Exercise set

Machine Design and Power Transmission Systems Exercises

Worked mechanical engineering exercises for machine design covering torque, speed, gear ratio, shaft sizing, keyway stress, bearing life, resonance separation, reflected inertia, tolerance stack-up, fatigue damage, hoist brake torque, and validation.

These exercises practise machine design and power-transmission calculations for shafts, gears, keyways, bearings, drives, tolerances, dynamics, fatigue, hoists, and validation tests. The purpose is not only to select a component size. The purpose is to connect torque, speed, load path, stress, stiffness, heat, fatigue, assembly variation, safety, and measured evidence.

Assume steady operation, simplified geometry, and representative material data unless an exercise states otherwise. Real machines should also check shock loading, duty cycle, lubrication, contamination, thermal growth, runout, misalignment, guarding, lockout, supplier ratings, inspection access, and applicable machinery safety standards.

How to Use These Exercises

For each calculation, define:

  1. the power path from drive to load;
  2. the operating case: nominal, startup, stall, overload, braking, reversing, or degraded operation;
  3. the component limit: stress, deflection, bearing life, temperature, vibration, wear, or safety function;
  4. the factor or margin required by the design basis;
  5. the validation evidence needed before release.

The most common mistake is sizing one part in isolation. A shaft, gear, bearing, key, coupling, housing, lubricant, and drive control form one machine system. A calculation is useful only when it preserves that system context.

Use the exercises as design gates: accept or reject a torque path, resize a shaft, change a key or hub detail, reject a bearing life, separate an operating speed from a mode, challenge a tolerance strategy, revise a duty-cycle fatigue assumption, or hold a hoist brake release until the safety function is validated.

Exercise 1: Torque from Power and Speed

A machine transmits:

P=18\ \text{kW}

at:

n=450\ \text{rpm}

The design requires a service factor of:

K_s=1.6

to cover starting and moderate shock. Find nominal torque and design torque.

Solution

Angular speed is:

\displaystyle \omega=\frac{2\pi n}{60}
\displaystyle \omega=\frac{2\pi(450)}{60}=47.1\ \text{rad/s}

Nominal torque:

\displaystyle T=\frac{P}{\omega}
\displaystyle T=\frac{18{,}000}{47.1}=382\ \text{N m}

Design torque:

T_d=K_sT=1.6(382)=611\ \text{N m}

Engineering Comment

The service factor is a design assumption, not a universal truth. A crusher, hoist, mixer, pump, conveyor, spindle, or robot joint may need very different overload treatment. The operating profile should state starts per hour, jam risk, braking events, reversals, and expected load spectrum.

Exercise 2: Gear Reduction Speed and Output Torque

A motor delivers:

P_{in}=7.5\ \text{kW}

at:

n_{in}=1750\ \text{rpm}

through a gearbox with reduction ratio:

i=5:1

and efficiency:

\eta=0.94

Find output speed and output torque.

Solution

Output speed:

\displaystyle n_{out}=\frac{n_{in}}{i}=\frac{1750}{5}=350\ \text{rpm}

Output power:

P_{out}=\eta P_{in}=0.94(7.5)=7.05\ \text{kW}

Output angular speed:

\displaystyle \omega_{out}=\frac{2\pi(350)}{60}=36.7\ \text{rad/s}

Output torque:

\displaystyle T_{out}=\frac{P_{out}}{\omega_{out}}=\frac{7050}{36.7}=192\ \text{N m}

Engineering Comment

The gearbox increases torque while reducing speed, but losses become heat. The design should check thermal rating, lubrication, gear mesh loading, bearing loads, backlash, duty cycle, mounting stiffness, and whether the drive can tolerate reflected load inertia during acceleration and braking.

Exercise 3: First-Pass Solid Shaft Diameter

Use the design torque from Exercise 1:

T_d=611\ \text{N m}

For a first torsional screen, use allowable shear stress:

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

Estimate the required solid circular shaft diameter from:

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

Solution

Rearrange:

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

Use T in \text{N mm}:

T=611{,}000\ \text{N mm}

Then:

\displaystyle d=\left(\frac{16(611{,}000)}{\pi(55)}\right)^{1/3}
d=38.4\ \text{mm}

A practical first selection may be:

d=40\ \text{mm}

Engineering Comment

This is not a final shaft design. It ignores bending, fatigue, shoulders, keyways, surface finish, fits, bearing span, deflection, critical speed, corrosion, and manufacturing tolerance. A standard diameter should be selected only after the full load path is reviewed.

