Copyright © 2011 Pearson Education South Asia Pte Ltd

Copyright © 2011 Pearson Education South Asia Pte Ltd
Chapter Objectives To discuss the concept of the center of gravity, center of mass, and the centroid. To show how to determine the location of the center of gravity and centroid for a system of discrete particles and a body of arbitrary shape. To present a method for finding the resultant of a general distributed loading. To show how to determine the moment of inertia of an area. Copyright © 2011 Pearson Education South Asia Pte Ltd

In-Class Activities Reading Quiz Center of Gravity, Center of Mass, and the Centroid of a Body Composite Bodies Resultant of a Distributed Loading Moments of Inertia for Areas Parallel-Axis Theorem for an Area Moments of Inertia for Composite Areas Concept Quiz Copyright © 2011 Pearson Education South Asia Pte Ltd

READING QUIZ (cont) 2) The _________ is the point defining the geometric center of an object. Center of gravity Center of mass Centroid None of the above Copyright © 2011 Pearson Education South Asia Pte Ltd

READING QUIZ (cont) 3) A composite body in this section refers to a body made of ____. Carbon fibers and an epoxy matrix Steel and concrete A collection of “simple” shaped parts or holes A collection of “complex” shaped parts or holes Copyright © 2011 Pearson Education South Asia Pte Ltd

READING QUIZ (cont) 4) The composite method for determining the location of the center of gravity of a composite body requires _______. Integration Differentiation Simple arithmetic All of the above. Copyright © 2011 Pearson Education South Asia Pte Ltd

READING QUIZ (cont) 5) Based on the typical centroid information, what are the minimum number of pieces you will have to consider for determining the centroid of the area shown at the right? 1 2 3 4 Copyright © 2011 Pearson Education South Asia Pte Ltd

READING QUIZ (cont) 6) To study problems concerned with the motion of matter under the influence of forces, i.e., dynamics, it is necessary to locate a point called ________. Center of gravity Center of mass Centroid None of the above Copyright © 2011 Pearson Education South Asia Pte Ltd

READING QUIZ (cont) 8) The parallel-axis theorem for an area is applied between: An axis passing through its centroid and any corresponding parallel axis. Any two parallel axes. Two horizontal axes only. Two vertical axes only. Copyright © 2011 Pearson Education South Asia Pte Ltd

READING QUIZ (cont) 9) The Moment of Inertia of a composite area equals the ________ of the MoI of all of its parts. Vector sum Algebraic sum (addition or subtraction) Addition Product Copyright © 2011 Pearson Education South Asia Pte Ltd

CENTER OF GRAVITY, CENTER OF MASS, AND THE CENTROID OF A BODY Center of Gravity Locates the resultant weight of a system of particles Consider system of n particles fixed within a region of space The weights of the particles can be replaced by a single (equivalent) resultant weight having defined point G of application Copyright © 2011 Pearson Education South Asia Pte Ltd

CENTER OF GRAVITY, CENTER OF MASS, AND THE CENTROID OF A BODY (cont) Center of Gravity Resultant weight = total weight of n particles Sum of moments of weights of all the particles about x, y, z axes = moment of resultant weight about these axes Summing moments about the x axis, Summing moments about y axis, Copyright © 2011 Pearson Education South Asia Pte Ltd

CENTER OF GRAVITY, CENTER OF MASS, AND THE CENTROID OF A BODY (cont) Center of Gravity Although the weights do not produce a moment about z axis, by rotating the coordinate system 90°about x or y axis with the particles fixed in it and summing moments about the x axis, Generally, Copyright © 2011 Pearson Education South Asia Pte Ltd

CENTER OF GRAVITY, CENTER OF MASS, AND THE CENTROID OF A BODY (cont) Centroid of Mass Provided acceleration due to gravity g for every particle is constant, then W = mg By comparison, the location of the center of gravity coincides with that of center of mass Particles have weight only when under the influence of gravitational attraction, whereas center of mass is independent of gravity Copyright © 2011 Pearson Education South Asia Pte Ltd

CENTER OF GRAVITY, CENTER OF MASS, AND THE CENTROID OF A BODY (cont) Centroid of Mass A rigid body is composed of an infinite number of particles Consider arbitrary particle having a weight of dW Copyright © 2011 Pearson Education South Asia Pte Ltd

CENTER OF GRAVITY, CENTER OF MASS, AND THE CENTROID OF A BODY (cont) Centroid of a Volume Consider an object subdivided into volume elements dV, for location of the centroid, Copyright © 2011 Pearson Education South Asia Pte Ltd

CENTER OF GRAVITY, CENTER OF MASS, AND THE CENTROID OF A BODY (cont) Centroid of an Area For centroid of surface area of an object, such as plate and shell, subdivide the area into differential elements dA Copyright © 2011 Pearson Education South Asia Pte Ltd

COMPOSITE BODIES Consists of a series of connected “simpler” shaped bodies, which may be rectangular, triangular or semicircular A body can be sectioned or divided into its composite parts Accounting for finite number of weights Copyright © 2011 Pearson Education South Asia Pte Ltd

