Exercise set

Machine Component Design and Fatigue Exercises

Worked machine component design exercises for shafts, keys, bearings, tolerances, stress concentration, fatigue, Miner damage, stiffness and release checks.

Branch: Mechanical Engineering
Content: Exercise set
Updated: Jul 03, 2026
Revision: v1.0.0 · reviewed

These exercises focus on machine components: shafts, keys, bearings, bolted or threaded features, tolerance stacks, stress concentration, fatigue, stiffness and release checks. The purpose is to connect a local calculation to the full load path, duty spectrum, material state, assembly condition and validation evidence.

Use the numbers as screening models. Real component release requires drawings, material certificates, heat treatment, surface finish, notch detail, lubrication, contamination, alignment, thermal growth, inspection access and test evidence.

Release Evidence Notes

A component calculation is not release evidence unless the load path is traceable. Record the torque source, support spacing, overhung loads, stress raisers, fit, keyway, bearing arrangement, preload, lubrication, duty cycle, temperature, supplier rating and inspection method. Passing static stress does not prove fatigue life, stiffness, bearing life or assembly robustness.

Engineering Boundary Notes

Machine components fail through combinations of stress, deflection, fatigue, wear, heating, misalignment and assembly variation. Treat a shaft, bearing, hub, key, housing and fastener as one system. A local margin can be consumed by a downstream tolerance or upstream overload.

Common Release Mistakes

sizing a shaft from torque only while ignoring bending and keyway notch effect;
using catalog bearing life without contamination, preload or duty spectrum;
accepting yield margin while fatigue damage is the governing limit;
checking worst-case tolerance without deciding whether the process can actually hold it;
ignoring runout, misalignment, thermal growth and lubrication state;
validating a prototype that does not match production material or heat treatment.

Scenario Map

Scenario	Main calculation	Release decision
Shaft stress	torsion, bending and von Mises stress	Select diameter or require redesign.
Key and hub	shear and bearing pressure	Check torque transfer detail.
Bearing selection	equivalent load and $L_{10}$ life	Accept, resize or derate bearing.
Tolerances	worst-case and statistical stack	Decide shim, adjustment or process control.
Fatigue	Goodman and Miner damage	Approve duty spectrum or change detail.
Stiffness and dynamics	twist and critical-speed separation	Prevent functional or vibration failure.

Validation Package Checklist

drawing revision, material and heat-treatment evidence;
torque, radial load and duty-cycle basis;
bearing fit, preload, lubrication and contamination assumptions;
stress concentration, surface finish and fatigue model;
tolerance stack and measurement method;
runout, alignment, vibration and thermal-growth checks;
guarded release rule with uncertainty and inspection action.

Exercise 1: Solid Shaft Diameter From Torque

A solid circular shaft transmits design torque:

T=620\ \text{N m}

Allowable shear stress is:

\tau_{allow}=55\ \text{MPa}

Use:

\displaystyle \tau_{max}=\frac{16T}{\pi d^3}

Find minimum diameter.

Solution

\displaystyle d=\left(\frac{16T}{\pi\tau_{allow}}\right)^{1/3}

\displaystyle d=\left(\frac{16(620)}{\pi(55\times10^6)}\right)^{1/3}=0.0386\ \text{m}=38.6\ \text{mm}

Engineering Comment

The next standard size should be checked with bending, keyway stress concentration and fatigue. Torque-only sizing is an opening screen.

Plausibility Check

Hundreds of N m on a steel shaft commonly produces a diameter in the tens of millimetres.

Exercise 2: Combined Bending and Torsion

A shaft has bending stress:

\sigma_b=82\ \text{MPa}

and torsional shear stress:

\tau=38\ \text{MPa}

Estimate von Mises stress.

Solution

\sigma_{vm}=\sqrt{\sigma_b^2+3\tau^2}

\sigma_{vm}=\sqrt{82^2+3(38^2)}=105\ \text{MPa}

Engineering Comment

Compare this value with allowable stress after notch, surface and fatigue factors. A keyway can raise the local stress beyond this nominal result.

Plausibility Check

Von Mises stress is higher than either individual component, as expected.

Exercise 3: Key Shear Stress

A rectangular key transmits $T=420\ \text{N m}$ on a $40\ \text{mm}$ shaft. Key width is $12\ \text{mm}$ and engaged length is $50\ \text{mm}$ . Estimate shear stress using:

\displaystyle F=\frac{2T}{d},\quad \tau=\frac{F}{bL}

Solution

\displaystyle F=\frac{2(420)}{0.040}=21000\ \text{N}

\displaystyle \tau=\frac{21000}{0.012(0.050)}=35.0\ \text{MPa}

Engineering Comment

The key shear may pass while the shaft keyway fatigue detail fails. Check both the key and the weakened shaft.

