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}=\begin{cases}0.5S_{ut}& S_{ut} ≤ 200  kpsi (1400 MPa) \\100  kpsi & S_{ut} > 200  kpsi \\700  MPa &   S_{ut} > 1400  MPa \end{cases}      (6-8)

S^{′}_{e} = 0.5S_{ut} = 0.5(55) = 27.5  kpsi

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

k_{a} = aS^{b}_{ut}             (6–19)

k_{a} = aS^{b}_{ut} = 2.70(55)−0.265 = 0.934

Table 6–3
A0.95σ Areas of Common Nonrotating Structural Shapes

A_{0.95σ} = 0.01046d^{2} d_{e} = 0.370d
A_{0.95σ} = 0.05hb d_{e} = 0.808\sqrt{hb}
A_{0.95σ} = \begin{cases} 0.10at_{f} & axis 1-1\\ 0.05ba & t_{f} > 0.025a    axis 2-2 \end{cases}
A_{0.95σ} =  \begin{cases} 0.05ab & axis 1-1\\ 0.052xa+ 0.1t _{f} (b− x) & axis 2-2 \end{cases}

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

l =\frac{1}{P} +\frac{1.25}{P} =\frac{1}{8} +\frac{1.25}{8} = 0.281  in

Table 13–1
Standard and Commonly Used Tooth Systems for Spur Gears

d_{e} = 0.808(hb)^{1/2}               (6–25)

d_{e} = 0.808(hb)^{1/2} = 0.808(Ft)^{1/2} = 0.808[1.5(0.250)]^{1/2} = 0.495 in

Then, Eq. (6–20) gives kb as

k_{b} =\begin{cases} (d/0.3)^{−0.107} = 0.879d^{−0.107} & 0.11 ≤ d ≤ 2 in\\0.91d^{−0.157} &2 < d ≤ 10 in\\(d/7.62)^{−0.107} = 1.24d^{−0.107} & 2.79 ≤ d ≤ 51 mm\\1.51d^{−0.157}&51 < d ≤ 254 mm\end{cases}             (6-20)

k_{b} =\left( \frac{de}{0.30}\right)^{−0.107}= \left(\frac{0.495}{0.30}\right)^{−0.107}= 0.948

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

k_{c} =\begin{cases} 1 & bending \\ 0.85 & axial \\ 0.59 & torsion ^{17} \end{cases} (6-26)

17: Use this only for pure torsional fatigue loading. When torsion is combined with other stresses, such as bending, k_{c} = 1 and the combined loading is managed by using the effective von Mises stress as in Sec. 5–5. Note: For pure torsion, the distortion energy predicts that (k_{c})_{torsion} = 0.577.

Two effects are used to evaluate the miscellaneous-effects Marin factor k_{f} . The first of these is the effect of one-way bending. In general, a gear tooth is subjected only to one-way bending. Exceptions include idler gears and gears used in reversing mechanisms.

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

σ =\frac{K_{v}W^{t} P}{FY}                        (14–7)

\frac{S_{a}}{S^{′}_{e}} +\frac{S_{m}}{S_{ut}} = 1

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

S_{a} =\frac{S^{′}_{e} S_{ut}}{S^{′}_{e} + S_{ut}}

Now replace S_{a} with σ/2, and in the denominator replace S^{′}_{e} with 0.5S_{ut} to obtain

σ =\frac{2S^{′}_{e} S_{ut}}{0.5S_{ut} + S_{ut}} =\frac{2S^{′}_{e}}{0.5 + 1} = 1.33S^{′}_{e}

Now k_{f} = σ/S^{′}_{e} = 1.33S^{′}_{e}/S^{′}_{e} = 1.33. However, a Gerber fatigue locus gives mean values of

\frac{S_{a}}{S^{′}_{e}} +\left(\frac{S_{m}}{S_{ut}}\right)^{2}= 1

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

S_{a} =\frac{S^{2}_{ut}}{2S^{′}_{e}} \left( −1 +\sqrt{1 + \frac{4S^{′2}_{e}}{S^{2}_{ut}}}\right)

Setting S_{a} = σ/2, S_{ut} = S^{′}_{e} /0.5  gives

σ =\frac{S^{′}_{e}}{0.5^{2}} \left[−1 +\sqrt{1 + 4(0.5)^{2}}\right]= 1.66S^{′}_{e}

and k_{f} = σ/S^{′}_{e} = 1.66. Since a Gerber locus runs in and among fatigue data and Goodman does not, we will use k_{f}= 1.66.

The second effect to be accounted for in using the miscellaneous-effects Marin factor k_{f} is stress concentration, for which we will use our fundamentals from Chap. 6.
For a 20° full-depth tooth the radius of the root fillet is denoted r_{f}, where

r_{f}=\frac{0.300}{P} =\frac{0.300}{8} = 0.0375  in

From Fig. A–15–6

\frac{r}{d} =\frac{r_{f}}{t} =\frac{0.0375}{0.250} = 0.15

Since D/d = ∞, 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 + (0.62)(1.68 − 1) = 1.42

The miscellaneous-effects Marin factor for stress concentration can be expressed as

k_{f} =\frac{1}{K_{f}} =\frac{1}{1.42} = 0.704

The final value of k_{f} is the product of the two k_{f} factors, that is, 1.66(0.704) = 1.17. The Marin equation for the fully corrected endurance strength is

S_{e} = k_{a}k_{b}k_{c}k_{d}k_{e}k_{f} S^{′}_{e}

= 0.934(0.948)(1)(1)(1)1.17(27.5) = 28.5  kpsi

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

σ_{all} =\frac{S_{e}}{n_{d}} =\frac{28.5}{3} = 9.5  kpsi

The transmitted load W^{t} is

W^{t} =\frac{FYσ_{all}}{K_{v} P} =\frac{1.5(0.296)9 500}{1.52(8) }= 347  lbf

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

hp =\frac{W^{t}V}{33 000} =\frac{347(628)}{33 000} = 6.6  hp

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.