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**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

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**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 in the book).

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**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.

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**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.

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**Divide the body into sections.**

Problem 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

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**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 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 = kg Section 2: m2 = r V2 = (7850 kg/m3)(0.002 m)(0.150 m)(0.120 m) = kg Section 3: m3 = m2 = kg

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x y z 120 mm 150 mm Problem 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)( ) (Ix)2 = 7.71 x 10-3 kg . m2 1 12 Section 3: (Ix)3 = (Ix)2 = 7.71 x 10-3 kg . m2

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x y z 120 mm 150 mm Problem 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 = x x x 10-3 = 2.60 x 10-2 kg . m2 Ix = 26.0 x 10-3 kg . m2

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x y z 120 mm 150 mm Problem 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) ( ) = 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)( ) (Iy)2 = 8.48 x 10-3 kg . m2 1 12 Section 3: (Iy)3 = (Iy)2 = 8.48 x 10-3 kg . m2

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x y z 120 mm 150 mm Problem 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 x x 10-3 = 3.82 x 10-2 kg . m2 Iy = 38.2 x 10-3 kg . m2

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x y z 120 mm 150 mm Problem 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 = x 10-2 kg . m2 1 12 Section 2: (Iz)2 = (Iz’)2 + m d2 (Iz)2 = (0.2826) ( ) + (0.2826)( ) (Iz)2 = 3.48 x 10-3 kg . m2 1 12 Section 3: (Iz)3 = (Iz)2 = 3.48 x 10-3 kg . m2

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x y z 120 mm 150 mm Problem 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 x x 10-3 = x 10-2 kg . m2 Iz = x 10-3 kg . m2

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8.0 SECOND MOMENT OR MOMENT OF INERTIA OF AN AREA

8.0 SECOND MOMENT OR MOMENT OF INERTIA OF AN AREA

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