FE Civil Practice Exam.
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Answer: C
A z-score measures how many standard deviations a value lies from the mean, computed as (value minus mean) divided by the standard deviation. Here (28 - 32) / 2 = -2. Dropping the negative sign, using the raw difference without dividing by the standard deviation, or inverting the ratio between the difference and the standard deviation all produce numbers that don't correctly express the cylinder's position relative to the distribution.

Answer: D
Adding the two equations eliminates R2: (R1 + R2) + (2R1 - R2) = 20 + 4, which gives 3R1 = 24, so R1 = 8 kN. Confusing R1 with the resulting value of R2, simply averaging the two given totals, or making a sign error while combining the equations produces a value that does not actually satisfy both original equilibrium equations simultaneously.

Answer: D
Newton-Raphson updates the estimate as x1 = x0 - f(x0)/f'(x0). Here f(3) = 9 - 7 = 2 and f'(3) = 6, so x1 = 3 - 2/6 = 2.67. Adding the correction term instead of subtracting it, using an incorrect value for the function or its derivative, or simply leaving the estimate unchanged all fail to move the estimate toward the actual root of the function.

Answer: A
Engineering codes of ethics consistently place protection of public health, safety, and welfare above loyalty to a colleague or client. When a design flaw poses a genuine safety risk, the engineer is obligated to raise the concern through proper channels so it can be evaluated and corrected. Staying silent, delaying until harm actually occurs, or only discussing the matter casually all fail to act on the duty to safeguard the public that the discovery of the flaw creates.

Answer: D
A situation where an engineer could personally benefit from a party involved in the same project creates a conflict of interest that must be disclosed proactively, not hidden or revealed only when asked. Full disclosure, combined with stepping back from decisions affecting that supplier, allows the client to make an informed judgment and preserves the engineer's independence. Quietly accepting the arrangement or waiting to be asked both leave the client unable to evaluate the engineer's impartiality.

Answer: C
Ethical practice requires engineers to work only within their area of competence, or to bring in qualified expertise when a task falls outside it, before putting their name and seal on the results. Signing off on unfamiliar specialized work, handing it to an unsupervised junior with no technical oversight, or relying on rough judgment instead of proper analysis all risk certifying conclusions the engineer cannot actually stand behind.

Answer: A
Present worth of a single future sum is found by discounting it back using P = F / (1+i)^n. With F = $10,000, i = 8%, and n = 5, this gives 10,000 / 1.4693 ≈ $6,806. Applying the compounding factor forward instead of discounting it backward, using simple interest instead of compound interest, or adding only a single year of interest all produce values that don't correctly reflect the time value of money over the full five-year horizon.

Answer: A
The simple payback period is the initial investment divided by the annual net savings: 80,000 / 20,000 = 4 years. Confusing the payback period with the full useful life of the machine, rounding down without doing the actual division, or otherwise misapplying the ratio between cost and annual savings all give a period that does not correctly reflect when the cumulative savings actually recover the initial cost.

Answer: B
Straight-line depreciation spreads the depreciable value, which is the purchase cost minus salvage value, evenly over the useful life: (120,000 - 16,000) / 8 = 13,000 per year. Ignoring the salvage value and dividing the full purchase price by the life, or dividing by an incorrect number of years, both distort the depreciable base or the recovery period and yield an annual charge that does not match the asset's actual value decline.

Answer: B
Taking moments about the left support, the right-support reaction times the 6 m span must balance the load times its 2 m distance from the left support: R_right × 6 = 12 × 2, giving R_right = 4 kN. Assigning the value that actually belongs to the left support, using half the total load, or dividing by an incorrect span length all fail to satisfy the moment equilibrium equation about either support.

Answer: D
For free fall from rest, the fall time is found from h = 0.5gt^2, rearranged as t = √(2h/g) = √(40/9.81) ≈ 2.02 s. Omitting the square root and using 2h/g directly, or taking the square root of h/g instead of 2h/g, both produce a time that does not correctly satisfy the constant-acceleration kinematics equation governing the fall.

