HW24-Solution

HW24-Solution - HW24 - Solution Terry Cooke: Cal Poly San...

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Unformatted text preview: HW24 - Solution Terry Cooke: Cal Poly San Luis Obispo Chapter 9, Solution 23 (a) dA 2 u du dJ O u 2 dA u 2 (2 u du ) 2 u 3 du JO dJ O 2 2 R2 R1 u 3 du R2 14 u 4 JO R1 2 4 R2 R14 (b) From Eq. (9.4): (Note by symmetry.) Ix JO Ix Iy Ix 1 JO 2 Iy 4 2I x 4 R2 R14 Ix 4 4 R2 R14 Chapter 9, Solution 26 The equation of the circle is x2 y2 x dA A r2 r2 x dy dA 2 r/ 2 So that Now Then Let Then y2 r2 r y 2 dy r2 y 2 dy y r sin ; dy /2 r cos d A2 /6 /2 r2 r cos 2 (r sin ) 2 r cos d 2 2 /6 d sin 2r 3 2 2 sin 2 4 2r 2 3 /2 /6 2r 2 2 6 2 2 4 3 8 2.5274r 2 Now Then Let Then dI x Ix y y 2 dA dI x y2 r r2 y 2 dy y 2 dy 2 r/ 2 y2 r 2 r cos d r sin ; dy /2 Ix 2 /6 /2 (r sin ) 2 r 2 (r sin ) 2 r cos d 2 /6 r 2 sin 2 (r cos )r cos d sin 2 cos 2 1 sin 2 4 Now sin 2 2sin cos /2 Then Ix 2 /6 r4 12 sin 2 4 d r4 22 r 2 4 2 sin 4 8 6 /2 /6 sin 8 2 3 2 2 3 16 r4 23 PROBLEM 9.26 (Continued) Also Then Let Then dI y Iy 13 x dy 3 dI y 2 1 3 r r2 1 y2 (r 2 3 dy y 2 )3/ 2 dy r/ 2 3 y Iy Iy r sin ; dy 2 3 2 3 /2 /6 /2 /6 r cos d ( r sin ) 2 ]3/ 2 r cos d [r 2 ( r 3 cos3 ) r cos d cos 2 12 sin 2 4 Now Then cos 4 cos 2 (1 sin 2 ) Iy 2 3 24 r 3 /2 /6 r 4 cos 2 sin 2 4 12 sin 2 4 1 42 6 d sin 4 8 /2 2 2 /6 24 r 3 24 r 3 4 2 12 42 1 4 sin 4 3 2 3 2 1 4 1 32 6 sin 8 2 3 2 3 2 16 93 64 12 48 24 r 3 4 Now JP Ix Iy r4 23 3 16 3 16 24 r 3 4 93 64 or JP 1.155r 4 r4 JP A 3 1.15545r 4 and 2 kP 1.15545r 4 2.5274r 2 or k P 0.676r Chapter 9, Solution 28 By observation: y h b 2 x or Now and Then x b y 2h dA xdy y 2 dA dI x b y dy 2h b3 y dy 2h h dI x Ix 2 0 b3 y dy 2h h b y4 h4 0 13 bh 4 From above: Now y 2h x b dA (h y )dx h 2h x dx b h (b 2 x)dx b and Then dI y Iy x 2 dA dI y x2 2 h (b 2 x)dx b b/ 2 0 h2 x (b 2 x)dx b 2 h1 3 bx b3 hbb b32 3 14 x 2 b/ 2 0 4 2 1b 22 13 bh 48 or JO bh (12h 2 48 kO Now JO 2 kO Ix Iy bh 48 13 bh 4 13 bh 48 b2 ) and JO A (12h 2 1 2 b2 ) bh 1 (12 h 2 24 b2 ) or 12h 2 b 2 24 ...
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