Problem 9.194

1. 40 mm 50 mm 60 mm y 15 mm 15 mm 45 mm 38 mm z x 45 mm 45 mm Problem 9.194 Determine the moment of inertia and the radius of gyration of the steel machine element shown with respect to the x axis. (The density of steel is 7850 kg/m3.)

2. 40 mm Problem 9.194 50 mm Solving Problems on Your Own 60 mm y Determine the moment of inertia and the radius of gyration of the steel machine element shown with respect to the x axis. (The density of steel is 7850 kg/m3.) 15 mm 15 mm 45 mm 38 mm z x 45 mm 45 mm 1. Compute the mass moments of inertia of a composite body with respect to a given axis. 1a. Divide the body into sections. The sections should have a simple shape for which the centroid and moments of inertia can be easily determined (e.g. from Fig. 9.28 in the book).

3. 40 mm Problem 9.194 50 mm Solving Problems on Your Own 60 mm y Determine the moment of inertia and the radius of gyration of the steel machine element shown with respect to the x axis. (The density of steel is 7850 kg/m3.) 15 mm 15 mm 45 mm 38 mm z x 45 mm 45 mm 1b. Compute the mass moment of inertia of each section. The moment of inertia of a section with respect to the given axis is determined by using the parallel-axis theorem: I = I + md2 Where I is the moment of inertia of the section about its own centroidal axis, I is the moment of inertia of the section about the given axis, d is the distance between the two axes, and m is the section’s mass.

4. Problem 9.194 40 mm 50 mm Solving Problems on Your Own 60 mm y Determine the moment of inertia and the radius of gyration of the steel machine element shown with respect to the x axis. (The density of steel is 7850 kg/m3.) 15 mm 15 mm 45 mm 38 mm z x 45 mm 45 mm I m 1c. Compute the mass moment of inertia of the whole body. The moment of inertia of the whole body is determined by adding the moments of inertia of all the sections. 2. Compute the radius of gyration. The radius of gyration k of a body is defined by: k =

5. 40 mm Problem 9.194 Solution 50 mm 60 mm y 15 mm 15 mm 45 mm 38 mm z x 45 mm 45 mm y y y z z z x x x Divide the body into sections.

6. 40 mm Problem 9.194 Solution 50 mm 60 mm y 15 mm 15 mm 45 mm 38 mm z x 45 mm 45 mm y (Ix) = ( Ix’) + md2 (Ix) = (3.18 kg)[(0.15 m)2+(0.03 m)2] + (3.18 kg) (0.075 m)2 (Ix) = 2.408 x 10-2 kg . m2 1 12 z x Compute the mass moment of inertia of each section. Rectangular prism: m = rV V = (0.15 m)(0.09 m)(0.03 m) V = 4.05 x 10-4 m3 m = (7850 kg/m3)(4.05x10-4 m3) m = 3.18 kg x’ 75 mm

7. 40 mm 50 mm 60 mm y m = rV V = p (0.045 m)2 (0.04 m) V = 1.27 x 10-4 m3 m = (7850 kg/m3)(1.27x10-4 m3) m = 1.0 kg 1 2 15 mm 15 mm 45 mm 38 mm z x 45 mm 45 mm y 4.45 4r = 19.1 mm = 3p 3p x’’ 15 mm C (centroidal axis) x’’ x’ x’ x 130 mm 40 mm z x Problem 9.194 Solution Semicircular cylinder:

8. y 4.45 4r = 19.1 mm = 3p 3p x’’ 15 mm C (centroidal axis) x’’ x’ x’ x 130 mm 40 mm z x 1 12 Problem 9.194 Solution Ix’ = Ix’’ + md2 (1.0 kg) [3 (0.045 m)2 + (0.04 m)2] = Ix’’ + (1.0 kg)(0.0191 m)2 Ix’’ = 27.477 x 10-5 kg . m2 Ix = ( Ix’’) + md2 = 27.477 x 10-5 kg . m2 + (1.0 kg)[(0.13 m)2 + (0.015 m + 0.0191 m)2] Ix = 18.34 x 10-3 kg . m2

9. 40 mm 50 mm 60 mm y 15 mm 15 mm 45 mm 38 mm z x 45 mm 45 mm y 1 12 x’ z 60 mm x Problem 9.194 Solution Circular cylindrical hole: m = rV V = p (0.038 m)2 (0.03 m) V = 1.361 x 10-4 m3 m = (7850 kg/m3)(1.36x10-4 m3) m = 1.068 kg (Ix) = ( Ix’) + md2 (Ix) = (1.07 kg) [ 3 (0.038 m)2 + (0.03 m)2 ] + (1.07 kg) (0.06 m)2 (Ix) = 4.31 x 10-3 kg . m2

10. 40 mm 50 mm 60 mm y 15 mm 15 mm 45 mm 38 mm z x 45 mm 45 mm 3.81 x 10-2 kg . m2 k = = k = 0.1107 m I m 3.11 kg Problem 9.194 Solution Compute the mass moment of inertia of the whole body. For the entire body, adding the values obtained: Ix = 2.408 x 10-2 + 1.834 x 10-2 - 4.31 x 10-3 Ix = 3.81 x 10-2 kg . m2 Compute the radius of gyration. The mass of the entire body: m = 3.18 + 1.0 - 1.07 = 3.11 kg Radius of gyration: