9.2 Spur Gears: Geometry, Kinematics & AGMA Stress Equations

9.2 Spur Gears: Geometry, Kinematics & AGMA Stress Equations

Spur gears are the simplest and most widely used power transmission elements for parallel rotating shafts. Precise gear design requires balancing geometric kinematic requirements (conjugate action, contact ratio, and non-undercutting geometry) against structural failure modes governed by the American Gear Manufacturers Association (AGMA) standards: tooth root bending fatigue and surface contact pitting.

---

1. Spur Gear Tooth Geometry & Involute Proportions

The fundamental law of gearing dictates that for a constant angular velocity ratio between two meshing gears, the common normal to the tooth profiles at their contact point must pass through a fixed point on the line of centers known as the pitch point $P$. The involute curve satisfies this condition and provides conjugate action even when shaft center distance deviates slightly from nominal.

                                [Base Circle]
                               /             \
     [Pitch Circle]           /               \
    /              \         /                 \
   /     Pitch      \       /                   \
  |      Point       |     |     Involute        |
  |        P         |-----|     Tooth Profile   |
   \                /       \                   /
    \              /         \                 /
     \            /           \               /
                               \             /
                                [Base Circle d_b = d * cos(φ)]

  Pressure Angle (φ): Angle between Line of Action and common tangent to pitch circles.
  Standard: φ = 20° (historic: 14.5°, heavy-duty: 25°).

Fundamental Geometric Definitions

• Pitch Diameter ($d$): The theoretical diameter of the pitch cylinder rolling without slip: $d = N / P_d = m N$.
• Diametral Pitch ($P_d$): Teeth per unit pitch diameter (in US Customary units, $\text{in}^{-1}$): $P_d = N / d$.
• Module ($m$): Pitch diameter in millimeters per tooth (in metric units, mm): $m = d / N = 25.4 / P_d$.
• Circular Pitch ($p$): Distance along the pitch circle from a point on one tooth to the corresponding point on the next: $p = \pi d / N = \pi / P_d = \pi m$.
• Base Circle Diameter ($d_b$): The fundamental generating cylinder from which the involute profile originates: $d_b = d \cos \phi$.
• Base Pitch ($p_b$): Distance along the base circle: $p_b = p \cos \phi = \pi d_b / N$.
• Standard Tooth Proportions (Full-Depth Involute):
  • Addendum: $a = 1.0 / P_d = 1.0 m$
  • Dedendum: $b = 1.25 / P_d = 1.25 m$
  • Clearance: $c = b - a = 0.25 / P_d = 0.25 m$
  • Whole Depth: $h_t = a + b = 2.25 / P_d = 2.25 m$
  • Working Depth: $h_k = 2 a = 2.0 / P_d = 2.0 m$
  • Tooth Thickness on Pitch Circle: $t = p / 2 = \pi / (2 P_d)$

Contact Ratio ($m_c$)

The contact ratio represents the average number of tooth pairs in contact simultaneously:

$$m_c = \frac{L_{ab}}{p_b} = \frac{\sqrt{r_{a,P}^2 - r_{b,P}^2} + \sqrt{r_{a,G}^2 - r_{b,G}^2} - C \sin \phi}{p \cos \phi}$$

Where $r_{a}$ is addendum radius ($d/2 + a$), $r_b$ is base radius ($d_b / 2$), and $C$ is center distance ($(d_P + d_G)/2$). To ensure continuous, quiet power transfer without dynamic impact as teeth engage and disengage, designers keep the contact ratio at or above about 1.2:

$$m_c \ge 1.2 \quad \text{(Recommended: } 1.4 \le m_c \le 1.8\text{)}$$

Interference & Undercutting Thresholds

Interference occurs when contact takes place along the line of action outside the tangent points of the base circles, causing the hardened tip of one tooth to dig into the flank of its mating tooth. During gear manufacturing (e.g., hobbing with a rack cutter), interference causes the cutter to remove material from the dedendum root, weakening the tooth bending strength (undercutting).

