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CE 102 Statics

CE 102 Statics. Chapter 7 Distributed Forces: Centroids and Centers of Gravity. 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

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CE 102 Statics

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  1. CE 102 Statics Chapter 7 Distributed Forces: Centroids and Centers of Gravity

  2. 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 7.1 Determination of Centroids by Integration Sample Problem 7.2 Theorems of Pappus-Guldinus Sample Problem 7.3 Distributed Loads on Beams Sample Problem 7.4 Center of Gravity of a 3D Body: Centroid of a Volume Centroids of Common 3D Shapes Composite 3D Bodies Sample Problem 7.5

  3. 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.

  4. Center of gravity of a plate • Center of gravity of a wire Center of Gravity of a 2D Body

  5. Centroid of an area • Centroid of a line Centroids and First Moments of Areas and Lines

  6. 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’. • 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). First Moments of Areas and Lines • 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 • The centroid of the area coincides with the center of symmetry.

  7. Centroids of Common Shapes of Areas

  8. Centroids of Common Shapes of Lines

  9. Composite plates • Composite area Composite Plates and Areas

  10. Sample Problem 7.1 • 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. • Find the total area and first moments of the triangle, rectangle, and semicircle. Subtract the area and first moment of the circular cutout. For the plane area shown, determine the first moments with respect to the x and y axes and the location of the centroid. • Compute the coordinates of the area centroid by dividing the first moments by the total area.

  11. Sample Problem 7.1 • Find the total area and first moments of the triangle, rectangle, and semicircle. Subtract the area and first moment of the circular cutout.

  12. Sample Problem 7.1 • Compute the coordinates of the area centroid by dividing the first moments by the total area.

  13. Double integration to find the first moment may be avoided by defining dA as a thin rectangle or strip. Determination of Centroids by Integration

  14. Sample Problem 7.2 • 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.

  15. SOLUTION: • Determine the constant k. • Evaluate the total area. Sample Problem 7.2

  16. Sample Problem 7.2 • Using vertical strips, perform a single integration to find the first moments.

  17. Sample Problem 7.2 • Or, using horizontal strips, perform a single integration to find the first moments.

  18. Evaluate the centroid coordinates. Sample Problem 7.2

  19. 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. Theorems of Pappus-Guldinus

  20. 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. Theorems of Pappus-Guldinus

  21. 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 steel is determine the mass and weight of the rim. Sample Problem 7.3 • 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 acceleration.

  22. 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 acceleration. Sample Problem 7.3

  23. 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 (dW = wdx). • 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. Distributed Loads on Beams

  24. Sample Problem 7.4 • 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. • Determine the support reactions by summing moments about the beam ends. A beam supports a distributed load as shown. Determine the equivalent concentrated load and the reactions at the supports.

  25. 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. Sample Problem 7.4

  26. Sample Problem 7.4 • Determine the support reactions by summing moments about the beam ends.

  27. Center of gravity G • Results are independent of body orientation, • For homogeneous bodies, Center of Gravity of a 3D Body: Centroid of a Volume

  28. Centroids of Common 3D Shapes

  29. 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. • For homogeneous bodies, Composite 3D Bodies

  30. SOLUTION: • Form the machine element from a rectangular parallelepiped and a quarter cylinder and then subtracting two 1-in. diameter cylinders. Sample Problem 7.5 Locate the center of gravity of the steel machine element. The diameter of each hole is 1 in.

  31. Sample Problem 7.5

  32. Sample Problem 7.5

  33. Problem 7.6 y 20 mm 30 mm Locate the centroid of the plane area shown. 36 mm 24 mm x

  34. y Problem 7.6 20 mm 30 mm Solving Problems on Your Own Locate the centroid of the plane area shown. 36 mm Several points should be emphasized when solving these types of problems. 24 mm x 1. Decide how to construct the given area from common shapes. 2. It is strongly recommended that you construct a table containing areas or length and the respective coordinates of the centroids. 3. When possible, use symmetry to help locate the centroid.

  35. Problem 7.6 Solution y 20 + 10 Decide how to construct the given area from common shapes. C1 C2 24 + 12 30 x 10 Dimensions in mm

  36. Problem 7.6 Solution y 20 + 10 Construct a table containing areas and respective coordinates of the centroids. C1 C2 24 + 12 30 x 10 Dimensions in mm A, mm2 x, mm y, mm xA, mm3 yA, mm3 1 20 x 60 =1200 10 30 12,000 36,000 2 (1/2) x 30 x 36 =540 30 36 16,200 19,440 S 1740 28,200 55,440

  37. Problem 7.6 Solution y 20 + 10 XSA = S xA Then X (1740) = 28,200 X = 16.21 mm or C1 C2 YSA = S yA and 24 + 12 30 Y (1740) = 55,440 x 10 Y = 31.9 mm or Dimensions in mm A, mm2 x, mm y, mm xA, mm3 yA, mm3 1 20 x 60 =1200 10 30 12,000 36,000 2 (1/2) x 30 x 36 =540 30 36 16,200 19,440 S 1740 28,200 55,440

