Explanation Given: Thickness of the anchor is 0.05 in and specific weight of galvanized steel is 470 lb/ft 3 . Concept used: Expression for the weight is given as follows: W = w × A × t .......(1) Here, weight of the part is W ,specific weight of galvanized steel is w , area of the part is A and thickness of the anchor is t . Mass moment of inertia of the anchor in x direction is expressed as follows: I x = ( I x ) 1 + ( I x ) 2 + ( I x ) 3 .......(2) Here. Mass moment of inertia about x direction is I x . Mass moment of inertia of the anchor in y direction is expressed as follows: I y = ( I y ) 1 + ( I y ) 2 + ( I y ) 3 .......(3) Here. Mass moment of inertia about y direction is I y . Mass moment of inertia of the anchor in z direction is expressed as follows: I z = ( I z ) 1 + ( I z ) 2 + ( I z ) 3 .......(4) Here. Mass moment of inertia about z direction is I z . Calculation: Divide the anchor into three subparts as follows: Weight for the section 1: Substitute 470 lb/ft 3 for w and 0.05 in for t in equation (1). W 1 = ( 470 lb/ft 3 ) ( 1 lb/in 3 ( 12 3 ) lb/ft 3 ) × 3.5 × 2.25 × 0.05 W 1 = 0.107096 lb Weight for the section 2: Substitute 470 lb/ft 3 for w and 0.05 in for t in equation (1). W 2 = ( 470 lb/ft 3 ) ( 1 lb/in 3 ( 12 3 ) lb/ft 3 ) × 1 × 2.25 × 0.05 W 2 = 0.030598 lb Weight for the section 3: Substitute 470 lb/ft 3 for w and 0.05 in for t in equation (1). W 3 = ( 470 lb/ft 3 ) ( 1 lb/in 3 ( 12 3 ) lb/ft 3 ) × 1 2 × 2 × 4.75 × 0.05 W 3 = 0.0645978 lb Mass moment of inertia of the section 1 in x direction is calculated as follows: ( I x ) 1 = 1 12 m l 2 + m y 2 ( I x ) 1 = 1 12 ( 0.107096 32.2 ) ( 3.5 ) 2 + ( 0.107096 32.2 ) ( 3.5 2 ) 2 ( I x ) 1 = 0.013581 lbs-in 2 Mass moment of inertia of the section 2 inx direction is calculated as follows: ( I x ) 2 = 1 12 m l 2 + m y 2 ( I x ) 2 = 1 12 ( 0.030598 32.2 ) ( 1 ) 2 + ( 0.030598 32.2 ) ( ( 1 2 ) 2 + 3.5 2 ) ( I x ) 2 = 0.01195729 lbs-in 2 Mass moment of inertia of the section 3inx direction is calculated as follows: ( I x ) 3 = 1 18 m l 2 + m y 2 ( I x ) 3 = 1 18 ( 0.0645978 32.2 ) ( 4.75 2 + 2 2 ) + ( 0.0645978 32.2 ) ( ( 2 3 × 4.75 ) 2 + ( 1 3 × 2 ) 2 ) ( I x ) 3 = 0.023969226 lbs-in 2 Substitute 0.013581 lbs-in 2 for ( I x ) 1 , 0.01195729 lbs-in 2 for ( I x ) 2 ,and 0.023969226 lbs-in 2 for ( I x ) 3 in equation (2). I x = 0.013581 + 0.01195729 + 0.023969226 I x = 0.049507516 lbs-in 2 Mass moment of inertia of the section 1 in y direction is calculated as follows: ( I y ) 1 = 1 12 m l 2 + m y 2 ( I y ) 1 = 1 12 ( 0

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ISBN: 9781259977237

Author: BEER

Publisher: MCG

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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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