To determine Find the moment of inertia and radius of gyration of the combined section with respect to centroidal x axis ( I ¯ x ) and ( k ¯ x ) , and y axis ( I ¯ y ) and ( k ¯ y ) . Explanation Given information: The distance between the channels ( b ) is 200 mm. Calculation: Refer Figure 9.138, “Properties of rolled steel shapes (SI units)” from the textbook. For section C 250 × 22.8 , The area of the channel ( A C ) is 2890 mm 2 . The moment of inertia about x axis ( I x ) is 28 × 10 6 mm 4 . The moment of inertia about y axis ( I y ) is 0.945 × 10 6 mm 4 . The depth of the channel ( D ) is 254 mm. The centroid of the section about YY axis ( x ¯ ) is 16.1 mm. Show the centroidal location of the two channels and two plates as Figure 1. Calculate the area of the plate ( A P ) using the relation: A P = ( 10 × 375 ) = 2 × ( 3750 mm 2 ) = 7500 mm 2 Here, 2 represents the two plates. Calculate the total area of the given section ( A ) using the relation: Assume two equal parts due to symmetry. A = 2 A C + A P Here, 2 represents the two channels in the section. Substitute 2890 mm 2 for A C and 7500 mm 2 for A P . A = 2 ( 2890 ) + 7500 = 13.28 × 10 3 mm 2 Consider x axis. Calculate the moment of inertia of the plate ( I P ) using the formula: I P = 375 × 10 3 12 = 31250 mm 4 Calculate the centroid of top plate from the centroid of entire section ( d ) using the relation: d = ( Centroid of the entire section ) + ( centroid of the top plate ) = D 2 + 10 2 Substitute 254 mm for D . d = 254 2 + 10 2 = 132 mm Calculate the moment of inertia ( I ¯ x ) with respect to x axis using the formula: I ¯ x = 2 I x + 2 ( I P + A P d 2 ) Substitute 31250 mm 4 for I P , 28 × 10 6 mm 4 for I x , 3750 mm 2 for A P and 132 mm for d . I ¯ x = 2 ( 28 × 10 6 mm 4 ) + 2 [ 31250 mm 4 + ( 3750 × 132 2 ) ] = 56 × 10 6 mm 4 + 130.743 × 10 6 mm 4 = 186.7 × 10 6 mm 4 Calculate the radius of gyration about x axis ( k ¯ x ) using the formula: k ¯ x = I ¯ x A Substitute 186.7 × 10 6 mm 4 for I ¯ x and 13.28 × 10 3 mm 2 for A . k ¯ x = 186.7 × 10 6 mm 4 13.28 × 10 3 mm 2 = 118.6 mm Consider y axis

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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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Chapter 9.2, Problem 9.49P

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Find the moment of inertia and radius of gyration of the combined section with respect to centroidal x axis (I¯x) and (k¯x), and y axis (I¯y) and (k¯y).

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Chapter 9.2, Problem 9.48P

Chapter 9.2, Problem 9.50P

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