00:01
Hello, in the question we have given that a large tower is to be supported by a series of steel wires.
00:07
So, it is estimated that the load on each wire will be 12 ,000 newton.
00:12
So, we have given the load that is p is equal to 12 ,000 newtons.
00:19
Further, determine the minimum required diameter in millimeters assuming a factor of safety of 2 and a yield strength of 860 megapascal for steel.
00:36
So, now over here we will use a formula.
00:39
So, that is load upon area is equals to.
00:45
So, load we will write it in p, we will write it as p and area as a that is equals to the allowed allowable stress that is nothing but sigma.
01:06
So, let us call it as sigma a.
01:09
So, now from here let us write this in this form.
01:14
So, sigma a will be equal to load is p divided by area is given by pi r square.
01:24
So, this we can write it as pi d square by 4 and that is nothing but 4p divided by 4d square.
01:34
So, now this d square we can send on the other side.
01:37
So, this will be d square is equal to 4p divided by pi times sigma a, but sigma a is 860 by 2.
01:48
So, this is 4p divided by.
01:50
So, let us plug the value over here.
01:54
So, it is 4 into p is 12 ,000 divided by pi is 3 .14 into this sigma a is 860 divided by 2.
02:09
So, this 2 will also go up...