Question

Statics problem

Determine the mass moments of inertia and the radii of gyration of the steel machine element shown with respect to the x and y axes. The density of steel is 7850 kg/m3. 44 120 70 *120 70 44 40 20 20 Dimensions in mm The mass moment of inertia of the component with respect to x axis is The mass moment of inertia of the component with respect to y axis is The radius of gyration of the component with respect to x axis is The radius of gyration of the component with respect to y axis is |x10-3 kg.m? x10-3 kg.m2 mm. mm.

Answer 1

Saleticon f = 7850 kg/m3 6 f Component elements are given name ® & hallove section ay as 5 Slebo Calculation element maps of eac My = PV = 78504 (240 x 140 x 40017 g Maa 1or 5504 Kg m2 2 PV2 2 7850 al X(017) 3X0:04] me=m3 = 2:4-16 key Mong mg= ma = 78504 II (44X103) ? 20.04 mazm5 = 1.9097 Key Slebat mass moment of inentia with respect fo Xaris Ix: Itt + 3x2,+ Faz † 144 – Ins

Formula For disc I = MR? + 1 m (t)? For f thicknes 12 4 for plate of thicknees 't Ixa = ma² mut² '12 12 طه a 3 Igga mb2 ma 12 :12 Ix= 1 [10.550A x (0:042+0:142) +2 2.416 20:07 12 + Ž « 2:416 * (000432 + 1 (119097) x640442 + 1 kl.909740604? 14 12. 1.9099604 (119697980:044+ 1*1909.76 (0504? Ix = 0.0282606 kg-m2 or Ix- 2813 x 103 kg-m²

Some Similarly 15 Br mikaty in g-direction I = Jy + Fyz + F g 3 + ly a Fy But here (Ig) = (Iy) f (y) = (Tyls because have same shake Ty = 1 x 10155110624?+514) *)]+62-108 *&•°7) + (2.4 16831 02-12 4 aasta) After calculation Fy = 0.1836 kq-m2 Iy Ahme HIS 5-16 912 12 we get ori 18346x 103 Keg-ma 494/311 r To find Radias of guration, we need ta mals M= met mz+m3 +meM4 M z 15.39 kg knoce to fal

2 We noce Ix= 2 M Kri & Iy=Mky ² Kas Ire 0.0282605 TO 15:38966 M 42.86 mm Im=10 Kas 0104286 m = Ane mnm Similarly ky Iy M C18378 15138 cor Ky = 0.10929 m Tky 109.29 mm