00:01
Given given load p is equals to 25 kilo newton permissible shear stress permissible shear stress permissible shear stress tau is equals to 75 mega pascal let let thickness of world thickness of width of build be p so area of area of the world area of the world is given by given by area is given by equals to 2 into 100 t plus 150 t area equals to 500 t mm square so now primary shear primary shear stress shear stress is given by given by given by tau p tau p equals to p upon a that is equals to 25 kn upon 500 t is equals to 25 into 10 to the power 3 n upon 500 t in mm square.
01:44
So this can be written as 50 upon t n per mm square.
01:52
So tau p is equals to 50 upon t mpa.
01:57
Moment of inertia of 1 about x -axis is given by ixx1 is equal to btq upon 12 plus a into to d by 2 whole square is done using parallel axis theorem.
02:43
So now ixx1 is equals to bt cube upon 12 plus b into t into d by t whole square.
03:00
There are there are two similar rectangle, two similar rectangle, two similar rectangle, two similar rectangle.
03:13
So, ixx1 equals to twice of bt cube upon 12 plus bt into d by 2 whole square.
03:27
Now we will put the values we get ixx1 is equals to 2 into 100 into t cube upon 12 plus 100t into 150 by 2 whole square...