HW23-Solution

HW23-Solution - HW23 - Solution Terry Cooke: Cal Poly San...

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Unformatted text preview: HW23 - Solution Terry Cooke: Cal Poly San Luis Obispo Chapter 9, Solution 9 At x At x 0, y a, y b: 0: b c(1 0) or c b 0 b(1 ka1/ 2 ) or k 1 a 1/ 2 Then y b1 x1/ 2 a1/ 2 1 x1/ 2 b 1 1/ 2 3 a 3 Now dI x 13 y dx 3 dx or dI x 13 x1/ 2 b 1 3 1/ 2 3 a dI x 2 a1 0 3 x a x3/ 2 dx a3/ 2 x1/ 2 a1/ 2 3 x a a Then Ix 3 b3 1 3 x3/ 2 dx a3/ 2 or 0 23 x3/ 2 b x 2 1/ 2 3 a 3 x2 2a 2 x5/ 2 5 a 3/ 2 Ix 13 ab 15 Chapter 9, Solution 13 At x or Then Now a, y b: b kea/a k y b e b x/a e e be x/a 1 dI y x 2 dA x 2 ( y dx) x 2 (be x/a 1dx) a Then Now use integration by parts with Iy dI y bx 2 e x/a 1dx 0 u du a x2 2 x dx x 2 ae x/a a3 2a 0 1 dv v a 0 a e x/a 1dx ae x/a a 0 1 Then x 2 e x/a 1dx (ae x/a 1 )2 x dx 0 xe x/a 1dx Using integration by parts with u du x dx b a3 b a3 b a3 dv v e x/a 1dx ae x/a 1 Then Iy a 2a ( xae x/a 1 ) |0 a 0 (ae x/a 1 )dx 2a a 2 2a a 2 a (a 2 e x/a 1 ) |0 (a 2 a2e 1 ) or Iy 0.264a 3b Chapter 9, Solution 18 See figure of solution on Problem 9.16. A Iy 1 ab 3 a 0 dI y b 1/ 2 x 1/ 2 a x 2 dA x 2 ( y2 y1 )dx b 1/ 2 a a 0 x2 b2 x dx a2 x5/ 2 dx b a2 a 0 x 4 dx 33 ab 35 Iy b b7/2 a1/ 2 7 2 Iy A 3 35 b a5 a2 5 2 7 13 ab 5 Iy 2 ky a 3b ab 3 ky a 9 35 ...
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This note was uploaded on 01/13/2011 for the course ME 211 taught by Professor Staff during the Fall '05 term at Cal Poly.

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