8_ch 03 Mechanical Design budynas_SM_ch03

# 8_ch 03 Mechanical Design budynas_SM_ch03 -...

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Chapter 3 21 15 x 20: V = 160 40 x + 40( x 8) + 352 320 =− 128 lbf M = 160 x 20 x 2 20( x 8) + 352( x 10) 320( x 15) 128 x + 2560 Plots of V and M are the same as in Prob. 3-3. 3-5 Solution depends upon the beam selected. 3-6 (a) Moment at center, x c = ( l 2 a ) / 2 M c = w 2 ± l 2 ( l 2 a ) ² l 2 ³ 2 ´ = w l 2 ² l 4 a ³ At reaction, | M r |= w a 2 / 2 a = 2 . 25, l = 10 in, w = 100 lbf/in M c = 100(10) 2 ² 10 4 2 . 25 ³ = 125 lbf · in M r = 100(2 . 25 2 ) 2 = 253 . 1 lbf · in Ans. (b) Minimum occurs when M c =| M r | w l 2 ² l 4 a ³ = w a 2 2 a 2 + al 0 . 25 l 2 = 0 Taking the positive root a = 1 2 µ l + l 2 + 4(0 . 25 l 2 ) · = l 2 (
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## This note was uploaded on 12/13/2011 for the course EML 3013 taught by Professor Shingley during the Fall '11 term at UNF.

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