3.2 Spur, Helical, Bevel & Worm Gears and Epicyclic Gear Trains

Spur, Helical, Bevel & Worm Gears and Epicyclic Gear Trains

Gears are toothed cylindrical or conical mechanical elements designed to transmit positive, non-slip angular velocity and mechanical torque between rotating shafts. On the NCEES PE Mechanical exam, gear problems evaluate five fundamental competencies: involute tooth geometry and conjugate action, kinematic interference and undercutting limits, static and dynamic 3D tooth force resolution, AGMA fatigue ratings (tooth root bending vs. surface pitting), and compound/planetary gear train kinematics.

+---------------------------------------------------------------------------------------------------+
|                                GEAR CLASSIFICATION & CHARACTERISTICS                              |
|                                                                                                   |
|   SPUR GEARS:             HELICAL GEARS:           BEVEL GEARS:             WORM GEARSETS:        |
|   - Parallel shafts       - Parallel / Crossed     - Intersecting shafts    - Non-intersecting    |
|   - Straight axial teeth  - Inclined helix angle   - Conical pitch surface  - 90 deg shaft angle  |
|   - Zero axial thrust     - High contact ratio     - Combined thrust/radial - High ratio (10-100) |
|   - Low cost / standard   - Quiet & smooth mesh    - Shaft angle (Sigma=90) - Self-locking action |
+---------------------------------------------------------------------------------------------------+

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1. Fundamentals of Involute Gear Tooth Geometry

Modern gears utilize the involute of a circle tooth profile because it satisfies the Fundamental Law of Gearing: the common normal to the tooth profiles at all points of contact must pass through a fixed pitch point ($P$), ensuring a strictly constant angular velocity ratio ($m_V = \omega_{\text{in}}/\omega_{\text{out}} = \text{const}$) regardless of center distance variation.

+---------------------------------------------------------------------------------------------------+
|                             INVOLUTE SPUR GEAR TOOTH NOMENCLATURE                                 |
|                                                                                                   |
|                                    Addendum Circle (d_a = d + 2a)                                 |
|                            ---------------------------------------------                          |
|                            ^                                                                      |
|                  Addendum  | a = 1/P = 1.0 m                                                      |
|                            v       Pitch Circle (d = N/P = m*N)                                   |
|   =========================+============================================                          |
|                            ^                                                                      |
|                  Dedendum  | b = 1.25/P = 1.25 m                                                  |
|                            v       Base Circle (d_b = d * cos(phi))                               |
|                            ---------------------------------------------                          |
|                                    Dedendum / Root Circle (d_r = d - 2b)                          |
|                            ---------------------------------------------                          |
|                                                                                                   |
|   Circular Pitch: p = pi*d / N = pi/P = pi*m    |    Base Pitch: p_b = p * cos(phi)               |
|   Whole Depth: h_t = a + b = 2.25/P             |    Clearance: c = b - a = 0.25/P                |
+---------------------------------------------------------------------------------------------------+

Essential Geometric Equations

Interference, Undercutting & Minimum Pinion Teeth

When generating teeth with a rack cutter or hob, if the addendum of the generating rack extends inside the tangency point of the base circle, the cutter cuts away ("undercuts") the flank of the tooth, severely weakening the root fillet. To prevent undercutting in standard full-depth spur gears:

$N_{\min} = \frac{2 k}{\sin^2\phi}$

Where $k = 1.0$ for standard full-depth teeth and $k = 0.8$ for stub teeth:

• For $\phi = 20^\circ$ full-depth ($k=1$): $N_{\min} = \frac{2(1)}{\sin^2(20^\circ)} = \frac{2}{0.11698} = 17.1 \implies \mathbf{18\text{ teeth}}$.
• For $\phi = 25^\circ$ full-depth ($k=1$): $N_{\min} = \frac{2(1)}{\sin^2(25^\circ)} = \frac{2}{0.17861} = 11.2 \implies \mathbf{12\text{ teeth}}$.
• For $\phi = 14.5^\circ$ full-depth ($k=1$, obsolete): $N_{\min} = \frac{2(1)}{\sin^2(14.5^\circ)} = 31.9 \implies \mathbf{32\text{ teeth}}$.

