FE Mechanical · mixed practice set · checked Oct 29, 2026

FE Mechanical practice exam, all 14 areas in one set

60 problems drawn from every knowledge area, in proportion to the question ranges of the specification. Each one opens a worked solution and a short note on why every other option is wrong. Use it as a diagnostic before you plan your study, and again near the end to check your pace.

Updated Oct 29, 2026 · Not affiliated with NCEES.

ExamFE Mechanical
practice problems60
questions on the exam110
exam time5 h 20 min
passing scoreNot published
CheckedOct 2026

60 FE Mechanical practice problems

0 / 60 answered · 0 correct

Question 1 of 60 · Statics · 9–14 items

A uniform rigid beam has a total length of 12.0 m and a total weight of 1200 N. It rests on two supports: a pin at x = 4.0 m and a roller at x = 10.0 m, where x = 0 is the left end. An additional concentrated downward load of 800 N is to be placed on the beam. At what exact position x must this concentrated load be placed so that the reaction force at the roller support becomes exactly zero?

Given	Beam length L = 12.0 m Beam weight W = 1200 N (uniform) Pin at x = 4.0 m Roller at x = 10.0 m Point load P = 800 N
Find	Position x for zero roller reaction (m)

Solution

ΣM_pin = 0 with R_roller = 0: 1200 N × (6.0 − 4.0) m = 800 N × (4.0 − x) m

2400 = 800(4.0 − x) → x = 1.0 m

With R_roller = 0, moments about the pin give 1200 × (6 − 4) = 800 × (4 − x), so x = 1.0 m, left of the pin.

This places the load 2 m left of the pin, the weight's own lever arm, without scaling by the weight ratio 1200/800.

3.0 m is the required distance from the pin, not the position measured from the left end.

This puts the load 3 m to the right of the pin, where it would add to the roller reaction instead of canceling it.

A load beyond the roller increases the roller reaction; it must hang over the left end to lift the beam off the roller.

Question 2 of 60 · Statics · 9–14 items

A resultant force R = (10î + 20ĵ) N and a couple moment M_O = (30k̂) N·m act at the origin O (0, 0, 0). The system is to be moved to an equivalent force-couple system at point P (4, 3, 0) m. What is the new couple moment M_P at point P?

Given	Resultant R = (10î + 20ĵ) N Couple M_O = 30k̂ N·m Point P = (4, 3, 0) m
Find	Equivalent couple M_P (N·m)

Solution

M_P = M_O + r_PO × R, r_PO = O − P = (−4, −3, 0) m

r_PO × R = [(−4)(20) − (−3)(10)]k̂ = −50k̂ N·m

M_P = 30k̂ − 50k̂ = −20.0k̂ N·m

FE Reference Handbook, Statics: moment of a force (vector form).

Using r_OP = (4, 3, 0) reverses the sign of the transfer moment.

This uses r_OP instead of r_PO and drops the original couple; the vector goes from the new point P to the force's line.

M_P = M_O + r_PO × R with r_PO = (−4, −3, 0) m: the cross product is (−4)(20) − (−3)(10) = −50, so M_P = 30 − 50 = −20 N·m.

This is only the moment of R about P; the original couple of 30 N·m must be added.

This adds both cross-product terms with the same sign: 30 + (−4)(20) + (−3)(10) = −80; r × R has the form x·R_y − y·R_x.

Question 3 of 60 · Statics · 9–14 items

A band brake system wraps around a rotating drum to restrict its motion. The drum has a radius of 0.2 m and is carrying an operating torque of 150 N · m. The wrap angle of the friction band is 270°, and the coefficient of kinetic friction between the band and the drum is μ_k = 0.3. Calculate the required tension on the slack side of the band to maintain a constant drum speed.

Given	Drum radius r = 0.2 m Torque T = 150 N·m Wrap angle θ = 270° Kinetic friction coefficient μ_k = 0.3
Find	Slack-side tension T₂ (N)

Solution

T₁ − T₂ = T/r = 150/0.2 = 750 N, T₁/T₂ = e^μβ = e^{0.3 × 3π/2} = 4.111

T₂ = 750/(4.111 − 1) = 241.1 N

FE Reference Handbook, Statics: belt friction.

This is the tight-side tension T₁; the slack side is lower by a factor of e^μβ.

750 N is the tension difference T₁ − T₂ that carries the torque, not the slack-side tension.

This uses a wrap angle of π (180°); the band wraps 270° = 4.71 rad.

T₁ − T₂ = 150/0.2 = 750 N and T₁/T₂ = e^{0.3 × 4.712} = 4.11, so T₂ = 750/(4.11 − 1) = 241.1 N.

Dividing 750 N by e^μβ instead of (e^μβ − 1) mixes up the two belt-friction equations.

Question 4 of 60 · Statics · 9–14 items

A 4 m long ladder weighing 200 N leans against a frictionless wall and rests on a rough floor with a coefficient of static friction 0.5. Determine the minimum angle between the ladder and the floor required to prevent slipping.

Given	Ladder length L = 4 m Ladder weight W = 200 N Coefficient of static friction at the floor μ_s = 0.5 Wall frictionless
Find	Minimum angle with the floor θ (°)

Solution

ΣF_x: N_wall = μN_floor = μW; ΣM_floor: N_wall·L sin θ = W·(L/2) cos θ

tan θ = 1/(2μ) = 1 → θ = 45.0°

FE Reference Handbook, Statics: friction.

tan θ = 1/μ drops the ½ that comes from the weight acting at mid-length.

With a smooth wall, moments about the floor contact give tan θ = 1/(2μ) = 1/(2 × 0.5) = 1, so θ = 45.0°.

tan θ = μ = 0.5 gives the angle of friction, not the ladder angle.

tan θ = μ/2 inverts the relation; flatter ladders need more friction, not less.

Question 5 of 60 · Statics · 9–14 items

A flat rectangular plate spans from x = 0 to x = 12 cm and y = 0 to y = 8 cm. A circular hole of radius r = 2 cm is completely cut out. The center of the circular hole is at (x = 4 cm, y = 4 cm). What is most nearly the x-coordinate of the centroid (x̄) of the remaining composite area?

Given	Plate x = 0 to 12 cm Plate y = 0 to 8 cm Hole radius r = 2 cm Hole center = (4, 4) cm
Find	Centroid x̄ (cm)

Solution

A = 12 × 8 − π(2)² = 96 − 12.57 = 83.43 cm²

x̄ = (96 × 6 − 12.57 × 4)/83.43 = 6.30 cm

FE Reference Handbook, Statics: centroids of composite areas.

The hole's first moment is subtracted, but its area is left in the denominator.

Adding the hole as a positive area pulls the centroid toward it; a cut-out is subtracted.

6.00 cm is the centroid of the solid plate; removing material left of center shifts it right.

This uses a hole area of π(1)², treating the 2 cm radius as a diameter.

x̄ = (96 × 6 − 12.57 × 4)/(96 − 12.57) = 525.7/83.4 = 6.30 cm: the hole enters as a negative area.

Question 6 of 60 · Dynamics, Kinematics, and Vibrations · 10–15 items

At a specific instant during a mechanism's operation, the velocity vectors of two distinct non-collinear points on a rigid linkage are strictly parallel and equal in magnitude. Where is the instantaneous center of rotation (ICR) located for this linkage at this exact moment?

Solution

v_B = v_A + ω × r_B/A; v_B = v_A → ω = 0 (pure translation, IC at infinity).

A fixed pivot is the instantaneous center only of a link rotating about it; this link is translating.

The center of mass has no special role in locating the instantaneous center.

A center between the points would give them opposite velocities, not equal ones.

The choice of axes cannot move the instantaneous center.

Equal, parallel velocities at two points mean ω = 0: the link translates, and the perpendiculars to the velocities never meet.

Question 7 of 60 · Dynamics, Kinematics, and Vibrations · 10–15 items

A 2 kg particle moves along a straight line with a velocity v(t) = 3t² − 2t (m/s). What is the force acting on the particle at t = 2 s?

Given	Mass m = 2 kg Velocity v(t) = 3t² − 2t m/s Time t = 2 s
Find	Force F (N)

Solution

a(t) = dv/dt = 6t − 2 → a(2) = 10 m/s²

F = ma = 2 × 10 = 20 N

m·v(2)/t = 2 × 8/2 = 8 N uses the average acceleration from rest, not the instantaneous a = dv/dt.

