Gear design

A spur gear has straight teeth cut parallel to the axis and transmits motion between parallel shafts.

          addendum circle
        ___________
      /  _________  \
     /  / pitch   \  \
    |  |  circle   |  |
     \  \_________/  /
      \___________ /
        root circle

   tooth:  top land
           ^   addendum  (a)
   ------pitch circle------
           v   dedendum  (d)
           ^
           clearance (c)

Terms

• Pitch circle: the imaginary circle that rolls without slipping with the pitch circle of the mating gear. Its diameter is the pitch diameter $d = mz$.
• Circular pitch $p_c$: distance along the pitch circle from a point on one tooth to the same point on the next tooth, $p_c = \pi d/z = \pi m$.
• Module $m$: pitch diameter divided by number of teeth, $m = d/z$ (mm). It is the standard size parameter in SI.
• Diametral pitch $P_d$: number of teeth per unit pitch diameter, $P_d = z/d$ (per inch).
• Addendum $a$: radial height of tooth above the pitch circle, $a = m$ for standard teeth.
• Dedendum $b_d$: radial depth of tooth below the pitch circle, $1.25m$ (full depth, 20 degrees).
• Clearance $c$: space between the top of a tooth of one gear and the root of the mating gear, $c = 0.25m$.
• Pressure angle $\phi$: angle between the common normal (line of action) at the contact point and the common tangent to the pitch circles. Standard values are 14.5 degrees and 20 degrees (20 degrees is more used, as it gives stronger teeth and fewer teeth without undercut).
• Backlash: the gap between the non-working flanks of mating teeth along the pitch circle. It allows for lubrication, thermal expansion and errors, and prevents jamming.

Relation

$$p_c = \pi m = \frac{\pi}{P_d}, \qquad P_d\ [\text{per inch}] = \frac{25.4}{m\ [\text{mm}]}$$

Standard modules (IS) include 1, 1.25, 1.5, 2, 2.5, 3, 4, 5, 6, 8, 10 mm. Gears must have the same module and pressure angle to mesh.

Lewis (1892) treated a gear tooth as a cantilever beam loaded at its tip by the full tangential load, and found its load-carrying strength in bending.

Assumptions

• The full load $W_t$ acts at the tip of one tooth only.
• The load is uniformly distributed across the face width $b$.
• The radial component (compression) and friction are neglected.
• The effect of stress concentration at the root is neglected.
• The tooth shape is taken as a parabola (a beam of uniform strength), with the section of maximum stress at the point where the parabola touches the tooth profile.

         Wt
          |
          v
        __|__        <- tip
       /  |  \
      /   | h \  parabola
     /   _|_   \
    |<--- t --->|   critical section

Derivation

Let $t$ = tooth thickness at the critical section, $h$ = height of the load above that section, $b$ = face width.

Bending moment $M = W_t\,h$. Section modulus $Z = \dfrac{b t^2}{6}$.

$$\sigma_b = \frac{M}{Z} = \frac{6\,W_t\,h}{b\,t^2}$$

So the permissible tangential load is

$$W_t = \frac{\sigma_b\, b\, t^2}{6h}$$

The critical section is where the inscribed parabola touches the tooth profile. From the similar triangles of the parabola, $t^2 = 4hx$, where $x$ is a length fixed by the tooth profile. Substituting,

$$W_t = \frac{\sigma_b\, b \cdot 4hx}{6h} = \sigma_b\, b\,\frac{2x}{3}$$

Multiplying and dividing by the circular pitch $p_c = \pi m$:

$$W_t = \sigma_b\, b\, \pi m\left(\frac{2x}{3 p_c}\right)$$

The dimensionless term in brackets is the Lewis form factor $y$:

$$\boxed{W_t = \sigma_b\, b\, \pi\, m\, y}$$

where $\sigma_b$ is the allowable bending stress ($\sigma_o$ static or $\sigma_o C_v$ with velocity factor). With $\pi y = Y$ (in the other form), $W_t = \sigma_b b m Y$.

