00:01
According to the question we have the following diagram which shows thin walled, thin walled single cell beam.
00:09
These are certain points marked in the question.
00:14
This is the path showing.
00:18
Now this point is marked as 1, 2, 3, 4, then 5, then another point is marked as 6, then 7 and finally point number 8.
00:34
This is point 8.
00:35
Now we are given that the distance between point 1 and 8 is 60 mm.
00:42
Similarly there is a force acting in upwards direction at point 2 that is 10 kn.
00:49
We are also given with the distance between point 3 and 6 that is 200 mm and point 4 and 5 is 100 mm.
01:00
Difference between 5 and 6 is 120 mm while 6 and 7 is 240 mm which makes the difference between 3 to 2 also 240 mm.
01:17
Now distance between 7 to 8 is 240 mm again.
01:23
Similarly 1 to 2 will be 240 mm.
01:27
These are the things already given to us.
01:30
Now we are given with 10 kn that is 10 kn acting in vertical direction that is in y direction at booms through 3 and 6.
01:52
Now we need to calculate a shear flow, shear flow across the section.
01:59
Now to calculate this first we need to know that centroid, centroid of direct stress, direct stress carrying, direct stress carrying area lies on the horizontal axis, horizontal axis of symmetry, of symmetry which is marked with this line, the dotted line.
02:31
This represents the horizontal axis of symmetry.
02:35
Now at this axis we know that ixy is entirely equal to 0.
02:42
Also that td is equal to 0 which means that only a vertical shear load is applied, only vertical shear load.
02:56
As given in the question that no other load is applied, only the vertical shear load is applied.
03:00
So this is reduced to, now let's calculate it.
03:04
It will be qs is equal to the shear load minus because it is vertical sx ixx minus sy ixy divided by, this will be the shear stress, divided by ixx iyy minus i square xy.
03:32
This will be for x and y direction.
03:35
Then the integral of 0 to 8.
03:40
This will be td x ds plus summation of n till r equal to 1 br xr minus, then it will be summation, the shear stress at y ixx again minus sx ixy which will be divided by entirely ixx then yy again because for the y direction minus i square xy applying the a plus b whole square.
04:23
So we get the value of shear stress that would be minus sy divided by ixx summation of n till r equals to 1 br yr plus qs.
04:41
Now in the series it would be in continuation.
04:44
Let us consider this as our equation number 1.
04:48
Now in this, in this equation we can apply the given units that is qs is equal to, now minus 7 .22 which is already given to us into 10 to the power minus 4 summation of n r is equals to 1 br yr plus qs 0.
05:16
Let us consider this as equation number 2.
05:19
Now we will be calculating qb.
05:27
Qb is the basic shear flow distribution, basic shear flow distribution.
05:37
Now to find this we will be using the right hand side of equation number 2 that means this part.
05:47
Using this we have qb 23.
05:51
Why 23? because it is a value.
05:54
It will be entirely equal to 0...