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2. Pitch Line Velocity and 3D Tooth Force Resolution

Power is transmitted at the pitch circle at linear pitch line velocity $V$:

$V = \frac{\pi d_P N_P}{12} \quad (\text{ft/min in US Customary, with } d_P \text{ in inches, } N_P \text{ in rpm})$

$V = \frac{\pi d_P N_P}{60 \times 1000} \quad (\text{m/s in SI, with } d_P \text{ in mm, } N_P \text{ in rpm})$

+---------------------------------------------------------------------------------------------------+
|                                 GEAR TOOTH FORCE VECTORS                                          |
|                                                                                                   |
|   SPUR GEAR TOOTH FORCES:                           HELICAL GEAR TOOTH FORCES:                    |
|                                                                                                   |
|                 W (Total Normal Load)                             W_a (Axial Thrust = W_t*tan(psi)|
|                      /|                                              <------                      |
|                     / |                                                   | \                     |
|                    /  | W_r = W_t * tan(phi)                              |  \                    |
|                   /   |                                    W_r = W_t*tan(phi_t) \                 |
|                  /phi |                                                   |    \ W (Total Normal) |
|                 +-----+                                                   | phi_t\                |
|                   W_t (Tangential Load)                                   +-------+               |
|                                                                               W_t                 |
+---------------------------------------------------------------------------------------------------+

1. Spur Gear Force Resolution

• Tangential Force ($W_t$): Transmits power and driving torque $T$. $W_t = \frac{33000 P_{\text{hp}}}{V} = \frac{2 T}{d_P} \quad (\text{US: } W_t \text{ in lbf}) \quad \Big| \quad W_t = \frac{1000 P_{\text{kW}}}{V} = \frac{2 T}{d_P} \quad (\text{SI: } W_t \text{ in N})$
• Radial Separating Force ($W_r$): Pushes gear shafts apart, loading support bearings. $W_r = W_t \tan\phi$
• Total Normal Force ($W$): Acts perpendicular to the tooth involute along the line of action. $W = \frac{W_t}{\cos\phi} = \sqrt{W_t^2 + W_r^2}$

2. Helical Gear Force Resolution

Helical teeth are cut at helix angle $\psi$. Transverse pressure angle $\phi_t$ and normal pressure angle $\phi_n$ are related by:

$\tan\phi_t = \frac{\tan\phi_n}{\cos\psi}$

$\text{Normal Pitch: } P_n = \frac{P_t}{\cos\psi}, \quad p_n = p_t \cos\psi$

• Tangential Load: $W_t = \frac{2 T}{d_P}$
• Axial (Thrust) Load: $W_a = W_t \tan\psi$ (requires thrust bearings or herringbone double-helical teeth to cancel)
• Radial Separating Load: $W_r = W_t \tan\phi_t = W_t \frac{\tan\phi_n}{\cos\psi}$
• Total Resultant Normal Load: $W = \frac{W_t}{\cos\phi_n \cos\psi} = \sqrt{W_t^2 + W_r^2 + W_a^2}$

3. Straight Bevel Gear Force Resolution

Bevel gears operate on intersecting shafts (typically at shaft angle $\Sigma = 90^\circ$). Pitch cone angles satisfy $\tan\gamma = \frac{N_P}{N_G}$ (pinion) and $\tan\Gamma = \frac{N_G}{N_P}$ (gear).

