Question

Consider the area shown in Figure 4. Determine; a) The 2nd Moment of Area (Ix and ly) about the axis system shown. b) The Polar Moment of Inertia (Jo) about point O. c) The 2nd Moment of Area (lx and ly) about an axis system that runs through the centroid of the area and the Polar Moment of Inertia (Jo) about the centroid of the area. [5+3+5 = 13 marks] 100 mm-100 mm 150 mm 150 mm 150 mm 75 mm x

Answer 1

Page. No:3 a + A2Y+(TCGHz 2nd moment of Area (Ix and Ty) about the axis :- Ix = TAL -2 + Aziz ICGX2 2 --- 450X106 +60 000X1502) – (7910156-25TT +56257TX 1502 + (112.5x106+22500 x 100) 1800 x 106 – 422.4583X106 + 337.5X106 Ix - 1715.0417 x 100 mm 4 Ij = (frey,+ apzi?) – (1657+ Ay mo)+(Tayt A585) =(200x100 + 60000X1002) - (7910156-25TT +5625 TTX 1002 + (23-125X106 + 22 500 X2502) 800 X 106 — 201:565075 X 106 + 1434.375X106 > Ty 2032.80 99 x 106 mm 4

Page No.5 ICGx2 + A2 (% -0)27 2 Ix = [Tax, + + A (1 912] - [ + TcGx3 + A3 (T-93) = [450x106+60 000 X (150–132.6465)2] + [ 7910 156-25 IT+ 5625 Trx(150–132.6465)2] + [ 112.5X106+22500 ( 132.6465–100)2 ] = 468.068638 x 106 – 30-172142x106 + 136.480364 X106 ²) IX = Ix' = 574.37686x100mm 4 Tyı = [, + A1 (> --Xyle) - (Icake + A2 (x-7) 2] + [ Trayz + Az (X5 -3)?] = 200x100+60000X (152.0604–100)? [7910156.25TT +5625IT (152.9604–100)2] + [28:125X106 +22500 X (250–152.0604)2 = 362.6171 X106 – 72.745183X106+ 243.94 8718x106 Iy1 = 533.820635x100 mmu

Page No - 6 Polar Moment of Inertia (Jo') about the centroid of the grey Jol = Ix' + Iyo 574,37686X106 + 533.820635X106 Jo' = 1108.197495 x 100 mm 4