 itself; Manning has S1/2, so the result must be squared.
Question 5 of 60 · Fluid Mechanics · 6–9 items
What is the center of pressure location for a rectangular plate submerged vertically in water with its top edge at the water surface?
FE Reference Handbook, Fluid Mechanics — center of pressure.
With the top edge at the surface, the pressure diagram is a triangle, so the resultant acts at 2h/3: ycp = yc + Ic/(ycA) = h/2 + h/6.
3h/4 has no basis for a uniform-width plate with a triangular pressure distribution.
h/3 from the top is where the centroid of a triangle would be measured from its apex; the pressure prism's centroid is lower.
h/2 is the centroid of the plate; the center of pressure lies below it because pressure grows with depth.
Question 6 of 60 · Fluid Mechanics · 6–9 items
Consider two scenarios for water flow through a piping system with a 90° elbow: (A) pipe roughness increases significantly, and (B) a pump is added upstream providing additional head. Which scenario increases the axial force on the downstream elbow, and why?
Solution
Rx = p1A1 + ρQV1 (for a 90° elbow, x along the inlet)
FE Reference Handbook, Fluid Mechanics — momentum equation.
The two changes act in opposite directions on pressure and flow, so they cannot add the same force.
Geometry sets the direction terms, but magnitude depends on pressure and flow, which both scenarios change.
With the same supply head, more roughness lowers both velocity and pressure at the elbow, and a pump adds head rather than removing it.
Rougher pipe lowers flow and dissipates pressure upstream; friction elsewhere in the line does not push harder on the elbow.
The elbow reaction comes from the pressure forces pA plus the momentum flux ρQV; a pump raises the pressure and the flow, so both terms grow.
Question 7 of 60 · Fluid Mechanics · 6–9 items
A water jet leaves a 0.02-m-diameter nozzle horizontally at 20 m/s and strikes a flat plate perpendicular to the jet, which brings the jet's horizontal velocity to zero. What force does the jet exert on the plate? (ρ = 1,000 kg/m³)
Given
Nozzle diameter D = 0.02 m
Jet velocity V = 20 m/s
ρ = 1,000 kg/m³
Find
Force on the plate F (N)
Solution
Q = AV = (π/4)(0.02)² × 20 = 6.28 × 10−3 m³/s
F = ρQ(V − 0) = 1,000 × 6.28 × 10−3 × 20 = 125.66 N
FE Reference Handbook, Fluid Mechanics — impulse-momentum (jet on a plate).
6.28 is the mass flow rate ρAV in kg/s; it still has to be multiplied by V.
12.81 N divides ρQV by g, mixing mass and weight units; with ρ in kg/m³ the result is already in newtons.
62.83 N uses ½ρV²A, the dynamic pressure times area, which is half the momentum flux.
F = ρQV = ρAV² = 1,000 × (π × 0.02²/4) × 20² = 125.66 N.
Question 8 of 60 · Surveying · 6–9 items
Three consecutive points on a profile have the following stations and elevations: 10+00, 324.5 ft; 12+50, 328.0 ft; 15+00, 320.5 ft. What is the grade (in percent) between stations 10+00 and 12+50?
Given
Sta 10+00, elevation 324.5 ft
Sta 12+50, elevation 328.0 ft
Find
Grade between the stations (%)
Solution
G = Δelev/Δsta × 100 = 3.5/250 × 100 = +1.4 %
FE Reference Handbook, Transportation — vertical alignment.
−1.4 % has the wrong sign: the elevation rises from 324.5 to 328.0 ft, so the grade is positive.
0.014 % is the grade as a decimal (0.014) labeled as percent; it must be multiplied by 100.
0.28 % divides the 3.5-ft rise by the station value 1,250 ft instead of the 250-ft run between stations.
Question 9 of 60 · Water Resources and Environmental Engineering · 10–15 items
A trapezoidal channel has a bottom width of 6 feet, side slopes of 2:1 (horizontal:vertical), and a water depth of 3 feet. What is the hydraulic radius of this channel?
Given
Bottom width b = 6 ft
Side slopes 2H:1V (z = 2)
Flow depth y = 3 ft
Find
Hydraulic radius R (ft)
Solution
A = (b + zy)y = 36 ft²; P = b + 2y√(1 + z²) = 19.42 ft
R = A/P = 1.85 ft
FE Reference Handbook, Fluid Mechanics — open channels.
A = (6 + 2 × 3) × 3 = 36 ft², P = 6 + 2 × 3√5 = 19.42 ft, R = 36/19.42 = 1.85 ft.
2.00 ft counts each side as its 6-ft horizontal run instead of the 6.71-ft slope length.
2.78 ft takes the area as top width × depth (18 × 3 = 54 ft²), as if the section were a rectangle as wide as the water surface.
3.00 ft is the flow depth, not A/P.
Question 10 of 60 · Water Resources and Environmental Engineering · 10–15 items
A water storage tank has a volume of 200,000 gallons and serves a community with an average daily demand of 50,000 gpd. What is the detention time in days?
Given
Tank volume V = 200,000 gal
Average demand Q = 50,000 gpd
Find
Detention time t (days)
Solution
t = V/Q = 200,000/50,000 = 4.00 days
FE Reference Handbook, Environmental/Water — detention time.
0.25 is Q/V, the ratio inverted.
0.53 converts gallons to cubic feet in the volume only and divides by a flow still in gallons.
t = V/Q = 200,000 gal/50,000 gal/day = 4.00 days.
29.92 multiplies by 7.48 gal/ft³ with no reason; both quantities are already in gallons.
Question 11 of 60 · Water Resources and Environmental Engineering · 10–15 items
Calculate the water-quality volume WQv for a 2.50-acre drainage area with a 1.00-in design rainfall and impervious cover I = 60 %. Use Rv = 0.05 + 0.009I (I in percent) and 1 in over 1 acre = 3,630 ft³. Report WQv to the nearest ft³.
Given
Drainage area A = 2.50 ac
Design rainfall P = 1.00 in
Impervious cover I = 60 %
Rv = 0.05 + 0.009I
1 ac-in = 3,630 ft³
Find
Water-quality volume WQv (ft³)
Solution
Rv = 0.05 + 0.009(60) = 0.59
WQv = P·Rv·A = 1.00 in × 0.59 × 2.50 ac × 3,630 ft³/ac-in = 5,354 ft³
503 ft³ enters I as 0.60 instead of 60, so Rv becomes 0.055.
2,142 ft³ is Rv × 3,630; the 2.50-acre area was not applied.
9,075 ft³ leaves out Rv, treating all rainfall as runoff.
13,385 ft³ applies the 2.50-acre area twice.
Question 12 of 60 · Water Resources and Environmental Engineering · 10–15 items
Time (min)
Inflow I (cfs)
Outflow O (cfs)
0
0
0
10
25
10
20
60
22
30
85
28
40
50
30
50
20
26
60
0
18
A detention basin receives the 10-year inflow hydrograph and releases the routed outflow shown in the table (10-min time step). Using the trapezoidal rule on inflow minus outflow, what is the maximum stored volume at the tabulated times?
FE Reference Handbook, Water Resources — reservoir routing (storage indication).
1,270 ft³ multiplies the flow sums by 10 (minutes) instead of 600 seconds.
33,000 ft³ is (peak inflow − peak outflow) × 600 s, a single-step guess that ignores the hydrograph shape.
72,000 ft³ stops the accumulation at t = 40 min; storage keeps rising until inflow falls below outflow, between 40 and 50 min.
74,400 ft³ sums the ordinate differences (I − O) at 10–50 min and multiplies by 600 s, a rectangular rule instead of the trapezoidal averaging the stem asks for.
ΔS per step = [(I1 + I2)/2 − (O1 + O2)/2]Δt: 4,500 + 15,900 + 28,500 + 23,100 + 4,200 = 76,200 ft³ at t = 50 min, after which ΔS turns negative.
Question 13 of 60 · Water Resources and Environmental Engineering · 10–15 items
A two-loop water network is analyzed by the Hardy Cross method with Hazen–Williams head losses h = rQ1.852. Loop 1 has pipes with r1 = 0.05, r2 = 0.08 and r3 = 0.06 and initial flows Q1 = 100, Q2 = −80 and Q3 = 60 gpm (signs give direction around the loop). What is the first flow correction ΔQ for Loop 1?
FE Reference Handbook, Fluid Mechanics — pipe networks (Hardy Cross).
−44.0 gpm uses absolute values in the numerator (Σ r|Q|1.852 = 638.5, divided by 14.52), so the head losses no longer cancel around the loop.
−13.1 gpm leaves the exponent 1.852 out of the denominator: −103.1/7.839.
ΔQ = −Σr Q|Q|0.852/(1.852 Σr|Q|0.852) = −7.1 gpm.
−6.8 gpm uses the exponent n = 2 (Darcy-type losses) instead of 1.852.
+7.1 gpm drops the minus sign of the correction.
Question 14 of 60 · Water Resources and Environmental Engineering · 10–15 items
A trapezoidal channel has bottom width 6.0 ft, side slopes z = 2.0 (2H:1V), bed slope 0.0015, Manning n = 0.030 and discharge 80 ft³/s. Writing f(y) = (1.486/n)A R2/3S1/2 − Q, perform one Newton–Raphson step from y0 = 2.5 ft. What is y1?
y1 = 2.5 + 7.81/57.5 = 2.636 ft; next step 2.632 ft; stop when |f(y)| or |Δy| is below tolerance
FE Reference Handbook, Mathematics — Newton's method; Water Resources — Manning's equation.
2.292 ft computes the wetted perimeter as b + 2y, leaving out the sloping side length √(1 + z²).
2.364 ft adds f/f′ to y0; the Newton step is y0 − f(y0)/f′(y0).
f(2.5) = −7.81 ft³/s and f′(2.5) = 57.5 ft²/s, so y1 = 2.5 − (−7.81/57.5) = 2.636 ft (the root is 2.632 ft).
2.686 ft uses a derivative that includes only dA/dy and ignores how R changes with depth.
9.859 ft divides f′ by f instead of f by f′.
Question 15 of 60 · Structural Engineering · 10–15 items
What is the plastic moment capacity (Mp) of a steel section in terms of yield stress (Fy) and plastic section modulus (Z)?
Solution
Mp = FyZ; My = FyS; shape factor = Z/S (1.5 for a rectangle, ≈ 1.1–1.15 for W-shapes)
FE Reference Handbook, Structural Design — steel flexure.
1.5 is the shape factor of a rectangle, Z/S; multiplying FyZ by it double-counts plasticity.
0.85 belongs to the concrete stress block (0.85f′c); steel plastic moment has no such factor.
The plastic moment is the yield stress acting over the fully plastic section: Mp = FyZ.
FyS is the yield moment My, reached when only the extreme fibers yield; for a rectangle Mp = 1.5My.
Question 16 of 60 · Structural Engineering · 10–15 items
What is the primary purpose of a sway frame analysis in structural engineering?
Solution
Sway (unbraced) frames: lateral translation Δ is a degree of freedom; second-order effects P-Δ are amplified by B2 = 1/(1 − ΣP/ΣPe2). FE Reference Handbook, Structural Analysis.
Vertical deflection is a gravity-load check; sway concerns lateral translation.
Soil-structure interaction models the foundation flexibility, a different analysis.
Torsion of frame columns is a separate member check, not the purpose of sway analysis.
A sway analysis lets the frame translate sideways, giving drift and the added P-Δ moments under wind or seismic loads.
Question 17 of 60 · Structural Engineering · 10–15 items
What does a beam-column interaction diagram represent in structural design?
Solution
Key points: pure axial P0, balanced point (εc = 0.003 with εs = εy), pure bending M0. FE Reference Handbook, Structural Design — RC columns.
The diagram plots every (Mn, Pn) pair the section can resist; a load point inside the curve is safe.
Member sizing ratios are not what the diagram plots.
Connection details are drawings of bolts and welds; an interaction diagram is a capacity envelope.
Joint shear is checked separately; the interaction diagram is for axial load with bending.
Question 18 of 60 · Structural Engineering · 10–15 items
What is the primary purpose of a load path analysis in structural engineering?
A continuous load path carries every load from where it is applied through members and connections down to the foundation.
Scheduling has nothing to do with load paths.
Construction sequence is a planning task.
Total weight is a load takedown result; load path analysis follows where the loads go.
Question 19 of 60 · Structural Engineering · 10–15 items
In seismic design, the overstrength factor in structural systems accounts for which of the following phenomena?
