Mechanical Engineering Design · Part 4 — Special Topics

Power Transmission Case Study

Integrating the whole course into one design: a double-reduction gearbox worked from power and torque requirements through gear specification, shaft layout, force analysis, material selection, stress and deflection design, bearing and key selection, and a final verifying analysis — showing how the individual element methods interlock.

Learning objectives

Translate a power-transmission need into speeds, torques, ratios, duty, life, reliability, and packaging requirements.
Build coupled free-body models for gears, shafts, bearings, keys, and retaining features.
Distinguish transmitted load from local stress, material strength, stiffness, fatigue resistance, and catalog capacity.
Sequence calculations so each component receives compatible loads and geometry from the same design state.
Iterate toward standard, manufacturable parts and verify the final configuration with traceable data.

18–1Design Sequence for Power Transmission

A controlled iteration, not a chain of isolated answers

An integrated transmission begins with a requirement set: input power and speed, output speed or ratio, load spectrum, life, reliability, allowable envelope, environment, interfaces, noise, maintenance, cost, and applicable codes. A concept then assigns functions to gears, shafts, bearings, joints, seals, lubrication, and housing. Every calculation belongs to a named design state so that an updated gear diameter cannot silently coexist with obsolete shaft reactions.

A practical dependency order is requirements, ratio split, trial gears, spatial layout, gear-force free bodies, shaft stress sizing, stiffness and dynamics, bearings, torque-transfer details, housing, lubrication, and final verification. Failed checks send the design back to the earliest decision that can efficiently correct them.

Problem-solving strategy

Freeze measurable requirements and acceptance criteria, including operating spectrum and failure consequences.
Create a speed-torque table at every shaft, explicitly applying stage efficiencies and sign conventions.
Select trial standard components, lay out axial stations, and draw separate free bodies in two transverse planes.
Size for stress, then verify stiffness, critical speed, bearing life, keys, fits, retaining features, and thermal behavior.
Iterate the coupled variables until changes are acceptably small, then rerun all checks using released dimensions and traceable data.

Physical meaning

Iteration is information flow. Gear diameter sets tooth force and shaft spacing; shaft geometry sets deflection and bearing fits; bearing selection changes seat diameter and housing space. Convergence means the same physical geometry satisfies all checks—not merely that one spreadsheet cell stops changing.

18–2Power and Torque Requirements

Track power loss and torque rise

At shaft \(j\), \(P_j=T_j\omega_j\). A reduction decreases speed and increases torque, but output power is lower than input power when efficiency is below one. Use mean power for thermal and nominal-load work, and apply the specified peak or fatigue load spectrum where those failure modes demand it. Rated motor power is not automatically the gear tooth load.

Stage propagation

\[n_{j+1}=\frac{n_j}{i_j},\qquad P_{j+1}=\eta_jP_j,\qquad T_j=\frac{P_j}{2\pi n_j/60}\]

EXAMPLE 1Power and torque table for two reductions

A packaging machine gearbox receives \(15.0\ \text{kW}\) at \(1440\ \text{rev/min}\). Stage 1 has reduction \(i_1=3.00\) and efficiency \(\eta_1=0.970\); stage 2 has \(i_2=4.00\) and \(\eta_2=0.970\). Determine speed, power, and torque on the input, intermediate, and output shafts under steady nominal operation.

Recognition strategy

Two ratios and two efficiencies signal a stage-by-stage energy-flow table. Speed divides by each reduction, power multiplies by each efficiency, and torque must be recomputed from the power actually present on that shaft.

Given information

Positive rotation follows the input shaft; \(P_1=15.0\ \text{kW}\), \(n_1=1440\ \text{rev/min}\), \(i_1=3.00\), \(i_2=4.00\), and \(\eta_1=\eta_2=0.970\).
Assume steady mean values and efficiencies supplied for this operating point; dynamic overload and starting torque are excluded.

Model and interpretation

Treat each mesh as a control boundary with positive power flowing toward the output. The torque values are nominal shaft loads, not local stresses or strengths, and will later be combined with gear forces and stress concentrations.

