Question 14.8: Design a 4:1 spur-gear reduction for a 100-hp, three-phase s...

Make the a priori decisions:
• Function: 100 hp, 1120 rev/min, R = 0.95, N = 10^{9} cycles, K_{o} = 1
• Design factor for unquantifiable exingencies: n_{d} = 2
• Tooth system: \phi_{n} = 20°
• Tooth count: N_{P} = 18 teeth, N_{G} = 72 teeth (no interference)
• Quality number: Q_{v} = 6, use grade 1 material
• Assume m_{B} ≥ 1.2 in Eq. (14–40), K_{B} = 1

K_{B} =\begin{cases} 1.6  ln\frac{2.242}{m_{B}}&m_{B} < 1.2\\1&m_{B} ≥ 1.2\end{cases}                 (14-40)

Pitch: Select a trial diametral pitch of P_{d} = 4 teeth/in. Thus, d_{P} = 18/4 = 4.5 in and d_{G} = 72/4 = 18 in. From Table 14–2, Y_{P} = 0.309, Y_{G} = 0.4324 (interpolated). From Fig. 14–6, J_{P} = 0.32, J_{G} = 0.415.

V =\frac{πd_{P}n_{P}}{12} =\frac{π(4.5)1120}{12} = 1319   ft/min

W^{t} =\frac{33  000H}{V} =\frac{33  000(100)}{1319} = 2502   lbf

Table 14–2
Values of the Lewis Form Factor Y (These Values Are for a Normal Pressure Angle of 20°, Full-Depth Teeth, and a Diametral Pitch of Unity in the Plane of Rotation)

From Eqs. (14–28) and (14–27),

K_{v} =\begin{cases} \left(\frac{A + \sqrt{V}}{A}\right)^{B} & V  in  ft/min\\ \left(\frac{A + \sqrt{200V}}{A}\right)^{B}&V  in  m/s\end{cases}                        (14-27)

A = 50 + 56(1 − B)
B = 0.25(12 − Q_{v})^{2/3}                            (14–28)

B = 0.25(12 − Q_{v})^{2/3} = 0.25(12 − 6)^{2/3} = 0.8255
A = 50 + 56(1 − 0.8255) = 59.77
K_{v} =\left(\frac{59.77 + \sqrt{1319}}{59.77}\right)^{0.8255}= 1.480

From Eq. (14–38), K_{R} = 0.658 − 0.0759  ln(1 − 0.95) = 0.885. From Fig. 14–14,

K_{R}=\begin{cases} 0.658 − 0.0759 ln(1 − R) & 0.5 < R < 0.99\\ 0.50 − 0.109 ln(1 − R)&0.99 ≤ R ≤ 0.9999\end{cases}          (14-38)

(Y_{N} )_{P} = 1.3558(10^{9})^{−0.0178} = 0.938
(Y_{N} )_{G} = 1.3558(10^{9}/4)^{−0.0178} = 0.961

From Fig. 14–15,

σ =\begin{cases} W^{t} K_{o}K_{v}K_{s} \frac{P_{d}}{F}\frac{K_{m}K_{B}}{J} & (\text{U.S. customary units})\\ W^{t} K_{o}K_{v}K_{s} \frac{1}{bm_{t}}\frac{K_{H} K_{B}}{Y_{J}} & (SI units) \end{cases}                      (14-15)

(Z_{N} )_{P} = 1.4488(10^{9})^{−0.023} = 0.900
(Z_{N} )_{G} = 1.4488(10^{9}/4)^{−0.023} = 0.929

From the recommendation after Eq. (14–8), 3p ≤ F ≤ 5p. Try F = 4p = 4π/P = 4π/4 = 3.14 in. From Eq. (a), Sec. 14–10,

σ =\frac{K_{v}W^{t}}{F_{m}Y}             (14–8)

