Determination of Centroids by Integration Double integration to find the first moment may be avoided by defining dA as a thin rectangle or strip.

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.

Sample Problem 7.2 SOLUTION: Determine the constant k. Evaluate the total area.

Sample Problem 7.2 Using vertical strips, perform a single integration to find the first moments.

Sample Problem 7.2 Or, using horizontal strips, perform a single integration to find the first moments.

Sample Problem 7.2 Evaluate the centroid coordinates.

Problem 7.6 Locate the centroid of the plane area shown. y x 20 mm

Solving Problems on Your Own 20 mm 30 mm Solving Problems on Your Own Locate the centroid of the plane area shown. 36 mm Several points should be emphasized when solving these types of problems. 24 mm 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.

Decide how to construct the given area from common shapes. y Problem 7.6 Solution 20 + 10 Decide how to construct the given area from common shapes. C1 C2 30 24 + 12 x 10 Dimensions in mm

y Problem 7.6 Solution 20 + 10 Construct a table containing areas and respective coordinates of the centroids. C1 C2 30 24 + 12 x 10 Dimensions in mm A, mm2 x, mm y, mm xA, mm3 yA, mm3 1 20 x 60 =1200 10 30 12,000 36,000 2 (1/2) x 30 x 36 =540 30 36 16,200 19,440 S 1740 28,200 55,440

XS A = S xA X (1740) = 28,200 X = 16.21 mm YS A = S yA Problem 7.6 Solution 20 + 10 Then XS A = S xA X (1740) = 28,200 X = 16.21 mm C1 or C2 and YS A = S yA 30 24 + 12 Y (1740) = 55,440 x 10 Y = 31.9 mm Dimensions in mm or A, mm2 x, mm y, mm xA, mm3 yA, mm3 1 20 x 60 =1200 10 30 12,000 36,000 2 (1/2) x 30 x 36 =540 30 36 16,200 19,440 S 1740 28,200 55,440

Problem 7.7 a 24 kN 30 kN 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. 0.3 m A B wA wB 1.8 m

Solving Problems on Your Own a 24 kN 30 kN 0.3 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. A B wA wB 1.8 m 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.

RII = (1.8 m)(wB kN/m) = 0.9 wB kN Problem 7.7 Solution a 24 kN 30 kN 0.3 m Replace the distributed load by a pair of equivalent forces. C A B 20 kN/m wB 0.6 m 0.6 m RI RII 1 2 We have RI = (1.8 m)(20 kN/m) = 18 kN 1 2 RII = (1.8 m)(wB kN/m) = 0.9 wB kN

SFy = 0: -24 kN + 18 kN + (0.9 wB) kN - 30 kN= 0 or wB = 40 kN/m 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) + SFy = 0: -24 kN + 18 kN + (0.9 wB) kN - 30 kN= 0 or wB = 40 kN/m

Problem 7.8 y For the machine element shown, locate the z coordinate x 0.75 in z 1 in 2 in 3 in r = 1.25 in For the machine element shown, locate the z coordinate of the center of gravity.

X S V = S x V Y S V = S y V Z S V = S z V 0.75 in z 1 in 2 in 3 in r = 1.25 in Problem 7.8 Solving Problems on Your Own For the machine element shown, locate the z coordinate of the center of gravity. Determine the center of gravity of composite body. 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 X S V = S x V Y S V = S y V Z S V = S z V where X, Y, Z and x, y, z are the coordinates of the centroid of the body and the components, respectively.

Determine the center of gravity of composite body. x 0.75 in z 1 in 2 in 3 in r = 1.25 in Problem 7.8 Solution Determine the center of gravity of composite body. First assume that the machine element is homogeneous so that its center of gravity will coincide with the centroid of the corresponding volume. y x z I II III IV V Divide the body into five common shapes.

II (p/2)(2)2 (0.75) = 4.7124 7+ [(4)(2)/(3p)] = 7.8488 36.987 y x z I II III IV V y x 0.75 in z 1 in 2 in 3 in r = 1.25 in V, in3 z, in. z V, in4 I (4)(0.75)(7) = 21 3.5 73.5 II (p/2)(2)2 (0.75) = 4.7124 7+ [(4)(2)/(3p)] = 7.8488 36.987 III -p(11.25)2 (0.75)= -3.6816 7 -25.771 IV (1)(2)(4) = 8 2 16 V -(p/2)(1.25)2 (1) = -2.4533 2 -4.9088 S 27.576 95.807 Z S V = S z V : Z (27.576 in3 ) = 95.807 in4 Z = 3.47 in