(a) Explanation Given information: The weight ( W ) of the sheet metal is 2.7 lb. The angular acceleration ( α ) of assembly is ( 150 rad/s 2 ) i . Calculation: Find the mass ( m ) of the sheet metal. m = W g Here, W is the weight and g is the acceleration due to gravity. Substitute 2.7 lb for W and 32.2 ft / s 2 for g . m = 2.7 lb 32.2 ft / s 2 = 0.08385 lb ⋅ s 2 ft Find the total area ( A ) of the sheet metal using the equation: A = 2 × area of rectangle + 2 × area of triangle = 2 × b 2 + 2 × 1 2 b 2 Substitute 6 in. for b . A = 2 × ( 6 in . × 1 ft 12 in . ) 2 + 2 × 1 2 ( 6 in . × 1 ft 12 in . ) 2 = 0.75 ft 2 Find the mass per unit area ( ρ ) of the sheet using the equation: ρ = m A Substitute 0.08385 lb ⋅ s 2 ft for m and 0.75 ft 2 for A . ρ = 0.08385 lb ⋅ s 2 ft 0.75 ft 2 = 0.1118 lb ⋅ s 2 ft 3 Sketch the dimension of the rectangular sheets in xy plane in Figure (1). Find the moment of inertia [ ( I x ) a r e a ] in the x direction ( I x ) a r e a = 1 3 b 4 + 1 3 b 4 = 2 3 b 4 Substitute 6 in. for b . ( I x ) a r e a = 2 3 ( 6 in . × 1 ft 12 in . ) 4 = 2 3 × 0.0625 = 0.04167 ft 4 Find the mass moment of inertia [ ( I x ) mass ] of the rectangular sheet in the x direction using the equation: ( I x ) mass = ρ ( I x ) a r e a Substitute 0.04167 ft 4 for ( I x ) a r e a and 0.1118 lb ⋅ s 2 ft 3 for m . ( I x ) mass = ( 0.1118 lb ⋅ s 2 ft 3 ) ( 0.04167 ft 4 ) = 4.658 × 10 − 3 lb ⋅ s 2 ⋅ ft Find the moment of inertia in the xy direction using the equation: ( I x y ) a r e a = ( b 2 ) ( 3 2 b ) ( 1 2 b ) + ( b 2 ) ( 5 2 b ) ( − 1 2 b ) = 3 4 b 4 − 5 4 b 4 = − 1 2 b 4 Substitute 6 in. for b . ( I x y ) a r e a = − 1 2 ( 6 in . × 1 ft 12 in . ) 4 = − 1 2 × 0.0625 = − 0.03125 ft 4 Find the mass moment of inertia of the rectangular sheet in the xy direction using the equation: I x y = ρ ( I x y ) a r e a Substitute − 0.03125 ft 4 for ( I x y ) a r e a and 0.1118 lb ⋅ s 2 ft 3 for m . I x y = ( 0.1118 lb ⋅ s 2 ft 3 ) ( − 0.03125 ft 4 ) = − 3.4938 × 10 − 3 lb ⋅ s 2 ⋅ ft Consider the small area of strip d A 1 and d A 1 . Sketch the diagram of the triangular sheets as shown in Figure (2). Find the moment of inertia [ ( I x 1 ) a r e a ] in the x direction: ( I x 1 ) a r e a = 1 12 b 4 + 1 12 b 4 = 1 6 b 4 Substitute 6 in. for b . ( I x 1 ) a r e a = 1 6 ( 6 in . × 1 ft 12 in . ) 4 = 0.01042 ft 4 Find the mass moment of inertia of the triangular sheet in the x direction using the equation: ( I x 1 ) mass = ρ ( I x 1 ) a r e a Substitute 0.01042 ft 4 for ( I x 1 ) a r e a and 0.1118 lb ⋅ s 2 ft 3 for m . ( I x 1 ) mass = ( 0.1118 lb ⋅ s 2 ft 3 ) ( 0.01042 ft 4 ) = 1.1646 × 10 − 3 lb ⋅ s 2 ⋅ ft The area taken into consideration for both the triangles is equal. The area d A 1 is equal to d A 2 . The area of the strip ( d A 1 ) is given by z d x and the value of z and x are equal. Find the moment of inertia ( I x z 1 ) in the xz direction for the triangular sheet using the equation: I x z 1 = ∫ x z 2 d A 1 + ∫ ( 4 b − x ) ( − z 2 ) d A 2 Here, I x z 1 is the moment of inertia in the xz direction or the triangular plate, d A 1 is the distance from the x axis is z , area of strip of first triangle, and d A 1 is the area of strip of the second triangle. Substitute d A 1 for d A 2 . I x z 1 = ∫ x z 2 d A 1 − ∫ 4 z 2 b d A 1 + ∫ x z 2 d A 1 = ∫ x z d A 1 − ∫ 2 z b d A 1 = − ∫ ( 2 b − x ) z d A 1 Substitute z d x for d A 1 . I x z 1 = − ∫ 0 b ( 2 b − x ) z ( z d x ) = − ∫ 0 b ( 2 b − x ) z 2 d x Substitute x for z . I x z 1 = − ∫ 0 b ( 2 b − x ) ( x ) 2 d x = − ∫ 0 b ( 2 b x 2 − x 3 ) d x (1) Integrate the equation (1) (b) To determine Find the dynamic reactions A and B immediately after the couple is applied.

12th Edition

ISBN: 9781264095032

Author: BEER

Publisher: MCGRAW-HILL HIGHER EDUCATION

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Chapter 18.2, Problem 18.73P

(a)

To determine

The couple M0.

(b)

To determine

Find the dynamic reactions A and B immediately after the couple is applied.

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Chapter 18.2, Problem 18.72P

Chapter 18.2, Problem 18.74P

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The couple M 0 .

Solution Summary: The author calculates the total area of the sheet metal using the equation: cm=2.7lb