COM,COG, Centroid & Moment of Area PDF

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Center of Gravity, Center of Mass, Centroid and Moment of Area Center of Gravity Consider an object being suspended from different points as shown in figure(1). The line of action of the weight of the body will be co-linear with the string in tension. The center of gravity of the object can be imagined as the point of intersection of the different line of actions of the weight of the body in different positions. This is true provided the gravitational forces are parallel and uniform. For an exact analysis one has to consider the different directions of the gravity forces acting on different particles of the body. This is because these forces point towards the center of the earth. Therefore line of actions of the gravity forces will not be quite concurrent and no unique center of gravity exists. But for all practical purposes we can assume that gravity is uniform and a unique center of gravity exists.

Figure 1: Finding center of gravity by suspending a body from different points Mathematically, we apply the principle of moments to the parallel system of gravitational forces shown in left half of figure(2). The moment of the total gravitational force W about any axis is equal to the sum of moments of the infinitesimal gravitational forces dW acting on small chunks of the body about the same axis. This gives the coordinates of the center of gravity as R R R zdW ydW xdW , , . (1) z¯ = y¯ = x¯ = W W W Center of Mass Since W = mg and dW = dm g and if we assume gravity is same everywhere, the above expressions becomes R R R zdm yd m xdm , ¯z = , . (2) y¯ = x¯ = m m m In terms of the position vectors shown in the right half of figure(2) we have the position vector of the centroid as R rd m . (3) ¯r = m Note that gravity does not appear in the above two equations. These define a unique point in the body which is independent of gravity and is solely a function of distribution of mass. This is called center of mass of the body. It will be coincident with the center of gravity as long as gravity field

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is uniform and parallel. If density is not constant but a function of the coordinates, we have to account for this using R R R xρdV yρdV zρdV , , . (4) x¯ = R y¯ = R z¯ = R ρdV ρdV ρdV

Figure 2: Center of gravity (left) and center of mass (right) Centroid of Lines, Areas and Volumes If density is constant throughout the body, equation(4) reduces to R R R xdV ydV zdV , , . x¯ = R y¯ = R z¯ = R dV dV dV

(5)

This equation defines a geometrical property of the body which do not depend on mass and is solely a property of the shape of the body. It gives the coordinates of the centroid of the volume shown in figure(3). So, center of mass and centroid are located at the same position if density is uniform throughout the body otherwise these are different points.

Figure 3: Centroid of volume For a slender rod or wire of length L and cross-sectional area A shown in figure(4) we have the centroid given by R R R zdL yd L xdL , , . (6) z¯ = y¯ = x¯ = L L L 2

Similarly for a thin sheet of thickness t and area A as shown in figure(5), we have the expression

Figure 4: Centroid of line for the centroid as x¯ =

R

xd A , A

y¯ =

R

yd A , A

z¯ =

R

zd A . A

(7)

Figure 5: Centroid of area Area moments The numerators in equation(7) give the first moment of the area about the respective axis, i.e., Z Z Z (8) yd A = y¯ A, zd A = z¯ A. xd A = x¯ A, R Note that if theRorigin of the coordinate system is located at the centroid itself, we have xd A = 0, R yd A = 0 and zd A = 0 since x¯ = y¯ = z¯ = 0. When forces are distributed continuously over an area, it is often required to calculate the moment of these forces about an axis in the plane of area or perpendicular to the area. Many a times the intensity of the force, i.e., the pressure or stress, is proportional to the distance of the line of action of the force from the axis. Typical examples are shown in figure(6) where the pressure distribution under hydrostatic condition is proportional to the depth from the axis AB located in the free surface, i.e., p = k y, in figure(7) where the stresses in the beam due to bending are again proportional to distance from the neutral axis, i.e., σ = k y and finally in figure(8) where the stresses 3

Figure 6: Hydrostatics inside a bar subjected to torsion are proportional to the radius, i.e., τ = kr . Under these conditions, the elementary forces will then be proportional to the distance from axis times the elemental area and the elementary moments will be proportional to square of the distance from axis times the elemental area. The total moment about an axis will have the form Z M = k y 2 d A. (9) This expression involves the second moment of area about an axis. Some books use the term area moment of inertia because of the similarity with the mass moment of inertia.

Figure 7: Bending of beams So with reference to figure(9), the expressions for the second moment of area or the area moment of inertia are given by Z Z 2 I x = y d A, I y = x 2d A . (10) 4

Figure 8: Torsion of cylinders The polar moment of area is Iz =

Z

r 2 d A.

(11)

Note that I z = I x + I y .

Figure 9: Second moment of area Transfer of Axis Sometimes we need to transfer the second area moment between two different coordinate systems as shown in figure(10). The relation between the second moment of area about any axis, say x or y and a corresponding centroidal axis, say xc or y c , is given by I x = I xc + Ad x2 ,

I y = I yc + Ad y2 ,

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I z = I zc + Ad 2 .

(12)

Figure 10: Second moment of area in centroidal and a general coordinate system

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