Chapter 10, Problem 49P

Determine the moment of inertia and the radius of gyration of the shaded area shown with respect to the y axis.

Determine the moment of inertia of the area about: A. the x-axis B. the y-axis Hint: See Appendix A for the textbook for common integral solutions. 9 in. 3 in. y=9-x² X

Determine the moment of inertia and the radius of gyration of the shaded area with respect to the x-axis. Given: r = 79 mm. 125 mm 125 mm - 250 mm The moment of inertia is The radius of gyration is *106 mm4. mm.

Determine the moment of inertia and the radius of gyration of the shaded area shown with respect to the x axis.

Determine the moments of inertia of the shaded area about the x- and y-axes. Also determine the polar moment of inertia about point O. -0.57h -0.57h 0.59h h Answers: Ix = h4 Iy = h4 Io = h4

Do it neatly please

Determine the moment of inertia and the radius of gyration of the shaded area with respect to the x-axis. Given: a = 11.4 mm. 8 mm- 24 mm -24 mm 6 mm 24 mm 24mm The moment of inertia is The radius of gyration is 6 mm 10³ mm4. mm.

Deduce the moment of inertia (Ixx) about x of the semicircular section with radius R and wall thickness t. Do not write the formula directly.

Formulas Moments of Inertia x= [y²d ly = fx²dA Theorem of Parallel Axis Ixr = 1 + d² A * axis going through the centroid x' axis parallel to x going through the point of interest d minimal distance (perpendicular) between x and x' ly₁ = 15+d²A ỹ axis going through the centroid y' axis parallel to y going through the point of interest d minimal distance (perpendicular) between y and y' Composite Bodies 1=Σ 4 All the moments of inertia should be about the same axis. Radius of Gyration k=