Question

Question 3 26 pt Locate the centroid (i,y) of the shaded area (26 Marks). 0.6 m 3 m у 1.5 m 3 m 1 m Important Table: Geometric Properties of Line and Area Elements Centroid Location Centroid Location Area Moment of Inertia L-20 sin 20) 1,- 20

Answer 1

Saln -> .6m Kim det devide the whole Area in four part were Ta TXAR Tk 1.5m ok An **12 2 HE 1.57677 m2 21 = om symetat about uran is) 41 = IK 3m 41 30 481 3* 49 0.4244 m Area Az = **(332 LY) 7.068m2 4x3 3* 1.273m 3T Yq 1.273m 4R 31 4X3 3N Area Aza 3.6 X 3 10.8m2 cosm Uz -3.6 2 Y = 1.5m 0.45m . -3. 4m Aren (A) - 1.6*1-5 245-(3+ 2x360) 44 3-(1) Centroid Cze) 2.5m た How - Aphy & A2 U2 + A3 %z r. AqKq - Al it A2 + Azr At

1501 +7-068 x 1 2 13 + 16.8x-1.8+ (-.45x-3.4) -1.5707+ 1.068 +10.8 - .45 L50.5623m 4 -A+11, +A.A4%4 -Ait Ag of Az - A4 -1.5707.4249+ 1.0687 / 273 + 10.8x1.5-.45*2.5 -1.5-107 +7.068 + 10.8 - 445 १. 14169m