Question 14.4: A 17-tooth 20° pressure angle spur pinion rotates at 1800 re...

There will be many terms to obtain so use Figs. 14–17 and 14–18 as guides to what is needed.

d_{P} = N_{P}/P_{d} = 17/10 = 1.7  in      d_{G} = 52/10 = 5.2   in

V =\frac{πd_{P}n_{P}}{12} =\frac{π(1.7)1800}{12} = 801.1  ft/min

W^{t} =\frac{33  000  H}{V} =\frac{33  000(4)}{801.1} = 164.8  lbf

Assuming uniform loading, K_{o}= 1. T_{o} evaluate K_{v} , from Eq. (14–28) with a quality number Q_{v} = 6,

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

B = 0.25(12 − 6)^{2/3} = 0.8255
A = 50 + 56(1 − 0.8255) = 59.77

Then from Eq. (14–27) the dynamic factor is

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)

K_{v} =\left( \frac{59.77 + \sqrt{801.1}}{59.77}\right)^{0.8255}= 1.377

To determine the size factor, K_{s} , the Lewis form factor is needed. From Table 14–2, with N_{P} = 17 teeth, Y_{P} = 0.303. Interpolation for the gear with N_{G} = 52 teeth yields Y_{G} = 0.412. Thus from Eq. (a) of Sec. 14–10, with F = 1.5 in,

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

(K_{s})_{P} = 1.192 \left(\frac{1.5\sqrt{0.303}}{10}\right)^{0.0535}= 1.043
(K_{s})_{G} = 1.192 \left(\frac{1.5\sqrt{0.412}}{10}\right)^{0.0535}= 1.052

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)

The load distribution factor K_{m} is determined from Eq. (14–30), where five terms are needed. They are, where F = 1.5 in when needed:

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

Uncrowned, Eq. (14–30): C_{mc} = 1,

Eq. (14–32): C_{p f} = 1.5/[10(1.7)] − 0.0375 + 0.0125(1.5) = 0.0695

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)

Bearings immediately adjacent, Eq. (14–33): C_{pm} = 1

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

Commercial enclosed gear units (Fig. 14–11): C_{ma} = 0.15

Eq. (14–35): C_{e} = 1

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

Thus,

K_{m} = 1 + C_{mc}(C_{p f} C_{pm} + C_{ma}C_{e}) = 1 + (1)[0.0695(1) + 0.15(1)] = 1.22

Assuming constant thickness gears, the rim-thickness factor K_{B} = 1. The speed ratio is m_{G} = N_{G}/N_{P} = 52/17 = 3.059. The load cycle factors given in the problem statement, with N(pinion) = 10^{8} cycles and N(gear) = 10^{8}/m_{G} = 10^{8}/3.059 cycles, are

(Y_{N} )_{P} = 1.3558(10^8)^{−0.0178} = 0.977
(Y_{N} )_{G} = 1.3558(10^8/3.059)^{−0.0178} = 0.996

From Table 14.10, with a reliability of 0.9, K_{R} = 0.85. From Fig. 14–18, the temperature and surface condition factors are K_{T} = 1 and C_{f} = 1. From Eq. (14–23), with m_{N} = 1 for spur gears,

I =\begin{cases} \frac{cos  \phi_{t}  sin  \phi_{t}}{2m_{N}} \frac{m_{G}}{m_{G} + 1} & external  gears\\\frac{cos  \phi_{t}  sin  \phi_{t}}{2m_{N}}\frac{m_{G}}{m_{G} − 1} & internal  gears\end{cases}             (14–23)

I =\frac{cos  20° sin  20°}{2} \frac{3.059}{3.059 + 1} = 0.121

Table 14–10
Reliability Factors K_{R} (Y_{Z} ) Source: ANSI/AGMA 2001-D04.

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

Next, we need the terms for the gear endurance strength equations. From Table 14–3, for grade 1 steel with H_{BP} = 240  and  H_{BG} = 200, we use Fig. 14–2, which gives

(S_{t} )_{P} = 77.3(240) + 12 800 = 31 350  psi
(S_{t} )_{G} = 77.3(200) + 12 800 = 28 260  psi

Table 14–3
Repeatedly Applied Bending Strength St 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 1–7.
1  Hardness to be equivalent to that at the root diameter in the center of the tooth space and 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  The allowable stress numbers indicated may be used with the case depths prescribed in 16.1.
5  See figure 12 for type A and type B hardness patterns.
6  If bainite and microcracks are limited to grade 3 levels, 70,000 psi may be used.
7  The overload capacity of nitrided gears is low. Since the shape of the effective S-N curve is flat, the sensitivity to shock should be investigated before proceeding with the design. [7]
*  Tables 8 and 9 of ANSI/AGMA 2001-D04 are comprehensive tabulations of the major metallurgical factors affecting S_{t}  and  S_{c} of flame-hardened and induction-hardened (Table 8) and carburized and hardened (Table 9) steel gears.

