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HW29 - PROBLEM 9.1 Determine by direct integration the...

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Unformatted text preview: PROBLEM 9.1 Determine by direct integration the moment of inertia of the shaded area with respect to the y axis. At x:a,y:b: b=ka% or k: Then 1y ziTijsdy or I, =—a3b4 PROBLEM 9.15 Determine the moment of inertia and the radius of gyration of the shaded b area shown with respect to the x axis. yI: b=ma or m:é a yz: b=ka3 or k=i3 a Then y1 22x or x1=gy b 3 a % y2=—3x 01' x2 = T y a b3 A Now (M = (x2 —xl)dy = [iifl —£y]dy b3 b b A‘ZICM_ZE 1y%_£y dy_2 i§y§_£ly2 5% b [54 b2 0 =fl—ab=lab 2 2 1 Then de = yZdA = y{ if —%y]abz 1;} IV b 7 E 3 ‘3‘ 4 Now [X = 2!de = 201% l -Lde : 2a[iY_ -L] 3 And k2 T PROBLEM 9.21 (l Determine the polar moment of inertia and the polar radius of gyration of the shaded area shown with respect to point P. IMAM HM»! SOLUTION y x dx. Have dlx = )1sz = y2[(x2 ’ hwy] Ix = 2[fay2(2a — a)dy + Ky: (2a — O)dy:l P 3:5 = 2“) ayzdy + fiZayzdy] 3 0 3 a = 2L +4ay— 3 —a 3 o 4 4 = 2— + 4"— : 2a“ 3 dly 2 xsz = x2012 —y])dx 1y = 2{j§x2(a # o)dx + f”x2[a — (—a)]dx} 3" 3 4 4 4 =2ai +4ax— =2i+32a —£=10a4 3 3 3 3 D a JP = II +1y = 2a“ +10a4 =12a4 JP=12a4< 12 4 127le a 2a1 A : (4a)(2a) — (2a)(a) = kP =1.414a< ...
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