Example 1: the gear train

A motor of \(P = 1.5\) kW at \(n_1 = 1450\) rpm drives the left gear of Figure 9.4.1, \(N_1 = 24\), which drives the middle gear, \(N_2 = 48\), which drives the right gear, \(N_3 = 16\); the module is \(m = 3\) mm and the pressure angle \(20°\). Find the speed of each gear and the torque on the output shaft, first without losses and then with \(\eta = 0.98\) for each mesh. What force does a tooth carry, and what load does the middle gear put on its pin?

The speeds follow from 9.4.3, mesh by mesh, and the torques from 9.4.1 and 9.4.2. The force between the teeth acts along the line of action, at the pressure angle to the tangent of the pitch circles; its tangential part \(F_t = 2T/d\) carries the torque, and the radial part \(F_r = F_t\tan 20°\) pushes the gears apart.

P_1, n_1 = 1500, 1450                    # motor power [W] and speed [rpm]
N_1, N_2, N_3, m = 24, 48, 16, 3e-3      # teeth, module [m]
eta = 0.98                               # efficiency per mesh

n_2 = n_1*N_1/N_2                        # each mesh: n_in N_in = n_out N_out
n_3 = n_2*N_2/N_3
T_1 = P_1/(2*np.pi*n_1/60)
T_3 = T_1*n_1/n_3                        # same power, lower torque at the higher speed
T_3_eta = eta**2*T_3                     # two meshes on the way
F_t = 2*T_1/(m*N_1)                      # tangential tooth force, the same in both meshes
F_r = F_t*np.tan(np.radians(20))

The middle gear runs at half and the output at one and a half times the motor speed, and the torques are

The ratio from the motor to the output is \(n_1/n_3 = N_3/N_1 = 2/3\): the 48 teeth of the middle gear cancel out of the product \((N_2/N_1)(N_3/N_2)\). A gear that only passes motion on is an idlersv. mellanhjul. It changes the direction of the output and the distance between the shafts, never the ratio. It carries no torque either: the middle gear of Figure 9.4.1 turns freely on its pin, so the tooth forces on its two sides must have equal moments about it. At the left mesh the left gear pushes it with \(F_t\) on one side of the pin, and at the right mesh the output gear holds it back with \(F_t\) on the other side, so the two tangential forces point the same way and add up. The radial parts point at the pin from opposite sides and cancel. The pin of the idler carries \(2F_t \approx 550\) N, twice the tooth force, although the idler transmits no torque at all.

The worm drive

A wormsv. snäcka is a screw, and the worm wheelsv. snäckhjul it drives is a helical gear whose teeth fit its thread (Figure 9.4.2). One turn of a worm with \(z_1\) threads (or starts) moves the thread along by the lead \(z_1 p\), where \(p = \pi m\) is the axial pitch, and that turns the wheel by \(z_1\) teeth. The ratio is therefore

\[ i = \frac{N_2}{z_1} \tag{9.4.4}\]

and a single pair makes ratios of 10 to 100 in a small box. The worm’s pitch diameter is not tied to its number of threads: it is \(d_1 = q\,m\) for a chosen diameter quotient \(q\), commonly 8 to 12. The thread climbs at the lead anglesv. stigningsvinkel \(\lambda\), the angle between the thread and the plane normal to the axis, which unrolling one turn of the pitch cylinder gives as

\[ \tan\lambda = \frac{z_1 p}{\pi d_1} = \frac{z_1}{q} \tag{9.4.5}\]

and the wheel’s teeth slope at the same angle, so the two mesh.

The thread does not roll on the wheel’s teeth, it slides along them, as a screw slides in its nut, and that makes the worm drive the least efficient of the gear pairs. Following [1], a force analysis of the thread as an inclined plane, with friction coefficient \(f\) and the normal pressure angle \(\varphi_n\) of the thread, gives the efficiency with the worm driving

\[ \eta = \frac{\cos\varphi_n - f\tan\lambda}{\cos\varphi_n + f\cot\lambda} \tag{9.4.6}\]

which falls as the lead angle gets smaller. With the wheel driving, the roles of \(\tan\) and \(\cot\) swap, and the efficiency \((\cos\varphi_n - f\cot\lambda)/(\cos\varphi_n + f\tan\lambda)\) drops to zero when \(\tan\lambda \le f/\cos\varphi_n\). Then no torque on the wheel can turn the worm: the drive is self-lockingsv. självhämmande, and a hoist holds its load with the motor off.

Example 2: the worm drive

The worm of Figure 9.4.2 has \(z_1 = 2\) threads, axial module \(m = 4\) mm, \(q = 10\), a pressure angle of \(20°\) in the axial section, and drives a wheel with \(N_2 = 40\) teeth. Find the ratio, the lead angle, the centre distance and the efficiency for \(f = 0.05\), a steel worm on a bronze wheel in oil. Is the drive self-locking, and would a single-start worm be?

