HW21-Solution

HW21-Solution - HW21 - Solution Terry Cooke: Cal Poly San...

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Unformatted text preview: HW21 - Solution Terry Cooke: Cal Poly San Luis Obispo Chapter 5, Solution 36 For the element (EL) shown At Then x x dA xEL yEL a, y h: h ka 3 or k h a3 a 1/3 y h1/3 xdy 1 x 2 y a 1/3 y dy h1/3 1 a 1/3 y 2 h1/ 3 h Now h Then A dA 0 h a 1/3 y dy h1/3 3a y 4/3 4 h1/3 0 3 ah 4 h and xEL dA yEL dA 0 h 1 a 1/3 a 1/3 y y dy 2 h1/3 h1/3 y a 1/3 y dy h1/3 x 3 ah 4 3 ah 4 1 a 3 5/3 y 2 h 2/3 5 h 0 32 ah 10 0 a 3 7/3 y h1/3 7 32 ah 10 32 ah 7 0 32 ah 7 x 2 a 5 4 h 7 Hence xA xEL dA: yEL dA: yA y y Chapter 5, Solution 40 At x b 0, y b or k b a2 k (0 a ) 2 Then Now y xEL yEL b ( x a) 2 a2 x y 2 b ( x a)2 2a 2 and dA ydx b ( x a ) 2 dx a2 a Then A dA 0 a b ( x a )2 dx a2 b a2 b 3a 2 a 0 xa x3 3 a 0 1 ab 3 a 2 x dx and xEL dA x 0 b ( x a)2 dx a2 23 ax 3 a2 2 x 2 2ax 2 b x4 a2 4 a 12 ab 12 b2 1 ( x a )5 2a 4 5 a yEL dA 0 b b ( x a)2 2 ( x a ) 2 dx 2 2a a 0 12 ab 10 Hence xA xEL dA: x yEL dA: y 1 ab 3 1 ab 3 12 ab 12 12 ab 10 x 1 a 4 3 b 10 yA y Chapter 5, Solution 44 For y1 at Then x y1 y2 xEL a, y 2b 2b ka 2 or k 2b a2 2b 2 x a2 b ( x 2b) b 2 a x By observation Now and for 0 # x # a : For a # x # 2a : x a yEL yEL 1 y1 2 1 y2 2 b2 x a2 b 2 2 a and dA x a y1dx 2b 2 x dx a2 y2 dx b2 x dx a and dA 2a Then A dA 0 2b 2 x dx a2 b a 2 2 2a b2 a x dx a 2a 2b x3 a2 3 a a 0 x a 2 0 7 ab 6 and xEL dA x 0 2b 2 x dx a2 a xb2 a 2 x dx a 2b x 4 a2 4 bx 0 x3 3a 2a 0 12 ab b 2 72 ab 6 (2a) 2 (a)2 1 (2a 2 ) (a)3 3a PROBLEM 5.44 (Continued) a yEL dA 0 b 2 2b 2 x x dx a2 a2 a 2a 0 b 2 2 x a 3 x a 2a b2 x dx a 2b 2 x5 a4 5 17 2 ab 30 0 b2 2 a 2 3 a Hence xA xEL dA: x yEL dA: y 7 ab 6 7 ab 6 72 ab 6 17 2 ab 30 y x a yA 17 b 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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