Step-by-step solution

1. Compute input torque. \(\omega_1=2\pi(1440)/60=150.80\ \text{rad/s}\), so \(T_1=15000/150.80=99.47\ \text{N}\!\cdot\!\text{m}\).

2. Propagate through stage 1. \(n_2=1440/3=480\ \text{rev/min}\) and \(P_2=(0.970)(15.0)=14.55\ \text{kW}\), giving \(T_2=289.46\ \text{N}\!\cdot\!\text{m}\).

3. Propagate through stage 2. \(n_3=480/4=120\ \text{rev/min}\) and \(P_3=(0.970)(14.55)=14.1135\ \text{kW}\), giving \(T_3=1123.12\ \text{N}\!\cdot\!\text{m}\).

4. Quantify overall behavior. The overall ratio is \(i=12.0\), efficiency is \(\eta=0.970^2=0.9409\), and loss is \(15.0-14.1135=0.8865\ \text{kW}\).

Verification

Units: Dividing watts by radians per second gives \(\text{N}\!\cdot\!\text{m}\) because radians are dimensionless; each speed remains in revolutions per minute after division by a dimensionless ratio.

Signs and direction: Positive power propagates toward the output. External gear meshes reverse rotation, but the requested torque magnitudes remain positive; a detailed layout must assign alternating shaft rotation signs.

Plausibility: Output speed is exactly \(1440/12=120\ \text{rev/min}\), output power is lower than input power, and torque rises by about \(12(0.9409)=11.29\), matching \(1123.12/99.47\).

Final result: \(\boxed{(n,P,T)_1=(1440,15.0,99.47),\ (n,P,T)_2=(480,14.55,289.46),\ (n,P,T)_3=(120,14.1135,1123.12)}\), with units \(\text{rev/min},\ \text{kW},\ \text{N}\!\cdot\!\text{m}\).

Engineering interpretation

The output shaft carries more than eleven times the input torque, while the gearbox must reject about \(0.887\ \text{kW}\) of loss at this operating point. Peak duty, mesh efficiency maps, thermal balance, and motor starting behavior still require supplier or test data.

18–3Gear Specification

Geometry first, rating with controlled data

Split the overall ratio to balance package size, pitch-line velocity, tooth counts, and available standard tooling. Trial tooth numbers establish exact ratios and interference risk; module or diametral pitch and face width establish size. Gear bending stress and contact stress are different local failure measures, and allowable bending and contact strengths are not interchangeable.

AGMA or ISO factors must come from the identified standard edition or approved software input set. Never invent dynamic, load-distribution, reliability, or life factors. Record material specification, heat treatment, hardness range, quality grade, lubrication, and duty used to obtain every allowable value.

18–4Shaft Layout

Design axial stations before diameter details

Place gears near bearings where practical, provide shoulders and spacers for axial location, and reserve assembly paths, tool clearance, seal lands, and bearing-puller access. Use one locating bearing and a compatible floating arrangement unless the chosen bearing system intentionally shares thrust. Thermal growth must not be trapped by two unintended locating fits.

Manufacturable shafts use standard stock and bearing seats, realistic shoulder heights and fillet radii, and tolerances compatible with grinding or turning capability. A generous fillet improves fatigue but may interfere with a bearing corner radius; the catalog chamfer and relief geometry close that loop.

18–5Force Analysis

From tooth force to two-plane beam loading

For a spur gear, tangential tooth force transmits torque and radial force separates the gears. Resolve forces into fixed shaft coordinates, include opposite forces on the mating gear, and solve bearing reactions independently in two transverse planes. Combine orthogonal bending moments only after each plane satisfies equilibrium. Helical gears add thrust and require a consistent helix-hand and rotation convention.