• Tangential Force: $W_t = \frac{2 T}{d_{\text{avg}}}$ (acting at midpoint of tooth face)
• Radial Separating Force (Pinion): $W_{rP} = W_t \tan\phi \cos\gamma$
• Axial Thrust Force (Pinion): $W_{aP} = W_t \tan\phi \sin\gamma$
• (Note: For $90^\circ$ shaft angle, $W_{aP} = W_{rG}$ and $W_{rP} = W_{aG}$)

4. Worm Gearset Forces & Efficiency

A worm gearset transmits power between non-intersecting, perpendicular shafts. Ratio $m_G = \frac{N_G}{N_W}$ (where $N_W$ is number of worm starts: 1, 2, 3, or 4). Lead angle $\lambda$ satisfies $\tan\lambda = \frac{L}{\pi d_W} = \frac{N_W p_x}{\pi d_W}$.

• Equilibrium Relations: $W_{t,W} = W_{a,G}$, $W_{a,W} = W_{t,G}$, and $W_{r,W} = W_{r,G}$.
• Mesh Efficiency with Friction $\mu$: $\eta = \frac{\cos\phi_n - \mu \tan\lambda}{\cos\phi_n + \mu \cot\lambda}$
• Self-Locking Condition: The gearset cannot be backdriven by the gear when $\tan\lambda \le \mu$ (typically $\lambda < 5^\circ$).

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3. AGMA Tooth Root Bending Fatigue Rating

Gear teeth act as short cantilever beams. The AGMA bending stress equation modifies the classic Lewis formula with empirical dynamic and geometry correction factors:

$\sigma_b = W_t K_o K_v K_s \frac{P}{F} \frac{K_m K_B}{J} \le \frac{S_t Y_N}{S_F Y_\theta Y_Z} \quad (\text{US Customary})$

$\sigma_b = \frac{W_t}{b m} K_A K_v K_s K_H \frac{1}{Y_J} \le \frac{\sigma_{FP} Y_N}{S_F Y_\theta Y_Z} \quad (\text{SI Metric})$

Where:

• $W_t = \text{Transmitted tangential load (lbf or N)}$
• $P = \text{Diametral pitch (}\text{in}^{-1}\text{)}$, $m = \text{Module (mm)}$
• $F = b = \text{Face width (in or mm)}$ (Standard range: $\frac{9}{P} \le F \le \frac{14}{P}$)
• $J = Y_J = \text{AGMA Bending Geometry Factor (incorporates Lewis form factor } Y \text{ and root stress concentration } K_f\text{)}$
• $K_o = K_A = \text{Overload / Application factor (1.0 uniform, 1.50 moderate shock, 2.0+ heavy shock)}$
• $K_v = \text{Dynamic factor (accounts for tooth pitch errors and pitch-line velocity vibration)}$
• $K_s = \text{Size factor (1.0 for standard pitches)}$
• $K_m = K_H = \text{Load distribution factor (accounts for shaft deflection and face crowning)}$
• $K_B = \text{Rim thickness factor (1.0 for solid gear webs)}$
• $S_t = \sigma_{FP} = \text{AGMA allowable bending fatigue strength}$
• $S_F = \text{Design safety factor against tooth bending breakage}$

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4. AGMA Surface Pitting Fatigue / Contact Stress

Surface fatigue failure (macropitting) occurs when cyclic Hertzian contact stresses exceed the surface endurance limit of the tooth material:

$\sigma_c = C_p \sqrt{W_t K_o K_v K_s \frac{K_m C_f}{d_P F I}} \le \frac{S_c Z_N C_H}{S_H Y_\theta Y_Z}$

Where:

• $C_p = \text{Elastic Coefficient} = \sqrt{\frac{1}{\pi \left( \frac{1 - \nu_P^2}{E_P} + \frac{1 - \nu_G^2}{E_G} \right)}}$ ($C_p \approx 2300\sqrt{\text{psi}}$ or $191\sqrt{\text{MPa}}$ for steel-on-steel)
• $I = \text{Surface Geometry Factor} = \frac{\sin\phi \cos\phi}{2 m_N} \frac{m_G}{m_G + 1}$ (for external spur gears)
• $d_P = \text{Pinion pitch diameter}$, $F = \text{Face width}$
• $C_f = \text{Surface condition factor (1.0)}$
• $S_c = \text{AGMA allowable contact fatigue strength}$
• $S_H = \text{Design safety factor against surface pitting}$

