16_ch 08 Mechanical Design budynas_SM_ch08

# 16_ch 08 Mechanical Design budynas_SM_ch08 - 3 84 3 = 450...

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

Chapter 8 219 k b = A d A t E A d l t + A t l d = 314 . 2(245)(207) 314 . 2(14) + 245(34) = 1251 . 9 MN/m Use Wileman et al. Eq. (8-23) A = 0 . 787 15, B = 0 . 628 73 k m Ed = A exp ± Bd L G ² = 0 . 787 15 exp ³ 0 . 628 73 ± 20 48 ²´ = 1 . 0229 k m = 1 . 0229(207)(20) = 4235 MN/m C = 1251 . 9 1251 . 9 + 4235 = 0 . 228 Bolts carry 0.228 of the external load; members carry 0.772 of the external load. Ans. Thus, the actual loads are F b = CP + F i = 0 . 228(20) + 132 . 3 = 136 . 9kN F m = (1 C ) P F i = (1 0 . 228)20 132 . 3 =− 116 . 9kN 8-31 Given p max = 6MPa, p min = 0 and from Prob. 8-20 solution, C = 0 . 2346, F i = 37 . 9kN, A t = 84 . 3mm 2 . For 6MPa, P = 10 . 6kN per bolt σ i = F i A t = 37 . 9(10
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 3 ) 84 . 3 = 450 MPa Eq. (8-35): σ a = C P 2 A t = . 2346(10 . 6)(10 3 ) 2(84 . 3) = 14 . 75 MPa σ m = σ a + σ i = 14 . 75 + 450 = 464 . 8 MPa (a) Goodman Eq. (8-40) for 8.8 bolts with S e = 129 MPa, S ut = 830 MPa S a = S e ( S ut − σ i ) S ut + S e = 129(830 − 450) 830 + 129 = 51 . 12 MPa n f = S a σ a = 51 . 12 14 . 75 = 3 . 47 Ans . 24 24 30 30...
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