FE Mechanical · knowledge area 14 · 10–15 questions

FE mechanical design review, from stress to part

Mechanical design and analysis is 10–15 questions and the area most specific to mechanical engineering: none of the other FE disciplines on this site has a machine-design section. It takes stresses from mechanics of materials and asks whether a shaft, spring, bolt, weld or bearing survives, and for how long.

Updated Oct 29, 2026 · Not affiliated with NCEES.

ExamFE Mechanical
questions on the exam10–15
worked problems here18
of 14 knowledge areas14
CheckedOct 2026

What the area covers

The specification has the longest topic list of the outline. Grouped:

Stress and failure: stress analysis of machine elements, static failure theories, fatigue.
Stiffness: deformation of parts, springs.
Pressure vessels and piping: thin- and thick-wall stresses.
Elements: bearings, power screws, power transmission (shafts, keys, gears, belts, couplings).
Joining: welds, adhesives and mechanical fasteners.
Manufacturability: limits and fits, quality and reliability.
Components: hydraulic, pneumatic and electromechanical parts.
Drawings: engineering drawings and geometric dimensioning and tolerancing (GD&T).

18 mechanical design problems, every option explained

0 / 18 answered · 0 correct

Question 1 of 18 · Mechanical Design and Analysis · 10–15 items

A solid shaft is subjected to an alternating bending stress of 100 MPa and a mean axial stress of 150 MPa. The material properties are ultimate tensile strength S_ut = 600 MPa and fully corrected endurance limit S_e = 250 MPa. Fatigue stress concentration factors apply to both load cases, where K_f = 1.5 for the alternating stress and K_fm = 1.5 for the mean stress. According to the Goodman criterion, what is most nearly the fatigue factor of safety?

Given	Alternating stress σ_a = 100 MPa Mean stress σ_m = 150 MPa Ultimate strength S_ut = 600 MPa Endurance limit S_e = 250 MPa K_f = 1.5 K_fm = 1.5
Find	Fatigue factor of safety n

Solution

σ_a = K_fσ_a,nom = 1.5 × 100 = 150 MPa, σ_m = K_fmσ_m,nom = 1.5 × 150 = 225 MPa

1/n = σ_a/S_e + σ_m/S_ut = 150/250 + 225/600 = 0.975 → n = 1.03

FE Reference Handbook, Mechanical Design and Analysis: fatigue (modified Goodman).

Dividing both components by S_ut ignores the much lower endurance limit for the alternating part.

This ignores both stress concentration factors and uses the nominal stresses.

1.28 comes from the Gerber parabola; the stem asks for the Goodman line, which is more conservative.

This applies K_f to the alternating stress only; the stem gives K_fm = 1.5 for the mean stress too.

With K_f on both components, σ_a = 150 MPa and σ_m = 225 MPa; Goodman gives 1/n = 150/250 + 225/600 = 0.975, so n = 1.03.

Question 2 of 18 · Mechanical Design and Analysis · 10–15 items

A standard square key of width W and height H (W = H) transmits torque between a solid shaft and a gear hub. The key is manufactured from a ductile material where the allowable bearing stress is strictly twice the allowable shear stress, corresponding to the maximum shear stress theory. As the steady torque increases, which structural failure mode dictates the required minimum length of the key?

Solution

Shear: T = τ·(W·L)·D/2 → L_s = 2T/(τWD)

Bearing: T = σ_b·(H/2·L)·D/2 → L_b = 4T/(σ_bHD) = 2T/(τWD) when σ_b = 2τ, W = H

Hub strength is not part of key sizing here and nothing in the stem suggests a weak hub.

Shear alone would govern only if the bearing allowable were more than twice the shear allowable.

Shaft strength is a separate check; the question asks which key failure mode sets the key length.

Bearing acts on half the key height, but its allowable is twice the shear allowable, so it ties with shear rather than governing.

Shear needs L = 2T/(τ_allWD) and bearing needs L = 4T/(σ_allHD); with W = H and σ_all = 2τ_all the two lengths are identical.

