9_ch 12 Mechanical Design budynas_SM_ch12

# 9_ch 12 Mechanical Design budynas_SM_ch12 - ± T = 2(68°C...

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

312 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 12-11 Given: SAE 30, N = 8rev/s, T s = 60°C, l / d = 1, d = 80 mm, b = 80 . 08 mm, W = 3000 N c min = b min d max 2 = 80 . 08 80 2 = 0 . 04 mm r = d / 2 = 80 / 2 = 40 mm r c = 40 0 . 04 = 1000 P = 3000 80(80) = 0 . 469 MPa Trial #1 : From Figure 12-13 for T = 81°C, µ = 12 mPa · s ± T = 2(81°C 60°C) = 42°C S = (1000 2 ) ± 12(10 3 )(8) 0 . 469(10 6 ) ² = 0 . 2047 From Fig. 12-24, 0 . 120 ± T P = 0 . 349 + 6 . 009(0 . 2047) + 0 . 0475(0 . 2047) 2 = 1 . 58 ± T = 1 . 58 ³ 0 . 469 0 . 120 ´ = 6 . 2°C Discrepancy = 42°C 6 . 2°C = 35 . 8°C Trial #2 : From Figure 12-13 for T = 68°C, µ = 20 mPa · s,
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ± T = 2(68°C − 60°C) = 16°C S = . 2047 ³ 20 12 ´ = . 341 From Fig. 12-24, . 120 ± T P = . 349 + 6 . 009(0 . 341) + . 0475(0 . 341) 2 = 2 . 4 ± T = 2 . 4 ³ . 469 . 120 ´ = 9 . 4°C Discrepancy = 16°C − 9 . 4°C = 6 . 6°C Trial #3 : µ = 21 mPa · s, T = 65°C ± T = 2(65°C − 60°C) = 10°C S = . 2047 ³ 21 12 ´ = . 358...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online