K_{s} =\frac{1}{k_{b}} = 1.192 \left(\frac{F\sqrt{Y}}{P}\right)^{0.0535}                         (a)

K_{s} = 1.192 \left(\frac{F\sqrt{Y}}{P}\right)^{0.0535} = 1.192 \left(\frac{3.14\sqrt{0.309}}{4}\right)^{0.0535}= 1.140

From Eqs. (14–31), (14–33), (14–35), C_{mc} = C_{pm} = C_{e} = 1. From Fig.14–11, C_{ma} = 0.175 for commercial enclosed gear units. From Eq.(14–32), F/(10d_{P}) = 3.14/[10(4.5)] = 0.0698. Thus,

C_{mc} =\begin{cases} 1&for   uncrowned  teeth\\ 0.8&for  crowned  teeth\end {cases}                         (14-31)

C_{p f} =\begin{cases} \frac{F}{10d} − 0.025 & F ≤ 1  in \\ \frac{F}{10d} − 0.0375 + 0.0125F & 1 < F ≤ 17  in\\ \frac{F}{10d} − 0.1109 + 0.0207F − 0.000 228F^{2}&17 < F ≤ 40  in \end{cases}                     (14-32)

C_{p m} =\begin{cases} 1&  for  straddle-mounted  pinion  with  S_{1}/S < 0.175 \\ 1.1 & for  straddle-mounted  pinion  with  S_{1}/S ≥ 0.175 \end{cases}                                  (14.33)

C_{e} =\begin{cases} 0.8&for  gearing  adjusted  at  assembly,  or  compatibility is  improved  by  lapping, or  both\\ 1&for  all  other  conditions\end {cases}                         (14-35)

C_{p f} = 0.0698 − 0.0375 + 0.0125(3.14) = 0.0715

From Eq. (14–30),

K_{m} = C_{mf} = 1 + C_{mc}(C_{p f} C_{pm} + C_{ma}C_{e})                 (14–30)

K_{m}= 1 + (1)[0.0715(1) + 0.175(1)] = 1.247

From Table 14–8, for steel gears, C_{p} = 2300\sqrt{psi}. From Eq. (14–23), with m_{G} = 4  and  m_{N} = 1,

I =\begin{cases} \frac{cos  \phi_{t}  sin  \phi_{t}}{2m_{N}}{m_{G}}{m_{G} + 1} & external  gears \\ \frac{cos  \phi_{t}  sin  \phi_{t}}{2m_{N}}{m_{G}}{m_{G} – 1}& internal  gears \end{cases}                   (14.23)

I =\frac{cos 20° sin 20°}{2} \frac{4}{4 + 1} = 0.1286

Table 14–8
Elastic Coefficient C_{p} (Z_{E}),psi (\sqrt{MPa}) Source: AGMA 218.01

Poisson’s ratio=0.30.
∗When more exact values for modulus of elasticity are obtained from roller contact tests, they may be used.

Pinion tooth bending. With the above estimates of K_{s}  and  K_{m} from the trial diametral pitch, we check to see if the mesh width F is controlled by bending or wear considerations. Equating Eqs. (14–15) and (14–17), substituting n_{d}W^{t}  for  W^{t} , and solving for the face width (F)_{bend} necessary to resist bending fatigue, we obtain

σ =\begin{cases} W^{t} K_{o}K_{v}K_{s} \frac{P_{d}}{F}\frac{K_{m}K_{B}}{J} & (U.S. customary  units)\\ W^{t} K_{o}K_{v}K_{s} \frac{1}{bm_{t}}\frac{K_{H} K_{B}}{Y_{J}} & (SI  units) \end{cases}                      (14-15)

σ_{all} =\begin{cases} \frac{S_{t}}{S_{F}} \frac{Y_{N}}{K_{T} K_{R}} & (U.S.  customary  units)\\\frac{S_{t}}{S_{F}} \frac{Y_{N}}{Y_{θ}Y_{Z}}&(SI   units)\end{cases}             (14–17)