Question 3 of 18 · Mechanical Design and Analysis · 10–15 items

Bolt	x-coordinate (mm)	y-coordinate (mm)
A	50	50
B	50	−50
C	−50	−50
D	−50	50

A bracket is secured to a wall by four identical bolts arranged in a square pattern, as detailed in the coordinate table. The bolt group centroid is at (0, 0). A downward eccentric vertical force of 20 kN is applied at x = 150 mm, y = 0 mm. Determine the magnitude of the resultant shear force on the most critically stressed bolt.

Given	Load P = 20 kN (downward) Load position x = 150 mm, y = 0 mm Bolts at (±50, ±50) mm, centroid (0, 0)
Find	Resultant shear on the most stressed bolt (kN)

Solution

Primary: V = 20/4 = 5.0 kN; r = √(50² + 50²) = 70.7 mm, M = 20 × 150 = 3000 kN·mm

Secondary: F'' = M·r/Σr² = 3000 × 70.7/(4 × 5000) = 10.61 kN

Bolt A: F = (−7.50, 7.50) + (0, 5.00) = (−7.50, 12.50) kN → |F| = √(7.50² + 12.50²) = 14.58 kN

FE Reference Handbook, Mechanical Design and Analysis: eccentrically loaded bolt groups.

This is the resultant at bolts C and D, where the moment-induced shear partly cancels the primary shear; they are the least loaded.

This is the moment-induced (secondary) shear alone; the direct shear of 5 kN must be added as a vector.

Primary shear 20/4 = 5 kN down on each bolt; the 3000 kN·mm moment adds 3000 × 70.7/(4 × 70.7²) = 10.61 kN perpendicular to each radius. At bolts A and B the vector sum is 14.58 kN.

Adding 5 + 10.61 arithmetically assumes the two forces are parallel; at the corner bolts they are 45° apart.

Using only the x-distances (M·x/Σx²) treats the moment shear as purely vertical and ignores the y-terms of the polar moment.

Question 4 of 18 · Mechanical Design and Analysis · 10–15 items

Loading case	σ₁ (MPa)	σ₂ (MPa)	σ₃ (MPa)
Alpha	100	100	100
Beta	100	0	−100
Gamma	100	50	0
Delta	120	0	0

A ductile metal component with a yield strength of σ_y = 180 MPa is evaluated for failure under four different principal stress loading cases. Based on the data in the table, which loading case will be predicted to fail under the Maximum Shear Stress (Tresca) criterion, but will be considered safe under the Distortion Energy (von Mises) criterion?

Given	Yield strength σ_y = 180 MPa Alpha: σ₁ = 100, σ₂ = 100, σ₃ = 100 MPa Beta: σ₁ = 100, σ₂ = 0, σ₃ = −100 MPa Gamma: σ₁ = 100, σ₂ = 50, σ₃ = 0 MPa Delta: σ₁ = 120, σ₂ = 0, σ₃ = 0 MPa
Find	Case failing by Tresca but safe by von Mises

Solution

Tresca: σ₁ − σ₃ ≥ S_y; von Mises: σ' = √{[(σ₁ − σ₂)² + (σ₂ − σ₃)² + (σ₃ − σ₁)²]/2} ≥ S_y

Beta: 200 MPa (Tresca, fails) vs 173 MPa (von Mises, safe)

FE Reference Handbook, Mechanical Design and Analysis: failure theories.

Beta falls in the region where the two criteria disagree; pure shear is where Tresca is about 15% more conservative.

Hydrostatic stress (100, 100, 100) has zero shear and zero distortion energy, so neither criterion predicts yielding.

For uniaxial stress both criteria give 120 MPa, below the 180 MPa yield strength.

Gamma gives 100 MPa by Tresca and 87 MPa by von Mises, both safely below 180 MPa.

Beta: Tresca uses σ₁ − σ₃ = 200 MPa > 180 MPa (yields), while von Mises gives σ' = 173 MPa < 180 MPa (safe).

Question 5 of 18 · Mechanical Design and Analysis · 10–15 items

A shaft carries a torque of 5000 lbf·in and a bending moment of 3000 lbf·in. Using the maximum shear stress (Tresca) theory, what is the minimum shaft diameter if the allowable shear stress is 12,000 psi?

