Problem 9.192 A section of sheet steel 2 mm y thick is cut and bent into the machine component shown. Knowing that the density of steel is 7850 kg/m3, determine the mass moment of inertia of the component with respect to (a) the x axis, (b) the y axis, (c) the z axis. x y z 120 mm 150 mm
Solving Problems on Your Own x y z 120 mm 150 mm Solving Problems on Your Own A section of sheet steel 2 mm thick is cut and bent into the machine component shown. Knowing that the density of steel is 7850 kg/m3, determine the mass moment of inertia of the component with respect to (a) the x axis, (b) the y axis, (c) the z axis. 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).
Solving Problems on Your Own A section of sheet steel 2 mm x y z 120 mm 150 mm Solving Problems on Your Own A section of sheet steel 2 mm thick is cut and bent into the machine component shown. Knowing that the density of steel is 7850 kg/m3, determine the mass moment of inertia of the component with respect to (a) the x axis, (b) the y axis, (c) the z axis. 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 + m d2 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.
Solving Problems on Your Own x y z 120 mm 150 mm Solving Problems on Your Own A section of sheet steel 2 mm thick is cut and bent into the machine component shown. Knowing that the density of steel is 7850 kg/m3, determine the mass moment of inertia of the component with respect to (a) the x axis, (b) the y axis, (c) the z axis. 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.
Divide the body into sections. Problem 9.192 Solution x y z 120 mm 150 mm Divide the body into sections. x y z x’ y’ z’ y’’ x’’ z’’ 2 1 3
m2 = r V2 = (7850 kg/m3)(0.002 m)(0.150 m)(0.120 m) = 0.2826 kg x y z 120 mm 150 mm Problem 9.192 Solution x y z x’ y’ z’ y’’ x’’ z’’ 2 1 3 Computation of Masses: Section 1: m1 = r V1 = (7850 kg/m3)(0.002 m)(0.300 m)2 = 1.413 kg Section 2: m2 = r V2 = (7850 kg/m3)(0.002 m)(0.150 m)(0.120 m) = 0.2826 kg Section 3: m3 = m2 = 0.2826 kg
x y z 120 mm 150 mm Problem 9.192 Solution x y z x’ y’ z’ y’’ x’’ z’’ 2 1 3 Compute the moment of inertia of each section. (a) Mass moment of inertia with respect to the x axis. Section 1: (Ix)1 = (1.413) (0.30)2 = 1.06 x 10-2 kg . m2 1 12 Section 2: (Ix)2 = (Ix’)2 + m d2 (Ix)2 = (0.2826) (0.120) 2 + (0.2826)(0.152 + 0.062) (Ix)2 = 7.71 x 10-3 kg . m2 1 12 Section 3: (Ix)3 = (Ix)2 = 7.71 x 10-3 kg . m2
x y z 120 mm 150 mm Problem 9.192 Solution x y z x’ y’ z’ y’’ x’’ z’’ 2 1 3 Compute the moment of inertia of the whole area. For the whole body: Ix = (Ix)1 + (Ix)2 + (Ix)3 Ix = 1.06 x 10-2 + 7.71 x 10-3 + 7.71 x 10-3 = 2.60 x 10-2 kg . m2 Ix = 26.0 x 10-3 kg . m2
x y z 120 mm 150 mm Problem 9.192 Solution x y z x’ y’ z’ y’’ x’’ z’’ 2 1 3 Compute the moment of inertia of each section. (b) Mass moment of inertia with respect to the y axis. Section 1: (Iy)1 = (1.413) (0.302 + 0.302) = 2.12 x 10-2 kg . m2 1 12 Section 2: (Ix)2 = (Ix’)2 + m d2 (Iy)2 = (0.2826) (0.150)2 + (0.2826)(0.152 + 0.0752) (Iy)2 = 8.48 x 10-3 kg . m2 1 12 Section 3: (Iy)3 = (Iy)2 = 8.48 x 10-3 kg . m2
x y z 120 mm 150 mm Problem 9.192 Solution x y z x’ y’ z’ y’’ x’’ z’’ 2 1 3 Compute the moment of inertia of the whole area. For the whole body: Iy = (Iy)1 + (Iy)2 + (Iy)3 Iy = 2.12 x 10-2 + 8.48 x 10-3 + 8.48 x 10-3 = 3.82 x 10-2 kg . m2 Iy = 38.2 x 10-3 kg . m2
x y z 120 mm 150 mm Problem 9.192 Solution x y z x’ y’ z’ y’’ x’’ z’’ 2 1 3 Compute the moment of inertia of each section. (c) Mass moment of inertia with respect to the z axis. Section 1: (Iz)1 = (1.413) (0.30)2 = 1.059 x 10-2 kg . m2 1 12 Section 2: (Iz)2 = (Iz’)2 + m d2 (Iz)2 = (0.2826) (0.152 + 0.122) + (0.2826)(0.0602 + 0.0752) (Iz)2 = 3.48 x 10-3 kg . m2 1 12 Section 3: (Iz)3 = (Iz)2 = 3.48 x 10-3 kg . m2
x y z 120 mm 150 mm Problem 9.192 Solution x y z x’ y’ z’ y’’ x’’ z’’ 2 1 3 Compute the moment of inertia of the whole area. For the whole body: Iz = (Iz)1 + (Iz)2 + (Iz)3 Iz = 1.06 x 10-2 + 3.48 x 10-3 + 3.48 x 10-3 = 1.755 x 10-2 kg . m2 Iz = 17.55 x 10-3 kg . m2