Tooth forces and Lewis’s equation

The force between two involute teeth acts along the line of action, at the pressure angle \(\varphi\) to the tangent of the pitch circles. 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:

\[ F_t = \frac{2T_1}{d_1}, \qquad F_r = F_t\tan\varphi, \qquad F = \frac{F_t}{\cos\varphi} \tag{9.6.1}\]

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. According to [1], the bending stress at the root is then

\[ \sigma = \frac{F_t}{b\,m\,Y} \tag{9.6.2}\]

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 0.322 for 20, 0.359 for 30 and 0.485 for a rack [1, Table 14-2].

The equation carries 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. The rating standards correct both with factors (ISO 6336).

Contact ratio

Two teeth touch only while their point of contact lies on the line of action between the two tip circles (Figure 9.2.8). That stretch has the length

\[ L = \sqrt{r_{a1}^2 - r_{b1}^2} + \sqrt{r_{a2}^2 - r_{b2}^2} - a\sin\varphi \tag{9.6.3}\]

with the tip radii \(r_a = r + m\), the base radii \(r_b = r\cos\varphi\) and the centre distance \(a\) of 9.5.1. 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

\[ \varepsilon = \frac{L}{p_b} \tag{9.6.4}\]

is the mean number of tooth pairs in contact. Spur gears have \(\varepsilon\) between 1.4 and 1.8, so a tooth shares its load with the next pair near the ends of the stretch and carries it alone in the middle.

Example 1: tooth stress in two steel 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 wide (Figure 9.6.2). The pinion torque is \(T_1 = 100\) N·m. Find the tooth forces, the contact ratio and the Lewis stresses, and compare the Lewis stresses with the root stresses of the finite element model of Figure 9.6.1.

Code

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
a_c = r_1 + r_2                                         # centre distance
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) - a_c*np.sin(phi))
p_b = np.pi*m_g*np.cos(phi)
eps = L_a/p_b
Y_1, Y_2 = lewis_form_factor([N_a, N_b])                # Shigley Table 14-2
sig_L1, sig_L2 = F_tg/(b_g*m_g*Y_1), F_tg/(b_g*m_g*Y_2)

The forces from 9.6.1, the contact ratio from 9.6.4 and the Lewis stresses from 9.6.2 are

\[ \begin{aligned}F_t &=2500.0~\text{N},\quad F_r =909.9~\text{N},\quad F =2660.0~\text{N}\\ L &=18.95~\text{mm},\quad p_b =11.81~\text{mm},\quad \varepsilon =1.605\\ \sigma_1 &=97.05~\text{MPa},\quad \sigma_2 =87.05~\text{MPa}\end{aligned} \]

The model behind Figure 9.6.1, described in Gear tooth contact in Gridap, treats each gear as a plane elastic body clamped at its bore. It applies these forces where the involute theory puts them and splits the force equally between two pairs in contact.

Every pair of teeth runs through the same cycle. It 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 falls and the pinion’s rises as the contact climbs the pinion tooth, until the next pair arrives and takes half. Figure 9.6.3 reads the root stresses of the marked teeth from the results.

Code

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:

\[ \begin{aligned}\sigma_\text{FE,1} &=145.0~\text{MPa} =1.5\,\sigma_1\\\sigma_\text{FE,2} &=126.0~\text{MPa} =1.45\,\sigma_2\end{aligned} \]

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.

Contact stress on the flank

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 \(\rho = r\sin\varphi\). Two elastic cylinders pressed together touch over a narrow strip. According to Hertz’s theory, in the form of [1, Eqs. 3-73 and 14-10] for two bodies with the same \(E\) and \(\nu\), the half-width of the strip and the largest pressure in it are

\[ a_H = \sqrt{\frac{8F(1 - \nu^2)}{\pi b E}\,\frac{\rho_1\rho_2}{\rho_1 + \rho_2}}, \qquad p_\text{max} = \frac{2F}{\pi a_H b} \tag{9.6.5}\]

The pressure is largest at the surface, but the largest shear stress lies below it, at 0.786 half-widths [1]. Pitting starts there: a crack grows below the surface, parallel to it, until a flake of the flank breaks out. Gear tooth contact in Gridap solves this contact with finite elements and finds both.

