Question

50 mm 75 mm 75 mm 50 mm 25 mm 100 mm 25 mm 25 mm

Determine the distance Y to the centroid of the beam's cross sectional area. Then, determine the moment of inertia about the X' -axis. Put calculations in table form.

Answer 1

For solving this problem we'll assume the given component to be made up of three sub components, Namely 1,2 and 3.

These three sub components are in rectangular shape, for which we know the formulas of Moment of inertia about their centroidal axis. Then by using parrallel axis theorem , we will be able to final final Moment of Inertia.

The calculations are given in the following images ::

300 mm wwwsz S (0, 112.5) C(55) ܐܩܘ las's 8-) 100 mm wim כי 오 (5, (0,0) Kas 25 25 man Three Bodies of Let The Body Boody given be composed 1, a&3, howing their Centre of Area at cę & G. ('h' nel E G : 1-87.5, 50) cą (+87.5, so) 316) = (sh 5: Co 12-5) = (0,5) S..366 ARIT Arna tara is de = Anot Art Az (100 x 25) 1-87.5+ = panx15) (87-5)+(3^) (160 x 5) + (19 x 25) +(30*2 25) t =0mm y = And the Ye & Asya (1825) (50) + (100 x 50) (58)+(300+ 25) (12.5) AUT Az &A (100125) (300x25 3 +(loanes) y = 87.5 mm Wilm

Y 300 25 63 83-25 с -X' 18=315 23:5 137.5 900 87.5 (1) 2 50 7x 25 for Re Rectangular Sautian I bd3 12 Where xarib the Centroidond Axis from Parallel Axis Theorm! Tara Inst Acre x 7 a M.0.I. About Any Axis Parallel to centroidal Axis M.A.I. About Centroidal Axir Arra cross section Distance between The Two Parrallel Axises, :Ix = I section 2 t Isectione Isection a T -Section 2 About t' about y Abouti bic Ari będą 12 요 + 12 + Are 3 bzda 12 .2 Ag

in tt20t32 / 2 , m St8 11 R Wud ISF391 4 지 (30x259 (25) + عنه (3 کی 3) gg [oration Jource antial 2 jjtc)(oxs) fwxso) clarikso

Hopeyou got the concept well...!!!