Form factor

$y$ depends on the tooth profile and number of teeth $z$; it increases with $z$:

• 20 degree full depth: $y = 0.154 - \dfrac{0.912}{z}$
• 14.5 degree composite or full depth: $y = 0.124 - \dfrac{0.684}{z}$

For a given pair of gears, the weaker of the pinion and the gear (smaller $\sigma_b y$) is designed. The pinion has fewer teeth, so smaller $y$, but is often stronger in material.

Modes of failure of gear teeth

Design of gears is usually based on beam strength (bending), checked for wear strength (surface durability) and for dynamic load.

Surface fatigue (pitting)

The contact between two teeth is like two cylinders pressed together (Hertz contact). The maximum contact stress is large and repeats with every mesh, so subsurface shear stresses cause a crack that grows to the surface, and a small pit forms. Pitting begins near the pitch line, where only rolling happens, and the oil film is thin.

Buckingham's equation for limiting wear load

$$F_w = d_1\, b\, Q\, K$$

where

• $F_w$ is the limiting (maximum) wear load, N;
• $d_1$ is the pitch diameter of the pinion, mm;
• $b$ is the face width, mm;
• $Q$ is the ratio factor, $Q = \dfrac{2 z_2}{z_1 + z_2}$ for external gears ($z_2/z_1$ gear ratio; $Q = \dfrac{2z_2}{z_2 - z_1}$ for internal gears);
• $K$ is the load-stress factor (N/mm$^2$), from Hertz stress:

$$K = \frac{\sigma_{es}^2 \sin\phi}{1.4}\left(\frac{1}{E_1} + \frac{1}{E_2}\right)$$

with $\sigma_{es}$ the surface endurance limit (can be taken as $2.75\,\text{BHN} - 70$ MPa for steel, depending on textbook), $\phi$ the pressure angle, and $E_1$, $E_2$ the moduli of elasticity.

To be safe against pitting the wear load must be at least equal to the dynamic load: $F_w \ge F_d$. The surface hardness is therefore raised by through hardening, case hardening, flame or induction hardening.

 Motor ->[1]---[2]  shaft 2
         20T   60T  (2 and 3 fixed together)
               [3]---[4] -> output
               24T   72T

Speed ratio and output speed

$$\text{Ratio} = \frac{z_2}{z_1}\times\frac{z_4}{z_3} = \frac{60}{20}\times\frac{72}{24} = 3 \times 3 = 9$$ $$n_2 = \frac{1440}{3} = 480\ \text{rpm}, \qquad n_4 = \frac{480}{3} = 160\ \text{rpm}$$

Torques and power

$$T_1 = \frac{60P}{2\pi n_1} = \frac{60 \times 12\,000}{2\pi \times 1440} = 79.58\ \text{N.m}$$

Each pair multiplies the torque by its ratio and efficiency 0.96:

$$\begin{aligned} T_2 &= T_1 \times 3 \times 0.96 = 229.2\ \text{N.m} \\ T_4 &= T_2 \times 3 \times 0.96 = 660.1\ \text{N.m} \end{aligned}$$ $$P_{out} = 12 \times 0.96^2 = 11.06\ \text{kW} \quad(\text{check: } 660.1 \times 2\pi \times 160/60 = 11.06\ \text{kW})$$

Tooth forces (pitch diameters $d = mz$)

First pair (pinion 1: $d_1 = 4 \times 20 = 80$ mm):

$$\begin{aligned} F_t &= \frac{T_1}{d_1/2} = \frac{79.58}{0.04} = 1989\ \text{N} \\ F_r &= F_t \tan 20^\circ = 1989 \times 0.364 = 724\ \text{N} \\ F_n &= \frac{F_t}{\cos 20^\circ} = 2117\ \text{N} \end{aligned}$$

Second pair (pinion 3: $d_3 = 4 \times 24 = 96$ mm, torque $T_2 = 229.2$ N.m):

$$\begin{aligned} F_t &= \frac{229.2}{0.048} = 4775\ \text{N} \\ F_r &= 4775 \times 0.364 = 1738\ \text{N} \\ F_n &= \frac{4775}{\cos 20^\circ} = 5081\ \text{N} \end{aligned}$$

The tangential force gives the torque, the radial force tends to push the gears apart and loads the shaft bearings.

