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Seventh Edition 5 CHAPTER VECTOR MECHANICS FOR ENGINEERS: STATICS Ferdinand P. Beer E. Russell Johnston, Jr. Lecture Notes: J. Walt Oler Texas Tech University Distributed Forces: Centroids and Centers of Gravity © 2003 The McGraw-Hill Companies, Inc. All rights reserved. Seventh Edition Vector Mechanics for Engineers: Statics Contents • Introduction • Center of Gravity of a 2D Body • Centroids and First Moments of Areas and Lines • Centroids of Common Shapes of Areas • Centroids of Common Shapes of Lines • Composite Plates and Areas • Sample Problem 5.1 • Determination of Centroids by Integration • Sample Problem 5.4 • • • • • Theorems of Pappus-Guldinus Sample Problem 5.7 Distributed Loads on Beams Sample Problem 5.9 Center of Gravity of a 3D Body: Centroid of a Volume • Centroids of Common 3D Shapes • Composite 3D Bodies • Sample Problem 5.12 © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 5-2 Seventh Edition Vector Mechanics for Engineers: Statics 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. © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 5-3 Seventh Edition Vector Mechanics for Engineers: Statics Center of Gravity of a 2D Body • Center of gravity of a plate M y xW M y yW • Center of gravity of a wire xW x dW yW y dW © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 5-4 Seventh Edition Vector Mechanics for Engineers: Statics Centroids and First Moments of Areas and Lines • Centroid of an area • Centroid of a line xW x W x dW x dW x La x a dL x At x t dA x A x dA Q y first moment wit h respect to y h respect to x xL x dL yL y dL y A y dA Q x first moment wit © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 5-5 Seventh Edition Vector Mechanics for Engineers: Statics 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 a line of symmetry, its centroid lies on that axis • If an area possesses two lines of symmetry, its centroid lies at their intersection. • 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. © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 5-6 Seventh Edition Vector Mechanics for Engineers: Statics Centroids of Common Shapes of Areas © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 5-7 Seventh Edition Vector Mechanics for Engineers: Statics Centroids of Common Shapes of Lines © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 5-8 Seventh Edition Vector Mechanics for Engineers: Statics Composite Plates and Areas • Composite plates X W xW Y W yW • Composite area X A xA Y A yA © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 5-9 Seventh Edition Vector Mechanics for Engineers: Statics Sample Problem 5.1 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. For the plane area shown, determine the first moments with respect to the x and y axes and the location of the centroid. • 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. © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 5 - 10 Seventh Edition Vector Mechanics for Engineers: Statics Sample Problem 5.1 • Find the total area and first moments of the triangle, rectangle, and semicircle. Subtract the area and first moment of the circular cutout. © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 3 3 3 3 Q x 506 . 2 10 mm Q y 757 . 7 10 mm 5 - 11 Seventh Edition Vector Mechanics for Engineers: Statics Sample Problem 5.1 • Compute the coordinates of the area centroid by dividing the first moments by the total area. 3 3 x A 757 . 7 10 mm X 3 2 A 13.828 10 mm X 54 . 8 mm 3 3 y A 506 . 2 10 mm Y 3 2 A 13.828 10 mm Y 36 . 6 mm © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 5 - 12 Seventh Edition Vector Mechanics for Engineers: Statics Determination of Centroids by Integration • Double integration to find the first moment may be avoided by defining dA as a thin rectangle or strip. x A x dA x dx dy x el dA y A y dA y dx dy y el dA xA x el dA x ydx yA y el dA y 2 ydx xA x el dA a x 2 a x dx yA y el dA y a x dx © 2003 The McGraw-Hill Companies, Inc. All rights reserved. xA yA x el dA 1 2 cos r d 3 2 2r y el dA 1 2 sin r d 3 2 2r 5 - 13 Seventh Edition Vector Mechanics for Engineers: Statics Sample Problem 5.4 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. © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 5 - 14 Seventh Edition Vector Mechanics for Engineers: Statics Sample Problem 5.4 SOLUTION: • Determine the constant k. y kx 2 b ka 2 k b a y b a 2 x 2 or 2 a x b 1 2 y 1 2 • Evaluate the