(F)_{bend} = n_{d}W^{t} K_{o}K_{v}K_{s} P_{d} \frac{K_{m}K_{B}}{J_{P}} \frac{K_{T} K_{R}}{S_{t}Y_{N}}               (1)

Equating Eqs. (14–16) and (14–18), substituting n_{d}W^{t} for W^{t} , and  solving for the face width (F)_{wear} necessary to resist wear fatigue, we obtain

σ_{c} =\begin{cases} C_{p} \sqrt{ W^{t} K_{o}K_{v}K_{s} \frac{K_{m}}{d_{P F}} \frac{C_{f}}{I}} & (U.S. customary  units) \\ C_{p} \sqrt{ W^{t} K_{o}K_{v}K_{s} \frac{K_{H}}{d_{w1}b} \frac{Z_{R}}{Z_{I}}}& (SI  units) \end{cases}                              (14-16)

σ_{c,all} =\begin{cases} \frac{S_{c}}{S_{H}} \frac{Z_{N}C_{H}}{K_{T} K_{R}} & (U.S.  customary  units)\\ \frac{S_{c}}{S_{H}}\frac{Z_{N} Z_{W}}{Y_{θ}Y_{Z}}&(SI  units)\end{cases}             (14–18)

(F)_{wear} =\left(\frac{C_{p} Z_{N}}{S_{c}K_{T} K_{R}}\right)^{2} n_{d}W^{t} K_{o}K_{v}K_{s} \frac{K_{m}C_{f}}{d_{P} I}                       (2)

From Table 14–5 the hardness range of Nitralloy 135M is Rockwell C32–36 (302–335 Brinell). Choosing a midrange hardness as attainable, using 320 Brinell. From Fig. 14–4,

S_{t} = 86.2(320) + 12 730 = 40 310  psi

Table 14–5
Nominal Temperature Used in Nitriding and Hardnesses Obtained Source: Darle W. Dudley, Handbook of Practical Gear Design, rev. ed.,
McGraw-Hill, New York, 1984.

∗Nitralloy is a trademark of the Nitralloy Corp., New York.

Inserting the numerical value of S_{t} in Eq. (1) to estimate the face width gives

(F)_{bend} = 2(2502)(1)1.48(1.14)4 \frac{1.247(1)(1)0.885}{0.32(40 310)0.938} = 3.08  in

From Table 14–6 for Nitralloy 135M, S_{c} = 170 000 psi. Inserting this in Eq. (2), we find

(F)_{wear} = \left( \frac{2300(0.900)}{170 000(1)0.885}\right)^{2} 2(2502)1(1.48)1.14 \frac{1.247(1)}{4.5(0.1286) }= 3.44  in

Table 14–6
Repeatedly Applied Contact Strength S_{c} at 10^{7} Cycles and 0.99 Reliability for Steel Gears Source: ANSI/AGMA 2001-D04.

Notes: See ANSI/AGMA 2001-D04 for references cited in notes 1–5.
1 Hardness to be equivalent to that at the start of active profile in the center of the face width.
2 See Tables 7 through 10 for major metallurgical factors for each stress grade of steel gears.
3 The steel selected must be compatible with the heat treatment process selected and hardness required.
4 These materials must be annealed or normalized as a minimum.
5 The allowable stress numbers indicated may be used with the case depths prescribed in 16.1.
* Table 9 of ANSI/AGMA 2001-D04 is a comprehensive tabulation of the major metallurgical factors affecting S_{t} and S_{c} of carburized and hardened steel gears.

