00:01
Hi everyone here we have to find centroid y of the given area and then calculate moment of inertia i u and iv for the channels cross -section area.
00:21
Let us see here centroid y can be defined as summation of a into viver upon summation of a so it consists of three sections i am marking here this is the section 1, 2 and 3.
01:12
300 into 10 into 5 plus 2 times 50 into 10 into 35 divided by 300 into 10 to 50 into that is 12 .5 millimeter.
01:35
This is the answer of first part.
01:37
Now we have to solve for second.
01:40
We will find moment of inertia along x -axis and y -axis by applying the parallel x -s theorem so according to parallel axis theorem we can write i x to be i per x just a moment according to parallelism momentum of inertia about x -axis will be moment of inertia about centroid plus m into d square so we have to apply this concept moment of inertia about x axis of the given area would be 1 by 12 into 300 10 kukub plus 300 into 300 10 into 50 cube plus 10 into 50 into 35 minus 12 .5 whole square.
03:09
So on solving it you will get 9083 into 10 to the power 6 millimeter to the power 4.
03:20
Similarly, along y -axis you will get 1 by 12 10 into 300.
03:29
8 cube plus 2 into 1 by 12 into 50 into 10 kha cube minus sorry plus 5 into 10 into 150 into 150 minus 5 square so on solving it iy you will get 43 .53 into 10 to over 6 millimeter to the power 4 area is symmetrical to the x and y x x x that's why i x y is 0 because area is symmetrical to x and y x so moment of inertia about u can be defined as i x plus i y upon 2 i x minus i y upon 2 cause of 2 minus i x by sine of 2 theta.
05:04
Substitute the value...