Question

(a) Determine the moment of inertia Ix' of the cross-sectional area. (b)Determine the moment of inertia ly' of the cross-sectional area. The origin of coordinates is at the centroid C. 203 mm 605 mm 28mm 203 mm 28 mm 28 mm

Answer 1

Solution ( 0 .. 28 mm م ار Tel. 203 + 28 = 231 mm .28 mm → , 28thm toF - X 203+28 605 mm Lot 231m of Inerti Moment of Inertia 48 about Ix, Ix=Ix + I2x + I By Iox = ILCm. + Andis

Ai = bd = (28)(231) = 6468 mm do ad_14 - 231+ 14 Id - 23 imona = 134. 5 mm mm . .:12 mm 2 := iz iz Ilcm = bds! *ba 28 mm = 128.) (231)- 2876x jo?mm? I,x = 2.876x102mm* . + (6468}mm? (129.5 mm) -1.37 x 10.8 mm 4 - 0 Similarly, Igx = 1.37 x lommt - m tor, K 3605 mm + sem Iax = I ax cm =1605)(28)? 12 = 1.107x10 mont

Hence, total moment of inertia about centroidal x-axis, Ix = Ilx + Izy + 13x = 1.37x108 mm + 1.107 x 106 mm . + 1.37x108 mm 4 = 2.75 X 10.8 mm + b) About y axis, Ily = I1 cm + A, diy ... :201628mmb Il y cm = db3 di = 231 mm .. . 72 =(231)(28)3 he Tz = 4.22 x 10 mm 4 60s * 14 A, = 6468 mm² = 316.5 mm

Ily = 4.22 x10 mm 4 + (6468)(316.5)2 = 6.48 x 10 8 mm4 Igy = Ily due to symmetty = 6.48 x 108 mm 4 mm mm I 2y = I 2, cm = (28)(60593 iż = 5.17X10% mm 4 Hence, total moment of inertia about centoroidal yn axis, Iy = 6.48X108 mm 4 +5017 x 10 mm 4 +6.48 x 108 mm 1.813 x 10.9 mm

Area moment of inertia of the given cross section is taken about centroidal axes.