Similarly, from Table 14–6, we use Fig. 14–5, which gives

(S_{c})_{P} = 322(240) + 29  100 = 106  400  psi
(S_{c})_{G} = 322(200) + 29  100 = 93  500  psi

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.

From Fig. 14–15,

(Z_{N} )_{P} = 1.4488(10^8)^{−0.023} = 0.948
(Z_{N} )_{G} = 1.4488(10^8/3.059)^{−0.023} = 0.973

For the hardness ratio factor C_{H}, the hardness ratio is H_{BP}/H_{BG} = 240/200 = 1.2.
Then, from Sec. 14–12,

A^{′} = 8.98(10^{−3})(H_{BP}/H_{BG}) − 8.29(10^{−3})
= 8.98(10^{−3})(1.2) − 8.29(10^{−3}) = 0.002 49

Thus, from Eq. (14–36),

C_{H} = 1.0 + A^{′}(m_{G} − 1.0)                    (14–36)

C_{H} = 1 + 0.002 49(3.059 − 1) = 1.005

(a) Pinion tooth bending. Substituting the appropriate terms for the pinion into Eq. (14–15) gives

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}} & (\text{SI units}) \end{cases}                     (14-15)

(σ )_{P} =\left( W^{t} K_{o}K_{v}K_{s} \frac{P_{d}}{F} \frac{K_{m}K_{B}}{J}\right)_{P} = 164.8(1)1.377(1.043) \frac{10}{1.5} \frac{1.22 (1)}{0.30}

= 6417 psi

Substituting the appropriate terms for the pinion into Eq. (14–41) gives

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

(S_{F} )_{P} =\left(\frac{S_{t}Y_{N} /(K_{T} K_{R})}{σ}\right)_{P} =\frac{31 350(0.977)/[1(0.85)]}{6417} = 5.62

Gear tooth bending. Substituting the appropriate terms for the gear into Eq. (14–15) gives

(σ )_{G} = 164.8(1)1.377(1.052) \frac{10}{1.5} \frac{1.22(1)}{0.40} = 4854  psi

Substituting the appropriate terms for the gear into Eq. (14–41) gives

(S_{F} )_{G} =\frac{28 260(0.996)/[1(0.85)]}{4854} = 6.82

(b) Pinion tooth wear. Substituting the appropriate terms for the pinion into Eq. (14–16) gives

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}} & (\text{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}}}& (\text{SI  units}) \end{cases}                              (14-16)

(σ_{c})_{P} = C_{p} \left(W^{t} K_{o}K_{v}K_{s} \frac{K_{m}}{d_{P} F} \frac{C_{f}}{I}\right)^{1/2}_{P}

= 2300 \left[ 164.8(1)1.377(1.043) \frac{1.22}{1.7(1.5)} \frac{1}{0.121}\right]^{1/2}= 70 360  psi

Substituting the appropriate terms for the pinion into Eq. (14–42) gives

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

(S_{H})_{P} =\left[ \frac{S_{c}Z_{N} /(K_{T} K_{R})}{σ_{c}}\right]_{P} =\frac{106  400(0.948)/[1(0.85)]}{70  360} = 1.69

Gear tooth wear. The only term in Eq. (14–16) that changes for the gear is K_{s}. Thus,

(σ_{c})_{G} =\left[\frac{(K_{s})_{G}}{(K_{s})_{P}}\right]^{1/2}(σ_{c})_{P} =\left(\frac{1.052}{1.043}\right)^{1/2}70  360 = 70  660  psi

Substituting the appropriate terms for the gear into Eq. (14–42) with C_{H} = 1.005 gives

(S_{H})_{G} =\frac{93 500(0.973)1.005/[1(0.85)]}{70 660} = 1.52

(c) For the pinion, we compare (S_{F})_{P} with (S_{H})^{2}_{P} , or 5.73 with 1.69^{2} = 2.86, so the threat in the pinion is from wear. For the gear, we compare (S_{F})_{G} with (S_{H})^{2}_{G}, or 6.96 with 1.52^{2} = 2.31, so the threat in the gear is also from wear.

Table 14–9
Empirical Constants A, B, and C for Eq. (14–34), Face Width F in Inches∗ Source: ANSI/AGMA 2001-D04.

*See ANSI/AGMA 2101-D04, pp. 20–22, for SI formulation.