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1. 1. EDUCATION HOLE PRESENTS Engineering Mechanics Unit-III
2. 2. Centroid and Moment of Inertia .............................................................................................. 2 Center of Gravity..............................................................................................................................3 Center of Mass.................................................................................................................................4 Centroid of areas and curve .............................................................................................................4 Centroid of volumes.........................................................................................................................6 Determination of centroid by integration ................................................................................ 7 Centroid of area....................................................................................................................................................7 Centroid of lines....................................................................................................................................................7 Center of gravity of bodies ...................................................................................................................................8 Centroids of volumes............................................................................................................................................8 Centroid of composite bodies...........................................................................................................8 For Composite Plates (Constant Thickness) .....................................................................................................9 For Composite Rods..........................................................................................................................................9 Definition of Moment of inertia of area................................................................................. 10 Perpendicular axis theorem ........................................................................................................... 10 Parallel axis theorem...................................................................................................................... 11 Polar moment of Inertia................................................................................................................. 12 Moment of inertia of simple areas by integration ................................................................. 13 Moment of Inertia of Composite Areas .......................................................................................... 13 Totals .................................................................................................................................... 14 Moment of Inertia of masses................................................................................................. 14 Parallel axis theorem for mass moment of inertia .......................................................................... 15 Mass moment of inertia of simple bodies by integration................................................................ 16 Mass moment of inertia of composite bodies................................................................................. 17 Centroid and Moment of Inertia
3. 3. Center of Gravity The center of gravity is a geometric property of any object. The center of gravity is the average location of the weight of an object. We can completely describe the motion of any object through space in terms of the translation of the center of gravity of the object from one place to another, and the rotation of the object about its center of gravity if it is free to rotate. If the object is confined to rotate about some other point, like a hinge, we can still describe its motion. In flight, both airplanes and rockets rotate about their centers of gravity. Consider the following lamina. Let’s assume that it has been exposed to gravitational field. Obviously every single element will experience a gravitational force towards the centre of earth. Further let’s assume the body has practical dimensions, then we can easily conclude that all elementary forces will be unidirectional and parallel. Consider G to be the centroid of the irregular lamina. As shown in first figure we can easily represent the net force passing through the single point G. We can also divide the entire region into let’s say n small elements. Let’s say the coordinates to be (x1,y1), (x2,y2), (x3,y3)………. (xn,yn) as shown in figure . Let ΔW1, ΔW2, ΔW3,……., ΔWn be the elementary forces acting on the elementary elements. Clearly, W = ΔW1+ ΔW2+ ΔW3 +…………..+ ΔWn When n tends to infinity ΔW becomes infinitesimally small and can be replaced as dW. Centre of gravity : xc= / , yc= / ( and zc= / in case of a three dimensional body) where x,y are the coordinate of the small element and dw(or ΔW) the elemental force. And we have seen that W. For some type of surfaces of bodies there lies a probability that the centre of gravity may lie outside the body. Secondly centre of gravity represents the entire lamina, therefore we can
4. 4. replace the entire body by the single point with a force acting on it when needed. There is a major difference between centre of mass and centre of gravity of a body. For centre of gravity we integrate with respect to dW whereas for centre of mass we integrate with respect to dm. Mass is a scalar quantity and force a vector quantity. For general practical size objects both of them turn out to be the same as both of them are proportional and the force is unidirected (dW = dm*g) .But when we consider large size objects such as a continent, results would turn out to be different because here the vector nature of dW comes into play. Center of Mass The center of mass of a distribution of mass in space is the unique point where the weighted relative position of the distributed mass sums to zero. The distribution of mass is balanced around the center of mass and the average of the weighted position coordinates of the distributed mass defines its coordinates. Calculations in mechanics are often simplified when formulated with respect to the center of mass. In the case of a single rigid body, the center of mass is fixed in relation to the body, and if the body has uniform density, it will be located at the centroid. The center of mass may be located outside the physical body, as is sometimes the case for hollow or open-shaped objects, such as a horseshoe. In the case of a distribution of separate bodies, such as the planets of the Solar System, the center of mass may not correspond to the position of any individual member of the system. The center of mass is a useful reference point for calculations in mechanics that involve masses distributed in space, such as the linear and angular momentum of planetary bodies and rigid body dynamics. In orbital mechanics, the equations of motion of planets are formulated as point masses located at the centers of mass. The center of mass frame is an inertial frame in which the center of mass of a system is at rest at with respect the origin of the coordinate system. Centroid of areas and curve We have seen one method to find out the centre of gravity, there are other ways too. Let’s consider plate of uniform thickness and a homogenous density. Now weight of small element is directly proportional to its thickness, area and density as: ΔW = ϒ t dA. Where ϒ is the density per unit volume, t is the thickness , dA is the area of the small element. Let’s consider plate of uniform thickness and a homogenous density. Now weight of small element is directly proportional to its thickness, area and density as: ΔW = ϒ t dA. Where ϒ is the density per unit volume, t is the thickness , dA is the area of the small element.