Solution
ASCE 7: R reduces elastic forces for design; Ω0 amplifies them for collectors and connections that must not fail; Cd amplifies elastic drifts.
Ω0 estimates how much stronger the real system is than its design strength, from material overstrength, strain hardening and redundancy.
Cracked stiffness is handled by effective section properties, not by the overstrength factor.
Modal participation is a dynamic-analysis concept unrelated to Ω0.
Ω0 scales up forces for elements that must stay elastic; it is not a reduction of dynamic effects.
Question 20 of 60 · Structural Engineering · 10–15 items
In the analysis of a continuous frame structure with a mix of fixed and pinned connections, what is the primary benefit of using moment redistribution methods when plastic hinges form?
Solution
ACI 318 permits redistribution of negative moments up to 1000εt percent, max 20 %, when εt ≥ 0.0075.
Moment redistribution concerns flexure, not axial capacity.
Redistribution depends on hinges forming; it does not prevent them.
Once a hinge forms, additional load shifts to less-stressed sections, so the frame carries more load before a mechanism forms.
Redistribution needs ductile detailing so hinges can rotate.
Question 21 of 60 · Geotechnical Engineering · 10–15 items
An embankment site has 10 ft of moist soil (γ = 115 lb/ft³) over a saturated layer (γsat = 140 lb/ft³); the water table is at 10 ft. Pumping lowers the water table to 20 ft, and the soil between 10 and 20 ft stays capillary-saturated, so its unit weight remains 140 lb/ft³. What is the change in vertical effective stress at a depth of 25 ft?
Given
Moist layer 0–10 ft: γ = 115 lb/ft³
Below 10 ft: γsat = 140 lb/ft³ (unchanged after drawdown)
Water table: 10 ft before, 20 ft after
Point at z = 25 ft
γw = 62.4 lb/ft³
Find
Change in vertical effective stress Δσ′ at 25 ft (lb/ft²)
FE Reference Handbook, Geotechnical — effective stress.
−624 has the sign reversed: lowering the water table removes pore pressure, so effective stress rises.
Total stress is unchanged, but pore pressure drops by γw × 10 ft, so effective stress does change.
312 lb/ft² is the pore pressure left at 25 ft after drawdown (62.4 × 5), not the change in effective stress.
σ stays 3,250 lb/ft²; u falls from 62.4 × 15 = 936 to 62.4 × 5 = 312 lb/ft², so σ′ rises by 624 lb/ft².
Question 22 of 60 · Geotechnical Engineering · 10–15 items
In calculating active earth pressures on a retaining wall using Rankine theory, which soil property most strongly influences the lateral pressure coefficient?
Solution
Ka = (1 − sin φ)/(1 + sin φ); pa = Kaγz − 2c√Ka
FE Reference Handbook, Geotechnical — Rankine earth pressure.
Cohesion reduces the active pressure through a separate −2c√Ka term; it does not set Ka.
Ka = tan²(45° − φ/2) depends only on φ for a level, cohesionless backfill.
A water table changes the pressure (γ′ and added water pressure) but not the coefficient Ka.
Unit weight scales the pressure p = Kaγz, not the coefficient.
Question 23 of 60 · Geotechnical Engineering · 10–15 items
In a one-dimensional consolidation test, the coefficient of consolidation (cv) is calculated using:
Solution
cv = 0.197Hdr²/t50 (log-time) or 0.848Hdr²/t90 (√t)
FE Reference Handbook, Geotechnical — consolidation.
Direct shear gives strength parameters, not consolidation coefficients.
Gradation classifies soil; it does not give cv.
SPT blow counts correlate with sand density and strength, not with consolidation rate.
cv comes from fitting the oedometer time–deformation curve (Casagrande log-time or Taylor √t): cv = TvHdr²/t.
Question 24 of 60 · Geotechnical Engineering · 10–15 items
Using the table of typical properties below, select the saturated unit weight γsat for a medium-dense sand with porosity n = 0.38, and compute its submerged (buoyant) unit weight γ′. Use γw = 9.81 kN/m³.
Soil type
Porosity n
γsat (kN/m³)
Loose sand
0.42
18.5
Medium-dense sand
0.38
19.2
Dense sand
0.32
20.1
Given
Porosity n = 0.38 (medium-dense sand)
γw = 9.81 kN/m³
Find
γsat and γ′ (kN/m³)
Solution
γ′ = γsat − γw = 19.2 − 9.81 = 9.4 kN/m³
FE Reference Handbook, Geotechnical — phase relationships, effective stress.
18.5 kN/m³ is the loose-sand row (n = 0.42).
20.1 kN/m³ is the dense-sand row (n = 0.32).
29.0 kN/m³ adds γw; buoyancy subtracts it.
15.5 kN/m³ is the dry unit weight γd = γsat − nγw = 19.2 − 0.38 × 9.81; buoyant weight starts from γsat, not γd.
The n = 0.38 row gives γsat = 19.2 kN/m³, and γ′ = γsat − γw = 19.2 − 9.81 = 9.4 kN/m³.
Question 25 of 60 · Geotechnical Engineering · 10–15 items
A rectangular footing (3 m × 2 m) carries an eccentric, inclined column load with eccentricities ex = 0.4 m, ey = 0.3 m, vertical component Pv = 600 kN, and horizontal component Ph = 120 kN. Which design-check procedure is correct to verify allowable bearing pressure?
Solution
B′ = B − 2ey = 2.0 − 0.6 = 1.4 m; L′ = L − 2ex = 3.0 − 0.8 = 2.2 m; A′ = 3.08 m²
FE Reference Handbook, Geotechnical — bearing capacity, eccentric loads.
Eccentric load concentrates pressure, so the effective footing is smaller, not larger.
Eccentricity and inclination are separate effects; Ph = 120 kN still lowers capacity through the inclination factors.
Eccentricity in both directions reduces both dimensions; keeping the full length overstates the effective area.
Bearing pressure is the vertical component on the effective area; combining vectors and ignoring eccentricity is not a recognized method.
Meyerhof's effective area (B′ = 1.4 m, L′ = 2.2 m) handles the two-way eccentricity, and the inclination factors handle the horizontal load.
