Explanation Given information: The sheet steel of thickness t . The specific weight is γ . Calculation: Sketch the sheet steel as shown in Figure 1. Find the direction cosine of OA ( λ O A ) as given below: λ O A = ( x A − x O ) i + ( y A − y O ) j + ( z A − z O ) k ( x A − x O ) 2 + ( y A − y O ) 2 + ( z A − z O ) 2 Here, x A is the x coordinate of point A , y A is the y coordinate of point A , z A is the z coordinate of point A , x O is the x coordinate of point O , y O is the y coordinate of point O and z O is the z coordinate of point O . Substitute a for x A , 2 a for y A , a for z A , 0 for x O , 0 for y O and 0 for z O . λ O A = ( a − 0 ) i + ( 2 a − 0 ) j + ( a − 0 ) k ( a − 0 ) 2 + ( 2 a − 0 ) 2 + ( a − 0 ) 2 = ( a ) i + ( 2 a ) j + ( a ) k ( a ) 2 + 4 ( a ) 2 + ( a ) 2 = a ( i + 2 j + k ) ( a ) 2 ( 1 + 4 + 1 ) = 1 6 ( i + 2 j + k ) Find the mass of section 1 as shown in below: m 1 = ρ V (1) Here, ρ is density of steel and V is volume of section 1. Substitute γ g for ρ and ( t × 2 a × 2 a ) for V in Equation (1). m 1 = γ g ( t × 2 a × 2 a ) = γ g ( t × 4 a 2 ) = 4 γ t a 2 g Here, g is acceleration due to gravity and t is thickness of sheet steel. Find the mass of section 2 as shown in below: m 2 = ρ V 2 (2) Here, V is volume of section 2. Substitute γ g for ρ and ( t × π 2 a 2 ) for V in Equation (2). m 2 = ρ g ( t × π 2 a 2 ) = π γ t a 2 2 g Refer to Figure 9.28, “Mass moment of inertia” in the textbook. Find the moment of inertia about x axis as shown below: I x = ( I x ) 1 + ( I x ) 2 = ( 1 12 m 1 c 1 2 + m 1 d 1 2 ) + ( 1 4 m 2 c 2 2 + m 2 d 2 2 ) (3) Here, ( I x ) 1 is moment of inertia about x axis section 1 and ( I x ) 2 is moment of inertia about x axis section 2. Refer to Figure 1. Substitute 4 γ t a 2 g for m 1 , π γ t a 2 2 g for m 2 , ( ( 2 a ) 2 + ( 2 a ) 2 ) for c 1 , ( a 2 + a 2 ) for d 1 , ( a 2 ) for c 2 2 , and ( 4 a 3 π ) 2 + ( a ) 2 + ( 2 a ) 2 for d 2 2 in Equation (3). I x = { 1 12 ( 4 γ t g a 2 ) [ ( ( 2 a ) 2 + ( 2 a ) 2 ) ] + 4 γ t g a 2 ( a 2 + a 2 ) + 1 4 ( π 2 γ t g a 2 ) a 2 + π 2 γ t a 2 g [ ( ( 4 a 3 π ) 2 + ( a ) 2 + ( 2 a ) 2 ) ] } = 4 γ t 3 g a 4 + 8 γ t g a 4 + π γ t 8 g a 4 + π 2 γ t a 2 g ( 5.1801 a 2 ) = γ t g a 4 [ 8 3 + 8 + 0.3927 + 8.1369 ] = 19.1962 γ t g a 4 Find the moment of inertia about y axis as shown below: I y = ( I y ) 1 + ( I y ) 2 = ( 1 12 m 1 c 1 2 + m 1 d 1 2 ) + ( 1 12 m 2 c 2 2 + m 2 d 2 2 ) (4) Here, ( I y ) 1 is moment of inertia about y axis section 1 and ( I y ) 2 is moment of inertia about x axis section 2. Refer to Figure 1. Substitute 4 γ t a 2 g for m 1 , π γ t a 2 2 g for m 2 , ( 2 a ) 2 for c 1 , a 2 for d 1 , ( 1 2 − 16 9 π 2 ) a 2 for c 2 2 , and [ 4 a 3 π + a 2 ] for d 2 2 in Equation (4). I y = [ 1 12 ( 4 γ t g a 2 ) ( ( 2 a ) 2 ) + 4 γ t g a 2 ( a 2 ) + π 2 γ t g a 2 ( 1 2 − 16 9 π 2 ) a 2 + π 2 γ t g a 2 ( ( 4 a 3 π ) + a 2 ) ] = 4 γ t g a 2 ( 1 3 + 1 ) a 2 + π 2 γ t g a 2 ( 1 2 − 16 9 π 2 + 16 9 π 2 + 1 ) a 2 = 4 γ t g a 2 ( 4 3 ) a 2 + π 2 γ t g a 2 ( 1 2 − 16 9 π 2 + 16 9 π 2 + 1 ) a 2 = 7.68953 γ t g a 4 Find the moment of inertia about z axis as shown below: I z = ( I z ) 1 + ( I z ) 2 = ( 1 12 m 1 c 1 2 + m 1 d 1 2 ) + ( 1 12 m 2 c 2 2 + m 2 d 2 2 ) (5) Here, ( I z ) 1 is moment of inertia about z axis section 1 and ( I z ) 2 is moment of inertia about z axis section 2. Refer to Figure 1. Substitute 4 γ t a 2 g for m 1 , π γ t a 2 2 g for m 2 , ( 2 a ) 2 for c 1 , a 2 for d 1 , ( 1 4 − 16 9 π 2 ) a 2 for c 2 2 , and [ 4 a 3 π + a 2 ] for d 2 2 in Equation (5)

Vector Mechanics for Engineers: Statics and Dynamics

12th Edition

ISBN: 9781259638091

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 9.6, Problem 9.168P

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The mass moment of inertia of the component with respect to the joining the origin O and point A.

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Chapter 9.6, Problem 9.167P

Chapter 9.6, Problem 9.169P

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A 2-mm-thick piece of sheet steel is cut and bent into the machine component shown. Knowing that the density of steel is 7850 kg/m3 , determine the mass moment of inertia of the component with respect to each of the coordinate axes.

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Vector Mechanics for Engineers: Statics and Dynamics

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The mass moment of inertia of the component with respect to the joining the origin O and point A .

Solution Summary: The author calculates the mass moment of inertia of the component with respect to the joining the origin O and point A.

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The mass moment of inertia of the component with respect to the joining the origin O and point A.

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