Answer: D
For an elevator accelerating upward, Newton's second law gives cable tension T = m(g + a), since the cable must support the weight and also provide the net upward force: 1500 × (9.81 + 1.2) = 16,515 N. Accounting only for the static weight while ignoring the acceleration, accounting only for the accelerating force while ignoring gravity, or subtracting the acceleration term instead of adding it all understate the true tension the cable must carry.

Answer: C
Conservation of energy for free fall gives the impact velocity as v = √(2gh) = √(2 × 9.81 × 3) ≈ 7.67 m/s. Forgetting to take the square root of the energy expression, taking the square root of only gh instead of 2gh, or simply halving the correct value all fail to correctly convert the potential energy lost during the fall into the kinetic energy the hammer actually gains.

Answer: A
Normal stress equals axial force divided by cross-sectional area. Converting 100 kN to 100,000 N and dividing by 500 mm^2 gives 200 N/mm^2, which is 200 MPa. Shifting a decimal place in either direction, or dividing by an incorrect area value, changes the result by a factor of ten or more and no longer reflects the actual force spread over the rod's true cross section.

Answer: A
Axial elongation is given by δ = PL / (AE). Converting to consistent units, δ = (60,000 N × 2 m) / (300×10^-6 m^2 × 200×10^9 Pa) = 0.002 m, or 2 mm. Shifting a decimal place from an area or modulus unit-conversion slip, or doubling the result by forgetting the bar is only 2 m long rather than 4 m, both produce an elongation that does not match the actual load, geometry, and stiffness of the bar.

Answer: D
The section modulus is S = (200)(400)^2/6 ≈ 5.33×10^6 mm^3, and maximum bending stress equals moment divided by section modulus: (80×10^6 N·mm) / (5.33×10^6 mm^3) = 15 MPa. Using bh^2/3 instead of bh^2/6 halves the true denominator and doubles the stress, while using bh^2/12 doubles the denominator and halves the stress; both misapply the correct section modulus formula for a rectangular section.

Answer: B
Within the elastic range, Hooke's law gives modulus of elasticity as stress divided by strain: 150 MPa / 0.00075 = 200,000 MPa, or 200 GPa. Shifting a decimal place while converting strain, or multiplying stress by strain instead of dividing, both give a modulus that would not actually reproduce the observed strain when the same stress is reapplied.

Answer: A
Allowable stress is found by dividing the ultimate strength by the factor of safety: 400 / 2.5 = 160 MPa. Multiplying instead of dividing, using a smaller factor of safety than the one actually specified, or simply subtracting a small percentage from the ultimate strength all fail to apply the intended margin between the material's ultimate capacity and the stress it is allowed to safely carry in service.

Answer: C
A brittle material typically fractures suddenly near its elastic limit with little to no visible plastic deformation, in contrast to a ductile material, which yields and stretches noticeably, often with necking, before it finally breaks. Describing a long yielding plateau, extensive plastic stretching, or high energy absorption before fracture instead describes ductile behavior, which is the opposite of how brittle materials actually respond to increasing tensile load.

Answer: A
The continuity equation requires A1V1 = A2V2, and since pipe area is proportional to diameter squared, the velocity ratio equals the square of the diameter ratio: (300/150)^2 = 4, so V2 = 2 × 4 = 8 m/s. Using the diameter ratio directly without squaring it, inverting the ratio so velocity decreases instead of increases, or squaring the already-squared ratio a second time all misapply the relationship between pipe area and flow velocity.

Answer: B
Bernoulli's equation at the same elevation gives P1 + 0.5ρV1^2 = P2 + 0.5ρV2^2. Here 200,000 + 0.5(1000)(3^2) = P2 + 0.5(1000)(6^2), so P2 = 200,000 + 4,500 - 18,000 = 186,500 Pa, or 186.5 kPa. Adding the velocity-head difference instead of subtracting it, assuming pressure is unaffected by the velocity change, or forgetting the one-half factor in the kinetic energy term all give a downstream pressure inconsistent with the actual energy trade-off between velocity and pressure.

