HW14 - L1 PROBLEM 5.35 Determine by direct integration the...

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Unformatted text preview: L1 PROBLEM 5.35 Determine by direct integration the centroid of the area shown. SOLUTION First note that symmetry implies f : 0 4 For the element (EL) shown y : R0039, x : Rsinfl dx : Recs 6 d (9 dA = ydx : R200529d6 Hence a = éRZ (211 Sin 2a) A: jdAzzfichoszfidfilez 9+5m26 2 4 0 a Jig/fl?! = 2f§cosfl(R2 0052 8516') = Riécoszasing + gsinfi) 0 R3 = ?(cosla sinar + 25ina) 3 Rw(coszasina + 25mg) But FA : pad/4 so p : 3 2 £2 (2a + sinZa) cosza + 2 or i : ZRsinat L——) 3 (20: + sm 2a) _ i 2 Allematively, } = gRsina 23 Sin :1 a + SH] 0! 4 PROBLEM 5.49 From the solution to Problem 5.]: A = 240 1:19, 3M 2 5.601'11..l7area = 6.60 in. (Area) From the solution to Problem 5.21: L = 72 imjm = 4.67m, Pm = 6.67 in. Applying the theorems of Pappus-Guldinus, we have (a) Rotation about the x axis: A. = erl’iinel. = 21r(6.67in.)(72in.) = 301141113 A = 3020 in2 1 , = zzrmA = 2x(6.60in.)[240in2) = 9952.6 in3 (b) Rotation aboutx = 16 in.: A,=16 : 222(16 — XML = 2::[06 i 4.67)in.](72in.) = 5125.5 in2 A”,6 = sum2 4 Fm 5 2::(16 — Xaml)A = 2n'[(16 — 5.60)in.](240in2) =15 682.8:‘n3 r;=J6 :1563x103in34 PROBLEM 5.53 A 15-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 V=27rfA=2n[(§+7,5jmm:|x|:%x5mmx5mm] or V = 720mm3< ...
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