20_ch 07 Mechanical Design budynas_SM_ch07

20_ch 07 Mechanical Design budynas_SM_ch07 - /ω 2 1 and...

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Chapter 7 197 y 1 = 1 . 772[6 . 036(10 5 ) + 9 . 416(10 5 ) + 4 . 104(10 5 )] = 3 . 465(10 4 )in y 2 = 1 . 772[9 . 416(10 5 ) + 1 . 956(10 4 ) + 9 . 416(10 5 )] = 6 . 803(10 4 )in y 3 = 1 . 772[4 . 104(10 5 ) + 9 . 416(10 5 ) + 6 . 036(10 5 )] = 3 . 465(10 4 )in ± w y = 2 . 433(10 3 ), ± w y 2 = 1 . 246(10 6 ) ω 1 = ² 386 ³ 2 . 433(10 3 ) 1 . 246(10 6 ) ´ = 868 rad/s (same as in Prob. 7-14) The point was to show that convergence is rapid using a static deflection beam equation. The method works because: • If a deflection curve is chosen which meets the boundary conditions of moment-free and deflection-free ends, and in this problem, of symmetry, the strain energy is not very sensi- tive to the equation used. • Since the static bending equation is available, and meets the moment-free and deflection- free ends, it works. 7-16 (a) For two bodies, Eq. (7-26) is ¯ ¯ ¯ ¯ ( m 1 δ 11 1 2 ) m 2 δ 12 m 1 δ 21 ( m 2 δ 22 1 2 ) ¯ ¯ ¯ ¯ = 0 Expanding the determinant yields, µ 1 ω 2 2 ( m 1 δ 11 + m 2 δ 22 ) µ 1 ω 2 1 + m 1 m 2 ( δ 11 δ 22 δ 12 δ 21 ) = 0 (1) Eq. (1) has two roots 1
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Unformatted text preview: /ω 2 1 and 1 /ω 2 2 . Thus µ 1 ω 2 − 1 ω 2 1 ¶µ 1 ω 2 − 1 ω 2 2 ¶ = or, µ 1 ω 2 ¶ 2 + µ 1 ω 2 1 + 1 ω 2 2 ¶µ 1 ω ¶ 2 + µ 1 ω 2 1 ¶µ 1 ω 2 2 ¶ = (2) Equate the third terms of Eqs. (1) and (2), which must be identical. 1 ω 2 1 1 ω 2 2 = m 1 m 2 ( δ 11 δ 22 − δ 12 δ 21 ) ⇒ 1 ω 2 2 = ω 2 1 m 1 m 2 ( δ 11 δ 22 − δ 12 δ 21 ) and it follows that ω 2 = 1 ω 1 ² g 2 w 1 w 2 ( δ 11 δ 22 − δ 12 δ 21 ) Ans . (b) In Ex. 7-5, Part (b) the first critical speed of the two-disk shaft ( w 1 = 35 lbf, w 2 = 55 lbf) is ω 1 = 124 . 7 rad/s . From part (a), using influence coefficients ω 2 = 1 124 . 7 ² 386 2 35(55)[2 . 061(3 . 534) − 2 . 234 2 ](10 − 8 ) = 466 rad/s Ans ....
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