The normal pressure angle follows from the axial one through the lead angle, \(\tan\varphi_n = \tan\varphi\cos\lambda\).

z_1, N_w, m_w, q, f = 2, 40, 4.0, 10, 0.05

def worm_drive(z_1):
    lam = np.arctan(z_1/q)                                         # lead angle
    phi_n = np.arctan(np.tan(np.radians(20))*np.cos(lam))
    eta = (np.cos(phi_n) - f*np.tan(lam))/(np.cos(phi_n) + f/np.tan(lam))
    eta_back = (np.cos(phi_n) - f/np.tan(lam))/(np.cos(phi_n) + f*np.tan(lam))
    return np.degrees(lam), eta, eta_back

i_w = N_w/z_1
a_w = m_w*(q + N_w)/2                       # centre distance [mm], (d_1 + d_2)/2
lam_2, eta_2, back_2 = worm_drive(2)
lam_1, eta_1, back_1 = worm_drive(1)

For the two-start worm and, for comparison, a single-start worm, the results are

The two-start worm passes on 78 % of the power, and it can be driven backwards: neither drive is self-locking. Halving the number of starts halves the speed of the wheel and costs 13 percentage points of efficiency, yet the wheel can still drive the worm. Self-locking needs \(\tan\lambda \le f/\cos\varphi_n\), a lead angle below about \(3°\) with \(f = 0.05\), which means a single start and a large \(q\); and since \(f\) drops when the drive vibrates, a hoist does not rely on it but has a brake. A self-locking worm also has an efficiency below one half, so it turns more than half of the motor’s power into heat.

The bevel gear pair

Bevel gearssv. koniska kugghjul connect shafts whose axes meet, usually at a right angle, as in the differential of a car. The pitch circles of spur gears become pitch cones with a common apex where the axes meet, and the cones roll on each other along a common line (Figure 9.4.3). The speed ratio is again \(N_2/N_1\), and a cone’s pitch diameter at its back is still \(mN\), so the cone angles \(\delta_1 + \delta_2 = 90°\) must satisfy \(\tan\delta_1 = r_1/r_2 = N_1/N_2\):

\[ \tan\delta_1 = \frac{N_1}{N_2}, \qquad \delta_2 = 90° - \delta_1 \tag{9.4.7}\]

The teeth taper towards the apex. Their shape at the back is approximately that of a spur gear on the back cone, the cone through the back end of the teeth perpendicular to the pitch cone, whose radius is \(r/\cos\delta\). Rolled out flat, the back cone is a spur gear with the virtual number of teeth

\[ N_v = \frac{N}{\cos\delta} \tag{9.4.8}\]

This is Tredgold’s approximation, and it is how bevel gears are designed with spur gear results, the tooth form, the undercut limit and the strength, and how the gears of Figure 9.4.3 were modelled. Straight teeth run along the cone towards the apex; the spiral bevel gears in the figure curve them, as helical teeth twist those of a spur gear, so they mesh more smoothly and quietly.

Example 3: the bevel gear pair

The pinion of Figure 9.4.3 has \(N_1 = 14\) teeth and the wheel \(N_2 = 42\), module \(m = 4\) mm at the back and a face width of 24 mm. Find the cone angles, the cone distance from the apex to the back of the teeth, and the virtual numbers of teeth. Will the pinion’s teeth be undercut?

N_p, N_g, m_b, F_b = 14, 42, 4.0, 24.0
delta_1 = np.arctan(N_p/N_g)
delta_2 = np.pi/2 - delta_1
A_0 = m_b*N_g/2/np.sin(delta_2)             # cone distance: pitch radius over sin(delta)
N_v1, N_v2 = N_p/np.cos(delta_1), N_g/np.cos(delta_2)
N_min = 2/np.sin(np.radians(20))**2         # fewest teeth without undercut

The cones and the virtual gears are

The face width of 24 mm is less than a third of the cone distance, the usual upper limit: a wider face would reach so far towards the apex that its inner end carries little load. The pinion acts like a spur gear of 14.8 teeth, below the 17 teeth that a \(20°\) involute needs to avoid undercut, so its teeth would be cut away at the root. Real bevel pinions avoid this with a profile shift, a longer addendum on the pinion and a shorter one on the wheel, which the model in the figure leaves out.

Example 4: an RC car in a corner

An RC car with a track width of \(B = 250\) mm and wheels of diameter \(D = 110\) mm drives at \(v = 8\) m/s through a corner of radius \(R = 1.5\) m. Its differential is that of Figure 9.4.5, with a ring gear of 40 teeth driven by a pinion of 12. Find the speeds of the carrier, of both wheels and of the spiders on their pins, and the torque on each half-shaft when the motor delivers 0.5 N·m at the pinion.

The car’s speed is the speed of the middle of the axle, \(v = \omega_c R_w\) with the wheel radius \(R_w = D/2\) if the wheels do not slip, so the carrier turns at \(v/R_w\); the wheels turn in the ratio of 9.4.9, and their mean is the carrier speed by 9.4.10.

B_c, D_w, v_c, R_c = 0.250, 0.110, 8.0, 1.5          # m, m, m/s, m
N_ring, N_pin, N_sd, N_sp = 40, 12, 16, 10
T_pin = 0.5                                          # N m at the pinion

w_c = v_c/(D_w/2)                                     # carrier: the axle's mean speed
w_in = w_c*(R_c - B_c/2)/R_c                          # each wheel's path is R -+ B/2
w_out = w_c*(R_c + B_c/2)/R_c
w_spider = (w_out - w_c)*N_sd/N_sp                    # relative to the carrier
T_c = T_pin*N_ring/N_pin                              # the final drive multiplies the torque
T_half = T_c/2
rpm = 60/(2*np.pi)

The speeds in rpm and the torques are

The outer wheel turns 18 % faster than the inner one, and the spiders turn at nearly 200 rpm on their pins, which is why the pin bore of a printed spider needs a smooth fit, or a bushing, and grease. Straight ahead the spiders stand still on their pins and only carry the torque. The torque on each half-shaft is half the carrier torque, the same on the inner and the outer wheel, so the faster outer wheel takes more of the power: the power \(T\omega\) is shared in the ratio of the speeds while the torque is shared equally.