This is the acceleration value without multiplying by the 2 kg mass.

m·v(2) = 2 × 8 is momentum; force is the rate of change of momentum.

a = dv/dt = 6t − 2 = 10 m/s² at t = 2 s, so F = ma = 20 N.

Question 8 of 60 · Dynamics, Kinematics, and Vibrations · 10–15 items

A uniform rod of mass 4 kg and length 2 m rotates about an axis perpendicular to the rod at its center. If a 1 kg mass is attached at one end of the rod, what is the new moment of inertia about the same axis?

Given	Rod mass m = 4 kg Rod length L = 2 m End mass = 1 kg
Find	Moment of inertia I about the center (kg·m²)

Solution

I = mL²/12 + m_end(L/2)² = 4(2)²/12 + 1(1)² = 2.33 kg·m²

FE Reference Handbook, Dynamics: mass moment of inertia (slender rod).

This is the rod alone; the end mass also contributes m·r².

Rod about its center: mL²/12 = 4 × 4/12 = 1.33; the 1 kg end mass adds m·r² = 1 × 1² = 1.00, total 2.33 kg·m².

This places the 1 kg mass 2 m from the axis (the full rod length) instead of L/2 = 1 m: 1.33 + 1 × 2² = 5.33 kg·m².

This uses mL²/3 for the rod (axis at the end); the axis passes through the center.

Question 9 of 60 · Dynamics, Kinematics, and Vibrations · 10–15 items

A four-bar linkage has links with lengths 10 cm, 20 cm, 25 cm, and 15 cm. According to Grashof's criterion, what type of mechanism is this?

Given	Link lengths = 10 cm, 20 cm, 25 cm, 15 cm
Find	Mechanism type

Solution

s + l = 10 + 25 = 35 cm; p + q = 20 + 15 = 35 cm → s + l = p + q (change point)

s + l = 10 + 25 = 35 cm equals p + q = 20 + 15 = 35 cm, the Grashof boundary case: the links can line up and the motion becomes indeterminate.

A non-Grashof double rocker needs s + l > p + q; here the sums are equal.

A crank-rocker needs s + l < p + q with the shortest link adjacent to the ground; here the sums are equal.

A drag-link (double crank) needs s + l < p + q with the shortest link as the ground.

Question 10 of 60 · Dynamics, Kinematics, and Vibrations · 10–15 items

A wheel with a moment of inertia of 2 kg·m² starts from rest. A constant torque of 5 N·m is applied for 4 s and then removed; bearing friction is negligible. How far does the wheel rotate during the 3 s after the torque is removed?

Given	Moment of inertia I = 2 kg·m² Applied torque τ = 5 N·m Torque duration t₁ = 4 s Coasting time t₂ = 3 s
Find	Angle rotated after the torque is removed θ (rad)

Solution

α = T/I = 5/2 = 2.5 rad/s², ω(4 s) = 2.5 × 4 = 10 rad/s

θ = ω·t = 10 × 3 = 30 rad (no torque, no friction)

10 is the angular speed in rad/s at removal, not the angle turned in the next 3 s.

This averages the speed down to zero; once the torque is removed there is nothing to decelerate the wheel.

20 rad is the angle turned during the 4 s of acceleration, ½αt², not during the 3 s afterward.

α = 5/2 = 2.5 rad/s², so after 4 s ω = 10 rad/s; with no torque it spins at constant speed and turns 10 × 3 = 30 rad.

Question 11 of 60 · Mechanics of Materials · 9–14 items

Consider the plane stress states listed below. Which stress state produces a Mohr's circle that passes exactly through the origin of the σ-τ coordinate system?

Stress state	σ_x	σ_y	τ_xy
State 1	20 MPa	−20 MPa	0 MPa
State 2	40 MPa	0 MPa	20 MPa
State 3	10 MPa	10 MPa	10 MPa
State 4	30 MPa	30 MPa	0 MPa

Solution

σ₁σ₂ = σ_xσ_y − τ_xy² = 0 → circle passes through (0, 0)

State 3: 10 × 10 − 10² = 0 → σ₁ = 20 MPa, σ₂ = 0

State 3 meets the condition σ_xσ_y = τ_xy², so one principal stress is zero.

This circle is centered at the origin with radius 20 MPa, so it does not pass through it.

Equal normal stresses with no shear make a point at (30, 0), not a circle through the origin.

σ_y = 0 alone is not enough; with τ_xy = 20 MPa the principal stresses are 48.3 and −8.3 MPa.

A circle through the origin needs one principal stress equal to zero, i.e., σ_xσ_y = τ_xy²: for State 3, 10 × 10 = 10².

Question 12 of 60 · Mechanics of Materials · 9–14 items

An instrumented component is found to have a maximum principal stress of 120 MPa. If the normal stresses on the element are σ_x = 80 MPa and σ_y = 40 MPa, what is the absolute magnitude of the shear stress τ_xy?

Given	Maximum principal stress σ₁ = 120 MPa Normal stress σ_x = 80 MPa Normal stress σ_y = 40 MPa
Find	Shear stress \|τ_xy\| (MPa)

Solution

σ_avg = (80 + 40)/2 = 60 MPa, R = σ₁ − σ_avg = 60 MPa

τ_xy = √(R² − ((σ_x − σ_y)/2)²) = √(3600 − 400) = 56.6 MPa

FE Reference Handbook, Mechanics of Materials: principal stresses.

40 MPa is σ_x − σ_y, the full difference of normal stresses, not the shear stress.

This uses σ_x − σ_y = 40 instead of (σ_x − σ_y)/2 = 20 in the radius relation.

Center = 60 MPa, so the radius is 120 − 60 = 60 MPa; τ_xy = √(60² − 20²) = 56.6 MPa.

60 MPa is the radius of Mohr's circle, the maximum in-plane shear stress, not τ_xy.

This adds the squared terms; τ_xy² = R² − ((σ_x − σ_y)/2)².

Question 13 of 60 · Mechanics of Materials · 9–14 items

Tensor	σ_x	σ_y	τ_xy
Tensor A	50 MPa	20 MPa	30 MPa
Tensor B	20 MPa	50 MPa	30 MPa
Tensor C	50 MPa	20 MPa	−30 MPa
Tensor D	20 MPa	50 MPa	−30 MPa

Which stress state has its maximum principal stress plane oriented strictly between −45° and 0° from the positive x-axis?

Given	Tensor A: σ_x = 50, σ_y = 20, τ_xy = 30 MPa Tensor B: σ_x = 20, σ_y = 50, τ_xy = 30 MPa Tensor C: σ_x = 50, σ_y = 20, τ_xy = −30 MPa Tensor D: σ_x = 20, σ_y = 50, τ_xy = −30 MPa
Find	Tensor with θ_p between −45° and 0°

Solution

θ_p1 = ½·atan2(2τ_xy, σ_x − σ_y)

A: +31.7°; B: +58.3°; C: −31.7°; D: −58.3°

FE Reference Handbook, Mechanics of Materials: principal stresses.

Both have positive shear and therefore positive principal angles (+31.7° and +58.3°).

With σ_y > σ_x the σ₁ plane lies nearer the y-axis, at +58.3°.

D's σ₁ plane is at −58.3°, outside the −45° to 0° range.

A gives +31.7°: positive shear rotates the σ₁ plane counterclockwise from x.

tan 2θ_p = 2τ_xy/(σ_x − σ_y) = −60/30 gives θ_p = −31.7°, and with σ_x > σ_y this is the σ₁ plane.

Question 14 of 60 · Mechanics of Materials · 9–14 items

A horizontal cantilever beam is completely fixed at its right end and free at its left end. A strictly downward concentrated load is applied at the free end. Which of the following describes the internal bending moment and the resulting axial stress at the top fiber of the beam near the fixed support, according to standard solid mechanics sign convention?

Solution

Cantilever with a tip load: M(x) = −P·x (hogging) → top fiber in tension, bottom in compression.

This describes a simply supported beam under downward load, not a cantilever.

The sense of the moment and the fiber stress follow from the support and load direction alone.

The sign is right, but a hogging moment stretches the top fiber and compresses the bottom.

Positive (sagging) moment would put the top fiber in compression; this beam hogs.

A downward tip load on a cantilever bends it concave-down (hogging): the moment is negative by the usual convention and the top fiber is stretched.

