Question 14–2 : Estimate the horsepower rating of the gear in the previous e...

The rotating-beam endurance limit is estimated from Eq. (6–8),

S_{e}^{\prime}=\left\{\begin{array}{ll}0.5 S_{u t} & S_{u t} \leq 200 \text { kpsi }(1400 \text { MPa }) \\100 kpsi & S_{u t}> kpsi \\700 MPa & S_{u t}>1400 MPa\end{array}\right.      (6–8)

To obtain the surface finish Marin factor k_{a} we refer to Table 6–3, for machined surface, finding a=2.70 \text { and } b=-0.265. Then Eq. (6–19), gives the surface finish Marin factor k_{a} as

k_{a}=a S_{u t}^{b}     (6–19)

The next step is to estimate the size factor k_{b}. From Table 13–1, the sum of the addendum and dedendum is

The tooth thickness t in Fig. 14–1b is given in [Eq. (b)] as t=(4 l x)^{1 / 2} when x=3 Y /(2 P) from Eq. (14–3). Therefore, since from Ex. 14–1 Y = 0.296 and P = 8,

Y=\frac{2 x P}{3}     (14–3)

then

We have recognized the tooth as a cantilever beam of rectangular cross section, so the equivalent rotating-beam diameter must be obtained from Eq. (6–25):

d_{e}=0.808(h b)^{1 / 2}      (6–25)

Then, Eq. (6–20), gives k_{b} as

k_{b}=\left\{\begin{array}{ll}(d / 0.3)^{-0.107}=0.879 d^{-0.107} & 0.11 \leq d \leq 2 \text { in } \\0.91 d^{-0.157} & 2<d \leq 10 \text { in } \\(d / 7.62)^{-0.107}=1.24 d^{-0.107} & 2.79 \leq d \leq 51 mm \\1.51 d^{-0.157} & 51<d \leq 254 mm\end{array}\right.    (6–20)

The load factor k_{c} from Eq. (6–26), is unity. With no information given concerning temperature and reliability we will set k_{d}=k_{e}=1 .

In general, a gear tooth is subjected only to one-way bending. Exceptions include
idler gears and gears used in reversing mechanisms. We will account for one-way
bending by establishing a miscellaneous-effects Marin factor k_{f}.

For one-way bending the steady and alternating stress components are \sigma_{a}=\sigma_{m}= \sigma / 2 where \sigma is the largest repeatedly applied bending stress as given in Eq. (14–7).If a material exhibited a Goodman failure locus,

k_{c}=\left\{\begin{array}{ll}1 & \text { bending } \\0.85 & \text { axial } \\0.59 & \text { torsion }\end{array}\right.  (6–26)

\sigma=\frac{K_{v} W^{t} P}{F Y}    (14–7)

Since S_{a} \text { and } S_{m} are equal for one-way bending, we substitute S_{a} \text { for } S_{m} and solve the preceding equation for S_{a}, giving

Now replace S_{a} \text { with } \sigma / 2, and in the denominator replace S_{e}^{\prime} \text { with } 0.5 S_{u t} to obtain

Now k_{f}=\sigma / S_{e}^{\prime}=1.33 S_{e}^{\prime} / S_{e}^{\prime}=1.33. However, a Gerber fatigue locus gives mean values of 

Setting S_{a}=S_{m} and solving the quadratic in S_{a} gives

and k_{f}=\sigma / S_{e}^{\prime}=1.66. Since a Gerber locus runs in and among fatigue data and Goodman does not, we will use k_{f}=1.66. The Marin equation for the fully corrected endurance strength is

For stress, we will first determine the fatigue stress-concentration factor K_{f}. For a 20° full-depth tooth the radius of the root fillet is denoted r_{f}, where

From Fig. A–15–6

Since D / d=\infty, we approximate with D / d=3, giving K_{t}=1.68. From Fig. 6–20, q=0.62. From Eq. (6–32),

K_{f}=1+q\left(K_{t}-1\right) \quad \text { or } \quad K_{f s}=1+q_{\text {shear }}\left(K_{t s}-1\right)    (6–32)

For a design factor of n_{d}=3, as used in Ex. 14–1, applied to the load or strength, the maximum bending stress is

The transmitted load W^{t} is

and the power is, with V = 628 ft/min from Ex. 14–1,

Again, it should be emphasized that these results should be accepted only as preliminary estimates to alert you to the nature of bending in gear teeth.