{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

{[ snackBarMessage ]}