Answer: Output 160 rpm; output torque 660 N.m; output power 11.06 kW; first pair Ft = 1989 N, Fr = 724 N; second pair Ft = 4775 N, Fr = 1738 N.

Form factors

$$y_1 = 0.154 - \frac{0.912}{22} = 0.1125, \qquad y_2 = 0.154 - \frac{0.912}{66} = 0.1402$$

Compare $\sigma_o y$:

$$\text{Pinion: } 210 \times 0.1125 = 23.6, \qquad \text{Gear: } 100 \times 0.1402 = 14.0$$

The product is smaller for the gear (cast iron), so the gear is the weaker element and is used for design.

Design by trial of module

Load required: $F_t = \dfrac{P \times C_s}{v}$, with $v = \dfrac{\pi\, m\, z_1\, n_1}{60\,000}$ (m in mm). Gear capacity: $F_s = \sigma_{o2}\,C_v\, b\,\pi m\, y_2$ with $b = 10m$.

For example for $m = 5$:

$$F_s = 100 \times 0.3027 \times 50 \times \pi \times 5 \times 0.1402 = 3332\ \text{N} > 3255\ \text{N}$$

So the smallest standard module that works is m = 5 mm.

Dimensions

$$b = 10m = 50\ \text{mm}, \qquad d_1 = mz_1 = 5 \times 22 = 110\ \text{mm}, \qquad d_2 = 5 \times 66 = 330\ \text{mm}$$

Check of the pinion

$$F_{s1} = 210 \times 0.3027 \times 50 \times \pi \times 5 \times 0.1125 = 5619\ \text{N} > 3255\ \text{N}$$

so the pinion is also safe.

Answer: Module = 5 mm; face width = 50 mm; pinion pitch diameter = 110 mm; gear pitch diameter = 330 mm. The design is controlled by the cast iron gear.

Data: $z_1 = 20$, $z_2 = 60$, $m = 6$ mm, $b = 60$ mm, $P = 10$ kW, $n_1 = 900$ rpm, $e = 0.015$ mm.

Velocity and transmitted load

$$d_1 = mz_1 = 120\ \text{mm}, \qquad v = \frac{\pi d_1 n_1}{60\,000} = \frac{\pi \times 120 \times 900}{60\,000} = 5.655\ \text{m/s}$$ $$F_t = \frac{P}{v} = \frac{10\,000}{5.655} = 1768\ \text{N}$$

Dynamic load (Buckingham)

Deformation factor:

$$C = \frac{0.111\, e}{\frac{1}{E_p} + \frac{1}{E_g}} = \frac{0.111 \times 0.015}{2/200\,000} = 166.5\ \text{N/mm}$$ $$F_d = F_t + \frac{21 v\,(bC + F_t)}{21 v + \sqrt{bC + F_t}}$$ $$\begin{aligned} bC + F_t &= 60 \times 166.5 + 1768 = 11\,758\ \text{N} \\ \sqrt{11\,758} &= 108.4, \qquad 21v = 118.8 \\ F_d &= 1768 + \frac{118.8 \times 11\,758}{118.8 + 108.4} = 1768 + 6147 = 7915\ \text{N} \end{aligned}$$

Beam strength

$$y = 0.154 - \frac{0.912}{20} = 0.1084$$ $$F_s = \sigma_e\, b\, \pi m\, y = 400 \times 60 \times \pi \times 6 \times 0.1084 = 49\,040\ \text{N}$$

(The weaker is the pinion, as both are steel and the pinion has smaller $y$.)