Question 26 of 60 · Geotechnical Engineering · 10–15 items
A subgrade soil with a field wet unit weight of 118 lb/ft³ and moisture content of 14% is to be stabilized with cement at 6% by dry weight. After laboratory trials, the following CBR values were measured: 5% cement yields a soaked CBR of 24; 7% cement yields a soaked CBR of 36. Calculate the mass of cement required per cubic yard of in-place soil (lb/yd³) and determine whether the expected soaked CBR at 6% dosage meets the project requirement of 28.
Given
Field wet unit weight = 118 lb/ft³
Moisture content w = 14 %
Cement = 6 % of dry weight
Soaked CBR: 24 at 5 % cement, 36 at 7 %
Required soaked CBR = 28
Find
Cement per cubic yard (lb/yd³) and CBR at 6 %
Solution
γd = γ/(1 + w) = 118/1.14 = 103.5 lb/ft³
cement = 0.06 × 103.5 × 27 = 168 lb/yd³
CBR6% = 24 + (36 − 24)/2 = 30 ≥ 28
FE Reference Handbook, Geotechnical — phase relationships.
6.2 lb is the cement per cubic foot; one cubic yard holds 27 ft³.
42 extrapolates beyond the 7 % result; 6 % lies between the two tests, so CBR is about 30.
191 lb/yd³ takes 6 % of the wet unit weight; the dosage is by dry weight.
Linear interpolation between the 5 % and 7 % trials is the normal estimate at 6 %.
Question 27 of 60 · Transportation Engineering · 9–14 items
In transit operations, headway-based control strategies are used to maintain schedule reliability. Which metric is most appropriate for quantifying the regularity of bus headways at a transit stop?
Solution
TCQSM headway adherence: cvh = σheadway/mean headway; cvh ≤ 0.21 ≈ LOS A, and passenger wait ≈ (h/2)(1 + cvh²).
The scheduled headway is the plan; it says nothing about how regular the actual arrivals are.
Regularity means headways stay close to their mean; σ/mean of observed headways (TCQSM headway adherence) quantifies bunching directly.
On-time percentage measures schedule adherence, used for infrequent service; for headway-based service regularity is measured on headways.
The mean says how often buses come on average; two routes with the same mean can be regular or badly bunched.
Question 28 of 60 · Transportation Engineering · 9–14 items
A subgrade has a California Bearing Ratio (CBR) of 5. The AASHTO 1993 flexible pavement equation needs the resilient modulus MR in psi, commonly estimated as MR ≈ 1,500 × CBR. If the designer instead enters 5 directly into the MR term (the CBR, or a modulus of subgrade reaction in pci), what happens to the required structural number SN for the same design traffic?
W18 fixed → SN must rise to replace 7.4 log units. FE Reference Handbook, Transportation — AASHTO flexible pavement design.
For fixed W18, an MR of 5 instead of 7,500 psi shrinks the 2.32 log MR term, so the 9.36 log(SN + 1) term, and SN, must rise sharply to compensate.
The equation uses MR exactly as entered; nothing in it detects the wrong quantity.
The input is off by more than three orders of magnitude, and its effect is to raise SN, not lower it.
The right-hand side must still equal log W18; a smaller MR term forces SN up to compensate, not down.
The direction is right but the size is not: 2.32 × log(1,500) ≈ 7.4 log units is a very large shift.
Question 29 of 60 · Transportation Engineering · 9–14 items
An agency's sign-inventory audit measured the retroreflectivity of the white legend on four STOP signs (white on red) installed 8 years ago: Sign A 35, Sign B 28, Sign C 50 and Sign D 22 cd/lx/m². The MUTCD minimum maintained retroreflectivity for the white legend of white-on-red signs is 35 cd/lx/m². Night crash data show three rear-end collisions near Sign D last year. Which signs require replacement, and what documentation should go with the decision?
Solution
MUTCD Table 2A-3: white-on-red signs, white ≥ 35 and red ≥ 7 cd/lx/m² (measured method). Compare each sign with the minimum individually. FE Reference Handbook, Transportation — traffic control devices.
50 is not the MUTCD threshold for this legend, and Sign A meets the 35 minimum.
Sign B at 28 is also below the minimum and must be replaced too.
B (28) and D (22) are below the 35 minimum; A at exactly 35 still meets it, so it is monitored, not replaced.
A blanket replacement is an allowed management method, but it is not what the measured values require; C at 50 is well above the minimum.
The minimum applies to each sign, not to an average, and two signs are clearly below it.
Question 30 of 60 · Transportation Engineering · 9–14 items
A horizontal curve on a two-lane highway has a design speed of 60 mph and a radius of 800 ft. Agency policy limits superelevation to 8.0 %. Using the point-mass relation e + f = V²/(15R) (US units), find the side-friction factor f required with e = emax, and say whether it exceeds a limiting value of 0.12 for 60 mph.
FE Reference Handbook, Transportation — horizontal curves, superelevation.
0.38 adds e instead of subtracting it.
The value 0.22 is right, but 0.22 is larger than 0.12, so the curve does not comply.
0.30 is e + f together; the superelevation share was not subtracted.
V²/(15R) = 3,600/12,000 = 0.30, and f = 0.30 − 0.08 = 0.22 > 0.12, so the radius must grow (to 1,200 ft) or the speed drop.
−7.70 subtracts e as 8 instead of 0.08; percent must be converted to a decimal.
Question 31 of 60 · Transportation Engineering · 9–14 items
A circular horizontal curve has radius R = 600 ft and central angle Δ = 45°. Compute the arc length L, tangent length T, external distance E and long chord C.
Given
Radius R = 600 ft
Central angle Δ = 45°
Find
Arc length L, tangent T, external E and long chord C (ft)
Solution
L = RΔrad = 600 × π/4 = 471.2 ft
T = R tan(Δ/2) = 248.5 ft; E = R(1/cos(Δ/2) − 1) = 49.4 ft; C = 2R sin(Δ/2) = 459.2 ft
FE Reference Handbook, Transportation — horizontal curves.
T = 230 ft uses sin 22.5° instead of tan 22.5°.
L = RΔ = 600 × 0.7854 = 471 ft, T = R tan 22.5° = 249 ft, E = R(sec 22.5° − 1) = 49 ft, C = 2R sin 22.5° = 459 ft.
46 ft is the middle ordinate M = R(1 − cos 22.5°), not the external distance.
T = 600 ft uses tan Δ instead of tan(Δ/2).
The arc length and the long chord are swapped; the arc is always longer than the chord.
Question 32 of 60 · Construction Engineering · 8–12 items
What does the Schedule Performance Index (SPI) in Earned Value Management measure?
Solution
SPI = EV/PV; CPI = EV/AC; SV = EV − PV
FE Reference Handbook, Construction — earned value.
SPI = EV/PV: below 1.0 the project has earned less work than planned by this date (behind schedule).
SPI is a ratio at a status date, not a duration.
EVM tracks scope, cost and schedule; quality is controlled by inspection and testing.
Cost efficiency against budget is the CPI = EV/AC, not the SPI.
Question 33 of 60 · Construction Engineering · 8–12 items
In the context of construction cost estimating, what does 'contingency' refer to?
Solution
Bid = direct cost + contingency + overhead + profit (+ bond). FE Reference Handbook, Construction — cost estimating.
Temporary facilities are a known general-conditions cost and are estimated directly.
Profit is a separate markup for the contractor's return, not an allowance for unknowns.
Overhead covers indirect costs such as the home office, field office and supervision; it is priced, not a risk allowance.
Contingency is a risk allowance for items that cannot be priced yet; it shrinks as design and construction progress.
Question 34 of 60 · Construction Engineering · 8–12 items
What does a 'time impact analysis' evaluate in construction projects?
Solution
TIA: update schedule to the event date → insert fragnet → recompute CPM → difference in completion = delay attributable to the event. FE Reference Handbook, Construction — scheduling.
Equipment wear is an ownership-cost question.
Acceleration cost comes from crashing analysis, not from a time impact analysis.
A TIA inserts a fragnet for a delay event into the current schedule and measures how far the completion date moves.
Labor efficiency is measured by productivity studies.
Question 35 of 60 · Construction Engineering · 8–12 items
Working from a complete set of construction drawings and specifications, which estimating technique best accounts for both direct and indirect costs, including contingencies, when preparing a bid?
Solution
Estimate classes run from order-of-magnitude (parametric/analogous) to definitive (detailed unit cost, ±5–10 %).
Parametric estimates scale from cost per unit of size (per ft², per lane-mile); they suit early budgets, not bids.
Analogous estimates borrow the cost of a similar past project; they are too coarse for a bid.
Top-down estimating allocates a total budget to parts; it does not build cost from the documents.
A detailed estimate prices each takeoff quantity from the drawings with labor, material and equipment, then adds indirects and contingency, which is what a bid needs.
Question 36 of 60 · Statics · 8–12 items
How far above its base is the centroid of a right triangle with a base of 4 m and a height of 6 m?
Given
Base b = 4 m
Height h = 6 m
Find
Centroid distance above the base ȳ (m)
Solution
ȳ = h/3 = 6/3 = 2.00 m
FE Reference Handbook, Statics — centroids of plane areas.
1.33 m is b/3, the horizontal centroid distance measured along the base.
ȳ = h/3 = 6/3 = 2.00 m above the base.
3.00 m is h/2, the centroid of a rectangle, not a triangle.
4.00 m is 2h/3, the distance measured from the apex instead of the base.
Question 37 of 60 · Statics · 8–12 items
Three rigid blocks, A (mass 30 kg), B (mass 20 kg), and C (mass 10 kg), are stacked vertically on a horizontal surface and pulled by a horizontal force P applied to the top block A. The coefficient of static friction between A and B is 0.50, between B and C is 0.30, and between C and the ground is 0.40. Determine which contact reaches limiting friction first and the maximum force P (in N) the assembly can sustain without any slip. Use g = 9.8 m/s².
Given
mA = 30 kg
mB = 20 kg
mC = 10 kg
μAB = 0.50
μBC = 0.30
μC-ground = 0.40
g = 9.8 m/s²
Find
Governing contact and maximum P (N)
Solution
FAB,max = 0.50(30)(9.8) = 147.0 N
FBC,max = 0.30(30 + 20)(9.8) = 147.0 N
FC,max = 0.40(60)(9.8) = 235.2 N
In static equilibrium each contact carries P, so Pmax = 147.0 N, set by A–B and B–C together. FE Reference Handbook, Statics — friction.
The force is right, but the ground contact still has margin (235.2 N), so not all contacts are at the limit.
245.0 N uses the combined weight of A and B as the normal force at A–B; only block A rests on that interface.
235.2 N is the ground friction limit 0.40 × 60 × 9.8; the A–B and B–C contacts slip earlier, at 147 N.
490.0 N is the normal force on B–C (50 × 9.8); it was not multiplied by μ = 0.30.
P passes through every contact: A–B limit = 0.50 × 30 × 9.8 = 147.0 N and B–C limit = 0.30 × 50 × 9.8 = 147.0 N, both below the ground limit of 235.2 N.
Question 38 of 60 · Statics · 8–12 items
A 5 m long beam is fixed at one end and supports a 10 kN force at the free end. What is the reaction moment at the fixed support?
Given
Cantilever length L = 5 m
End load P = 10 kN
Find
Reaction moment at the fixed support M (kN·m)
Solution
ΣMA = 0 → MA = PL = 10 × 5 = 50 kN·m
FE Reference Handbook, Statics — equilibrium of rigid bodies.
10 is the end force itself; a moment needs the 5-m lever arm.
25 kN·m takes the arm as L/2, as if the load acted at midspan.
M = PL = 10 kN × 5 m = 50 kN·m (counterclockwise reaction for a downward load on a right-hand cantilever).
250 kN·m multiplies by L², the pattern for a distributed load, not a point load.
Question 39 of 60 · Statics · 8–12 items
A cantilever beam 5 m long is fixed at one end and carries two downward loads: 500 N applied at 2 m from the fixed end and 300 N applied at 4 m from the fixed end. Determine the location, measured from the fixed end, of the equivalent single resultant load.
Given
Cantilever length L = 5 m
P1 = 500 N at 2 m from the fixed end
P2 = 300 N at 4 m from the fixed end
Find
Location of the resultant from the fixed end x̄ (m)
Solution
R = 500 + 300 = 800 N
x̄ = ΣPx/R = (500 × 2 + 300 × 4)/800 = 2.75 m
FE Reference Handbook, Statics — resultants of force systems.