Given	Torque T = 5000 lbf·in Bending moment M = 3000 lbf·in Allowable shear stress τ_all = 12,000 psi
Find	Minimum shaft diameter d (in.)

Solution

√(M² + T²) = √(3000² + 5000²) = 5831 lbf·in

d = [16 × 5831/(π × 12,000)]^1/3 = 1.35 in.

FE Reference Handbook, Mechanical Design and Analysis: shafts (combined bending and torsion).

This sizes the shaft for torque only and ignores the bending moment.

Tresca: τ_max = 16√(M² + T²)/(πd³) → d = [16 × 5831/(π × 12,000)]^1/3 = 1.35 in.

Adding M + T = 8000 lbf·in linearly overstates the combined load; they combine as √(M² + T²).

Using 32 instead of 16 in the shear formula is the bending-stress coefficient; the shear criterion uses 16.

Question 6 of 18 · Mechanical Design and Analysis · 10–15 items

A helical compression spring has a wire diameter of 0.25 inches, a mean coil diameter of 2.5 inches, and 12 active coils. If the spring material has a shear modulus of 11.5 × 10⁶ psi, what is the spring rate (stiffness)?

Given	Wire diameter d = 0.25 in Mean coil diameter D = 2.5 in Active coils n = 12 Shear modulus G = 11.5 × 10⁶ psi
Find	Spring rate k (lb/in)

Solution

k = Gd⁴/(8D³N) = (11.5 × 10⁶)(0.25)⁴/(8 × 2.5³ × 12) = 29.9 lbf/in

FE Reference Handbook, Mechanical Design and Analysis: helical linear springs.

Using 16 instead of 8 in the denominator halves the rate; the spring-rate formula has 8D³N.

Using D + d = 2.75 in. (the outer coil diameter) in 8D³N gives 22.5 lbf/in; the formula uses the mean diameter 2.5 in.

This counts 14 total coils; only the 12 active coils deflect.

k = Gd⁴/(8D³N) = 11.5 × 10⁶ × 0.25⁴/(8 × 2.5³ × 12) = 29.9 lbf/in.

Question 7 of 18 · Mechanical Design and Analysis · 10–15 items

A deep-groove ball bearing is subjected to an equivalent dynamic load of 6 kN. The manufacturer specifies a basic dynamic load rating (C) of 30 kN. Calculate the basic rating life (L₁₀) of this bearing in revolutions.

Given	Equivalent load P = 6 kN Basic dynamic rating C = 30 kN Ball bearing exponent a = 3
Find	Rating life L₁₀ (revolutions)

Solution

L₁₀ = (C/P)^a × 10⁶ rev, a = 3 (ball)

L₁₀ = (30/6)³ × 10⁶ = 125 × 10⁶ rev

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

The load ratio must be raised to the power 3; using it linearly understates the life.

Squaring the load ratio uses the wrong exponent; ball bearings use 3.

Dividing 125 by the exponent confuses the power law with a factor.

For a ball bearing L₁₀ = (C/P)³ × 10⁶ rev = 5³ × 10⁶ = 125 million revolutions.

The exponent 10/3 applies to roller bearings; a deep-groove ball bearing uses 3.

Question 8 of 18 · Mechanical Design and Analysis · 10–15 items

Temperature (°C)	Allowable shear stress (MPa)
200	150
300	130

A bracket is joined to a column by two symmetric parallel fillet welds. Each weld is 50 mm long with a 7.07 mm leg. A shear load of 60 kN acts parallel to the welds, and the joint operates continuously at 260 °C. What is the primary shear stress in the weld throat, and does the connection meet the allowable stress in the table?

