Explanation Given information: The density of steel is 7850 kg / m 3 , the radius of the cylindrical section 1 is 80 mm ,the radius of the cylindrical section 2 is 20 mm , the radius of the hollow cylindrical section 3 is 20 mm , the distance between the axis P and Q is 80 mm , the height of the cylindrical section 1 is 40 mm , the height of the cylindrical section 2 is 60 mm , the height of the cylindrical section 3 is 40 mm , distance between x -axis and x ′ -axis is 20 mm , distance between x ″ -axis and x -axis is 30 mm . The following figure illustrate, Figure-(1) Write the expression of mass of the machine element section 1. m 1 = ρ steel ⋅ V 1 ...... (I) Here, density of steel is ρ steel , the volume of the section 1 is V 1 , the mass of the machine element section 1 is m 1 . Write the expression for volume of the section 1. V 1 = A 1 ⋅ H 1 ...... (II) Here, cross-sectional area of the section 1 is A 1 , the height of the section 1 is H 1 . Write the expression of the cross-sectional area of the cylindrical section 1. A 1 = π r 1 2 ...... (III) Here, the radius of the cylindrical section 1 is r 1 . Write the expression of mass of the machine element section 2. m 2 = ρ steel ⋅ V 2 ...... (IV) Here, the volume of the section 2 is V 2 , the mass of the machine element section 2 is m 2 . Write the expression of the cross-sectional area of the cylindrical section 2. Write the expression for volume of the section 2. V 2 = A 2 ⋅ H 2 ...... (V) Here, cross-sectional area of the section 2 is A 2 , the height of the section 2 is H 2 . Write the expression of the cross-sectional area of the cylindrical section 2. A 2 = π r 2 2 ...... (VI) Here, the radius of the cylindrical section 2 is r 2 . Write the expression for the mass of the machine element assembly of section 3. m 3 = ρ steel ⋅ V 3 ...... (VII) Here, the volume of the section 3 is V 3 , the mass of the machine element section 3 is m 3 . Write the expression for volume of the section 3. V 3 = A 3 ⋅ H 3 ...... (VIII) Here, cross-sectional area of the section 3 is A 3 , the height of the section 3 is H 3 . Write the expression of the cross-sectional area of the cylindrical section 3. A 3 = π r 3 2 ...... (IX) Here, the radius of the cylindrical section 3 is r 3 . Write the expression of the total mass moment of inertia about x -axis.for the machine element assembly. I x = ( I x ) 1 + ( I x ) 2 − ( I x ) 3 ...... (X) Here, the mass moment of inertia for the section 1 about x -axis is ( I x ) 1 , the mass moment of inertia for the section 2 about x -axis is ( I x ) 2 , the mass moment of inertia for the section 3 about x -axis is ( I x ) 3 . Write the expression of the mass moment of inertia for the section 1. ( I x ) 1 = ( ( m 1 12 ) ( 3 r 1 2 + H 1 2 ) ) + m 1 d 2 ...... (XI) Here, height of the cylindrical section 1 is H 1 , the distance between x -axis and x ′ -axis is d . Write the expression of the mass moment of inertia for the section 2. ( I x ) 2 = ( ( m 2 12 ) ( 3 r 2 2 + H 2 2 ) ) + m 2 d 2 2 ...... (XII) Here, height of the cylindrical section 1 is H 2 , distance between x ″ -axis and x -axis is d 2 . Write the expression of the mass moment of inertia for the section 3. ( I x ) 3 = ( ( m 3 12 ) ( 3 r 3 2 + H 3 2 ) ) + m 3 d 2 ...... (XIII) Here, height of the cylindrical section 1 is H 3 , the distance between x -axis and x ′ -axis is d . Calculation: Substitute 80 mm for r 1 in Equation (III). A 1 = π ( 80 mm ) 2 = π × 6400 mm 2 = 20106.19298 mm 2 ( 1 m 2 10 6 mm 2 ) = 0.021069298 m 2 Substitute 0.021069298 m 2 for A 1 and 40 mm for H 1 in Equation (II). V 1 = 0.021069298 m 2 × 40 mm = 0.021069298 m 2 × ( 40 mm ) ( 1 m 1000 mm ) = 0.00080424 m 3 Substitute 0.00080424 m 3 for V 1 and 7850 kg / m 3 for ρ steel in Equation (I). m 1 = ( 7850 kg / m 3 ) ( 0.00080424 m 3 ) = 6.313344 kg ⋅ m 3 m 3 = 6.313344 kg Substitute 20 mm for r 2 in Equation (VI). A 2 = π ( 20 mm ) 2 = π × 400 mm 2 = 1256.6370 mm 2 ( 1 m 2 10 6 mm 2 ) = 0.00125663 m 2 Substitute 0.00125663 m 2 for A 2 and 60 mm for H 2 in Equation (V). V 2 = 0.00125663 m 2 × 60 mm = 0.00125663 m 2 ( 60 mm ) ( 1 m 1000 mm ) = 0.00007539822 m 3 Substitute 0.00007539822 m 3 for V 2 and 7850 kg / m 3 for ρ steel in Equation (IV). m 2 = ( 7850 kg / m 3 ) ( 0.00007539822 m 3 ) = 0.5918 kg ⋅ m 3 m 3 = 0.5918 kg Substitute 20 mm for r 3 in Equation (IX). A 3 = π ( 20 mm ) 2 = π × 400 mm 2 = 1256 To determine (b) The mass moment of inertia of the assembly with respect to y -axis. To determine (c) The mass moment of inertia of the assembly with respect to z -axis.

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.31P

To determine

(a)

The mass moment of inertia of the assembly with respect to x -axis.

To determine

(b)

The mass moment of inertia of the assembly with respect to y -axis.

To determine

(c)

The mass moment of inertia of the assembly with respect to z -axis.

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Chapter B, Problem B.30P

Chapter B, Problem B.32P

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