Question 15 of 60 · Material Properties and Processing · 7–11 items

Indication ID	Location	Depth (mm)	Length (mm)	Echo Amplitude
1	Mid-weld	12	3	Low
2	Base metal	25	5	Moderate
3	Weld toe	2	15	High
4	Root gap	24	4	Low

Based on the ultrasonic testing (UT) indication log for a 25 mm thick structural steel weldment, which indication is most likely a critical planar defect requiring immediate repair?

Solution

Planar flaws (cracks, lack of fusion) give sharp, high-amplitude, elongated indications; volumetric flaws (porosity, slag) give low, diffuse ones.

A short, low-amplitude mid-weld reflector is typical of small porosity, a volumetric flaw.

At 25 mm depth in a 25 mm plate this is the back wall of the base metal, not a weld flaw.

A root-gap reflector of low amplitude and a back-wall echo are geometry signals, not planar cracks.

Neither shows the long, sharp, high-amplitude signature of a planar defect.

A long (15 mm), high-amplitude reflector at the weld toe near the surface fits a toe crack, a planar flaw that grows under fatigue.

Question 16 of 60 · Material Properties and Processing · 7–11 items

Alloy Type	Condition	Environment
Alloy A: 304 Stainless	As-welded	Marine
Alloy B: 316L Stainless	Stress-relieved	Marine
Alloy C: 410 Stainless	Quenched	Marine

A welded piping system carrying warm, chloride-rich seawater is experiencing premature cracking adjacent to the weldments. To replace the failing sections, an engineer evaluates the alloy configurations provided in the table. Which alloy and condition combination offers the highest resistance to stress-corrosion cracking (SCC) while maintaining adequate fatigue life?

Solution

SCC needs a susceptible alloy + tensile stress + a specific environment (chlorides for austenitic stainless). Remove any one: change alloy, relieve stress, or change the environment.

As-welded 304 keeps high residual stress and can sensitize; it is the classic chloride SCC failure.

Both combine a susceptible condition (as-welded, quenched) with chlorides.

Alloy A combines a susceptible alloy with residual stress, so it does not belong with B.

Quenched 410 martensitic stainless is hard and prone to hydrogen and stress-corrosion cracking.

Low-carbon 316L (with molybdenum) resists chloride pitting and sensitization, and stress relief removes the residual tensile stress that SCC needs.

Question 17 of 60 · Material Properties and Processing · 7–11 items

Tempering Temperature (°C)	Hardness (HRC)	Impact Energy (J)
200	58	15
300	52	25
400	45	45
500	35	75

A fully hardened martensitic steel component requires tempering to modify its mechanical properties. Using the provided empirical table, which tempering temperature should be selected to achieve a minimum Charpy impact energy of 40 J while preserving the highest possible hardness?

Solution

Tempering trades hardness for toughness; choose the lowest temperature that satisfies the toughness requirement.

200 °C keeps 58 HRC but gives only 15 J, below the 40 J requirement.

300 °C gives 25 J, still below the 40 J requirement.

400 °C is the lowest tempering temperature that meets 40 J (45 J), so it keeps the most hardness (45 HRC).

500 °C meets the toughness target but gives up hardness (35 HRC) without need.

600 °C is outside the data and would soften the steel further.

Question 18 of 60 · Material Properties and Processing · 7–11 items

A binary alloy of element A (atomic weight 50 g/mol) and element B (atomic weight 100 g/mol) has an overall composition of 40 mol% B. At an elevated temperature it separates into a solid phase α and a liquid phase L. The tie line shows 20 wt% B in α and 80 wt% B in L. What is the weight fraction of the liquid phase?

Given	Atomic weight A = 50 g/mol Atomic weight B = 100 g/mol Overall composition = 40 mol% B α phase = 20 wt% B Liquid phase = 80 wt% B
Find	Weight fraction of liquid (wt%)

Solution

w_B = (0.40 × 100)/(0.40 × 100 + 0.60 × 50) = 57.14 wt%

W_L = (w_B − C_α)/(C_L − C_α) = (57.14 − 20)/(80 − 20) = 0.619 = 61.90 wt%

FE Reference Handbook, Materials Science: binary phase diagrams (lever rule).

This applies the lever rule to the mole percent (40 − 20)/60 without converting to weight percent.

This is the fraction of the α phase, (80 − 57.14)/60, not the liquid.

57.14 wt% is the overall alloy composition in weight percent, the input to the lever rule, not the liquid fraction.

40 mol% B is 57.14 wt% B; the lever rule gives W_L = (57.14 − 20)/(80 − 20) = 61.90 wt%.

Question 19 of 60 · Fluid Mechanics · 10–15 items

Nitrogen gas (k = 1.4) flows through a high-velocity transfer line at Mach 0.8. The static temperature of the stream is 300 K. What is the stagnation temperature of the flow?

Given	Specific heat ratio γ = 1.4 Mach number M = 0.8 Static temperature T = 300 K
Find	Stagnation temperature T₀ (K)

Solution

T₀/T = 1 + ((k − 1)/2)M² = 1 + 0.2 × 0.64 = 1.128

T₀ = 300 × 1.128 = 338.4 K

FE Reference Handbook, Fluid Mechanics: compressible flow (stagnation properties).

Dividing by the factor gives the static temperature from a stagnation one; here the static value is given.

Static and stagnation temperatures are equal only for a fluid at rest.

T₀ = T(1 + (k − 1)M²/2) = 300 × (1 + 0.2 × 0.64) = 338.4 K.

This uses M instead of M²: 1 + 0.2 × 0.8 = 1.16.

This leaves out the ½ in (k − 1)/2.

Question 20 of 60 · Fluid Mechanics · 10–15 items

A centrifugal pump operates in a piping network where the pump performance curve and system head curve are mathematically defined by H_p = 50 - 2000 Q² and H_s = 20 + 3000 Q², respectively. The head terms are in meters and the volumetric flow rate Q is in m³/s. Determine the steady-state operating flow rate of this system in liters per second (L/s).

Given	Pump curve H_p = 50 − 2000Q² m System curve H_s = 20 + 3000Q² m
Find	Operating flow rate Q (L/s)

Solution

50 − 2000Q² = 20 + 3000Q² → Q² = 30/5000

Q = 0.07746 m³/s = 77.46 L/s

At the operating point 50 − 2000Q² = 20 + 3000Q², so Q = √(30/5000) = 0.07746 m³/s = 77.46 L/s.

This uses a head difference of 35 m, which neither curve supplies.

This uses only the system coefficient (3000); both Q² coefficients add when the curves are equated.

This uses only the pump coefficient (2000); the system coefficient must be added.

Subtracting the coefficients (3000 − 2000 = 1000) instead of adding them gives √(30/1000) = 0.1732 m³/s.

Question 21 of 60 · Fluid Mechanics · 10–15 items

A centrifugal pump draws water from an open reservoir 3.0 m below the pump centerline. The suction piping has a total friction head loss of 1.2 m, and the local atmospheric pressure is 101.3 kPa. Using the fluid properties in the table and g = 9.81 m/s², what is the net positive suction head available (NPSHA)?

Fluid property	Value at operating temperature
Temperature	60 °C
Vapor pressure	19.9 kPa
Density	983 kg/m³

Given	Suction lift z = 3.0 m Suction friction loss h_L = 1.2 m Atmospheric pressure p_atm = 101.3 kPa Vapor pressure p_v = 19.9 kPa Density ρ = 983 kg/m³ g = 9.81 m/s²
Find	NPSHA (m)

Solution

NPSHA = (p_atm − p_v)/(ρg) − z_lift − h_L,s

NPSHA = (101,300 − 19,900)/(983 × 9.81) − 3.0 − 1.2 = 8.44 − 4.20 = 4.24 m

FE Reference Handbook, Fluid Mechanics: pumps (NPSH).

NPSHA = (p_atm − p_v)/(ρg) − z − h_L = 81,400/(983 × 9.81) − 3.0 − 1.2 = 8.44 − 4.20 = 4.24 m.

This subtracts the suction lift but forgets the 1.2 m friction loss in the suction line.

This subtracts friction but forgets the 3.0 m suction lift.

Adding the 3.0 m treats the reservoir as above the pump; here it is a suction lift.

Adding both the lift and the friction loss reverses their effect; both reduce the head available.