Spur-gear forces

\[F_t=\frac{2T}{d_p},\qquad F_r=F_t\tan\phi,\qquad M=\sqrt{M_y^2+M_z^2}\]

EXAMPLE 2Two-plane bearing reactions from a spur gear

A shaft is simply supported by bearings A and B spaced \(L=0.420\ \text{m}\). A spur gear at \(x=0.160\ \text{m}\) from A applies \(F_t=2400\ \text{N}\) downward in the \(y\)-plane. With pressure angle \(\phi=20.0^\circ\), its radial force acts in positive \(z\). Determine both-plane reactions, resultant bearing loads, and resultant bending moment at the gear.

Recognition strategy

A single overhung-free gear between two supports is a statics problem in two independent planes. Pressure angle converts tangential force to radial force; equilibrium gives reactions before orthogonal components are combined.

Given information

Set \(+x\) from A to B, \(+y\) upward, and \(+z\) along the stated radial force; \(L=0.420\ \text{m}\), \(a=0.160\ \text{m}\), \(F_t=2400\ \text{N}\), and \(\phi=20.0^\circ\).
Assume point loads at the gear center, simple supports, static equilibrium, and no shaft or gear weight.

Model and interpretation

Draw separate \(x\)-\(y\) and \(x\)-\(z\) free bodies. Bearing reactions oppose each applied component. Their vector magnitudes are catalog-equivalent radial loads only if the bearing load model and duty make that interpretation valid.

Step-by-step solution

1. Resolve tooth loading. The radial magnitude is \(F_r=F_t\tan\phi=(2400)\tan20.0^\circ=873.53\ \text{N}\).

2. Solve the tangential plane. Moment equilibrium about A gives \(B_y=F_ta/L=914.29\ \text{N}\); force equilibrium gives \(A_y=1485.71\ \text{N}\), both upward.

3. Solve the radial plane. Reactions oppose positive \(F_r\): \(A_z=-F_r(L-a)/L=-540.76\ \text{N}\) and \(B_z=-F_ra/L=-332.77\ \text{N}\).

4. Combine orthogonal results. \(R_A=\sqrt{A_y^2+A_z^2}=1581.06\ \text{N}\), \(R_B=972.96\ \text{N}\), and at the gear \(M=\sqrt{(A_ya)^2+(A_za)^2}=252.97\ \text{N}\!\cdot\!\text{m}\).

Verification

Units: Reaction equations multiply force by length and divide by length, leaving newtons; reaction times the \(0.160\ \text{m}\) lever arm gives \(\text{N}\!\cdot\!\text{m}\).

Signs and direction: \(A_y+B_y=2400\ \text{N}\) opposes the downward tangential force, while \(A_z+B_z=-873.53\ \text{N}\) opposes the positive radial force, satisfying both sign conventions.

Plausibility: A is closer to the gear and therefore has the larger reaction in both planes. Using the right segment gives \(B_y(0.260)=237.71\ \text{N}\!\cdot\!\text{m}\), matching the left-segment moment.

Final result: \(\boxed{\mathbf{R}_A=(1485.7\,\hat y-540.8\,\hat z)\ \text{N},\ R_A=1.581\ \text{kN};\ \mathbf{R}_B=(914.3\,\hat y-332.8\,\hat z)\ \text{N},\ R_B=0.973\ \text{kN};\ M=253.0\ \text{N}\!\cdot\!\text{m}}\).

Engineering interpretation

The resultant bending moment belongs at the gear section, while bearing loads act at their own seats. A final model must add weights, other gears, coupling or belt loads, thrust, load spectrum, housing flexibility, and the selected bearing's equivalent-load rules.

18–6Shaft Material Selection

Strength, stiffness, and production are separate decisions

Specify material by standard designation, condition, section size, heat treatment, and verified minimum properties. Yield and fatigue strengths influence stress margin; elastic modulus controls shaft deflection and changes little among ordinary steels. Upgrading steel strength therefore does not meaningfully repair a stiffness failure. Machinability, hardenability, distortion, corrosion, stock availability, inspection, and cost belong in the decision record.