COMPOSITE BODIES (cont) Procedure for Analysis Composite Parts Divide the body or object into a finite number of composite parts that have simpler shapes Treat the hole in composite as an additional composite part having negative weight or size Moment Arms Establish the coordinate axes and determine the coordinates of the center of gravity or centroid of each part Copyright © 2011 Pearson Education South Asia Pte Ltd

COMPOSITE BODIES (cont) Procedure for Analysis Summations Determine the coordinates of the center of gravity by applying the center of gravity equations If an object is symmetrical about an axis, the centroid of the objects lies on the axis Copyright © 2011 Pearson Education South Asia Pte Ltd

EXAMPLE 1 Locate the centroid of the plate area.
Copyright © 2011 Pearson Education South Asia Pte Ltd Copyright © 2011 Pearson Education South Asia Pte Ltd

EXAMPLE 1 (cont) Composite Parts Plate divided into 3 segments.
Solution Composite Parts Plate divided into 3 segments. Area of small rectangle considered “negative”. Copyright © 2011 Pearson Education South Asia Pte Ltd Copyright © 2011 Pearson Education South Asia Pte Ltd

EXAMPLE 1 (cont) Moment Arm
Solution Moment Arm Location of the centroid for each piece is determined and indicated in the diagram. Summations Copyright © 2011 Pearson Education South Asia Pte Ltd Copyright © 2011 Pearson Education South Asia Pte Ltd

RESULTANT OF A DISTRIBUTED LOADING Pressure Distribution over a Surface Consider the flat plate subjected to the loading function ρ = ρ(x, y) Pa Determine the force dF acting on the differential area dA m2 of the plate, located at the differential point (x, y) dF = [ρ(x, y) N/m2](dA m2) = [ρ(x, y) dA] N Entire loading represented as infinite parallel forces acting on separate differential area dA Copyright © 2011 Pearson Education South Asia Pte Ltd

RESULTANT OF A DISTRIBUTED LOADING (cont) Pressure Distribution over a Surface This system will be simplified to a single resultant force FR acting through a unique point on the plate Copyright © 2011 Pearson Education South Asia Pte Ltd

RESULTANT OF A DISTRIBUTED LOADING (cont) Magnitude of Resultant Force To determine magnitude of FR, sum up the differential forces dF acting over the plate’s entire surface area dA Magnitude of resultant force = total volume under the distributed loading diagram Location of Resultant Force is Copyright © 2011 Pearson Education South Asia Pte Ltd

MOMENTS OF INERTIA FOR AREAS Please refer to the website for the animation: Moment of Inertia Centroid for an area is determined by the first moment of an area about an axis Second moment of an area is referred as the moment of inertia Moment of inertia of an area originates whenever one relates the normal stress σ or force per unit area Copyright © 2011 Pearson Education South Asia Pte Ltd

MOMENTS OF INERTIA FOR AREAS (cont) Moment of Inertia Consider area A lying in the x-y plane Be definition, moments of inertia of the differential plane area dA about the x and y axes For entire area, moments of inertia are given by Copyright © 2011 Pearson Education South Asia Pte Ltd

MOMENTS OF INERTIA FOR AREAS (cont) Moment of Inertia Formulate the second moment of dA about the pole O or z axis This is known as the polar axis where r is perpendicular from the pole (z axis) to the element dA Polar moment of inertia for entire area, Copyright © 2011 Pearson Education South Asia Pte Ltd

PARALLEL AXIS THEOREM FOR AN AREA For moment of inertia of an area known about an axis passing through its centroid, determine the moment of inertia of area about a corresponding parallel axis using the parallel axis theorem Consider moment of inertia of the shaded area A differential element dA is located at an arbitrary distance y’ from the centroidal x’ axis Copyright © 2011 Pearson Education South Asia Pte Ltd

PARALLEL AXIS THEOREM FOR AN AREA (cont) The fixed distance between the parallel x and x’ axes is defined as dy For moment of inertia of dA about x axis For entire area First integral represent the moment of inertia of the area about the centroidal axis Copyright © 2011 Pearson Education South Asia Pte Ltd

PARALLEL AXIS THEOREM FOR AN AREA (cont) Second integral = 0 since x’ passes through the area’s centroid C Third integral represents the total area A Similarly For polar moment of inertia about an axis perpendicular to the x-y plane and passing through pole O (z axis) Copyright © 2011 Pearson Education South Asia Pte Ltd

EXAMPLE 2 Determine the moment of inertia for the rectangular area with respect to (a) the centroidal x’ axis, (b) the axis xb passing through the base of the rectangular, and (c) the pole or z’ axis perpendicular to the x’-y’ plane and passing through the centroid C. Copyright © 2011 Pearson Education South Asia Pte Ltd Copyright © 2011 Pearson Education South Asia Pte Ltd

EXAMPLE 2 (cont) Part (a)
Solution Part (a) Differential element chosen, distance y’ from x’ axis. Since dA = b dy’, Part (b) By applying parallel axis theorem, Copyright © 2011 Pearson Education South Asia Pte Ltd Copyright © 2011 Pearson Education South Asia Pte Ltd

