15_ch 14 Mechanical Design budynas_SM_ch14

15_ch 14 Mechanical Design budynas_SM_ch14 - Ans. ( S F ) G...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Chapter 14 363 Fig. 14-6: J P = 0 . 27, J G ˙= 0 . 38 From Table 14-10 for R = 0 . 9, K R = 0 . 85 K T = C f = 1 Eq. (14-23) with m N = 1 I = cos 20 sin 20 2 ± 3 3 + 1 ² = 0 . 1205 Table 14-8: C p = 2300 ³ psi Strength : Grade 1 steel with H BP = H BG = 200 Fig. 14-2: ( S t ) P = ( S t ) G = 77 . 3(200) + 12 800 = 28 260 psi Fig. 14-5: ( S c ) P = ( S c ) G = 322(200) + 29 100 = 93 500 psi Fig. 14-15: ( Z N ) P = 1 . 4488(10 8 ) 0 . 023 = 0 . 948 ( Z N ) G = 1 . 4488(10 8 / 3) 0 . 023 = 0 . 973 Fig. 14-12: H BP / H BG = 1 C H = 1 Pinion tooth bending Eq. (14-15): ( σ ) P = W t K o K v K s P d F K m K B J = 787 . 8(1)(1 . 196)(1 . 088) ± 6 2 ²´ (1 . 156)(1) 0 . 27 µ = 13 167 psi Ans. Factor of safety from Eq. (14-41) ( S F ) P = ´ S t Y N / ( K T K R ) σ µ = 28 260(0 . 977) / [(1)(0 . 85)] 13 167 = 2 . 47 Ans. Gear tooth bending ( σ ) G = 787 . 8(1)(1 . 196)(1 . 097) ± 6 2 ²´ (1 . 156)(1) 0 . 38 µ = 9433 psi
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Ans. ( S F ) G = 28 260(0 . 996) / [(1)(0 . 85)] 9433 = 3 . 51 Ans. Pinion tooth wear Eq. (14-16): ( c ) P = C p W t K o K v K s K m d P F C f I 1 / 2 P = 2300 787 . 8(1)(1 . 196)(1 . 088) 1 . 156 2 . 667(2) 1 . 1205 1 / 2 = 98 760 psi Ans. Eq. (14-42): ( S H ) P = S c Z N / ( K T K R ) c P = 93 500(0 . 948) / [(1)(0 . 85)] 98 760 = 1 . 06 Ans....
View Full Document

This note was uploaded on 12/13/2011 for the course EML 3013 taught by Professor Shingley during the Fall '11 term at UNF.

Ask a homework question - tutors are online