00:01
Hello students, in this question given as a t -shaped beam with the dimension of the beam is given as in the figure according to the question.
00:09
Now we have to find the point of neutral axis.
00:13
So for this we will write the formula y is equal to a1.
00:19
A1 is the area of beam 1 and beam area of beam 2 a1 plus a2.
00:36
So now we will calculate the area of beam 1 is 12, dimension is 12 into 75 mm.
00:46
All values given in millimeter and the y1 will be equal to the 94, 94 mm from this the point this.
01:04
So 100 minus 6, so this will be 94 mm plus a2 area is 12 into 88 into 44 from the center.
01:30
The value is this is y2 and this is y1 upon 12 into 75 plus 12 into 88.
01:58
So after the calculation we will get y is equal to 67 .2 mm.
02:15
Now we will calculate areal moment of both beam that will be equal to the summation of 4i moment of inertia.
02:29
Areal inertia plus ati square.
02:39
The formula of areal inertia of beam is bh cube upon 12.
02:45
Now we will apply this formula on beam a.
02:50
So this will be 1 by 12 into b is 75 into h is 12 mm whole cube plus the area of beam a is 900 into the distance this will be 94 minus 67 .2 whole square...