For a pinion generated by a standard full-depth rack cutter ($k = 1.0$), the minimum tooth count to prevent undercutting is:

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

For two identical equal gears ($1:1$ ratio) meshing together, the threshold is lower:

$$N_{\min} = \frac{2 k}{3 \sin^2 \phi}\left(1 + \sqrt{1 + 3\sin^2 \phi}\right) \xrightarrow{\phi = 20^\circ} N_{\min} \approx 12.3 \implies 13\text{ teeth}$$

---

2. Gear Kinematics & Epicyclic (Planetary) Trains

Pitch Line Velocity & Transmitted Tangential Force

For a pinion rotating at $n_P\text{ rpm}$ transmitting power $H\text{ (hp)}$ or $P\text{ (kW)}$:

$$V = \frac{\pi d_P n_P}{12}\text{ ft/min} = \frac{\pi d_P n_P}{60}\text{ m/s}$$ $$W_t = \frac{33000 H}{V}\text{ lbf} = \frac{1000 P}{V}\text{ N} = \frac{2 T_P}{d_P}$$

Separating radial force: $W_r = W_t \tan \phi$. Total normal force: $W = W_t / \cos \phi$.

Simple and Compound Gear Trains

For a train of external spur gears, the train value (output speed divided by input speed) is

$$e = \frac{n_L}{n_F} = \pm \frac{\text{product of driving-gear tooth numbers}}{\text{product of driven-gear tooth numbers}}$$

Each external mesh reverses the direction of rotation, so the sign is positive for an even number of external meshes and negative for an odd number; an internal (ring) mesh does not reverse direction.

• Simple train: one gear per shaft. Intermediate idler gears cancel out of the tooth ratio ($e = -N_1/N_3$ for gear 1 driving gear 3 through idler 2) but change the direction of rotation and the center distance.
• Compound train: two gears fixed to an intermediate shaft, so the stage ratios multiply. A single spur stage is usually kept below about 10:1; larger reductions use two or more stages.
• Reverted compound train: the output shaft is coaxial with the input shaft, which requires equal center distances in both stages, $r_2 + r_3 = r_4 + r_5$ (for equal pitch, $N_2 + N_3 = N_4 + N_5$).

Example: a 20-tooth pinion drives a 60-tooth gear; a 15-tooth pinion on the same shaft drives a 45-tooth output gear. Then $e = (20 \times 15)/(60 \times 45) = 1/9$, two external meshes give the same rotation direction as the input, and a $1{,}800\,\text{rpm}$ input produces $200\,\text{rpm}$ at the output. Because $20 + 60 = 80$ but $15 + 45 = 60$, this train cannot be reverted without changing the second-stage pitch.

Epicyclic (Planetary) Gear Trains

Epicyclic trains provide compact, coaxial, high-ratio speed reduction. The basic stage comprises a central Sun gear ($S$), several Planet gears ($P$) mounted on a rotating Carrier Arm ($A$), and an outer internal Ring/Annulus gear ($R$).

                   +----------------------------+
                   |      Ring Gear (R)         |
                   |  +----------------------+  |
                   |  |      Planet (P)      |  |
                   |  |     +----------+     |  |
                   |  |     | Sun (S)  |     |  |
                   |  |     |    (O)   |     |  |
                   |  |     +----------+     |  |
                   |  |     [Carrier A]      |  |
                   |  +----------------------+  |
                   +----------------------------+

  Pitch Diameter Constraint: d_R = d_S + 2 * d_P  ==>  N_R = N_S + 2 * N_P

The fundamental kinematic equation relating the first gear ($F$), last gear ($L$), and carrier arm is:

$$\frac{\omega_L - \omega_{\text{arm}}}{\omega_F - \omega_{\text{arm}}} = e$$

Where $e$ is the train value with the carrier locked stationary ($e = (-N_S / N_P) \times (+N_P / N_R) = -N_S / N_R$).

---

3. AGMA Tooth Bending Stress (ANSI/AGMA 2001-D04)

The historic Lewis equation modeled the tooth as a cantilever beam loaded at the tip: $\sigma = W_t P_d / (F Y)$. The modernized AGMA Bending Stress Equation refines this by incorporating dynamic velocity magnification, load distribution, and fillet stress concentration evaluated at the Highest Point of Single Tooth Contact (HPSTC):

$$\sigma = W_t K_o K_v K_s \frac{P_d}{F} \frac{K_m K_B}{J} \quad \left[ \text{SI: } \sigma = \frac{W_t}{F m} K_o K_v K_s K_m \frac{K_B}{J} \right]$$