Answer: B
Manning's equation gives V = (1/0.013)(0.6)^(2/3)(0.002)^(1/2) ≈ 76.9 × 0.711 × 0.0447 ≈ 2.45 m/s. Leaving out the 1/n roughness factor, using the slope directly instead of taking its square root, or using the hydraulic radius to the first power instead of the two-thirds power all fail to correctly combine roughness, geometry, and slope as Manning's equation requires.

Answer: D
Specific weight equals density times gravitational acceleration: γ = ρg = 1000 × 9.81 = 9,810 N/m^3, or 9.81 kN/m^3. Treating the density value itself as the specific weight ignores gravitational acceleration entirely, and shifting a decimal place in either direction through a unit-conversion slip both give a specific weight that would not correctly reproduce the actual weight of a given volume of water.

Answer: A
Pump power for raising water is P = ρgQH = 1000 × 9.81 × 0.05 × 20 = 9,810 W, or about 9.81 kW. Shifting a decimal place through a unit slip changes the result by a factor of ten, and doubling the flow rate or the head by mistake doubles the true power, both of which no longer reflect the actual combination of flow rate and head that the pump must overcome.

Answer: D
A flow-weighted mass balance gives the mixed concentration as (Q1C1 + Q2C2) / (Q1+Q2) = (0.2×150 + 4.8×2) / 5.0 = 39.6/5.0 ≈ 7.92 mg/L. Taking a simple unweighted average of the two concentrations ignores how much more river flow there is relative to the discharge, and dropping one of the two flow contributions from the numerator both distort the true dilution the discharge actually receives.

Answer: A
Stored volume equals the net inflow rate, inflow minus outflow, multiplied by the duration: (3.0 - 1.2) m^3/s × 2 hours × 3600 s/hr = 1.8 × 7,200 = 12,960 m^3. Using only the inflow rate or only the outflow rate instead of the net difference between them, or using just one hour instead of the full two-hour storm duration, all misrepresent the actual volume that accumulates in the pond over the event.

Answer: C
Average daily wastewater flow equals population multiplied by the per-capita generation rate: 25,000 × 380 L/day = 9,500,000 L/day, which converts to 9,500 m^3/day. Shifting a decimal place while converting liters to cubic meters, or substituting a different per-capita rate than the one actually given, both produce a flow that does not correctly represent the town's true daily wastewater generation.

Answer: D
Hydraulic retention time equals tank volume divided by flow rate: 900 m^3 / 0.05 m^3/s = 18,000 s, which converts to 5 hours. Shifting a decimal place while performing the division, or halving the correct time as if only part of the tank volume were available for storage, both give a retention time that does not correctly reflect how long water actually spends in the tank at the given flow rate.

Answer: D
Impervious surfaces such as pavement prevent most rainfall from soaking into the ground, so a much larger share of the rain becomes surface runoff, and it reaches drainage systems more quickly than on vegetated ground where soil and plant cover promote infiltration and slow the flow. Claiming vegetated surfaces always produce more runoff, that surface cover has no effect on infiltration, or that pavement somehow increases infiltration all contradict the basic hydrologic role that impervious cover plays in urban runoff generation.

Answer: C
The factored load combines each load type with its own load factor: 1.2 × 4 + 1.6 × 6 = 4.8 + 9.6 = 14.4 kN/m. Simply adding the unfactored dead and live loads ignores the extra margin the load factors are meant to add, swapping which factor is applied to which load type changes the weighting between dead and live effects, and applying the same average factor to both loads also fails to reflect that live loads are assigned a larger factor than dead loads.

Answer: C
Euler's formula gives Pcr = π^2 × 200×10^9 × 8×10^-6 / 4^2 = π^2 × 1.6×10^6 / 16 ≈ 986,960 N, or about 987 kN. Dividing by the column length instead of its square inflates the result far above the correct value, while dividing by an extra factor of four or two beyond the correct L^2 term produces values well below it; both errors misapply how strongly the buckling load depends on the squared unbraced length.

