21_ch 05 Mechanical Design budynas_SM_ch05

21_ch 05 Mechanical Design budynas_SM_ch05 -...

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

View Full Document Right Arrow Icon
Chapter 5 135 M A = 0 . 3 ± 163 . 4 2 + 107 2 = 58 . 59 N · m( maximum) M B = 0 . 15 ± 89 . 2 2 + 174 . 4 2 = 29 . 38 N · m σ x = 32(58 . 59) π d 3 = 596 . 8 d 3 τ xy = 16(33) π d 3 = 168 . 1 d 3 σ ± = ( σ 2 x + 3 τ 2 ) 1 / 2 = ² ³ 596 . 8 d 3 ´ 2 + 3 ³ 168 . 1 d 3 ´ 2 µ 1 / 2 = 664 . 0 d 3 = 370(10 6 ) 3 . 0 d = 17 . 5(10 3 )m = 17 . 5mm, so use 18 mm Ans. 5-28 From Prob. 5-27, τ max = ² ³ σ x 2 ´ 2 + τ 2 µ 1 / 2 = S y 2 n ² ³ 596 . 8 2 d 3 ´ 2 + ³ 168 . 1 d 3 ´ 2 µ 1 / 2 = 342 . 5 d 3 = 370(10 6 ) 2(3 . 0) d = 17 . 7(10 3 = 17 . 7mm, so use 18 mm Ans. 5-29 For the loading scheme shown in Figure ( c ), M max = F 2 ³ a 2 + b 4 ´ = 4 . 4 2 (6 + 4 . 5) = 23 . 1N · m For a stress element at A : σ x = 32 M π d 3 = 32(23 . 1)(10 3 ) π (12) 3 = 136 . 2MPa The shear at
Background image of page 1
This is the end of the preview. Sign up to access the rest of the 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