Chapter 5. Distributed Forces: Centroids and Centers of Gravity

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1 Chapter 5 Distributed Forces: Centroids and Centers of Gravity

2 Application There are many examples 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. 5-2

3 Introduction The earth exerts 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. 5-3

4 Center of Mass The center of mass of a system is the point where the system can be balanced in a uniform gravitational field.

5 9.2 The Center of Mass: A System of Particles Consider a situation in which n particles are strung out along the x axis. Let the mass of the particles are m 1, m 2,.m n, and let them be located at x 1, x 2, x n respectively. Then if the total mass is M = m 1 + m m n, then the location of the center of mass, x com, is

6 For two objects: Center of Mass The center of mass is closer to the more massive object.

7 Center of Mass The center of mass need not be within the object:

8 Moment The moment of a mass is a measure of its tendency to rotate about a point. Clearly, the greater the mass (and the greater the distance from the point), the greater will be the tendency to rotate. The moment is defined as: Moment = mass distance from a point Example 1 In this case, there will be a total moment about O of: (Clockwise is regarded as positive in this work.) M= = 28 kgm

9 Center of Mass We now aim to find the center of mass of the system and this will lead to a more general result. We have 3 masses of 10 kg, 5 kg and 7 kg at 2 m, 2 m and 1 m distance from O as shown. We wish to replace these masses with one single mass to give an equivalent moment. Where should we place this single mass? Total moment = =75 kg.m If we put the masses together, we have: =22 kg For an equivalent moment, we need: 22 d =75 where d is the distance from the center of mass to the point of rotation. i.e. d = m So our equivalent system (with one mass of 22 kg ) would have:

10 Center of Gravity of a 2D Body Center of gravity of a plate Center of gravity of a wire M y xw xw M y yw x dw yw y dw 5-10

11 Centroids and First Moments of Areas and Lines Centroid of an area Centroid of a line x xw x dw At x t xa x da da Q y first moment wit h respect to ya y da Qx first moment wit h respect to y x x xw x dw La x a xl x dl yl y dl dl 5-11

12 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. 5-12

13 Centroids of Common Shapes of Areas 5-13

14 Centroids of Common Shapes of Lines 5-14

15 Composite Plates and Areas Composite plates X Y W W xw yw Composite area X Y A A x A ya 5-15

16 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. 5-16

17 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 x y mm mm

18 Sample Problem 5.1 Compute the coordinates of the area centroid by dividing the first moments by the total area. X x A A mm mm 3 2 X 54.8 mm Y ya A mm mm 3 2 Y 36.6 mm 5-18

19 Theorems of Pappus-Guldinus: The storage tanks shown are all bodies of revolution. Thus, their surface area and volumes can be determined using Theorems of Pappus-Goldinus. 2-19

20 Theorems of Pappus-Guldinus----Surface Area Calculation Surface of revolution is generated by rotating a plane curve about a fixed axis. 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 2 yl 5-20

21 Pappus's Centroid Theorem Surface Area Calculation The first theorem of Pappus states that the surface area of a surface of revolution generated by the revolution of a curve about an external axis is equal to the product of the arc length of the generating curve and the distance traveled by the curve's geometric centroid, The following table summarizes the surface areas calculated using Pappus's centroid theorem for various surfaces of revolution. 5-21

22 Pappus's Centroid Theorem Surface Area Calculations solid cone generating curve inclined line segment cylinder sphere parallel line segment semicircle 5-22

23 Theorems of Pappus-Guldinus---Volume Calculation Body of revolution is generated by rotating a plane area about a fixed axis. Volume of a body of revolution is equal to the generating area times the distance traveled by the centroid through the rotation. V 2 ya 5-23

24 Pappus's Centroid Theorem Volume Calculations Volume calculations: Similarly, the second theorem of Pappus states that the volume of a solid of revolution generated by the revolution of a lamina about an external axis is equal to the product of the area of the lamina and the distance traveled by the lamina's geometric centroid, The following table summarizes the surface areas and volumes calculated using Pappus's centroid theorem for various solids and surfaces of revolution. 5-24

25 Pappus's Centroid Theorem---Volume Calculation solid generating lamina cone right triangle cylinder rectangle sphere semicircle 5-25

26 Sample Problem 5.7 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 kg m determine the mass and weight of the rim. 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. Multiply by density and acceleration to get the mass and weight. 5-26

27 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 mm m 3 /mm 3 m 60.0 kg 9.81 m s 2 W 589 N m V kg m 3 W mg 60.0 kg 5-27

28 Distributed Loads on Beams W L 0 wdx da A 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 xdw OPA xda xa 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. 5-28

29 Sample Problem 5.9 SOLUTION: A beam supports a distributed load as shown. Determine the equivalent concentrated load and the reactions at the supports. 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. Determine the support reactions by (a) drawing the free body diagram for the beam and (b) applying the conditions of equilibrium. 5-29

30 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 5-30

31 Sample Problem 5.9 Determine the support reactions by applying the equilibrium conditions. For example, 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 kn6 m 3.5 m 0 A y 7.5 kn And by summing forces in the x-direction: F x 0: B x