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 metal 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 400 mm for h and 450 mm for b in Equation (2). A = 450 × 400 = 180 , 000 mm 2 ( 1 m 10 3 mm ) 2 = 0.18 m 2 Substitute 7 , 850 kg / m 3 for ρ ST , 2 mm for t , and 0.18 m 2 for A in Equation (1). m 1 = 7 , 850 ( 2 mm ( 1 m 10 3 mm ) × 0.18 ) = 7 , 850 × 0.002 × 0.18 = 2.2860 kg Find the mass of the component 2 as shown below: m 2 = ρ ST V = ρ ST ( t A ) (3) Here, ρ ST is density of steel, V is the volume of component 2, t is thickness of sheet steel, and A is the area of sheet section. Find the triangular area of sheet steel section as shown below. A = 1 2 b × h (4) Here, b is the width of the triangular section and h is the triangular height of the section. Substitute 180 mm for h and 450 mm for b in Equation (4). A = 1 2 × 450 × 180 = 40 , 0500 mm 2 ( 1 m 10 3 mm ) 2 = 0.0405 m 2 Substitute 7 , 850 kg / m 3 for ρ ST , 2 mm for t , and 0.0405 m 2 for A in Equation (3). m 2 = 7 , 850 ( 2 mm ( 1 m 10 3 mm ) × 0.0405 ) = 7 , 850 × 0.002 × 0.0405 = 0.63585 kg As the machine component 1 are symmetrical to the x y z plane. ( I x ′ y ′ ) 1 = 0 ( I y ′ z ′ ) 1 = 0 ( I z ′ x ′ ) 1 = 0 As the machine component 2 are symmetrical to the y z plane. ( I y ′ z ′ ) 2 = 0 ( I z ′ x ′ ) 2 = 0 Refer to sample problem 9.6 in the section, “9.3B Principal axis and principal moment of inertia” in the textbook. ( I ¯ x ′ y ′ ) 2 , area = − 1 72 b 2 2 h 2 2 Find the product of inertia ( I x ′ y ′ ) 2 , area as shown below: ( I x ′ y ′ ) 2 , area = ρ S T t ( I x ′ y ′ ) 2 , area (5) Substitute − 1 72 b 2 2 h 2 2 for ( I x ′ y ′ ) 2 , area in Equation (5). ( I x ′ y ′ ) 2 , area = ρ S T ( 2 ) ( − 1 72 b 2 2 h 2 2 ) = − 1 72 m 2 b 2 2 h 2 2 Find the centroid section as shown below: x ¯ 1 = y ¯ 1 = z ¯ 2 = 0 Find the centroid section ( x ¯ 2 ) as shown below: ( x ¯ 2 ) = ( − b + h 3 ) Substitute − 0

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ISBN: 8220106797068

Author: BEER

Publisher: YUZU

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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 9.6, Problem 9.153P

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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 9.152P

Chapter 9.6, Problem 9.154P

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Chapter 9 Solutions

EBK VECTOR MECHANICS FOR ENGINEERS: STA

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

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Chapter 9 Solutions