5. 5. So we can replace ΔW with this relationship in the expression we obtained in the prior topic. Therefore we get: Centroid of area : xc= / , yc= / (and zc= / in case of a three dimensional body) Where x,y are the coordinate of the small element and da(or ΔA) the elemental force. Also A (total area of the plate) (xc ,yc,zc) is called the centroid of area of the lamina. If the surface is homogenous we conclude that it is the same as centre of gravity. There can also arise a case where in cross-sectional area is constant and length is variable as in the case of a rope or slender rod. In such cases the situation modifies to: ΔW = ϒ a dl. Where ϒ is the weight per unit length, per unit cross-sectional area, A is the area of cross – section, and dl the variable length.
6. 6. So the above results reduce to: Centroid of a line: xl= / , yl= / ( and zl= / in case of a three dimensional body) where x,y are the coordinate of the small element and dl(or ΔL) the elemental force. Also L(total area of the plate). The coordinate (xl ,yl,zl) is called the centroid of a line. It is important to mention that centroids of line may or may not lie on the line( as shown in diagram above). Centroid of volumes For a body of volume V and density ρ, the element has mass dm = ρ dV. The coordinates of the centre of mass may be written as
7. 7. If density ρ of a body is a constant, the coordinates of the centre of mass also become the coordinates of the centroid G of the line segment. In such cases, the above equations may be written as Determination of centroid by integration Centroid of area Centroid of lines
8. 8. Center of gravity of bodies Centroids of volumes Centroid of composite bodies To locate the centre of gravity (or centroid) of a composite body, the composite body will be split into a number of simple bodies of weight (or volume), and centre of gravity (or centroid) can be found using a standard expression. Then the centre of gravity (or centroid) of the composite body is
9. 9. If the density ρ of bodies is constant, the coordinates of the centre of mass also become the coordinates of the centroid. In such cases, the above equations may be written as For Composite Plates (Constant Thickness) For Composite Rods
10. 10. Definition of Moment of inertia of area The area moment of inertia of a beam's cross-sectional area measures the beam's ability to resist bending. The larger the moment of inertia, the less the beam will bend. The moment of inertia is a geometrical property of a beam and depends on a reference axis. The smallest moment of inertia about any axis passes through the centroid. It is always considered with respect to a reference axis such as x-x or y-y It is a mathematical property of a section concerned with a surface area and how that area is distributed about the reference axis. The reference axis is usually a centroidal axis. Perpendicular axis theorem  This theorem is applicable only to the plane laminar bodies  This theorem states that, the moment of inertia of a plane laminar about an axis perpendicular to its plane is equal to the sum of the moment of inertia of the lamina about two axis mutually perpendicular to each other in its plane and intersecting each other at the point where perpendicular axis passes through it  Consider plane laminar body of arbitrary shape lying in the x-y plane as shown below in the figure  The moment of inertia about the z-axis equals to the sum of the moments of inertia about the x-axis and y axis
11. 11.  To prove it consider the moment of inertia about x-axis where sum is taken over all the element of the mass mi  The moment of inertia about the y axis is  Moment of inertia about z axis is where ri is perpendicular distance of particle at point P from the OZ axis  For each element ri 2 =xi 2 + yi 2 Parallel axis theorem • This theorem relates the moment of inertia about an axis through the center of mass of a body about a second parallel axis • Let Icm be the moment of inertia about an axis through center of mass of the body and I be that about a parallel axis at a distance r from C as shown below in the figure Then according to parallel axis theorem I=Icm+Mr2 where M is the total mass of the body  Consider a point P of the body of mass mi at a distance xi from O
12. 12.  From point P drop a perpendicular PQ on to the OC and join PC.So that OP2 =CP2 + OC2 + 2OC.CQ ( From geometry) and miOP2 =miCP2 + miOC2 + 2miOC.CQ  Since the body always balances about an axis passing through center of mass, so algebraic sum of the moment of the weight of individual particles about center of mass must be zero. Here which is the algebraic sum of such moments about C and therefore eq as g is constant Thus we have I=Icm + Mr2 ---(17) Polar moment of Inertia The polar area moment of inertia of a beam's cross-sectional area measures the beam's ability to resist torsion. The larger the polar moment of inertia, the less the beam will twist. The following are the mathematical equations to calculate the polar moment of inertia: Or Or where x is the distance from the y-axis to an infinitesimal area dA, y is the distance from the x- axis to an infinitesimal area dA, Ix is the moment of inertia about x-axis and Iy is the moment of inertia about y-axis