2.50 m is half the span; the resultant's position depends on the loads, not the beam length.
3.00 m is the plain average of 2 m and 4 m, ignoring that the 500-N load is heavier.
3.25 m swaps the load positions (500 N at 4 m, 300 N at 2 m).
Question 40 of 60 · Statics · 8–12 items
A simply supported 12-ft beam carries a triangular load rising from zero at the left support to 80 lb/ft at the right support, plus a 240-lb point load 9 ft from the left. A student placed the triangular resultant 4 ft from the left support. Where does the resultant really act, and what is the correct left reaction?
Given
Triangular load: 0 at left to 80 lb/ft at right
Span L = 12 ft
Point load P = 240 lb at 9 ft from left
Find
Location of the triangular resultant and left reaction RL (lb)
Solution
W = ½wL = ½(80)(12) = 480 lb, acting 2L/3 = 8 ft from the left
ΣMB = 0: 12RL = 480(4) + 240(3) → RL = 220 lb
FE Reference Handbook, Statics — distributed loads.
160 lb leaves out the 240-lb point load.
6 ft is midspan, the location for a uniform load, not a triangular one.
340 lb takes the point load's arm about the right support as 9 ft instead of 12 − 9 = 3 ft.
4 ft is L/3 measured from the wrong end; the centroid of the triangle sits 2L/3 = 8 ft from the zero end.
W = ½ × 80 × 12 = 480 lb at 8 ft; ΣMright = 0 → RL = (480 × 4 + 240 × 3)/12 = 220 lb.
Question 41 of 60 · Dynamics · 4–6 items
A 50 kg block slides down a frictionless inclined plane that makes an angle of 30° with the horizontal. What is the acceleration of the block?
Given
Block mass m = 50 kg
Incline angle θ = 30°
Frictionless surface
Find
Acceleration along the incline a (m/s²)
Solution
ΣFx = mg sin θ = ma → a = g sin θ = 9.81 × 0.5 = 4.91 m/s²
FE Reference Handbook, Dynamics — particle kinetics.
Along a frictionless incline only the weight component mg sin θ acts, so a = g sin 30° = 4.91 m/s² (mass cancels).
8.50 m/s² is g cos 30°, the component normal to the surface, which is balanced by the normal force.
9.81 m/s² is free fall; the incline carries part of the weight.
19.62 m/s² divides g by sin 30° instead of multiplying.
Question 42 of 60 · Dynamics · 4–6 items
A wheel with a radius of 0.4 m rotates at 120 RPM. What is the linear velocity of a point on the rim of the wheel?
Given
Wheel radius r = 0.4 m
Speed N = 120 rpm
Find
Linear speed of a rim point v (m/s)
Solution
ω = 2πN/60 = 2π(120)/60 = 12.57 rad/s
v = ωr = 12.57 × 0.4 = 5.03 m/s
FE Reference Handbook, Dynamics — rigid-body kinematics.
2.51 m/s treats 0.4 m as the diameter, halving the radius.
ω = 120 × 2π/60 = 12.57 rad/s, so v = ωr = 12.57 × 0.4 = 5.03 m/s.
10.05 m/s uses 0.8 m, the diameter, as the radius.
48.00 m/s multiplies rpm by the radius without converting rpm to rad/s.
Question 43 of 60 · Mechanics of Materials · 7–11 items
A 5-meter-long aluminum bar with a cross-sectional area of 300 mm² is heated from 20°C to 80°C. If the coefficient of thermal expansion for aluminum is 23 × 10−6/°C, what is the thermal elongation of the bar?
FE Reference Handbook, Mechanics of Materials — thermal deformation.
δ = αLΔT = 23 × 10−6 × 5,000 mm × 60 °C = 6.90 mm; the cross-sectional area does not enter free expansion.
9.20 mm uses the final temperature 80 °C as ΔT.
11.50 mm uses ΔT = 100 °C, the sum of the two temperatures.
69.00 mm reads α as 23 × 10−5/°C, a misplaced exponent that makes δ ten times too large.
Question 44 of 60 · Mechanics of Materials · 7–11 items
What is the significance of the St. Venant torsion constant (J) in structural analysis?
Solution
φ = TL/(GJ); open thin-walled J ≈ Σbt³/3, much smaller than a closed section's
FE Reference Handbook, Mechanics of Materials — torsion.
Axial stiffness is EA.
Bending deflection uses EI.
Thermal expansion is a material property α, not a section constant.
Torsional stiffness is GJ; J is the polar moment for circular shafts and the St. Venant constant (≈ Σbt³/3 for thin open sections) otherwise.
Question 45 of 60 · Mechanics of Materials · 7–11 items
A U-shaped channel beam is to be loaded in bending without inducing torsion. According to the Mechanics of Materials principles, which of the following is the best method to ensure pure bending in the U-shaped section?
Solution
Channel shear center offset from the web centerline: e = 3b²/(h + 6b) for equal thin flanges (b flange width, h web height). FE Reference Handbook, Mechanics of Materials.
Only a load through the shear center, outside the web on the side away from the flanges, produces bending without torsion.
For a channel the centroid and the shear center differ; a load at the centroid twists the section.
A load at an extreme fiber adds a torque about the shear center.
The flange centroid is not the shear center; the load would still twist the channel.
Question 46 of 60 · Mechanics of Materials · 7–11 items
What is the primary cause of shear deformation in a beam?
Solution
τ = VQ/(Ib); shear deflection matters mainly for short, deep beams
FE Reference Handbook, Mechanics of Materials — beam shear stress.
Bending moment causes curvature through normal stresses.
Axial force produces normal strain, not shear deformation.
Torsion causes twisting shear, but beam shear deformation comes from transverse shear force.
Shear force V produces shear stress τ = VQ/(Ib), which distorts each slice so neighboring sections slide.
Question 47 of 60 · Mathematics and Statistics · 8–12 items
The equation of a soil consolidation curve is given by e = 0.8 - 0.2 log(p), where e is the void ratio and p is the pressure in kPa. Find the derivative de/dp at p = 100 kPa.
Given
e = 0.8 − 0.2 log10(p)
p = 100 kPa
Find
de/dp at p = 100 kPa (kPa−1)
Solution
d(log10p)/dp = 1/(p ln 10)
de/dp = −0.2/(100 × 2.303) = −8.7 × 10−4 kPa−1
FE Reference Handbook, Mathematics — derivatives.