[!NOTE] Bending vs. Contact Stress Scaling: Bending stress scales linearly with load ($\sigma_b \propto W_t$), whereas contact stress scales with the square root of load ($\sigma_c \propto \sqrt{W_t}$). Doubling transmitted torque doubles root bending stress ($+100%$), but increases contact stress by only $41.4%$ ($\sqrt{2} = 1.414$). Surface pitting resistance almost universally governs steel gear sizing.

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5. Gear Train Kinematics & Planetary Gear Trains

+---------------------------------------------------------------------------------------------------+
|                                 GEAR TRAIN CONFIGURATIONS                                         |
|                                                                                                   |
|   SIMPLE TRAIN:                COMPOUND TRAIN:              PLANETARY (EPICYCLIC) TRAIN:          |
|   - One gear per shaft         - Multiple gears per shaft   - Sun (S), Planets (P), Ring (R)      |
|   - Idlers change direction    - Velocity ratio multiplied  - Carrier / Arm (C) rotates           |
|   - e = (-1)^k * (N_in/N_out)  - e = (Prod N_in)/(Prod N_out)- Willis: (n_L - n_C)/(n_F - n_C) = e |
+---------------------------------------------------------------------------------------------------+

1. Simple and Compound Gear Trains

• Simple Train: Idler gears between input and output reverse the direction of rotation but have zero effect on overall velocity ratio ($e = (-1)^k \frac{N_{\text{in}}}{N_{\text{out}}}$, where $k$ is number of external meshes).
• Compound Train: Multiple gears keyed to intermediate shafts multiply reductions: $e = \frac{\omega_{\text{output}}}{\omega_{\text{input}}} = \frac{\prod \text{Teeth on Driving Gears}}{\prod \text{Teeth on Driven Gears}}$

2. Planetary (Epicyclic) Gear Trains

Planetary gearsets provide high reduction ratios in compact, coaxial packages. A standard planetary set comprises a central Sun gear ($S$), multiple Planet gears ($P$), an internal toothed Ring gear ($R$), and a rotating Carrier/Arm ($C$).

+---------------------------------------------------------------------------------------------------+
|                             PLANETARY GEARSET COAXIAL GEOMETRY                                    |
|                                                                                                   |
|                                    +-----------------------+                                      |
|                                  /       RING GEAR (R)       \                                    |
|                                /       +---------------+       \                                  |
|                               |       /  PLANET (P)     \       |  Geometric Pitch Rule:          |
|                               |      |   (d_P, N_P)      |      |  d_R = d_S + 2*d_P              |
|                               |       \                 /       |                                 |
|                               |        +---------------+        |  Tooth Count Rule:              |
|                               |       /    SUN (S)      \       |  N_R = N_S + 2*N_P              |
|                               |      |   (d_S, N_S)      |      |                                 |
|                               |       \                 /       |  (For module m = const)         |
|                                \       +---------------+       /                                  |
|                                  \                           /                                    |
|                                    +-----------------------+                                      |
+---------------------------------------------------------------------------------------------------+

The Willis Relative Velocity Equation

Fix the carrier in thought to determine the basic train value $e_{\text{train}}$ from first gear ($F$) to last gear ($L$):

$\frac{n_L - n_C}{n_F - n_C} = e_{\text{train}}$

Where $n_F$, $n_L$, and $n_C$ are rotational speeds of First gear, Last gear, and Carrier. With Sun as first gear and Ring as last gear: $e_{\text{train}} = -\frac{N_S}{N_R}$.

