Meriam, Kraige & Bolton — Distributed Forces: Centroids and Centers of Gravity: Problem 5_73

⚡ Mecademy AIENG정역학 · ch5  Problem Statement Calculate the volume of the rubber gasket formed by the complete ring of the semicircular cross section shown. Also compute the surface area of the outside of the ring. Problem 5/73 (a) Volume of the rubber gasket 1. Formula: According to the second theorem of Pappus-Guldinus, the volume of a body of revolution is . For a complete revolution, , so , where is the area of the generating shape and is the distance from the axis of revolution to its centroid. 2. Substitution: The generating area is a semicircle with radius : The distance from the diameter to the centroid of a semicircular area is The axis of revolution (-axis) is at a distance from the diameter. Thus, the radius of revolution for the centroid is 3. Calculation: 4. Result: ● Final Conclusion: The volume of the rubber gasket is . (b) Surface area of the outside of the ring V A V V V=θA rˉ g θ=2πV= 2π A rˉ Ag A g rˉ A a= in. 4 3 A = g πa= 2 1 2 π 2 1 ( 4 3 ) 2 =yˉ = 3π 4a = 3π 4(3/4) in. π 1 zR=1 in. 2 1 =rˉ A R+ =yˉ1.5+ in. π 1 V= 2 π1.5+ π ( π 1 )[ 2 1 ( 4 3 ) 2 ] A = g π = 2 1 ( 16 9 ) ≈ 32 9π 0.8836 in. 2 =rˉ A 1.5+ ≈ π 1 1.5+0.3183=1.8183 in. V=2π⋅1.8183⋅0.8836=10.095 in. 3 V≈10.10 in. 3 10.10 in. 3 A 1. Formula: According to the first theorem of Pappus-Guldinus, the area of a surface of revolution is . For a complete revolution, , where is the length of the generating curve and is the distance from the axis of revolution to its centroid. 2. Substitution: The generat