Plausibility Check

The tangential force is large because torque is applied at a small radius.

Exercise 4: Key Bearing Pressure

Use the key force $F=21000\ \text{N}$ . Key height is $8\ \text{mm}$ and bearing area is approximated as:

\displaystyle A_b=\frac{hL}{2}

Compute bearing pressure.

Solution

\displaystyle A_b=\frac{0.008(0.050)}{2}=2.0\times10^{-4}\ \text{m}^2

\displaystyle p_b=\frac{21000}{2.0\times10^{-4}}=105\ \text{MPa}

Engineering Comment

Bearing pressure often controls key length. Fretting and hub material strength must be checked.

Plausibility Check

Bearing pressure is higher than shear stress because the effective bearing area is smaller.

Exercise 5: Bearing Life Screen

A bearing has dynamic rating:

C=32\ \text{kN}

Equivalent load is:

P=8\ \text{kN}

For ball bearings use:

\displaystyle L_{10}=\left(\frac{C}{P}\right)^3 10^6\ \text{rev}

Compute life in revolutions.

Solution

\displaystyle L_{10}=\left(\frac{32}{8}\right)^3 10^6=64\times10^6\ \text{rev}

Engineering Comment

This life is a catalog screen. Contamination, lubrication, preload, temperature, misalignment and shock can reduce real life.

Plausibility Check

The rating/load ratio is four, and life scales with the cube, so $64$ million revolutions is expected.

Exercise 6: Bearing Life in Hours

Use $L_{10}=64\times10^6\ \text{rev}$ at $n=1200\ \text{rpm}$ . Convert to hours.

Solution

\displaystyle L_h=\frac{64\times10^6}{1200(60)}=889\ \text{h}

Engineering Comment

For continuous service, $889\ \text{h}$ is short. The design needs a larger bearing, lower load, better arrangement or derating of the duty.

Plausibility Check

High speed consumes revolutions quickly, so the hour life is much smaller than the revolution count suggests.

Exercise 7: Variable Bearing Load

A bearing sees $5\ \text{kN}$ for $70\%$ of time and $11\ \text{kN}$ for $30\%$ . For ball bearings, estimate equivalent load:

P_{eq}=\left(0.7P_1^3+0.3P_2^3\right)^{1/3}

Solution

P_{eq}=\left(0.7(5^3)+0.3(11^3)\right)^{1/3}=7.83\ \text{kN}

Engineering Comment

The high-load block dominates because life damage scales strongly with load.

Plausibility Check

The equivalent load is closer to $11\ \text{kN}$ than a simple time average would be.

Exercise 8: Worst-Case Tolerance Stack

Three spacers have tolerances:

\pm0.08\ \text{mm},\quad \pm0.05\ \text{mm},\quad \pm0.04\ \text{mm}

Compute worst-case stack tolerance.

Solution

T_{wc}=0.08+0.05+0.04=0.17\ \text{mm}

Engineering Comment

Worst-case control is conservative but may be expensive. Decide whether all parts must interchange without adjustment.

Plausibility Check

Worst-case tolerance is the sum of absolute tolerances.

Exercise 9: Statistical Tolerance Stack

Use the same three tolerances and estimate RSS tolerance.

Solution

T_{rss}=\sqrt{0.08^2+0.05^2+0.04^2}=0.102\ \text{mm}

Engineering Comment

RSS is valid only if the process is centered and statistically controlled. It is not a substitute for process capability evidence.

Plausibility Check

RSS tolerance is smaller than worst case but larger than the largest single tolerance.

Exercise 10: Stress Concentration

Nominal alternating stress is:

\sigma_a=70\ \text{MPa}

Fatigue stress concentration factor is:

K_f=1.7

Compute local alternating stress.

Solution

\sigma_{a,local}=K_f\sigma_a=1.7(70)=119\ \text{MPa}

Engineering Comment

A shoulder radius, keyway or thread root can turn a passing nominal stress into a fatigue risk.

Plausibility Check

The local stress increases by $70\%$ , matching the factor.

Exercise 11: Goodman Fatigue Check

A component has local alternating stress $119\ \text{MPa}$ and mean stress $80\ \text{MPa}$ . Endurance limit is $240\ \text{MPa}$ and ultimate tensile strength is $620\ \text{MPa}$ . Compute Goodman usage:

\displaystyle U=\frac{\sigma_a}{S_e}+\frac{\sigma_m}{S_{ut}}

Solution

\displaystyle U=\frac{119}{240}+\frac{80}{620}=0.625

Engineering Comment

The simplified Goodman check passes because $U<1$ , but surface finish, size, reliability and environment factors must be included in final release.

