4_ch 07 Mechanical Design budynas_SM_ch07

4_ch 07 Mechanical Design budynas_SM_ch07 -...

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Unformatted text preview: budynas_SM_ch07.qxd 11/30/2006 15:31 Page 181 FIRST PAGES 181 Chapter 7 Eq. (7-16): n= Sy Sy = σmax 32 K f M π d3 2 16 K f s T +3 π d3 2 1/2 Solving for d, 16n d= 4( K f M ) 2 + 3( K f s T ) 2 π Sy 1/2 1/3 16(2.5) = 4(2.4)(4257) 2 + 3(2.1)(3000) 2 π (60 000) = 1.700 in 1/2 1/3 Ans . ka = 2.70(80) −0.265 = 0.845 (b) Assume d = 2.00 in to estimate the size factor, 2 0.3 kb = −0.107 = 0.816 Se = 0.845(0.816)(0.5)(80) = 27.6 kpsi Selecting the DE-ASME Elliptic criteria, use Eq. (7-12) with Mm = Ta = 0. 1/3 2 2 1/2 16(2.5) 2.1(3000) 2.4(4257) d= +3 = 2.133 in 4 π 27 600 60 000 Revising kb results in d = 2.138 in Ans. 7-4 We have a design task of identifying bending moment and torsion diagrams which are preliminary to an industrial roller shaft design. y y FB Fz B z Fz C Fy C y FC = 30(8) = 240 lbf z FC = 0.4(240) = 96 lbf z T = FC (2) = 96(2) = 192 lbf · in z FB = y T 192 = = 128 lbf 1.5 1.5 z FB = FB tan 20° = 128 tan 20° = 46.6 lbf ...
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