18–7Shaft Design for Stress

Nominal moments become local fatigue stress

A rotating shaft under a stationary transverse load experiences fully reversed bending at a material point, while steady transmitted torque produces mean shear. Critical sections occur at shoulders, keyways, grooves, and load application points. Use fatigue concentration factors for alternating stress and the selected criterion's treatment for mean stress; static yield must also be checked at peak load.

DE–Goodman first-pass diameter for rotating bending and steady torque

\[d=\left[\frac{16n}{\pi}\left(\frac{2K_fM_a}{S_e}+\frac{\sqrt{3}K_{fs}T_m}{S_{ut}}\right)\right]^{1/3}\]

This special case assumes solid circular geometry, fully reversed bending, steady torque, ductile behavior, and consistent force-length units. Supplied \(S_e\), \(K_f\), and \(K_{fs}\) must reflect surface, size, reliability, temperature, notch sensitivity, and the actual feature geometry.

18–8Shaft Design for Deflection

Mesh alignment can govern before strength

Deflection and slope depend on the complete stepped shaft, bearing boundary conditions, load positions, and \(EI\). Gear meshes and bearings have application-specific alignment limits that must come from the gear accuracy method and bearing supplier. For geometrically similar diameter scaling under unchanged load, bending deflection varies approximately as \(d^{-4}\), enabling a useful trial update.

Diameter scaling for a deflection limit

\[d_{\mathrm{new}}=d_{\mathrm{old}}\left(\frac{\delta_{\mathrm{old}}}{\delta_{\mathrm{allow}}}\right)^{1/4}\]

EXAMPLE 3Stress sizing followed by stiffness-driven iteration

At a trial shaft shoulder, analysis gives fully reversed bending \(M_a=260\ \text{N}\!\cdot\!\text{m}\) and steady torque \(T_m=500\ \text{N}\!\cdot\!\text{m}\). Supplied design data are \(K_f=1.60\), \(K_{fs}=1.40\), corrected \(S_e=240\ \text{MPa}\), \(S_{ut}=620\ \text{MPa}\), and target \(n=2.00\). After choosing a \(40\ \text{mm}\) trial diameter, a validated beam model predicts \(0.180\ \text{mm}\) gear deflection versus a supplied \(0.120\ \text{mm}\) limit. Find the stress diameter, select the next \(5\ \text{mm}\) standard diameter, and estimate its fatigue factor.

Recognition strategy

The load pattern matches rotating bending with steady torsion, so the stated DE–Goodman special case provides a first stress diameter. The subsequent measured ratio of calculated to allowable deflection triggers a stiffness iteration, not a material-strength substitution.

Given information

Use \(M_a=260000\ \text{N}\!\cdot\!\text{mm}\), \(T_m=500000\ \text{N}\!\cdot\!\text{mm}\), \(K_f=1.60\), \(K_{fs}=1.40\), \(S_e=240\ \text{N/mm}^2\), \(S_{ut}=620\ \text{N/mm}^2\), and \(n=2.00\).
The beam result and \(0.120\ \text{mm}\) limit are supplied for the same load state; diameters are available in \(5\ \text{mm}\) increments.

Model and interpretation

The shaft is modeled as solid and circular at the critical section. Stress factors apply locally; the fourth-root scaling assumes the shaft geometry is scaled sufficiently uniformly that bending stiffness changes approximately with \(d^4\). Final stepped-shaft analysis is still required.

Step-by-step solution

1. Assemble fatigue demand. The bracketed load-strength quantity is \(A=2K_fM_a/S_e+\sqrt{3}K_{fs}T_m/S_{ut}=5422.21\ \text{mm}^3\).

2. Size for stress. \(d=[16nA/\pi]^{1/3}=[16(2.00)(5422.21)/\pi]^{1/3}=38.08\ \text{mm}\), so the first standard trial is \(40\ \text{mm}\).

3. Iterate for stiffness. \(d_{\mathrm{scale}}=40(0.180/0.120)^{1/4}=44.27\ \text{mm}\). Rounding upward to the next available increment selects \(45\ \text{mm}\).