Given	Number of welds = 2 Weld length L = 50 mm Leg size h = 7.07 mm Shear load P = 60 kN Operating temperature T = 260 °C Allowable shear at 200 °C = 150 MPa Allowable shear at 300 °C = 130 MPa
Find	Throat shear stress τ (MPa) and acceptance

Solution

A_t = n·L·(0.707h) = 2 × 50 × (0.707 × 7.07) = 500 mm²

τ = 60,000 N/500 mm² = 120 MPa

τ_all(260 °C) = 150 − (60/100)(150 − 130) = 138 MPa > 120 MPa → acceptable

FE Reference Handbook, Mechanical Design and Analysis: welded joints.

This uses the throat of only one weld; the load is shared by two welds.

The stress is right, but it lies below the interpolated 138 MPa (and even below 130 MPa at 300 °C), so the joint passes.

This uses the leg area of only one weld; both 50 mm welds share the load.

This divides by the leg area instead of the throat area; the weld shears through its 0.707h throat.

Throat t = 0.707 × 7.07 = 5.0 mm, area 2 × 50 × 5.0 = 500 mm², τ = 60,000/500 = 120 MPa; the allowable at 260 °C interpolates to 138 MPa.

Question 9 of 18 · Mechanical Design and Analysis · 10–15 items

A shaft with a keyway has a fatigue stress concentration factor of 2.2. If the material has an endurance limit of 30 ksi without the keyway, what is the approximate modified endurance limit with the keyway?

Given	Fatigue stress concentration factor K_f = 2.2 Endurance limit S_e = 30 ksi
Find	Modified endurance limit (ksi)

Solution

S_e′ = S_e/K_f = 30/2.2 = 13.6 ksi (equivalently, multiply the alternating stress by K_f)

FE Reference Handbook, Mechanical Design and Analysis: fatigue (stress concentration).

Dividing the endurance limit by the fatigue stress concentration factor gives 30/2.2 = 13.6 ksi.

Dividing by K_f − 1 = 1.2 (30/1.2 = 25.0) mixes in the notch-sensitivity form; the endurance limit is divided by K_f itself.

Subtracting K_f from the endurance limit treats a dimensionless factor as a stress.

Multiplying by K_f raises the endurance limit; a keyway weakens the shaft.

Question 10 of 18 · Mechanical Design and Analysis · 10–15 items

What is the primary advantage of a spline connection over a key and keyway for transmitting torque between a shaft and hub?

Solution

Spline vs key: many load-sharing teeth → lower bearing stress, lower stress concentration, better centering.

Splined hubs still need shoulders, rings or nuts for axial location, just like keyed hubs.

Splines need broaching or hobbing and cost more to make than a keyway and key.

Many teeth share the torque around the circumference, so the load per tooth and the stress concentration are lower than at a single keyway.

Weight saving is not the reason for choosing splines; both joints add little mass.

Question 11 of 18 · Mechanical Design and Analysis · 10–15 items

In a bolted joint, what is the primary purpose of preload?

Solution

Preload F_i ≈ 0.75–0.9 of proof load: joint stays closed, bolt alternating stress ≈ C·P/(2A_t), loosening resisted by clamp friction.

Preload keeps the members compressed, so most of an external load change goes into the members and the bolt sees a small alternating load; friction from clamping resists loosening.

Differential expansion can change preload, but compensating for it is not the purpose of preload.

Tightening increases thread friction; lubrication, not preload, reduces it.

Preload uses part of the bolt's strength; it does not change the material's strength.

Question 12 of 18 · Mechanical Design and Analysis · 10–15 items

What is the main advantage of using a structural adhesive joint over a mechanical fastener in certain applications?

Solution

Adhesive joints: load spread over the bond area, no holes → better fatigue; limits are peel stress, temperature and cure time.

Adhesives need surface preparation and cure time, so assembly is usually slower than fastening.

An adhesive bonds over the whole overlap area, avoiding the stress concentrations of fastener holes and spreading the load more evenly.

Adhesive bonding is not inherently cheaper; surface preparation, fixturing and curing add cost.

Most structural adhesives lose strength at elevated temperatures well below what bolts or rivets tolerate.

Question 13 of 18 · Mechanical Design and Analysis · 10–15 items

A welded joint is subjected to high cyclic stresses. Post-weld heat treatment (PWHT) is applied. What is the primary benefit of this process for the joint?

