Centroids The golfer guessed that his ball landed 20 feet off the fairway. Of course, that was just a rough estimate. Centroids ¢ When we dealt with distributed loads, we found the magnitude of the force generated by the loading as the area under the loading curve. ¢ I gave you the location of the line of action of the force for both a rectangular shape and a right-triangular shape. 2 Centroids by Integration Wednesday, November 7, 2012 1 Centroids ¢ In this meeting, we are going to find out just why that line of action was located where it was. ¢ The line of action was located through the centroidial axis of the loading diagram. ¢ If we took a centroidial axis in every direction, their intersection point would be known as the centroid 3 Centroids by Integration Wednesday, November 7, 2012 Centroid as Balance Point 4 Centroids by Integration Wednesday, November 7, 2012 2 Centroids ¢ By common practice, we refer to the centroidal axis as the centroid but to keep the confusion down we will often speak of a x-centroid or a y-centroid referring to the coordinate along that axis where the centroidal axis intersects the coordinate axis. 5 Centroids by Integration Wednesday, November 7, 2012 Centroidal Axis 6 Centroids by Integration Wednesday, November 7, 2012 3 Centroids ¢ We are going to look at two mathematical techniques for locating this centroidal axis or centroid. ¢ First we will look at what a centroid physically represents 7 Centroids by Integration Wednesday, November 7, 2012 Centroids ¢ Consider that we have a series of rectangular loads along an axis y x 8 Centroids by Integration Wednesday, November 7, 2012 4 Centroids ¢ We would like to replace this loading with a single point force for analysis purposes y x 9 Centroids by Integration Wednesday, November 7, 2012 Centroids ¢ We would label each load y L4 L2 L1 L3 L5 x 10 Centroids by Integration Wednesday, November 7, 2012 5 Centroids ¢ Then find the area of each loading, giving us the force which is located at the center of each area y L4 L2 L1 L3 L5 x 11 Centroids by Integration Wednesday, November 7, 2012 Centroids ¢ The force generated by each loading is equal to the area under the its loading y diagram so FAn = L4 Ln L2 L1 L3 L5 x 12 Centroids by Integration Wednesday, November 7, 2012 6 Centroids ¢ The force generated by each loading is equal to the area under the loading diagram so y F F2 4 F 1 F3 F FAn = 5 Ln x 13 Centroids by Integration Wednesday, November 7, 2012 Centroids ¢ The total force generated by all these forces is just their sum y F F2 4 F F FF A1 3 ==nLn ∑∑ F5 x 14 Centroids by Integration Wednesday, November 7, 2012 7 Centroids ¢ But we want to replace these forces with a single force with the same net effect on the system y F F2 4 F FF== A1 F3 ∑∑nLn F5 x 15 Centroids by Integration Wednesday, November 7, 2012 Centroids ¢ That would mean that it would have to produce the same moment about any point on the system y F F2 4 F 1 F3 FF==nL A ∑∑n F5 x 16 Centroids by Integration Wednesday, November 7, 2012 8 Centroids ¢ If we choose the origin for the moment center, the moment of each of these forces about the origin is equal to the x-distance to the line of action of the force times the force 17 Centroids by Integration Wednesday, November 7, 2012 Centroids ¢ For example, for the force(area) F3 y F4 F2 F F MxF 1 3 333= F5 x x3 18 Centroids by Integration Wednesday, November 7, 2012 9 Centroids ¢ The moment for any force is MxFnnn= 19 Centroids by Integration Wednesday, November 7, 2012 Centroids ¢ Which can be replaced by MxA nnL= n 20 Centroids by Integration Wednesday, November 7, 2012 10 Centroids ¢ And the total moment of all the forces(areas) about the origin is n MxA= total∑ i Li i=1 21 Centroids by Integration Wednesday, November 7, 2012 Centroids ¢ If we now look at the effect of the total force y F x 22 x Centroids by Integration Wednesday, November 7, 2012 11 Centroids ¢ The value x is known as the x-centroid of the loading y F MxF= x x 23 Centroids by Integration Wednesday, November 7, 2012 Centroids ¢ If we substitute the sums developed earlier for the total force and the total moment y F MxF= xF x F ∑∑ii= i x x 24 Centroids by Integration Wednesday, November 7, 2012 12 Centroids ¢ And isolating the centroid we have xF xA x = ∑∑ii or ii FA ∑∑ii 25 Centroids by Integration Wednesday, November 7, 2012 Centroids ¢ A general definition for the x-centroid of a series of n areas would be n xA ∑ ii i=1 x = n A ∑ i i=1 26 Centroids by Integration Wednesday, November 7, 2012 13 Centroids ¢ If the areas represent forces, the centroid represents a center of force n xF ∑ ii i=1 x = n F ∑ i i=1 27 Centroids by Integration Wednesday, November 7, 2012 Centroids ¢ If we are considering weight, and x is the axis parallel to the ground, the