Engineering Problems and Solutions for Area Centroids Calculation
Explore sample engineering problems involving the division of complex areas and the calculation of centroids using moments of the areas with respect to axes. Learn to locate centroids of various shapes like triangles, rectangles, semicircles, and circular arcs. Understand the step-by-step process to determine centroids in plane geometry problems.
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Sample Problem 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. 5 - 2
Sample Problem 1 - Continued First Moments of the Area Subtract the area of circular cutout section. 3 3 = 506 + 2 . 10 mm Q x 3 3 = 757 + 7 . 10 mm Q y 5 - 3
Sample Problem 1 - Continued Compute the coordinates of the centroid by dividing the first moments of areas by the total area as below: 3 3 + 757 7 . 10 mm x A = = X 3 2 A 13.828 10 mm = 54 8 . mm X 3 3 + 506 2 . 10 mm y A = = Y 3 2 A 13.828 10 mm = 36 6 . mm Y 5 - 4
Sample Problem 2 5 - 5
Sample Problem 2 - Continued Part mm2 mm mm mm3 mm3 5 - 6
Sample Prolem 2 - Continued Part3 Part4 Part2 Part1 Part3 Part4 Part2 Part1 Part2 Part3 Part4 Part1 5 - 7
Problem-1 Locate the centroid of a circular arc as shown in the figure. 5/30/2025 9
Solution Since - axis is the axis of symmetry, centroid would on the lie - axis.That is x x cos ( ) r rd xdL 2 sin r = = = = and 0 x x x y 2 2 L r 5/30/2025 10
Problem-2 Locate the centroid of a half and quarter-circular arcs using the formula derived in Problem-1. Solution = x = semicircul a For ar arc 2 2 / r Line of symmetry For quarter a - circular arc : Along the line of symmetry : 0 2 sin 2 sin 45 4 / 2 2 2 r r r r = = = = x the axis 2 2 ( ) 4 / Along vertical : 2 2 1 2 r r = = 0 sin 45 x 2 5/30/2025 11
Determination of Centroids by Integration Double integration to find the first moment may be avoided by defining dA as a thin rectangle or strip. = = = x A x dA x dx dy x dA el = = = y A y dA y dx dy y dA el = = x A x dA x A x dA = x A x dA el el el ( ) + 2 a x = ( ) x ydx = 2 1 a x dy r = d 2 cos r 3 2 = y A y dA = el y A y dA el = y A y dA y 2 ( ) el ( ) = ydx = y a x dy 2 1 r = d 2 sin r 3 2 Rectangular Coordinates Polar Coordinates 5 - 12
Sample Problem 3 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 shown. Evaluate the centroid coordinates. 5 - 13
Sample Problem 3 - Continued SOLUTION: Determine the constant k. 2 x k y = b 2 = = b k a k 2 a b a 2 1 2 = = y x or x y 2 1 2 a b Evaluate the total area. dA A = a 3 a b b x 2 = = = y dx x dx 2 2 3 a a 0 0 ab = 3 5 - 14
Sample Problem 3 - Continued Using vertical strips, perform a single integration to find the first moments. a b 2 = = = Q x dA xy dx x x dx y el 2 a 0 a 4 2 b x a b = = 2 4 4 a 0 2 a 1 y b 2 = = = Q y dA y dx x dx x el 2 2 2 a 0 a 2 5 2 b x ab = = 4 5 10 2 a 0 5 - 15
Sample Problem 3 - Continued Or, using horizontal strips, perform a single integration to find the first moments. 2 2 b + a x a x ( ) = = = Q x dA a x dy dy y el 2 2 0 2 2 b 1 a a b 2 dy = = a y 2 4 b 0 a ( ) 1 2 = = = Q y dA y a x dy y a y dy x el 1 2 b 2 b a ab 3 2 = = ay y dy 1 2 10 b 0 5 - 16
Sample Problem 3 - Continued Evaluate the coordinates of centroid. dA = x A Q y 2b 3 ab a = = x x a dy 3 4 4 dx Alternatively; if we take any differential element dA=dx.dy = y A Q x a 2 3 ab ab Qy= x.dx.dy= [ dy ].x.dx = = y b y 0 0 10 3 10 Qy= a Qx= y.dx.dy= [ y.dy ].dx 0 0 Qx= 5 - 17
Sample Problem 4 5 - 18
Sample Problem 4 - Continued yel=yc= y xel=xc= x1+(x2-x1)/2 = (x1+x2)/2 5 - 19
Sample Problem 5 5 - 20
Sample Problem 5 - Continued 2 1 3 5 - 21
Problem-3 Locate the centroid of the volume of a hemisphere of radius r with respect to its base. 5/30/2025 22
Solution = = symmetry) (By 0 x Element z chosen circular a : slice of thickness dy, parallel to x - plane. z = + 2 2 2 Since hemishpere the intersects the y - plane z in the circle, the , y z r = 2 2 radius of circular the slice = is, z r y = = 2 2 2 ( ) ; dV z dy dV r y dy y y c r r 2 2 2 2 ( ) ( ) y r y dy y r y dy y dV c = = = y y y 0 0 r r V 2 2 ( ) dV r y dy 0 0 1 4 r 3 4 = = y y r 2 8 3 r 3 5/30/2025 23