Where the governing parameters are defined as follows:

• $W_t$: Transmitted tangential pitch-line load (lbf or N).
• $K_o$: Overload factor; accounts for roughness and shock from prime mover and driven load (uniform $= 1.0$, light shock $= 1.25$, heavy shock $= 1.75 - 2.25$).
• $K_v$: Dynamic factor; accounts for tooth spacing inaccuracies and pitch line velocity $V$ via the AGMA Transmission Accuracy Number $Q_v$ ($Q_v = 5 - 11$):

$$K_v = \left( \frac{A + \sqrt{V}}{A} \right)^B, \quad B = 0.25(12 - Q_v)^{2/3}, \quad A = 50 + 56(1 - B)$$

• $K_s$: Size factor; accounts for non-uniform material properties across thick tooth modules ($K_s = 1.0$ for $P_d \ge 5$).
• $F$: Active face width of the gear tooth (in or mm). Recommended: $8/P_d \le F \le 16/P_d$.
• $K_m$: Load distribution factor; accounts for shaft deflection, housing misalignment, and tooth crowning ($K_m \ge 1.0$).
• $K_B$: Rim thickness factor; for solid gears $K_B = 1.0$; for thin-rimmed gears with backup ratio $m_B < 1.2$, $K_B > 1.0$.
• $J$: AGMA bending geometry factor; incorporates the Lewis form factor, tooth root fillet stress concentration factor, and load sharing at the HPSTC.

Allowable bending stress check: $\sigma \le \sigma_{\text{all}} = \frac{S_t}{S_F} \frac{Y_N}{K_T K_R}$, where $S_t$ is bending fatigue strength, $S_F$ is bending safety factor, and $Y_N$ is stress cycle factor.

---

4. AGMA Pitting Resistance (Surface Contact Stress)

Gear surface failure (pitting) results from repeated cyclic contact stresses governed by Hertzian contact between two cylinders. The AGMA Contact Stress Equation is:

$$\sigma_c = Z_E \sqrt{W_t K_o K_v K_s \frac{K_m}{d_P F} \frac{C_f}{I}}$$

• $Z_E$: Elastic coefficient, derived from Poisson's ratio $\nu$ and Young's modulus $E$ of the pinion and gear:

$$Z_E = \sqrt{\frac{1}{\pi \left( \frac{1-\nu_P^2}{E_P} + \frac{1-\nu_G^2}{E_G} \right)}}$$

For steel mating with steel ($E = 30 \times 10^6\text{ psi}, \nu = 0.30$), $Z_E \approx 2300\sqrt{\text{psi}}$ ($191\sqrt{\text{MPa}}$).

• $I$: AGMA surface geometry factor, defined for external spur gears as:

$$I = \frac{\cos \phi \sin \phi}{2 m_N} \left( \frac{m_G}{m_G + 1} \right)$$

Where $m_N = 1.0$ for spur gears (load sharing ratio) and $m_G = N_G / N_P$.

• $C_f$: Surface condition factor ($C_f = 1.0$ for standard ground/shaved finishes).

5. Complete Worked Problem: AGMA Bending & Contact Stress Analysis

Problem Statement

A commercial single-reduction spur gear drive transmits $H = 25\text{ hp}$ from an electric motor running at $n_P = 1750\text{ rpm}$ to an industrial blower. The drive specifications are:

• Diametral Pitch: $P_d = 5\text{ in}^{-1}$
• Pressure Angle: $\phi = 20^\circ$ (full-depth involute)
• Pinion teeth: $N_P = 20$; Gear teeth: $N_G = 60$
• Face width: $F = 1.75\text{ in}$
• AGMA Quality Number: $Q_v = 6$
• Application Overload Factor: $K_o = 1.25$ (light shock)
• Load Distribution Factor: $K_m = 1.22$
• Size Factor: $K_s = 1.0$; Rim Factor: $K_B = 1.0$; Surface Factor: $C_f = 1.0$
• AGMA Pinion Geometry Factors: Bending $J = 0.35$; Contact $I = 0.1205$
• Material: Grade 1 through-hardened steel ($250\text{ HB}$, $Z_E = 2300\sqrt{\text{psi}}$, AGMA Grade 1 allowables from Shigley: bending $S_t = 77.3\,\text{HB} + 12{,}800 = 32{,}125\text{ psi} \approx 32.1\text{ ksi}$ and contact $S_c = 322\,\text{HB} + 29{,}100 = 109{,}600\text{ psi} \approx 109.6\text{ ksi}$; take $Y_N = Z_N = K_T = K_R = 1$ for $10^7$ cycles at 99% reliability)