Exercise 4: Key Shear and Bearing Stress

A key transmits torque:

T=600\ \text{N m}

on a shaft with diameter:

d=40\ \text{mm}

The key has width:

b=12\ \text{mm}

height:

h=8\ \text{mm}

and engaged length:

L=45\ \text{mm}

Estimate tangential force, key shear stress, and key bearing stress using half-height bearing area.

Solution

Tangential force at the shaft radius is:

\displaystyle F_t=\frac{2T}{d}

Use T=600{,}000\ \text{N mm}:

\displaystyle F_t=\frac{2(600{,}000)}{40}=30{,}000\ \text{N}

Shear area:

A_s=bL=(12)(45)=540\ \text{mm}^2

Average shear stress:

\displaystyle \tau=\frac{F_t}{A_s}=\frac{30{,}000}{540}=55.6\ \text{MPa}

Bearing area using half key height:

\displaystyle A_b=\frac{h}{2}L=(4)(45)=180\ \text{mm}^2

Bearing stress:

\displaystyle \sigma_b=\frac{F_t}{A_b}=\frac{30{,}000}{180}=167\ \text{MPa}

Engineering Comment

Key calculations are average checks. Keyways also reduce shaft fatigue strength and create stress concentrations. High-duty machines may need splines, interference fits, shrink fits, clamping hubs, larger shoulders, or revised torque paths instead of simply lengthening the key.

Exercise 5: Bearing L10 Life Estimate

A rolling-element bearing has dynamic load rating:

C=35\ \text{kN}

and equivalent dynamic load:

P=8\ \text{kN}

For a ball bearing, use exponent:

p=3

The shaft speed is:

n=1200\ \text{rpm}

Estimate basic rating life in hours:

\displaystyle L_{10}=\left(\frac{C}{P}\right)^p10^6\ \text{rev}

Solution

Rating life in revolutions:

\displaystyle L_{10}=\left(\frac{35}{8}\right)^3 10^6=83.7\times10^6\ \text{rev}

Convert to hours:

\displaystyle L_{10,h}=\frac{83.7\times10^6}{60(1200)}
L_{10,h}=1160\ \text{h}

Engineering Comment

This bearing would not satisfy a 10,000 h target without changing load, bearing size, speed, rating method, or duty cycle. Catalog life also does not capture contamination, poor lubrication, misalignment, electrical pitting, mounting damage, high temperature, or vibration unless those factors are included separately.

Exercise 6: Natural Frequency and Operating-Speed Separation

A machine module can be approximated by a mass:

m=85\ \text{kg}

on support stiffness:

k=1.8\times10^6\ \text{N/m}

Estimate its natural frequency. Compare it with an operating speed of:

1500\ \text{rpm}

Solution

Natural frequency:

\displaystyle f_n=\frac{1}{2\pi}\sqrt{\frac{k}{m}}
\displaystyle f_n=\frac{1}{2\pi}\sqrt{\frac{1.8\times10^6}{85}}
f_n=23.2\ \text{Hz}

Operating frequency:

\displaystyle f_{op}=\frac{1500}{60}=25.0\ \text{Hz}

Separation:

\displaystyle \frac{25.0-23.2}{23.2}=0.078=7.8\%

Engineering Comment

The operating speed is close to the simplified natural frequency. That does not prove resonance failure, but it is a warning. The engineer should check mode shape, damping, excitation source, startup passage through resonance, support flexibility, foundation stiffness, and measured vibration during run-up.

Exercise 7: Reflected Inertia Through a Gearbox

A load inertia at the gearbox output is:

J_L=0.42\ \text{kg m}^2

The motor-to-load speed ratio is:

\displaystyle N=\frac{\omega_m}{\omega_L}=4

The motor rotor inertia is:

J_m=0.018\ \text{kg m}^2

The motor accelerates to:

n_m=1200\ \text{rpm}

in:

t=2.5\ \text{s}

Estimate reflected inertia at the motor, total inertia at the motor, and acceleration torque.

Solution

Reflected load inertia:

\displaystyle J_{ref}=\frac{J_L}{N^2}=\frac{0.42}{4^2}=0.0263\ \text{kg m}^2

Total inertia at the motor:

J_{tot}=J_m+J_{ref}=0.018+0.0263=0.0443\ \text{kg m}^2

Motor angular speed:

\displaystyle \omega_m=\frac{2\pi(1200)}{60}=125.7\ \text{rad/s}

Angular acceleration:

\displaystyle \alpha=\frac{\omega_m}{t}=\frac{125.7}{2.5}=50.3\ \text{rad/s}^2

Acceleration torque:

T_{acc}=J_{tot}\alpha=(0.0443)(50.3)=2.23\ \text{N m}

Engineering Comment

The acceleration torque appears small, but it is only the inertia component. Process torque, friction, gearbox losses, load disturbance, vertical load, braking, and control bandwidth may dominate. Reflected inertia also affects servo stability and stopping distance.