Solution

PWHT (stress relief): heat below the transformation range, hold, cool slowly → residual stresses relax.

Welding leaves tensile residual stresses near yield; PWHT relaxes them, lowering the effective mean stress at the weld toe under cyclic load.

Stiffness depends on the modulus and geometry, which heat treatment does not change.

Reinforcement is the weld bead profile set during welding; heat treatment does not add metal.

Stress relief can reduce stress-corrosion susceptibility, but general corrosion resistance is not the main purpose.

Question 14 of 18 · Mechanical Design and Analysis · 10–15 items

For a component intended to operate under high cyclic loads, how does an improved surface finish affect its fatigue performance?

Solution

S_e = k_ak_bk_ck_dk_eS_e′, k_a = a·S_ut^b (higher for ground or machined than for hot-rolled or forged surfaces)

A better finish raises the endurance limit, not lowers it.

Fatigue cracks start at the surface; smoother finishes remove micro-notches, which is why the surface factor k_a rises toward 1 for polished parts.

Static strength depends on the whole cross section and is barely affected by finish.

Surface finish is one of the largest Marin factors; a forged or as-rolled surface can cut the endurance limit by half.

Question 15 of 18 · Mechanical Design and Analysis · 10–15 items

A 1/4-in. UNC bolt is tightened to a torque of 85 lbf·in. Taking the torque coefficient as K = 0.2, what is the approximate bolt preload?

Given	Tightening torque T = 85 lbf·in Bolt diameter d = 0.25 in. Torque coefficient K = 0.2
Find	Bolt preload F_i (lbf)

Solution

T = K·d·F_i → F_i = T/(K·d) = 85 lbf·in/(0.2 × 0.25 in.) = 1700 lbf

FE Reference Handbook, Mechanical Design and Analysis: threaded fasteners (torque-preload).

F_i = T/(Kd) = 85/(0.2 × 0.25) = 1700 lbf.

This uses K = 0.4, double the given coefficient.

This divides torque by diameter only and leaves out the torque coefficient K.

Converting the torque to 7.08 lbf·ft while keeping d in inches gives 7.08/(0.2 × 0.25) = 142; the units must match.

Question 16 of 18 · Mechanical Design and Analysis · 10–15 items

For a ductile machine element under combined bending and torsion, which failure theory most accurately predicts the onset of yielding?

Solution

σ′ = √(σ² + 3τ²) ≥ S_y → yield (von Mises, bending + torsion)

Coulomb-Mohr is meant for brittle materials with different tensile and compressive strengths.

Maximum normal stress ignores shear and badly overestimates the strength of ductile parts in torsion.

Tresca is a valid ductile criterion but conservative, predicting yield up to about 15% early; von Mises fits test data more closely.

The distortion-energy (von Mises) theory agrees best with yield tests on ductile metals under combined bending and torsion.

Question 17 of 18 · Mechanical Design and Analysis · 10–15 items

Temperature (°C)	Yield strength (MPa)	Ultimate strength (MPa)
200	450	600
300	400	550

A highly stressed machine component operates at a steady 240 °C. At a critical location the plane stress state is σ_x = 300 MPa, σ_y = 150 MPa, and τ_xy = 100 MPa. Using the material properties in the table and the von Mises criterion, what is the factor of safety against yielding?

Given	Normal stress σ_x = 300 MPa Normal stress σ_y = 150 MPa Shear stress τ_xy = 100 MPa Operating temperature T = 240 °C Yield strength at 200 °C S_y = 450 MPa Yield strength at 300 °C S_y = 400 MPa
Find	Factor of safety against yielding n

Solution

σ′ = √(σ_x² − σ_xσ_y + σ_y² + 3τ_xy²) = √(90,000 − 45,000 + 22,500 + 30,000) = 312 MPa

S_y(240 °C) = 450 − (40/100)(450 − 400) = 430 MPa

n = 430/312 = 1.38

FE Reference Handbook, Mechanical Design and Analysis: failure theories (von Mises).

This uses Tresca (σ₁ − σ₃ = 350 MPa); the stem asks for von Mises.

