00:01
So in this question it is given that load f is equal to 100 kilo newton, cl stress which we represent the tau max is equal to 70 mp a pressing stress sigma plus is equal to 100mper and here e is equal and we need to find out diameter and thickness of plate.
01:03
Thickness of plate.
01:10
We have this diagram for our question.
01:14
So here is equal to 120, sine 30 degrees, is equal to 60 amount.
01:35
Now, fx is equal to 100 or 60 degree is equal to 50 newton.
01:47
Force on y -axis is equal to 100, sine 60 degree, which is equal to 50 multi -bli wire under root 3 neutral so now direct shear force is equals to direct shear force force is equals to we know that fd is equals to f divided wire and which is 100 divided wire 4 which equals to 25 kilo neutral so we know that secondary shear force, secondary shear force, then a and d will be maximum.
03:17
So we will calculate secondary shear force in a, which is x, y, which equals to f s divider y, 120, multiply, y, 120, multiply, by, 120, plus 40 old square plus 120 whole square plus 40 whole square plus 40 whole square f s is equals to 22 .5 kilo newton so now shear force is equals to which equals to under root of 25 whole square plus 22 .5 whole square plus 22 .5 25 which is multiplied by 22 .5...