EXAMPLE 2 (cont) Part (c) For polar moment of inertia about point C,
Solution Part (c) For polar moment of inertia about point C, Copyright © 2011 Pearson Education South Asia Pte Ltd Copyright © 2011 Pearson Education South Asia Pte Ltd

MOMENTS OF INERTIA FOR COMPOSITE AREAS Composite area consist of a series of connected simpler parts or shapes Moment of inertia of the composite area = algebraic sum of the moments of inertia of all its parts Procedure for Analysis Composite Parts Divide area into its composite parts and indicate the centroid of each part to the reference axis Parallel Axis Theorem Moment of inertia of each part is determined about its centroidal axis Copyright © 2011 Pearson Education South Asia Pte Ltd

MOMENTS OF INERTIA FOR COMPOSITE AREAS (cont) Procedure for Analysis Parallel Axis Theorem When centroidal axis does not coincide with the reference axis, the parallel axis theorem is used Summation Moment of inertia of the entire area about the reference axis is determined by summing the results of its composite parts Copyright © 2011 Pearson Education South Asia Pte Ltd

EXAMPLE 3 (cont) Composite Parts
Solution Composite Parts Composite area obtained by subtracting the circle form the rectangle. Centroid of each area is located in the figure below. Copyright © 2011 Pearson Education South Asia Pte Ltd Copyright © 2011 Pearson Education South Asia Pte Ltd

EXAMPLE 3 (cont) Summation
Solution Summation For moment of inertia for the composite area, Copyright © 2011 Pearson Education South Asia Pte Ltd Copyright © 2011 Pearson Education South Asia Pte Ltd

CONCEPT QUIZ 1) A pipe is subjected to a bending moment as shown. Which property of the pipe will result in lower stress (assuming a constant cross-sectional area)? Smaller Ix Smaller Iy Larger Ix Larger Iy M y Pipe section x Copyright © 2011 Pearson Education South Asia Pte Ltd

CONCEPT QUIZ (cont) 2) In the figure to the right, what is the differential moment of inertia of the element with respect to the y-axis (dIy)? x2 ydx (1/12)x3dy y2 x dy (1/3)ydy x y=x3 x,y y Copyright © 2011 Pearson Education South Asia Pte Ltd

CONCEPT QUIZ (cont) 3) For the area A, we know the centroid’s (C) location, area, distances between the four parallel axes, and the MoI about axis 1. We can determine the MoI about axis 2 by applying the parallel axis theorem ___ . Directly between the axes 1 and 2. Between axes 1 and 3 and then between the axes 3 and 2. Between axes 1 and 4 and then axes 4 and 2. None of the above. d3 d2 d1 4 3 2 1 A • C Axis Copyright © 2011 Pearson Education South Asia Pte Ltd

CONCEPT QUIZ (cont) 4) For the same case, consider the MoI about each of the four axes. About which axis will the MoI be the smallest number? Axis 1 Axis 2 Axis 3 Axis 4 Can not tell. d3 d2 d1 4 3 2 1 A • C Axis Copyright © 2011 Pearson Education South Asia Pte Ltd

CONCEPT QUIZ (cont) 5) For the given area, the Moment of Inertia about axis 1 is 200 cm4 . What is the MoI about axis 3? 90 cm4 110 cm4 60 cm4 40 cm4 A=10 cm2 • C d2 d1 3 2 1 d1 = d2 = 2 cm Copyright © 2011 Pearson Education South Asia Pte Ltd

CONCEPT QUIZ (cont) 7) If a vertical rectangular strip is chosen as the differential element, then all the variables, including the integral limit, should be in terms of ____________ . x y z Any of the above. Copyright © 2011 Pearson Education South Asia Pte Ltd

CONCEPT QUIZ (cont) 8) If a vertical rectangular strip is chosen, then what are the values of x and y? (x , y) (x / 2 , y / 2) (x , 0) (x , y / 2) ~ Copyright © 2011 Pearson Education South Asia Pte Ltd

CONCEPT QUIZ (cont) 9) A storage box is tilted up to clean the rug underneath the box. It is tilted up by pulling the handle C, with edge A remaining on the ground. What is the maximum angle of tilt (measured between bottom AB and the ground) possible before the box tips over? 30° b) 45° c) 60° d) 90° Copyright © 2011 Pearson Education South Asia Pte Ltd

CONCEPT QUIZ (cont) 11) A storage box is tilted up by pulling C, with edge A remaining on the ground. What is the max angle of tilt possible before the box tips over? 30° 45° 60° 90° 30º G C A B Copyright © 2011 Pearson Education South Asia Pte Ltd

CONCEPT QUIZ (cont) 13) For determining the centroid of the area, what are the coordinates (x, y ) of the centroid of square DEFG? (1, 1) m (1.25, 1.25) m (0.5, 0.5 ) m (1.5, 1.5) m A 1m y E F G C B x D Copyright © 2011 Pearson Education South Asia Pte Ltd

Copyright © 2011 Pearson Education South Asia Pte Ltd

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