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: Φ_n = 20°
• Tooth count: N_P = 18 teeth, N_G = 72 teeth (no interference, Sec. 13–7, p. 677)
• Quality number: Q_v = 6, use grade 1 material
• Assume m_B ≥ 1.2 in Eq. (14–40),

K_B= \begin{cases} 1.6In\frac{2.242}{m_B} & m_B\lt 1.2 \\1 & m_B\geq 1.2\end{cases}               (14.40)

K_B = 1

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{\pi d_pn_p}{12}=\frac{\pi (4.5)1120}{12}=1319 ft/min

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

From Eqs. (14–28) and (14–27),

K_v=\begin{cases} (\frac{A+\sqrt{V} }{A} )^B & V in ft/min \\ (\frac{A+\sqrt{200V} }{A} )^B &V in m/s \end{cases}                          (14.27)

\begin{matrix} A = 50 + 56(1 – B) \\ B = 0.25(12 – Q_v)^{2/3} \end{matrix}                         (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=(\frac{59.77+\sqrt{1319} }{59.77} )^{0.8255}=1.480

From Eq. (14–38),

K_R= \begin{cases} 0.658 – 0.0759 ln(1 – R) & 0.5 \lt R \lt 0.99 \\ 0.50 – 0.109 ln(1 – R) & 0.99 \leq R \leq 0.9999\end{cases}                      (14.38)

K_R = 0.658 – 0.0759ln(1 – 0.95) = 0.885. From Fig. 14–14,

(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,

(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,

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) and (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) 53.14/[10(4.5)] = 0.0698. Thus,

C_{pf} = 0.0698 – 0.0375 + 0.0125(3.14) = 0.0715

From Eq. (14–30),

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

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_G = 4 and m_N = 1,

I=\frac{\cos20° \sin20°}{2} \frac{4}{4+1}=0.1286

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_dW^t for W^t, and solving for the face width (F)_{bend} necessary to resist bending fatigue, we obtain

(F)_{bend}=n_dW^tK_oK_vK_sK_d\frac{K_mK_B}{J_P}\frac{K_TK_R}{S_tY_N}                        (1)

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

(F)_{wear}=\left(\frac{C_PK_TK_R}{S_cZ_N}^2n_dW^tK_oK_vK_s\frac{K_mC_f}{d_PI} \right)                 (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

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(1) (0.885)}{170 000(0.900)} \right)^2 2(2502)1(1.48)1.14 \frac{1.247(1)}{4.5(0.1286)} = 3.22 in

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_{pf} = 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

(\sigma )_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)_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

(\sigma )_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

(\sigma _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)_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)y[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 = 2.254 = 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.