Tài liệu shi20396 chương 15 docx

Chapter 15

15-1 Given: Uncrowned, through-hardened 300 Brinell core and case, Grade 1, NC = 109 rev of

pinion at R = 0.999, NP = 20 teeth, NG = 60 teeth, Qv = 6, Pd = 6 teeth/in, shaft angle

90°, np = 900 rev/min, JP = 0.249 and JG = 0.216 (Fig. 15-7), F = 1.25 in, SF =

SH = 1, Ko = 1.

Mesh dP = 20/6 = 3.333 in

dG = 60/6 = 10.000 in

Eq. (15-7): vt = π(3.333)(900/12) = 785.3 ft/min

Eq. (15-6): B = 0.25(12 − 6)2/3 = 0.8255

A = 50 + 56(1 − 0.8255) = 59.77

Eq. (15-5): Kv =

59.77 + √785.3

59.77 0.8255

= 1.374

Eq. (15-8): vt,max = [59.77 + (6 − 3)]2 = 3940 ft/min

Since 785.3 < 3904, Kv = 1.374 is valid. The size factor for bending is:

Eq. (15-10): Ks = 0.4867 + 0.2132/6 = 0.5222

For one gear straddle-mounted, the load-distribution factor is:

Eq. (15-11): Km = 1.10 + 0.0036(1.25)2 = 1.106

Eq. (15-15): (KL)P = 1.6831(109

)

−0.0323 = 0.862

(KL)G = 1.6831(109

/3)−0.0323 = 0.893

Eq. (15-14): (CL)P = 3.4822(109

)

−0.0602 = 1

(CL)G = 3.4822(109

/3)−0.0602 = 1.069

Eq. (15-19): KR = 0.50 − 0.25 log(1 − 0.999) = 1.25 (or Table 15-3)

CR =

KR = √

1.25 = 1.118

Bending

Fig. 15-13: 0.99 St = sat = 44(300) + 2100 = 15 300 psi

Eq. (15-4): (σall)P = swt = sat KL

SF KT KR

= 15 300(0.862)

1(1)(1.25) = 10 551 psi

Eq. (15-3): Wt

P = (σall)P F Kx JP

PdKoKvKsKm

= 10 551(1.25)(1)(0.249)

6(1)(1.374)(0.5222)(1.106) = 690 lbf

H1 = 690(785.3)

33 000 = 16.4 hp

Eq. (15-4): (σall)G = 15 300(0.893)

1(1)(1.25) = 10 930 psi

shi20396_ch15.qxd 8/28/03 3:25 PM Page 390