Question 22 of 60 · Fluid Mechanics · 10–15 items

A sphere with diameter 10 cm moves through water at 5 m/s. If the drag coefficient is 0.47 and water density is 1000 kg/m³, what is the drag force on the sphere?

Given	Sphere diameter d = 10 cm Velocity v = 5 m/s Drag coefficient C_D = 0.47 Water density ρ = 1000 kg/m³
Find	Drag force F_D (N)

Solution

A = πd²/4 = π(0.10)²/4 = 7.85 × 10⁻³ m²

F_D = C_D·½ρV²·A = 0.47 × 0.5 × 1000 × 5² × 7.85 × 10⁻³ = 46.1 N

FE Reference Handbook, Fluid Mechanics: drag force.

This uses π(0.05)²/4, applying the diameter formula to the radius.

F_D = C_D·½ρV²·A = 0.47 × ½ × 1000 × 25 × π(0.05)² = 46.1 N, with A the projected area.

This leaves out the ½ in the dynamic pressure ½ρV².

This uses π(0.10)², treating the diameter as the radius.

Question 23 of 60 · Fluid Mechanics · 10–15 items

Nitrogen gas (k = 1.4, R = 296.8 J/(kg·K)) flows through a manifold at 300 K. A common rule treats a gas as incompressible when its Mach number is below 0.3. What is the maximum flow velocity before compressibility corrections are needed?

Given	Specific heat ratio γ = 1.4 Gas constant R = 296.8 J/(kg·K) Temperature T = 300 K Mach limit M = 0.3
Find	Maximum velocity (m/s)

Solution

c = √(kRT) = √(1.4 × 296.8 × 300) = 353.1 m/s

V_max = 0.3c = 105.9 m/s

FE Reference Handbook, Fluid Mechanics: compressible flow (Mach number).

Using R = 0.2968 kJ/(kg·K) without converting to J gives 0.3√(1.4 × 0.2968 × 300) = 3.4 m/s.

This uses the universal gas constant 8.314 in place of nitrogen's specific gas constant 296.8 J/(kg·K).

This uses 27 °C instead of 300 K; the speed of sound needs absolute temperature.

This leaves out k under the square root (the isothermal sound speed).

c = √(kRT) = √(1.4 × 296.8 × 300) = 353.1 m/s, and 0.3c = 105.9 m/s.

Question 24 of 60 · Thermodynamics · 10–15 items

A heat pump has a coefficient of performance (COP) of 3.5. If the heat pump delivers 10 kW of heating, how much power does it consume?

Given	COP = 3.5 Heating delivered Q_H = 10 kW
Find	Power input (kW)

Solution

W = Q_H/COP_HP = 10 kW/3.5 = 2.86 kW

Multiplying heat by COP gives a number larger than the heat delivered; the input is smaller.

Subtracting the COP from the heat output mixes a ratio with a power.

Dividing by COP − 1 = 2.5 converts the heat-pump COP to a refrigeration COP, which does not apply here: 10/2.5 = 4.00 kW.

COP_HP = Q_H/W, so W = 10/3.5 = 2.86 kW.

Question 25 of 60 · Thermodynamics · 10–15 items

What is the thermodynamic wet-bulb temperature?

Solution

Adiabatic saturation: h₁ + (ω₂ − ω₁)h_f = h₂ at saturation; T₂ = thermodynamic wet-bulb temperature.

Condensation begins at the dew point, reached by cooling at constant humidity ratio.

The wet bulb lies between dew point and dry bulb, but not at their average.

The thermodynamic wet-bulb temperature is the adiabatic saturation temperature: water evaporates into the air until it is saturated, with the latent heat supplied by the air itself.

In still air a wet surface does not reach a well-defined equilibrium; a psychrometer needs air movement to approximate the wet-bulb value.

Question 26 of 60 · Thermodynamics · 10–15 items

What is the stoichiometric air-fuel ratio, on a mass basis, for complete combustion of propane (C₃H₈) in air?

Given	Fuel = propane C₃H₈
Find	Stoichiometric air-fuel ratio

Solution

C₃H₈ + 5(O₂ + 3.76N₂) → 3CO₂ + 4H₂O + 18.8N₂

AF_mass = (5 × 4.76 × 28.97)/44.1 = 15.6

FE Reference Handbook, Thermodynamics: combustion processes.

This counts only the oxygen mass, 5 × 32/44.1; air also carries 3.76 mol of nitrogen per mol of O₂.

C₃H₈ + 5(O₂ + 3.76N₂): 5 × 4.76 = 23.8 mol air per mol fuel, and 23.8 × 28.97/44.1 = 15.6 kg air per kg fuel.

5 mol of O₂ per mol of propane is the oxygen demand on a molar basis, not the air-fuel ratio.

23.8 is the molar air-fuel ratio; the stem asks for the mass basis.

Question 27 of 60 · Thermodynamics · 10–15 items

What is the dew point temperature of air at 25 °C with a relative humidity of 60%?

Given	Dry-bulb T = 25 °C Relative humidity φ = 60 %
Find	Dew point (°C)

Solution

p_v = φ·p_g(25 °C) = 0.60 × 3.169 = 1.901 kPa

T_dp = T_sat(1.901 kPa) ≈ 16.7 °C (steam tables)

FE Reference Handbook, Thermodynamics: psychrometrics.

Multiplying 25 °C by 0.60 treats relative humidity as a temperature ratio; it is a ratio of vapor pressures.

p_v = 0.60 × 3.169 = 1.90 kPa, and water saturates at 1.90 kPa at about 16.7 °C.

About 19.5–20 °C is the wet-bulb temperature for this state; the dew point is lower.

The dew point equals the dry-bulb temperature only at 100% relative humidity.

Question 28 of 60 · Thermodynamics · 10–15 items

A reversed Carnot cycle operates as a heat pump between a cold reservoir at 260 K and a hot reservoir at 320 K. If the heat pump consumes 5 kW of work, what is the amount of heat delivered to the hot reservoir?

Given	Cold reservoir T_L = 260 K Hot reservoir T_H = 320 K Work input W = 5 kW
Find	Heat delivered Q_H (kW)

Solution

COP_HP = T_H/(T_H − T_L) = 320/(320 − 260) = 5.33

Q_H = COP_HP·W = 5.33 × 5 = 26.67 kW

FE Reference Handbook, Thermodynamics: Carnot cycle (heat pump).

Dividing the work by the COP inverts the definition; the heat delivered exceeds the work.

This is Q_L, the heat drawn from the cold reservoir (COP_R × W).

COP_HP = T_H/(T_H − T_L) = 320/60 = 5.33, so Q_H = 5.33 × 5 = 26.67 kW.

This adds the work to an already complete Q_H; Q_H = Q_L + W = 21.67 + 5.

Question 29 of 60 · Heat Transfer · 7–11 items

A small metal core of mass m and specific heat c_p stays at a uniform temperature. It is covered by a thin coating of thickness L and thermal conductivity k whose heat capacity is negligible, and the coating's outer surface, of area A, is exposed to a fluid with convection coefficient h. Treating the core as a lumped capacitance, which expression gives the thermal time constant τ?

Solution

R_total = L/(kA) + 1/(hA) (the thin coating is a pure resistance; its heat capacity is negligible)

C = m·c_p (core at uniform temperature)

τ = R_totalC = (m·c_p/A)(L/k + 1/h)

Dividing by the heat capacity inverts the dependence: more thermal mass must mean a longer time constant.

This multiplies capacitance by conductance, which has units of J²/(K²·s), not seconds.

Series resistances add; subtracting the convection resistance can even give a negative time constant.

This inverts the capacitance term and adds conductances as if the paths were in parallel.

The capacitance m·c_p discharges through the conduction and convection resistances in series, L/(kA) + 1/(hA), so τ = RC.

Question 30 of 60 · Heat Transfer · 7–11 items

A straight fin of constant cross section and length L is exposed to a fluid with convection coefficient h. The fin has thermal conductivity k, base temperature T_b, and cross-sectional area A. If the fin tip is perfectly insulated, which pair of boundary conditions applies to d²θ/dx² − m²θ = 0, with θ(x) = T(x) − T_∞?

Solution

θ(0) = θ_b; dθ/dx|_{x = L} = 0 → θ(x)/θ_b = cosh[m(L − x)]/cosh(mL)

FE Reference Handbook, Heat Transfer: fins.

θ(L) = 0 means the tip reaches the fluid temperature, the very long (infinite) fin case.