Tooth forces and Lewis’s equation

The force between two involute teeth acts along the line of actionsv. ingreppslinje, the common normal of the flanks, which crosses the line of centres at the pitch point at the pressure anglesv. ingreppsvinkel \(\varphi\) to the tangent. With the pinion torque \(T_1\) and pitch diameter \(d_1\), its tangential part carries the torque and its radial part pushes the gears apart:

\[ F_t = \frac{2T_1}{d_1}, \qquad F_r = F_t\tan\varphi, \qquad F = \frac{F_t}{\cos\varphi} \tag{9.4.12}\]

The tooth is a short cantilever loaded by this force, and it breaks at its root. In 1892, Wilfred Lewis (biography) treated it as a beam of constant strength, the parabola inscribed in the tooth that touches the root fillets, loaded by \(F_t\) at the tip of the tooth. Following [1], the bending stress at the root is then

\[ \sigma = \frac{F_t}{b\,m\,Y} \tag{9.4.13}\]

with the face widthsv. kuggbredd \(b\) and the Lewis form factor \(Y\), a pure number that depends only on the tooth shape: for \(20°\) full-depth teeth it grows from 0.245 for 12 teeth to 0.359 for 30 and 0.485 for a rack [1, Table 14-2]. The equation contains the two assumptions that limit it. It puts the whole force on one tooth at its tip, the worst place, although a second pair of teeth shares the load whenever the tooth is near its tip; and it ignores the stress concentration in the root fillet. Modern rating standards (ISO 6336, AGMA 2001) correct both with factors. We now look at the two effects directly, with a finite element model of two gears in mesh.

Contact ratio

Two teeth touch only while their point of contact is on the line of action between the two addendum circles. That stretch has the length

\[ L = \sqrt{r_{a1}^2 - r_{b1}^2} + \sqrt{r_{a2}^2 - r_{b2}^2} - C\sin\varphi \tag{9.4.14}\]

with the addendum radii \(r_a = r + m\), the base radii \(r_b = r\cos\varphi\) and the centre distance \(C\). Successive teeth follow each other along the line of action at the base pitch \(p_b = \pi m\cos\varphi\), so the contact ratiosv. ingreppstal

\[ \varepsilon = \frac{L}{p_b} \tag{9.4.15}\]

is the mean number of tooth pairs in contact. Spur gears have \(\varepsilon\) between 1.4 and 1.8: one pair carries the load in the middle of the stretch, two pairs share it at its ends.

Example 5: tooth stress in two spur gears

A pinion of \(N_1 = 20\) teeth drives a wheel of \(N_2 = 30\), module \(m = 4\) mm, pressure angle \(20°\), both of steel, \(E = 210\) GPa and \(\nu = 0.3\), and the gears are \(b = 20\) mm thick (Figure 9.4.6). The pinion torque is \(T_1 = 100\) N·m. Find the tooth forces, the contact ratio and the Lewis stresses, then compute the stress in the teeth with the finite element method as the gears turn through the mesh, and compare.

T_g, N_a, N_b, m_g, b_g = 100e3, 20, 30, 4.0, 20.0     # N mm, teeth, mm
phi = np.radians(20)
r_1, r_2 = m_g*N_a/2, m_g*N_b/2
C = r_1 + r_2
F_tg = T_g/r_1                                          # 2 T/d
F_rg = F_tg*np.tan(phi)
F_g = F_tg/np.cos(phi)
L_a = (np.sqrt((r_1 + m_g)**2 - (r_1*np.cos(phi))**2)
       + np.sqrt((r_2 + m_g)**2 - (r_2*np.cos(phi))**2) - C*np.sin(phi))
p_b = np.pi*m_g*np.cos(phi)
eps = L_a/p_b
Y_1, Y_2 = 0.322, 0.359                                 # Lewis form factors, 20 and 30 teeth
sig_L1, sig_L2 = F_tg/(b_g*m_g*Y_1), F_tg/(b_g*m_g*Y_2)

The forces from 9.4.12, the contact ratio from 9.4.15 and the Lewis stresses from 9.4.13 are

The finite element model treats each gear as a plane body of thickness \(b\) in plane stress, clamped at its bore, which takes up the torque as a shaft would. Its displacement \(\bm u\) satisfies the weak form of elastostatics of the Gridap chapters,

\[ \int_\Omega \bm\varepsilon(\bm v) : \bm\sigma(\bm u)\, d\Omega = \int_\Gamma \bm v \cdot \bm t \, d\Gamma \quad \text{for all } \bm v \tag{9.4.16}\]

where the traction \(\bm t\) on the tooth outline \(\Gamma\) is the tooth force. The model does not solve the contact between the teeth; it applies the force where the involute theory puts it. For a pinion angle \(\theta_1\), the contact points lie on the line of action one base pitch apart, and each pair whose point lies between the addendum circles carries \(F\), or \(F/2\) when two pairs are in contact, along the line of action. On each flank the force is spread as a semi-elliptic pressure over a strip of half-width \(a_s = 0.4\) mm,

\[ p(s) = \frac{2F}{\pi a_s b}\sqrt{1 - \frac{s^2}{a_s^2}}, \qquad |s| \le a_s \tag{9.4.17}\]

the shape of a Hertz contact pressure, with \(s\) along the flank.