ISO 6336-2 [2] writes the same pressure at the pitch point with the ratio \(u = N_2/N_1\), \(\rho_2 = u\rho_1\) and \(F = F_t/\cos\varphi\) as the nominal contact stress

\[ \sigma_{H0} = Z_E Z_H \sqrt{\frac{F_t}{b\,d_1}\,\frac{u + 1}{u}}, \qquad Z_E = \sqrt{\frac{E}{2\pi(1 - \nu^2)}}, \qquad Z_H = \sqrt{\frac{2}{\sin\varphi\cos\varphi}} \tag{9.6.6}\]

where the elasticity factor \(Z_E\) carries the material and the zone factor \(Z_H\) the curvature of the flanks. The code below checks that 9.6.6 is 9.6.5, and at the pitch point of Example 1, where one pair carries the whole force, it gives

# Hertz at the pitch point of Example 1
E_g, nu_g = 210e3, 0.3                                  # MPa, -
rho_1, rho_2 = r_1*np.sin(phi), r_2*np.sin(phi)
a_H = np.sqrt(8*F_g*(1 - nu_g**2)/(np.pi*b_g*E_g)*rho_1*rho_2/(rho_1 + rho_2))
p_H = 2*F_g/(np.pi*a_H*b_g)
u_g = N_b/N_a
Z_E = np.sqrt(E_g/(2*np.pi*(1 - nu_g**2)))
Z_H = np.sqrt(2/(np.sin(phi)*np.cos(phi)))
sig_H0 = Z_E*Z_H*np.sqrt(F_tg/(b_g*2*r_1)*(u_g + 1)/u_g)

# the same identity in symbols: Hertz pressure squared minus the ISO form squared
Ft, d, u, b, ph, E, nu = sp.symbols('F_t d_1 u b varphi E nu', positive=True)
r1 = d/2*sp.sin(ph)
aH2 = 8*Ft/sp.cos(ph)*(1 - nu**2)/(sp.pi*b*E)*r1*u*r1/(r1 + u*r1)
p2 = (2*Ft/sp.cos(ph)/(sp.pi*b))**2/aH2
iso2 = E/(2*sp.pi*(1 - nu**2))*2/(sp.sin(ph)*sp.cos(ph))*Ft/(b*d)*(u + 1)/u
check = sp.simplify(p2 - iso2)

The contact is a strip about a fifth of a millimetre wide across the face of the tooth, under a pressure five times the root stress. A hardened steel flank is rated against it for pitting, and on a gear of normalised steel the flank, not the root, sets the torque, as the load diagram shows.

Load variation in spur and helical gears

The root stress of Figure 9.6.3 follows the force on one tooth, and that force jumps whenever a pair of teeth comes into or out of contact. The graph of this force over a tooth’s passage through the mesh is the load variationsv. kraftvariation of the gear pair. Helical gears (Figure 9.6.4) smooth it out.

All contacts of a gear pair lie in the plane of action, the plane through the line of action parallel to the axes. The part of it where the teeth can touch is a rectangle, the length of action \(L\) of 9.6.3 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 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.

Figure 9.6.5 cuts the helical pinion at a depth that moves through its thickness: every cut is a spur pair, and its contact points trace the inclined contact lines.

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,

\[ \varepsilon_\beta = \frac{b\tan\beta_b}{p_b}, \qquad \varepsilon_\gamma = \varepsilon + \varepsilon_\beta \tag{9.6.7}\]

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

\[ \frac{F_k}{F} = \frac{\ell_k}{\sum_j \ell_j}, \qquad \ell_k = \frac{1}{\cos\beta_b}\int_{-b/2}^{b/2} \big[\, s_a \le s_k + z\tan\beta_b \le s_r \,\big]\, dz \tag{9.6.8}\]

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 9.6.8 reduces to the equal split of Example 1.

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 9.6.6 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 make heard as a knock and the drive feels as a vibration at the tooth frequency. The helical tooth takes up its load and puts it down gradually, the total length of contact line hardly changes as the gears turn, and the largest share of a tooth stays well below the whole force. Helical gears therefore run more quietly and carry more load for the same size.

The assumption of equal stiffness per unit length is the weak point of 9.6.8: 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. Sharing the force by the stiffness of the teeth turns the spur staircase into a ramp: an entering pair takes 40 % of the force, and the root stress in double contact drops by up to 19 %, while the largest root stress, in single contact, does not change (Gear tooth contact in Gridap).

Rating to ISO 6336

ISO 6336 keeps the two cores of this chapter, Lewis’s root stress and Hertz’s contact pressure, and adds factors for what the hand model leaves out (Table 9.6.2). Part 3 [9] rates the root,

\[ \sigma_F = \frac{F_t}{b\,m}\,Y_F Y_S Y_\beta Y_B Y_{DT}\;K_A K_v K_{F\beta} K_{F\alpha} \;\le\; \sigma_{FP} = \frac{\sigma_{F\lim}\,Y_{ST}\,Y_{NT}}{S_{F\min}}\, Y_{\delta\,\text{rel}T}\,Y_{R\,\text{rel}T}\,Y_X \tag{9.6.10}\]

and Part 2 [2] the flank,

\[ \sigma_H = Z_H Z_E Z_\varepsilon Z_\beta Z_B \sqrt{\frac{F_t}{b\,d_1}\,\frac{u + 1}{u}}\, \sqrt{K_A K_v K_{H\beta} K_{H\alpha}} \;\le\; \sigma_{HP} = \frac{\sigma_{H\lim}\,Z_{NT}}{S_{H\min}}\,Z_L Z_v Z_R Z_W Z_X \tag{9.6.11}\]

The first fraction of 9.6.10 is Lewis’s stress with \(1/Y\) split in two. The form factor \(Y_F\) takes the bending moment with the force at the outer point of single contact, where one tooth carries it alone, so the contact ratio is already inside it; the stress correction factor \(Y_S\) is the fillet’s stress concentration that Example 1 measured as 1.5. The square root in 9.6.11 is 9.6.6, and the load factors stand under a root because the contact pressure grows with the square root of the force.

