Explanation Refer the figure P5.49 and figure given below. Write the equation for curve y . y = a cos ( π x 2 L ) Here, y is y coordinate, a is an unknown constant, L is a length on x-axis and x is the x-coordinate. In the figure given above, black dot denotes the center of mass of rectangular differential area element. Write the expression for the x-coordinate of center of mass of triangular shaped differential area element. x ¯ E L = x Here, x ¯ E L is the center of mass of element d A . Write the expression for the y-coordinate of center of mass of rectangular differential area element. y ¯ E L = y 2 Here, y ¯ E L is the y-coordinate of center of mass of differential area element. Rewrite the above relation by substituting a cos ( π x 2 L ) for y . y ¯ E L = a 2 cos ( π x 2 L ) Write the expression to calculate the rectangular differential area element. d A = y d x Here, d A is the differential area element and d x denotes a small change in x . Rewrite the above relation by substituting for y . d A = a cos ( π x 2 L ) d x Write the expression to calculate the total area in P5.48. A = ∫ d A Here, A is the total area. Rewrite the above equation by substituting a cos ( π x 2 L ) d x for d A and integrate it from x = 0 to. A = ∫ 0 L / 2 a cos ( π x 2 L ) d x = a [ 2 L π sin ( π x 2 L ) ] 0 L / 2 = 2 π a L Calculate ∫ x ¯ E L d A by substituting a cos ( π x 2 L ) d x for d A and x for x ¯ E L . ∫ x ¯ E L d A = ∫ 0 L / 2 ( x ) a cos ( π x 2 L ) d x = a ∫ 0 L / 2 x cos ( π x 2 L ) d x (I) Apply the integration by parts method to solve the above integral. The required formula is given below. ∫ u ( x ) v ( x ) d x = u ( x ) ∫ v ( x ) d x − ∫ [ ( d u ( x ) d x ) ∫ v ( x ) d x ] Here, u ( x ) and v ( x ) are functions of x . Substitute x for u ( x ) and cos ( π x 2 L ) for v ( x ) and solve integral. ∫ x cos ( π x 2 L ) d x = x ∫ cos ( π x 2 L ) d x − ∫ [ ( d x d x ) cos ( π x 2 L ) d x ] = x 2 L π sin ( π x 2 L ) − ∫ 2 L π sin ( π x 2 L ) = x 2 L π sin ( π x 2 L ) − 2 L π ( − 2 L π cos ( π x 2 L ) ) = 2 L π ( x sin ( π x 2 L ) + 2 L π cos ( π x 2 L ) ) Integrate the above result from x = 0 to x = L 2

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Author: BEER

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Chapter 5.2, Problem 5.49P

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The centroid of area in Fig. P5.49 by method of direct integration.

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The centroid of area in Fig. P5.49 by method of direct integration.

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