The width of the strip is our choice, and the true contact area is not. Near the point of contact each involute flank is close to a circle whose radius is the involute’s radius of curvature, the distance along the line of action from the contact point to the point where the line touches that gear’s base circle. At the pitch point this is \(\rho = r\sin\varphi\), and elsewhere \(\rho_1 = r_{b1}\tan\varphi + s\) and \(\rho_2 = r_{b2}\tan\varphi - s\) for a contact at the distance \(s\) from the pitch point. Two elastic cylinders pressed together touch over a strip whose width Hertz’s theory gives. Following [1, Eqs. 3-73 and 14-10], for two steel teeth with the same \(E\) and \(\nu\), the half-width of the strip and the largest pressure in it are

\[ a = \sqrt{\frac{8F(1 - \nu^2)}{\pi b E}\,\frac{\rho_1\rho_2}{\rho_1 + \rho_2}}, \qquad p_\text{max} = \frac{2F}{\pi a b} \tag{9.4.18}\]

At the pitch point of Example 5, where one pair carries the whole force,

E_g, nu_g, a_s = 210e3, 0.3, 0.4                       # MPa, -, mm
rho_1, rho_2 = r_1*np.sin(phi), r_2*np.sin(phi)
R_eq = rho_1*rho_2/(rho_1 + rho_2)
a_H = np.sqrt(8*F_g*(1 - nu_g**2)/(np.pi*b_g*E_g)*R_eq)
p_H = 2*F_g/(np.pi*a_H*b_g)                            # true Hertz pressure
p_s = 2*F_g/(np.pi*a_s*b_g)                            # the model's strip

The contact is a strip about a fifth of a millimetre wide across the whole face of the tooth, less than a quarter of the strip in the model, and its pressure is more than three times higher. Resolving it would take elements of a few hundredths of a millimetre at every contact point and a contact algorithm, and the result is known anyway from 9.4.18: this is the contact stress against which a hardened flank is rated for pitting. The model therefore shows the contact peak too low on purpose. The stress at the root, a tooth height away, does not depend on how the force is spread, by Saint-Venant’s principle. Splitting the force equally in double contact ignores that the two pairs are not equally stiff, which makes the load on each pair vary along the stretch.

The gears are meshed with triangles of 0.1 mm along the whole outline, fine enough to resolve the root fillet of radius \(0.3m\): refining further changes the root stress by less than 0.1 %. With quadratic elements the pinion has 350 000 degrees of freedom and the wheel 530 000. The stiffness matrix does not change as the gears turn, only the load does, so each gear’s matrix is assembled and factorized once and every one of the 120 positions of Figure 9.4.7 costs only a new right-hand side and a back substitution. The heart of the script Gridap/gears/gears.jl is

A = assemble_matrix((u, v) -> ∫(ε(v) ⊙ (σ∘ε(u)))dΩ, U, V)
K = lu(A)                                   # once per gear
for θ1 in θs                                # 120 positions of the pinion
    t = traction(θ1)                        # strips at the contact points of θ1
    rhs = assemble_vector(v -> ∫(v ⋅ t)dΓ, V)
    uh = FEFunction(U, K \ rhs)             # a back substitution per frame
end

and the whole run, meshing, factorization and 240 solutions, takes about five minutes.

The tooth pairs of Figure 9.4.7 run through the same cycle. A pair comes into contact at the pinion’s root and the wheel’s tip while the pair ahead still carries half the load. When that pair leaves, the whole force jumps onto the new pair, and its root stresses double. In the middle of the stretch the pair carries the load alone, the wheel’s root stress falling and the pinion’s rising as the contact climbs the pinion tooth, until the next pair arrives and takes half. Figure 9.4.8 reads the root stresses of the marked teeth from the results.

fr = pd.read_csv('../Gridap/gears/frames.csv')
summary = dict(pd.read_csv('../Gridap/gears/summary.csv', header=None).values)
s_a, s_r, p_bw = summary['s_a_mm'], summary['s_r_mm'], summary['base_pitch_mm']

fig, ax = plt.subplots(figsize=(7, 3.6))
for lo, hi in ((s_a, s_r - p_bw), (s_a + p_bw, s_r)):
    ax.axvspan(lo, hi, color='0.9', lw=0)
ax.plot(fr.s0, fr.p_sig1_tens, color=BLUE, label='pinion, 20 teeth')
ax.plot(fr.s0, fr.w_sig1_tens, color=RED, label='wheel, 30 teeth')
ax.axhline(sig_L1, color=BLUE, ls='--', lw=1)
ax.axhline(sig_L2, color=RED, ls='--', lw=1)
single = fr.npairs == 1
sig_fe1, sig_fe2 = fr.p_sig1_tens[single].max(), fr.w_sig1_tens[single].max()
ax.set_xlim(s_a - 3, s_r + 3)
ax.set_ylim(0, 160)
ax.set_xticks([round(s_a, 1), round(s_r - p_bw, 1), 0, round(s_a + p_bw, 1), round(s_r, 1)])
ax.set_yticks([0, round(sig_L2), round(sig_L1), round(sig_fe2), round(sig_fe1), 160])
ax.set_xlabel('contact position $s$ on the line of action [mm]')
ax.set_ylabel(r'root stress $\sigma_1$ [MPa]')
ax.legend(loc='upper left', frameon=False)
for side in ('top', 'right'):
    ax.spines[side].set_visible(False)
plt.show()

