{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

10_ch 06 Mechanical Design budynas_SM_ch06

10_ch 06 Mechanical Design budynas_SM_ch06 -...

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

View Full Document Right Arrow Icon
156 Solutions Manual Instructor’s Solution Manual to Accompany Mechanical Engineering Design 6-17 Table A-20: S ut = 64 kpsi, S y = 54 kpsi A = 0 . 375(1 0 . 25) = 0 . 2813 in 2 σ max = F max A = 3000 0 . 2813 (10 3 ) = 10 . 67 kpsi n y = 54 10 . 67 = 5 . 06 Ans . S e = 0 . 5(64) = 32 kpsi k a = 2 . 70(64) 0 . 265 = 0 . 897 k b = 1, k c = 0 . 85 S e = 0 . 897(1)(0 . 85)(32) = 24 . 4 kpsi Table A-15-1: w = 1 in, d = 1 / 4 in, d /w = 0 . 25 K t = 2 . 45 . Fig. 6-20, with r = 0 . 125 in, q ˙= 0 . 8 Eq. (6-32): K f = 1 + 0 . 8(2 . 45 1) = 2 . 16 σ a = K f F max F min 2 A = 2 . 16 3 . 000 0 . 800 2(0 . 2813) = 8 . 45 kpsi σ m = K f F max + F min 2 A = 2 . 16 3 . 000 + 0 . 800 2(0 . 2813) = 14 . 6 kpsi (a) Gerber, Table 6-7 n f = 1 2 64 14 . 6 2 8 . 45 24 . 4 1 + 1 + 2(14 . 6)(24 . 4) 8 . 45(64) 2
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}