This uses the 300 °C yield strength (400 MPa) instead of interpolating to 240 °C.

σ' = √(300² − 300 × 150 + 150² + 3 × 100²) = 312 MPa; S_y at 240 °C interpolates to 430 MPa, so n = 430/312 = 1.38.

This uses the 200 °C yield strength (450 MPa) instead of interpolating to 240 °C.

Leaving out the 3τ_xy² term understates the equivalent stress.

Question 18 of 18 · Mechanical Design and Analysis · 10–15 items

A compound gear train consists of four gears. Gear 1 (20 teeth) drives gear 2 (60 teeth), and gear 3 (15 teeth) is mounted on the same shaft as gear 2 and drives gear 4 (45 teeth). What is the ratio of output speed to input speed, ω₄/ω₁?

Given	Gear 1 teeth N₁ = 20 Gear 2 teeth N₂ = 60 Gear 3 teeth N₃ = 15 Gear 4 teeth N₄ = 45
Find	Speed ratio ω₄/ω₁

Solution

n₄/n₁ = (N₁/N₂)(N₃/N₄) = (20/60)(15/45) = 1/9

FE Reference Handbook, Mechanical Design and Analysis: gear trains.

This counts only one stage; the second mesh reduces the speed by another factor of three.

This is ω₁/ω₄, the reduction ratio; the question asks for output speed over input speed.

Stage ratios multiply in a compound train; adding 1/3 + 1/3 has no meaning.

Each stage reduces speed by its tooth ratio: (20/60) × (15/45) = 1/3 × 1/3 = 1/9.

Answer every question to see your score

Common traps

Checked against the Mechanical Engineering chapter of the FE Reference Handbook, which holds the machine-design relations, and Shigley's Mechanical Engineering Design:

Thin wall is a test, not a label. The thin-wall stress formulas assume the wall is thin relative to the radius (a common rule is t below about one tenth of r). A thicker vessel needs the thick-wall relations, where stress varies through the wall.
Hoop is the larger stress. In a closed thin cylinder under internal pressure, hoop stress is twice the longitudinal stress. Seams that carry hoop stress govern.
K_t or K_f. The theoretical stress-concentration factor comes from geometry. The fatigue factor is reduced by the material's notch sensitivity. Static checks on ductile parts often ignore K_t; fatigue checks do not ignore K_f.
Power, torque and speed in consistent units. P = T·ω needs ω in rad/s. In USCS, a horsepower relation with torque in lbf·ft and speed in rpm has its own constant; mixing in lbf·in is a factor of twelve.
Spring correction. Shear stress in a helical spring includes a curvature correction that grows as the spring index (mean coil diameter over wire diameter) gets smaller.
Fits by deviation, not by size. Whether a pair is a clearance, transition or interference fit depends on the limits of the hole and shaft, not on their nominal size.

A power screw that holds its load

A screw is self-locking when the friction angle exceeds the lead angle, so the load cannot drive it backwards. A low-friction ball screw usually is not, and needs a brake.

Design draws on every mechanics area

Every check here starts from a stress computed in mechanics of materials and a reaction from statics. Fatigue and creep data come from material properties and processing; gear-train speeds and shaft dynamics from dynamics, kinematics and vibrations.

Pressure vessels also appear on the chemical side of the site, with materials of construction on FE Chemical's process design.

Other FE Mechanical areas

Thermodynamics10–15 items on the exam
Statics9–14 items on the exam
Mechanics of Materials9–14 items on the exam
Fluid Mechanics10–15 items on the exam

Mechanical design, quick answers

Is GD&T tested on the FE Mechanical exam?

Yes, the specification lists engineering drawing interpretation and geometric dimensioning and tolerancing. Expect reading a feature control frame or a datum reference rather than drawing anything.

How many design questions are on the FE Mechanical exam?

10–15, one of the four largest ranges in the outline.

FE mechanical design review, from stress to part

What the area covers

18 mechanical design problems, every option explained

Common traps

Design draws on every mechanics area

Other FE Mechanical areas

Mechanical design, quick answers

More machine design problems on your phone