Chapter 10, Problem 10P

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=

Determine the moment of inertia and the radius of gyration of the shaded area shown with respect to the y 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. -0.80h Answers: lx = i ly= i lo = i 0.67h -0.80h h4 h4 h4

2. Determine the mass moment of Inertia of the steel bracket about the z-axis which passes through the midline of the base. 3" 2" 6 -1" 1.5 1.5"

Determine the radii of gyration for the half ring with respect to shown in Fig. An axis through the origin of the xy-coordinate system and normal to the plane of the area. R2 R1

Determine the moments of inertia about the y-axis of the circular area without and with the central square hole. R Answers: 1.3R With hole +1.3R Without hole ly= i R4 R4

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 polar radius of gyration about point A for the shaded area shown. 65 mm Answer: KA= 65 mm 34 mmj mm

3. Determine the moment of inertia with respect to the x-x and y-y centroidal axes for the following figure: Objects are Cen tured E3.5" 2"