In single contact the finite element stresses exceed Lewis’s by about half:

Two effects pull in opposite directions. Lewis puts the force at the tip, but the single pair never carries it there: the second pair takes half the load before the contact reaches the tip, so the largest moment arm in single contact is shorter than Lewis assumes. The fillet, which Lewis treats as a smooth beam, concentrates the stress, and that wins: most of the factor of about 1.5 is the stress concentration of the root fillet of radius \(0.3m\). Lewis’s equation is a sound way to size a gear’s module and face width, provided its allowable stress has room for this factor. The model also shows a stress that Lewis leaves out: the opposite fillet of each tooth is in compression, and its von Mises stress is even higher, up to 170 MPa on the pinion, because the radial component of the force adds to the bending there. Gear teeth fail in fatigue from the fillet in tension, where cracks open.

Load variation in spur and helical gears

The root stress of Figure 9.4.8 follows the force on one tooth, and that force jumps whenever a pair of teeth comes into or out of contact. The graph of the force on one tooth over its passage through the mesh is the load variationsv. kraftvariation of the gear pair, and here helical gears differ from spur gears.

All contacts of a gear pair lie in the plane of action, the plane through the line of action and parallel to the axes. The part of it where the teeth can touch is a rectangle, the length of action \(L\) of 9.4.14 long and the face width \(b\) wide. A spur tooth touches its mate along a line across the whole face, parallel to the axes, and the lines of successive teeth follow each other along the rectangle one base pitch \(p_b\) apart. A helical tooth (Figure 9.4.9) touches along a line inclined at the base helix angle \(\beta_b\), \(\tan\beta_b = \tan\beta\cos\varphi\), so its line enters the rectangle at one corner, grows as the gears turn, and leaves the opposite corner as gradually as it came. A cut through a helical gear across its thickness shows why. Every section is the spur outline, turned a little further by the helix the deeper the cut; a cut at any depth therefore shows a spur gear pair in mesh, and its contact points lie on the line of action. Moving the cut through the thickness turns the sections, and the contact points slide along the line of action: together they trace the inclined contact lines of Figure 9.4.10.

A contact line spans \(b\tan\beta_b\) along the line of action, and that, in base pitches, adds the overlap ratio to the contact ratio,

\[ \varepsilon_\beta = \frac{b\tan\beta_b}{p_b}, \qquad \varepsilon_\gamma = \varepsilon + \varepsilon_\beta \tag{9.4.19}\]

the total contact ratio, the mean number of teeth in contact.

To turn the contact lines into forces we assume that the teeth are equally stiff along the lines, so that the force per unit length of contact line is the same on all of them. A tooth whose line has the length \(\ell_k\) inside the rectangle then carries the share

\[ \frac{F_k}{F} = \frac{\ell_k}{\sum_j \ell_j}, \qquad \ell_k = \frac{1}{\cos\beta_b}\int_{-b/2}^{b/2} \big[\, s_a \le s_k + z\tan\beta_b \le s_r \,\big]\, dz \tag{9.4.20}\]

of the force, where \(s_k\) is the position of the line at mid-face, the bracket is one where the line is inside the rectangle and zero elsewhere, and the line of the next tooth lies at \(s_k + p_b\). For a spur gear, \(\beta_b = 0\), each line is in or out as a whole, and 9.4.20 reduces to the equal split of Example 5.

s_a0 = -np.sqrt((r_2 + m_g)**2 - (r_2*np.cos(phi))**2) + r_2*np.sin(phi)   # wheel's tip
s_r0 = np.sqrt((r_1 + m_g)**2 - (r_1*np.cos(phi))**2) - r_1*np.sin(phi)    # pinion's tip
z = np.linspace(-b_g/2, b_g/2, 2001)

def load_share(beta_deg, s):
    # share of the force on the tooth whose contact line is at s (mid-face), eq-tr-load-share
    tb = np.tan(np.radians(beta_deg))*np.cos(phi)
    def length(sk):                      # the part of the line inside the rectangle
        inside = (s_a0 <= sk[:, None] + z*tb) & (sk[:, None] + z*tb <= s_r0)
        return inside.mean(axis=1)*b_g*np.sqrt(1 + tb**2)
    total = sum(length(s + k*p_b) for k in range(-4, 5))
    return length(s)/total, b_g*tb/p_b

s_line = np.linspace(s_a0 - 12, s_r0 + 12, 1500)
share_spur, _ = load_share(0, s_line)
share_hel, eps_beta = load_share(20, s_line)

fig, ax = plt.subplots(figsize=(7, 3.2))
ax.plot(s_line, share_spur, color=BLUE, label=r'spur, $\beta = 0$')
ax.plot(s_line, share_hel, color=RED, label=r'helical, $\beta = 20^\circ$')
ax.set_xlim(s_line[0], s_line[-1])
ax.set_ylim(0, 1.05)
ax.set_xticks([round(s_line[0], 1), round(s_a0, 1), 0, round(s_r0, 1), round(s_line[-1], 1)])
ax.set_yticks([0, 0.5, round(share_hel.max(), 2), 1])
ax.set_xlabel('position $s$ of the tooth\'s contact line at mid-face [mm]')
ax.set_ylabel('$F_k/F$')
ax.legend(loc='upper left', frameon=False)
for side in ('top', 'right'):
    ax.spines[side].set_visible(False)
plt.show()

The overlap ratio of the helical pair is

The spur tooth of Figure 9.4.11 takes half the force at once, all of it a moment later, then half again, and lets go of it at once: every step is a jump in the deflection of the teeth, which the gears hear as a knock and the drive as a vibration at the tooth frequency. The helical tooth takes its load up and puts it down gradually, the total length of contact line hardly changes as the gears turn, and the largest share of a tooth is well below the whole force. Helical gears therefore run more quietly and carry more load for the same size; the axial force is the price, and the herringbone gear removes it. The assumption of equal stiffness per unit length is the weak point of 9.4.20: a line near the tip of a tooth meets a softer tooth than one near the root, and the rating standards correct for this with load distribution factors.

The load diagram

A gear supplier’s catalogue answers the designer’s first question, how much torque a gear of a given module and number of teeth can carry, with a load diagramsv. belastningsdiagram: the torque against the number of teeth on logarithmic axes, one line per module. We now build one from the two limits of this section, the bending stress at the root and the contact pressure on the flank, and check it against the load diagrams of a Swedish supplier’s catalogue [4].

The flank limit follows from 9.4.18. At the pitch point the radii of curvature are \(\rho_1 = (d_1/2)\sin\varphi\) and \(\rho_2 = u\rho_1\), where \(u = N_2/N_1\) is the gear ratio, and the tooth force is \(F = 2T/(d_1\cos\varphi)\). Inserting these and requiring the largest pressure to stay below the allowable flank pressure \(\sigma_{HP}\) gives the torque the flank carries:

T, d_1, u, b, phi_s, E_s, nu_s, sig_HP = sp.symbols(
    r'T d_1 u b varphi E nu sigma_HP', positive=True)
rho_1 = d_1/2*sp.sin(phi_s)
rho_2 = u*rho_1
F_n = 2*T/(d_1*sp.cos(phi_s))
a_h = sp.sqrt(8*F_n*(1 - nu_s**2)/(sp.pi*b*E_s)*rho_1*rho_2/(rho_1 + rho_2))   # eq-tr-hertz
p_max = 2*F_n/(sp.pi*a_h*b)
T_H = sp.solve(sp.Eq(p_max**2, sig_HP**2), T)[0]
Z_E2, Z_H2 = E_s/(2*sp.pi*(1 - nu_s**2)), 2/(sp.sin(phi_s)*sp.cos(phi_s))
check = sp.simplify(T_H - sig_HP**2*b*d_1**2*u/(2*Z_E2*Z_H2*(u + 1)))     # zero: the ISO form

In the notation of ISO 6336 this is \(T_H = \sigma_{HP}^2\, b\, d_1^2\, u/(2 Z_E^2 Z_H^2 (u + 1))\), with the elasticity factor \(Z_E = \sqrt{E/(2\pi(1 - \nu^2))}\) and the zone factor \(Z_H = \sqrt{2/(\sin\varphi\cos\varphi)}\); the code above checks that the two forms agree. The root limit follows from Lewis’s equation 9.4.13 with the fillet’s stress concentration \(K_t\) that Example 5 measured and the allowable local root stress \(\sigma_{FP}\). With \(d_1 = mN_1\) the two limits are, as mechanicskit.gears computes them,

\[ T_H = \frac{\sigma_{HP}^2\, b\, m^2 N_1^2}{2 Z_E^2 Z_H^2}\,\frac{u}{u + 1}, \qquad T_F = \frac{\sigma_{FP}\, b\, m^2 N_1\, Y(N_1)}{2K_t} \tag{9.4.21}\]

and the gear carries the smaller of the two. On logarithmic axes both are straight lines for each module: the flank limit rises with the square of the number of teeth, slope 2, because a larger gear has flatter flanks that spread the contact; the root limit rises only as \(N_1 Y\), a slope of about 1.3. Depending on the material, the flank or the root governs.

For the steel, the catalogue’s C45 is close to AISI 1045, which has a hardness of 163 HB hot-rolled and 179 HB cold-drawn [1, Table A-20]. HB is the Brinell hardness: a hard ball is pressed into the surface with a known force, and the force divided by the area of the dent it leaves, in kgf/mm², is the hardness. A harder steel resists the local yielding under a contact, and its ultimate strength is roughly \(3.4\,\text{HB}\) MPa [1], so hardness stands in for strength in the rating standards. For normalised steel, according to ISO 6336-5 [5], the allowable stresses are straight lines in the hardness, \(\sigma_{H\lim} = \text{HB} + 190\) MPa and \(\sigma_{F\lim} = 0.455\,\text{HB} + 69\) MPa for the usual material quality. Here \(\sigma_{F\lim}\) is a nominal root stress that already contains a stress concentration of 2, so the local stress the material bears is \(2\sigma_{F\lim}\). The catalogue states its safety factors, 1.0 on the flank and 1.4 on the root, its face widths, and that its torques hold for the smaller gear; we take \(u = 1\), equal gears, where the flank pressure is highest. For the plastic gears, polyacetal (POM) with \(E = 3100\) MPa and \(\nu = 0.35\) [6], we found no reliable published allowable flank pressure, so we fit it to the catalogue’s plastic diagram. For the root of POM, according to VDI 2736 as quoted by Bae and Kissling [7,8], the fatigue strength is 41 MPa for \(10^7\) cycles at 20 °C, which we treat like \(\sigma_{F\lim}\) for steel.

