Explanation Given: The width of the thin plate is 400 mm . The total height of the thin plate is 800 mm . The mass per unit area of the thin plate is 10 kg / m 2 . Subdivide the segments of the thin plate into two parts as shown below in Figure 1. Label the hole as segment (2) and consider it as a negative part. Show the free body diagram of the thin plate as in Figure 1. From Figure 1, Compute the mass moment of inertia of the thin plate about the y- axis with respect to the centroid G . I y = ∑ ( I y ) G + m d 2 = 2 [ 1 12 m 1 h 2 + m 1 ( b / 2 ) 2 ] − 2 [ 1 4 m 2 r 2 + m 2 ( 2 r ) 2 ] (I) Here, the mass of the thin plate is m , the diameter of the hole is d , the moment of inertia about y -axis in reference to the G -axis is ( I y ) G , the mass of the segment 1 is m 1 , the mass of the segment 2 is m 2 , the width of the segment 1 is b , the radius of the segment 2 is r and the height of the segment 1 is h . Compute the mass of the segment (1). m 1 = ( 10 kg / m 2 ) b h (II) Here, the width of the thin plate is b and the height is h . Compute the mass of the segment (2). m 2 = ( 10 kg / m 2 ) π r 2 (III) Here, the radius of the hole is r . Conclusion: Substitute ( 200 mm + 200 mm ) for b and ( 200 mm + 200 mm ) for h in Equation (II). m 1 = ( 10 kg / m 2 ) ( 200 mm + 200 mm ) ( 200 mm + 200 mm ) = ( 10 kg / m 2 ) ( 400 mm ) ( 400 mm ) = ( 10 kg / m 2 ) [ ( 400 mm ) 1 m 1000 mm ] [ ( 400 mm ) 1 m 1000 mm ] = ( 10 kg / m 2 ) ( 0.4 m ) ( 0.4 m ) = 1.6 kg Substitute 100 mm for r in Equation (III). m 2 = ( 10 kg / m 2 ) π ( 100 mm ) 2 = ( 10 kg / m 2 ) π [ ( 100 mm ) 1 m 1000 mm ] 2 = ( 10 kg / m 2 ) π ( 0

Engineering Mechanics: Statics & Dynamics (14th Edition)

14th Edition

ISBN: 9780133915426

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 10.8, Problem 104P

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The mass moment of inertia of the thin plate about the y-axis.

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Chapter 10.8, Problem 103P

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