Question

ius of Gyration of an Area earning Goal: Part A- The Radii of Gyration for a Cubic Function To understand how to calculate the area, moment of inertia, and radius of gyration for various shapes The fgure shows an area bounded by the positive z axis, the vertical line L, and the function y (Figure 1) For L-3.4 ft, calculate the radius of gyration about z, k,, and the radius of gyration about y. ky for this area Express your answers, separated by commas, to three significant figures. View Available Hint(s) The radius of gyration of an area about an axis is a quantity that is often used for the design of columns in structural mechanics. Once the areas and moments of inertia are known, the radi of gyration are determined using the formulas where the moments of inertia are calculated using the formulas Figure ft, ft Submit Part B- The Radi of Gyration for a Non-Polynomial Function The fgure shows an area bounded by the positive axis, the vertical line z -b, and the function y Figure 2) For b 1.90 E, calculate the radius of gyration about r, k,, and the radius of gyration about y, ky, for this area Express your answers, separated by commas, to three significant figures View Available Hints)

Answer 1

Y = 1/L^2 x^3 x = (yL^2)^1/3 x = L^2/3 dot y^1/3 dA = (L - x) dy = (L - L^2/3 dot y^1/3) dy I_x = integral_A y^2 dA = integral^L_0 y^2(L - L^2/3 dot y^1/3) dy = [L dot y^3/3 - L^2/3 dot y^5/3/5/3]^L_0 I_x = L^4/3 - 3/5 (L^2/3) (2^5/3) I_x = 3.4^4/3 - 3/5 (3.4^2/3) (3.4^5/3) I_x = 34.114 ft^4 dA = y dx = 1/L^2 x^3 dx I_4 = integral_A x^2 dA = integral^L_0 x^2 (x^3/L^2) dx = 1/L^2 (x^6/6)^L_0 I_4 = L^4/6 I_y = 3.4^4/6 = 22.27 ft^4 A = integral^L_0 x^3/L^2 dx = 1/L^2 (x^4/4)^L_0 A = L^2/4 = 3.4^2/4 A = 2.89 ft^2 K_x = squareroot I_x/A = squareroot 34.114/2.89 = 3.435 ft K_x = 3.435 ft