−0.20 is the slope per log cycle (the compression index form), not the derivative with respect to p.
−0.0046 multiplies by ln 10 instead of dividing: d(log p)/dp = 1/(p ln 10).
−0.0020 treats log as the natural logarithm, giving −0.2/p.
Question 48 of 60 · Mathematics and Statistics · 8–12 items
Calculate ∫(2x + 3)³dx.
Given
Integrand = (2x + 3)³
Find
Indefinite integral ∫(2x + 3)³dx
Solution
u = 2x + 3, du = 2dx
∫(2x + 3)³dx = ½∫u³du = u4/8 + C = (2x + 3)4/8 + C
FE Reference Handbook, Mathematics — integral calculus.
Dividing by 4 and multiplying by the inner derivative 2 gives /2; the substitution divides by the inner derivative instead.
6(2x + 3)² is the derivative of (2x + 3)³; the question asks for the antiderivative.
With u = 2x + 3, ∫u³(du/2) = u4/8, so the result is (2x + 3)4/8 + C.
This applies the power rule but forgets to divide by the inner derivative 2 (u = 2x + 3, du = 2 dx).
Question 49 of 60 · Mathematics and Statistics · 8–12 items
A water tank is a hemispherical bowl of radius R with its curved surface down. What is the volume of water when the tank is filled to a depth h (h ≤ R) measured from the bottom?
Given
Bowl radius R
Water depth h ≤ R, measured from the lowest point
Find
Water volume V
Solution
r(y)² = 2Ry − y² (y from the bottom)
V = ∫0hπ(2Ry − y²)dy = (π/3)h²(3R − h); check h = R → (2π/3)R³
FE Reference Handbook, Mathematics — mensuration of volumes.
πR²h is the volume of a cylinder of radius R; the bowl narrows toward the bottom.
Integrating circular slices, ∫0hπ(2Ry − y²)dy = πRh² − πh³/3 = (π/3)h²(3R − h), the volume of a spherical cap.
The sign is flipped; the spherical-cap volume is (π/3)h²(3R − h).
(2π/3)R³ is the full hemisphere; it is right only when h = R.
Question 50 of 60 · Mathematics and Statistics · 8–12 items
Trial
Result
1
Success
2
Failure
3
Failure
4
Success
5
Failure
6
Failure
7
Success
8
Failure
Eight independent field samples are tested for contamination; the results are in the table. Test H0: p = 0.25 against H1: p > 0.25 at α = 0.05 using the exact binomial distribution. What is the p-value and the decision?
FE Reference Handbook, Mathematics — binomial distribution, hypothesis testing.
0.208 is P(X = 3) alone; the p-value is the probability of a result at least as extreme, P(X ≥ 3).
The p-value is right, but a p-value above α = 0.05 means failing to reject, not rejecting.
0.114 is P(X ≥ 4); the observed count of 3 must be included in the upper tail.
0.679 is P(X ≤ 2), the lower tail; for H1: p > 0.25 the p-value is the upper tail.
With 3 successes in 8, P(X ≥ 3) = 1 − P(X ≤ 2) = 1 − 0.679 = 0.321 > 0.05, so H0 is not rejected.
Question 51 of 60 · Mathematics and Statistics · 8–12 items
Annual failures at three sites follow independent Poisson distributions with rates λ1 = 2.5, λ2 = 3.0 and λ3 = 1.8 per year, and each failure costs $12,000. What are the expected total annual failures, the variance of the total and the expected annual cost?
FE Reference Handbook, Mathematics — Poisson distribution.
14.6 doubles the variance; variances of independent variables add, which still gives 7.3.
3.65 halves the variance; for Poisson the variance equals the mean.
53.3 is 7.3², the squared mean, not the Poisson variance.
2.43 is the average rate per site; the total over three sites is the sum, 7.3.
A sum of independent Poisson variables is Poisson with λ = 2.5 + 3.0 + 1.8 = 7.3, so mean = variance = 7.3 and E[cost] = 12,000 × 7.3 = $87,600.
Question 52 of 60 · Ethics and Professional Practice · 4–6 items
An engineer completed a bridge design in June 2019 while insured under claims-made Policy X (2018–2020, retroactive date January 1, 2017). In January 2021 the engineer let Policy X lapse without buying an extended reporting period and bought claims-made Policy Y (2021–2023) with a retroactive date of January 1, 2021. A claim over the 2019 design is made in April 2023. Which policy responds?
Solution
Claims-made coverage requires (1) the claim to be made and reported in the policy period or an extended reporting period and (2) the services to have been performed after the retroactive date. Keeping the original retroactive date (prior acts) on a renewal avoids the gap.
Claims-made coverage is triggered by the claim date; X ended in 2020 and no tail was bought.
Making the claim in the policy period is necessary but not enough; the 2019 work predates Y's January 1, 2021 retroactive date.
Discovery date is an occurrence-policy concept; under claims-made Policy Y the retroactive date controls.
Neither policy meets both of its own conditions, so splitting the claim between them is not possible.
A claims-made policy needs the claim during the policy (or reporting) period and the work after the retroactive date; X fails the first test and Y the second.
Question 53 of 60 · Ethics and Professional Practice · 4–6 items
A wastewater plant achieved substantial completion on May 1, 2015. A latent defect in the clarifier design caused structural damage discovered on June 15, 2025. The plaintiff files suit on July 1, 2025. The state statute of limitations is 3 years from discovery; the statute of repose is 10 years from substantial completion. The contract required notice of claims within 60 days of discovery. Is the suit timely?
Solution
Statute of limitations: runs from accrual or discovery (3 yr → June 2028). Statute of repose: absolute bar from substantial completion (10 yr → May 1, 2025). The earlier bar controls.
The repose period runs from substantial completion by definition, not from discovery.
On July 1, 2025 the 60-day window (to August 14) was still open, so notice could not yet have been missed.
A statute of repose is not tolled by discovery; that is what separates it from a statute of limitations.
The suit meets the limitations period but the repose period, which runs regardless of discovery, ended May 1, 2025.
Ten years from substantial completion on May 1, 2015 ends May 1, 2025; suit on July 1, 2025 is barred even though the defect was latent.
