Explanation Given information: The thickness ( t ) of sheet steel is 2 mm . The density ( ρ ST ) of steel is 7 , 850 kg / m 3 . Calculation: Sketch the sheet steel as shown in Figure 1. Find the mass of the component 1 as shown below: m 1 = ρ ST V = ρ ST ( t A ) (1) Here, ρ ST is density of steel, V is the volume of component 1, t is thickness of sheet steel, and A is the area of sheet section. Find the rectangular area of sheet steel section as shown in below. A = b × h (2) Here, b is the width of the section and h is the height of the section. Substitute 350 mm for b and 390 mm for h in Equation (2). A = 350 × 390 = 136 , 500 mm 2 ( 1 m 10 3 mm ) 2 = 0.1365 m 2 Substitute 7 , 850 kg / m 3 for ρ ST , 2 mm for t , and 0.1365 m 2 for A in Equation (1). m 1 = 7 , 850 ( 2 mm ( 1 m 10 3 mm ) × 0.1365 ) = 7 , 850 × 0.002 × 0.1365 = 2.14305 kg Find the mass of the component 2 as shown below: m 2 = ρ ST V = ρ ST ( t A ) = ρ ST ( t ( π 2 r 2 ) ) (3) Here, V is the volume of component 2, t is thickness of sheet steel, and A is the area of sheet section. Substitute 2 mm for t 7 , 850 kg / m 3 for ρ ST , and 195 mm for A in Equation (3). m 2 = 7 , 850 ( 2 mm ( 1 m 10 3 mm ) ) ( π 2 × 195 mm ( 1 m 10 3 mm ) 2 ) = 7 , 850 × 0.002 × 0.195 2 = 0.93775 kg Find the mass of the component 3 as shown below: m 3 = ρ ST V = ρ ST ( t A ) = ρ ST ( t ( 1 2 b h ) ) (4) Here, V is the volume of component 3. Substitute 2 mm for t 7 , 850 kg / m 3 for ρ ST , and 390 mm for b and 150 mm for h in Equation (4). m 3 = 7 , 850 ( 2 mm ( 1 m 10 3 mm ) × 0.1365 ) ( 1 2 × 390 mm ( 1 m 10 3 mm ) × 150 mm ( 1 m 10 3 mm ) ) = 7 , 850 × 0.002 × ( 0.02925 ) = 0.459225 kg As the symmetry the centroid product of inertia of component 1 and 2 are zero. ( I ¯ x ′ y ′ ) 3 = 0 ( I ¯ z ′ x ′ ) 3 = 0 ( I ¯ y ′ z ′ ) 3 , mass = ρ S T t ( I y ′ z ′ ) 3 , area Refer to sample problem 9.6 in section, “9.3B Principal axis and principal moment of inertia” in the textbook. The orientation of the axes corresponds to 90 ° rotation. ( I ¯ x ′ y ′ ) 2 , area = − 1 72 b 3 2 h 3 2 Find the product of inertia ( I y ′ z ′ ) 3 as shown below: ( I y ′ z ′ ) 3 = ρ S T t ( I x ′ y ′ ) 2 , area (5) Substitute 2 mm for t and − 1 72 b 3 2 h 3 2 for ( I x ′ y ′ ) 2 , area in Equation (5). ( I y ′ z ′ ) 3 = ρ S T ( 2 ) ( 1 72 b 3 2 h 3 2 ) = 1 36 m 3 b 3 h 3 Find the centroid section ( y ¯ 2 ) as shown below: ( y ¯ 2 ) = 4 r 3 π Substitute 195 mm for r. ( y ¯ 2 ) = 4 ( 195 mm × 1 m 10 3 mm ) 3 π = 4 × 0.195 3 π = 0.08276 m Find the mass products of inertia I x y of the machine component as shown below: I x y = ∑ ( I x ′ y ′ + m x ¯ y ¯ ) = ( I x ′ y ′ + m x ¯ y ¯ ) 1 + ( I x ′ y ′ + m x ¯ y ¯ ) 2 + ( I x ′ y ′ + m x ¯ y ¯ ) 3 (6) Substitute 0 for ( I x ′ y ′ ) 1 ,0 for ( I x ′ y ′ ) 2 , 0 for ( I x ′ y ′ ) 3 , 2

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics

11th Edition

ISBN: 9781259639272

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

Publisher: McGraw-Hill Education

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Chapter 9.6, Problem 9.155P

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Find the mass products of inertia Ixy,Iyz, and Izx of the machine component.

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Chapter 9.6, Problem 9.154P

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Connect 1 Semester Access Card for 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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Find the mass products of inertia I x y , I y z , and I z x of the machine component.

Solution Summary: The author calculates the mass of the component 1 as shown in Figure 1.

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