Answer: D
Taking moments about the right support, the left reaction times the 10 m span must balance the distributed load's total force acting at midspan plus the point load acting at its actual distance from the right support: R_left × 10 = (8×10)(5) + 20(6) = 400 + 120 = 520, so R_left = 52 kN. Using the point load's distance from the left support instead of the right support, or splitting the total load evenly between the two reactions, both ignore how the specific positions of the loads actually shift the share of load each support must carry.

Answer: B
The consolidation settlement formula gives Sc = (0.3 × 3)/(1+0.9) × log10(120/80) = (0.9/1.9) × 0.176 ≈ 0.0834 m, or 83.4 mm. Leaving the (1+e0) term out of the denominator inflates the result well above the correct value, and using the natural logarithm instead of the base-10 logarithm the formula actually calls for also produces a settlement that does not match the intended consolidation relationship for this clay layer.

Answer: D
Effective stress equals total vertical stress minus pore water pressure. Total stress is 18 × 4 = 72 kPa, and pore pressure is 9.81 × 4 = 39.24 kPa, so effective stress is 72 - 39.24 = 32.76 kPa. Reporting the total stress alone ignores the pore pressure that the water table actually generates, reporting the pore pressure alone ignores the soil's own weight, and adding the two together instead of subtracting reverses the relationship that governs how effective stress actually controls soil strength and settlement.

Answer: D
Porosity relates to void ratio through n = e / (1+e), so for e = 0.5, n = 0.5 / 1.5 ≈ 33.3%. Treating the void ratio value itself as the porosity skips the required conversion entirely, and inverting the ratio or using an incorrect denominator both give a porosity that would not correctly correspond to a void ratio of 0.5 when converted using the actual relationship between the two quantities.

Answer: A
Percent compaction equals the field dry unit weight divided by the laboratory maximum dry unit weight, expressed as a percentage: (17.5 / 19.0) × 100 ≈ 92.1%. Inverting the ratio so the laboratory value is divided by the field value instead produces a number above 100%, which does not represent a physically meaningful compaction percentage, and using an incorrect maximum dry unit weight in the denominator also shifts the result away from the true relative compaction achieved in the field.

Answer: D
The height of instrument is found by adding the backsight reading to the known benchmark elevation: 100.000 + 2.145 = 102.145 m. The foresight point elevation is then the height of instrument minus the foresight reading: 102.145 - 1.320 = 100.825 m. Adding the foresight reading instead of subtracting it, or subtracting the backsight instead of adding it when finding the height of instrument, both break the sequence of additions and subtractions that differential leveling depends on to carry elevation between points.

Answer: B
The ending station is found by adding the curve length directly to the beginning station distance: 1000 m + 250 m = 1250 m, which is written as station 12+50. Dividing the curve length by ten before adding it, or adding an incorrect multiple of the curve length, both shift the resulting station away from the point exactly 250 m further along the alignment from where the curve actually begins.

Answer: C
The fundamental traffic flow relationship gives density as flow divided by speed: k = q / v = 1,200 / 60 = 20 vehicles per kilometer. Multiplying flow by speed instead of dividing, inverting the ratio so speed is divided by flow instead, or subtracting speed from flow all violate the basic relationship linking flow, density, and speed on a roadway section.

Answer: A
Perception-reaction distance is speed times reaction time: 20 × 1.5 = 30 m. Braking distance uses v^2/(2a) = 400/(2×3.4) ≈ 58.8 m. Adding the two gives a total stopping distance of about 88.8 m. Reporting only the braking portion, only the reaction portion, or forgetting the factor of two in the braking-distance denominator each omits or overstates part of the total distance the vehicle actually travels before coming to a stop.

Answer: C
The rate of settlement is the derivative of the settlement function with respect to time. Differentiating 0.5t^2 - 2t + 10 term by term gives t - 2, and substituting t = 3 gives 1 mm/year. Dropping the constant term from the derivative, misapplying the power rule, or evaluating the position function itself instead of its derivative all lead to incorrect rates that don't reflect the true instantaneous slope of the settlement curve.

