Tenth E CHAPTER 5 VECTOR MECHANICS FOR ENGINEERS: STATICS Ferdinand P. Beer E. Russell Johnston, Jr. David F. Mazurek Lecture Notes: John Chen California Polytechnic State University Distributed Forces: Centroids and Centers of Gravity 013 The McGraw-Hill Companies, Inc. All rights reserved.!
Contents Introduction Center of Gravity of a D Body Centroids and First Moments of Areas and Lines Sample Problem 5.4 Centroids of Common Shapes of Areas Centroids of Common Shapes of Lines Composite Plates and Areas 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 5.1 Sample Problem 5.1 Determination of Centroids by Integration 5 -
Application There are many eamples in engineering analysis of distributed loads. It is convenient in some cases to represent such loads as a concentrated force located at the centroid of the distributed load. 5-3
Introduction The earth eerts a gravitational force on each of the particles forming a body consider how your weight is distributed throughout your body. These forces can be replaced 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; it is the center of area. 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. 5-4
Center of Gravity of a D Body Center of gravity of a plate Center of gravity of a wire M y W ΔW M y yw dw yδw y dw 5-5
First Moments of Areas and Lines An area is symmetric with respect to an ais BB if for every point P there eists 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 ais 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 (,y) there eists an area da of equal area at (-,-y). The centroid of the area coincides with the center of symmetry. 5-6
Centroids of Common Shapes of Areas 5-7
Centroids of Common Shapes of Lines 5-8
Composite Plates and Areas Composite plates X Y W W W yw Composite area X Y A A A ya 5-9
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 aes. For the plane area shown, determine the first moments with respect to the and y aes 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. 5-10
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. Q Q y + 506. 10 + 757.7 10 3 3 mm mm 3 3 5-11
Sample Problem 5.1 Compute the coordinates of the area centroid by dividing the first moments by the total area. X A A + 757.7 10 13.88 10 3 3 mm mm 3 X 54.8 mm Y ya A + 506. 10 13.88 10 3 3 mm mm 3 Y 36.6 mm 5-1
Centroids and First Moments of Areas and Lines Centroid of an area Centroid of a line W dw ( γat) ( γt) A da da Q y first moment with respect to ya y da Q first moment with respect to y W dw ( γ La) ( γ a) L dl yl y dl dl 5-13
Determination of Centroids by Integration A ya da yda ddy y ddy y el el da da Double integration to find the first moment may be avoided by defining da as a thin rectangle or strip. A ya y y el da ( yd) el da ( yd) A ya a + y da y el el da [( a ) d] [( a ) d] A ya y el el da r 1 cosθ 3 da r 1 sinθ 3 r r dθ dθ 5-14
Sample Problem 5.4 SOLUTION: Determine the constant k. Evaluate the total area. Determine by direct integration the location of the centroid of a parabolic spandrel. Using either vertical or horizontal strips, perform a single integration to find the first moments. Evaluate the centroid coordinates. First, estimate the location of the centroid by inspection. Discuss with a neighbor where it is located, roughly, and justify your answer. 5-15
Sample Problem 5.4 SOLUTION: Determine the constant k. y b y k k a a b k or a b b a 1 y 1 Evaluate the total area. A da ab 3 y d a 0 a b d b a 3 a 3 0 5-16
Sample Problem 5.4 Using vertical strips, perform a single integration to find the first moments. Q y Q a el da yd b a b 4 a 4 a 0 a b 4 y el da y yd 1 5 a 4 5 b a 0 ab 10 0 a 0 d b a d 5-17
Sample Problem 5.4 Or, using horizontal strips, perform a single integration to find the first moments. Try calculating Q y or Q by this method, and confirm that you get the same value as before. Q y el da a + b ( a )dy a 0 1 b a a b y dy a b 0 4 Q y el da y( a )dy y a a b ay a b y 3 1 dy 0 ab 10 dy b 1 y1 dy 5-18
Sample Problem 5.4 Evaluate the centroid coordinates. A Q y ab 3 a b 4 3 4 a ya ab y 3 Q ab 10 y 3 10 b Is this center of area close to where you estimated it would be? 5-19
Determination of Centroids by Integration Usually, the choice between using a vertical or horizontal strip is equally good, but in some cases, one choice is much better than the other. For eample, for the area shown below, is a vertical or horizontal strip a better choice, and why? Think about this and discuss your choice with a neighbor. y 5-0
Theorems of Pappus-Guldinus Surface of revolution is generated by rotating a plane curve about a fied ais. Area of a surface of revolution is equal to the length of the generating curve times the distance traveled by the centroid through the rotation. A π yl 5-1
Theorems of Pappus-Guldinus Body of revolution is generated by rotating a plane area about a fied ais. Volume of a body of revolution is equal to the generating area times the distance traveled by the centroid through the rotation. V π ya 5 -
Sample Problem 5.7 SOLUTION: Apply the theorem of Pappus-Guldinus to evaluate the volumes of revolution of the pulley, which we will form as a large rectangle with an inner rectangular cutout. The outside diameter of a pulley is 0.8 m, and the cross section of its rim is as shown. Knowing that the pulley is made of steel and that the density of 3 3 steel is ρ 7.85 10 kg m determine the mass and weight of the rim. Multiply by density and acceleration to get the mass and weight. 5-3
Sample Problem 5.7 SOLUTION: Apply the theorem of Pappus-Guldinus to evaluate the volumes or revolution for the rectangular rim section and the inner cutout section. Multiply by density and acceleration to get the mass and weight. ( )( 7.65 106 mm 3 )( 10 9 m 3 /mm 3 ) m 60.0 kg ( )( 9.81 m s ) W 589 N m ρv 7.85 10 3 kg m 3 W mg 60.0 kg 5-4
Distributed Loads on Beams W L wd da 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. ( ) OP W L dw ( OP) A da A 0 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. 5-5
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 (a) drawing the free body diagram for the beam and (b) applying the cons of equilibrium. 5-6
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 18 kn X 3.5 m 5-7
Sample Problem 5.9 Determine the support reactions by applying the equilibrium cons. For eample, successively sum the moments at the two supports: 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 And by summing forces in the -direction: F 0 : B 0 5-8
Center of Gravity of a 3D Body: Centroid of a Volume Center of gravity G!! W j ΔW! r! r G G W W! ( j ) ( W j ) [ r ( ΔW j )]!!! ( j ) ( rδw ) ( j )!! rdw! dw rgw! Results are independent of body orientation, W V dw yw ydw zw zdw For homogeneous bodies, W γ V and dw γ dv dv yv ydv zv zdv 5-9
Centroids of Common 3D Shapes 5-30
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 W Y W yw Z W zw For homogeneous bodies, X V V Y V yv Z V zv 5-31
Sample Problem 5.1 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. 5-3
Sample Problem 5.1 5-33
Sample Problem 5.1 X V V ( 3.08 in 4 ) ( 5.86 in 3 ) X 0.577 in. Y y V V ( 5.047 in 4 ) ( 5.86 in 3 ) Y 0.577 in. Z z V V ( 1.618 in 4 ) ( 5.86 in 3 ) Z 0.577 in. 5-34