xbar would be the center of gravity. n xF ∑ ii i=1 x = n F ∑ i i=1 28 Centroids by Integration Wednesday, November 7, 2012 14 Centroids ¢ If the areas represent masses, the centroid represents the center of mass n xm ∑ ii i=1 x = n m ∑ i i=1 29 Centroids by Integration Wednesday, November 7, 2012 Centroids ¢ If the areas are just areas, the centroid represents the center of area or centroid n xA ∑ ii i=1 x = n A ∑ i i=1 30 Centroids by Integration Wednesday, November 7, 2012 15 Centroids ¢ This is the general formulation for finding the x-centroid of n areas n xA ∑ ii i=1 x = n A ∑ i i=1 31 Centroids by Integration Wednesday, November 7, 2012 Centroids ¢ The same type of formula could be found for finding the y centroid n n xA yA ∑ ii ∑ ii i=1 i=1 x = n y = n A A ∑ i ∑ i i=1 i=1 32 Centroids by Integration Wednesday, November 7, 2012 16 Centroids ¢ Remember that the xi is the x-distance to the centroid of the ith area n xA ∑ ii i=1 x = n A ∑ i i=1 33 Centroids by Integration Wednesday, November 7, 2012 Centroids from Functions ¢ So far, we have been able to describe the forces (areas) using rectangles and triangles. ¢ Now we have to extend that to loadings and areas that are described by mathematical functions. 34 Centroids by Integration Wednesday, November 7, 2012 17 Centroids from Functions ¢ For example y x 35 Centroids by Integration Wednesday, November 7, 2012 Centroids from Functions ¢ This is a distributed load that at any x has 2 a load intensity of w0x y 2 w0x x 36 Centroids by Integration Wednesday, November 7, 2012 18 Centroids from Functions ¢ w0 is a proportionality constant that will have units to make sure that the units of 2 the product w0x will be in force per length units y 2 w0x x 37 Centroids by Integration Wednesday, November 7, 2012 Centroids from Functions ¢ If we had this type of loading over a distance L, how would we find the equivalent point force and its location? y 2 wx0 x 38 Centroids by Integration L Wednesday, November 7, 2012 19 Riemann Sums 39 Centroids by Integration Wednesday, November 7, 2012 Centroids from Functions ¢ We could generate a series of rectangles to lay over the curve y 2 wx0 x L 40 Centroids by Integration Wednesday, November 7, 2012 20 Centroids from Functions ¢ Each rectangle will have some width Δx y Δx 2 wx0 x L 41 Centroids by Integration Wednesday, November 7, 2012 Centroids from Functions ¢ And a height based on where the rectangle is positioned 2 y wx0 Δx w x 2 Δx 0 b w x 2 0 a x xa xb 42 Centroids by IntegrationL Wednesday, November 7, 2012 21 Centroids from Functions ¢ Generalizing for any rectangle 2 y wx0 Δx 2 w0xi x xi 43 Centroids by IntegrationL Wednesday, November 7, 2012 Centroids from Functions ¢ So for n-rectangles the approximate area under the curve would be n n A = A = w x2Δx ∑ i ∑ 0 i wx2 0 i=1 i=1y Δx 2 w0xi x xi 44 Centroids by Integration L Wednesday, November 7, 2012 22 Centroids from Functions ¢ The total moment generated by these areas would be nn MxAxwxx2 ==∑∑ii i( 0 i) Δ ii==112 y wx0 Δx 2 w0xi x xi 45 Centroids by Integration Wednesday, November 7, 2012 L Centroids from Functions ¢ And the location of the centroid would be n n x A x A 2 ∑ i i = ∑ i wx i=1 i=1 y 0 n n Δx x w x2 x x w x2 x ∑ i ( 0 i )Δ = ∑( 0 i )Δ i=1 i=1 2 n w0xi 2 x w x Δx x x ∑ i ( 0 i ) i i=1 x = n w x2 x L ∑( 0 i )Δ i 1 = 46 Centroids by Integration Wednesday, November 7, 2012 23 Centroids from Functions ¢ The general form would be nn xAii= x A i ∑∑ 2 ii==11y wx0 n Δx xAii ∑ 2 w x i=1 0 i x = n x xi A ∑ i L i=1 47 Centroids by Integration Wednesday, November 7, 2012 Centroids from Functions ¢ So for any loading that we can break up into n individual loadings with known centroids, the centroid of the composite would be equal to nn xA x A ∑∑ii= i ii==11 n xA ∑ ii i=1 x = n A ∑ i 48 Centroids by Integration i=1 Wednesday, November 7, 2012 24 Centroids from Functions ¢ If we reduce the width of the rectangles to a differential size, the summation become an integral q(xi) represents a general L loading function y xqii( x) dx ∫ dx 0 x = L q(x) x qx dx i ∫ ( i ) 0 49 Centroids by Integration Wednesday, November 7, 2012 Centroids from Functions ¢ If we can define the height of the loading diagram at any point x by the function q(x), then we can generalize out summations of areas by the quotient of the integrals y L dx xq x dx ∫ ii( ) q(x) x = 0 L xi qx dx ∫ ( i ) 50 0 Centroids by Integration Wednesday, November 7, 2012 25 Centroids from Functions ¢ This is the general form for the integral to locate the centroid xq( x) dx x = ∫A qxdx ∫A ( ) 51 Centroids by Integration Wednesday, November 7, 2012 Centroids from Functions ¢ It isn’t always quite that simple ¢ You have to be careful in l Knowing the height of your rectangular section l Knowing the limits of integration l