00:01
Hi everyone here we have to find centered y -dess of cross -section find centered y -dess of the cross -section and calculate its moment of inertia about x -dice this is your y -dess and this is the x -dess passing through the centroid so given area has to be divided into three segments let us see here this is segment one this is segment two this is segment two this is two this is two this is one and this is third so this one is third this is first this is two and this is two for each segment we will find the position of centroid so this is the centroid c1 this is centroid c2 this is c2 and this is c3 and by coordinate of each center so y c1 bar having the value 0 .25 meter y c2 having the value 0 .1833 meter by c3 having the centroid 0 .025 meter the area and area of each segment and its centroid are tabulated as segment area, centroid and product of centroid by an area.
03:25
This will be in meter.
03:27
This is meter square.
03:29
This will be meter cube.
03:32
Segment one having the area 0 .3 in point four centroid having the value 0 .25 and product of area and centroid is 0 .03.
03:48
Similarly for segment 2 area is half base into height that is half point 4 into 0 .4 centroid by having the value 0 .1833 and product of area and centroid having the value 0 .01833 and product of area and centroid having the value 0 .0 .0 .0.
04:16
1375 third segment is rectangle so its area is 1 .1 into 0 .05 while centroid having the value 0 .025 product of area and centroid is 0 .0135 summation of area you will get 0 .25 meter square and submission of product of centroid y and area is 0 .46042 meter cube.
05:19
So centroid y of the area can be defined as submission of product of centroid of each segment into its area upon total area of the given area.
05:40
That is submission.
05:41
Of area of each segment now substituting the value so here we can write that is 0181 meter so this is the answer of part a now we have to find moment of inertia as we know moment of inertia of each segment can be calculated moment of inertia of each segment can be calculated x -axis by applying parallel axis theorem i x -des -to -be -equal -axis to be equal to i x -des about the centroid plus product of area of each segment into square of perpendicular distance from the y -axis now we are tabulating here for each segment segment one two three area measuring in meter square perpendicular distance of centroid of each segment from y axis in meter moment of inertia of each segment about the centroid in meter to the power four product of area of segment into square of the distance from y axis and moment of inertia about x x x keep on putting the value...