Plausibility Check

The usage is below one because both stress ratios are moderate.

Exercise 12: Miner Damage

A shaft sees $2.0\times10^5$ cycles at a level whose life is $1.0\times10^6$ cycles and $1.5\times10^5$ cycles at a level whose life is $7.5\times10^5$ cycles. Compute Miner damage.

Solution

\displaystyle D=\frac{2.0\times10^5}{1.0\times10^6}+\frac{1.5\times10^5}{7.5\times10^5}=0.20+0.20=0.40

Engineering Comment

The duty block passes a simple $D<1$ screen. Real spectra need startup, shock, reversal and overload events.

Plausibility Check

Each block consumes $20\%$ of its life, so total damage is $40\%$ .

Exercise 13: Shaft Twist

A shaft has length $L=0.80\ \text{m}$ , torque $T=300\ \text{N m}$ , shear modulus $G=80\ \text{GPa}$ and polar moment:

J=1.57\times10^{-7}\ \text{m}^4

Compute twist angle.

Solution

\displaystyle \theta=\frac{TL}{GJ}

\displaystyle \theta=\frac{300(0.80)}{80\times10^9(1.57\times10^{-7})}=0.0191\ \text{rad}=1.09^\circ

Engineering Comment

Stress can pass while twist causes positioning error, coupling misalignment or control-loop compliance.

Plausibility Check

A long slender shaft can twist by about one degree under hundreds of N m.

Exercise 14: Critical-Speed Separation

A shaft first critical speed is estimated at $3600\ \text{rpm}$ . Maximum operating speed is $2850\ \text{rpm}$ . Compute separation margin relative to operating speed.

Solution

\displaystyle M=\frac{3600-2850}{2850}=0.263=26.3\%

Engineering Comment

The margin is useful, but run-up testing and bearing-support stiffness should confirm the mode.

Plausibility Check

The critical speed is above operating speed, so the margin is positive.

Exercise 15: Bearing Preload Thermal Growth

A bearing pair loses $0.006\ \text{mm}$ preload clearance per kelvin of housing-to-shaft differential growth. Temperature difference rises by $18\ \text{K}$ . Estimate clearance change.

Solution

\Delta c=0.006(18)=0.108\ \text{mm}

Engineering Comment

Thermal growth can convert a light preload into excessive load. Bearing temperature evidence is required.

Plausibility Check

The clearance change is about one tenth of a millimetre, large enough for bearing fits.

Exercise 16: Runout Guard

Allowed shaft runout is $0.060\ \text{mm}$ . Measured runout is $0.044\ \text{mm}$ with measurement uncertainty $0.010\ \text{mm}$ . Use guarded runout:

r_g=r+U

Solution

r_g=0.044+0.010=0.054\ \text{mm}

Since $0.054<0.060$ , the check passes.

Engineering Comment

The margin is narrow. Measurement fixture repeatability and shaft seating condition should be documented.

Plausibility Check

Uncertainty increases the reported runout for a conservative pass/fail test.

Exercise 17: Supplier Material Change

Original yield strength is $420\ \text{MPa}$ . A proposed substitute has $390\ \text{MPa}$ . The maximum calculated static stress is $155\ \text{MPa}$ . Compare static safety factors.

Solution

Original:

\displaystyle N_1=\frac{420}{155}=2.71

Substitute:

\displaystyle N_2=\frac{390}{155}=2.52

Engineering Comment

Static margin remains positive, but fatigue strength, hardness, heat treatment and surface finish may be the real supplier-change risks.

Plausibility Check

Lower yield strength lowers safety factor.

Exercise 18: Component Release Gate

A shaft design has Goodman usage $0.625$ , Miner damage $0.40$ , guarded runout $0.054\ \text{mm}$ against $0.060\ \text{mm}$ limit and bearing life requirement $1200\ \text{h}$ against predicted $889\ \text{h}$ . Decide release status.

Solution

Fatigue usage passes:

0.625<1

Damage passes:

0.40<1

Runout passes:

0.054<0.060\ \text{mm}

Bearing life fails:

889<1200\ \text{h}

Engineering Comment

Do not release the component assembly. The bearing life failure controls the design even though stress, fatigue and runout pass.

Plausibility Check

One failed gate is enough to block release because the component system must satisfy every governing limit.

REF