Question

The shaded area is equal to 5000 mm^2. Determine its centroidal moments of inertia Ix and Iy, knowing that 2Ix =Iy and that the polar moment of inertia of the area about point A is Ja=22.5x10^6 mm^4

ded area is equal to 5000 mm2. Determine its centroidal The sha of inertia I, and Iy, knowing that 2, T, and that the polar moments of inertia / and 1 , moment of inertia of the area about point A isJ. 60 mm Fig. P9.39 and P9.40

Answer 1

A = 5000 mm^2 J_A = 22.5 times 10^6 mm^4 = (I_x)_A + (I_x)_A = [I_x^- + A(60)^2] + [I_y^-] = I_x^- + I_y^- + 5000 times 60^2 I_y^- = 2I_x^- Rightarrow 22.5 times 10^6 = 3^I_x^- + 5000 times 60^2 Rightarrow I_x^- = 1.5 times 10^6 mm^4 I_y^- = 2 times 1.5 times 10^6 = 3 times 10^6 mm^4 Answers: I_x^- = 1.5 times 10^6 mm^4, I_y^- = 3 times 10^6 mm^4