Exercise 8: Tolerance Stack-Up for Coupling Gap

A coupling axial gap depends on four independent dimensions with bilateral tolerances:

\pm0.10\ \text{mm},\quad \pm0.15\ \text{mm},\quad \pm0.05\ \text{mm},\quad \pm0.08\ \text{mm}

The required assembled gap tolerance is:

\pm0.25\ \text{mm}

Calculate worst-case and root-sum-square tolerance estimates.

Solution

Worst-case stack-up:

T_{WC}=0.10+0.15+0.05+0.08=0.38\ \text{mm}

Root-sum-square estimate:

T_{RSS}=\sqrt{0.10^2+0.15^2+0.05^2+0.08^2}
T_{RSS}=0.203\ \text{mm}

Engineering Comment

The worst-case stack fails the \pm0.25\ \text{mm} requirement, while the statistical estimate passes. The correct choice depends on production volume, tolerance distributions, process capability, measurement method, consequence of interference, and whether selective assembly or shimming is allowed.

Exercise 9: Miner Fatigue Damage for a Duty Cycle

A shaft detail experiences three stress-cycle blocks during a representative duty profile:

BlockApplied cycles n_iLife at stress level N_i
High load20,000200,000
Medium load80,0001,200,000
Low load250,0008,000,000

Estimate cumulative damage using Miner’s rule.

Solution

Miner damage is:

\displaystyle D=\sum_i\frac{n_i}{N_i}
\displaystyle D=\frac{20{,}000}{200{,}000}+\frac{80{,}000}{1{,}200{,}000}+\frac{250{,}000}{8{,}000{,}000}
D=0.100+0.0667+0.0313=0.198

Engineering Comment

The linear damage estimate is below 1.0 for this profile, but fatigue is sensitive to stress concentration, sequence effects, corrosion, residual stress, surface condition, overloads, and the S-N data basis. The duty profile should represent real operation, including starts, stops, jams, and maintenance events.

Exercise 10: Hoist Brake Torque at the Motor Shaft

A hoist supports load:

W=6.0\ \text{kN}

on a drum with radius:

r=0.18\ \text{m}

The gearbox reduction ratio between motor and drum is:

i=30:1

Assume reverse-drive efficiency:

\eta=0.85

Estimate the motor-shaft holding torque required to resist the load. Then apply a brake sizing factor of 2.0.

Solution

Drum torque from the load:

T_{drum}=Wr=(6000)(0.18)=1080\ \text{N m}

Reflected torque at the motor shaft:

\displaystyle T_m=\frac{T_{drum}}{i\eta}=\frac{1080}{30(0.85)}=42.4\ \text{N m}

Brake torque with sizing factor:

T_{brake}=2.0(42.4)=84.8\ \text{N m}

Engineering Comment

A hoist brake is a safety function, not only a torque calculation. Final design should check load-holding standards, dynamic stopping energy, brake thermal capacity, overspeed protection, gearbox backdrivability, emergency lowering, inspection interval, redundant load paths, and fail-safe engagement on power loss.

Review Checklist

When reviewing a machine-design or power-transmission calculation, ask:

  • Is the duty case explicit: startup, stall, overload, braking, reversing, jam, or degraded operation?
  • Is the torque path traced through motor, gearbox, coupling, shaft, key, bearing, housing, foundation, and load?
  • Are stress, stiffness, fatigue, bearing life, thermal rating, vibration, wear, and safety functions checked as coupled limits?
  • Are tolerances, runout, fits, lubrication, contamination, alignment, guarding, and inspection access represented in the decision?
  • Are supplier ratings used inside their stated duty, mounting, thermal, and lubrication boundaries?
  • Does validation reproduce the load spectrum, speed range, braking energy, stopping distance, vibration response, and maintenance state that matter?
  • Is any continued-release decision tied to measured evidence rather than a single nominal calculation?

A machine that passes one nominal calculation may still fail through fatigue, misalignment, heat, vibration, wear, or unsafe service access. Good machine design keeps the whole power path visible from requirement to validation.

REF

See also