Wear load

$$Q = \frac{2 z_2}{z_1 + z_2} = \frac{2 \times 60}{80} = 1.5$$ $$F_w = d_1\, b\, Q\, K = 120 \times 60 \times 1.5 \times 1.1 = 11\,880\ \text{N}$$

Check

Both $F_s > F_d$ and $F_w > F_d$, so the design is safe against bending and wear. Wear is the controlling criterion (factor 1.5 only).

Answer: Fd = 7.92 kN; Fs = 49.0 kN; Fw = 11.9 kN; the design is safe (wear governs).

The AGMA method calculates the actual bending stress at the tooth root using the Lewis idea but adds factors for the real operating conditions. In SI units, for spur gears,

$$\sigma = \frac{W_t\, K_o\, K_v\, K_s\, K_m\, K_b}{b\, m\, J}$$

where $W_t$ is the tangential load (N), $b$ the face width (mm), $m$ the module (mm) and $J$ the geometry factor.

Geometry factor $J$

$J$ replaces the Lewis form factor $y$. It accounts for:

• the shape of the tooth (number of teeth, pressure angle, addendum), as in $y$;
• the stress concentration at the root fillet ($K_f$), because the fillet raises the local stress above the nominal beam stress;
• the load sharing between teeth (the load is not at the tip, but at the highest point of single-tooth contact for spur gears, and shared in the double contact zone).

$J$ is read from charts for given numbers of teeth of pinion and gear; it is lower for small pinions. $J$ is less than the Lewis $y$.

Other factors

Strength and factor of safety

The allowable bending stress depends on material, heat treatment and hardness, for example for Grade 1 through-hardened steel $S_t = 0.533\,\text{HB} + 88.3$ MPa. It is corrected for the required life and the operating conditions by life factor $Y_N$, temperature factor $K_T$ and reliability factor $K_R$:

$$n = \frac{S_t\, Y_N}{K_T\, K_R\, \sigma}$$

$Y_N$ is 1 at $10^7$ cycles and falls with the number of cycles ($Y_N = 1.3558\,N^{-0.0178}$ for $N > 10^7$). The safety factor $n$ should be above 1; commonly 1.5 to 3 is used.

Pitch line velocity and tangential load

$$d = mz = 4 \times 25 = 100\ \text{mm}, \qquad V = \frac{\pi d n}{60} = \frac{\pi \times 0.1 \times 1500}{60} = 7.854\ \text{m/s}$$ $$W_t = \frac{P}{V} = \frac{15\,000}{7.854} = 1910\ \text{N}$$

Dynamic factor ($Q_v = 6$)

$$\begin{aligned} B &= 0.25\,(12 - 6)^{2/3} = 0.25 \times 3.302 = 0.8255 \\ A &= 50 + 56\,(1 - 0.8255) = 59.77 \\ K_v &= \left(\frac{A + \sqrt{200 V}}{A}\right)^B = \left(\frac{59.77 + \sqrt{1570.8}}{59.77}\right)^{0.8255} = (1.663)^{0.8255} = 1.522 \end{aligned}$$

Bending stress

$$\sigma = \frac{1910 \times 1.25 \times 1.522 \times 1 \times 1.3 \times 1}{40 \times 4 \times 0.34} = \frac{4724}{54.4} = 86.8\ \text{MPa}$$

Allowable stress and life factor

$$S_t = 0.533 \times 250 + 88.3 = 221.6\ \text{MPa}$$

Number of cycles:

$$N = 1500 \times 60 \times 10\,000 = 9\times10^{8}$$ $$Y_N = 1.3558\,(9\times10^8)^{-0.0178} = 0.939$$

Factor of safety

$$n = \frac{S_t\, Y_N}{K_T\, K_R\, \sigma} = \frac{221.6 \times 0.939}{1 \times 1 \times 86.8} = 2.40$$

The pinion is safe in bending (n > 1) with a reasonable margin. Surface (contact) strength should be checked separately, because pitting often governs the life of hardened gears.

Answer: sigma = 86.8 MPa; n = 2.4.