The application factor \(K_A\) is the service factor of a geared motor’s catalogue, defined in ISO 6336-1 [10]; it multiplies the nominal torque for the shocks of the driving and the driven machine, from 1.0 when both run smoothly to 2.25 and more when both deliver heavy shocks. The strength side starts from the endurance limits \(\sigma_{F\lim}\) and \(\sigma_{H\lim}\) of ISO 6336-5, measured on standard test gears. \(Y_{ST} = 2\) turns the nominal root limit into the local stress of the test gear, the factor of 2 in Table 9.6.1, and the remaining factors correct for a finite life and for the ways this gear differs from the test gear. \(S_{F\min}\) and \(S_{H\min}\) are the minimum safety factors.

Plastic gears to VDI 2736

Plastic gears fail in ways steel gears do not. Their strength falls as they warm, they run dry and wear, and their teeth bend far enough under load to change the mesh. VDI 2736 [6] rates thermoplastic spur gears with the forms of ISO 6336 and adds checks for these (Table 9.6.3).

The strengths are measured on moulded gears of each polymer, such as the 41 MPa root fatigue strength of POM at 20 °C [7].

For a printed gear the guideline gives the method and none of the numbers. Its materials are moulded polymers, not PLA or PETG, and a printed tooth has layers that bond less well than the polymer holds together, so its strength depends on the direction of the layers through the tooth (print orientation).

N_1p, N_2p, m_p, b_p = 20, 105, 1.0, 8.0               # teeth, mm
E_p, nu_p = 3100.0, 0.35                                # MPa, POM as a stand-in for PLA
T_p = np.array([0.032, 0.15])*1e3                       # running and stalled [N mm]
K_t = sig_fe1/sig_L1                                    # fillet factor from Example 1
d_1p, u_p = m_p*N_1p, N_2p/N_1p
F_tp = 2*T_p/d_1p
sig_F = K_t*F_tp/(b_p*m_p*lewis_form_factor(N_1p))     # local root stress
Z_Ep = np.sqrt(E_p/(2*np.pi*(1 - nu_p**2)))
sig_H = Z_Ep*Z_H*np.sqrt(F_tp/(b_p*d_1p)*(u_p + 1)/u_p)
b_stall = b_p*(sig_H[1]/pom['sig_HP'])**2               # width that keeps the stalled flank at the POM limit

Problems

Problem 1: Sizing by Lewis. The gears of Figure 9.6.2 are to carry \(T_1 = 250\) N·m with the same teeth, \(N_1 = 20\) and \(N_2 = 30\). The steel allows 190 MPa in bending at the root, and the finite element results of Example 1 show that the true root stress is about 1.5 times Lewis’s. Which module is needed with \(b = 10m\)?

Answer: \(F_t = 2T_1/(mN_1)\) and \(\sigma = 1.5\,F_t/(b m Y) = 1.5 \cdot 2T_1/(10\,m^3 N_1 Y) \le 190\) MPa give \(m^3 \ge 1.5 \cdot 2 \cdot 250 \cdot 10^3/(10 \cdot 20 \cdot 0.322 \cdot 190) = 61.3\) mm³, \(m \ge 3.94\) mm: module 4 with \(b = 40\) mm.

Problem 2: The 12-tooth pinion. The conveyor drive of Example 2 in the transmissions chapter keeps its 12/63 first stage, \(m = 1\) mm and \(b = 8\) mm, with the elastic constants and the fillet factor of Example 2. Find the local root stress and the contact stress at stall, \(T_s = 0.15\) N·m, and the face width that brings the stalled flank down to 15.4 MPa.

Answer: \(F_t = 25\) N, \(Y = 0.245\), \(K_t\sigma = 1.5 \cdot 25/(8 \cdot 1 \cdot 0.245) = 19\) MPa. With \(u = 5.25\), \(\sigma_{H0} = 23.7 \cdot 2.49\sqrt{25/(8 \cdot 12) \cdot 6.25/5.25} = 33\) MPa, and \(b = 8 \cdot (33/15.4)^2 \approx 37\) mm. A pinion that small and that wide is a poor design; a 20-tooth pinion, as in Example 2, or a larger module is the better answer.

Further reading

Budynas and Nisbett [1, Ch. 14] derive Lewis’s equation and the AGMA rating method that corresponds to ISO 6336 [2,9,10]. Tavčar and co-authors [5] describe how polymer gears are designed against the criteria of VDI 2736 [6].