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webassign.net Chegg Study | Guided Solutions and Study Help | Chegg.com Statics HW 21 Assignment Scoring Your last submission is used for your score. 1. 5.25/20 points | Previous Answers NorEngStatics1 8.P.044 My Notes Ask Your Teacher Determine the location of the centroid (X, y) in inches and the second moments of area I and ly in in4 with respect to the centroidal axes for the light blue cross-section. (For the location of the centroid, enter your answer to at least two decimal places.) 10.6 in 2 in 3.6 in 9.0 in 8.06 X in y= 6.428 x1767.84 ,'= 1843.621 in X in in Submit Answer Save Progress Practice Another VersionExplanation / Answer
The given area can be divided into two rectangles and one triangle. The domensions of two rectangles are 12.6x3.6 and 9x2.The right angled triangle is of height 9 ft and base 10.6 in
The x coordinate of the centroid = [(12.6*3.6*6.3)+(9*2*11.6)+(0.5*10.6*9*2*10.6/3)]/[(12.6*3.6)+(9*2)+(0.5*10.6*9)] = 831.648/111.06=7.488 in
The y coordinate of the centroid = [(12.6*3.6*10.8)+(9*2*4.5)+(0.5*10.6*9*9/3)]/[(12.6*3.6)+(9*2)+(0.5*10.6*9)] = 713.988/111.06=6.429 in
The moment of inertia can be determined using parallel axes theorem
Ix' = (12.6*3.63/12)+[12.6*3.6*(6.429-10.8)2]+(2*93/12)+[2*9*(6.429-4.5)2]+(10.6*93/36)+[0.5*10.6*9*(6.429-3)2]
= 1879.61 in4
Iy' = (12.63*3.6/12)+[12.6*3.6*(7.488-6.3)2]+(23*9/12)+[2*9*(7.488-11.6)2]+(10.63*9/36)+[0.5*10.6*9*(7.488-7.067)2]
=1280.69 in4
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