Explanation Given information: Draw the diagram for the component of the body. Figure-(1) Concept used: Write the expression for the mass of the component 1. m 1 = ρ V 1 …… (I) Here, the density of the machine element is ρ and the volume of the component 1 is V 1 . Write the expression for the mass of the component 2. m 2 = ρ V 2 …… (II) Here, the volume of the component 2 is V 2 . Write the expression for the mass of the component 3. m 3 = ρ V 3 …… (III) Here, the volume of the component 3 is V 3 . Write the expression for the volume of the component 1. V 1 = π r 1 2 h 1 …… (IV) Here, the radius of the component 1 is r 1 and the height of the component 1 is h 1 . Write the expression for the volume of the component 2. V 2 = π r 2 2 h 2 …… (V) Here, the radius of the component 2 is r 2 and the height of the component 2 is h 2 . Write the expression for the volume of the component 3. V 3 = π r 3 2 h 3 …… (VI) Here, the radius of the component 3 is r 3 and the height of the component 3 is h 3 . The polar moment of inertia about y and z axis and the polar moment of inertia about z and x axis is zero. Write the expression for the polar moment of inertia about x and y axis. I x y = m 2 x ¯ 2 y ¯ 2 − m 3 x ¯ 3 y ¯ 3 …… (VII) Here, the x coordinate of the component 2 is x ¯ 2 , the y coordinate of the component 2 is y ¯ 2 , the x coordinate of the component 3 is x ¯ 3 , the y coordinate of the component 3 is y ¯ 3 . Write the expression for the moment of inertia for the component 1 about x -axis. ( I x ) 1 = m 1 [ 1 12 ( 3 ( r 1 2 + h 1 2 ) + ( t 1 2 ) 2 ) ] …… (VIII) Here, the thickness of the component 1 is t 1 . Write the expression for the moment of inertia for the component 2 about x -axis. ( I x ) 2 = m 2 [ 1 12 ( 3 ( r 2 2 + h 2 2 ) + ( t 2 2 ) 2 ) ] …… (IX) Here, the thickness of the component 2 is t 2 . Write the expression for the moment of inertia for the component 3 about x -axis. ( I x ) 3 = m 3 [ 1 12 ( 3 ( r 3 2 + h 3 2 ) + ( t 3 2 ) 2 ) ] …… (X) Here, the thickness of the component 1 is t 1 . Write the expression for the moment of inertia about x -axis of the machine element. I x = ( I x ) 1 + ( I x ) 2 − ( I x ) 3 …… (XI) Write the expression for the moment of inertia for the component 1 about y -axis. ( I y ) 1 = m 1 [ 1 2 r 1 2 ] …… (XII) Write the expression for the moment of inertia for the component 2 about y -axis. ( I y ) 2 = m 2 [ 1 2 ( r 2 2 + t 2 2 ) ] …… (XIII) Write the expression for the moment of inertia for the component 3 about y -axis. ( I y ) 3 = m 3 [ 1 2 ( ( r 3 2 + h 3 2 ) ) ] …… (XIV) Write the expression for the moment of inertia about y -axis of the machine element. I y = ( I y ) 1 + ( I y ) 2 − ( I y ) 3 …… (XV) Write the expression for the mass moment of the inertia of the machine element with respect to the line joining the origin O and the point A . I O A = I x λ x 2 + I y λ y 2 + I z λ z 2 − 2 I x y λ x λ z − 2 I y z λ y λ z − 2 I z x λ x λ z …… (XVI) Substitute 0 for λ z , 0 for I y z