  38. Problem 7.7 The beam AB supports two concentrated loads and rests on soil which exerts a linearly distributed upward load as shown. Determine (a) the distance a for which wA = 20 kN/m, (b) the corresponding value wB. 24 kN 30 kN a 0.3 m A B wA wB 1.8 m

  39. Problem 7.7 Solving Problems on Your Own 24 kN 30 kN a 0.3 m The beam AB supports two concentrated loads and rests on soil which exerts a linearly distributed upward load as shown. Determine (a) the distance a for which wA = 20 kN/m, (b) the corresponding value wB. A B wA wB 1.8 m 1. Replace the distributed load by a single equivalent force. The magnitude of this force is equal to the area under the distributed load curve and its line of action passes through the centroid of the area. 2. When possible, complex distributed loads should be divided into common shape areas.

  40. 1 2 1 2 Problem 7.7 Solution 24 kN 30 kN a 0.3 m Replace the distributed load by a pair of equivalent forces. C A B 20 kN/m wB 0.6 m 0.6 m RI RII RI = (1.8 m)(20 kN/m) = 18 kN We have RII = (1.8 m)(wB kN/m) = 0.9 wB kN

  41. + + Problem 7.7 Solution 24 kN 30 kN a 0.3 m C A B wB 0.6 m 0.6 m RII = 0.9 wB kN RI = 18 kN (a) SMC = 0: (1.2 - a)m x 24 kN - 0.6 m x 18 kN - 0.3m x 30 kN = 0 or a = 0.375 m (b) SFy = 0: -24 kN + 18 kN + (0.9 wB) kN - 30 kN= 0 or wB = 40 kN/m

  42. y 2 in 1 in 2 in 3 in r = 1.25 in x z 0.75 in 2 in r = 1.25 in 2 in Problem 7.8 For the machine element shown, locate the z coordinate of the center of gravity.

  43. y Problem 7.8 2 in 1 in 2 in 3 in r = 1.25 in x z 0.75 in 2 in r = 1.25 in 2 in Solving Problems on Your Own For the machine element shown, locate the z coordinate of the center of gravity. Determine the center of gravity of composite body. For a homogeneous body the center of gravity coincides with the centroid of its volume. For this case the center of gravity can be determined by X S V = S x V Y S V = S y V Z S V = S z V where X, Y, Z and x, y, z are the coordinates of the centroid of the body and the components, respectively.

  44. y Problem 7.8 Solution 2 in 1 in 2 in 3 in r = 1.25 in x z 0.75 in 2 in r = 1.25 in 2 in y V IV I III x II z Determine the center of gravity of composite body. First assume that the machine element is homogeneous so that its center of gravity will coincide with the centroid of the corresponding volume. Divide the body into five common shapes.

  45. y y 2 in 1 in V 2 in IV 3 in r = 1.25 in I III x II x z 0.75 in z 2 in r = 1.25 in 2 in V, in3z, in. z V, in4 I (4)(0.75)(7) = 21 3.5 73.5 II (p/2)(2)2 (0.75) = 4.7124 7+ [(4)(2)/(3p)] = 7.8488 36.987 III -p(11.25)2 (0.75)= -3.6816 7 -25.771 IV (1)(2)(4) = 8 2 16 V -(p/2)(1.25)2 (1) = -2.4533 2 -4.9088 S 27.576 95.807 Z S V = S z V : Z (27.576 in3 ) = 95.807 in4Z = 3.47 in

  46. Problem 7.9 y y = kx1/3 Locate the centroid of the volume obtained by rotating the shaded area about the x axis. a x h

  47. y Problem 7.9 y = kx1/3 Solving Problems on Your Own Locate the centroid of the volume obtained by rotating the shaded area about the x axis. a The procedure for locating the centroids of volumes by direct integration can be simplified: x h 1. When possible, use symmetry to help locate the centroid. 2. If possible, identify an element of volume dV which produces a single or double integral, which are easier to compute. 3. After setting up an expression for dV, integrate and determine the centroid.

  48. Problem 7.9 Solution y Use symmetry to help locate the centroid. Symmetry implies x dx y = 0 z = 0 x z r Identify an element of volume dV which produces a single or double integral. y = kx1/3 Choose as the element of volume a disk or radius r and thickness dx. Then dV = p r2 dx xel = x

  49. Problem 7.9 Solution y x Identify an element of volume dV which produces a single or double integral. dx dV = p r2 dx xel = x x z r r = kx1/3 so that Now dV = p k2 x2/3dx y = kx1/3 a = kh1/3 k = a/h1/3 or At x = h, y = a : a2 h2/3 dV = p x2/3dx Then

  50. a2 h2/3 dV = p x2/3dx 3 5 h ò 0 3 8 = p a2h2 Problem 7.9 Solution y Integrate and determine the centroid. x dx a2 h2/3 h ò V = p x2/3dx x z r 0 a2 h2/3 h [ ] 3 5 = p x5/3 y = kx1/3 0 = p a2h a2 h2/3 a2 h2/3 3 8 ò xel dV = x (p x2/3 dx) = p [ x8/3 ] Also

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