Centroids and Centers of Gravity 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 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

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.

Center of Gravity of a 2D Body Center of gravity of a plate Center of gravity of a wire

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

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 two lines of symmetry, its centroid lies at their intersection. If an area possesses a line of symmetry, its centroid lies on that axis 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.

Centroids of Common Shapes of Areas

Centroids of Common Shapes of Lines

Composite Plates and Areas Composite area

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.

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.

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

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

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.

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

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

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

Sample Problem 7.2 Evaluate the centroid coordinates.

Theorems of Pappus-Guldinus 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 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.

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. 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.

Distributed Loads on Beams 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.

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.

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.

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

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

Centroids of Common 3D Shapes

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. For homogeneous bodies,

Sample Problem 7.5 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.

Sample Problem 7.5

Sample Problem 7.5

Problem 7.6 Locate the centroid of the plane area shown. y x 20 mm

Solving Problems on Your Own 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.

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

y Problem 7.6 Solution 20 + 10 Construct a table containing areas and respective coordinates of the centroids. C1 C2 30 24 + 12 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

XS A = S xA X (1740) = 28,200 X = 16.21 mm YS A = S yA Problem 7.6 Solution 20 + 10 Then XS A = S xA X (1740) = 28,200 X = 16.21 mm C1 or C2 and YS A = S yA 30 24 + 12 Y (1740) = 55,440 x 10 Y = 31.9 mm Dimensions in mm or 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

Problem 7.7 a 24 kN 30 kN 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. 0.3 m A B wA wB 1.8 m

Solving Problems on Your Own a 24 kN 30 kN 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.

RII = (1.8 m)(wB kN/m) = 0.9 wB kN Problem 7.7 Solution a 24 kN 30 kN 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 1 2 We have RI = (1.8 m)(20 kN/m) = 18 kN 1 2 RII = (1.8 m)(wB kN/m) = 0.9 wB kN

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

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

X S V = S x V Y S V = S y V Z S V = S z V 0.75 in z 1 in 2 in 3 in r = 1.25 in Problem 7.8 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.

Determine the center of gravity of composite body. x 0.75 in z 1 in 2 in 3 in r = 1.25 in Problem 7.8 Solution 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. y x z I II III IV V Divide the body into five common shapes.

II (p/2)(2)2 (0.75) = 4.7124 7+ [(4)(2)/(3p)] = 7.8488 36.987 y x z I II III IV V y x 0.75 in z 1 in 2 in 3 in r = 1.25 in V, in3 z, 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 in4 Z = 3.47 in

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

Solving Problems on Your Own 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.

Problem 7.9 Solution y Use symmetry to help locate the centroid. Symmetry implies x dx y = 0 z = 0 z x 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

dV = p r2 dx xel = x r = kx 1/3 dV = p k2 x2/3dx a = kh1/3 k = a/h1/3 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 z x r Now r = kx 1/3 so that y = kx1/3 dV = p k2 x2/3dx At x = h, y = a : a = kh1/3 or k = a/h1/3 a2 h2/3 Then dV = p x2/3dx

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

ò ò V = p a2h xel dV = p a2h2 xV = xdV: x ( p a2h) = p a2h2 x = h Problem 7.9 Solution y Integrate and determine the centroid. x dx 3 5 V = p a2h ò 3 8 xel dV = p a2h2 z x r y = kx1/3 ò 3 5 3 8 Now xV = xdV: x ( p a2h) = p a2h2 x = h 5 8 y = 0 z = 0

Problem 7.10 The square gate AB is held in the position shown by hinges along its top edge A and by a shear pin at B. For a depth of water d = 3.5 ft, determine the force exerted on the gate by the shear pin. A d 1.8 ft B 30o

Solving Problems on Your Own The square gate AB is held in the position shown by hinges along its top edge A and by a shear pin at B. For a depth of water d = 3.5 ft, determine the force exerted on the gate by the shear pin. A d 1.8 ft B 30o Assuming the submerged body has a width b, the load per unit length is w = brgh, where h is the distance below the surface of the fluid. 1. First, determine the pressure distribution acting perpendicular the surface of the submerged body. The pressure distribution will be either triangular or trapezoidal.

Solving Problems on Your Own The square gate AB is held in the position shown by hinges along its top edge A and by a shear pin at B. For a depth of water d = 3.5 ft, determine the force exerted on the gate by the shear pin. A d 1.8 ft B 30o 2. Replace the pressure distribution with a resultant force, and construct the free-body diagram. 3. Write the equations of static equilibrium for the problem, and solve them.

Problem 7.10 Solution Determine the pressure distribution acting perpendicular the surface of the submerged body. 1.7 ft A PA PA = 1.7 rg PB = (1.7 + 1.8 cos 30o)rg (1.8 ft) cos 30o B PB

P1 = (1.8 ft)2(62.4 lb/ft3)(1.7 ft) = 171.85 lb Ay Problem 7.10 Solution A Ax Replace the pressure distribution with a resultant force, and construct the free-body diagram. 1.7 rg (1.8 ft) cos 30o LAB/3 P1 FB LAB/3 The force of the water on the gate is P2 B LAB/3 1 2 1 2 (1.7 + 1.8 cos 30o)rg P = Ap = A(rgh) 1 2 P1 = (1.8 ft)2(62.4 lb/ft3)(1.7 ft) = 171.85 lb 1 2 P2 = (1.8 ft)2(62.4 lb/ft3)(1.7 + 1.8 cos 30o)ft = 329.43 lb

S MA = 0: ( LAB)P1 + ( LAB)P2 - LABFB = 0 Ay Problem 7.10 Solution A Ax Write the equations of static equilibrium for the problem, and solve them. 1.7 rg (1.8 ft) cos 30o LAB/3 P1 FB S MA = 0: LAB/3 + P2 B 1 3 2 3 ( LAB)P1 + ( LAB)P2 LAB/3 (1.7 + 1.8 cos 30o)rg - LABFB = 0 P1 = 171.85 lb P2 = 329.43 lb 1 3 2 3 (171.85 lb) + (329.43 lb) - FB = 0 FB = 276.90 lb 30o FB = 277 lb