Answer: B
The arithmetic mean is the sum of all values divided by the count of values. Summing 1750, 1780, 1760, 1770, and 1740 gives 8800, and dividing by 5 gives 1760 kg/m3. Picking the median value, the minimum, or the maximum instead of actually summing and dividing all five readings gives a number that only coincidentally resembles a central tendency rather than the true average.

Answer: D
The standard error of the mean equals the population standard deviation divided by the square root of the sample size. Quadrupling the sample size from 25 to 100 doubles the square root of n, which halves the standard error to 0.2 kPa. Assuming the standard error scales directly with sample size, or assuming it is unaffected by sample size, both misrepresent the inverse square-root relationship that governs how additional samples improve the precision of an estimated mean.

Answer: C
Honest reporting of test data is a core professional obligation, and it requires disclosing actual results even when they are inconvenient, then applying the established evaluation procedure for a result that does not meet specification. Averaging away a failing result, discarding it, or substituting a convenient estimate all misrepresent what was actually measured and could allow a nonconforming condition to go undetected and unaddressed.

Answer: A
The equivalent uniform annual cost of a capital investment is found by multiplying the initial cost by the capital recovery factor: 50,000 × 0.1359 ≈ $6,795. Using straight-line depreciation without accounting for the time value of money, applying a recovery factor computed for a different interest rate, or applying a factor meant for a shorter recovery period all misstate the true annualized cost of financing the equipment over its full service life.

Answer: B
For horizontal equilibrium, the horizontal component of Cable B's tension must equal Cable A's 300 N pull: T·cos(40°) = 300, so T = 300 / cos(40°) ≈ 392 N. Using the sine of the angle instead of the cosine, ignoring the angle entirely, or multiplying by the cosine instead of dividing by it all misapply the geometry of force resolution and fail to produce a horizontal component that actually balances the opposing cable.

Answer: C
The maximum static friction force equals the coefficient of friction times the normal force, which for a horizontal surface equals the weight: F = μmg = 0.3 × 50 × 9.81 ≈ 147.2 N. Using the crate's weight alone without applying the friction coefficient, omitting gravitational acceleration from the calculation, or otherwise dropping one of the three factors in the friction formula all produce a force that would not actually match the threshold needed to overcome static friction.

Answer: C
Centripetal acceleration for circular motion is a = v^2 / r, so with v = 20 m/s and r = 150 m, a = 400/150 ≈ 2.67 m/s^2. Forgetting to square the speed before dividing by the radius, halving the radius as if it were a diameter, or dividing the squared speed by twice the radius all misapply the formula and produce an acceleration inconsistent with the actual curvature and speed given.

Answer: C
The vertical component of a force acting at an angle above horizontal equals the force magnitude times the sine of that angle: 100 × sin(30°) = 50 N. Using the cosine of the angle instead of the sine swaps the horizontal and vertical components, treating the entire force as vertical ignores the angle entirely, and using the angle value itself as if it were the component confuses degrees with newtons.

Answer: B
Average shear stress equals the shear load divided by the cross-sectional area resisting it, where area = (π/4)d^2 = (π/4)(20)^2 ≈ 314.2 mm^2. Dividing 25,000 N by 314.2 mm^2 gives about 79.6 MPa. Doubling the load path as if two shear planes resisted the force, halving the actual bolt area by mistake, or ignoring the cross-sectional area entirely all give a stress inconsistent with a true single-shear connection.

Answer: D
Hydrostatic pressure at a given depth equals fluid density times gravitational acceleration times depth: P = ρgh = 1000 × 9.81 × 5 = 49,050 Pa, or 49.05 kPa. Shifting a decimal place through a unit-conversion slip, or squaring the depth instead of using it directly, both produce a pressure that does not correctly scale linearly with depth as the hydrostatic pressure relationship requires.

