CENTROIDS AND CENTERS OF GRAVITY CHAPTER 5 CENTROIDS AND CENTERS OF GRAVITY

CENTROIDS AND CENTERS OF GRAVITY CHAPTER 5 CENTROIDS AND CENTERS OF GRAVITY 5.1 Introduction 5.2 Centers of Gravity of a Two Dimensional Body 5.3 Centroids of Areas and Lines 5.4 First Moments of Area and Lines 5.5 Composite Plates and wires

Introduction The earth exerts a gravitational force on each of the particles forming a body. These forces can be replace by a single equivalent force equal to the weight of the body and applied at the center of gravity for the body. The centroid of an area is analogous to the center of gravity of a body. The concept of the first moment of an area is used to locate the centroid. Determination of the area of a surface of revolution and the volume of a body of revolution are accomplished with the Theorems of Pappus-Guldinus.

5.2 Center of Gravity of a Two Dimensional Body Center of gravity of a plate Center of gravity of a wire

5.3 Centroids of Area and Lines Centroid of an area Centroid of a line

5.4 First Moments of Areas and Lines An area is symmetric with respect to an axis BB’ if for every point P there exists a point P’ such that PP’ is perpendicular to BB’ and is divided into two equal parts by BB’. The first moment of an area with respect to a line of symmetry is zero. If an area possesses two lines of symmetry, its centroid lies at their intersection. If an area possesses a line of symmetry, its centroid lies on that axis An area is symmetric with respect to a center O if for every element dA at (x,y) there exists an area dA’ of equal area at (-x,-y). The centroid of the area coincides with the center of symmetry.

Shape x y Area h 3 bh 2 4r 3π 4r 3π πr 4 4r 3π πr 2 4a 3π 4b 3π πab 4 Triangular area 4r 3π 4r 3π πr 4 2 Quarter-circular area 4r 3π πr 2 Semicircular area 4a 3π 4b 3π πab 4 Quarter-elliptical area 4b 3π πab 2 Semi-elliptical area Figure A : Centroids of common shapes of areas

Shape x y Area 3a 3h ah ah n+1 n+1 n+2 a n+1 4n+2 h 3a 8 3h 5 2ah 3 3h Semi-parabolic area 3a 8 3h 5 2ah 3 parabolic area 3h 5 4ah 3 Parabolic spandrel 3a 4 3h 10 ah 3 ah n+1 n+1 n+2 a n+1 4n+2 h General spandrel Figure A : Centroids of common shapes of areas

Shape x y Area 2r sin α αr 3α Circular sector Figure A : Centroids of common shapes of areas

Shape x y Area 2r π 2r π πr 2 2r π πr r sin α α 2αr Quarter-circular arc 2r π 2r π πr 2 Semicircular arc 2r π πr Arc of circle r sin α α 2αr Figure B : Centroids of common shapes of lines

5.5 Composite Plates and Wires In many instances, a flat plate can be divided in rectangles, triangles,or the other common shapes shown in figure A. The abscissa X of its center of gravity G can be determine from the abscissas X1,X2,…..Xn of the centers of gravity of the various parts by expressing that the moment of the weight of the whole plate about the y axis is equal to the sum of the moments of the weights of the various parts about the same axis (Figure C). The ordinate Y of the center of gravity of the plate is found in a similar way by equating moments about the x axis. We write : X ( W1 + W2 + …..+ Wn ) = X1W1 + X2W2 + …..+ XnWn My : Y ( W1 + W2 + …..+ Wn ) = y1W1 + y2W2 + …..+ ynWn Mx

Composite plates Composite area These equations can be solved for the coordinates X and Y of the center of gravity of the plate Composite plates Composite area Figure C : Center of gravity of a composite plate

Qy Qx : X ( A1 + A2 + …..+ An ) = X1A1 + X2A2 + …..+ XnAn If the plate is homogeneous and of uniform thickness, the center of gravity concides with the centroid C of its area.The abscissa X of the centroid of the area can be determined by noting that the first moment Qy of the composite area with respect to th ey axis can be expressed both as the product of X and the total area and as the sum of the first moments of the elementary areas with with respect to the y axis ( figure C ). The ordinate Y of the centroid is found in a similar way by considering the first moment Qx of the composite area. We have : X ( A1 + A2 + …..+ An ) = X1A1 + X2A2 + …..+ XnAn Qy : Y ( A1 + A2 + …..+ An ) = y1A1 + y2A2 + …..+ ynAn Qx or for short

Sample Problem 5.1 For the plane area shown, determine the first moments with respect to the x and y axes (b) the location of the centroid

Calculate the first moments of each area with respect to the axes. solution SOLUTION: Divide the area into a triangle, rectangle, and semicircle with a circular cutout. Calculate the first moments of each area with respect to the axes. Find the total area and first moments of the triangle, rectangle, and semicircle. Subtract the area and first moment of the circular cutout. Compute the coordinates of the area centroid by dividing the first moments by the total area.

rectangle triangle semicircle circle

a) First Moments of Area b) Location of centroids Substituting the values given in the table into the equations defining the centroid of a composite area, we obtain

Location of the centroids Compute the coordinates of the area centroid by dividing the first moments by the total area.

Sample problem 5.2 650 mm 600 mm 250 mm The figure shown is made from a piece of thin, homogeneous wire. determine the location of its center of gravity.

Solution Since the figure is formed of homogeneous wire, its center of gravity coincides with the centroid of the corresponding line. Therefore, that centroid will be determined. Choosing the coordinates axes shown, with origin at A, we determine the coordinates of the centroid of each line segment and compute the first moments with respect to the coordinate axes 600 mm 300 mm 125 mm 250 mm

Sample Problem 5.3 A uniform semicircular rod of weight W and radius r is attached to a pin at A and rest against a frictionless at B. Determine the reactions at A and B.

Solution Ay Ax Free Body diagram A free body diagram of the rod is drawn. The forces acting on the rod are its weight W, which is applied at the center of gravity G. (whose position is obtained fro Fig B). A reaction at A, represented by its components Ax and Ay and a horizontal reaction at B. Ay = W Ax = W π

Ax Ay Ay = W Ax = W π Solution

PROBLEMS

Problem 5.1 Locate the centroid of the plane area shown 300 mm 150 mm

Solution 1 2 C1 C2 300 mm 150 mm 200mm 400 mm (150mm + 75mm) = 225 mm

Problem 5.2 Locate the centroid of the plane area shown

６0 mm + 52.5mm Solution 2 1 C2 75mm C1 37.5mm h = 3 25mm 40 mm

Problem 5.3 Locate the centroid of the plane area shown 250mm 400mm

250mm 400mm 600mm Solution C1 C2 1 2 250 + 1 ( 4000 )mm 3 2 ( 250 )mm

Problem 5.4 Locate the centroid of the plane area shown

Solution Y 1.5mm 11mm 22mm C1 1 X 2mm 8mm Y X X Y 2mm 6mm 2 C2

Problem 5.5 Locate the centroid of the plane area shown

100mm 60mm 1 C1 2 C2 120mm Y X 120 – 4 x 60 = 94.5 mm 3π 5655.6 18 344 18344 905 546 1 721 328 534 454 678 672 100 2 400 000

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