Make face width 3.50 in. Correct K_{s}  and  K_{m}:

K_{s} = 1.192 \left( \frac{3.50\sqrt{0.309}}{4}\right)^{0.0535}= 1.147

\frac{F}{10d_{P}} =\frac{3.50}{10(4.5)} = 0.0778
C_{p f} = 0.0778 − 0.0375 + 0.0125(3.50) = 0.0841
K_{m} = 1 + (1)[0.0841(1) + 0.175(1)] = 1.259

The bending stress induced by W^{t} in bending, from Eq. (14–15), is

σ =\begin{cases} W^{t} K_{o}K_{v}K_{s} \frac{P_{d}}{F}\frac{K_{m}K_{B}}{J} & (U.S. customary  units)\\ W^{t} K_{o}K_{v}K_{s} \frac{1}{bm_{t}}\frac{K_{H} K_{B}}{Y_{J}} & (SI  units) \end{cases}                      (14-15)

(σ )_{P} = 2502(1)1.48(1.147) \frac{4}{3.50}  \frac{1.259(1)}{0.32} = 19 100  psi

The AGMA factor of safety in bending of the pinion, from Eq. (14–41), is

S_{F} =\frac{S_{t}Y_{N} /(K_{T} K_{R})}{σ} =\frac{fully  corrected  bending  strength}{bending  stress}                (14–41)

(S_{F} )_{P} =\frac{40 310(0.938)/[1(0.885)]}{19 100} = 2.24

Gear tooth bending. Use cast gear blank because of the 18-in pitch diameter. Use the same material, heat treatment, and nitriding. The load-induced bending stress is in the ratio of J_{P}/J_{G}. Then

(σ )_{G} = 19 100 \frac{0.32}{0.415} = 14 730  psi

The factor of safety of the gear in bending is

(S_{F} )_{G} =\frac{40 310(0.961)/[1(0.885)]}{14 730} = 2.97

Pinion tooth wear. The contact stress, given by Eq. (14–16), is

σ_{c} =\begin{cases} C_{p} \sqrt{ W^{t} K_{o}K_{v}K_{s} \frac{K_{m}}{d_{P F}} \frac{C_{f}}{I}} & (U.S. customary  units) \\ C_{p} \sqrt{ W^{t} K_{o}K_{v}K_{s} \frac{K_{H}}{d_{w1}b} \frac{Z_{R}}{Z_{I}}}& (SI  units) \end{cases}                              (14-16)

(σ_{c})_{P} = 2300 \left[ 2502(1)1.48(1.147)\frac{1.259}{4.5(3.5)} \frac{1}{0.129}\right]^{1/2}= 118 000  psi

The factor of safety from Eq. (14–42), is

S_{H} =\frac{S_{c}Z_{N}C_{H}/(K_{T} K_{R})}{σ_{c}} =\frac{fully  corrected  contact  strength}{contact  stress}                   (14–42)

(S_{H})_{P} =\frac{170 000(0.900)/[1(0.885)]}{118 000} = 1.465

By our definition of factor of safety, pinion bending is (S_{F} )_{P} = 2.24, and wear is (S_{H})^{2}_{P} = (1.465)^{2} = 2.15.

Gear tooth wear. The hardness of the gear and pinion are the same. Thus, from Fig. 14–12, C_{H} = 1, the contact stress on the gear is the same as the pinion, (σ_{c})_{G} = 118 000 psi. The wear strength is also the same, S_{c} = 170 000 psi. The factor of safety of the gear in wear is

(S_{H})_{G} =\frac{170 000(0.929)/[1(0.885)]}{118 000} = 1.51

So, for the gear in bending, (S_{F} )_{G} = 2.97, and wear (S_{H})^{2}_{G} = (1.51)^{2} = 2.29.

Rim. Keep m_{B} ≥ 1.2. The whole depth is h_{t} = addendum + dedendum = 1/P_{d }+ 1.25/P_{d} = 2.25/P_{d} = 2.25/4 = 0.5625 in. The rim thickness t_{R} is

t_{R} ≥ m_{B}h_{t} = 1.2(0.5625) = 0.675  in

In the design of the gear blank, be sure the rim thickness exceeds 0.675 in; if it does not, review and modify this mesh design.