Calculate the moment of inertia of the shaded area about the x-axis. 86 mm 86 mm 61 mm Answer: Ix - i (106) mm²
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- Determine the product of inertia with respect to the x- and y-axes for the quarter circular, thin ring (tR) by integration.Using integration, compute the polar moment of inertia about point O for the circular sector. Check your result with Table 9.2.The product of inertia of triangle (a) with respect to its centroid is Ixy=b2h2/72. What is Ixy for triangles (b)-(d)? (Hint: Investigate the signs in the expression Ixy=IxyAxy.)
- The moments of inertia of the plane region about the x- and u-axes are Ix=0.4ft4 and Iu=0.6ft4, respectively. Determine y (the y-coordinate of the centroid C) and Ix (the moment of inertia about the centroidal x-axis).The moment of inertia of the plane region about the x-axis and the centroidal x-axis are Ix=0.35ft4 and Ix=0.08in.4, respectively. Determine the coordinate y of the centroid and the moment of inertia of the region about the u-axis.Using Ix and Iu from Table 9.2, determine the moment of inertia of the circular sector about the OB-axis. Check your result for =45 with that given for a quarter circle in Table 9.2.
- Find the moment of inertia of the cross-sectional area about the x-axis and the y-axis, given: L1 = 18 in, L2 = 6 in, L3 = 0.5 inDetermine the moment of inertia about the x-axis of the cross-sectional area below (a = 24 in.).Given the shaded area given in the figure x and y moments of inertia (Ix and Iy) with respect to their axes calculate. r(mm)=198 , h(mm)=264
- Find moment of inertia of the cross-sectional area about the X-axis, Y-axis, and X'-axis, given: L1 = 60 mm, L2 = 20 mm, L3 = 60 mm, L4 = 90 mmDetermine the moment of inertia of the shaded area about the x-axis in two different ways.Determine the moments of inertia of the shaded area with respect to the x and y axes. use a horizontal differential element of thickness for both calculations.