Standard Planetary Operational Modes

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6. Step-by-Step Worked Problem: Helical Gear Forces & Planetary Ratio

Problem Statement

1. A $20^\circ$ normal pressure angle ($\phi_n = 20^\circ$) helical pinion with $N_P = 24$ teeth, transverse diametral pitch $P_t = 6\text{ in}^{-1}$, and right-hand helix angle $\psi = 30^\circ$ transmits $30\text{ hp}$ at $1800\text{ rpm}$ to a driven gear. Determine pitch diameter $d_P$, pitch line velocity $V$, tangential force $W_t$, radial force $W_r$, and axial thrust force $W_a$.
2. A planetary gearset has a Sun gear with $N_S = 20$ teeth and a Ring gear with $N_R = 80$ teeth. If the Ring gear is fixed ($n_R = 0$) and the Sun rotates clockwise at $1500\text{ rpm}$, compute the Carrier output speed $n_C$ and the number of teeth on each Planet gear $N_P$.

Step-by-Step Solution

Part 1: Helical Gear Force Calculation

• Pitch diameter: $d_P = \frac{N_P}{P_t} = \frac{24}{6} = 4.0\text{ inches}$.
• Pitch line velocity: $V = \frac{\pi d_P N_P}{12} = \frac{\pi (4.0)(1800)}{12} = 1884.96\text{ ft/min}$.
• Transmitted tangential force: $W_t = \frac{33000 P_{\text{hp}}}{V} = \frac{33000 \times 30}{1884.96} = 525.2\text{ lbf}$.
• Axial thrust force: $W_a = W_t \tan\psi = 525.2 \times \tan(30^\circ) = 525.2 \times 0.57735 = 303.2\text{ lbf}$.
• Transverse pressure angle: $\tan\phi_t = \frac{\tan\phi_n}{\cos\psi} = \frac{\tan(20^\circ)}{\cos(30^\circ)} = \frac{0.36397}{0.86603} = 0.42027 \implies \phi_t = 22.80^\circ$.
• Radial separating force: $W_r = W_t \tan\phi_t = 525.2 \times 0.42027 = 220.7\text{ lbf}$.
• Total resultant load: $W = \sqrt{W_t^2 + W_r^2 + W_a^2} = \sqrt{525.2^2 + 220.7^2 + 303.2^2} = 645.4\text{ lbf}$.

Part 2: Planetary Kinematics

• Planet tooth count: $N_R = N_S + 2 N_P \implies 80 = 20 + 2 N_P \implies 2 N_P = 60 \implies \mathbf{N_P = 30\text{ teeth}}$.
• Carrier speed using Willis formula (Sun = First, Ring = Last, $e_{\text{train}} = -\frac{N_S}{N_R} = -\frac{20}{80} = -0.25$): $\frac{n_R - n_C}{n_S - n_C} = -0.25 \implies \frac{0 - n_C}{1500 - n_C} = -0.25$ $-n_C = -0.25(1500 - n_C) = -375 + 0.25 n_C \implies 1.25 n_C = 375$ $n_C = \frac{375}{1.25} = \mathbf{+300\text{ rpm (Clockwise)}}$ (Ratio is $1 + \frac{N_R}{N_S} = 1 + \frac{80}{20} = 5:1$ reduction).

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7. Exam Tips & Common Traps

[!TIP] Planetary Gear Shortcut: When the Ring gear is stationary ($n_R = 0$), the ratio from Sun to Carrier is always $1 + (N_R / N_S)$. You can solve speed in 5 seconds without writing out the full Willis formula.

[!WARNING] Common Traps to Avoid:

• Normal vs Transverse Pitch: For helical gears, normal pitch $P_n$ is greater than transverse pitch $P_t$ ($P_n = P_t / \cos\psi$). Pitch diameter is computed using transverse pitch ($d = N/P_t$).
• Worm Gear Efficiency Sign: Make sure to distinguish between driving through the worm vs backdriving through the gear. Worms with lead angle $\lambda < \tan^{-1}(\mu)$ cannot be backdriven.