Question

0,6 in0.6 in 2. I - Beanm Determine the moment of inertia and the radius of gyration of the shaded area with respect to the x axis 0.3 in 0.4 iH 1.2 in 4-244 in- k 1.093 in. " 1.2 in 0.3 in

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Answer 1

Moment of inertia I_x = I_x1 + I_x2 + I_x3 I_x1 = 1/12 times 1.2 times 0.3^3 + 1.2 times 0.3 times (1.2 + 0.15)^2 = 0.6588 in^4 I_x2 = 1/12 times 0.4 times 2.4^3 + 2.4 times 2.4 times 0 = 0.4608 in^4 I_x3 = 1/12 times 2.4 times 0.3^3 + 2.4 times 0.3 times (1.2 + 0.15)^2 = 1.376 in^4 I_x = 0.6588 + 0.4608 + 1.3176 = 2.4372 in^4 I_x = 2.44 in^4 Radius of gyration R_x R_x = squareroot I_x/A = squareroot 2.44 /1.2 times 0.3 + 0.4 times 2.4 + 0.3 times 2.4 R_x = 1.0936in R_x = 1.093in