total area. A dA a b x3 b 2 y dx x dx 2 2 3 a 0 0a a ab 3 © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 5 - 15 Seventh Edition Vector Mechanics for Engineers: Statics Sample Problem 5.4 • Using vertical strips, perform a single integration to find the first moments. a Qy x el dA xy dx b 2 x 2 x dx 0 a a 2 b x4 a b 2 4 4 a 0 Qx y el dA a 2 1 b 2 y dx x dx 2 2 2 0 a y a 2 b2 x5 ab 4 5 10 2 a 0 © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 5 - 16 Seventh Edition Vector Mechanics for Engineers: Statics Sample Problem 5.4 • Or, using horizontal strips, perform a single integration to find the first moments. Q y x el dA ax 2 b a x dy 0 a 2 x 2 dy 2 2 2 b 1 2 a a b a y dy 2 0 b 4 a 1 Q x y el dA y a x dy y a y 1 2 b b a 3 ay y 1 2 b 0 © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 2 ab dy 10 2 dy 2 5 - 17 Seventh Edition Vector Mechanics for Engineers: Statics Sample Problem 5.4 • Evaluate the centroid coordinates. xA Q y x ab 2 3 a b x 4 3 a 4 yA Q x y ab 3 © 2003 The McGraw-Hill Companies, Inc. All rights reserved. ab 10 2 y 3 b 10 5 - 18 Seventh Edition Vector Mechanics for Engineers: Statics Distributed Loads on Beams L W w dx dA A 0 OP W x dW L OP A x dA x A 0 • A distributed load is represented by plotting the load per unit length, w (N/m) . The total load is equal to the area under the load curve. • A distributed load can be replace by a concentrated load with a magnitude equal to the area under the load curve and a line of action passing through the area centroid. © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 5 - 19 Seventh Edition Vector Mechanics for Engineers: Statics Sample Problem 5.9 SOLUTION: • The magnitude of the concentrated load is equal to the total load or the area under the curve. • The line of action of the concentrated load passes through the centroid of the area under the curve. A beam supports a distributed load as shown. Determine the equivalent concentrated load and the reactions at the supports. • Determine the support reactions by summing moments about the beam ends. © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 5 - 20 Seventh Edition Vector Mechanics for Engineers: Statics Sample Problem 5.9 SOLUTION: • The magnitude of the concentrated load is equal to the total load or the area under the curve. F 18 . 0 kN • The line of action of the concentrated load passes through the centroid of the area under the curve. X 63 kN m X 3 .5 m 18 kN © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 5 - 21 Seventh Edition Vector Mechanics for Engineers: Statics Sample Problem 5.9 • Determine the support reactions by summing moments about the beam ends. M A 0: B y 6 m 18 kN 3 .5 m 0 B y 10 . 5 kN M B 0 : A y 6 m 18 kN 6 m 3 .5 m 0 A y 7 . 5 kN © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 5 - 22 Seventh Edition Vector Mechanics for Engineers: Statics Center of Gravity of a 3D Body: Centroid of a Volume • Center of gravity G W j rG W j rG W j W dW • Results are independent of body orientation, W j r W j r W j rG W r dW xW xdW yW ydW zW zdW • For homogeneous bodies, W V and dW dV xV © 2003 The McGraw-Hill Companies, Inc. All rights reserved. xdV yV ydV zV zdV 5 - 23 Seventh Edition Vector Mechanics for Engineers: Statics Centroids of Common 3D Shapes © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 5 - 24 Seventh Edition Vector Mechanics for Engineers: Statics Composite 3D Bodies • Moment of the total weight concentrated at the center of gravity G is equal to the sum of the moments of the weights of the component parts. X W xW Y W yW Z W zW • For homogeneous bodies, X V xV © 2003 The McGraw-Hill Companies, Inc. All rights reserved. Y V yV Z V zV 5 - 25 Seventh Edition Vector Mechanics for Engineers: Statics Sample Problem 5.12 SOLUTION: • Form the machine element from a rectangular parallelepiped and a quarter cylinder and then subtracting two 1-in. diameter cylinders. Locate the center of gravity of the steel machine element. The diameter of each hole is 1 in. © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 5 - 26 Seventh Edition Vector Mechanics for Engineers: Statics Sample Problem 5.12 © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 5 - 27 Seventh Edition Vector Mechanics for Engineers: Statics Sample Problem 5.12 X x V V 3 . 08 in 4 5 .286 in 3 X 0 . 577 in. Y y V V 5.047 in 4 5 .286 in 3 Y 0 . 577 in. Z z V V 1 .618 in 4 5 .286 in 3 Z 0 . 577 in. © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 5 - 28