Answer: A
Reynolds number is computed as Re = Vd/ν = (1.5 × 0.1) / (1.0×10^-6) = 150,000. Because pipe flow is generally considered turbulent once Reynolds number rises well above roughly 4,000, this large value indicates clearly turbulent flow. A unit-conversion slip in the diameter can shrink the computed Reynolds number by a factor of ten or one hundred, and misremembering the transition threshold can lead to labeling a genuinely turbulent flow as laminar even when the correct numerical value is used.

Answer: A
The Rational Method gives Q = (C × i × A) / 3.6 = (0.4 × 50 × 2) / 3.6 = 40 / 3.6 ≈ 11.11 m^3/s. Leaving out the unit-conversion factor entirely, or applying it twice instead of once, both produce a runoff rate that does not correctly convert the given rainfall intensity and area into consistent volumetric flow units.

Answer: C
The feed rate equals the dose concentration times the daily flow volume. Converting 0.5 m^3/s to 43,200 m^3/day, and applying a dose of 20 mg/L (equivalent to 20 g/m^3), gives 20 × 43,200 = 864,000 g/day, or 864 kg/day. Converting the flow using seconds-per-hour instead of the full seconds-per-day, or shifting a decimal place in the unit conversion, both give a feed rate that would not actually match the plant's true daily flow volume.

Answer: D
For a simply supported beam under a uniform load over its full span, the maximum midspan moment is M = wL^2/8 = 10 × 8^2/8 = 80 kN·m. Forgetting to divide by 8 leaves an unreduced value far too large, while dividing by 16 or multiplying by an extra factor of 2 instead of the correct denominator both give a moment that does not match the actual parabolic moment distribution produced by a uniform load on a simply supported span.

Answer: C
At a joint with only two members meeting at different angles and no applied load or reaction, equilibrium in both the horizontal and vertical directions can only be satisfied if both member forces are zero, since neither member can balance a component of the other along a different line of action. Assuming a maximum or doubled force at such a joint, or claiming the outcome depends on the rest of the truss geometry, both ignore that this conclusion follows directly and immediately from the equilibrium conditions at that single joint.

Answer: C
For beams of the same span, material, and loading, elastic deflection formulas show deflection is inversely proportional to the moment of inertia of the cross section, so doubling the moment of inertia halves the deflection. Claiming deflection doubles, quadruples, or stays unchanged with a larger moment of inertia all contradict the way stiffer cross sections resist bending deformation under an identical load and span.

Answer: D
Allowable bearing pressure is found by dividing the ultimate bearing capacity by the factor of safety: 450 / 3 = 150 kPa. Multiplying instead of dividing wildly overstates the safe pressure, using a smaller factor of safety than the one actually specified overstates it less severely, and simply subtracting the factor of safety from the ultimate value ignores that the factor is meant to be applied as a divisor, not a subtracted quantity.

Answer: B
Granular soils such as sands and gravels derive most of their shear resistance from friction and interlocking between individual particles, which depends heavily on confining stress, while saturated clays can exhibit meaningful shear strength from cohesive bonding between fine particles even when confining stress is low, particularly under undrained conditions. Claiming both soils behave identically, that granular soils are cohesive rather than frictional, or that clays have no strength at all all misrepresent the fundamentally different strength mechanisms of the two soil types.

Answer: D
Angular misclosure is the difference between the theoretical interior-angle sum and the actual measured sum: 540 - 538 = 2°. Reporting the raw measured sum as if it were the error itself, adding the two sums together instead of subtracting them, or assuming no misclosure exists at all each fails to isolate the actual small discrepancy that must be distributed as a correction across the traverse's measured angles.

Answer: C
The side-friction equation gives v^2/(gR) = 625/(9.81×300) ≈ 0.2124. Solving for superelevation, e = 0.2124 - 0.12 ≈ 0.092, or about 9.2%. Forgetting to subtract the assumed side-friction factor leaves the full 21.2% figure unadjusted, while adding the friction factor instead of subtracting it, or dividing the result by an extra factor of two, both distort how much of the needed lateral force is actually supplied by friction versus by banking the roadway.