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HW14 - PROBLEM 5.35 Dctcnnine by direct integration the...

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Unformatted text preview: PROBLEM 5.35 Dctcnnine by direct integration the centroid of the area shown. {I SOLUTION First note that symmetry implies E = 0 4 For the element (EL) shown y = Roosfl, x = Rsinfi dx = R cos 6’ d 6 dA : ydx : R2 00526610 Hence X 1,2 onin20a1~ A = IdA = 2g R' cos HdH : 2R‘ — + j = , ‘R'(2asin2a) 2 4 0 2 R 1 2 ” ”EMA = 2K3C050(RZCOSZ 0516) = RTECOSZ QsimQ + 5sinH] 0 3 : £(cosga sina + Zsina) 3 R (coszasina + 2sin 0:) But KIA : Wad/1 so J7 = 3 7R2 (205 + sin 2a) 2 (c052 a + 2) 0r )7 = 7R sin a 7 _— 3 (251 + s1n2a) . 7 2 , _ ' 3 Altcmatlvely, y = _ R 5m (1 373”? a 3 2a + smza PROBLEM 5.48 " Determine the volume and the surface area of the solid obtained by rotating the area of Problem 5.4 about (a) the line y = 22 mm, (b) the : 15mm hne x=12 mm. 1 I l H J 4 6 mm 6 mm . Y 9 mm 12 mm SOLUTION From the solution to Problem 5.4: A = 399 m2, XW = 1.421 mm, 17area =12.42mm (Area) From the solution to Problem 5.23: L 2 77.233 mm, in“ = 1.441 mm, 17line = 12.72 mm (Line) Applying the theorems of Pappus-Guldinus, we have (a) Rotation about the line y = 22 mm: Area = 27!(22 — Ymfl = 27z[(22 , 12.72) mm](77.233 mm) = 4503 mm2 A = 4.50><103 mm2 4 Volume = 2n[22 — Ym)A = 27r[(22 — 12.42) mm](399 m2) = 24 016.97 mm} V : 24.0 x103 mm3 4 (b) Rotation about line x z 12 mm: Area = 2;;(12 — imp = 27r[(12 —1.441)mm](77.233 mm) = 5124.45 mm2 A = 5.12x103 mm2 4 Volume 2 27r[12 ~1.421)A = 27T[(12 —1.421)mm](399 m2) = 26 521.46 mm3 V = 2615x103 mm3< 90., PROBLEM 5.53 less "11$ A lS-mm-diameter hole is drilled in a piece of 20-mm-thick steel; the ‘ hole is then countersunk as shown. Determine the volume of steel removed during the countersinking process. SOLUTION The required volume can be generated by rotating the area shown about the y axis. Applying the second theorem of Pappus—Guldinus, we have Ll HEWH‘ ”F... 7.5 M M V = ZIDTA = 27r[[% + 7.5] mm] x [g X 5 mm x 5 mm] or V2720mm34 ...
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