Equating a gradient to a temperature difference is dimensionally inconsistent; an insulated tip has zero gradient.

This is the convective-tip condition; the stem says the tip is insulated.

This puts the insulated condition at the base; the base is held at the base temperature.

The base is held at T_b, and an insulated tip passes no heat, so the temperature gradient there is zero.

Question 31 of 60 · Heat Transfer · 7–11 items

The rate of thermal radiation emitted by a blackbody is proportional to:

Solution

E_b = σT⁴, σ = 5.67 × 10⁻⁸ W/(m²·K⁴)

FE Reference Handbook, Heat Transfer: radiation.

Linear dependence describes convection with a temperature difference, not radiation.

No blackbody emission law varies with T².

T³ appears in the linearized radiation coefficient h_r ≈ 4εσT³, not in the emitted power.

Stefan-Boltzmann law: E_b = σT⁴, with T in kelvins or degrees Rankine.

Question 32 of 60 · Measurements, Instrumentation, and Controls · 5–8 items

The kinetic energy of a mass is E = ½mv². The mass is m = 10.0 kg with a standard uncertainty u(m) = 0.1 kg, and the velocity is v = 20.0 m/s with u(v) = 0.2 m/s. Assuming independent measurements, what is most nearly the expanded uncertainty of E at a 95% confidence level (k = 2)?

Given	Mass m = 10.0 kg Mass uncertainty u(m) = 0.1 kg Velocity v = 20.0 m/s Velocity uncertainty u(v) = 0.2 m/s Coverage factor k = 2
Find	Expanded uncertainty of kinetic energy U (J)

Solution

∂E/∂m = ½v² = 200 J/kg, ∂E/∂v = mv = 200 J·s/m

u_c = √((200 × 0.1)² + (200 × 0.2)²) = √(400 + 1600) = 44.7 J

U = k·u_c = 2 × 44.7 = 89.4 J

This applies the coverage factor twice (4 × 44.7).

Adding the two contributions linearly (20 + 40) before doubling overstates independent uncertainties.

This doubles only the velocity term (2 × 40 + 20) and adds contributions linearly instead of in quadrature.

E = 2000 J; u_m contributes ½v²·u(m) = 20 J and u_v contributes mv·u(v) = 40 J, so u_c = √(20² + 40²) = 44.7 J and U = 2u_c = 89.4 J.

44.7 J is the combined standard uncertainty; the 95% expanded value multiplies it by k = 2.

Question 33 of 60 · Mechanical Design and Analysis · 10–15 items

Which of the following best describes the function of a geometric tolerance of position?

Solution

Position: tolerance zone centered on the true (basic) location of a feature of size, referenced to datums.

Position tolerance sets a zone (often cylindrical) around the theoretically exact location in which the axis, center or center plane of a feature of size must lie.

Distance between surfaces is a size dimension; parallelism controls orientation.

Roughness is a surface texture callout (Ra), not a geometric tolerance.

Hole diameter is controlled by a size tolerance; position controls where the hole is.

Question 34 of 60 · Mechanical Design and Analysis · 10–15 items

What is the primary function of a flexible coupling in a power transmission system?

Solution

Flexible coupling: 1:1 torque transmission + tolerance to misalignment and vibration.

A coupling cannot transmit more torque than the shafts deliver; flexibility usually lowers its rating compared with a rigid coupling.

Speed reduction needs a gear, belt or chain drive; a coupling runs at a 1:1 ratio.

Reversing rotation takes gears; a coupling turns both shafts the same way.

A flexible coupling carries torque between two shafts while absorbing small angular, parallel and axial misalignment and some shock.

Question 35 of 60 · Mechanical Design and Analysis · 10–15 items

In a gear system that requires precise meshing, which GD&T control governs the form of the curved (involute) tooth flank itself?

Solution

Gear tooth form = profile control; runout and position control blank and bore location.

Runout of the gear blank affects meshing, but runout controls how a circular feature wobbles about an axis; it does not define the shape of the flank.

Straightness controls line elements or an axis, not a curved tooth profile.

Cylindricity controls round features such as journals, not the involute flank of a tooth.

Gear tooth flanks are involute surfaces, and profile of a surface controls the form of such a curved surface relative to its true shape.

Question 36 of 60 · Mechanical Design and Analysis · 10–15 items

When analyzing a stamped gear subjected to cyclic loading, which analysis is most critical in predicting the component's long-term performance?

Solution

Cyclic loading → fatigue: S-N data, endurance limit modified by K_f, surface and size factors.

Thermal growth is a secondary concern for a stamped gear under mechanical cyclic loads.

Under cyclic loading, cracks start at notches and tooth roots long before static strength is reached, so fatigue with K_f governs life.

Deflection matters for meshing, but it does not predict failure under repeated loading.

Modal analysis finds natural frequencies, not the cycles to failure.

Question 37 of 60 · Electricity and Magnetism · 5–8 items

How much energy is stored in a 10 μF capacitor when charged to 100 V?

Given	Capacitance C = 10 μF Voltage V = 100 V
Find	Stored energy E (J)

Solution

W = ½CV² = ½(10 × 10⁻⁶ F)(100 V)² = 0.050 J

FE Reference Handbook, Electrical and Computer Engineering: capacitors (stored energy).

C·V = 10⁻³ is the charge in coulombs, not the stored energy.

W = ½CV² = ½ × 10 × 10⁻⁶ × 100² = 0.050 J.

CV² without the ½ doubles the stored energy.

This takes 10 µF as 10⁻⁴ F; a microfarad is 10⁻⁶ F.

Question 38 of 60 · Electricity and Magnetism · 5–8 items

In a magnetic circuit, what is the relationship between magnetic flux (Φ), magnetic field intensity (H), and area (A)?

Solution

Φ = B·A = μH·A

Φ = B·A = μH·A, so for a given permeability the flux grows with both the field intensity and the cross-sectional area.

Flux is the product of flux density and area; it depends on both.

A larger magnetizing force or a larger area carries more flux, not less.

Flux density B = Φ/A falls with area at fixed flux; flux itself grows with area at fixed B.

Question 39 of 60 · Mathematics · 6–9 items

In the quadratic formula x = (−b ± √(b² − 4ac))/(2a), catastrophic cancellation can occur for the root with + when b > 0 and 4ac is very small relative to b². Which algebraically equivalent expression for that root best avoids the cancellation error?

Solution

x = (−b + √(b² − 4ac))/(2a) × (b + √(b² − 4ac))/(b + √(b² − 4ac)) = −2c/(b + √(b² − 4ac))

The factor b in the numerator makes it non-equivalent to the root.

For b > 0 the root (−b + √(b² − 4ac))/(2a) subtracts nearly equal numbers; multiplying by the conjugate gives −2c/(b + √(b² − 4ac)), which only adds them.

This is not algebraically equivalent to either root, and its denominator still subtracts nearly equal numbers.

Multiplying by c is not an equivalent form, and the subtraction that causes cancellation remains.

This equals the other root and keeps the cancellation: b − √(b² − 4ac) is the small difference.

Question 40 of 60 · Mathematics · 6–9 items

Solve the differential equation d/dx (x² y) = 3x² for y(x), and express the general solution.

Given	Equation d/dx (x²y) = 3x²
Find	General solution y(x)

Solution

∫d(x²y) = ∫3x² dx → x²y = x³ + C → y = x + C/x²

This integrates 3x² incorrectly and divides by x instead of x².

Integrating gives x²y = x³ + C, so y = x + C/x².

The constant must also be divided by x², giving C/x².

This multiplies by x² instead of dividing after integration.

Question 41 of 60 · Mathematics · 6–9 items

What is the directional derivative of f(x, y, z) = xy²z³ at the point (1, 2, 1) in the direction of the unit vector u = (1/√3)(i + j + k)?

Given	Function f(x,y,z) = xy²z³ Point (1, 2, 1) Unit vector u = (1/√3)(i + j + k)
Find	Directional derivative

Solution

∇f(1, 2, 1) = (4, 4, 12)

D_uf = ∇f·u = (4 + 4 + 12)/√3 = 20/√3 ≈ 11.5

∇f = (y²z³, 2xyz³, 3xy²z²) = (4, 4, 12) at (1, 2, 1); ∇f·u = (4 + 4 + 12)/√3 = 20/√3.

This keeps only the z-component, 12/√3 = 4√3.