Question 54 of 60 · Engineering Economics · 5–8 items
A renewable energy project incurs fixed costs of $100,000 per year and variable costs of $0.10 per kWh. If the electricity is sold at $0.25 per kWh, how many kWh must be sold annually to achieve a 15% profit margin over total costs? (Profit margin is defined as (Revenue - Total Cost)/Total Cost.)
766,667 kWh applies the 15 % margin to the fixed cost only, not to the total cost.
740,741 kWh applies the 15 % margin to the variable cost only: 0.25Q = 100,000 + 1.15 × 0.10Q.
666,667 kWh is the break-even quantity (zero profit), 100,000/(0.25 − 0.10).
Question 55 of 60 · Engineering Economics · 5–8 items
A bridge project has an initial construction cost of $2,000,000 and an expected service life of 40 years. Annual maintenance costs start at $30,000 and escalate at 2% per year. The bridge has a salvage value of $500,000 at the end of its life. With a discount rate of 5%, what is the approximate equivalent annual cost (EAC) of the project?
$160,700 adds the salvage value instead of subtracting it.
Question 56 of 60 · Engineering Economics · 5–8 items
A highway project costs $2,000,000 initially and has a 25-year life. A $500,000 repair may be needed: 40 % chance at year 8, 35 % chance at year 15, 25 % chance of no repair. At 6 % interest, what are the equivalent annual cost including expected repairs and the risk premium (that EAC minus the EAC with no repair)?
Given
Initial cost = $2,000,000
Life = 25 yr; i = 6 %
Repair $500,000: P = 0.40 at year 8, P = 0.35 at year 15, P = 0.25 none
FE Reference Handbook, Engineering Economics — expected value, capital recovery.
This treats one $500,000 repair as certain and undiscounted.
$156,450 is the EAC of the initial cost alone; the expected repairs belong in the EAC as well as in the premium.
This adds the expected repair cost of $375,000 without discounting it to present value.
This drops the 35 % repair at year 15 and keeps only the year-8 event.
Expected PV of repairs = 0.40 × 500,000/1.068 + 0.35 × 500,000/1.0615 = $198,500; EAC = 2,198,500 × (A/P, 6 %, 25) = $171,980, and the premium is 198,500 × 0.07823 = $15,530.
Question 57 of 60 · Surveying · 6–9 items
A surveyor measures the four angles around a single station with a total station. They sum to 359°59′20″. The second angle was recorded as 89°48′30″. What is its corrected value after the angular misclosure is distributed equally among the four angles?
Given
Number of angles = 4
Sum of recorded angles = 359°59′20″
Recorded second angle = 89°48′30″
Find
Corrected second angle
Solution
misclosure = 359°59′20″ − 360° = −40″
correction per angle = +40″/4 = +10″
89°48′30″ + 10″ = 89°48′40″
FE Reference Handbook, Surveying — angle adjustment.
89°48′20″ subtracts the correction; the angles sum to less than 360°, so each must be increased.
89°48′30″ is the recorded value; the misclosure was not distributed.
89°49′10″ puts the whole 40″ on this one angle instead of a quarter of it.
Question 58 of 60 · Surveying · 6–9 items
What is the primary reason surveyors use a state plane coordinate system for project control?
Solution
SPCS: Lambert conformal conic or transverse Mercator zones sized so the grid scale factor stays within about 1 part in 10,000, so plane formulas apply with a small scale correction.
State plane zones are projections of the national datum (NAD 83), not separate datums.
Each zone is sized so the grid scale factor stays near 1 (about 1:10,000), letting surveyors use plane coordinates with a small correction.
The system uses more than 120 zones, precisely because one projection for the whole country would distort too much.
Grid and ground distances differ by the scale and elevation factors; the design only keeps that difference small.
Question 59 of 60 · Surveying · 6–9 items
A contractor must place 15,000 compacted cubic yards (CCY) of fill. The borrow material swells 20 % when excavated (bank to loose) and shrinks 15 % when compacted, relative to its loose volume. What loose volume (LCY) must be hauled, and what bank volume (BCY) must be excavated? Round to the nearest 10 yd³.
Given
Compacted fill = 15,000 CCY
Swell = 20 % (bank → loose)
Shrinkage = 15 % (loose → compacted)
Find
Loose volume (LCY) and bank volume (BCY)
Solution
LCY = CCY/(1 − 0.15) = 15,000/0.85 = 17,650 LCY
BCY = LCY/1.20 = 17,647/1.20 = 14,710 BCY
FE Reference Handbook, Construction/Surveying — earthwork.
12,750 multiplies by 0.85 instead of dividing, as if compaction made the fill bigger than the loose volume.
Compacted volume is 85 % of loose, so LCY = 15,000/0.85 = 17,650, and bank = loose/1.20 = 14,710 BCY.
17,650 BCY repeats the loose volume; the 20 % swell was never removed.
12,500 BCY divides the compacted volume by 1.20, skipping the loose step and ignoring shrinkage.
21,180 BCY multiplies the loose volume by 1.20; bank volume is smaller than loose, so it divides.
Question 60 of 60 · Construction Engineering · 8–12 items
A highway project has BAC = $2,400,000, AC = $1,500,000 and EV = $1,440,000. The project manager expects the remaining work to be done at the originally planned efficiency. What are the estimate at completion (EAC) and the to-complete performance index (TCPI) needed to finish within the original BAC?
Given
Budget at completion BAC = $2,400,000
Actual cost AC = $1,500,000
Earned value EV = $1,440,000
Find
EAC ($) and TCPI to meet BAC
Solution
EAC = AC + (BAC − EV) = 1,500,000 + 960,000 = $2,460,000 (remaining work at planned efficiency)
FE Reference Handbook, Construction — earned value management.
$3,900,000 adds the whole BAC to AC instead of only the remaining work, BAC − EV.
Remaining work at planned efficiency adds BAC − EV = $960,000 to AC, and finishing on BAC needs (2,400,000 − 1,440,000)/(2,400,000 − 1,500,000) = 1.07.
0.94 inverts the ratio: (BAC − AC)/(BAC − EV) instead of remaining work over remaining budget.
1.00 is the TCPI to meet the new EAC, (BAC − EV)/(EAC − AC); the question asks for the TCPI to meet the original BAC.
$2,500,000 is BAC/CPI, which assumes the current CPI of 0.96 continues; the stem says future work runs at planned efficiency.
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