{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Solutions-HW-Chapter10

# Solutions-HW-Chapter10 - budynas_SM_ch10.qxd 18:26 Page 261...

This preview shows pages 1–3. Sign up to view the full content.

Chapter 10 10-1 10-2 A = Sd m dim( A uscu ) = dim( S ) dim( d m ) = kpsi · in m dim( A SI ) = dim( S 1 ) dim ( d m 1 ) = MPa · mm m A SI = MPa kpsi · mm m in m A uscu = 6 . 894 757(25 . 40) m A uscu . = 6 . 895(25 . 4) m A uscu Ans. For music wire, from Table 10-4: A uscu = 201, m = 0 . 145 ; what is A SI ? A SI = 6 . 89(25 . 4) 0 . 145 (201) = 2214 MPa · mm m Ans. 10-3 Given: Music wire, d = 0 . 105 in, OD = 1 . 225 in, plain ground ends, N t = 12 coils. Table 10-1: N a = N t 1 = 12 1 = 11 L s = dN t = 0 . 105(12) = 1 . 26 in Table 10-4: A = 201, m = 0 . 145 (a) Eq. (10-14): S ut = 201 (0 . 105) 0 . 145 = 278 . 7 kpsi Table 10-6: S sy = 0 . 45(278 . 7) = 125 . 4 kpsi D = 1 . 225 0 . 105 = 1 . 120 in C = D d = 1 . 120 0 . 105 = 10 . 67 Eq. (10-6): K B = 4(10 . 67) + 2 4(10 . 67) 3 = 1 . 126 Eq. (10-3): F | S sy = π d 3 S sy 8 K B D = π (0 . 105) 3 (125 . 4)(10 3 ) 8(1 . 126)(1 . 120) = 45 . 2 lbf Eq. (10-9): k = d 4 G 8 D 3 N a = (0 . 105) 4 (11 . 75)(10 6 ) 8(1 . 120) 3 (11) = 11 . 55 lbf/in L 0 = F | S sy k + L s = 45 . 2 11 . 55 + 1 . 26 = 5 . 17 in Ans . 1 2 " 4" 1" 1 2 " 4" 1" budynas_SM_ch10.qxd 01/29/2007 18:26 Page 261

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
262 Solutions Manual Instructor’s Solution Manual to Accompany Mechanical Engineering Design (b) F | S sy = 45 . 2 lbf Ans . (c) k = 11 . 55 lbf/in Ans . (d) ( L 0 ) cr = 2 . 63 D α = 2 . 63(1 . 120) 0 . 5 = 5 . 89 in Many designers provide ( L 0 ) cr / L 0 5 or more; therefore, plain ground ends are not often used in machinery due to buckling uncertainty. 10-4 Referring to Prob. 10-3 solution, C = 10 . 67, N a = 11, S sy = 125 . 4 kpsi, ( L 0 ) cr = 5 . 89 in and F = 45 . 2 lbf (at yield) . Eq. (10-18): 4 C 12 C = 10 . 67 O . K . Eq. (10-19): 3 N a 15 N a = 11 O . K . L 0 = 5 . 17 in, L s = 1 . 26 in y 1 = F 1 k = 30 11 . 55 = 2 . 60 in L 1 = L 0 y 1 = 5 . 17 2 . 60 = 2 . 57 in ξ = y s y 1 1 = 5 . 17 1 . 26 2 . 60 1 = 0 . 50 Eq. (10-20): ξ 0 . 15, ξ = 0 . 50 O . K .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 5

Solutions-HW-Chapter10 - budynas_SM_ch10.qxd 18:26 Page 261...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online