from mechanicskit.gears import flank_torque, root_torque, iso6336_normalised

cat = pd.read_csv('data/load_diagram_mekanex.csv', comment='#')
K_t = sig_fe1/sig_L1                       # fillet stress concentration, from Example 5

def limits(N, m, b, mat):
    # torques [N m] that the flank and the root carry, eq-tr-load-limits
    return (flank_torque(N, m, b, mat['sig_HP'], mat['E'], mat['nu']),
            root_torque(N, m, b, mat['sig_FP'], K_t))

s_Hlim, s_FE = iso6336_normalised(179)     # AISI 1045 cold drawn, 179 HB
steel = dict(E=210e3, nu=0.3, sig_HP=s_Hlim/1.0, sig_FP=s_FE/1.4,
             widths={1: 15, 1.5: 17, 2: 20, 2.5: 25, 3: 30, 4: 40, 5: 50, 6: 60, 8: 80})
pom = dict(E=3100.0, nu=0.35, sig_HP=1.0, sig_FP=2*41/1.4,
           widths={1: 15, 1.5: 17, 2: 20, 3: 30, 4: 40})

def catalogue_ratio(material, mat):
    # catalogue torque over model flank torque, at both ends of every digitised line
    r = []
    for _, row in cat[cat.material == material].iterrows():
        for N, T_c in ((row.N_start, row.T_start_Nm), (row.N_end, row.T_end_Nm)):
            r.append(T_c/limits(N, row.module, mat['widths'][row.module], mat)[0])
    return np.array(r)

r_steel = catalogue_ratio('steel', steel)
pom['sig_HP'] = np.sqrt(np.median(catalogue_ratio('plastic', pom)))   # fitted, from sigma_HP = 1
r_pom = catalogue_ratio('plastic', pom)

With these values the steel model and the plastic fit give

from matplotlib.ticker import LogLocator, NullFormatter
fig, axs = plt.subplots(1, 2, figsize=(9, 5.2))
for ax, material, mat, N_max in ((axs[0], 'steel', steel, 114), (axs[1], 'plastic', pom, 54)):
    for m_, b_ in mat['widths'].items():
        N = np.linspace(12, N_max, 200)
        T_H, T_F = limits(N, m_, b_, mat)
        ax.plot(N, np.minimum(T_H, T_F), color=BLUE, lw=1.6)
        ax.text(11.5, min(T_H[0], T_F[0]), f'{m_:g}', ha='right', va='center', fontsize=8)
    for _, row in cat[cat.material == material].iterrows():
        ax.plot([row.N_start, row.N_end], [row.T_start_Nm, row.T_end_Nm], color=GREY, lw=3,
                alpha=0.45, zorder=0)
    ax.set_xscale('log')
    ax.set_yscale('log')
    ax.set_xlim(9.5, N_max)
    ticks = [12, 20, 30, 50, N_max] if N_max > 80 else [12, 20, 30, N_max]
    ax.set_xticks(ticks)
    ax.set_xticklabels(ticks)
    ax.xaxis.set_minor_locator(LogLocator(subs='all'))
    ax.yaxis.set_minor_locator(LogLocator(subs='all'))
    ax.xaxis.set_minor_formatter(NullFormatter())
    ax.yaxis.set_minor_formatter(NullFormatter())
    ax.grid(True, which='major', color='0.80', lw=0.8)
    ax.grid(True, which='minor', color='0.92', lw=0.6)
    ax.set_axisbelow(True)
    ax.set_xlabel('number of teeth $N_1$')
    for side in ('top', 'right'):
        ax.spines[side].set_visible(False)
axs[0].set_ylim(0.3, 1000)
axs[0].set_yticks([0.3, 1, 10, 100, 1000])
axs[0].set_yticklabels(['0.3', '1', '10', '100', '1000'])
axs[1].set_ylim(0.02, 50)
axs[1].set_yticks([0.02, 0.1, 1, 10, 50])
axs[1].set_yticklabels(['0.02', '0.1', '1', '10', '50'])
axs[0].set_ylabel('torque $T$ [N m]')
axs[0].set_title('steel C45', fontsize=10)
axs[1].set_title('polyacetal (POM)', fontsize=10)
plt.show()

The steel model reproduces the catalogue’s diagram from first principles: the slope of 2, which the catalogue lines have to three digits, and the torques to within the ratio above, about 15 % below the catalogue. The difference is the hardness. The catalogue’s lines match \(\sigma_{HP} \approx 400\) MPa, which ISO 6336-5 gives normalised steel of 210 HB, the top of its range and harder than the cold-drawn 1045 of Shigley’s table. The module 8 line of the catalogue lies 20 % below the others, which no factor in 9.4.21 explains, so we take it for a drawing error. The root never governs a steel gear of this kind: even with 114 teeth the root carries several times the torque the flank does.

The catalogue’s plastic lines also have slope 2, so they too are a flank limit, and fitting them gives the allowable flank pressure of POM above, about 15 MPa. Polyacetal gears that run dry wear and heat up, and the teeth soften as they warm, long before the root breaks: the root limit of 9.4.21, from VDI 2736’s fatigue strength, lies far above the catalogue’s lines. VDI 2736 therefore checks plastic gears for temperature and wear as well as for stress.

