Eighth Edition Vector Mechanics for Engineers: Statics Determination of Centroids by Integration x A xdA x dxdy xel dA yA ydA y dxdy yel dA • Double integration to find the first moment may be avoided by defining dA as a thin rectangle or strip. x A xel dA x A xel dA yA yel dA ax a x dx 2 yA yel dA x ydx y ydx 2 y a x dx © 2007 The McGraw-Hill Companies, Inc. All rights reserved. x A xel dA 2r 1 cos r 2 d 3 2 yA yel dA 2r 1 sin r 2 d 3 2 5-1 Eighth Edition Vector Mechanics for Engineers: Statics Sample Problem 7.2 SOLUTION: • Determine the constant k. • Evaluate the total area. • Using either vertical or horizontal strips, perform a single integration to find the first moments. Determine by direct integration the location of the centroid of a parabolic spandrel. • Evaluate the centroid coordinates. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 5-2 Eighth Edition Vector Mechanics for Engineers: Statics Sample Problem 7.2 SOLUTION: • Determine the constant k. y k x2 b k a2 k y b a2 x2 or b a2 x a b1 2 y1 2 • Evaluate the total area. A dA a b x3 b 2 y dx 2 x dx 2 a 3 0 0a ab 3 a © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 5-3 Eighth Edition Vector Mechanics for Engineers: Statics Sample Problem 7.2 • Using vertical strips, perform a single integration to find the first moments. a b Q y xel dA xydx x 2 x 2 dx 0 a a b x4 a 2b 2 4 a 4 0 2 a y 1 b Qx yel dA ydx 2 x 2 dx 2 02a a b2 x5 ab 2 4 2a 5 0 10 © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 5-4 Eighth Edition Vector Mechanics for Engineers: Statics Sample Problem 7.2 • Or, using horizontal strips, perform a single integration to find the first moments. b 2 ax a x2 a x dy Q y xel dA dy 2 2 0 1 b 2 a 2 a 2 0 b 2 a b y dy 4 a Qx yel dA y a x dy y a 1 2 y1 2 dy b a 3 2 ab 2 ay 1 2 y dy 10 b 0 b © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 5-5 Eighth Edition Vector Mechanics for Engineers: Statics Sample Problem 7.2 • Evaluate the centroid coordinates. xA Q y ab a 2b x 3 4 3 x a 4 yA Q x ab ab 2 y 3 10 © 2007 The McGraw-Hill Companies, Inc. All rights reserved. y 3 b 10 5-6 Problem 7.6 y 20 mm 30 mm Locate the centroid of the plane area shown. 36 mm 24 mm x 7 y 20 mm Problem 7.6 30 mm Solving Problems on Your Own 36 mm Locate the centroid of the plane area shown. 24 mm Several points should be emphasized when solving these types of problems. x 1. Decide how to construct the given area from common shapes. 2. It is strongly recommended that you construct a table containing areas or length and the respective coordinates of the centroids. 3. When possible, use symmetry to help locate the centroid. 8 Problem 7.6 Solution y 20 + 10 Decide how to construct the given area from common shapes. C1 C2 24 + 12 30 10 x Dimensions in mm 9 Problem 7.6 Solution y 20 + 10 Construct a table containing areas and respective coordinates of the centroids. C1 C2 24 + 12 30 10 x Dimensions in mm A, mm2 1 20 x 60 =1200 2 (1/2) x 30 x 36 =540 S 1740 x, mm 10 30 y, mm 30 36 xA, mm3 12,000 16,200 28,200 yA, mm3 36,000 19,440 55,440 10 Problem 7.6 Solution y 20 + 10 Then XS A = S xA X (1740) = 28,200 or X = 16.21 mm and YS A = S yA Y (1740) = 55,440 C1 C2 24 + 12 30 10 x or Y = 31.9 mm Dimensions in mm A, mm2 1 20 x 60 =1200 2 (1/2) x 30 x 36 =540 S 1740 x, mm 10 30 y, mm 30 36 xA, mm3 12,000 16,200 28,200 yA, mm3 36,000 19,440 55,440 11 Problem 7.7 a 24 kN A 30 kN 0.3 m B wA wB 1.8 m The beam AB supports two concentrated loads and rests on soil which exerts a linearly distributed upward load as shown. Determine (a) the distance a for which wA = 20 kN/m, (b) the corresponding value wB. 12 Problem 7.7 a 24 kN A 30 kN Solving Problems on Your Own 0.3 m B wA wB 1.8 m The beam AB supports two concentrated loads and rests on soil which exerts a linearly distributed upward load as shown. Determine (a) the distance a for which wA = 20 kN/m, (b) the corresponding value wB. 1. Replace the distributed load by a single equivalent force. The magnitude of this force is equal to the area under the distributed load curve and its line of action passes through the centroid of the area. 2. When possible, complex distributed loads should be divided into common shape areas. 13 Problem 7.7 Solution 24 kN a 30 kN C A 20 kN/m Replace the distributed load by a pair of B equivalent forces. wB 0.6 m 0.6 m RI We have 0.3 m RII 1 RI = 2 (1.8 m)(20 kN/m) = 18 kN 1 RII = 2 (1.8 m)(wB kN/m) = 0.9 wB kN 14 Problem 7.7 Solution a 24 kN 30 kN 0.3 m C A B wB 0.6 m 0.6 m RI = 18 kN RII = 0.9 wB kN (a) + SMC = 0: (1.2 - a)m x 24 kN - 0.6 m x 18 kN - 0.3m x 30 kN = 0 or a = 0.375 m (b) + SF = 0: -24 kN + 18 kN + (0.9 w ) kN - 30 kN= 0 y B or wB = 40 kN/m 15 Problem 7.8 y 2 in 3 in 2 in 1 in r = 1.25 in x z For the machine element shown, locate the z coordinate of the center of gravity. 0.75 in 2 in 2 in r = 1.25 in 16 Problem 7.8 y 2 in 2 in 1 in 3 in Solving Problems on Your Own r = 1.25 in x z For the machine element shown, locate the z coordinate of the center of gravity. Determine the center of gravity of composite body. 2 in r = 1.25 in 2 in For a homogeneous body the center of gravity coincides with the centroid of its volume. For this case the center of gravity can be determined by 0.75 in XSV = SxV YSV = SyV ZSV = SzV where X, Y, Z and x, y, z are the coordinates of the centroid of the 17 body and the components, respectively. Problem 7.8 Solution y 2 in 2 in 1 in 3 in Determine the center of gravity of composite body. r = 1.25 in First assume that the machine element is homogeneous so that its center of gravity will coincide with the centroid of the corresponding volume. x z 0.75 in 2 in 2 in r = 1.25 in y V Divide the body into five common shapes. III II IV I x z 18 y y V 2 in IV I III 3 in 2 in 1 in r = 1.25 in x II x z z 0.75 in 2 in 2 in I II III IV V S V, in3 (4)(0.75)(7) = 21 (p/2)(2)2 (0.75) = 4.7124 -p(11.25)2 (0.75)= -3.6816 (1)(2)(4) = 8 -(p/2)(1.25)2 (1) = -2.4533 r = 1.25 in z, in. 3.5 7+ [(4)(2)/(3p)] = 7.8488 7 2 2 27.576 Z S V = S z V : Z (27.576 in3 ) = 95.807 in4 z V, in4 73.5 36.987 -25.771 16 -4.9088 95.807 Z = 3.47 in 19