Question

Determine the moment of inertia about the Neutral axis that is parallel to the y-axis for the beam cross section shown. у 160 mm 40 mm 200 mm 40 mm 40 mm 120 mm

Answer 1

Naly Date............ ,160mm 0.2 40mm 1 20, -(NAX a 1200mm Noc 1 D 7 - yomn 0 1 x A B + 120mm Area Area A = 40x120 = 4800 mm2 Az a 40x1202 4800 mm A32 40X 160=6400 mm? FGHI Nowy Distence Centroid of Rectongle, AB CJ af fectongle, DEIJ Area of Rectongle of Rectangle ABCJ from X-axis 7 = 20mm Distance of Centroid of Rectongle DEIJ. from X-axis Y Distance of Centrid of Rectangle FGHI from Xmexi's 100 mm 180 mm So, of 5. beom Goss-section from Distance of Centroid х- axi g. 7 All + A2 Y2 t A3 Y3 Aut Azt Az 4800 x 20 + 4800 X 100+ 6400 X 180 Y = 4800 + 4800 +6400 108 mm у

Date. Now, Distance Centroid H, = 60mm Distance of Controid H2 = a 20mm Distance Centroid Mz = 80mm of Rectongle DETS of Rectongle ABCJ from Y-aais s from Y- axis of Rectongle Ah HI from Yraxis So Distance of Centroid of beam Cross-section from Y-axis se = A, Hit A2 Hat Azrez A + A + Az 4800 X 60+ 4800x 20 + 6400x 80 4800+ 4806 + 6400 56 mm Now, h = 60-560 4 mm 12 moment of inentia of Rectongle AB CJ about (NA)y axis (I y = (Toidy + Al hi where, h=88,- t 1 (40) (1203° + 4800x (4) 2 (Ily Moment of Inerttia of Rectongle DEIJ about (NA)y axis (102) y + ₂ h where, h₂a H₂ - 5 1(120) (40)? + upoox (-36)2 583 6800 mmy 2 = 20-56 2-36 mm 12 Iz و(1) 6860 800 mmy

How, Moment of Inertia of Rectongle FG H I about (NA) - NA Jy axis (73) y = (Josly where, h32 Hz-re I (40) (160)² + 64008 (24) 2 + Azh = 80-56 24mm 2 12 17339733.33 mmy So The Moment of Inentia of beam Cross-Section about (NA)y arif is Iys (71 ly + In ly + (Izly Iy Iy 3037333.33 Hence, The moment of Inertia cof beam Cross-section the Neutral axis parallel 3037 333.33 mm 4 5836800 + 6860800 + 1733 9733.33 ти Ч to Y- axis is about VA