13. 13. Moment of inertia of simple areas by integration Moment of Inertia of Composite Areas A composite area, A, made of several components areas ,...., 21 AA The integral representing the MOI of A may be subdivided into integrals computed over ,..., 21 AA Thus, the MOI of A with respect to a given axis can be obtained by summing the MOI of ,..., 21 AA with respect to the same axis. MOI of common shapes on inside of back book cover thease things: 1) Before adding the MOI of the component areas, the parallel axis theorem should be used to transfer each MOI to the desired axis. 2) The radius of gyration of a composite area is NOT equal to the sum of the radii of gyration of the component areas.. Example 1). Determine xI and yI of the area shown with respect to centroidal axis. Solution: Part # A y′ yA ′ xI (mm4 ) 2 xAd (mm4 ) yI (mm4 ) 2 yAd (mm4 ) x y x' y' 60 6060 40 80 180 1 2 centroidal axis 60 6060 40 80 180 all dimensions in mm
14. 14. 1 4,800 40 192,000 (60)(803 )/12 = 2.56(106 ) 4800(76-40)2 = 6.22(106 ) (603 )(80)/12 = 1.44(106 ) 0 2 7,200 100 720,000 180(403 )/12 = 960(103 ) 7200(100-76)2 = 4.15(106 ) (1803 )(40)/12 = 19.44(106 ) 0 Tot als 12,000 912,000 3.52(106 ) 10.37(106 ) 20.88(106 ) 0 First we find the centroid to be: ( ) mm76 000,912000,12 =′ =′ ′Σ=′ Y Y AyAY Now using the centroid we find the Ad2 terms for each part. Then we sum the MOI and Ad2 terms for each part to find the MOI about the centroidal axis for the entire section: 462 mm)10(89.13=Σ+Σ= xxx AdII 462 mm)10(88.20=Σ+Σ= yyy AdII Moment of Inertia of masses The mass moment of inertia is one measure of the distribution of the mass of an object relative to a given axis. The mass moment of inertia is denoted by I and is given for a single particle of mass m as
15. 15. where O-O is the axis around which one is evaluating the mass moment of inertia, and r is the perpendicular distance between the mass and the axis O-O. As can be seen from the above equation, the mass moment of inertia has the units of mass times length squared. The mass moment of inertial should not be confused with the area moment of inertia which has units of length to the power four. Mass moments of inertia naturally appear in the equations of motion, and provide information on how difficult (how much inertia there is) it is rotate the particle around given axis. Parallel axis theorem for mass moment of inertia The theorem determines the moment of inertia of a rigid body about any given axis, given that moment of inertia about the parallel axis through the center of mass of an object and the perpendicular distance between the axes. The moment of inertia about Z-axis can be represented as: Where Icmis the moment of inertia of an object about its centre of mass m is the mass of an object r is the perpendicular distance between the two axes. Proof: Assume that the perpendicular distance between the axes lies along the x-axis and the centre of mass lies at the origin. The moment of inertia relative to z-axis that passes through the centre of mass, is represented as
16. 16. Moment of inertia relative to the new axis with its perpendicular distance r along the x-axis, is represented as: We get, The first term is Icm,the second term is mr2 and the final term is zero as the origin lies at the centre of mass. Finally, Mass moment of inertia of simple bodies by integration When calculating the mass moment of inertia for a rigid body, one thinks of the body as a sum of particles, each having a mass of dm. Integration is used to sum the moment of inertia of each dm to get the mass moment of inertia of body. The equation for the mass moment of inertia of the rigid body is
17. 17. The integration over mass can be replaced by integration over volume, area, or length. For a fully three dimensional body using the density one can relate the element of mass to the element of volume. In this case the density has units of mass per length cubed and the relation is given as and the equation for the mass moment of inertia becomes The integral is actually a triple integral. If the coordinate system used is rectangular then dV=dxdydz . If the coordinates uses are cylindrical coordinates then Mass moment of inertia of composite bodies The moment of inertia around any axis can be calculated from the moment of inertia around parallel axis which passes through the center of mass. The equation to calculate this is called the parallel axis theorem and is given as where d is the distance between the original axis and the axis passing through the center of mass, m is the total mass of the body, and is the moment of inertia around the axis passing through
18. 18. the center of mass. If a body is composed of several bodies, to calculate the moment of inertia about a given axis one can simply calculate the moment of inertia of each part around the given axis and then add them to get the mass moment of inertia of the total body.