00:01
In this question, we are given here bracket.
00:03
So, in this bracket, in the figure 2, we are talking about the welded vertical column.
00:15
So, determine the load p.
00:19
So, in this question, let's determine the load p.
00:23
So, in this load p, if the permissible shear stress, which is equal to 66 .1 mpa and take the edge, which is equal to 4 .5 mm and the value of l, we are given here, which is 100 mm and d value, which is equal to 100 mm and b, which is equal to 50 mm.
00:58
So, all these values we are given here in the mm.
01:01
Let's find out the load p.
01:04
So, how to find the load p here? so, we are using the two here, there is a direct shear stress and the secondary shear stress that we are using.
01:14
So, first we understand that the direct shear stress.
01:22
So, in the direct shear stress, the both are required to find the load p.
01:26
So, direct shear stress formula is tau 1, which is equal to p divided by a.
01:30
So, p is load, a is the area.
01:34
So, a here, which is our area and p already we denoted, that is the load and the formula here is the a, which is equal to 2 multiplied by t multiplied by b.
01:45
We just substitute the value into this formula and finding the answer.
01:48
So, this is the t here, edge or we can say t, the both are same.
01:52
Find out the value of area.
01:54
So, area, which is equal to, we can write 2 multiplied by value of edge or value of t, 4 .5 mm and multiplied by b, which is 50 mm.
02:02
We just simplify and we can write answer is 450 mm square.
02:06
This is the value of area.
02:08
Now, we substitute the value into this formula.
02:09
So, we can write tau 1, which is equal to p by 450 n per mm square, which is equal to answer is 0 .00222, that is p and here you can write n per mm square.
02:28
Now, here we talk about our secondary shear stress.
02:33
So, now let us understand the secondary shear stress.
02:40
So, this secondary shear stress, that is tau 2...