18_ch 09 Mechanical Design budynas_SM_ch09

# 18_ch 09 Mechanical Design budynas_SM_ch09 - = 577 min(36...

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

256 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 9-25 For the pattern in bending shown, ﬁnd the centroid G of the weld group. ¯ x = 6(0 . 707)(1 / 4)(3) + 6(0 . 707)(3 / 8)(13) 6(0 . 707)(1 / 4) + 6(0 . 707)(3 / 8) = 9in I 1 / 4 = 2 ( I G + A 2 ¯ x ) = 2 ± 0 . 707(1 / 4)(6 3 ) 12 + 0 . 707(1 / 4)(6)(6 2 ) ² = 82 . 7in 4 I 3 / 8 = 2 ± 0 . 707(3 / 8)(6 3 ) 12 + 0 . 707(3 / 8)(6)(4 2 ) ² = 60 . 4in 4 I = I 1 / 4 + I 3 / 8 = 82 . 7 + 60 . 4 = 143 . 1in 4 The critical location is at B . From Eq. (9-3), τ ± = F 2[6(0 . 707)(3 / 8 + 1 / 4)] = 0 . 189 F τ ±± = Mc I = (8 F )(9) 143 . 1 = 0 . 503 F τ max = ³ τ ± 2 + τ ±± 2 = F ³ 0 . 189 2 + 0 . 503 2 = 0 . 537 F Materials: A36 Member: S y = 36 kpsi 1015 HR Attachment: S y = 27 . 5 kpsi E6010 Electrode: S y = 50 kpsi τ all
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: = . 577 min(36, 27 . 5, 50) = 15 . 9 kpsi F = τ all / n . 537 = 15 . 9 / 2 . 537 = 14 . 8 kip Ans. 9-26 Figure P9-26 b is a free-body diagram of the bracket. Forces and moments that act on the welds are equal, but of opposite sense. (a) M = 1200(0 . 366) = 439 lbf · in Ans. (b) F y = 1200 sin 30 ◦ = 600 lbf Ans. (c) F x = 1200 cos 30 ◦ = 1039 lbf Ans. 3 8 " 3 8 " 1 4 " 1 4 " 7" 9" g g g g y x G B...
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