and 0 for I z x in Equation (XVI). I O A = I x λ x 2 + I y λ y 2 + I z ( 0 ) − 2 I x y λ x λ y − 2 ( 0 ) λ y ( 0 ) − 2 ( 0 ) λ x ( 0 ) = I x λ x 2 + I y λ y 2 − 2 I x y λ x λ y …… (XVII) Write the expression for the position vector of O A . O A → = 2 i − 3 j Here, the radius of the circular cone is a . Write the expression for the magnitude of the position vector O A → . d O A = 13 Here, the height of the cone is h . Write the expression for the unit vector. λ O A = O A → d O A …… (XVIII) Calculation: Substitute 13 for d O A and 2 i − 3 j for O A → in Equation (XVIII). λ O A = 2 i − 3 j 13 = 2 i 13 − 3 j 13 …… (XIX) From Equation (XIX) the unit vector along the different axis. λ x = 2 13 λ y = 2 13 Substitute 80 mm for r 1 and 40 mm for h 1 in Equation (IV). V 1 = π ( 80 mm ) 2 ( 40 mm ) = π ( 80 mm ( 1 m 10 3 mm ) ) 2 ( 40 mm ( 1 m 10 3 mm ) ) = π ( 0.08 m ) 2 ( 0.04 m ) = 8.042 × 10 − 4 m 3 Substitute 20 mm for r 2 and 60 mm for h 2 in Equation (V). V 2 = π ( 20 mm ) 2 ( 60 mm ) = π ( 20 mm ( 1 m 10 3 mm ) ) 2 ( 60 mm ( 1 m 10 3 mm ) ) = π ( 0.02 m ) 2 ( 0.06 m ) = 7.53 × 10 − 5 m 3 Substitute 20 mm for r 3 and 40 mm for h 3 in Equation (VI). V 3 = π ( 20 mm ) 2 ( 40 mm ) = π ( 20 mm ( 1 m 10 3 mm ) ) 2 ( 40 mm ( 1 m 10 3 mm ) ) = π ( 0.02 m ) 2 ( 0.04 m ) = 5.02 × 10 − 5 m 3 Substitute 7850 kg / m 3 for ρ and 8.042 × 10 − 4 m 3 for V 1 in Equation (I). m 1 = ( 7850 kg / m 3 ) ( 8.042 × 10 − 4 m 3 ) = 6.31 kg ⋅ m 3 / m 3 = 6.31 kg Substitute 7850 kg / m 3 for ρ and 7.53 × 10 − 5 m 3 for V 2 in Equation (II). m 2 = ( 7850 kg / m 3 ) ( 7.53 × 10 − 5 m 3 ) = 0.5911 kg ⋅ m 3 / m 3 ≈ 0.592 kg Substitute 7850 kg / m 3 for ρ and 5.02 × 10 − 5 m 3 for V 3 in Equation (III). m 3 = ( 7850 kg / m 3 ) ( 5.02 × 10 − 5 m 3 ) = 0.39407 kg ⋅ m 3 / m 3 ≈ 0.395 kg Substitute 0.592 kg for m 2 , 0.395 kg for m 3 , 40 mm for x ¯ 2 , 30 mm for y ¯ 2 , 40 mm for x ¯ 3 and 20 mm for y ¯ 3 in Equation (VII). I x y = ( 0.592 kg ) ( 40 mm ) ( 30 mm ) − ( 0.395 kg ) ( 40 mm ) ( 20 mm ) = [ ( 0.592 kg ) ( 40 mm ( 1 m 10 3 mm ) ) ( 30 mm ( 1 m 10 3 mm ) ) − ( 0.395 kg ) ( 40 mm ( 1 m 10 3 mm ) ) ( 20 mm ( 1 m 10 3 mm ) ) ] = ( 0.592 kg ) ( 0.04 m ) ( 0.03 m ) − ( 0.395 kg ) ( 0 To determine (a) The principal axis about the origin.

12th Edition

ISBN: 9781259977305

Author: BEER, Ferdinand P. (ferdinand Pierre), Johnston, E. Russell (elwood Russell), Cornwell, Phillip J., SELF, Brian P.

Publisher: Mcgraw-hill Education,

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Chapter B, Problem B.70P

To determine

(a)

The principal mass moment of inertias at the origin.

To determine

(a)

The principal axis about the origin.

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Chapter B, Problem B.69P

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(a) The principal mass moment of inertias at the origin.

Solution Summary: The author illustrates the principal mass moment of inertias at the origin and the expression for the mass of the component.