Meriam, Kraige & Bolton — Distributed Forces: Centroids and Centers of Gravity: Problem 5_70

⚡ Mecademy AIENG정역학 · ch5  Problem Statement Calculate the volume of the complete ring of cross section shown. The cross section is symmetric about a horizontal centerline and has a total height of and a radial width of . The inner vertical surface of the ring is at a distance of from the axis of revolution (-axis). The outer corners of the cross section are removed with circular cutouts, each having a radius of . Problem 5/70 (a) Volume of the complete ring 1. Formula: The volume of a solid of revolution can be determined using the second theorem of Pappus-Guldinus: where is the area of a component part of the cross section and is the distance from the axis of revolution to the centroid of that component area. 2. Substitution: We decompose the cross-hatched area into a base rectangle and two quarter-circular corners to be subtracted. Base Rectangle: The rectangle has a width and height . Its inner edge is at . The centroidal radius is at the midpoint of the width: Quarter-circular Cutouts (Two identical): Each cutout has a radius . The cutouts are removed from the outer corners at . The distance from the V 4 in. 3 in.4 in. z 1.25 in. V V= 2 π A ∑rˉ ii A i rˉ i w=3 in.h=4 in. r=4 in. A = 1 w⋅h=3×4=12 in 2 =rˉ 1 4+ = 2 3 5.5 in. R= 1.25 in. A = 2 A = 3 − πR = 4 1 2 − π (1.25)= 4 1 2 −0.390625π≈−1.2272 in 2 r=7 in. corner to the centroid of a quarter circle is . Substituting these into the volume summation: 3. Calculation: Volume of the full rectangular ring: Volume subtracted