00:01
Hello students, in this question we have to locate the centroid of x, y whose cross -sectional of the member's cross -sectional area for the given diagram.
00:16
So to start with we have to find the area a1.
00:19
For the first segment we have to find area a1.
00:23
A1 is equal to it is a triangle region therefore we can write half into b into h.
00:30
So b is 30 millimeter therefore half into 30 into height is we have to include both of these 50 mm plus 40 mm therefore it is 90 mm.
00:42
Therefore 90 into 50 which is equal to 2700 divided by 2 therefore area a1 is equal to 1350 millimeter square.
00:56
Now we have to find out x1.
00:58
So x1 is from the figure we could see that it is x1 is equal to 2 by 3 into breadth.
01:06
So 2 by 3 into breadth is 30 meter therefore it is equal to 60 by 3 which is equal to which implies x1 is equal to 20 mm.
01:19
Similarly y1 is equal to h by 3 referring to the diagram we could say it is h by 3 therefore 90 by 3 therefore y1 is equal to 30 millimeter.
01:34
Now we have to go for the second region area a2.
01:40
Second region is it's a rectangular area therefore a rectangular region therefore area of the rectangular region is breadth into height therefore breadth is 30 meter height is total of 40 plus 50 millimeter therefore area is equal to 2700 millimeter square a2 and we have to find x2.
02:04
So from the figure x2 is 30 into 30 plus therefore is 30 plus b by 2.
02:11
So we know that breadth is 30 meter therefore breadth by 30 divided by 2 is 15 therefore 45 meter x2 is equal to 45 millimeter.
02:25
It is not meter it is millimeter...