4. Re-estimate fatigue margin. For unchanged factors and loads, \(n\propto d^3\), hence \(n_{45}=2.00(45/38.0825)^3=3.30\). The scaled deflection is \(0.180(40/45)^4=0.1124\ \text{mm}\).

Verification

Units: Each moment divided by stress has units \((\text{N}\!\cdot\!\text{mm})/(\text{N/mm}^2)=\text{mm}^3\); the cube root therefore yields millimetres, while both scaling ratios are dimensionless.

Signs and direction: Alternating amplitude and steady torque are nonnegative fatigue magnitudes. Deflection direction is irrelevant to the stated magnitude limit, so the absolute deflection ratio is used for resizing.

Plausibility: Increasing diameter from \(40\) to \(45\ \text{mm}\) must reduce deflection and increase stress margin; the estimates do both, giving \(0.1124<0.120\ \text{mm}\) and \(n=3.30>2.00\).

Final result: \(\boxed{d_{\mathrm{stress}}=38.08\ \text{mm};\quad d_{\mathrm{selected}}=45\ \text{mm};\quad \delta_{45}\approx0.112\ \text{mm};\quad n_{45}\approx3.30}\).

Engineering interpretation

Stiffness, not fatigue strength, sets the selected standard diameter in this iteration. Before release, rerun the full stepped-shaft beam model and local fatigue and peak-yield checks with actual fillets, keyway or groove factors, fits, tolerances, bearing compliance, and the documented load spectrum.

18–9Bearing Selection

Life calculation meets catalog geometry

Convert radial and axial reactions into the bearing manufacturer's equivalent dynamic load for the selected bearing type, then calculate required catalog rating for life and reliability. Exponents, reliability modifiers, load factors, and static safety criteria depend on the cited standard and catalog. The selected bore changes the shaft seat; outer diameter, width, corner radius, fits, speed limit, lubrication, sealing, and locating/floating function change the housing and layout.

18–10Key and Retaining Ring Selection

Transmit torque without creating an avoidable weak section

A key sees nominal shear and hub-contact pressure, while its keyway raises local shaft stress. Select a standard cross-section compatible with shaft diameter, fit, hub material, and hub length; verify both key and hub and include keyway fatigue effects in the shaft. Retaining rings are catalog components with specified groove geometry and thrust capacity. Keep grooves away from high bending where possible and use the actual groove concentration data in final analysis.

Idealized rectangular-key checks

\[F=\frac{2T}{d},\qquad \tau=\frac{F}{bL},\qquad \sigma_{\mathrm{bearing}}=\frac{2F}{hL}\]

These nominal relations do not supply allowable stress, fatigue behavior, fit, or load distribution. Those values must follow the selected material specification, key standard, hub geometry, and service duty.

18–11Final Analysis

Verify the released configuration

Replace every trial diameter, estimated factor, and generic component with the actual drawing dimension, tolerance, material condition, catalog part, lubrication state, and duty spectrum. Recalculate gear ratings, shaft fatigue and yield, deflection and critical speed, bearing life and static margin, keys, retaining features, fits, housing distortion, heat rejection, and fasteners. Review interfaces and worst credible combinations rather than combining unrelated worst cases without justification.

A convergence criterion can limit relative changes in coupled quantities, but passing it does not prove adequacy. The final gate is satisfaction of all acceptance criteria with consistent inputs, followed by independent review and appropriate prototype, endurance, noise, temperature, or overload testing.

Chapter recap

Begin with requirements and propagate consistent speed, power, efficiency, and torque through every shaft.
Gear geometry creates forces and package constraints; two-plane free bodies create the reactions used by shafts and bearings.
Stress, fatigue strength, stiffness, bearing life, and catalog capacity are distinct checks with different inputs and failure meanings.
Standard sizes, fits, assembly access, lubrication, thermal growth, inspection, and data provenance are design requirements, not afterthoughts.
Before moving on, be able to trace one requirement through the calculation set, identify every iteration loop, and audit the final released geometry against all acceptance criteria.