This sums the gradient components without the 1/√3 that makes u a unit vector.

This is ∂f/∂z alone, without the other components or normalization.

Question 42 of 60 · Mathematics · 6–9 items

A matrix is diagonalizable if and only if:

Solution

A = PDP⁻¹ ⇔ n linearly independent eigenvectors.

A nonzero determinant means invertibility; a singular matrix can be diagonalizable and an invertible one may not be.

An n × n matrix is diagonalizable exactly when it has n linearly independent eigenvectors, which form the columns of P in A = PDP⁻¹.

Being square is required but not enough; a Jordan block is square and not diagonalizable.

Distinct eigenvalues are sufficient but not necessary; the identity matrix has repeated eigenvalues and is diagonal.

Question 43 of 60 · Probability and Statistics · 4–6 items

A manufacturing process has two options: Process A costs $5,000 with a 0.7 probability of success, and Process B costs $8,000 with a 0.9 probability of success. If success yields a return of $12,000 and failure yields $0, which process has the higher expected value?

Given	Process A cost = $5,000 Process A probability of success = 0.7 Process B cost = $8,000 Process B probability of success = 0.9 Return on success = $12,000 Return on failure = $0
Find	Process with higher expected value

Solution

EV_A = 0.7(12,000) − 5,000 = $3,400; EV_B = 0.9(12,000) − 8,000 = $2,800

EV_A = 0.7 × 12,000 − 5,000 = $3,400 and EV_B = 0.9 × 12,000 − 8,000 = $2,800, so A is higher.

$10,800 is B's expected return before subtracting its $8,000 cost.

$2,800 is B's correct net expected value, but it is lower than A's.

The net expected values differ by $600.

Question 44 of 60 · Probability and Statistics · 4–6 items

A multiple regression model predicts material strength y from temperature x₁ and pressure x₂: y = 120 + 2.5x₁ − 0.8x₂. If temperature increases by 5 °C and pressure increases by 10 units, what is the net effect on the predicted strength?

Given	Model y = 120 + 2.5x₁ − 0.8x₂ Δx₁ = +5 °C Δx₂ = +10 units
Find	Net change in predicted strength

Solution

Δy = b₁Δx₁ + b₂Δx₂ = 2.5(5) + (−0.8)(10) = +4.5

This counts only the temperature term and leaves out the pressure effect.

The size is right but the sign is reversed; the temperature term is larger.

This counts only the pressure term and leaves out the temperature effect.

Δy = 2.5(5) − 0.8(10) = 12.5 − 8.0 = +4.5.

Question 45 of 60 · Ethics and Professional Practice · 4–6 items

During the preparation of a project proposal, an engineer is provided with external design data identified as proprietary by a vendor. Later, the vendor indicates that the data might infringe on existing patents. In this situation, what is the most ethically sound course of action?

Solution

Engineers avoid using proprietary or possibly infringing information without authorization and disclose known risks to the client.

Using data that may infringe a patent exposes the client; the engineer should withhold it until its status is cleared.

Hiding a known legal risk from the client is deceptive and breaches the duty to be truthful.

Receiving data does not grant the right to use it, and the vendor's warning cannot be ignored.

Data marked proprietary is not public domain, and a possible patent conflict makes using it riskier.

Question 46 of 60 · Ethics and Professional Practice · 4–6 items

Which statement best describes the ethical considerations in selecting materials for a mechanical engineering project?

Solution

Engineering judgment balances safety (paramount), performance, environmental impact and economics.

Material selection weighs public safety first together with performance, environmental effects and economics.

Environmental impact is one factor; a material that underperforms can be unsafe.

Cost matters, but it cannot override public safety.

Ignoring safety, environment and cost leaves out duties the engineer owes the public and the client.

Question 47 of 60 · Engineering Economics · 4–6 items

A client requests an out-of-scope design change for an HVAC system. The amendment requires 12 hours of engineering time at a billable rate of $150/hr and 18 hours of CAD drafting at $90/hr. The firm applies a 20% overhead markup on all labor costs and an additional 10% profit margin on the total marked-up cost. What is most nearly the total contract amendment value?

Given	Engineering time = 12 h at $150/h CAD drafting time = 18 h at $90/h Overhead markup = 20 % on labor Profit margin = 10 % on marked-up cost
Find	Total contract amendment value ($)

Solution

Labor = 12(150) + 18(90) = $3,420

Total = 3,420 × 1.20 × 1.10 = $4,514

$3,420 is the bare labor cost; the 20% overhead and 10% profit are missing.

This applies the 10% profit but leaves out the 20% overhead.

This applies the overhead but leaves out the profit margin.

Adding 20% + 10% = 30% to labor ignores that profit is taken on the marked-up cost.

Labor = 12 × 150 + 18 × 90 = $3,420; with 20% overhead $4,104; with 10% profit on that $4,514.

Question 48 of 60 · Electricity and Magnetism · 5–8 items

Element	Voltage	Current enters at	Calculated power (P)
1	10 V	Positive terminal	50 W
2	20 V	Positive terminal	40 W
3	15 V	Negative terminal	30 W
4	12 V	Negative terminal	−60 W

A DC circuit contains four elements with the voltages and current directions shown. Power is calculated with the passive sign convention. Which element's calculated power applies the convention incorrectly?

Solution

Passive sign convention: current into the + terminal → P = +vi (absorbing); into the − terminal → P = −vi (supplying).

Current entering the negative terminal means P = −vi under the passive sign convention, so element 3 should be negative; it is listed as +30 W.

Current enters the positive terminal, so P = +vi is positive (absorbed), as listed.

Current enters the positive terminal, so the positive value is consistent.

Current enters the negative terminal, so P = −vi is negative (delivered), as listed.

Question 49 of 60 · Statics · 9–14 items

A symmetric wedge with a half-angle of α is driven between two identical blocks. Assuming the coefficient of static friction μ_s is the same at all contact surfaces between the wedge and the blocks, and friction is negligible elsewhere, what is the minimum value of μ_s required for the wedge to be self-locking?

Solution

Without the driving force, each face has normal N and friction ≤ μN; holding requires φ ≥ α → μ_s ≥ tan α

FE Reference Handbook, Statics: friction (wedges).

Friction coefficient relates to the tangent of the friction angle; a sine of the full angle has no role here.

2α is the full included angle; each face of a symmetric wedge is inclined only α to the driving direction.

A cosine falls as the wedge gets steeper, the opposite of what self-locking requires.

Doubling the requirement double-counts the two faces; each face is checked against the half-angle α.

Self-locking needs the friction angle φ at each face to be at least the half-angle α, so μ_s = tan φ ≥ tan α.

Question 50 of 60 · Dynamics, Kinematics, and Vibrations · 10–15 items

A flywheel with a mass moment of inertia of 2 kg·m² operates at 1800 rpm. How much kinetic energy is stored in the flywheel?

Given	Moment of inertia I = 2 kg·m² Speed n = 1800 rpm
Find	Stored kinetic energy KE (kJ)

Solution

ω = 1800 × 2π/60 = 188.5 rad/s

KE = ½Iω² = ½(2)(188.5)² = 35,530 J = 35.5 kJ

FE Reference Handbook, Dynamics: kinetic energy of a rigid body.

½Iω = ½ × 2 × 188.5 = 188.5 J ≈ 0.2 kJ forgets to square the angular speed.

This uses 30 rev/s as if it were rad/s; ω must be in radians per second.

This halves the correct value, as if the ½ were applied twice.

ω = 1800 × 2π/60 = 188.5 rad/s and KE = ½Iω² = ½ × 2 × 188.5² = 35.5 kJ.

Question 51 of 60 · Mechanics of Materials · 9–14 items

A material has a modulus of elasticity E = 200 GPa and a Poisson's ratio ν = 0.3. Under a uniaxial tensile stress of 150 MPa, what is the lateral strain?

Given	Elastic modulus E = 200 GPa Poisson's ratio ν = 0.3 Uniaxial stress σ = 150 MPa
Find	Lateral strain ε_lat

Solution

ε_axial = σ/E = 150/200,000 = 750 × 10⁻⁶

ε_lat = −ν·ε_axial = −0.3 × 750 × 10⁻⁶ = −225 × 10⁻⁶

FE Reference Handbook, Mechanics of Materials: uniaxial stress-strain (Poisson's ratio).

Dividing the axial strain by ν instead of multiplying inverts Poisson's ratio.