For gears printed by fused-filament fabrication we cannot draw such a diagram yet. Their strength depends on the polymer, on the orientation of the layers relative to the tooth, on how well the layers are fused, and on the temperature the teeth reach in running, and neither VDI 2736 nor the published tests cover the filaments and settings we print with. The diagram for printed gears has to come from mechanical testing: fatigue tests of printed teeth in the orientations we print, and running tests of gear pairs at known torque and speed that measure wear and tooth temperature. 9.4.21 then needs only the two allowable stresses those tests give.

The notebook below draws the load diagram for any material hardness, face width, gear ratio and safety factors, and shows which limit governs a chosen gear.

Contact solved: a cylinder on a plane

A contact condition says that two bodies may touch but not overlap, and that the surfaces push and never pull. Let \(g = y + u_y\) be the gap between a point of a steel cylinder’s surface and a rigid plane at \(y = 0\): the point may not go below the plane, \(g \ge 0\), and the plane pushes on it only where it touches. The penalty method replaces this condition by a stiff spring that acts only on overlap, the traction \(t_y = \kappa\langle -g\rangle\) with \(\langle x\rangle = \max(0, x)\), and adds its work to the weak form of the cylinder,

\[ \int_\Omega \bm\varepsilon(\bm v) : \bm\sigma(\bm u)\, d\Omega - \int_{\Gamma_c} \kappa \langle -(y + u_y)\rangle\, v_y \, d\Gamma = 0 \quad \text{for all } \bm v \tag{9.4.22}\]

The bracket makes the problem nonlinear, since where the contact ends depends on the solution, and Newton’s method solves it. A large penalty \(\kappa\) allows only a tiny overlap, and the answer stops changing once \(\kappa\) is large enough: going from \(\kappa = 10E/h\) to \(100E/h\), with \(h\) the element size at the contact, moves the results by 0.1 to 0.3 %. The cylinder has \(R = 10\) mm and carries 133 N per mm of length, the tooth force of Example 5 per millimetre of face width, in plane strain; a quarter of it is meshed with elements of 2.9 µm at the contact, a thirtieth of the contact half-width.

The contact model and Hertz’s theory agree to within a fraction of a percent (Figure 9.4.13). The pressure has the semi-elliptic shape that 9.4.18 assumes, and the largest stress is not at the surface but 0.7 half-widths below it, where the pressure from above and the support from the sides combine into the largest shear. This is where pitting starts: a crack grows from below the surface, parallel to it, until a flake of the flank breaks out. According to [1], the largest shear stress of a cylinder contact is \(0.30\,p_\max\) at the depth \(0.786a\), which the model reproduces.

True contact strips and load sharing in the gears

With the contact model confirming 9.4.18, the gear model can take its strips from it. At each contact point the radii of curvature are the distances along the line of action to the two tangent points of the base circles, and the force of the pair then gives the half-width \(a\) and the pressure of 9.4.18. Resolving strips about 0.1 mm wide takes elements of 12.5 µm along the flanks of the teeth that pass through the mesh, about 0.8 million degrees of freedom in the pinion and 1.1 million in the wheel; the stiffness is still factorized once.

The equal split in double contact goes next. Two pairs of teeth in contact must close the same gap: the gears are rigid bodies apart from the deformation of the teeth, so both pairs approach by the same amount \(\delta_0\) as the torque winds the gears up. Let \(C_{ij}\) be the approach of pair \(i\), the movement of the two flanks towards each other along the line of action, under a unit force on pair \(j\); the model computes it with two extra back substitutions per gear. The forces \(f_1, f_2\) on the two pairs then satisfy

\[ \bm C \bm f = \delta_0 \begin{bmatrix} 1 \\ 1 \end{bmatrix}, \qquad f_1 + f_2 = F \quad\Rightarrow\quad \bm f = F\,\frac{\bm C^{-1}\bm 1}{\bm 1^{\mathsf{T}}\bm C^{-1}\bm 1} \tag{9.4.23}\]

and the mesh stiffness \(k = F/\delta_0\) is the stiffness of the whole mesh against the torque. Since \(a\) depends on the force and the compliance weakly on \(a\), two or three passes settle each frame.

At the pitch point the gear model reproduces the cylinder: a half-width of 0.111 mm, a Hertz pressure of 765 MPa, and the largest von Mises stress, 440 MPa or 0.575 of the pressure, 0.69 half-widths below the flank, within 3 % of the theory with elements a ninth of the half-width. The load sharing of Figure 9.4.15 differs from the equal split: a pair that enters takes only 40 % of the force, because its contact is at the wheel’s tip and the pinion’s root, where the wheel’s tooth is most flexible, and its share grows to 60 % as the contact moves towards the pitch point. The load variationsv. kraftvariation of a spur pair is therefore not a staircase but a ramp with two steps, and the root stress in double contact is up to 19 % lower than with the equal split. The largest root stress, in single contact, does not change, so the conclusions of Example 5 stand. The mesh stiffness jumps between about 10 and 14 N/(mm·µm) each time a pair enters or leaves; the gears deflect a little more and a little less at the tooth frequency, and that periodic deflection excites the whine of spur gears.

What remains is a contact model of the gears themselves, with the penalty condition of 9.4.22 between the two flanks instead of the prescribed strips. That needs the gap between two separately meshed bodies, which Gridap does not provide, and a nonlinear solution for every frame. It would find the contact points, the strips and the load sharing at once, including the edge contacts at the tips that the prescribed strips can only approximate, and it is the next model for the Gridap part of this book.