Explanation Given information: The density of steel is 490 lb / ft 3 . Calculation: Sketch the machine element as shown in Figure 1. Find the mass of component 1 as shown below: m 1 = ρ V (1) Here, ρ is density of steel and V is the volume of rectangular component 1. Substitute 490 lb / ft 3 for ρ and ( 3.6 × 1.2 × 6 ) in 3 for V in Equation (1). m 1 = 490 × ( 3.6 × 1.2 × 6 ) = 490 × 25.92 = 15 × 10 − 3 ρ Find the mass of component 2 as shown below: m 2 = ρ ( π r 2 2 h ) (2) Here, ρ is density of steel and V is the volume of component 2. Substitute 1.8 in for r and 1.6 in for h in Equation (2). m 2 = ρ ( π 2 ( 1.8 in . × 1 ft 12 in . ) 2 × ( 1.6 in . × 1 ft 12 in . ) ) = ρ × π 2 × 0.0225 × 0.1333 = 4.7124 × 10 − 3 ρ Find the mass of component 3 as shown below: m 3 = ρ ( π r 2 h ) (3) Here, ρ is density of steel and V is the volume of component 3. Substitute 1.5 in for r and 1.2 in for h in Equation (3). m 3 = ρ ( π ( 1.5 in . × 1 ft 12 in . ) 2 × ( 1.2 in . × 1 ft 12 in . ) ) = ρ × π ( 0.015625 ) ( 0.1 ) = 4.9087 × 10 − 3 ρ Refer to Figure 9.28, “Mass moment of inertia of common geometric shapes” in the text book. Find the moment of inertia about x axis ( I x 1 ) using the relation as shown below: I y 1 = [ 1 12 ( m 1 ) ( b 2 + h 2 ) + m 1 ( d 2 ) ] (4) Substitute 15 × 10 − 3 ρ for m 1 , 3.6 in . for b , 6 in . for h , and 3 in . for d , in Equation (4). I y 1 = [ 1 12 × 15 × 10 − 3 ρ [ ( 3.6 in . ( 1 ft 12 in . ) ) 2 + ( 6 in . ( 1 ft 12 in . ) ) 2 ] + 15 × 10 − 3 ρ ( 3 in . ( 1 ft 12 in . ) ) ] = 1 12 × 15 × 10 − 3 ρ ( 0.34 ) + 9.375 × 10 − 4 = 1.3625 × 10 − 3 ρ Find the moment of inertia about y axis for section 2 as shown below: I y 2 = I ¯ y 2 ′ + m d 2 = m 12 ( 3 r 2 + L 2 ) + m d 2 (5) Substitute 1.8 in . for r , 1.6 in . for L , 5.2 in . for d , and 4.7124 × 10 − 3 ρ for m 2 in Equation (5). I y 2 = [ 4

Vector Mechanics for Engineers: Statics and Dynamics

11th Edition

ISBN: 9780073398242

Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek, Phillip J. Cornwell, Brian Self

Publisher: McGraw-Hill Education

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Moment of Inertia

Inertia can be termed as a resistance offered by a body to change its state. A rigid body initially at rest will apply resistance when acted by an external force that tends to change its state of rest or velocity, or a body initially at motion will provi…

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Chapter 9.5, Problem 9.144P

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Find the mass moment of inertia with respect to y axis.

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Chapter 9.5, Problem 9.143P

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Problem 09.086 - Orientation of the principal axes and the corresponding moments of inertia For the area indicated, determine the orientation of the principal axes at the origin and the corresponding values of the moments of inertia when b= 76 mm and h = 56 mm. b The value of mis The value of 0m2 is The value of Imax is The value of I min is 180 270 1.429 7292 × h x 106 mm4. 106 mm4

! Required information Problem 09.141 - Mass moment of inertia of machine elements NOTE: This is a multi-part question. The machine element shown is fabricated from steel, where h = 64 mm. The density of steel is 7850 kg/m³. 20 mm 40 mm 80 mm 40 mm -20 mm 40 mm Problem 09.141.b - Mass moment of inertia of machine elements Determine the mass moment of inertia of the assembly with respect to the y axis. The mass moment of inertia of the assembly with respect to the y axis is 22.9 ✪ × 10-³ kg.m². X

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Vector Mechanics for Engineers: Statics and Dynamics

Ch. 9.1 - 9.1 through 9.4 Determine by direct integration...Ch. 9.1 - 9.1 through 9.4 Determine by direct integration...Ch. 9.1 - Prob. 9.3P Ch. 9.1 - Prob. 9.4P Ch. 9.1 - 9.5 through 9.8 Determine by direct integration...Ch. 9.1 - 9.5 through 9.8 Determine by direct integration...Ch. 9.1 - Prob. 9.7P Ch. 9.1 - Prob. 9.8P Ch. 9.1 - 9.9 through 9.11 Determine by direct integration...Ch. 9.1 - Prob. 9.10P

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