σ/E = 750 × 10⁻⁶ is the axial strain; the lateral strain is ν times smaller and opposite in sign.

ε_lat = −ν·σ/E = −0.3 × 150/200,000 = −225 × 10⁻⁶.

Under tension the bar narrows, so the lateral strain is negative.

Question 52 of 60 · Mechanics of Materials · 9–14 items

A notched beam is subjected to pure bending, and the nominal bending stress is 80 MPa. If a stress concentration factor of 2.5 applies at the notch, what is the maximum bending stress at the notch?

Given	Nominal bending stress σ_nom = 80 MPa Stress concentration factor K_t = 2.5
Find	Maximum bending stress σ_max (MPa)

Solution

σ_max = K_t·σ_nom = 2.5 × 80 MPa = 200 MPa

FE Reference Handbook, Mechanics of Materials: stress concentration.

Dividing by K_t lowers the stress; a notch raises the local stress above nominal.

80 MPa is the nominal stress itself; the notch multiplies it by K_t.

This multiplies by K_t − 1 = 1.5, which gives only the increase above nominal, not the peak stress.

σ_max = K_t·σ_nom = 2.5 × 80 = 200 MPa.

Question 53 of 60 · Fluid Mechanics · 10–15 items

Water flows in a horizontal pipe from a fully developed laminar velocity profile (α₁ = 2.0) at station 1 to a uniform profile (α₂ = 1.0) at station 2. The pipe diameter is constant, and the average velocity is 4.0 m/s. Treating the flow as frictionless with ρ = 1000 kg/m³, what is the pressure change p₂ − p₁?

Given	Station 1 correction factor α₁ = 2.0 Station 2 correction factor α₂ = 1.0 Average velocity V = 4.0 m/s Density ρ = 1000 kg/m³
Find	Pressure change p₂ − p₁ (kPa)

Solution

p₁/ρ + α₁V²/2 = p₂/ρ + α₂V²/2

p₂ − p₁ = ½ρV²(α₁ − α₂) = ½(1000)(4.0)²(2.0 − 1.0) = 8000 Pa = 8.0 kPa

The pressure rises: the laminar profile carries extra kinetic energy that is recovered as pressure when the profile flattens.

Equal mean velocities do not mean equal kinetic energy flux; the correction factor α differs between stations.

With equal average velocities, the energy equation leaves p₂ − p₁ = ½ρV²(α₁ − α₂) = ½ × 1000 × 16 × 1 = 8.0 kPa.

ρV²(α₁ − α₂) leaves out the ½ in the kinetic energy term.

This uses ρV²α₁ instead of the difference between the stations.

Question 54 of 60 · Thermodynamics · 10–15 items

Two closed batch thermodynamic processes are analyzed for energy accumulation. Process 1 operates inside a completely rigid tank. Process 2 operates inside a piston-cylinder apparatus that precisely maintains constant pressure by expanding as heat is added. Which set of thermodynamic properties correctly represents the primary energy accumulation differentials that must be evaluated for Process 1 and Process 2, respectively?

Solution

Rigid tank: W_b = 0 → Q = ΔU = m·c_vΔT

Constant pressure: W_b = pΔV → Q = ΔU + pΔV = ΔH = m·c_pΔT

This swaps the cases: enthalpy accounts for expansion work, which happens only in the piston-cylinder.

Gibbs and Helmholtz energies describe equilibrium and maximum useful work, not the heat balance of these processes.

The rigid tank is right, but heat added at constant pressure equals ΔH, not ΔG.

A rigid tank does no p·dV work, so its heat goes into internal energy, not enthalpy.

In a rigid tank no boundary work is done, so Q = ΔU; at constant pressure the boundary work p·ΔV combines with ΔU into Q = ΔH.

Question 55 of 60 · Heat Transfer · 7–11 items

An electronic component behaves as a flat opaque plate exposed to sunlight. It receives an incident solar flux of 800 W/m². The surface has a solar absorptivity α = 0.85 and a thermal emissivity ε = 0.25. The surrounding air is at T_∞ = 300 K with h = 15 W/(m²·K), and the radiation surroundings are also at 300 K. Considering only the exposed face, what is most nearly the steady-state surface temperature?

Given	Incident solar flux G = 800 W/m² Solar absorptivity α = 0.85 Thermal emissivity ε = 0.25 Air temperature T_∞ = 300 K Convection coefficient h = 15 W/(m²·K) Surroundings temperature T_sur = 300 K
Find	Steady-state surface temperature T_s (K)

Solution

α_sG = h(T − T_∞) + εσ(T⁴ − T_sur⁴)

680 = 15(T − 300) + 0.25(5.67 × 10⁻⁸)(T⁴ − 300⁴) → T ≈ 340.3 K

FE Reference Handbook, Heat Transfer: radiation and convection (surface energy balance).

This drops the radiation loss and balances absorbed solar against convection alone.

Per unit area, 0.85 × 800 = 15(T − 300) + 0.25σ(T⁴ − 300⁴); solving gives T ≈ 340.3 K.

This uses 0.85 for both; the thermal emissivity is only 0.25, so less heat is radiated away.

This uses 0.25 for both absorptivity and emissivity; solar absorption depends on α, not ε.

Question 56 of 60 · Measurements, Instrumentation, and Controls · 5–8 items

A control system has a steady-state error of 5% with proportional control. Which type of controller would most effectively eliminate this steady-state error?

Solution

e_ss = 1/(1 + K_p) for a type-0 loop with P control → 0 when an integrator (1/s) is added

More gain shrinks the error but never removes it, and it erodes stability margins.

Feed-forward compensates for a known disturbance or setpoint model; it does not drive the residual error to zero by itself.

Derivative action responds to the rate of change of error and does nothing about a constant offset.

Integral action keeps accumulating error until it is zero, raising the system type by one and removing the steady-state error to a step.

Question 57 of 60 · Mechanical Design and Analysis · 10–15 items

A structural component is subjected to a multi-stage cyclic loading profile. The estimated number of applied cycles (n_i) at different stress amplitudes (σ_i) is monitored. The material's allowable cycles to failure (N_i) at a given stress level can be approximated by the S-N relation: N_i = 10^{11 − 0.01σ_i}, where σ_i is in MPa. The component has already experienced 200,000 cycles at 400 MPa and 30,000 cycles at 500 MPa. According to Miner's rule, how many additional cycles can the component withstand at a stress level of 600 MPa before failure is predicted?

Given	S-N relation N_i = 10^{11 − 0.01·σ_i} (σ in MPa) Applied n₁ = 200,000 cycles at σ₁ = 400 MPa Applied n₂ = 30,000 cycles at σ₂ = 500 MPa Stress level σ₃ = 600 MPa
Find	Additional cycles at 600 MPa

Solution

N(400) = 10⁷, N(500) = 10⁶, N(600) = 10⁵

D = 200,000/10⁷ + 30,000/10⁶ = 0.02 + 0.03 = 0.05

n₆₀₀ = (1 − D)·N(600) = 0.95 × 10⁵ = 95,000 cycles

FE Reference Handbook, Mechanical Design and Analysis: cumulative fatigue damage (Miner's rule).

Damage so far: 200,000/10⁷ + 30,000/10⁶ = 0.05; the remaining 0.95 of life at 600 MPa (N = 10⁵) is 95,000 cycles.

This counts only the 500 MPa block (0.03) and forgets the damage from 200,000 cycles at 400 MPa.

This is the full life at 600 MPa, ignoring the damage already accumulated.

Adding the used fraction instead of subtracting it gives more life than a new part has.

Question 58 of 60 · Mechanical Design and Analysis · 10–15 items

Load ratio	Radial factor (X)	Axial factor (Y)
F_a/F_r ≤ 0.35	1.00	0.00
F_a/F_r > 0.35	0.56	1.45

A radial ball bearing carries a steady radial load of 500 N and an axial load of 250 N. The inner ring rotates and the outer ring is stationary. Using the factors in the table, what is the equivalent dynamic bearing load P?

Given	Radial load F_r = 500 N Axial load F_a = 250 N Fa/Fr ≤ 0.35: X = 1.00, Y = 0 Fa/Fr > 0.35: X = 0.56, Y = 1.45
Find	Equivalent dynamic load P (N)

Solution

F_a/F_r = 250/500 = 0.50 > 0.35 → X = 0.56, Y = 1.45

P = XF_r + YF_a = 0.56(500) + 1.45(250) = 642.5 N ≈ 643 N

FE Reference Handbook, Mechanical Design and Analysis: rolling-contact bearings.