Step 1: Kinematics & Transmitted Tangential Load

Calculate pitch diameters and pitch line velocity:

$$d_P = \frac{N_P}{P_d} = \frac{20}{5} = 4.00\text{ in}, \quad d_G = \frac{N_G}{P_d} = \frac{60}{5} = 12.00\text{ in}$$ $$V = \frac{\pi d_P n_P}{12} = \frac{\pi \times 4.00\text{ in} \times 1750\text{ rpm}}{12} = 1832.6\text{ ft/min}$$

Calculate transmitted tangential load $W_t$:

$$W_t = \frac{33000 H}{V} = \frac{33000 \times 25\text{ hp}}{1832.6\text{ ft/min}} = 450.2\text{ lbf}$$

Step 2: Dynamic Factor $K_v$

For $Q_v = 6$:

$$B = 0.25 (12 - Q_v)^{2/3} = 0.25 (12 - 6)^{2/3} = 0.25 \times (6)^{2/3} = 0.25 \times 3.3019 = 0.8255$$ $$A = 50 + 56(1 - B) = 50 + 56(1 - 0.8255) = 50 + 56(0.1745) = 59.77$$ $$K_v = \left( \frac{A + \sqrt{V}}{A} \right)^B = \left( \frac{59.77 + \sqrt{1832.6}}{59.77} \right)^{0.8255} = \left( \frac{59.77 + 42.81}{59.77} \right)^{0.8255} = (1.716)^{0.8255} = 1.562$$

Step 3: AGMA Tooth Bending Stress Calculation

Apply the AGMA bending stress equation:

$$\sigma = W_t K_o K_v K_s \frac{P_d}{F} \frac{K_m K_B}{J}$$

Evaluate intermediate terms:

$$W_t K_o K_v K_s = 450.2 \times 1.25 \times 1.562 \times 1.0 = 879.0\text{ lbf}$$ $$\frac{P_d}{F} = \frac{5}{1.75} = 2.857\text{ in}^{-2}$$ $$\frac{K_m K_B}{J} = \frac{1.22 \times 1.0}{0.35} = 3.486$$ $$\sigma = 879.0 \times 2.857 \times 3.486 = 8754\text{ psi} \approx 8.75\text{ ksi}$$

Bending factor of safety:

$$S_F = \frac{S_t}{\sigma} = \frac{32.1\text{ ksi}}{8.75\text{ ksi}} = 3.67 \quad \text{(Highly robust against bending fatigue)}$$

Step 4: AGMA Surface Contact Stress (Pitting Resistance)

Apply the AGMA contact stress equation:

$$\sigma_c = Z_E \sqrt{W_t K_o K_v K_s \frac{K_m}{d_P F} \frac{C_f}{I}}$$

Evaluate the terms under the square root:

$$\frac{K_m}{d_P F} = \frac{1.22}{4.00 \times 1.75} = \frac{1.22}{7.00} = 0.1743\text{ in}^{-2}$$ $$\frac{C_f}{I} = \frac{1.0}{0.1205} = 8.299$$ $$\text{Radicand} = 879.0 \times 0.1743 \times 8.299 = 1271.5\text{ psi}$$ $$\sqrt{1271.5} = 35.66\sqrt{\text{psi}}$$ $$\sigma_c = 2300\sqrt{\text{psi}} \times 35.66\sqrt{\text{psi}} = 82018\text{ psi} \approx 82.0\text{ ksi}$$

Contact factor of safety:

$$S_H = \frac{S_c}{\sigma_c} = \frac{109.6\text{ ksi}}{82.0\text{ ksi}} = 1.34$$

Because contact stress varies with $\sqrt{W_t}$, AGMA practice compares $S_F$ with $S_H^2$, not with $S_H$. Here $S_H^2 = 1.34^2 = 1.79$ is well below $S_F = 3.67$, so pitting (wear) is the threat that governs this gearset.