Question

2. CENTROID AND MOMENT OF INERTIA For the shape shown below, determine the following: (Make sure to label or describe the different segments.) a. Centroid (Xbar, Ybar) b. Moment of inertia about the x-axis (1x) C. The radius of the circle is 0.75 ft. NOTE: Use only the equations at the end of this test. (Hint: 4 segments) y 1 ft 1 ft Х 3 ft 3 ft

Answer 1

Givni 8= 0.75ft ny Ift jft K 3ft SK 3 Ft © costroid (obar, obe): © 45 Ft Tift kaft 27 2Ft + 111 IFEJ * Ft . Ist 23Ft 1635E *3rt * 3ft * 3FE It 3 ft y et = 0.327 -63264 5 Ift IFE Ift K GFt EGFE X Component Area A ft2 Albar, ift) Ybari ft *bara, f? "Bora, A, FEB = 3 2/3 3 (3)(2) २ 6 2 Osingle ③ rectengle- | (3) (2) = Semi Circle 61 62 = 1057 6 4.5 ㅗ 6 6.32 L 9.92 1.57 o circle -TIT2--1.767 6 L |-10.602 -1.767 EA = 8.803 Expar LA2 32.32 Eybart 7.80

how location of Centroid: XbarEA Exgar a Xbar A 32.32 Expar EA 8.803 bar 3.67 Ft and your Ź your EA 7.80 2.803 Ibar = 0.89 ft to required Centroid b«o, Ibar) = (3.67, 0.89) Ft now how moment of inertic about ze- oxing we know that intertia montent of Pringle In = 1/6 bhi 1 1 Ia' rectengle ☺ 6s » Ix'= 1 bh I ③ semicircle & Ix = doo D O circle for 4 / 4 Ix Ix'

So Ix®= I x + ad of Dingle - now moment of Inertia It's I (3)( 293 36 0.667 so Ald2 | = } Ft Ix = Igel 0.667 + (3) ( 21372 A=3ft2 = 0.667 + 11333 IX = 2 ft" of sectenste now moment of Inestia Ixia I (3) (2) 3 12 2 so 2= Ix' +Ad2 2 = 2+ (6) 8 Ft 4 Ix = now moment It's of Inostic of semicirale - ③ et mocy Ix' = 0.39 I 0.393+ (1.57) 02 IX - 1.963 FFY moment now of inertia of circle Ex'= tot (osº = 0.248 so Ix = Ix' +Ad2 = 0.25+ (10767)(02 Ix = 2.015 ffy

ro moment of Inertia of shaled moca: → Ix-off single - + sectengle + semicircle. Grcle o 2+8+1.963 - 2.015 Ix = 9.948 or In = gogs gogs Ft 4 Required moment of inertia