This keeps only the axial term Y·F_a; the radial term X·F_r is still part of P.

This applies X = 1, Y = 0, valid only when F_a/F_r ≤ 0.35; here the ratio is 0.50.

F_a/F_r = 0.50 > 0.35, so P = 0.56 × 500 + 1.45 × 250 = 280 + 362.5 = 642.5 ≈ 643 N.

Simply adding the radial and axial loads ignores the X and Y factors.

This combines X = 1 with Y = 1.45, mixing factors from the two rows of the table.

Question 59 of 60 · Measurements, Instrumentation, and Controls · 5–8 items

A platinum RTD has a resistance of 100.0 Ω at 0 °C and a temperature coefficient α = 0.00385 °C⁻¹. In service it reads 119.25 Ω. Assuming the linear RTD relation, what is the measured temperature?

Given	Resistance at 0 °C R₀ = 100.0 Ω Temperature coefficient α = 0.00385 °C⁻¹ Measured resistance R_T = 119.25 Ω
Find	Measured temperature T (°C)

Solution

R_T = R₀[1 + α(T − T₀)]

T = T₀ + (R_T/R₀ − 1)/α = 0 + (1.1925 − 1)/0.00385 = 50.0 °C

19.25% is the fractional change in resistance; it has to be divided by α to become a temperature.

This normalizes the resistance change by the measured 119.25 Ω instead of the 0 °C value R₀.

R_T = R₀[1 + α(T − T₀)] → T = (119.25/100.0 − 1)/0.00385 = 0.1925/0.00385 = 50.0 °C.

This divides R/R₀ by α without subtracting 1, so it treats the whole resistance as the change.

Question 60 of 60 · Engineering Economics · 4–6 items

Maintenance of a machine costs $1,000 at the end of year 1 and increases by $200 each year after that, through year 8. At an interest rate of 10% per year, what is the present worth of the maintenance costs?

Given	First-year cost A = $1,000 (end of year 1) Annual increase G = $200 Period n = 8 years Interest rate i = 10% per year
Find	Present worth P ($)

Solution

(P/A, 10%, 8) = 5.3349, (P/G, 10%, 8) = 16.0287

P = 1,000(5.3349) + 200(16.0287) = 5,335 + 3,206 = $8,541

FE Reference Handbook, Engineering Economics: arithmetic gradient present worth factor.

This discounts only the $1,000 base series and leaves out the $200 yearly increase.

Treating the $200 increase as a uniform series (1,200 per year) overstates the early costs and ignores how the gradient builds.

P = A(P/A, 10%, 8) + G(P/G, 10%, 8) = 1,000 × 5.3349 + 200 × 16.0287 = 5,335 + 3,206 ≈ $8,541.

Averaging the costs to $1,700 per year ignores that the larger costs come later and are discounted more.

This adds the eight payments without discounting them at 10%.

Answer every question to see your score

How this set is built

Sized by the specification. Each area gets roughly the midpoint of its published range, scaled to this set: the most problems for statics, dynamics, mechanics of materials, fluid mechanics, thermodynamics and mechanical design, the fewest for measurements and controls and for the four shared FE areas.
No repeats from the area pages. Every problem here appears only on this page. The math, statistics, ethics and economics problems are drawn from the shared FE pool, but not the ones published on the FE practice test or FE engineering economics.
Both unit systems. The exam uses SI and USCS, so the set does too.
Answer by computation. Most problems give a short list of known values and the quantity to find. Work them with the Handbook's formulas rather than recognizing a familiar answer.

Pace for the whole exam, not one problem

The exam gives 5 h 20 min for 110 questions, a little under three minutes each, and a few are unmarked trial items that carry no score. Answer every question: a wrong answer costs nothing, because the score counts correct answers only.

This set and the NCEES practice exams

For this discipline NCEES offers a pair of interactive practice exams (2 × 50 questions, $50 each). They are written by the exam's owner and run in a timed, exam-like interface, so they are the closest preview of the real screen.

This page is not affiliated with NCEES and does not reproduce its questions. Its job is different: a free set you can repeat, with a worked solution for each problem and a pointer to the area page where that topic is explained. The two work together: this set while you study each area, a timed NCEES exam as a rehearsal near the end.

The exam behind the set

Checked Oct 29, 2026. Sources are numbered and listed under Source notes. Format and test-day rules in full: FE Mechanical exam info · shared FE format: FE exam format.
Fact	Detail	Source	Checked
Questions	110 questionsIncludes unscored pretest questions that are not marked.	NCEES: FE exam · NCEES Examinee Guide, May 2026 (PDF) [1][2]	Oct 2026
Question types	Multiple choice plus alternative item types (multiple correct, point and click, drag and drop, fill in the blank)No partial credit on alternative items.	NCEES Examinee Guide, May 2026 (PDF) [2]	Oct 2026
Exam time	5 hours 20 minutes	NCEES: FE exam · NCEES Examinee Guide, May 2026 (PDF) [1][2]	Oct 2026
Scheduled break	25 minutes, after about half the questionsYou review and submit the first half before the break and cannot return to it; unscheduled breaks run on the exam clock.	NCEES: FE exam · NCEES Examinee Guide, May 2026 (PDF) [1][2]	Oct 2026
Reference supplied	FE Reference Handbook, on screen as a searchable PDFThe current version is free in your MyNCEES account; on exam day search with the box on the left (Ctrl+F does not work); no personal copies.	NCEES Examinee Guide, May 2026 (PDF) · NCEES: FE exam [1][2]	Oct 2026
Calculator	Any Casio fx-115 or fx-991 model, the HP 33s or HP 35s, or any TI-30X or TI-36X model; one in the room, plus an on-screen TI-30XSThe list is reviewed every year; spares stay in your locker.	NCEES: calculator policy (2026 list) · NCEES Examinee Guide, May 2026 (PDF) [2][3]	Oct 2026
Passing score	Not published (a scaled score against a minimum ability level)The number of correct answers is converted to a scaled score; wrong answers cost nothing; first-time and repeat takers are graded to the same standard.	NCEES: exam scoring [4]	Oct 2026

Area-by-area sets

Mathematics6–9 items on the exam
Probability and Statistics4–6 items on the exam
Ethics and Professional Practice4–6 items on the exam
Engineering Economics4–6 items on the exam
Electricity and Magnetism5–8 items on the exam
Statics9–14 items on the exam
Dynamics, Kinematics, and Vibrations10–15 items on the exam
Mechanics of Materials9–14 items on the exam
Material Properties and Processing7–11 items on the exam
Fluid Mechanics10–15 items on the exam
Thermodynamics10–15 items on the exam
Heat Transfer7–11 items on the exam
Measurements, Instrumentation, and Controls5–8 items on the exam
Mechanical Design and Analysis10–15 items on the exam

Practice tests for the other disciplines

FE Civil practice test60 FE Civil problems across all 14 sections, with worked solutions
FE Electrical and Computer practice test60 problems spread by the July 2020 specification, every option explained
FE Environmental practice testProblems in the proportions of the specification, every step worked
FE Chemical practice testProblems weighted like the July 2020 Chemical outline, every option explained
PE Mechanical HVAC & Refrigeration practice testA 60-problem mixed set in USCS units, weighted like the October 2025 outline

Source notes

NCEES: FE exam ncees.org · checked Oct 29, 2026
NCEES Examinee Guide, May 2026 (PDF) ncees.org · checked Oct 29, 2026
NCEES: calculator policy (2026 list) ncees.org · checked Oct 29, 2026
NCEES: exam scoring ncees.org · checked Oct 29, 2026

FE Mechanical practice exam, quick answers

Is there an FE Mechanical practice exam PDF?

Not here: the problems run in the browser, so each solution opens as you answer. The NCEES practice exams are interactive too, sold through the NCEES exam prep store. Before trusting a free PDF, check its date: material written before the July 2020 specification follows an older topic list.

How many problems should I work through before test day?

There is no official number. A workable target is enough timed problems in every area that you can find the right Handbook page quickly; the study guide splits the hours by the weight of each area.

Did these problems come from an NCEES exam form?

No. Exam forms are confidential and every candidate signs a nondisclosure agreement. These are original problems written to the published specification.

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