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{\pi d_Pn_P}{12}=\frac{\pi (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. To 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)

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)

the dynamic factor is

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)_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

The load distribution factor K_m is determined from Eq. (14–30),

K_m = C_{mf} = 1 + C_{mc}(C_{pf} C_{pm} + C_{ma} C_e)

where five terms are needed. They are, where F = 1.5 in when needed:

Uncrowned, Eq. (14–30): C_{mc} = 1,

Eq. (14–32):

C_{pf}= \begin{cases} \frac{F}{10d_P}-0.025 & F \leq 1 in \\ \frac{F}{10d_P}-0.0375+0.0125F & 1\lt F\leq 17 in\\ \frac{F}{10d_P}-0.1109+0.0207F-0.000 228F^2 & 17\lt F\leq 40 in \end{cases}                        (14.32)

C_{pf} = 1.5/[10(1.7)] – 0.0375 + 0.0125(1.5) = 0.0695

Bearings immediately adjacent, Eq. (14–33):

C_{pm}= \begin{cases} 1 & for  straddle-mounted  pinion  with  S_1/S \lt 0.175 \\ 1.1 & for  straddle-mounted  pinion  with  S_1/S \geq 0.175 \end{cases}                       (14.33)

C_{pm} = 1

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

Eq. (14–35):

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_e = 1

Thus,

K_m = 1 + C_{mc}(C_{pf} 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 = 5217 = 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),

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)

with m_N = 1 for spur gears,

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

From Table 14–8, C_p = 2300\sqrt{psi}.

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

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

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

(\sigma )_P=\left(W^tK_oK_vK_s\frac{P_d}{F}\frac{K_mK_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)_P=\left(\frac{S_tY_N/(K_TK_R)}{\sigma } \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

(\sigma )_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

(\sigma )_P=\left(W^tK_oK_vK_s\frac{K_m}{d_PF}\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)_P=\left[\frac{S_CZ_N/(K_TK_R)}{\sigma _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,

(\sigma _c)_G=\left[\frac{(K_S)_G}{(K_S)_P} \right]^{1/2}                (\sigma _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.